Jonathan M. Kane
Writing Proofs in Analysis
Writing Proofs in Analysis
Jonathan M. Kane
Writing Proofs in Analysis
123
Jonathan M. Kane Department of Mathematics University of Wisconsin - Madison Madison, WI, USA
ISBN 978-3-319-30965-1 ISBN 978-3-319-30967-5 (eBook) DOI 10.1007/978-3-319-30967-5 Library of Congress Control Number: 2016936668 © Springer International Publishing Switzerland 2016 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. Printed on acid-free paper This Springer imprint is published by Springer Nature The registered company is Springer International Publishing AG Switzerland
To the memory of Sylvan Burgstaller, Duane E. Anderson, and especially James L. Nelson who, at the University of Minnesota Duluth, taught me the fundamentals of writing proofs in analysis.
Acknowledgments
I wish to thank Natalya St. Clair for her excellent work creating the illustrations appearing in this textbook. She took my crude sketches and vague ideas and turned them into pleasing artwork and instructive diagrams. I also wish to thank Daniel M. Kane, Alan Gluchoff, Thomas Drucker, and Walter Stromquist for their insightful comments about the presentation, content, and correctness of the text.
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Preface
After learning to solve many types of problems such as those found in the first courses in Algebra, Geometry, Trigonometry, and Calculus, mathematics students are usually exposed to a “transition” course where they are expected to write proofs of various theorems. I taught such a course for a dozen years and was never satisfied with the textbooks available for that course. Although such textbooks often teach the fundamentals of logic (conditionals, biconditionals, negations, truth tables) and give some common proof strategies such as mathematical induction, the textbooks failed to teach what a student needs to be thinking about when trying to construct a proof. Many of these books present a great number of well-written proofs and then ask students to write proofs of similar statements in the hope that the students will be able to mimic what they have seen. Some of these books are also designed to be used as an introductory textbook in Analysis, Abstract Algebra, Topology, Number Theory, or Discrete Mathematics, and, as such, they concentrate more on explaining the fundamentals of those topic areas than on the fundamentals of writing good proofs. This Book Is Not Your Traditional Transition Textbook The goal of this book is to give the student precise training in the writing of proofs by explaining what elements make up a correct proof, by teaching how to construct an acceptable proof, by explaining what the student is supposed to be thinking about when trying to write a proof, and by warning about pitfalls that result in incorrect proofs. In particular, this book was written with the following directives: • Unlike many transition books which do not give enough instruction about how to write proofs, most of the proofs presented in this text are preceded by detailed explanations describing the thought process one goes through when constructing the proof. Then a good proof is given that incorporates the elements of that discussion. • For proofs that share the same general structure such as the proof of lim f .x/ D L x!a for various functions, proof templates are provided that give a generic approach to writing that type of proof. ix
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• Many transition books begin with several chapters covering an introduction to logic, set theory, cardinal numbers, and an axiomatic construction of the real numbers. I find that students do not appreciate the details of these discussions when these concepts are presented before they are needed to write a specific proof. For example, truth tables are very helpful in verifying the truth of a complex logical statement, but it is hard for students at that level to see the connection between the truth value of a complex statement and the formation of a proof. Therefore, I introduce many of these ideas as needed within the contexts of writing Analysis proofs and have kept the introductory material to a minimum. • Many books that propose to teach students to write proofs in Analysis get carried away with covering those great topics in Analysis and cut back on the proof writing instruction. The books may start out teaching about proofs, but after a few chapters of introduction, they assume that the students now understand everything they need to know about writing proofs, and the books concentrate entirely on the concepts of Analysis. This book covers plenty of Analysis and can be used as a textbook for a typical beginning Real Analysis course, but it never loses sight of the fact that its primary focus is about proof writing skills. Certainly, one can use this book for a beginning course in Real Analysis because it thoroughly covers the standard theorems, but as a first course in proof writing, it will succeed where others fail. If the students using this book have already had a thorough background in writing proofs, then this book could be used as a standard one-semester course in Real Analysis. Theses students might begin in Sect. 2.5 and, depending on their background, be expected to cover the material through Chaps. 6, 7, or 8. On the other hand, if the students are using this book both as an introduction to proof writing and an introduction to Analysis, then the textbook can be used for a two-semester course in Real Analysis and proof writing. The first semester might aim to cover the first five or six chapters, while the second semester aims to complete the book. For most of the topics, it is important that the chapters be covered in their prescribed order. Elements of later chapters do depend on the material covered in earlier chapters. Madison, Wisconsin, USA 2016
Jonathan M. Kane
Contents Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . List of Proof Templates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 What Are Proofs, and Why Do We Write Them? . . . . . . . . . . . . . . . . . . . . . . . . 1.1 What Is a Proof? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Why We Write Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 The Basics of Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 The Language of Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Conditional Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 Negation of a Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.3 Proofs of Conditional Statements . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Template for Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Proofs About Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Set Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.3 Proofs About Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.5 Proofs About Set Equality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Proofs About Even and Odd Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Definitions of Even and Odd Integers . . . . . . . . . . . . . . . . . . . . . 2.4.2 Proofs About Even and Odd Integers . . . . . . . . . . . . . . . . . . . . . 2.4.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Basic Facts About Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Ordered Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5.2 The Completeness Axiom and the Real Numbers . . . . . . . . 2.5.3 Absolute Value, the Triangle Inequality, and Intervals . . . 2.5.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Function, Domain, Codomain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.2 Surjection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.3 Injection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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2.6.4 Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 The Definition of Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Proving lim f .x/ D L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x!a 3.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 One-Sided Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Limits at Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Limit of a Sequence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Definition of Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 Arithmetic with Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Monotone Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.4 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.5 Limit of a Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.6 Limits of Monotone Sequences and Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.7 Cauchy Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6 Proving That a Limit Does Not Exist . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.1 Why a Limit Might Not Exist . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.2 Quantifiers and Negations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.3 Proving No Limit Exists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Accumulation Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 The Arithmetic of Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9.1 Limit of a Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9.2 Limit of a Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9.3 Limit of a Quotient. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9.4 Limit of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9.5 Other Types of Limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10 Other Limit Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10.1 The Limit of a Positive Function . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10.2 Uniqueness of Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10.3 The Squeezing Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10.4 Limits of Subsequences. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3.11
4
5
6
Liminf and Limsup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 The Definition of Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Proving the Continuity of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Compactness and the Heine–Borel Theorem . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Open Covers and Subcovers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 Proofs of the Heine–Borel Theorem . . . . . . . . . . . . . . . . . . . . . . 4.4.3 Uniform Continuity on Closed Bounded Intervals . . . . . . . 4.4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 The Arithmetic of Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Composition, Absolute Value, Maximum, and Minimum . . . . . . . . . . 4.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Other Continuity Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.1 Boundedness of Continuous Functions . . . . . . . . . . . . . . . . . . . 4.7.2 Obtaining Extreme Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.3 The Intermediate Value Property . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Discontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 The Definition of Derivative. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Differentiation and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Calculating Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 The Arithmetic of Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Chain Rule and Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Increasing Functions, Decreasing Functions, and Critical Points . . 5.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 L’Hopital’s Rule. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9 Intermediate Value Property and Limits of Derivatives . . . . . . . . . . . . . Riemann Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Area. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Cardinality of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Measure Zero. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Areas in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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6.5
7
8
Definition of Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Properties of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Integrable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 Step Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9 Integrals of Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.10 Characterization of Integrable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Convergence of Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Absolute and Conditional Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 The Arithmetic of Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Tests for Absolute Convergence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 Comparison Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.2 Ratio Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.3 Root Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.4 Integral Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Alternating Series Test. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 The Smallest Divergent Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 Rearrangement of Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7.1 Addition of Parentheses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7.2 Order of Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8 Cauchy Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Sequences of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Pointwise Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Monotone Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Series of Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.1 Absolute Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.2 Interval of Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.3 Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.4 Taylor’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
169 172 173 177 178 183 183 189 189 194 194 200 201 201 204 205 206 208 209 209 212 214 215 218 219 222 222 224 224 225 230 231 236 239 239 241 241 246 246 252 255 255 256 259 263
CONTENTS
8.5.5 Arithmetic of Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Fundamental Question of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Topology of the Real Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Interior, Exterior, and Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Open and Closed Sets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Unions and Intersections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Continuous Functions Applied to Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Closure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.7 Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Definition of Metric Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.1 Cauchy–Schwarz Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.2 Minkowski Inequality. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Examples of Metric Spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Topology of Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Limits in Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Continuous Functions on Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7 Homeomorphism. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.7.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.8 Connected Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.8.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.9 Compact Metric Spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.9.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.10 Complete Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.10.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.11 Contraction Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.11.1 Contraction Mapping Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.11.2 Picard’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
xv
265 266 267 269 269 274 274 278 278 282 282 285 285 288 288 291 291 293 295 295 297 297 298 298 299 305 306 307 308 311 311 314 315 316 316 317 317 322 323 327 327 327 329
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CONTENTS
10.11.3 Fractals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 10.11.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 Books for Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
List of Figures Fig. 1.1
Dividing the disk with the chords from n points . . . . . . . . . . . . . . . . . . . . . .
6
Fig. Fig. Fig. Fig. Fig.
List of implications for P ! Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .A [ B/c is equal to Ac \ Bc . . . . . . . . . . . . . . . . .p ........................... Showing the least upper bound of S is s D r . . . . . . . . . . . . . . . . . . . . . . . . Triangle inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Composition .f ı g/.x/ D z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15 26 37 39 43
2.1 2.2 2.3 2.4 2.5
Fig. 3.1 Fig. 3.2 Fig. 3.3 Fig. Fig. Fig. Fig. Fig. Fig. Fig.
3.4 3.5 3.6 3.7 3.8 3.9 3.10
Fig. 4.1 Fig. 4.2 Fig. Fig. Fig. Fig.
4.3 4.4 4.5 4.6
Fig. Fig. Fig. Fig. Fig. Fig. Fig.
4.7 4.8 4.9 4.10 4.11 4.12 4.13
lim f .x/ D L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
x!a
lim f .x/ D L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
x!a
lim f .x/ D L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
x!a
Graph of f .x/ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Approaching a limit as x ! 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proving bounded monotone sequences converge . . . . . . . . . . . . . . . . . . . . . f has no limit at x D 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Graph of sin 1x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Set with accumulation point a and isolated point b . . . . . . . . . . . . . . . . . . . Sequences approaching the lim sup and lim inf . . . . . . . . . . . . . . . . . . . . . . .
56 57 63 71 72 74 94
Continuity of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A function equal to 2x for rational x and x C 1 for irrational x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . f .x/ D 1x is not uniformly continuous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Heine–Borel Theorem first proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Heine–Borel Theorem second proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . y and z straddle one endpoint but remain in an interval of the open cover. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proving that a continuous function on Œa; b is bounded . . . . . . . . . . . . . . The maximum and minimum of a function f .x/ on an interval . . . . . . f passing through each y between f .c/ and f .d/ . . . . . . . . . . . . . . . . . . . . . . A function with a jump discontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Graph of sin 1x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Graphs of sgn.x/ and bxc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Graphs of functions with discontinuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
100 104 106 112 113 115 125 126 128 130 130 131 132
xvii
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LIST OF FIGURES
Fig. 4.14 Graph of Thomae’s function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Fig. Fig. Fig. Fig. Fig. Fig. Fig.
5.1 5.2 5.3 5.4 5.5 5.6 5.7
Slope of a Secant Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Tangent Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Restricting sin.x/ to get sin1 .x/ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Graph showing maxima and minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The proof of Rolle’s Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Point c where the tangent line is parallel to the secant line . . . . . . . . . . . x2 sin x12 and its derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
134 134 142 145 147 148 156
Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.
6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9
The union of countably many countable sets is countable . . . . . . . . . . . . Determining y using a diagonalization argument . . . . . . . . . . . . . . . . . . . . . Construction of the Cantor set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Covering a line segment with smaller and smaller squares . . . . . . . . . . . An 8 8 grid of rectangles overlaying a triangle . . . . . . . . . . . . . . . . . . . . . Approximating the area under a curve with narrowing rectangles . . . The step function s.x/ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Choosing j on .xj1 ; xj / . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The mean value theorem for integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
161 161 164 165 168 170 184 185 191
Fig. 7.1 Fig. 7.2 Fig. 7.3
Comparing the series with the integral in the Integral Test . . . . . . . . . . . 216 Converging to ln 2 with an alternating series . . . . . . . . . . . . . . . . . . . . . . . . . . 220 Rearranging terms to converge to L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
Fig. 8.1
The sequence of functions xn converging to a discontinuous function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . nC1 The sequence of functions jxj n converging to the function f .x/ D jxj . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A sequence of functions with integral 1 converging to the function f .x/ D 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A sequence of functions converging uniformly . . . . . . . . . . . . . . . . . . . . . . . If continuous function fn is close to f , then f .x/ is close to f .a/ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Fig. 8.2 Fig. 8.3 Fig. 8.4 Fig. 8.5 Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.
9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9
Interior, boundary, and exterior of a set. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x in @.@S/, y in @S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . An open set S, its boundary, and its complement Sc . . . . . . . . . . . . . . . . . . Showing boundaries of sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The union of open sets is an open set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Mapping sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The closure of a set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The sets Œ0; 1 and .4; 5/ are disconnected. . . . . . . . . . . . . . . . . . . . . . . . . . . . . The set C is a connected set. The set N is not a connected set. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Fig. 9.10 Graph of sin 1x with the y-axis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
240 240 241 242 243 270 273 275 278 279 284 287 292 293 293
Fig. 10.1 Metric distances in the plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296
LIST OF FIGURES
Fig. Fig. Fig. Fig. Fig. Fig. Fig. Fig.
10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9
Fig. Fig. Fig. Fig. Fig. Fig.
10.10 10.11 10.12 10.13 10.14 10.15
Euclidean distance is R2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . N.0; 1/ in the Euclidean, taxicab, and supremum metrics . . . . . . . . . . . . Some functions in CŒ0; 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proving that the union of open sets is open. . . . . . . . . . . . . . . . . . . . . . . . . . . . Limit of f W X ! Y as x approaches a is L . . . . . . . . . . . . . . . . . . . . . . . . . . . . Limit of a sequence in a metric space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A compact set of a metric space is closed and bounded . . . . . . . . . . . . . . Extrema of a continuous real-valued function on a compact set. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Continuous bijection on a compact metric space. . . . . . . . . . . . . . . . . . . . . . Enclosing a closed bounded set in a grid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Contraction mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Mandelbrot set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Stages in the forming of the Sierpinski triangle . . . . . . . . . . . . . . . . . . . . . . . Generation of a fractal fern. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
xix
300 301 302 307 308 310 318 320 322 326 328 333 333 339
List of Proof Templates Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proving A B for sets A and B . . . . . . . . . . . . . . . Proving A D B for sets A and B . . . . . . . . . . . . . . . Proving a function f is surjective . . . . . . . . . . . . . . Proving a function f is injective . . . . . . . . . . . . . . . Proving lim f .x/ D L . . . . . . . . . . . . . . . . . . . . x!a Proving a result using mathematical induction . . . . . . . Proving lim f .x/ does not exist . . . . . . . . . . . . . . . x!a Proving the function f is continuous at the point a . . . . . Proving the function f is uniformly continuous on the set A Proving is a metric space . . . . . . . . . . . . . .
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. 12 . 20 . 23 . 41 . 42 . 50 . 65 . 70 . 102 . 106 . 296
xx
Chapter 1
What Are Proofs, and Why Do We Write Them?
1.1 What Is a Proof? A statement in Mathematics is just a sentence which could be designated as true or false. The sentences “1 C 1 D 2” and “x D 4 implies x2 D 16” are true statements while “All rational numbers are positive” and “There is a real number x such that x2 C 5 D 2” are false statements. Some sentences like “Green is nice” or “Authenticity runs hot” are too ambiguous, a matter of opinion, or are just plain nonsense and cannot be said to be true or false, so mathematicians would not consider them to be statements. Mathematicians have a lot of words for kinds of statements including many that you have heard: definition, axiom, postulate, principle, conjecture, lemma, proposition, law, theorem, contradiction, and others. You are certainly familiar with the numbers you use for counting items: 1; 2; 3; 4, and so forth. Suppose you wish to investigate statements about these numbers to see which statements hold true for all of these numbers. This is an admirable mathematical pursuit, so how would you get started? Mathematicians know from experience that if you want to begin an investigation, you better start with definitions, that is, you better make some clear statements about the objects you are about to study, because there are examples of mathematicians running off to study something without first making clear what it is they are studying, and later running into problems because they have not been consistent about how they are treating these new objects. This happened, for example, when people investigated the concept of limit before a precise definition of limit was in place. OK, so perhaps you make some statements about the numbers with which you want to work so that you are confident that you understand the collection 1; 2; 3; 4; 5; : : : . What are you going to be able to do with these numbers? If you only know the names of these numbers and have a symbolic representation for each, there is not a great deal you can do with them. Perhaps you could get a collection of blocks and paint one number on each block. Then you could have fun rearranging these numbers just as you have seen done by countless children.
© Springer International Publishing Switzerland 2016 J.M. Kane, Writing Proofs in Analysis, DOI 10.1007/978-3-319-30967-5_1
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1 What Are Proofs, and Why Do We Write Them?
But more likely you are interested in investigating some properties of these numbers having to do with their order or how they behave when operated on by addition or multiplication. This, of course, would mean that you will need to make clear statements about addition and multiplication operations and a less than relationship, again, so that you do not run into problems later because you were being ambiguous. So, you might write definitions of addition, multiplication, and less than, and then make statements about how these operations behave such as a Commutative Law of Addition (mCn D nCm), a Distributive Law of Multiplication over Addition (a.b C c/ D ab C ac), and an Order Property of Addition (r < s implies r C t < s C t). These statements about how these defined quantities work are called axioms, postulates, or principles. They are statements that you accept as the guiding rules for how your mathematical objects behave and go beyond the definitions to describe and make precise just what the definitions are talking about. Once you have made definitions and laid out your axioms, you should have the tools necessary to begin an investigation of other properties. Suppose that someone looks at a few examples and notice that 1 C 9 D 10 and 10 is 2 times another number, 5. They then notice that 4C12 D 16, 3C147 D 150, and 1002C6 D 1008, and all of these results are also numbers equal to 2 times another number. This might lead them to make the statement that “if you add two natural numbers together, the result is always 2 times another number.” Such a statement would be called a conjecture, a statement whose truth has not yet been determined. Of course, you know that this statement is false and came about because the investigator had not yet considered enough examples. Once they stumble upon 5 C 8 D 13 and notice that 13 cannot be represented as 2 times another number, they will know that the statement does not hold true in every case. Other conjectures such as “for every natural number a, the number a2 3a C 12 is a multiple of 2” hold up to more scrutiny. At some point in your investigation you might see a convincing argument that this conjecture is, in fact, a true statement. Such a convincing argument is what is called a proof. Once it is known that a statement has a proof, it is known as a theorem, lemma, corollary, proposition, or law. So, a proof of a statement in mathematics is a convincing argument that establishes the truth of that statement. Some statements are very easily proved, and certainly mathematicians often set up axioms in order to make particular statements easy to prove. At first this may appear to be cheating or, at best, unproductive and uninteresting because it seems to defeat the purpose of establishing truth by dictating rules that make it trivial to establish the truth. But this is certainly not the case. It is common for mathematicians to have an intuitive idea about how a system should work before they feel that they understand it enough to set down formal definitions and axioms. Perhaps you wanted addition of all natural numbers a and b to satisfy a C b D b C a. Then it would make sense to include this rule among your axioms. The axioms are written with the idea of establishing enough structure so that the statements the mathematicians want to hold true can easily be proved. The richness of mathematics is that after assuring that the obvious can be proved from the axioms, there are many more results that can be proved that are not immediately obvious from the definitions and axioms,
1.1 What Is a Proof?
3
statements which might never have been apparent to those who set up the system in the first place. For example, Fermat’s Last Theorem (there are no natural numbers a, b, c, and n > 2 such that an C bn D cn ) is a statement about natural numbers which could only be conjectured after investigating a large number of examples, and stood as a conjecture for hundreds of years before a proof was provided. Occasionally, it is shown that a conjecture is independent of the axioms; that is, neither the truth nor the falseness of the statement follows from the axioms. Two famous examples are the statements about sets known as the Axiom of Choice and the Continuum Hypothesis which have been shown to be independent of the original axioms of Zermelo-Fraenkel Set Theory. The independence of such statements suggests that the axiom system is not rich enough in structure to establish the truth of these statements, and that if one chose to do so, those statements could be added to the list of axioms for the system. The Axiom of Choice or something equivalent to it, for example, is now usually listed along with the Zermelo-Fraenkel axioms. One certainly hopes that it is not possible to prove two contradictory statements about objects in a system. Such an occurrence would say that the axioms of the system were inconsistent, and this would require the axioms to be changed. After the original ground rules for Set Theory were established by Georg Cantor in the 1870s and 1880s, Bertrand Russell pointed out in 1901 a paradox (contradiction) that is a consequence of those rules. Now commonly known as Russell’s paradox, it stimulated a flurry of activity which resulted in the young field of Set Theory being put on a firm foundation (we hope) with the creation and adoption of the Zermelo–Fraenkel axioms. The language of a proof can vary depending on who is writing the proof and who is the intended reader. In other words, what makes a convincing argument may well depend on who it is that needs to be convinced. For example, if two experts in Functional Analysis are speaking to each other, one might prove a statement by saying “Oh, that’s just a consequence of the Hahn-Banach Theorem.” That proof might be sufficient since it completely describes the reasoning behind the statement in question due to the shared knowledge of the two experts. On the other hand, if one of these experts were speaking to a beginning mathematics graduate student, the proof would need to include far more detail in order for it to be a convincing argument. If the expert were speaking to a high school student, the proof might need to be a complete book that both introduces the needed concepts and explains many results needed to understand the proof. It is important to understand that there is a difference between knowing why a statement is true and knowing how to write a good proof of the statement. It is quite possible to learn a great deal of mathematics, to be able to solve many types of mathematical problems, and to understand why particular properties must hold without being able to write coherent proofs of these properties. It is analogous to a police detective who has gathered enough evidence to be convinced which of the many suspects has committed a particular crime, but it is quite another thing to have the criminal successfully prosecuted in a court of law resulting in the criminal’s conviction and eventual punishment for the crime. A student in Analysis needs to learn many strategies that can be brought to bear when writing proofs. Some of these
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1 What Are Proofs, and Why Do We Write Them?
strategies are methods or tricks that enter a student’s bag of tricks which can be employed later when solving problems or writing proofs. A student of proof writing needs to learn how to take those strategies and turn them into coherent proofs where the ideas are presented in a logical order, fill in all necessary details, and make clear to the proof reader exactly why the chosen strategies justify the needed result. This book talks about how you should go about writing proofs of the kinds of statements typically found in the branch of Mathematics called Analysis. The branches of mathematics are not precisely defined. After a new branch arises, some mathematicians begin to combine ideas from older branches with ideas from the new branch to form even newer areas of study. For example, there are branches called Algebra, Geometry, and Topology. During the twentieth century mathematicians began talking about Algebraic Topology, Algebraic Geometry, and Geometric Topology. Very roughly speaking, then, some of the branches of mathematics are • Set Theory: the study of sets, set operations, functions between sets, orderings of sets, and sizes of sets • Algebra: the study of sets upon which there are binary operations defined (such as addition or multiplication) and includes Group Theory, Ring Theory, Field Theory, and Linear Algebra • Topology: the study of continuous functions and properties of sets that are preserved by continuous functions • Analysis: the study of sets for which there is a measure of distance allowing for the definition of various limiting processes such as those found in the subjects of Calculus, Differential Equations, Functional Analysis, Complex Variables, Measure Theory, and many other areas. Other areas of study such as Applied Mathematics, Combinatorics, Geometry, Logic, Probability are considered by some mathematicians to be their own branch of mathematics or just as part of one or more of the above four branches. The exact designation is important to some mathematicians and not to others. Although mathematicians learn to write proofs in each of these branches of mathematics, one has to begin the learning process someplace. Many teachers feel that Analysis is a good area to start because students who have completed a study of Calculus will already be familiar with just about all of the theorems discussed in a beginning course in Analysis, and may already have an intuitive feeling for why these results hold. That does not mean that those same students can write convincing proofs of these theorems. It is the goal of this book to provide the training necessary so that a student can learn to write proofs of these and similar theorems. Undergraduate courses in Topology, Group Theory, Advanced Calculus, Graph Theory, and so forth generally present the beginning concepts in each of these fields and try to give students a feel for why the major results in the fields are true. Sometimes this involves having the students learn proofs of these results while other times it only involves a presentation of definitions and known results with the idea that the students will be able to take the why it is true and turn it into a proof themselves. This book is much more interested in turning known strategies into proofs than in introducing a wealth of new strategies.
1.2 Why We Write Proofs
5
Some arguments in Analysis follow a standard format or template. This book will present several templates for proofs as a tool for teaching how one might approach the writing of a proof. For example, one can learn to prove a statement of the form lim f .x/ D L by following a standard pattern. This book will display x!1 proof patterns by presenting proof templates, and for each template it will discuss proof examples showing how to use the template and the thought process needed to complete such proofs. After that, a student would be expected to produce similar proofs. There are other theorems in Analysis whose proofs involve the introduction of some clever idea which time has shown to be useful. Beginning students would not be expected to produce proofs using these new ideas on their own, so some of these proofs are presented in order to teach the new proof strategy. The experienced mathematician will have seen a large number of these clever proof techniques and can be expected to reuse these techniques when writing a proof of some new statement. Beginning students do not have this catalog of proof techniques from which to draw, so they are not expected to be able to write proofs for such a wide variety of statements. But one must start someplace when building up this catalog, and it is a goal of this book to get students started in the right direction.
1.2 Why We Write Proofs There are many reasons why mathematicians put a lot of weight on the writing of proofs. Here are some of the reasons. Determining Truth Research mathematicians use proofs to determine what mathematical statements are true. Although many statements in mathematics are obviously true, many remain unproved conjectures for long periods of time before being proved. When a conjecture stands unproved for many years, there is time for more mathematicians to learn about the statement, and the conjecture may attract a great deal of attention. When the conjecture is first stated, some may find it interesting, but finding a suitable proof may not appear to be a difficult problem until many people have tried unsuccessfully to find a proof. As this interesting statement remains a conjecture for a longer and longer period of time, the mathematical community realizes that the problem of finding a proof is much more involved than originally expected. This is exciting partly because a wider community of experts begin to wonder whether the statement under consideration is true and because it becomes clear that new techniques will be needed to find a suitable proof if, in fact, the statement can be proved at all. The problem of determining whether or not the mathematical statement is true takes on the same sort of interest that some people would take in the success of their favorite sports teams; sitting and waiting to see how they will fair in the upcoming contest. When a longstanding conjecture is finally proved, the announcement of the accomplishment will often be covered by the lay press giving mathematics an uncharacteristic brief period of pubic admiration. Perhaps you are familiar with some of these famous problems whose
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1 What Are Proofs, and Why Do We Write Them?
Fig. 1.1 Dividing the disk with the chords from n points
1 Point, 1 Region
2 Points, 2 Regions
3 Points, 4 Regions
4 Points, 8 Regions
5 Points, 16 Regions
6 Points, 31 Regions
resolution has alluded mathematicians for years (at least at the time of the writing of this text in January 2016): The Riemann Hypothesis, the Goldbach Conjecture, the Twin Prime Conjecture, the P versus NP Problem, and the Navier–Stokes Equations Existence and Smoothness Problem. During the last 40 years resolutions have been announced for several long-standing problems including the Four Color Theorem, The Bieberbach Conjecture (now called de Branges’s Theorem), Fermat’s Last Theorem, and the Poincaré Conjecture. Why do mathematicians expend so much effort trying to prove statements, some of which may seem obvious from the start? One reason is that mathematicians are very skeptical of statements that appear obvious, and rightfully so. There is a long history that includes mathematical statements which appear to be true which are eventually shown to be false. Even very clear patterns can be deceptively seductive. Take, for example, the following problem. Select a set of n points along the circumference of a circle, draw the chords between each pair of points, and find out the maximum number of regions into which these segments can divide the disk. Figure 1.1 shows the results for the first few values of n. Although from considering n D 1; 2; 3; 4; 5 it appears that the chords can divide the disk into 2n1 regions, this fails to be true when n D 6. With a bit more thinking n1 itnis not hard to see that2n could not be the correct answer. With n points there are chords and at most 4 intersections of two chords. This number of intersections 2 grows as a fourth-degree polynomial in n suggesting that the number of regions will
1.2 Why We Write Proofs
7
also grow as a fourth-degree polynomial in n. It would, therefore, be suspicious for the number of regions to grow at the exponential rate of 2n1 . Another well-known example comes from Number Theory. The function .x/ gives the number of positive prime integers less than or equal to the number x. The growth rate of this function has long been of central importance in Number Theory. The Prime Number Theorem says that the function grows at the same rate as the logarithmic integral Z
x
li.x/ D 0
dt : ln t
In fact, for many years it was thought that li.x/ > .x/ for all x > 0 because this holds for all small values of x which can be practically checked, for example, all x between 0 and 1024 . It has now been shown that li.x/ .x/ switches sign infinitely often, although only for extremely large values of x. It is apparent that sometimes seemingly very obvious patterns do not hold in every case, so mathematicians rely on proofs to convince themselves that the patterns do indeed hold in the general case. Testing Axiom Systems In the next chapter you will read about the writing of proofs for some very elementary facts in mathematics; so elementary that you may wonder why anyone would bother with these proofs. Clearly, it makes sense to begin any training in the writing of proofs with some very simple results that are easy to understand so that the student can feel confident about all the statements being made in the proofs. But these proofs are not being presented just because they are elementary. When one sets up a mathematical system by making definitions and determining axioms, it is usually with a particular application or example in mind. The desired result is that the new system will include the already partially understood application so that any new discoveries will immediately tell something new about the original application. Suppose someone sets up an axiom system for the real numbers, for example, but is not able to prove that addition of real numbers satisfies the commutative property. Since the commutative property is an important aspect of addition of real numbers, it would appear that the new axiom system does not have enough power to represent all that one would want to show about the real numbers. Perhaps the axiom system will need to be expanded to include an axiom about the commutativity of addition. Thus, if one cannot prove that the expected simple properties hold, then it says that something is missing from the axioms. So mathematicians write proofs to confirm that their axiom systems are representative of the applications they are trying to describe. Exhibiting Beauty There are no rules about what composers of music need to write, but many composers try to write in standardly accepted formats such as string quartets or symphonies because there are already organizations ready to perform such works and groups of people happy to listen to such works. Scholars of literature compare literary works by writing literary analysis, a form which holds a lot of meaning for those who read and write in that field. Although painters
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1 What Are Proofs, and Why Do We Write Them?
choose to make pictures of every sort of object or scene, real or imagined, most painters eventually try their hand at painting some of the standard subjects (still life, nudes, famous religious or historical depictions). Similarly, mathematicians write proofs partly because that is what mathematicians enjoy doing. Although many mathematicians make substantial contributions to the sciences, social sciences, and arts through the application of their mathematical skills, others live in a world of creating and discussing abstract concepts that have no immediate application to real world problems, or at least no application apparent to the mathematicians doing the research. To them, mathematics is studied as part of the humanities and is appreciated for its beauty. And much of the beauty of mathematics lies in the proofs of its theorems. One gets a great deal of pleasure reading a clever proof of a complicated result when the proof can be stated in just a few lines, especially if previous proofs of the same result were considerably longer and more difficult to understand. Many mathematicians like reading articles and attending conferences where they are exposed mainly to proofs of results, partly so that they can learn about new results, but more importantly so they can appreciate the techniques brought to bear to construct the proofs. Testing Students One should not underestimate the need to educate future mathematicians. A good way to test whether a student understands a particular result is to ask the student to present a proof of the result. The presentation of a proof shows a deep understanding of why the result is true and shows an ability to discuss many details about the objects involved. At the graduate school level in mathematics, most test problems require the student to produce a proof of a particular result. The student who has completed a study of Calculus is likely to have mastered basic skills in Algebra, Geometry, Trigonometry, and Elementary Functions. This is a good point in one’s studies to begin writing proofs. It should not be assumed that one can just begin writing proofs at this stage even if they have had years of experience watching teachers and authors present proofs to them any more than someone can be expected to sit down and begin playing the piano just because they have watched many other people present concerts using the instrument. In this book the reader will be taken through the construction of many proofs in a step-by-step manner that presents the thought process used to write the proofs. Some incorrect proofs are shown and explained so that the student can learn about common pitfalls to avoid. Some students dread the transition to writing proofs because they feel that they do not understand how to write proofs, and are leery of the day when they will be expected to produce what they cannot now do. But the ability to write good proofs is a skill no different from the ability to factor polynomials or integrate rational functions. There is no expectation that the beginner can produce a good proof, but every expectation that the beginner can learn.
Chapter 2
The Basics of Proofs
2.1 The Language of Proofs 2.1.1 Conditional Statements Most theorems concern mathematical objects x that satisfy a set of properties P, that is, P.x/ D the properties P hold for object x. The theorem may say that if P.x/ is true, then some additional properties Q.x/ must also be true. Such statements are called conditional statements and can be written P.x/ ! Q.x/. In the context of proving theorems, the “P.x/” portion of the statement is referred to as the hypothesis of the statement, and the “Q.x/” portion of the statement is referred to as the conclusion of the statement. The hypothesis of a conditional statement is often called the antecedent while the conclusion of the conditional statement is often called the consequent. For example, a well-known theorem is that all functions differentiable at a point are also continuous at that point. There are many equivalent ways to express this fact: • • • •
All functions differentiable at a point are also continuous at that point. If the function f is differentiable at a point, then f is continuous at that point. The function f is differentiable at a point only if f is continuous at that point. If the function f is not continuous at a point, then f is not differentiable at that point. • There are no functions f such that f is both differentiable at a point and discontinuous at that point. • The function f is differentiable at a point implies that f is continuous at that point. • The function f is differentiable at x ! f is continuous at x.
All of these statements assert that if a function f satisfies the hypothesis that it has a derivative at a point x, then f must also satisfy the conclusion that f is continuous at x. Note that the truth of a conditional statement, P.x/ ! Q.x/, suggests nothing about the truth of the statement Q.x/ ! P.x/ which is known as the converse © Springer International Publishing Switzerland 2016 J.M. Kane, Writing Proofs in Analysis, DOI 10.1007/978-3-319-30967-5_2
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of the conditional statement P.x/ ! Q.x/. Indeed, the converse of this theorem is the clearly false statement: “If the function f is continuous at a point, then f is differentiable at that point.” Certainly, there are functions f both continuous and differentiable at a point, but knowing that a function is continuous at a point does not allow one to conclude that it is differentiable at that point. The converse of a conditional statement is not logically equivalent to the original statement, but since the two statements are concerned with the same subject matter, mathematicians are often interested in the converse of a given conditional. If someone succeeds in proving a new theorem expressed as a conditional statement, you might wonder whether the converse of the statement could also be true. Sometimes the truth of the converse statement is a trivial matter because it is well known. But there are many examples where the converse does not hold in every case; that is, there are many known values of x where the converse statement “Q.x/ ! P.x/” is false. Other times, the converse statement is something that has been previously established. But very often, the truth of the converse statement remains an open question, and the proof of the original conditional statement may generate research interest in its converse. One of the equivalent forms of a conditional statement P.x/ ! Q.x/ is the statement “if Q.x/ is false, then P.x/ must be false.” This can be written as “:Q.x/ ! :P.x/” using the negation symbol : . This form of the statement is called the contrapositive of the original conditional statement. For example, the contrapositive of the statement discussed above is “If the function f is not continuous at a point, then f is not differentiable at that point.” Although logically equivalent to the original conditional statement, the contrapositive often gives you a different way to think about the statement, and you will often see a proof which is a proof of the contrapositive statement instead of a proof of the original conditional statement.
2.1.2 Negation of a Statement The negation of a statement is a statement with the opposite truth value of the original statement, that is, a statement which is false exactly when the original statement is true. For example, the negation of “n is an integer” is “n is not an integer.” The negation of the statement P.x/ is “not P.x/” or simply “:P.x/.” The conditional statement “P.x/ ! Q.x/” says that every time P.x/ holds it must be the case that Q.x/ also holds. The negation of this statement must, therefore, state that for at least one value of x, P.x/ is true and Q.x/ is false or “P.x/ and :Q.x/.” A proof by contradiction is a proof that assumes both that P.x/ and :Q.x/ are true, and derives a statement that must be false (known as a contradiction) showing that it is impossible to have P.x/ being true at the same time that Q.x/ is false. The well-known Pythagorean Theorem is a conditional statement: “If a right triangle has legs with lengths a and b and a hypotenuse with length c, then a2 C b2 D c2 .” The converse of the Pythagorean Theorem is also true: “If a triangle has sides with lengths a, b, and c satisfying a2 C b2 D c2 , then the triangle is a right triangle.” When a conditional statement, P.x/ ! Q.x/ and its converse
2.1 The Language of Proofs
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Q.x/ ! P.x/ are both true, the two statements can be combined into one as P.x/ ! Q.x/. This can also be stated as “P.x/ if and only if Q.x/.” Such statements are called biconditional statements. Thus, the Pythagorean Theorem and its converse could be combined into the single biconditional statement: “A triangle is a right triangle if and only if the triangle has side lengths a, b, and c satisfying a2 C b2 D c2 .”
2.1.3 Proofs of Conditional Statements Conditional statements often make assertions about a very large number of objects or even an infinite set of objects. Indeed, the statement about differentiable functions being continuous refers to infinitely many functions, and the Pythagorean Theorem refers to an infinite number of triangles. How, then, are you supposed to prove these results since you clearly cannot consider every case individually? A general approach to proving the conditional statement “P.x/ ! Q.x/” is to select a generic element x which could represent any object satisfying P.x/ and then to prove the statement Q.x/. Since a generic object x satisfying P.x/ has been shown to satisfy Q.x/, it follows that every object satisfying P.x/ must also satisfy Q.x/, and the result has been proved. This will be the format of most of the proofs you will ever write in analysis. If the statement “P.x/ ! Q.x/” is not true, it means that there is at least one value of x that makes “P.x/ ! Q.x/” a false statement. Such an x is called a counterexample to the statement, and exhibiting such a counterexample would be a way to prove that “P.x/ ! Q.x/” is false. A proof of “P.x/ ! Q.x/” is essentially an argument showing that no counterexamples exist. There are many phrases that occur so frequently when writing proofs, that mathematicians have developed a short hand notation for these phrases. There is little need to use these abbreviations within a textbook such as this or even in a journal article, but the short hand can be useful when writing out a proof by hand on paper or a blackboard. Here is a list of some of the commonly used symbols. Shorthand Symbols for Proofs 9 there exists
9Š there exists exactly one
8 for all
S suppose (or assume)
3 such that ! implies
! if an only if
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2 The Basics of Proofs
2.1.4 Exercises Perform the follows steps for each of the conditional statements in Exercises 1–6. A B C D E
identify the hypothesis and the conclusion. write the converse of the statement. decide whether or not the converse of the statement is true. write the contrapositive of the statement. write the negation of the statement.
1. If x D 1 and y D 1, then xy D 1. 2. If x is an integer, then 2x C 1 is also an integer. 3. f .x/ and g.x/ are both continuous at x D 0 only if f .x/ C g.x/ is continuous at x D 0. 4. xy D 0 if x D 0 or y D 0. 5. If xy 9y D 0 and y > 0, then x D 9. 6. A rectangle has area xy if two adjacent sides of the rectangle have lengths x and y. 7. Write the following without using shorthand symbols. (a) 9x 2 R 3 x C 4 D 2. (b) 8x 2 R 9y 2 R 3 x C y D 10.
2.2 Template for Proofs Many proofs can be written by following a simple formula or template that suggests guidelines to follow when writing the proof. Mathematicians reading a proof that follows a traditional template will find the proof easier to follow because there will be an expectation about what will be presented in the proof. For example, many proofs will follow the general template given here. TEMPLATE followed by many proofs • • • •
SET THE CONTEXT ASSERT THE HYPOTHESIS LIST IMPLICATIONS STATE THE CONCLUSION
To illustrate this template, consider this proof of a well-known theorem from elementary Algebra.
2.2 Template for Proofs
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PROOF (Quadratic Formula): For constants a, b, and p c, the quadratic b2 4ac b ˙ . polynomial ax2 C bx C c has roots given by x D 2a SET THE CONTEXT: Let a, b, and c be constants with a ¤ 0. ASSERT THE HYPOTHESIS: Suppose that x satisfies ax2 C bx C c D 0. LIST IMPLICATIONS: Since a ¤ 0, it follows that x2 C ba x C ac D 0. b2 b2 Then x2 C ba x C ac C 4a 2 D 4a2 . b 2 b2 C ac D 4a • Factoring shows that x C 2a 2. b 2 b2 c b2 4ac • Then x C 2a D 4a2 a D 4a2 .
• • • •
2
b 4ac • This means that x C 2a must be one of the two square roots of b 4a 2 . s p p 2 2 2 b ˙ b 4ac b 4ac b ˙ b 4ac D˙ , and x D . • So, x C D 2 2a 4a 2a 2a • STATE THE CONCLUSION: Thus, the p roots of the quadratic polynomial b ˙ b2 4ac ax2 C bx C c are given by x D . 2a
The proof template begins with the suggestion to “SET THE CONTEXT” which represents statements designed to tell the reader what is being assumed in the proof. This is usually a sentence or two telling the reader about the properties of the objects that will be encountered in the proof. It may also introduce which variables will appear in the proof and what kinds of objects they represent. So, in the given proof of the Quadratic Formula, the first line tells that the variables a, b, and c are going to represent known constants with a not being 0. Clearly, the fact that a is not 0 needs to be stipulated because if a D 0, the polynomial ax2 C bx C c would not be quadratic and would not have the proposed roots. Generally, you are not looking for a lengthy narrative here, and, in fact, brevity is a particularly cherished attribute of a proof. Saying what needs to be said, but only what needs to be said is usually best. Some authors who state a theorem and immediately follow the statement of the theorem with its proof will forgo setting the context at the beginning of the proof because the reader will have just seen the statement of the theorem and may not need to see a repeat of the context for that proof. For example, in the example proof, the statement of the theorem does introduce the constants a, b, and c and polynomial ax2 C bx C c, so some authors might just skip the first line of the proof. On the other hand, if the first line of the proof instead introduced the constants r, s, and t, the proof could have proceeded using these variables instead of a, b, and c. The same result would have been proved. So the “SET THE CONTEXT” of the proof makes the proof independent of the statement of the theorem being proved. Thus, for completeness, it is good to establish the habit of including the setting of the context at the beginning of each proof, at least until the student’s experience in proof writing has matured. Your choices of variables used to represent particular objects in the proof are not critically important to the structure or correctness of the proof, but there are
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certain variables that mathematicians associate with various uses, and sticking to these conventional choices simplifies the understanding of the proof because those variable choices bring with them a history of context that the reader will recognize. There are very few Algebra students who would recognize the Quadratic Formula p s ˙ s2 4rt if you gave them z D . Proofs about limits usually refer to the 2r variables and ı which represent small positive real numbers used in specific ways in the proof. Using these two variables in their traditional contexts makes the proofs easier to understand because the reader will expect these variables to play specific roles, just as they have in many other proofs the reader has seen. Seeing many examples of proofs will familiarize the novice proof writer with these traditional uses of variables. Suppose that the statement being proved indicates that every object satisfying the properties listed in the hypothesis of the theorem also satisfies some properties listed in the conclusion of the theorem. One generally structures a proof of such a statement by first selecting a generic object satisfying the properties listed in the hypothesis. The “ASSERT THE HYPOTHESIS” part of the proof is where the writer selects an arbitrary element satisfying the hypothesized properties. In the Quadratic Formula proof, it was assumed that x satisfied the quadratic equation ax2 C bx C c D 0. Other examples would be statements such as • • • • • •
Let n be any natural number bigger than 3. Let x be an element of set A. Let y be a root of the polynomial p.x/. Assume that the real valued function f has a zero at the point z. Suppose G and H are any two linesR that intersect at a point P. s Assume that the function f .s/ D 0 g.x/ dx is a differentiable function of s. In addition assume that 0 f .s/ 10 for all s 0.
It is possible that there are infinitely many objects which could play the role of the generically chosen object. But if an argument proves the result is true for this generic object, then the theorem will have been shown to hold for any object that could have played the role of the generic object, and, therefore, the theorem will have been proved for all objects satisfying the hypothesis. The Quadratic Formula proof addresses the one generic polynomial ax2 C bx C c and in doing so derives a formula that works for all quadratic polynomials including 5x2 17x C 126 and rx2 C sx C t. Often the reader of a proof will form a mental picture of the generic object being chosen. For example, after reading “Let n be any natural number bigger than 3,” the reader may think, “OK, how about n D 7?” As the proof progresses, the reader may take each statement of the proof and verify that it is valid and makes sense for their choice of n D 7. This helps the reader follow the logic of the proof and verifies that they are understanding what the proof is saying. The proof will be completed when it is shown that the generically chosen element satisfying the hypothesis of the theorem is, in fact, an element satisfying the conclusion of the theorem as stated in the “STATE THE CONCLUSION” part of the template. There will certainly need to be some statements placed between
2.2 Template for Proofs
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the original assertion of the hypothesis and the end of the proof that justify the conclusion of the theorem. Those statements make up the “LIST IMPLICATIONS” part of the template. In almost all cases, most of the body of the proof belongs to this list of statements. Each statement in the list should follow from definitions or be simple implications following from previous statements in the proof. In a wellwritten complete proof, the reader should easily see why each implication follows logically from other statements made earlier in the proof (Fig. 2.1). If an implication is not clear on its own, it will need some justification so the reader can follow the logic. The justification may just be a reminder of a key point made earlier in the proof (“as shown earlier, f is continuous at point a”) or a reminder of a well-known definition or theorem (“Since all continuous functions on the interval Œ0; 4 are R4 integrable there, it follows that f .x/dx exists.”) The given Quadratic Formula proof 0
contains six lines of implications. Each line follows easily from the line before using standard rules of Algebra, and any student familiar with the algebraic manipulations of equations will be able to understand these implications. In the fourth step of the b2 proof, the quantity 4a 2 is added to both sides of an equation. Although this step surely follows the rules of Algebra, it may not be clear to the reader of the proof why the step is important. As it turns out, this “completing the square” operation prepares for the factoring performed in the fifth step of the proof and is arguably the most clever step of the proof. A proof will often require a clever step such as this. The proof writer may have labored for years looking for the inspiration needed to find such a step, but the proof itself need only make clear the justification for what is being done and does not need to refer to the sweat that went into producing it. Some implications will be easy for the reader to follow without having to justify the step. Other statements may need some deeper explanation. Here is where the proof writer will need to consider the expertise of the target audience for the proof in order to decide how much detail to provide. How to make your proof “easy to follow” is only clear when you know for whom it is meant to be easy. For example, it b2 made sense to follow the line x2 C ba xC ac D 0 with the statement x2 C ba xC ac C 4a 2 D b2 4a2
because this just used the fact that you can add equal quantities to both sides of an equation to get a new equation that is equivalent. On the other hand, suppose you wish to combine a conditional statement on line 8 of a proof with the fact stated on line 15 of the proof in order to show that the hypothesis of that conditional is satisfied. This would allow the writer to state the conclusion of the conditional
Fig. 2.1 List of implications for P ! Q
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statement to get line 16 of the proof, but the reader may have to be reminded about which statements are being combined to get that conclusion. Sometimes the writer of a long or complicated proof will need to make a new definition or point out some new property that will be important later in the proof. Depending on the complexity of the new idea, the proof writer may want to include an example or two of objects satisfying the new definition or property. This will serve to help the reader understand the new concept or to verify that the reader is understanding the new concept. It is admirable to include such examples if the complexity of the proof can be made clearer. But in most other contexts, the proof should be kept short without the inclusion of unnecessary statements. If the intended readers are able to easily construct these examples on their own, then the examples should be left out of the proof. The remainder of this chapter will discuss proofs that follow this general proof template in contexts that the student should find easy to follow. It will also give an opportunity to present some definitions and notation that will be used in later chapters.
2.2.1 Exercises 1. If you were writing a proof of “All prime numbers greater than 2 are odd,” which of the following would be appropriate ways to begin the proof. (There may be more than one correct answer.) (a) (b) (c) (d) (e) (f) (g)
Let n be an odd prime number. Assume that all odd prime numbers are greater than 2. Let n be a prime number greater than 2. Assume that 2 is a prime number. Assume that n and k are integers with n > k > 2. The numbers 3, 5, 7, and 11 are prime numbers greater than 2. Let n be a number greater than 2 which is not prime.
2. If you were writing a proof of “The diagonals of a parallelogram bisect each other,” which of the following would be appropriate ways to begin the proof. (There may be more than one correct answer.) (a) (b) (c) (d) (e) (f)
Let ABCD be a parallelogram. Let ABCD be a quadrilateral whose diagonals bisect each other. Let ABCD be a parallelogram whose diagonals bisect each other. All rectangles are parallelograms. Assume that the diagonals of a parallelogram bisect each other. Assume that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.
2.3 Proofs About Sets
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3. If you were writing a proof of “Every cubic polynomial with real coefficients has at least one real root,” which of the following would be appropriate ways to begin the proof. (There may be more than one correct answer.) (a) Assume that every cubic polynomial with real coefficients has at least one real root. (b) Assume that p.x/ is a polynomial with at least one real root. (c) Assume that a, b, c, and d are real numbers with a ¤ 0, and let p.x/ D ax3 C bx2 C cx C d. (c) The polynomial x3 8 has exactly one real root at x D 2. (e) Let p.x/ be a cubic polynomial with real coefficients and q.x/ be a cubic polynomial with complex coefficients. (f) Let p.x/ be a cubic polynomial with real coefficients with real root r. Write an appropriate first sentence that would begin proofs of each of the following statements. 4. If m and n are relatively prime integers, then there exist integers x and y such that mx C ny D 1. 5. The three angle bisectors of any triangle intersect at a common point. 6. If a and b are real numbers with a b, and f is a function continuous on the closed interval Œa; b, then there is a real number M such that jf .x/j M for all x 2 Œa; b. u , ! u ! u .! 7. If ! v , and ! w are 3-dimensional vectors, then .! v / ! w D! v ! w /:
2.3 Proofs About Sets 2.3.1 Set Notation Most courses in mathematics discuss sets: sets of numbers, sets of points, sets of functions, sample spaces, and so forth. This should have given any Calculus student an intuitive understanding of sets. Many theorems in mathematics are statements about sets in disguise. For example, the statement that “If the function f is differentiable at a point, then f is continuous at that point” is equivalent to the statement “The set of functions differentiable at a point is a subset of the set of functions continuous at that point.” For the purposes of this text, it will be enough to define a set as a collection of elements. That is, elements are those objects that belong to sets, and the notation x 2 A says that x is an element of the set A, and x … A says that x is not an element of the set A. The set A is a subset of the set B, or A is contained in the set B, if each element of A is also an element of B in which case this fact is written as A B. Two sets, A and B, are equal if they have the same elements, that is, all the elements in the set A are in the set B, and all the elements in the set B are in the set A. Notationally, this says that A D B if and only if both A B and B A.
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There are many ways to express the contents of a set. One is to list the elements such as A D fa; b; cg or B D f1; 3; 5; 7; : : : g. Another way is to use set builder notation which states that the set consists of all elements satisfying a given property P.x/ and is written fx j P.x/g, or to emphasize that the elements of the set are also in set A, it is often written fx 2 A j P.x/g. Examples are fx j x > 0g, fy j y2 C3y2 > 7g, and ff j f is a function differentiable at x D 3g. Note that a set is determined by the elements that are in the set. Thus, f1; 2; 3g D f3; 2; 1g D f1; 2; 2; 3; 3; 3; 1; 2; 3g because all three of these sets contain exactly the same three elements. In some contexts, mathematicians will talk about a multiset which is an object similar to a set but allows elements of the collection to appear with different multiplicities. Thus, f1; 2; 3g and f1; 2; 2; 3; 3; 3; 1; 2; 3g would be different multisets even though, in the notation of sets, they are the same set. One special set is the empty set written as ; or fg and is the set that has no elements. In some contexts there is an understanding of a universal set, U, such that all other sets under consideration are subsets of U. For example, the sets A D f1; 2; 3g and B D f2; 4; 6; 8; : : : g can be thought of as subsets of the universal set U D f1; 2; 3; 4; : : : g. Take care not to confuse elements and subsets. Remember that sets are collections of elements and sets are subsets of other sets. It is possible that a set contains other sets as elements, but this would need to be explicitly clear from the definition of that set. It is correct to write 3 2 f1; 2; 3; 4; 5g, f1; 3; 5g f1; 2; 3; 4; 5g, and f1; 2g 2 f1; f1; 2g; f1; f1; 2ggg, but it is incorrect to write 3 f1; 2; 3; 4; 5g, f1; 2g 2 f1; 2; 3; 4; 5g, or ; 2 f1; 2; 3; 4; 5g. The student should be familiar with the following standard set operations. The union of sets A and B is A [ B D fx j x 2 A or x 2 Bg, and the intersection of sets A and B is A \ B D fx j x 2 A and x 2 Bg. When there is an understood universal set, U, it makes sense to refer to the complement of a set Ac D fx 2 U j x … Ag. It does not make sense to discuss the complement of a set if there is no understood universal set. For example, is f1; 2; 3gc D f4; 5; 6; 7; : : : g, or is it f: : : ; 3; 2; 1; 0; 4; 5; 6; 7; : : : g? For that matter, is your right shoe an element of f1; 2; 3gc ? The difference of two sets is AnB D fx 2 A j x … Bg, and, equivalently, if there is an understood universal set, AnB D A \ Bc . For example, if A D f1; 2; 3; 4; 5g and B D f2; 4; 6; 8g, then A [ B D f1; 2; 3; 4; 5; 6; 8g, A \ B D f2; 4g, AnB D f1; 3; 5g, and BnA D f6; 8g.
2.3.2 Exercises 1. Which of the following statements are true? (a) (b) (c) (d)
6 2 f1; 2; 3; : : : ; 10g. f3; 5g 2 f1; 2; 3; : : : ; 10g. ; 2 f1; 2; 3; : : : ; 10g. f6; 8g f1; 2; 3; : : : ; 10g.
2.3 Proofs About Sets
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(e) f2; 5; 5; 6g f1; 2; 3; : : : ; 10g. (f) f1; 2; 3; : : : ; 10g f1; 2; 3; : : : ; 10g. 2. Given A D f1; 3; 5; 7; 9; 11; 13g, B D f2; 3; 4; 5; 6g, and C D f1; 4; 7; 11; 14g evaluate each of the following expressions. (a) (b) (c) (d) (e) (f)
A[A A\B .A [ B/ \ C .B [ C/ \ A .A \ B/nC .AnB/ [ C
2.3.3 Proofs About Subsets There are many simple statements about sets which should be immediately obvious to students reading this text, but learning to write proofs for these types of statements will be instructive and useful in the proof writing discussed in the following chapters. Here are some of those simple statements that apply to all sets A, B, and C. Some Statements About All Sets A, B, and C A A [ B. A \ B A. An.B [ C/ .A [ B/nC. .A [ B/ \ C A [ .B \ C/. A [ B D B [ A, the Commutative Law of Union. A \ B D B \ A, the Commutative Law of Intersection. .A [ B/ [ C D A [ .B [ C/, the Associative Law of Union. .A\B/\C D A\.B\C/, the Associative Law of Intersection. A [ .B \ C/ D .A [ B/ \ .A [ C/, the Distributive Law of Union Over Intersection. • A \ .B [ C/ D .A \ B/ [ .A \ C/, the Distributive Law of Intersection Over Union. • .A [ B/c D Ac \ Bc , DeMorgan’s Laws. • .A \ B/c D Ac [ Bc , DeMorgan’s Laws.
• • • • • • • • •
The first four of these statements propose that one set is a subset of a second set. From the definition of subset, for A B to be true, it is required that for every x 2 A, x must also be in B. There is a standard template for proofs of statements of this form:
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TEMPLATE for proving A B for sets A and B • SET THE CONTEXT: State what is being assumed about the sets A and B. • ASSERT THE HYPOTHESIS: Let x 2 A. • LIST IMPLICATIONS: Use the properties of set A to show x belongs to set B. • STATE THE CONCLUSION: x 2 B. Therefore, by the definition of subset, A B. For example, how would one prove the statement “For all sets A and B, A A [ B?” Because this proof is supposed to apply to any sets A and B regardless of what properties they may possess, all that would be necessary for the “SET THE CONTEXT” part of the proof is a statement introducing to the reader the fact that the variables A and B will represent sets. Since A A [ B exactly when every element of A is also an element of A [ B, the “ASSERT THE HYPOTHESIS” part of the proof needs to select a generic element of the set A so that the proof can conclude that the generic element is an element of set A [ B. The first two lines of the proof read: Suppose that A and B are any two sets. Let x 2 A. The “LIST IMPLICATIONS” for this proof can be very short. It merely needs to show that the definition of “set union” implies that x is in the union A [ B. This completes the proof. PROOF: A A [ B. • • • • •
SET THE CONTEXT: Suppose that A and B are any two sets. ASSERT THE HYPOTHESIS: Let x 2 A. LIST IMPLICATIONS: Since x 2 A, it is true that x 2 A or x 2 B. By the definition of set union x 2 A [ B. STATE THE CONCLUSION: Therefore, by the definition of subset, A A [ B.
Do the statements of this proof have to appear in exactly this order using exactly these words? Of course not. There can be many variations in what makes up a good proof. But it does not hurt to review why these statements make a good proof. The first line Suppose that A and B are any two sets just makes it clear to the reader that the variables A and B can be used to represent any two sets. Here is where the reader of the proof may well mentally choose two sets so that when reading the remainder of the proof, the reader can verify that the statements make sense when applied to those two sets. The second line Let x 2 A is required because by the definition of “subset,” one must show that each element of A is also an element of A [ B, so selecting an arbitrary element of A is the natural way to do this. The next line Since x 2 A, it is true that x 2 A or x 2 B is just a statement of logic that says if statement p is true, then statement p or q is also true. Of course, this particular p or q statement is exactly the definition of x being a member of A [ B, which is exactly what is needed to complete the proof.
2.3 Proofs About Sets
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Could one have interchanged the third and fourth lines of this proof? Well, yes; the proof would be complete if that were done, but the fact that the definition of set union is invoked right after its conditions are verified makes the statements of the proof flow smoothly. The reader facing the definition of set union in line three might wonder why that definition is being shown at that point. By placing that statement as the fourth statement where the proof reader has just seen that x 2 A or x 2 B, the proof reader will immediately see that the definition of set union applies. Note that each of the five statements in the proof has been placed on a separate line in the display box. This has been done merely to facilitate the discussion about that proof. In practice, there is no requirement that these statements appear on a separate lines. The second statement about all sets is A \ B A. This can be proved using the same proof template as the first statement. Since this statement also applies to any two sets A and B, the first line of this proof will be the same as the first line of the previous proof. Because the assertion of the statement being proved is that A \ B is a subset of another set, the “ASSERT THE HYPOTHESIS” line of the proof would change to the assertion that x belongs to A \ B. After reading this second line, what does the proof reader know about x? Only that x belongs to the intersection of two sets. Thus, that only direction that the proof can proceed is to invoke the definition of set intersection to make the additional assertion that x 2 A and x 2 B. This is a statement of the form p and q, so logic allows the assertion that p is true, or, in this case, that x 2 A. This is the required “STATE THE CONCLUSION” statement, and the complete proof would be PROOF: A \ B A. • SET THE CONTEXT: Suppose that A and B are any two sets. • ASSERT THE HYPOTHESIS: Let x 2 A \ B. • LIST IMPLICATIONS: By the definition of set intersection, x 2 A and x 2 B. • Thus, x 2 A. • STATE THE CONCLUSION: Therefore, by the definition of subset, A \ B A. For a more substantial example, consider the third of the list of statements about sets An.B [ C/ .A [ B/nC. A proof of this statement will need to refer to the definition of set difference as well as the definitions of set union and subset. Since the statement being proved involves three sets, the “SET THE CONTEXT” part of the proof will need to refer to all three sets. The “ASSERT THE HYPOTHESIS” statement will need to select an arbitrary element from An.B [ C/. To emphasize that the choice of which variable to use is arbitrary, this time use y rather than x to represent the arbitrarily chosen element. Once it is known that y 2 An.B [ C/, the only property of y that can be used is the fact that y is a member of a set difference. Thus, this would be a good time to invoke the definition of set difference. That assures that y 2 A and y … .B [ C/. At that point one can use the definition of
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2 The Basics of Proofs
set union to conclude that since y … .B [ C/ that y … B and y … C. Now these facts can be combined to get the “STATE THE CONCLUSION” statement required by the proof template. The complete proof would be PROOF: An.B [ C/ .A [ B/nC. • SET THE CONTEXT: Suppose that A, B, and C are any three sets. • ASSERT THE HYPOTHESIS: Let y 2 An.B [ C/. • LIST IMPLICATIONS: By the definition of set difference, y 2 A and y … .B [ C/. • By the definition of set union y cannot be an element of either set B or set C, or it would be in B [ C. • Also by the definition of set union, since y 2 A, y is also a member of A [ B. • Now, y 2 .A [ B/ and y … C, so by the definition of set difference, y 2 .A [ B/nC. • STATE THE CONCLUSION: Therefore, by the definition of subset, An.B [ C/ .A [ B/nC.
2.3.4 Exercises Write proofs for each of the following statements. 1. For all sets A, B, and C, .A \ B/ \ C A \ C. 2. For all sets A, B, and C, .A \ B/ \ .A \ C/ B \ C. 3. For all sets A, B, and C, .AnB/ \ .AnC/ An.B \ C/.
2.3.5 Proofs About Set Equality Let A and B be sets. From the definition of set equality it follows that one can prove A D B by proving the two separate facts A B and B A. That suggests the following proof template for proving that two sets are equal.
2.3 Proofs About Sets
23
TEMPLATE for proving A D B for sets A and B • SET THE CONTEXT: Make statements about what is being assumed about sets A and B. PART 1: SHOW A B. • ASSERT THE HYPOTHESIS: Let x 2 A. • LIST IMPLICATIONS: Use the properties of set A to show x belongs to set B. • CONCLUDE PART 1: x 2 B. Therefore, by the definition of subset, A B. PART 2: SHOW B A. • ASSERT THE HYPOTHESIS: Let x 2 B. • LIST IMPLICATIONS: Use the properties of set B to show x belongs to set A. • CONCLUDE PART 2: x 2 A. Therefore, by the definition of subset, B A. • STATE THE CONCLUSION: Therefore, because A and B are subsets of each other, by the definition of set equality, A D B. Is it correct to use the same variable x in both parts of the above proof template? Yes, since the use of the variable x is only important in the context of showing A B or B A, there is little chance that the reader will be confused by these two uses of the same variable. On the other hand, there would be nothing incorrect about using the variable x to represent the element of set A in the first part of the proof and to use the variable y to represent the element of set B in the second part of the proof. Using the same variable has the advantage that it is used the same way in both parts of the proof, that is, to represent an element of a set that is being shown to also be an element of a second set. Is it correct that the variables A and B are used to represent the sets in both parts of the proof? Could, for example, the first part of the proof use sets A and B, and the second part of the proof use sets C and D? Here the answer is that it is very important to use the same variables in both parts of the proof. To show A D B it must be shown that A B and B A for the same pair of sets A and B. Showing A B and C D does not let one conclude that A D B. After introducing A and B in the “SET THE CONTEXT” part of the proof, it would be wrong to change the use of these variables later in the proof or to change which variables were representing the two sets. Consider how to write proofs of three of the example statements: • A [ B D B [ A, the Commutative Law of Union. • .A \ B/ \ C D A \ .B \ C/, the Associative Law of Intersection. • .A [ B/c D Ac \ Bc , DeMorgan’s Law. The first proof follows easily from the fact that in logic the statements p or q and q or p are equivalent. This leads to the proof
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2 The Basics of Proofs
PROOF: A [ B D B [ A. • SET THE CONTEXT: Suppose that A and B are any two sets. PART 1 A [ B B [ A • • • • •
ASSERT THE HYPOTHESIS: Let x 2 A [ B. LIST IMPLICATIONS: By the definition of set union, x 2 A or x 2 B. Thus, x 2 B or x 2 A. By the definition of set union x 2 B [ A. CONCLUDE PART 1: Hence, from the definition of subset, it follows that A [ B B [ A.
PART 2 B [ A A [ B • • • • •
ASSERT THE HYPOTHESIS: Now suppose that x 2 B [ A. LIST IMPLICATIONS: By the definition of set union, x 2 B or x 2 A. Thus, x 2 A or x 2 B. By the definition of set union x 2 A [ B. CONCLUDE PART 2: Hence, from the definition of subset, it follows that B [ A A [ B.
• STATE THE CONCLUSION: Therefore, because A [ B and B [ A are subsets of each other, by the definition of set equality A [ B D B [ A. Note that the “PART 1” and “PART 2” labels have been included in the above display as guides to the student, but they are not required elements of the proof itself. This proof can be shortened. Since the second part of the proof is identical to the first part of the proof except that the roles of the sets A and B are interchanged, one might save the reader from having to think through the details of the second half of the proof which are identical to the details of the first half. The proof could be written as PROOF: A [ B D B [ A. • • • • • • • •
Suppose that A and B are any two sets. Let x 2 A [ B. By the definition of set union, x 2 A or x 2 B. Thus, x 2 B or x 2 A. By the definition of set union x 2 B [ A. Hence, from the definition of subset, it follows that A [ B B [ A. Similarly, one can conclude that B [ A A [ B. Therefore, since A [ B and B [ A are subsets of each other, by the definition of set equality A [ B D B [ A.
In fact, the first half of the proof is the second half of the proof. The first half of the proof shows that A[B B[A for any two sets A and B. In particular, that proof applies when the roles of the two sets are interchanged; just let the variable A in the
2.3 Proofs About Sets
25
first part of the proof represent the set B from the second part of the proof, and let the variable B in the first part of the proof represent the set A from the second part of the proof. The Associative Law of Intersection refers to three sets and requires repeated use of the definition of “set intersection.” The definition is used to break down the statement x 2 .A \ B/ \ C into the three simple statements x 2 A, x 2 B, and x 2 C and then these facts are put back together to form the needed x 2 A\.B\C/. Again, the proof needs two parts. The result is PROOF: .A \ B/ \ C D A \ .B \ C/. • Suppose that A, B, and C are any three sets. PART 1 .A \ B/ \ C A \ .B \ C/ Let x 2 .A \ B/ \ C. By the definition of set intersection, x 2 .A \ B/ and x 2 C. Also, by the definition of set intersection, x 2 A and x 2 B. Thus, x 2 A, x 2 B, and x 2 C. Since x 2 B and x 2 C, by the definition of set intersection x 2 B \ C. Since x 2 A and x 2 B \ C, by the definition of set intersection x 2 A \ .B \ C/. • Hence, from the definition of subset, it follows that .A \ B/ \ C A\ .B \ C/.
• • • • • •
PART 2 A \ .B \ C/ .A \ B/ \ C Now, let x 2 A \ .B \ C/. By the definition of set intersection, x 2 A and x 2 B \ C. Also, by the definition of set intersection, x 2 B and x 2 C. Thus, x 2 A, x 2 B, and x 2 C. Since x 2 A and x 2 B, by the definition of set intersection x 2 A \ B. Since x 2 A \ B and x 2 C, by the definition of set intersection x 2 .A \ B/ \ C. • Hence, from the definition of subset, it follows that A\.B\C/ .A\B/\C. • Therefore, because .A \ B/ \ C and A \ .B [ C/ are subsets of each other, by the definition of set equality .A \ B/ \ C D A \ .B \ C/.
• • • • • •
The two DeMorgan’s Laws are useful because they tell how to simplify the complement of a set formed by a combination of unions and intersections of sets. Proving these laws can follow the template for showing set equality. The proofs will need to refer to the definitions of set union, set intersection, and set complement. The order in which these definitions are invoked follows from what is known at that point of the proof. For example, if you know that x 2 .A [ B/c , then the only way to make progress in the proof is to apply the definition of set complement because the only attribute known about the set is that it is the complement of some other set.
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Fig. 2.2 .A [ B/c is equal to Ac \ B c
A
B
(A ∪ B)C
A
B
AC ∩ BC
Yes, that other set is a union of two sets, but there is no way to use that information at this point of the proof because complementation was performed after the union was taken (Fig. 2.2). PROOF: .A [ B/c D Ac \ Bc . • Suppose that A and B are any two sets. PART 1 .A [ B/c Ac \ Bc Let x 2 .A [ B/c . By the definition of set complement, x … .A [ B/. If x 2 A or x 2 B, then x 2 A [ B which is false. Thus, x … A and x … B, so by the definition of set complement, x 2 Ac and x 2 Bc . • By the definition of set intersection x 2 Ac \ Bc . • Hence, from the definition of subset, it follows that .A [ B/c Ac \ Bc .
• • • •
PART 2 Ac \ Bc .A [ B/c Now, let x 2 Ac \ Bc . By the definition of set intersection, x 2 Ac and x 2 Bc . Thus, by the definition of set complement, x … A and x … B. If x 2 A [ B, then by the definition of union, it would follow that x 2 A or x 2 B which is false. • Thus, x … A [ B, and, by the definition of set complement, x 2 .A [ B/c . • Hence, from the definition of subset, it follows that Ac \ Bc .A [ B/c . • Therefore, because Ac \ Bc and .A [ B/c are subsets of each other, by the definition of set equality .A [ B/c D Ac \ Bc .
• • • •
2.3.6 Exercises Give that A, B, and C are sets, write proofs for each of the following statements. 1. A \ B D B \ A. 2. A \ .BnA/ D ;. 3. .AnB/ [ .BnA/ D .A [ B/n.A \ B/.
2.4 Proofs About Even and Odd Integers
4. 5. 6. 7.
27
.A [ B/ [ C D A [ .B [ C/. .A [ B/ \ C D .A \ C/ [ .B \ C/. .A \ B/ [ C D .A [ C/ \ .B [ C/. An.B [ C/ D .AnB/ \ .AnC/.
2.4 Proofs About Even and Odd Integers 2.4.1 Definitions of Even and Odd Integers You are already very familiar with the natural numbers, N D f1; 2; 3; 4; : : : g, which are sometimes called the counting numbers or whole numbers. By adding zero and the negative natural numbers to this set, one obtains the integers, Z D f: : : ; 3; 2; 1; 0; 1; 2; 3; : : : g. The natural numbers are often referred to as the positive integers. Much of a student’s first study of mathematics is concerned with these two sets of numbers. By a very young age most people are already familiar with even and odd integers and some of their properties. This section will construct proofs of some of these properties both because the student will feel very comfortable with the concepts and because it allows for the introduction of some basics about how to write proofs. Before proceeding with proofs, though, it is necessary that there is agreement on the definitions of even and odd integers. Indeed, there are many possible definitions of even integers: n 2 Z is an even integer if • the decimal representation of n has a ones digit equal to 0, 2, 4, 6, or 8. • n is either 0 or the prime factorization of n contains a factor of 2. • there is an integer k such that npD 2k. • in is a real number, where i D 1. • the number .1/n is positive. • 9n 1 .mod 10/. • sin. n / D 0. 2 • n2 2 Z. Which of these definitions should be used when writing proofs about even and odd integers? Actually, since all the definitions are equivalent, one could adopt any one of these definitions and then prove theorems that show that all the other definitions are equivalent to the chosen definition. This is not an unusual situation in mathematics, especially for a concept as elementary as even integers. But it turns out that one of these definitions is particularly well suited for writing proofs, and that is, n 2 Z is even if there is a k 2 Z such that n D 2k. This makes a useful
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definition because it provides a fairly easy way to check whether a given integer is even, and because knowing that a number n is even immediately gives you a number k for which n D 2k, and that is a powerful tool for proving facts about even integers. For this reason, this chosen definition is called the working definition, that is, it is the definition easiest to apply in the wide variety of contexts. It is the definition chosen from which all other properties of even numbers can be derived. A similar discussion could take place about how to define odd integers. The working definition is that n 2 Z is odd if there is a k 2 Z such that n D 2k C 1. There is a long list of facts you could prove about even and odd numbers. Facts About Even and Odd Integers • • • • • •
Every integer is either even or odd. No integer is both even and odd. n 2 Z is even if and only if n C 1 2 Z is odd. The sum of any two even integers is even. The sum of any two odd integers is even. The sum of an even integer and an odd integer is odd. • The product of two odd integers is odd. • The product of two integers is odd only if both of the factors are odd.
Together, the first two of these facts say that each integer is either even or odd but not both. This says that the sets of even and odd integers form a partition of Z, that is, the sets are disjoint and the union of the sets is all of Z. Some authors require that all the sets of a partition be nonempty as in the case with even and odd integers. So why is it that every integer is either even or odd? This depends on the Division Algorithm that states that if m; n 2 Z with n > 0, then there are unique q; r 2 Z with 0 r < n such that m D nq C r. In this case q is called the quotient of the division, and r is called the remainder of the division. Using the Division Algorithm, any integer m can be divided by 2 giving a quotient and remainder where the remainder is either 0 or 1. If the remainder is 0, then m D 2q for integer q implying that m is even, and if the remainder is 1, then m D 2q C 1 for integer q implying that m is odd.
2.4.2 Proofs About Even and Odd Integers How can these ideas be used to write a good proof of Every integer is either even or odd? First it is easier to reword the statement as If m 2 Z, then either m is even or m is odd. This is a conditional statement, so the natural way to begin a proof is to assume that the hypothesis of the statement is satisfied, that is, that m is an integer. Now apply the Division Algorithm to get the quotient q and remainder r guaranteed by the algorithm. Finally, the value of r shows that m either satisfies the
2.4 Proofs About Even and Odd Integers
29
definition of being an even integer or the definition of being an odd integer. The result would be PROOF: Ever integer is either even or odd. • Let m be an integer. • By the Division Algorithm there are integers q and r with 0 r < 2 such that m D 2q C r. • If r D 0, then m D 2q for integer q which means that m satisfies the definition for being even. • If r D 1, then m D 2q C 1 for integer q which means that m satisfies the definition for being odd. • Since r must be either 0 or 1, it follows that every integer is either even or odd. Next consider the how to prove the statement The sum of any two odd integers is even. The statement concerns the sum of any two odd integers, so the proof reader would expect the proof to consider two arbitrarily chosen odd integers. Once two odd integers are chosen, the definition of odd integer should be invoked because, at that point, that is the only information that is known about the two integers. Finally, a little algebra will help to show that the sum of these two odd integers satisfies the definition of even integer. Here is an attempt to write such a proof that makes several common proof writing errors. PROOF ATTEMPT: The sum of any two odd integers is even. The two integers are odd, so each has the form 2k C 1. The sum of these two integers is .2k C 1/ C .2k C 1/ D 4k C 2. k could be even or odd. The number 2 is even since it is 2 1. 4k is even since it is 2 2k. The sum of two even numbers is even, so the sum of 4k and 2 is an even number. • Therefore, the sum of two odd integers is always even.
• • • • • •
Here are some complaints about the above proof attempt. • The proof begins talking about two integers, but the proof reader has not yet been introduced to these integers and does not know what two integers are being discussed. The proof is missing a “SET THE CONTEXT” sentence to introduce the idea of starting with any two odd integers. • The proof uses the variable k without introducing what that variable represents. The proof requires that k be an integer, but the fact that k is an integer is not stated anywhere. As far as the proof reader knows, k could be any complex number. Later, the proof claims that 2k is an integer which is needed to show 4k is an even integer. Without knowing that k is an integer, it does not follow that 2k is also an integer. • The definition of odd integer allows you to take an odd integer and represent it as 2k C 1, where k is another integer. To apply this definition, then, the proof should
30
•
•
•
•
2 The Basics of Proofs
start with an odd integer, say m, and then represent it as 2kC1 rather than starting with 2k C 1. The subtle point is that one should start with odd integer and use its definition to move on to 2k C 1 rather than starting with 2k C 1 which jumps the gun. The reader of the proof could wonder whether 2k C 1 could represent a generic odd integer. Well, it can, but this takes some thought which can be avoided by starting with an odd integer m and then using the definition of odd to select the integer k such that m D 2k C 1. The definition of “odd integer” does refer to “2k C 1,” but it is more precise. It does not say “has the form.” It says that there is an integer k such that the odd number equals 2k C 1: It is a major error to allow both odd integers to equal 2k C 1 for the same number k. The only way this can happen is for the two odd integers themselves to be equal. Thus, this “proof” only applies to a small subset of cases where one adds two identical odd integers together such as 3 C 3 or 117 C 117. The statement k could be even or odd is certainly correct, but it does not contribute to the proof. It is a statement about items in the proof that is not part of the proof. Occasionally, one will make a definition as part of a long proof, and then give some examples to help the reader understand that definition. But if a statement is not needed either as a critical step in a proof or an important illustration to aid the understanding of the proof, then the statement should be left out of the proof because it distracts from the proof and complicates it. The statement The sum of two even numbers is even is correct, but it has not been proved yet, at least in this text, and is equivalent in difficulty to proving the corresponding statement about the sum of odd integers. Thus, it is not appropriate to use the result about sums of even integers to prove one about the sum of odd integers.
Considering these ideas, one can construct a better proof. PROOF: The sum of any two odd integers is even. • Let m and n be two odd integers. • From the definition of odd integer, there is an integer k1 such that m D 2k1 C 1 and an integer k2 such that n D 2k2 C 1. • Then m C n D .2k1 C 1/ C .2k2 C 1/ D 2.k1 C k2 C 1/. • Since k1 and k2 are integers, so is k1 C k2 C 1. • Thus, the sum m C n is equal to twice an integer, so by the definition of even integer, m C n is even. • Therefore, the sum of any two odd integers is always even. The form of this proof can be copied almost word for word to get a similar proof of the statement The product of two odd integers is odd.
2.5 Basic Facts About Real Numbers
31
PROOF: The product of any two odd integers is odd. • Let m and n be two odd integers. • From the definition of odd integer, there is an integer k1 such that m D 2k1 C 1 and an integer k2 such that n D 2k2 C 1. • Then mn D .2k1 C 1/.2k2 C 1/ D 4k1 k2 C 2k1 C 2k2 C 1 D 2.2k1 k2 C k1 C k2 / C 1. • Since k1 and k2 are integers, so is 2k1 k2 C k1 C k2 . • Thus, the product mn is equal to one more than twice an integer, so by the definition of odd integer, mn is odd. • Therefore, the product of any two odd integers is always odd.
2.4.3 Exercises Write proofs for each of the following statements. 1. 2. 3. 4. 5.
The sum of any two even integers is even. The product of an even integer and an odd integer is even. The difference of an even integer and an odd integer is odd. If the product of two integers is odd, then both of the integers must have been odd. The sum of any four consecutive integers is even.
2.5 Basic Facts About Real Numbers 2.5.1 Ordered Fields Many of the theorems of Calculus involve properties of the real numbers. Some of these properties are subtle, so it is essential to understand this important set of numbers. Already introduced are the sets of natural numbers, N, and the integers, Z. Also of importance is the set of rational numbers, Q D f mn j m; n 2 Z; n ¤ 0g. This definition comes with the understanding that the two rational numbers mn and ab are equal whenever mb D na. Thus, there are always infinitely many representations for each rational number. For all rational numbers r ¤ 0, one can always find relatively prime integers m and n with n > 0 such that r D mn . Together with an agreement to write the rational number 0 as 01 , each rational number has a unique lowest terms representation. The set of rational numbers is more than a set of fractions with integers for numerators and denominators. It also comes with the two binary operations of addition (C) and multiplication () and with the order relation less than (), greater than or equal to (), and less than or equal to () in the obvious ways, that is, r > s whenever s < r, r s whenever either r > s or r D s, and r s whenever either r < s or r D s. There are many other ordered fields, and it is constructive to consider how to justify the fifteen ordered field axioms for a different ordered field. For example, p the set T D fr C s 2 j r; s 2 Qg is an ordered field p using the usual p addition and multiplication operations. For two elements a C b 2 and c C d 2 in T, define p p p addition as .a C b 2/ C .c C d 2/ D .a C c/ C .b C d/ 2 and multiplication p p p as .a C b 2/ .c C d 2/ D .ac C 2bd/ C .ad C bc/ 2. To define the less than p p p relation you would want .a C b 2/ < .c C d 2/ whenever a c < .d b/ 2 p which can be checked by squaring both a c and .d b/ 2, although you will need topalso considerpthe signs of a c and d b. Thus, the definition becomes .a C b 2/ < .c C d 2/ if one of the following holds: • a c < 0 and 0 < d b, • 0 < a c, 0 < d b, and .a c/2 < 2.d b/2 , or • a c < 0, d b < 0, and .a c/2 > 2.d b/2 . It is fairly easy to check that T is an ordered field. The only field axiom which does not follow immediately from the properties of rational p numbers is the inverse axiom for multiplication. You should verify that for a C b 2 ¤ 0, its multiplicative inverse is a b p C 2 a2 2b2 a2 2b2 which is in T. The order axioms take more work to verify due to the complicated definition of less than. For example, to verify the less than relation p p works correctly 2, c C d 2, and with addition, one would begin with three elements of T, a C b p p p e C f p2 where it ispgiven that a C p b 2 < cpC d 2. One needs to compare .a C b 2/ C .e C f 2/ with .c C d 2/ C .e C f 2/. To do this, one compares the values of .a C e/ .c C e/ D a c and .d C f / .b C f / D d b. But this reduces to comparing p a c and p d b which are known to satisfy the correct conditions because a C b 2 < c C d 2 was given. Every ordered field satisfies a long list of simple properties that you will associate with facts learned in Arithmetic and Algebra. Here are some of those properties.
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2 The Basics of Proofs
Some Properties Obeyed By All Ordered Fields Let r; s; t all be elements of ordered field F . Then 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.
r 0 D 0. If r C t D s C t, then r D s. If r t D s t and t ¤ 0, then r D s. .r/ D r. If r ¤ 0, then 11 D r. r r D s if and only if r D s. r D .1/ r. .r/ C .s/ D .r C s/. .r/ .s/ D r s. If r < s and t < 0, then s t < r t. If r ¤ 0, then r2 > 0. 0 < 1. If 0 < r, then 0 < 1r . If 0 < r < s, then 0 < 1s < 1r . If n is any natural number and r > 1, then rn < rnC1 .
The reader may wish to prove some of these properties by applying the axioms. This book will not dwell on these proofs since the techniques used in proving them are not essential for writing most proofs in Analysis. Two simple proofs are given here as examples. PROOF: If r is any element of the field F, then r 0 D 0. Let r be an element of field F . Since 0 is the additive identity of F , 0 D 0 C 0. Then r 0 D r .0 C 0/. By the Distributive Law, r 0 D r 0 C r 0. By adding r 0 to each side of this equality, one gets 0 D r 0 r 0 D .r 0 C r 0/ r 0 D r 0 C .r 0 r 0/ D r 0 C 0 D r 0. • Therefore, for any r 2 F , r 0 D 0.
• • • • •
The next theorem essentially says that if .1/ r has the same properties as r, it must equal r.
2.5 Basic Facts About Real Numbers
35
PROOF: If r is any element of the field F, then r D .1/ r. Let r be an element of field F . Then .1/ r D = = = = = = =
.1/ r C 0 .1/ r C .r C r/ Œ.1/ r C r C r Œ.1/ r C 1 r C r Œ.1/ C 1 r C r 0 r C r 0 C r r
Additive Identity Additive Inverses Associative Law of Addition Multiplicative Identity Distributive Law Additive Inverses r0D0 Additive Identity
Therefore, .1/ r D r for every r 2 F . Note that every ordered field F will contain a copy of Q. This follows since 0; 1 2 F , and if n is a natural number in F , then n C 1 2 F . Thus, it follows by mathematical induction that n 2 F for all n 2 N. Moreover, since 0 < 1, it follows for each n 2 N that n D n C 0 < n C 1 showing that all natural numbers are distinct elements of F . The existence of the negatives of all numbers in F implies that the integers is a subset of F , and the existence of reciprocals implies that all of Q lies in F . There are fields which are not ordered fields, and some of them do not contain copies of Q. Indeed, there are finite fields as well as infinite fields that do not contain Q or even N.
2.5.2 The Completeness Axiom and the Real Numbers There are infinitely many ordered fields. The real numbers, R, is special because it includes every number that is considered a possible “distance from zero,” either positive, negative, or zero. An easy way to ensure that R contains every possible distance is to require it to satisfy the Completeness Axiom. This axiom considers nonempty subsets of an ordered field, F , (actually, any ordered set would do). A subset S F is said to be bounded above if there is an M 2 F such that all x 2 S satisfy x M. In this case, M is called an upper bound of S. Similarly, S F is bounded below by lower bound K 2 F if all x 2 S satisfy x K. If S F is both bounded above and bounded below, then S is said to be bounded. If M is an upper bound for a set S, and it is less than or equal to every upper bound of S, then M is the least upper bound of S. Similarly, if K is a lower bound for a set S, and it is greater than or equal to every lower bound of S, then K is the greatest lower bound p of S. For example, if S is the interval .1; 5 D fx j 1 x < 5g, then 10, 6, and 30
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2 The Basics of Proofs
are all upper bounds of S, but 5 is the least upper bound of S. Also, 2, 0, and 12 are all lower bounds of S, but 1 is the greatest lower bound of S. One often uses the notation l.u.b..S/ or sup.S/ to represent the least upper bound or supremum of S and g.l.b..S/ or inf.S/ to represent the greatest lower bound or infimum of S. Axioms for the Real Numbers The real numbers, R, is an ordered field that satisfies The Completeness Axiom: Every nonempty set S R which is bounded above has a least upper bound in R. Note, for example, that the set S D fx 2 Q j x2 < 7g is a nonempty subset of Q which is bounded above by 4, 3, and 2.7, but there is no element of Q which is a least upper bound of p S. The set of real numbers, though, does contain a least upper bound of S, namely 7. The Completeness Axiom is sometimes called the Least Upper Bound Principle. The Completeness Axiom comes up frequently in proofs about the real numbers to show that numbers with particular properties exist. For example, consider the two theorems, the Archimedian Principle and the Existence of Square Roots. Both of these theorems are easily understood, but they cannot be proved without using the Completeness Axiom. The Archimedian Principle states that for every real number r there is a natural number greater than r. It can be proved using a proof by contradiction. The proof makes the assumption that there is a real number greater than every natural number and uses this to derive a contradiction, a statement that is false. Because one cannot derive a false statement from a true statement, the assumption most recently made in the proof must be a false statement, and you can conclude that no real number exists that is greater than every natural number. PROOF (Archimedian Principle): If r 2 R, then there exists n 2 N such that r < n. • Suppose that there is an r 2 R such that r > n for every n 2 N. • Then the set N is a nonempty subset of R with an upper bound, so by the Completeness Axiom, N has least upper bound M. • Then M 1 < M, so M 1 is not an upper bound for N. • Thus, there is a k 2 N with the property that k > M 1. • But then k C 1 is also in N, yet k C 1 > .M 1/ C 1 D M where M is an upper bound for N. • This is a contradiction since no element of a set can be greater than an upper bound for that set. • Therefore, the assumption that r > n for every n 2 N must be false, and for every r 2 R there must be at least one n 2 N with n > r.
2.5 Basic Facts About Real Numbers
37
You may not have ever doubted that every nonnegative real number has a square root, but this is a fact that can be proved using the axioms for the real numbers. It is a nice application of both the Trichotomy Property and the Completeness Axiom. Given a positive real number, r, the proof constructs the set S D fx 2 R j x2 rg and then uses the Completeness Axiom to exhibit a value, s, equal to the least upper bound of S. Then it shows that s2 cannot be greater than r and cannot be less than r, so by the Trichotomy Property, s2 must equal r. In particular, the proof first assumes that s2 > r and shows that there is a number y > 0 such that the square of s y is also greater than r. This shows that s y is an upper bound for S which contradicts the fact that s is the least upper bound of S. The 2 proof magically suggests that y D s 4sr works. Where did this magical expression for y come from? It came from considering what property you would want such a y to have. If you want .s y/2 > r, this suggests that you want s2 2sy C y2 > r. This inequality is quadratic in y and has an unnecessarily messy solution. But one of the most important lessons about writing proofs in Analysis is that one can often be a little sloppy when trying to show that an inequality holds. Here, for example, rather than finding a y such that s2 2sy C y2 > r, it would be sufficient to find a y such that s2 2sy > r, because if s2 2sy > r, then certainly the needed s2 2sy C y2 > r also holds. The advantage of making this change is that the inequality s2 2sy > r 2 2 is very easy to solve for y yielding y < s 2sr . Thus, the value y D s 4sr ought to work fine, and, hence, the magic is demystified. Of course, there are many other possible values of y that would also have worked in this proof, but only one value for y is needed. After showing that s2 > r cannot be true, the proof assumes that s2 < r and shows that there is a number y > 0 such that the square of s C y is less than r. This shows that s C y is in S which contradicts the fact that s is an upper bound of 2 S. Again, the proof just suggests setting y D rs . Can you figure out where this 4s expression for y came from? Indeed, the calculation is similar to the one above. You need .s C y/2 r, so s2 C 2sy C y2 r. It is simpler if y2 could be replaced by 2sy. If you assume y 2s, it allows you to conclude y2 2sy so that .s C y/2 D s2 C 2sy C y2 s2 C 2sy C 2sy D s2 C 4sy. You then want a y that satisfies 2 gives the needed value of y (Fig. 2.3). Putting s2 C 4sy r. Thus, the value y D rs 4s these ideas together gives the following proof.
S
s2 – r s 4s
– s2 S s r 4s
Fig. 2.3 Showing the least upper bound of S is s D
p r
S
s
38
2 The Basics of Proofs
PROOF (Existence of Square Roots): If r 2 R and r 0, then there exists an s 2 R such that s2 D r. Let r 0 be a real number. If r D 0, then 02 D r and 0 satisfies the needed condition. So assume that r > 0. Let S D fx 2 R j x2 rg. S is nonempty since 0 2 S. S is bounded above by r C 1 since x > .r C 1/ implies x2 > r2 C 2r C 1 > r so x … S. • Thus, by the Completeness Axiom, S has a least upper bound s. • By the Trichotomy Property, either s2 > r, s2 < r, or s2 D r.
• • • • • •
2
If s2 > r, note that y D s 4sr > 0, and .s y/2 D s2 2sy C y2 > s2 2sy D 2 2 s2 2s s 4sr D s 2Cr > r. Because .s y/2 > r, it follows that s y < s is an upper bound of S. This contradicts the fact that s is the least upper bound of S. Therefore, s2 > r must be false. 2
If s2 < r, let y be the smaller of 2s and rs . Then y > 0, and 4s 2 2 2 2 2 2 D r. .s C y/ D s C 2sy C y s C 2sy C 2sy D s C 4sy s2 C 4s rs 4s Because .sCy/2 r, it follows that sCy 2 S and sCy > s. This contradicts the fact that s is an upper bound of S. Therefore, s2 < r must be false. • Thus, it must be true that s2 D r which proves that for every real number r 0 there is an s 2 R with s2 D r.
2.5.3 Absolute Value, the Triangle Inequality, and Intervals The concept that separates the area of Mathematics known as Analysis from other branches such as Algebra, Topology, Set Theory, and Combinatorics is the idea of distance. In the real numbers, one canmeasure distance by using the absolute x if x 0 value function which is defined as jxj D : For a real number x, the x if x < 0 absolute value of x can be thought of as the distance that x is from the real number 0. Note that for all x 2 R it holds that jxj x jxj. If k > 0, then ˇ the set fx j jxj < kg is the same ˇas the set fx j k < x < kg. Similarly, the set fx ˇ jxj > kg is the same as the set fx ˇ k > x or x > kg. The distance between two real numbers x and y can be defined as jx yj. Note that this distance is positive unless x D y. One property of the absolute value function used frequently in proofs in Analysis is the triangle inequality which states that for all x; y 2 R, jx C yj jxj C jyj. The name of this inequality comes from geometry where it is known that the sum of the
2.5 Basic Facts About Real Numbers
39
Fig. 2.4 Triangle inequality
y
x+y
x lengths of two sides of a triangle always exceeds the length of the third side of the triangle (Fig. 2.4). One simple proof of the triangle inequality is PROOF (Triangle Inequality): jx C yj jxj C jyj • Let x and y be elements of R. • Then jxj x jxj and jyj y jyj. Adding these inequalities yields .jxj C jyj/ x C y .jxj C jyj/. • This last inequality is equivalent to jx C yj jxj C jyj. A subset S contained in R is called connected if it has the property that for any two numbers in S, all the numbers between those two numbers are also in S. More precisely, S is connected if for all x; y 2 S with x < y, it follows that z 2 S for all z with x < z < y. Informally, this means that there are no holes in the set S. Another word for a connected set of real numbers is an interval. If a < b are real numbers, all of the following sets are intervals. Intervals of Real Numbers
; empty set fag D Œa; a single point fx j a < x < bg D .a; b/ open bounded interval fx j a x bg D Œa; b closed bounded interval fx j a x < bg D Œa; b/ bounded interval open on the right fx j a < x bg D .a; b bounded interval open on the left fx j a < xg D .a; 1/ open right infinite interval fx j x < bg D .1; b/ open left infinite interval fx j a xg D Œa; 1/ closed right infinite interval fx j x ag D .1; b closed left infinite interval R entire real line
40
2 The Basics of Proofs
2.5.4 Exercises 1. Show that for any real number x it follows that jxj C jx 6j 6. 2. Show that for any real number x, jx 1j C jx 3j 2. 3. Show that for any real numbers x and y it follows that jx2 C3x4yjCjx14yj .x C 1/2 . 4. Show that for any real numbers x and y, jx C yj C jx y 2j C j2x C 8j 10. 5. Show that for any real numbers x and y it follows that jxjCj3x5yjCj5x4yj jx C yj. 6. Show that the intersection of any two intervals is always an interval. 7. Under what conditions is the union of two intervals an interval?
2.6 Functions 2.6.1 Function, Domain, Codomain Intuitively, a function is a mapping that assigns to each point of some domain A a value that resides in some codomain B. This is usually written f W A ! B. More precisely, the function f is defined as a set of ordered pairs .x; y/ where each x resides in the domain A of f and each y resides in the codomain B of f , and for each x 2 A there is exactly one y 2 B such that .x; y/ 2 f . Since there is a unique ordered pair .x; y/ 2 f for each x 2 A, f associates or links the value of y to the value of x and allows one to write f .x/ D y.
2.6.2 Surjection The domain of the function f is exactly the set of all x that are first coordinates of the order pairs in f , that is, the domain is A D fx j .x; y/ 2 f g. The range of f is defined as the image of f , that is, the range is fy j .x; y/ 2 f g. Clearly, the codomain of f can be any set that contains the range of f . This can lead to some confusion since the codomain of f is not precisely defined. It is simply a convenience. When one defines a function f W R ! R, one means that f is defined for every real number, and that for any x 2 R, the value f .x/ also lies in R. This is the case whether or not R is the range of f or if the range of f is actually some proper subset of R. It could be difficult and unnecessary to calculate exactly which subset of R is the range of f , so it might be easier to just give the codomain as R and avoid the technicalities of figuring out just what values of R are in the range of f . For example, the function f .x/ D 3x6 15x4 C 12x3 C 25x2 32x C 14 maps the real numbers into the real numbers, but to find the range of f , you would need to find the minimum value of f . This minimum exists, but it may not be possible to give its value explicitly.
2.6 Functions
41
If the range of f W A ! B is actually all of B, we say that f maps A onto B, and f is called surjective, and f is called a surjection. Thus, one can prove that a function is surjective by showing that each element of the codomain is in the range. TEMPLATE for proving a function f is surjective • SET THE CONTEXT: Make a statement introducing f , its domain A, and its codomain B. • Select an arbitrary value y 2 B. • Exhibit a value x 2 A such that y D f .x/. • STATE THE CONCLUSION: Therefore, f is a surjection. Note that the crucial step in proving that a function is surjective is showing the existence of an x with f .x/ D y and verifying that the x is in the domain A of the function. For example, the function f .x/ D 5x2 C 1 is a surjection from the negative real numbers onto the interval .1; 1/. To prove this you would need to show that for each real number y > 1 there is a negative real number x for which f .x/ D y. But this just involves a simple algebraic manipulation. That is, if you need 5x2 C 1 D y, then you can solve to get 5x2 D y1 and x2 D y1 . Here one needs to be careful because 5 q y1 it is easy to continue by writing x D which always results in a positive value 5 for x. The q proof needs to exhibit a negative value for x, so it is important to set
. There is no need for the proof to display the steps of solving the x D y1 5 equation for x. The goal is to produce a value of x 2 A such that f .x/ D y; how you arrived at that x is not important. It may be interesting, but it is not an essential part of the proof, and, therefore, it should not be part of the proof. PROOF: The function f .x/ D 5x2 C 1 is a surjection from the negative real numbers onto the interval .1; 1/. • Let f .x/ D 5x2 C 1.
q . • For any y > 1 let x D y1 5
> 0, so • Because y > 1, y1 5 negative real number.
q
y1 5
is a positive real number and x is a
q 2 C 1 D y. • Moreover, f .x/ D 5x C 1 D 5 y1 C 1 D 5 y1 5 5 2
• Therefore, f is a surjection.
2.6.3 Injection The definition of function requires that each value x in the domain of f is found in exactly one ordered pair .x; y/ 2 f . The same does not have to hold for values in
42
2 The Basics of Proofs
the codomain, that is, one value y in the codomain could appear in many order pairs .x; y/ 2 f . For example, for the constant function f W R ! R given by f .x/ D 1 for all x 2 R, the value 1 appears as the second coordinate in all the ordered pairs of the function. If a function has the property that no value of y appears as the second coordinate of more than one ordered pair in f , then f is said to be injective or, less formally, that f is one-to-one. In this case the function f is called an injection. In such a case, one sees that f .x1 / D f .x2 / only if x1 D x2 . This gives a procedure for proving that a function is injective. TEMPLATE for proving a function f is injective • SET THE CONTEXT: Make a statement introducing f , its domain A, and its codomain B. • Assume that for two values x1 and x2 in A that f .x1 / D f .x2 /. • Show that x1 D x2 . • STATE THE CONCLUSION: Therefore, f is an injection. p For example, the function f .x/ D 4x C 7 maps the positive real numbers to the positive real numbers. It is not a surjection, but it is an injection. The proof would require that you show that f .x1 / D f .x2 / implies that x1 D x2 . Again, this is just an algebraic manipulation. p PROOF: The function f .x/ D 4x C 7 is an injection from the positive real numbers to the positive real numbers. p • Let f .x/ D 4x C 7. • Assume p that for positive p real numbers x1 and x2 , f .x1 / D f .x2 /. • Then 4x1 C 7 D 4x2 C 7. • Squaring yields 4x1 C 7 D 4x2 C 7, so 4x1 D 4x2 , and x1 D x2 . • Therefore, f is an injection. If a function f W A ! B is both surjective and injective, that is, if f is both one-toone and onto, then f is bijective, and f is called a bijection. In this case, f exhibits a one-to-one correspondence between the set A and the set B. Two functions f and g whose ranges are in the real numbers can be combined arithmetically. Specifically, one can define f C g, f g, fg, and gf in natural ways: .f C g/.x/ D f .x/ C g.x/,.f g/.x/ D f .x/ g.x/, .fg/.x/ D f .x/ g.x/, and, f .x/ for x such that g.x/ ¤ 0, gf .x/ D g.x/ . When functions f and g are combined in this way, the domain of the sum, difference, product, or quotient is assumed to be the intersection of the domain of f and the domain of g with the exception that the domain of gf also excludes values of x for which g.x/ D 0. Thus, the function p p f .x/ D x 4 is defined for all x 4, and g.x/ D 5 x is defined p the function p for all x 5. It follows that the function x 4 C 5 x is defined only for those x satisfying 4 x 5. Similarly, the function ff .x/ is only defined for x > 4 even .x/ though it is identically 1 for those x. That function has a natural extension to all real numbers.
2.6 Functions
43
g(x) y f(x) B
x
z A
f◦g
C
Fig. 2.5 Composition .f ı g/.x/ D z
2.6.4 Composition If g is a function assigning values in its domain A to values in its range contained in the set B, and if f is a function assigning values in its domain B to values in its range contained in the set C, then the composition of f with g is the function .f ı g/.x/ D f g.x/ which assigns to values in its domain A values in its range contained in set C (Fig. 2.5). The main reason for considering compositions is that it is often easiest to represent complicated functions as compositions of simpler 2x functions. For example, the function f .x/ D psinxC4 is clearly a quotient where the numerator is the composition of the function p x2 with the function sin x, and the denominator is the composition of the function x with the function x C 4. It is easily shown that if g W A ! B and f W B ! C are both surjective functions, then their composition, f ı g W A ! C, is also surjective. To prove this, you would follow the template for proving that a function is surjective. That requires that you select an arbitrary z 2 C and show that there is an x 2 A such that .f ı g/.x/ D z. Why might you use the variable z here rather than the variable y? Well, that allows you to think of g as mapping x to y, and f , in turn, mapping y to z. Faced with the statement f g.x/ D z, there is little you can do except to apply what you know about the function f , that is, that f is surjective. Because f is surjective, and z is in the codomain of f , you know that there is a y in the domain of f such that f .y/ D z. Can you find an x such that g.x/ D y? Of course y is in the domain of f which is the codomain of g. The function g is surjective, so there must be an x in the domain of g that maps onto y. These ideas give the following proof.
44
2 The Basics of Proofs
PROOF: If g W A ! B and f W B ! C are both surjective functions, then their composition f ı g W A ! C is also surjective. • • • • • •
Let g W A ! B and f W B ! C be two surjective functions. Let z 2 C. Then since f is a surjection from B to C, there is a y 2 B such that f .y/ D z. Since g is a surjection from A to B, there is an x 2 A such that g.x/ D y. Therefore, .f ı g/.x/ D f g.x/ D f .y/ D z. It follows that f ı g W A ! C is surjective.
It is also true that if g W A ! B and f W B ! C are both injective functions, then their composition, f ı g W A ! C, is also injective. You would prove this by following the template for proving that a function is injective. That is, you would assume that .f ı g/.x 1/ D .f ıg/.x2 / for some x1 and x2 in A. Again, what can you say if you know f g.x1 / D f g.x2 / ? All that you can do is apply what you know about the function f , that is, that f is injective. Since f is injective, you can conclude that g.x1 / D g.x2 /. Then because g is injective, you can conclude x1 D x2 , and you are done. PROOF: If g W A ! B and f W B ! C are both injective functions, then their composition f ı g W A ! C is also injective. • • • • • •
Let g W A ! B and f W B ! C be two injective functions. Assume that for some x1 and x2 inA, .f ı g/.x1 / D .f ı g/.x2 /. By the definition of composition f g.x1 / D f g.x2 / . Then since f is an injection, it follows that g.x1 / D g.x2 /. Since g is an injection, it follows that x1 D x2 . It follows that f ı g W A ! C is injective.
2.6.5 Exercises Write a proof for each of the following statements. 1. For each real number r there is a real number x such that x3 D r. 2. For each real number r 0 there is a real number x 0 such that x4 D r. 3. If n is an odd positive integer, then for each real number r there is a real number x such that xn D r. 4. If n is an even positive integer, then for each real number r 0 there is a real number x 0 such that xn D r. 5. If h W A ! B, g W B ! C, and f W C ! D are three functions, then .f ı g/ ı h D f ı .g ı h/. In other words, function composition is associative. (Hint: Show that both functions .f ı g/ ı h and f ı .g ı h/ give the same result when applied to an x 2 A.)
2.6 Functions
45
6. If h W A ! B, g W B ! C, and f W C ! D are three surjective functions, then their composition f ı g ı h is surjective. 7. If h W A ! B, g W B ! C, and f W C ! D are three injective functions, then their composition f ı g ı h is injective.
Chapter 3
Limits
3.1 The Definition of Limit In a typical Calculus course students develop an intuitive understanding of the concept of limit which, of course, is the central concept of Calculus and, indeed, the central concept of Analysis. In particular, if f is a function defined on an open interval containing a 2 R, then f has limit L at a if the values of f .x/ get closer and closer to L as x approaches a. In order to prove theorems about limits, one needs a rigorous definition of limit which makes clear what is meant by “closer and closer” and “approaches.” In Analysis the distance between two real numbers is measured by the absolute value of the difference of the two numbers. Thus, the ideas of “closer and closer” and “approaches” naturally involve statements about the absolute values of differences of two quantities. Consider the definition of lim f .x/ D L, where the function f is defined in an x!a
open interval in R containing the point a. This limit should give you a mental image similar to Fig. 3.1 where the graph of the function gets close to L as x approaches a. So, how can you quantify what “f .x/ is getting close to L” means? Is within 1 1 close? Is within 14 close? Is within 1000 close? Clearly, there needs to be a way to say “arbitrarily close” or “as close as one likes.” Analysts have found that a good way to express f .x/ getting arbitrarily close to L is to say that for any positive distance, jf .x/ Lj can be made to be less than that distance. Of course, jf .x/ Lj cannot be made to be negative, and it is not reasonable to require it to be zero since that would require f .x/ to actually equal L. Hence, one usually says that for any > 0, one can achieve jf .x/ Lj < . The use of the Greek letter (epsilon) is arbitrary, but the tradition of using in this context has been universal since Cauchy introduced its use in the early 1800s. Figure 3.2 shows a tolerance of a small around the limit value L. The goal is to show that the function f .x/ stays within that tolerance when x is close to a. In the figure you can see that for the values of x near a, the function f .x/ falls within the prescribed tolerance of L. You could find a small interval centered at a © Springer International Publishing Switzerland 2016 J.M. Kane, Writing Proofs in Analysis, DOI 10.1007/978-3-319-30967-5_3
47
48
3 Limits
Fig. 3.1 lim f .x/ D L x!a
L
a
y = f(x)
a
y = f(x)
Fig. 3.2 lim f .x/ D L x!a
L
Fig. 3.3 lim f .x/ D L x!a
L+ε L L–ε
ε ε y = f(x)
a
a–
a+
such that jf .x/ Lj < for all x in that interval. That is, there is a value ı > 0 such that every x satisfying jx aj < ı also satisfies jf .x/ Lj < as seen in Fig. 3.3. Again, the choice of the Greek letter ı (delta) is completely arbitrary, but the tradition of using ı in this context is universal. Note that for the function whose graph appears in Fig. 3.3 the value L is the limit of the function as x approaches a, and L also happens to be the value of f .x/ at x D a. You should recall that this sometimes happens (specifically when f is continuous at x D a), but that this is not a requirement. Indeed, one reason for discussing limits in the first place is because there is a need to evaluate the behavior of a function as x approaches a value a when the function fails to be defined at x D a. Thus, in general, one does not want to require jf .x/ Lj < for all x with jx aj less than some positive value ı since this would require jf .x/ Lj < at x D a. Instead, one excludes the need for the function to satisfy any conditions at all at x D a by saying that there is a positive ı such that f is within the desired tolerance of L for all x with 0 < jx aj < ı. Clearly, the value of ı must be chosen to be positive since no negative value would represent a distance and, ı D 0 would not result in a region around the number a satisfying jx aj < ı.
3.2 Proving lim f .x/ D L
49
x!a
Combining these ideas results in the following definition. Suppose that the function f is defined for all x in an open interval containing a 2 R except perhaps at x D a. Then the limit of f as x approaches a is L, lim f .x/ D L, means that for x!a every > 0 there exists a ı > 0 such that for every x satisfying 0 < jx aj < ı, it follows that jf .x/ Lj < . The power of this definition is the fact that the and ı are arbitrary positive numbers. For example, what if you knew that for every > 0 there were a ı > 0 such that whenever 0 < jx aj < ı, then jf .x/ Lj < 2? Would this be sufficient for showing lim f .x/ D L? The answer is yes, because the x!a
is arbitrary. Suppose that for any > 0 you can find a ı > 0 that will ensure that jf .x/ Lj < 2. Then since 2 is also a positive number, you can find a ı 0 > 0, likely smaller than ı, that will ensure that jf .x/ Lj < 2 2 D . The point here is that since was arbitrary, you can replace it with any positive number, including 2 .
3.1.1 Exercises Which of the following definitions is equivalent to the definition of lim f .x/ D L? x!a
1. For all 0 there is a ı 0 such that if 0 < jx aj < ı, then jf .x/ Lj < . 2. For all > 0 there is a ı > 0 such that if 0 < jx aj < 4ı , then jf .x/ Lj < 7. 3. For all > 0:001 there is a ı > 0:001 such that if 0 < jx aj < ı, then jf .x/ Lj < . 4. For all ı > 0 there is an > 0 such that if 0 < jx aj < ı, then jf .x/ Lj < . 5. There exists a ı > 0 such that for all > 0, if 0 < jxaj < ı, then jf .x/Lj < . 6. For all ı > 0 there is an > 0, such that if 0 < jx aj < , then jf .x/ Lj < ı. 7. For all > 1 there is a ı > 1, such that if 1 < jx aj C 1 < ı, then jf .x/ Lj C 1 < .
3.2 Proving lim f .x/ D L x!a
The definition of limit provides a formula by which one can construct a proof that a particular function f has a limit of L at the point a. The definition requires that for every > 0 there is a ı > 0 that satisfies certain properties. Thus, a proof of a limit must show that for every > 0 you can exhibit a ı > 0 which has the needed property. As with other proofs that some property holds for all elements of a set, the proof begins by selecting an arbitrary element of that set. In this case, one would select an arbitrary > 0. The goal is to present a value for ı > 0 such that every x satisfying 0 < jx aj < ı also satisfies jf .x/ Lj < . That suggests the following proof template.
50
3 Limits
TEMPLATE for proving lim f .x/ D L x!a
• SET THE CONTEXT: Make statements telling what is known about the function f and the numbers a and L. • SELECT AN ARBITRARY : Given > 0, • PROPOSE A VALUE FOR ı: let ı D . Here you would insert an appropriate value for ı. • SELECT AN ARBITRARY x: Select x such that 0 < jx aj < ı. • LIST IMPLICATIONS: Derive the result jf .x/ Lj < . • STATE THE CONCLUSION: Therefore, lim f .x/ D L. x!a
For example, consider proving that lim 2x 3 D 7. As stated in the template, x!5
the proof would begin with “Let f .x/ D 2x 3. Given > 0, : : :.” Your task is to determine a value of ı > 0 that ensures the inequality jf .x/ Lj < holds for all x with 0 < jx aj < ı. Since the function f is not constant, the choice of ı will surely depend on the value of . But how is this value of ı determined? A common approach is to work backwards from the final conclusion jf .x/ Lj < to see what value of ı is needed. In this example, f .x/ D 2x 3, a D 5, and L D 7. The value of jf .x/ Lj D j.2x 3/ 7j D j2x 10j D 2jx 5j. Note that this expression has a factor of x a, where a D 5. When finding the limit of a polynomial where L D f .a/, this will always be the case. For more complicated functions f , other properties of f will often allow you to write f .x/ L as an expression where x a is a factor. This makes it easier to determine a value of ı since the choice of ı restricts the size of jx aj which, in turn, will make jf .x/ Lj small, the desired result. So here, jf .x/ Lj D 2jx 5j. To force jf .x/ Lj to be less than some arbitrary > 0, it is, therefore, sufficient for 2jx 5j to be made less than . This is done by making jx 5j < 2 , and the needed value of ı is 2 . Note that it is stipulated that is positive, so ı D 2 is also greater than zero, a requirement of the definition of limit. Now a complete proof can be written by following the template. PROOF: lim 2x 3 D 7 x!5
• • • • •
Let f .x/ D 2x 3. Given > 0, let ı D 2 > 0. Select x such that 0 < jx 5j < ı D 2 . Then ı > jx5j implies > 2jx5j D j2x10j D j.2x3/7j D jf .x/7j. Therefore, lim 2x 3 D 7. x!5
Identifying an appropriate value for ı is very easy when f is linear as in the example above, but it can be trickier for other functions. Consider proving lim 3x2 D 48. In this example, f .x/ D 3x2 , a D 4, and L D 48. The value of
x!4
jf .x/ Lj D j3x2 48j D 3jx2 16j D 3jx C 4j jx 4j. As suggested above,
3.2 Proving lim f .x/ D L
51
x!a
it should not be surprising that this expression includes a factor of x a D x 4 which can be forced to be small by selecting a small value for ı. In particular, the proof will need to justify jf .x/ Lj < which means 3jx C 4j jx 4j < and jx 4j < 3jxC4j . Here is an attempt at a proof using this idea, but it falls short of being correct. PROOF ATTEMPT: lim 3x2 D 48 x!4
• Let f .x/ D 3x2 . • For any x, let ı D 3jxC4j . • Then 0 < jx 4j < ı D 3jx2 16j D jf .x/ 48j. • Therefore, lim 3x2 D 48.
, 3jxC4j
implies that > jx 4j 3jx C 4j D
x!4
Can you spot some of the errors in this proof? • The second line of the proof refers to the variable which has not yet been introduced in the proof. In particular, without having specified that > 0, one does not know that ı > 0 which is required by the definition of limit. The proof should include the phrase Given > 0. • The value of ı in the second line of the proof is undefined when x D 4. • The most serious error here is that the value of ı depends on the value chosen for x. The definition of limit requires that for every > 0 there is a ı > 0. That value of ı can depend on the value of but certainly cannot depend on x which has not yet been introduced in the definition. After ı is specified, the definition requires that a condition hold for all x satisfying 0 < jx aj < ı, and only then does the definition refer to values of x. Still one needs a value of ı which will be less than 3jxC4j for all the values of x considered in the proof. One way around this would be to find a value for ı which is less than 3jxC4j for every value of x. But this cannot be done because the expression gets arbitrarily small as x gets large. On the other hand, the value of x will be 3jxC4j restricted so that 0 < jx 4j < ı. Thus, unless ı is very large, x cannot wander too far away from a D 4, and 3jxC4j cannot get arbitrarily small. So how does one choose a ı which both ensures that jx C 4j does not grow too large and also makes jx 4j small? The technique is to select ı in two stages. First, to ensure that jx C 4j does not grow too large, restrict the value of ı so that x cannot wander too far from a D 4. Almost any restriction in the size of ı will work, so how about suggesting that ı not exceed 1? If ı 1, then when you choose an x with 0 < jx4j < ı, you will know that jx4j < 1 which is equivalent to 1 < x4 < 1 and, thus, 1 C 8 < .x 4/ C 8 < 1 C 8. That is, 7 < x C 4 < 9, and it follows that jx C 4j < 9. Here is another attempt at a proof that uses this idea. Unfortunately, it too has problems.
52
3 Limits
PROOF ATTEMPT: lim 3x2 D 48 x!4
• Let f .x/ D 3x2 . • Given > 0, let ı D 1. • Then for any x such that 0 < jx 4j < ı D 1, it follows that 1 < x 4 < 1 so 1 C 8 < .x 4/ C 8 < 1 C 8 and jx C 4j < 9. • Now let ı D 27 > 0. • Then 0 < jx 4j < ı D 27 implies that > jx 4j 27 > jx 4j 3jx C 4j D 2 j3x 48j D jf .x/ 48j. • Therefore, lim 3x2 D 48. x!4
The only problem with the above proof is in its use of the variable ı. In the second line of the proof ı is set to 1, and in the fourth line it is set to 27 . It does not make sense to set the value of ı equal to both of these values because, except in the rare case that D 27, the value of ı cannot be equal to both values at the same time. The solution is to choose one value for ı that satisfies two separate conditions. For example, you can first require that ı < 1. Then a choice of x with 0 < jx 4j < ı will guarantee that jx C 4j < 9. Then 3jxC4j > 39 D 27 . This suggests that you should select ı D 27 . But you also need ı 1. What happens if someone suggests that be some rather large number such as D 100? Then ı D 27 would not satisfy ı < 1. This is not a problem since one can always get away with selecting a positive value for ı that is smaller than needed. Thus, you can select ı to be the lesser of 1 and 27 . This choice is usually written as ı D min 27 ; 1 . Now you can put this all together to get a formal proof that is completely correct. PROOF: lim 3x2 D 48 x!4
Let f .x/ D 3x2 . Given > 0, let ı D min 27 ;1 . Select x such that 0 < jx 4j < ı. Since ı 1, it follows that jx4j < 1 and 1 < x4 < 1, so 7 < xC4 < 9. Thus, jx C 4j < 9. • Since ı 27 , it follows that jx 4j < 27 D 39 < 3jxC4j . 2 • Then ı > jx 4j implies > 3jx C 4j jx 4j D j3x 48j D jf .x/ 48j. • Therefore, lim 3x2 D 48.
• • • •
x!4
xC2 2 x!2 x C3xC2
As a third example, consider proving the limit lim
D 1. In this
example f .x/ D a D 2, and L D 1. Note that f .2/ is not defined even though the limit as x approaches 2ˇ exists. The proof ˇ of this limit must conclude ˇ ˇ xC2 with the inequality > jf .x/Lj D ˇ x2 C3xC2 .1/ˇ. As in the previous examples, xC2 , x2 C3xC2
xC2 it would be convenient if the expression x2 C3xC2 .1/ would contain a factor of x C 2 so that it could be made small by requiring x .2/ to be less than some ı.
3.2 Proving lim f .x/ D L
53
x!a
But this follows with some fairly straightforward algebra. Assuming that x ¤ 2, x2
xC2 xC2 1 1 C .x C 1/ xC2 .1/ D C1 D C1 D D : C 3x C 2 .x C 2/.x C 1/ xC1 xC1 xC1
ˇ ˇ ˇ ˇ ˇ D jx C 2j ˇ 1 ˇ would follow if jx C 2j never The needed inequality > ˇ xC2 xC1 xC1 exceeds jx C 1j which, in turn, would happen if ı < jx C 1j. Again, there is a problem because the choice of ı > 0 cannot depend on the value of x, yet jx C 1j can get arbitrarily close to zero as x gets close to 1. The strategy, then, would be to restrict the value of ı so that x could not get close to 1. If x is supposed to be close to 2, ı could be chosen so that it does not exceed 12 . Then, jx C 1j could not get smaller than 1 12 D 12 , and jx C 1j > 2 . You would not want ı to exceed either 2 or 12 . Thus, one can select ı D min 2 ; 12 . The complete proof follows. xC2 2 x!2 x C3xC2
PROOF: lim
D 1
xC2 Let f .x/ D x2 C3xC2 . Given > 0, let ı D min 2 ; 12 . Select x such that 0 < jx .2/j < ı. Since ı 12 , it follows that jx C 2j < 12 and 12 < x C 2 < 12 , so 32 < x C 1 < 12 . Thus, jx C 1j > 12 . • Since ı 2 , it follows that jx C 2j < 2 < jx C 1j. • Then ı > jx .2/j > 0 implies 2 > jx C 2j and 1 1 jx C 2j D jxC1j j1 C .x C 1/j D > 2jx C 2j > jxC1j ˇ ˇ 1 ˇ ˇˇ xC2 ˇ ˇ C 1ˇ D ˇ 2 .1/ˇ D jf .x/ .1/j.
• • • •
xC1
x C3xC2 xC2 2 x!2 x C3xC2
• Therefore, lim
D 1.
Clearly, at the point that you stipulate that ı should be less than 12 , you are making a rather arbitrary decision. What would have happened if you had chosen some other reasonable bound on the size of ı? For example, what if instead you only require ı < 34 ? This would also work, although that decision would affect the final choice of ı for now jx C 1j can get as small as 14 , and jxC2j could be as large as 3jxC1j 4jx C 2j. This suggests that you then select ı D min 4 ; 4 . This choice is no better or worse than the ı chosen earlier. When one makes such arbitrary decisions, it is good form to make a selection that does not lead to unnecessary arithmetic or algebraic complications because one does not want to make the proof any harder to read than necessary. Thus, it p would perfectly adequate but enormously awkward to select the bound on ı to be p5 . As long as the bound is less than 1, it will do the 1C 5
job of keeping jx C 1j bounded away from zero, but optimal choice.
p 5 p 1C 5
would certainly not be an
54
3 Limits
3.2.1 Exercises Write a proof of each of the following limits. 1. lim 35 x C 1 D 4 x!5
2. lim 5x 8 D 7 x!3
3. lim 2x2 D 18 x!3
4. lim 9x2 D 4 x! 23
5. lim 3x2 5x 7 D 1 x!1
6. lim x2 C 3x C 1 D 29 x!4
7. lim 2x3 D 16 x!2
6 D2 x!1 2xC5 xC4 lim 2 D x!8 x 10xC10
8. lim 9.
2
10. lim mx C b D ma C b x!a
11. lim ax2 D au2 x!u
12. lim ax2 C bx C c D au2 C bu C c x!u
3.3 One-Sided Limits The one-sided limits lim f .x/ D L and lim f .x/ D L are very similar to twox!aC
x!a
sided limits except that the value of x is only allowed to approach the real number a from one side. As a result, the definitions of these one-sided limits are very similar to the definition of limit with minor alterations that forces x to stay on one side of a. The definition of limit states that for a function f defined in a neighborhood of a, but not necessarily at a, the limit lim f .x/ D L means for every > 0 there exists a x!a
ı > 0 such that for every x satisfying 0 < jx aj < ı, it follows that jf .x/ Lj < . What is it about this definition that allows x to approach a from two sides? It is the inequality 0 < jx aj < ı that allows x to be either greater than or less than a since jx aj is positive in either case. By removing the absolute value function in this inequality and writing instead 0 < x a < ı, the choice of x becomes restricted to being a value greater than a, or writing instead 0 < a x < ı, the choice of x becomes restricted to being a value less than a. Thus, if f is a function defined for all x in an open interval with right end at a, then the limit of f at a from the left is L, lim f .x/ D L, means that for every > 0 there is a ı > 0 such that for every x x!a
satisfying 0 < a x < ı, it follows that jf .x/ Lj < . Similarly, if f is a function
3.3 One-Sided Limits
55
defined for all x in an open interval with left end at a, then the limit of f at a from the right is L, lim f .x/ D L, means that for every > 0 there is a ı > 0 such that x!aC
for every x satisfying 0 < x a < ı, it follows that jf .x/ Lj < . One-sided limits are particularly useful in cases where the function f behaves 1 differently on one side of a as on the other side such as the way e x behaves quite 1 differently as x approached 0 from the right where 1x is positive from how e x behaves as x approaches 0 from the left where 1x is negative. Similarly, the derivative of f .x/ D jxj has different limits as x approaches 0 from the right and from the left. There are also p cases where a function is not even defined for x on one side of a such as f .x/ D x which is not defined for x < 0. Proving the existence of one-sided limits is very similar to proving two-sided limits except that care must be taken to ensure that the value of x remains on one side of a. Take, for example, the limit lim 2x2 5x D 3. Here f .x/ D 2x2 5x, x!3C
a D 3, and L D 3. As with a proof of other limits earlier in the chapter, the proof needs to give a value for ı > 0 which will ensure > jf .x/Lj D j.2x2 5x/3j D j.2x C 1/.x 3/j. This will follow if jx 3j < ı j2xC1j for all suitable values of x. What is needed is the largest possible value of 2x C 1, but 2x C 1 is not bounded unless x is restricted to be close to 3. Thus, stipulate that ı be less than 1 which will ensure that x3 will be less than1, x will not exceed 4, and 2xC1 will not exceed 9. Then ı can be chosen to be min 9 ; 1 , and the proof can be written as follows. PROOF: lim 2x2 5x D 3 x!3C
Let f .x/ D 2x2 5x. Given > 0, let ı D min 9 ; 1 . Select x such that 0 < x 3 < ı. Since ı 1, it follows that 0 < x 3 < 1, 3 < x < 4, and j2x C 1j < 9. Then 0 < x 3 < ı implies 9 > x 3 and > 9.x 3/ > .2x C 1/.x 3/ D 2x2 5x 3 D jf .x/ 3j. • Therefore, lim 2x2 5x D 3. • • • • •
x!3C
Consider a function where its left limit differs from its right limit such as the function 5 7x if x < 1 f .x/ D : Then lim f .x/ D 2 while lim f .x/ D 1. Thus, x!1 x if x 1 x!1C while proving lim f .x/ D 2, it is important to use that fact that x < 1 as part x!1
of the proof since the required inequalities will not hold for x > 1 (Fig. 3.4). The following shows one possible proof.
56
3 Limits
Fig. 3.4 Graph of f .x/
PROOF: lim x!1
5 7x if x < 1 x if x 1
D 2
• • • • •
5 7x if x < 1 Let f .x/ D . x if x 1 Given > 0, let ı D 7 . Select x such that 0 < 1 x < ı D 7 . Then x < 1, so f .x/ D 5 7x. It follows that > 7.1 x/ D 5 7x .2/ D jf .x/ .2/j. Therefore, lim f .x/ D 2. x!1
In the third line of the proof, 0 < 1 x < ı ensures that x < 1 which, in turn, is needed to conclude that f .x/ D 5 7x and not f .x/ D x. The fact that x < 1 is also used in the fourth line of the proof to conclude that 5 7x .2/ D jf .x/ .2/j which follows because 5 7x .2/ is positive for all x < 1.
3.3.1 Exercises Write a proof of each of the following one-sided limits. 1. lim x2 C 4x D 21 x!3C
2. lim 8 3x D 1 x!3
3. lim x!2
4. lim
x!4C
5. lim x!2
6. lim
x!2C
x2 4 x2 3xC2
D4
x2 4x
D
2x2 7x4 8jx2j x2 4 8jx2j x2 4
D 2 D2
4 9
3.4 Limits at Infinity
57
3.4 Limits at Infinity The definitions given in the last two sections do not make sense when the real number that x approaches, a, is replaced by infinity. Infinity, of course, is not an element of the real numbers, R, but it does make sense to ask whether a function approaches a limit when x increases without bound, that is, as x approaches infinity. When one writes lim f .x/ D L, one is thinking that f .x/ is getting close to the real x!1 number L as x increases without bound. But it does not make sense to measure how close x is to infinity by choosing a ı > 0 so that when x is within ı of infinity, f .x/ is close to L. Since infinity is not a real number, one cannot measure the distance from the real number, x, to infinity, even less expect x to get within ı of infinity. So how does one quantify “getting closer to infinity?” The answer lies in the phrase “increases without bound” which suggests that for any bound, N, you could place on the size of x, the value of x can be made to be greater than that bound. Thus, instead of selecting a ı > 0 and requiring 0 < jx aj < ı, one chooses a number N 2 R and requires x > N. This allows the following definition. Suppose that the function f is defined for all x > K for some real number K. Then the limit of f as x approaches infinity is L, lim f .x/ D L, means that for every > 0 there exists x!1 an N 2 R such that for every x > N, it follows that jf .x/ Lj < (Fig. 3.5). Now consider how one might write a proof of a limit at infinity. For example, consider x x D 0. Here f .x/ D x2 C6 and L D 0. As with other limit proving the limit lim x2 C6 x!1
proofs, the goal is to arrange that jf .x/ Lj < for an chosen > 0. ˇ ˇ arbitrarily ˇ x ˇ Again, you can work backwards. Since jf .x/ Lj D ˇ x2 C6 ˇ, as long as x > 0, it ˇ ˇ ˇ x ˇ would follow that ˇ x2 C6 ˇ < xx2 D 1x . Thus, there is an expression, 1x , which is larger than jf .x/ Lj for all suitably large values of x. This will help because if you can assure that 1x is less than , it will follow that jf .x/ Lj is also less than . It would not have been helpful to exhibit an expression that was always less than jf .x/ Lj because making that expression small would not imply that jf .x/ Lj is small. Now, if x > 1 , it follows that 1x < suggesting that 1 is a suitable value for N. x x!1 x2 C6
PROOF: lim • • • • •
D0
x Let f .x/ D x2 C6 . Given > 0, let N D 1 . Select x such that x > N > 0. Then x > 1 implies > 1x D xx2 > x Therefore, lim x2 C6 D 0. x!1
Fig. 3.5 Approaching a limit as x ! 1
x x2 C6
ˇ ˇ ˇ x ˇ D ˇ x2 C6 0ˇ D jf .x/ 0j.
58
3 Limits
Note that it is important that the third step of the proof pointed out that N is positive. It is used in the fourth step when 1x is calculated, and this would not have been allowed if the value of x could have been zero. For a second example, consider proving lim 2xC5 D 2. Again, you can work x!1 x7 ˇ ˇˇ .2xC5/2.x7/ ˇˇ ˇ 19 ˇ ˇ 2xC5 backwards to get > jf .x/ Lj D ˇ x7 2ˇ D ˇ ˇ D ˇ x7 ˇ. From here x7 there are a number of ways to proceed. You can solve for x in the previous inequality to get x > 7 C 19 which gives a reasonable value for N. Another way would be to 19 is less than say that if x > 14, then x 7 < x 2x D 2x . In this case the fraction x7 19 38 38 , and it becomes clear that x > is sufficient. x D x 2 x This is an example demonstrating the enormous flexibility one sometimes has in writing proofs in analysis where you often need to prove an inequality which can be done in many ways. It is usually easier to prove an inequality involving a simple fraction rather than a complicated fraction, so you can use the strategy of replacing a fraction with a simpler fraction that is clearly larger, or in some cases, clearly smaller. Keep in mind that a ratio of positive values gets larger if its numerator gets larger or its denominator gets smaller. A complete proof can be written as follows. 2xC5 x!1 x7
PROOF: lim • • • • •
D2
Let f .x/ D 2xC5 . x7 Given > 0, let N D 7 C 19 . Select x such that x > N > 7. Then x > 7 C 19 implies x 7 > Therefore, lim 2xC5 D 2. x7
19
and >
19 x7
D
2xC5 x7
2 D jf .x/ 2j.
x!1
As in the previous proof it is important that x > 7 is pointed out in the third step of the proof because that fact is needed both to ensure that f .x/ is defined by assuring x 7 ¤ 0 and that x 7 is positive allowing the absolute value function to be introduced in the fifth step of the proof. With a slight adjustment of the definition of lim f .x/ D L, one gets a definition x!1
of lim f .x/ D L. This time rather than choosing an N and requiring jf .x/ Lj < x!1
for all x > N, one instead needs f .x/ to be within of L for those x < N. Thus, lim f .x/ D L means that for every > 0 there exists an N 2 R such that for all x!1 x < N it follows that jf .x/ Lj < . 2 D 3, one can identify an N such that x < N implies that To prove lim 6x2x2C5x 7 x!1 ˇ ˇ 2 ˇ ˇ 2 2 ˇ 6x C5x ˇ ˇ .6x C5x/3.2x2 7/ ˇ D 3j < by working backwards. That is, 3 j 6x2x2C5x ˇ ˇD ˇ ˇ 7 2x2 7 2x2 7 ˇ 5xC21 ˇ ˇ 2 ˇ. It would be nice to simplify this rather messy expression; something you 2x 7 can do as long as you do not introduce changes that prevent the final inequality from holding. In this case, the 7 term in the denominator of 5xC21 is an inconvenience, 2x2 7
3.4 Limits at Infinity
59
and it would be nice to remove it. Simply removing this negative term would make the absolute value of the fraction smaller when what is needed is to make the fraction larger. A strategy that does work is to take part of the 2x2 term, which grows very large as x goes to 1, and pair it with the 7 term. For example, 2x2 7 p can be written as x2 C .x2 7/. Because x2 7 is a positive value for all x < 7, removing it from the denominator makes the absolute value of the fraction greater. Also note that when x < 21 , the numerator j5x C 21j < 5jxj, and this happens for 10 ˇ 5xC21 ˇ p 5 5 ˇ ˇ all x < 7. It would then be sufficient for > jxj D 5jxj 2 > 2x2 7 or that x < x p as long as x < 7. A proof would be 6x2 C5x x!1 2x2 7
PROOF: lim
D3
6x2 C5x . 2x2 7
• Let f .x/ D
p • Given > 0, let N D min 7; 5 . p • Select x such that x < N 7. ˇ ˇ ˇ ˇ ˇˇ 5xC21 ˇˇ ˇ>ˇ 2 2 ˇD • Then x < N 5 implies > ˇ 5x ˇ D ˇ 5x x2 x C.x 7/ ˇ 2 ˇ ˇ ˇ ˇ .6x C5x/3.2x2 7/ ˇ ˇ 6x2 C5x ˇ D D jf .x/ 3j. 3 ˇ ˇ ˇ ˇ 2x2 7 2x2 7 6x2 C5x 2 x!1 2x 7
• Therefore, lim
D 3.
3.4.1 Exercises Find ways to justify each of the following inequalities that hold for large values of x. 1. 2. 3.
3x5 < 2x 2x2 4xC7 < 5x 2x2 6 2 5x C3xC1 < 10 x x3 x2 1
Write proofs for each of the following limits. 4 D0 x!1 xC4 3x9 lim D1 x!1 3xC4
4. lim 5.
9x2 D3 2 x!1 3x 10 3 x lim D 15 3 2 x!1 5x 2x 4
6. lim 7.
60
3 Limits
3.5 Limit of a Sequence 3.5.1 Definition of Sequence A sequence is just a function whose domain is the set of natural numbers, N. In this chapter the codomain of a sequence will be the real numbers, R, but you can have a sequence with any set serving as the codomain. Functions are usually referenced using the notation f .x/. But for sequences it is traditional to place the argument of a sequence in a subscript rather than within parentheses as in a1 ; a2 ; a3 ; : : : . The entire sequence is notated with angle brackets as in . Note that this is not the same as the set fa1 ; a2 ; a3 ; : : : g which is just the collection of the values taken on by the sequence, that is, the range of the function a W N ! R. For each n 2 N, an is called a term of the sequence, or specifically, the nth term of the sequence.
3.5.2 Arithmetic with Sequences As with any real-valued function, you can add, subtract, multiply, and divide sequences. The sum of sequences and is the sequence where, for each n 2 N, cn D an C bn . Similarly, one can define the difference of sequences and product of sequences as cn D an bn and cn D an bn , respectively. If the sequence has no terms equal to zero, then the quotient of sequence and is the sequence cn D abnn . Other arithmetic operations can be similarly defined. If f is any real-valued function with a domain that includes the range of the sequence , then it makes sense to define the sequence cn D f .an /. For example,pif pis the p 1; 3; 5; 7; : : : , then the sequence < an > is the sequence 1; 3; 5; 7; : : : .
3.5.3 Monotone Sequences A sequence is a monotone increasing sequence if a1 a2 a3 : : : , or in other words, for natural numbers i < j it follows that ai aj . Similarly, is a monotone decreasing sequence if a1 a2 a3 : : : , or for natural numbers i < j it follows that ai aj . A monotone sequence is a sequence that is either monotone increasing or monotone decreasing. If a monotone increasing sequence satisfies ai < aj for all natural numbers i < j, then it is a strictly monotone increasing sequence. Similarly, is strictly monotone decreasing if ai > aj for all natural numbers i < j. For example, the following sequences are monotone increasing:
3.5 Limit of a Sequence
61
• 1; 2; 3; : : : • 1; 1; 2; 3; 3; 4; 5; 5; : : : • 12 ; 23 ; 34 ; 45 ; : : : • 13 ; 23 ; 33 ; 43 ; : : : whereas the following sequences are monotone decreasing: • • • •
1; 0; 1; 2; 3; : : : 8; 4; 2; 1; 12 ; 14 ; : : : 0; 0; 0; 12 ; 12 ; 12 ; 1; 1; 1; 32 ; : : : 1 1 1 44 ; 55 ; 66 ; : : :
It is interesting to notice that every sequence of real numbers can be written as a sum of a monotone increasing sequence and a monotone decreasing sequence. In particular, if is a sequence of real numbers, define an increasing sequence and a decreasing sequence as follows. Let a1 D c1 and b1 D 0. Then for all n 2 N if cn cnC1 , define anC1 D cnC1 bn and bnC1 D bn , and if cn > cnC1 , define anC1 D an and bnC1 D cnC1 an . These definitions make it clear that cn D an C bn for each n 2 N. The sequence is increasing because cn cnC1 implies that anC1 an D .cnC1 bn / .cn bn / D cnC1 cn 0, and cn < cnC1 , implies an D anC1 . Similarly, is decreasing because cn > cnC1 implies that bnC1 bn D .cnC1 an / .cn an / D cnC1 cn < 0, and cn cnC1 implies bn D bnC1 . Thus, 1; 1; 2; 2; 3; 3; : : : can be written as the sum of the two sequences 1; 1; 4; 4; 9; 9; : : : and 0; 2; 2; 6; 6; 12; : : : .
3.5.4 Subsequences Intuitively, a subsequence of a sequence is a sequence whose terms include some of the terms of the sequence in the same order as they appear in the original sequence. Formally, if there is a strictly increasing sequence of natural numbers i W N ! N, then is a subsequence of the sequence . Thus, the sequence 1; 1; 2; 2; 3; 3; : : : has the following subsequences • 1; 2; 3; : : : • 1; 1; 3; 3; 5; 5; : : : • 2; 3; 5; 7; 11; : : : The sequence 1; 2; 2; 3; 3; 3; 4; 4; 4; 4; : : : is not a subsequence of 1; 1; 2; 2; 3; 3; : : : since there are no repeated values in the original sequence, so there can be no repeated values in any of its subsequences.
62
3 Limits
3.5.5 Limit of a Sequence The definition of the limit of a sequence is similar to that of the limit of a function as x ! 1 except that the function is only defined on the natural numbers. Thus, if is a sequence of real numbers, then the limit of the sequence is L, lim an D L, means that for all > 0 there is an N such that for every natural n!1 number n > N it follows that jan Lj < . A sequence that has limit L is said to converge to L and is said to be a convergent sequence. A sequence that does not converge is said to diverge and is said to be a divergent sequence. Except for slight notational changes, proving that a sequence has a particular limit involves the same type of work as proving that a function has a particular limit 2 as its variable approaches infinity. For example, the sequence an D 4n2nCnC2 has 2 7 limit 2. To prove this, given an > 0, you would need to exhibit a number N such that jan 2j < for all n > N. As writing about functions, ˇ ˇone can ˇ with ˇ proofs ˇ nC16 ˇ ˇ nC16n ˇ ˇ 4n2 CnC2 ˇ ˇ ˇ work backwards from jan 2j D ˇ 2n2 7 2ˇ D 2n2 7 ˇ n2 C.n2 7/ ˇ. If you stipulate thatˇ n ˇ 3, then n2 7 9 7 D 2 > 0 allowing you to conclude that ˇ D 17 which can easily be made less than by requiring n > 17 . jan 2j < ˇ 17n n n2 This is what is needed for the proof. PROOF: lim
n!1
• • • • •
4n2 CnC2 2n2 7
D2
2
Let an D 4n2nCnC2 2 7 . Given > 0, let N D max 3; 17 . Select an n > N. Since N 3, it follows that n2 > 9. Also, n > N gives n 17 . Thus, ˇ 2 ˇ ˇ 4n CnC2 ˇ ˇ 2n2 7 2ˇ D jan 2j. 4n2 CnC2 2 n!1 2n 7
• Therefore, lim
17 n
ˇ ˇ ˇˇ nC16n ˇˇ ˇ nC16 ˇ ˇ>ˇ 2 2 ˇ>ˇ 2 ˇD D ˇ 17n n2 n C.n 7/ 2n 7
D 2.
3.5.6 Limits of Monotone Sequences and Mathematical Induction A function f W A ! R is said to be bounded above if the set ff .x/ j x 2 Ag is bounded above, that is, if there exists an M 2 R such that f .x/ M for all x in the domain A of f . In this case M is an upper bound of f . Similarly, the function is said to be bounded below if the set ff .x/ j x 2 Ag is bounded below. A function that is both bounded above and bounded below is said to be bounded. Because a realvalued sequence is just a real-valued function whose domain is the natural numbers, N, these definitions apply to sequences as well.
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63
One of the most important properties of monotone sequences is that monotone increasing sequences that are bounded above must converge and monotone decreasing sequences that are bounded below must converge. Thus, bounded monotone sequences converge. If a monotone sequence does not converge, then its terms must continue to grow without bound and approach plus or minus infinity. So how would you prove that a monotone increasing sequence that is bounded above converges? When proving a limit of the form lim an D L, you can work with n!1
the inequality > jan Lj in order to find an appropriate value of N that allows you to use the definition of limit to complete the proof. But in this case, you do not have a general expression for the terms an , and you have not been given a value for L. Somehow you need to use the only known facts about , that is, the fact that the sequence is both monotone increasing and bounded, to come up with a candidate to serve as the limit, L, in the proof. The definition of a sequence being bounded above holds the key. That definition says that the sequence is bounded above if the set fan j n 2 Ng is bounded above, so there is a real number M which is greater than or equal to each term of the sequence. Will this M be the limit of the sequence? Well, not usually. If M is an upper bound for the sequence, then so are M C 1, M C 100, and M C 20;000. They are all upper bounds, but they cannot all be limits of the sequence. You should recognize that the terms of the sequence must get close to the limit, and the only upper bound of the set fan j n 2 Ng that the terms could get close to is the least upper bound of the set. Since fan j n 2 Ng is both nonempty and bounded above, the Completeness Axiom for the real numbers guarantees that such a least upper bound exists. This gives you a candidate for L. The proof will require you to show that for all n greater than some N, the terms of the sequence, , are within of L. How can this be arranged? Here is where you can use the fact that the sequence is monotone increasing because once you find a single term, an , that gets within of L, all the terms that come after this term in the sequence will necessarily have to be between an and L, so they also will be within of L. How do you find one term, an , within of L? This follows from the fact that L is a least upper bound of fan j n 2 Ng. Because L is the least upper bound, L being less than the least upper bound, L, is not an upper bound, so there must be an element of the set fan j n 2 Ng greater than L . This gives all the tools needed for the proof (Fig. 3.6).
L–
a1
a2
a3
a4
L
a5 … aN an
Fig. 3.6 Proving bounded monotone sequences converge
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3 Limits
So how would you write the proof? Certainly the proof would begin with selecting a generic sequence and making a statement about the properties the sequence is assumed to have, that is, its being monotone increasing and bounded above. Then, the proof would proceed to justify the existence of the least upper bound for the set of terms of the sequence; that will give you the target value of L. Then, as with most proofs about limits, it would select a value for > 0. Unlike the limit proofs earlier in this chapter, one cannot immediately state a value for N. The existence of N must be proved as discussed in the previous paragraph. Finally, the properties of the sequence can be brought together to show jan Lj < for all n > N. Here is one possible proof. PROOF: A monotone increasing sequence that is bound above converges. • Let be a monotone increasing sequence of real numbers that is bounded above. • Since the set of terms A D faj j j 2 Ng contains a1 , it is nonempty, and since it is bounded above, the Completeness Axiom guarantees that A has a least upper bound, L. • Given > 0, the number L is less than L. Since L is the least upper bound of A, L is not an upper bound of A. Thus, there is an N 2 N such that the term aN is in A and is larger than L . • Select an n > N. • Because is monotone increasing, an aN . Because L is an upper bound for A, an L. Therefore, L < aN an L, and jan Lj < j.L / Lj D . • This proves that the sequence has limit L and that converges. Note that the proof needs to refer to the sequence as well as a particular element of the sequence an . It could be confusing to the proof reader to use the variable n in both contexts here, especially since the sequence notation is used after the choice of a specific value of n is made. That is the reason the proof changed to using the variable j to refer to a generic term index. Then, it could refer to a specific term using index n without confusing the two uses. There is also a theorem stating that a monotone decreasing sequence that is bounded below converges. The proof of this is left as an exercise. As an illustration of the usefulness of the above result, consider a sequence p defined recursively by a1 D 2, and for n p 1, anC1 D an C 12. That is, p p p a1 D 2, a2 D a1 C 12 D 14, a3 D 14 C 12, and so forth. One can prove that this sequence converges by showing that the sequence is both monotone increasing and bounded above. Indeed, both of these facts can be established by mathematical induction. The reader is likely already familiar with proofs by mathematical induction, but this is an appropriate opportunity to review the method and its merits.
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65
Suppose the variable n represents any natural number, and there is a statement S.n/ that includes this variable as part of the statement. For example, the statement could be lim xn D an . Mathematical induction is a proof technique that uses the x!a
following proof template to show that S.n/ is true for all n greater than or equal to some base value b 2 N. TEMPLATE for using mathematical induction to prove the statement S.n/ is true for all natural numbers n b. • SET THE CONTEXT: The statement will be proved by mathematical induction on n for all n b. • PROVE S.b/: Prove that the statement is true when the variable n is equal to the base value, b. • STATE THE INDUCTION HYPOTHESIS: Assume that S.n/ is true for some natural number n D k b. • PERFORM THE INDUCTION STEP: Using the fact that S.k/ is true, prove that S.k C 1/ is true. • STATE THE CONCLUSION: Therefore, by mathematical induction, S.n/ is true for all natural numbers n b. It is important to understand that the technique of mathematical induction works. That is, if the statement S.b/ is true, and if the statement S.k/ ! S.k C 1/ is true, then, in fact, S.n/ must be true for all natural numbers n b. Certainly, S.b/ is true. Because S.b/ is true, and S.k/ ! S.k C 1/ is true for all k b, it follows that S.b/ ! S.bC1/, so S.bC1/ is true. Then S.bC1/ ! S.bC2/, S.bC2/ ! S.bC3/, and so forth, so the fact that S.n/ is true for all n b follows. The strength of mathematical induction is that it is often much easier to provide a proof for the one step S.k/ ! S.k C 1/ than it is to prove S.n/ in the general case. The reader has likely seen many statements proved by mathematical induction while studying Algebra, Calculus, or just about any other branch of mathematics. Mathematical induction is an excellent tool for proving that the previously introduced recursive sequence is both monotone increasing and bounded above. p p Clearly, a2 D 14 > 4 D 2 D a1 so a1 < a2 . Supposepthat for some p k 1 one has ak < akC1 . Then it follows that ak C12 < akC1 C12 so ak C 12 < akC1 C 12 which shows that akC1 < akC2 . Thus, by mathematical induction it follows that an < anC1 for all n, and the sequence is monotone increasing. Alsop clear is that ap1 D 2 < 4. p Suppose that for some k 1 that ak < 4. Then akC1 D ak C 12 < 4 C 12 D 16 D 4. Thus, by mathematical induction it follows that an < 4 for all n, and the sequence is bounded above. The limit of this sequence can be shown p to be 4. In particular, if the limit is L, one can conclude that a C 12 should be n p which should equal the limit of an which is also L. Thus, one converging to L C 12p would expect that L D L C 12. This equation has only one positive real solution, L D 4.
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3 Limits
3.5.7 Cauchy Sequences A Cauchy sequence is a sequence whose terms get close together. As with the definition of limit, the concept of “close” needs to be made precise. As with the definition of limit, “close” means that given any tolerance > 0, one can go out far enough in the sequence to ensure that all terms of the sequence beyond that point are within of each other. Thus, a sequence is Cauchy if for every > 0 there is an N such that if natural numbers m and n are both greater than N, then jam an j < . If a sequence of real numbers converges, then the sequence is Cauchy. The proof of this fact uses a strategy employed repeatedly in Analysis, that is, if two quantities are very close to the same value, then they must be very close to each other. This standard technique for proving that two quantities are close to each other involves the use of the triangle inequality. In particular, if lim aj D L, then for every > 0 j!1
there is an N such that if natural number n > N, then jan Lj < . Well then, certainly if m and n are both natural numbers greater than N, then both jam Lj < and jan Lj < . Adding these two inequalities together shows that jam Lj C jan Lj < C . The triangle inequality states that for any real numbers x and y, jxj C jyj jx C yj. Thus, 2 > jam Lj C jan Lj D jam Lj C jL an j j.am L/ C .L an /j D jam an j. Of course, the definition of Cauchy sequence requires you to show that jam an j is less than , not 2. But you have an enormous amount of flexibility when working with these types of inequalities, so you could have asked instead for an N such that for all natural numbers n greater than N, you have jan Lj less than 2 rather than less than . Thus, the proof could be as follows. PROOF: Every convergent sequence is Cauchy. • Let be a sequence of real numbers with lim aj D L. j!1
• Let > 0 be given. • From the definition of limit, there is a number N such that for all natural numbers j > N, it follows that jaj Lj < 2 . • Then for all natural numbers m and n greater than N, jam Lj < 2 and jan Lj < 2 , so D 2 C 2 > jam Lj C jan Lj D jam Lj C jL an j j.am L/ C .L an /j D jam an j. • This shows that the convergent sequence is Cauchy. Note that the converse of this theorem also holds. That is, any sequence of real numbers that is Cauchy is a convergent sequence. This result will be proved in Sect. 3.7. An important and useful consequence of the above theorem is its contrapositive: If a sequence is not Cauchy, then it does not converge. Often when one wants to show that a sequence does not converge, one shows that there is some > 0 such that for every N there are natural numbers m and n greater than N for which jam an j . Another important property of Cauchy sequences is that all Cauchy sequences are bounded. If the sequence is Cauchy, then there is a natural number N such that whenever m; n N, the difference jam an j < 1. The set fa1 ; a2 ; a3 ; : : : ; aN g
3.5 Limit of a Sequence
67
is a finite set, so it is bounded by some number, K. That is, jan j K for all n N. If m > N, then, since both N and m are greater than or equal to N, it follows that jam aN j < 1 from which it follows that jam j < jaN j C 1 K C 1. Then the sequence is necessarily bounded above by K C 1 and below by .K C 1/, and the sequence is bounded. A complete proof follows. PROOF: All Cauchy sequences are bounded. • Let be a Cauchy sequence. • Then there is a natural number N such that for all m; n N, jam an j < 1. • The set fa1 ; a2 ; a3 ; : : : ; aN g is a finite set, so there is a K such that the set is bounded above by K and bounded below by K. • Let m be any natural number. If m N, then jam j K. If m > N, then jam aN j < 1, so jam j D jam aN C aN j jam aN j C jaN j < 1 C K. • It follows that all terms of the sequence lie between .K C 1/ and K C 1, and, thus, the sequence is bounded. One consequence of the last two results is that since all convergent sequences are Cauchy, all convergent sequences are bounded. The concept of a Cauchy sequence is not only applied to sequences of numbers but also to much more general sequences such as sequences of vectors, sequences of functions, and sequences of linear operators. Of course, one would need a way to discuss distances between the terms of a sequence in these other contexts, but when that makes sense, the concept of a Cauchy sequence becomes important.
3.5.8 Exercises 1. Which of the following sequences are monotone? Which of them are bounded above? Which of them are bounded below? Which of them are bounded? (a) (b) (c) (d) (e) (f) (g)
an an an an an an an
D .1/n n D nC1 D 5n .1/n D 5n n D 1C.1/ nCn1 D 5 n.1/n D 1 12 13 1n
2. Write proofs of each of the following limits. 6n n!1 3nC1 lim 4n1 n!1 nC6
(a) lim (b)
(c) lim
D2 D4
n2 C2nC1
2 n!1 n 2n5
D1
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p 3. If a1 D 3 and an is defined recursively by anC1 D 3an C 10, show that the sequence converges. p 4. If a1 D 7 and an is defined recursively by anC1 D 3an C 4, show that the sequence converges. 5. Prove that a monotone decreasing sequence that is bounded below converges. 6. Let be any sequence. Prove that has a monotone subsequence. 7. Prove that if is a sequence such that L D lim a2n D lim a2nC1 , then the n!1 n!1 sequence converges to L. 8. Prove that if is a sequence that converges to L, then the sequence a1 ; a1 ; a2 ; a2 ; a3 ; a3 ; : : : also converges to L. 9. Prove that if is a sequence that converges to L, then the sequence a1 ; a2 ; a2 ; a3 ; a3 ; a3 ; a4 ; a4 ; a4 ; a4 ; : : : also converges to L.
3.6 Proving That a Limit Does Not Exist 3.6.1 Why a Limit Might Not Exist lim f .x/ D L means that if x is required to stay close to a, then f .x/ will stay close
x!a
to L. So what does it mean for lim f .x/ not to exist? Intuitively, it could mean that x!a
in every neighborhood of a there are values of x for which f .x/ is close to one value L1 and other values of x for which value L2 . That is what f .x/ is close to another 4x 5 if x < 2 happens with the function f .x/ D as x approaches 2. For some 10 2x if x 2 values of x near 2, f .x/ is close to 3, and for some values of x near 2, f .x/ is close to 6. Thus, the limit does not exist. Another well-known example is f .x/ D sin 1x which oscillates wildly as x approaches zero, and in every neighborhood of 0, the function takes on all values in the interval Œ1; 1 infinitely often. Another way for the limit not to exist is for the values of f .x/ to grow without bound and approach xC3 infinity or negative infinity such as what happens to f .x/ D .x5/ 2 as x approaches 5. One can write a proof showing that a particular function has no limit at x D a, but before discussing how to do this, it is worth taking a close look at the definition of limit.
3.6.2 Quantifiers and Negations To say that a function f has a limit at x D a is to say that there exists a real number L such that for all > 0 there is a ı > 0 such that for every x, 0 < jx aj < ı implies jf .x/ Lj < . This definition is actually a fairly complicated statement. At the heart of it is the conditional statement “0 < jx aj < ı implies jf .x/ Lj < .”
3.6 Proving That a Limit Does Not Exist
69
But this is an open statement, that is, even though the function f and the limit point a are supposedly known, the statement contains variables x, L, , and ı, all of which are unknown. Thus, this open statement does not have any truth value until these four variables have been stipulated. They are stipulated with four phrases: “there is a real number L,” “for all > 0,” “there is a ı > 0,” and “for every x.” These four phrases are called quantifications of the variables because they indicate for which values of the variables the following statement must hold. Two of the phrases use the existential quantifier “there exists.” It indicates that there is at least one value of the variable that will make the following statement true. The other two phrases use the universal quantifier “for all.” It indicates that every possible value of that variable will make the following statement true. So • The statement “there exists a real number L such that for all > 0 there is a ı > 0 such that for every x, 0 < jx aj < ı implies jf .x/ Lj < ” begins with the existential quantifier “there exists a real number L,” and the entire statement is true if, in fact, there is a value of the variable L that makes the following statement true, that is, “for all > 0 there is a ı > 0 such that for every x, 0 < jx aj < ı implies jf .x/ Lj < .” • The statement “for all > 0 there is a ı > 0 such that for every x, 0 < jx aj < ı implies jf .x/ Lj < ” begins with the universal quantifier “for all > 0,” and the entire statement is true if, in fact, every possible positive value of the variable makes the following statement true, that is, “there is a ı > 0 such that for every x, 0 < jx aj < ı implies jf .x/ Lj < .” • The statement “there is a ı > 0 such that for every x, 0 < jx aj < ı implies jf .x/ Lj < ” begins with the existential quantifier “there is a ı > 0,” and the entire statement is true if, in fact, there is a positive value of the variable ı that makes the following statement true, that is, “for every x, 0 < jx aj < ı implies jf .x/ Lj < .” • The statement “for every x, 0 < jx aj < ı implies jf .x/ Lj < ” begins with the universal quantifier “for every x,” and the entire statement is true if, in fact, every possible value of the variable x makes the following statement true, that is, “0 < jx aj < ı implies jf .x/ Lj < .” A proof that no limit exists must prove the negation of the statement that says that a limit does exist, so it is important that one can generate the negation of a statement that contains quantifiers such as this one does. The logic of doing this is not hard to follow. Suppose the P.y/ is a statement that depends on the value of a variable y. Then the universally quantified statement “for every y, P.y/” says that P.y/ is true for every possible value of y. The negation of “for every y, P.y/” must be that it is false that every value of y makes P.y/ true, so there must be at least one y that makes P.y/ a false statement. This means that the negation of “for every y, P.y/” is the statement “there is a y such that :P.y/.” To negate a universally quantified statement, change the universal quantifier to an existential quantifier and negate the statement that follows. What if the original statement is an existentially quantified statement such as “there is a y such that P.y/?” This statement says that some value of y makes
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P.y/ true. The negation of this statement must be that no value of y makes P.y/ true which is to say that every value of y makes P.y/ a false statement. This means that the negation of “there is a y such that P.y/” is the statement “for all y, :P.y/.” To negate an existentially quantified statement, change the existential quantifier to a universal quantifier and negate the statement that follows. The statement that f has a limit at x D a is a statement that has an existential quantifier followed by a universal quantifier followed by an existential quantifier followed by a universal quantifier followed by a conditional statement. To prove that f does not have a limit at x D a requires a proof of the negation of that statement. From the previous discussion it is now clear that to get the negation of the statement that f has a limit at a, you must flip the two existential quantifiers to universal quantifiers, flip the two universal quantifiers to existential quantifiers, and end with the negation of the conditional statement. The result is “for all real numbers L there is an > 0 such that for all ı > 0 there is an x such that 0 < jx aj < ı and jf .x/ Lj .”
3.6.3 Proving No Limit Exists Getting back to writing a proof that a limit does not exist, the proof would need to show that for every real number L there is an > 0 such that for every ı > 0 there is an x within ı of a such that jf .x/ Lj . This is often done by exhibiting an x1 and an x2 within ı of a such that f .x1 / and f .x2 / are so far apart that they could not both be within of any L. That suggests the following template for proving that a particular limit does not exist. TEMPLATE for proving lim f .x/ does not exist x!a
• SET THE CONTEXT: Make statements about what is known about the function f and the number a. • SELECT AN ARBITRARY LIMIT L: Given L 2 R, • PROPOSE A VALUE FOR : let D . Here you would insert a value for . • SELECT AN ARBITRARY ı > 0: Select ı > 0. • SELECT VALUES FOR x1 AND x2 : Let x1 D and x2 D . Note that 0 < jx1 aj < ı, 0 < jx2 aj < ı, and jf .x1 /f .x2 /j 2. You would have selected appropriate x1 and x2 in such a way that jf .x1 / f .x2 /j exceeds 2. • LIST IMPLICATIONS: Assume that jf .x1 / Lj < and jf .x2 / Lj < . Then 2 D C > jf .x1 / Lj C jf .x2 / Lj D jf .x1 / Lj C jL f .x2 /j jf .x1 / L C L f .x2 /j D jf .x1 / f .x2 /j. • STATE THE CONTRADICTION: This shows that 2 > jf .x1 / f .x2 /j which is a contradiction. • STATE THE CONCLUSION: Thus, it cannot hold that both jf .x1 / Lj < and jf .x2 / Lj < , and the limit does not exist.
3.6 Proving That a Limit Does Not Exist
71
4x 5 if x < 2 For example, consider the limit of f .x/ D as x approaches 10 2x if x 2 2. Here the limit from the left is 3, and the limit from the right is 6. Thus, no matter how close x is supposed to be to 2, there will be values x1 and x2 within that required tolerance where f .x1 / is close to 3 and f .x2 / is close to 6. If f .x1 / and f .x2 / are both supposed to be within of some limit L, then it will follow that f .x1 / and f .x2 / will have to be within 2 of each other. Again, you employ the technique of showing that two quantities close to the same value must be close to each other. In particular, if x1 is chosen to be less than 2, f .x1 / will be less than 3. If x2 is chosen to be between 2 and 2 12 , f .x2 / will be greater than 5. In this case it would be impossible to have f .x1 / and f .x2 / within 2 of each other, and, therefore, it would be impossible to have them both within D 1 of some limit L. This suggests that you will get a contradiction if you set D 1. Indeed, if a ı > 0 is chosen, you can let x1 D 2 2ı (that is, less than 2 but within ı of 2), and let x2 D min 2 C 2ı ; 2 C 12 (that is, greater than 2 but within ı of 2 and not so large that f .x/ is less than 5). The point of all of this is that now, no matter what value is chosen for L, f .x1 / and f .x2 / are more than 2 apart, so how could they both be within 1 of L? Specifically, if jf .x1 / Lj < 1 and jf .x2 / Lj < 1, it follows from the triangle inequality that 2 D 1 C 1 > jf .x1 / Lj C jf .x2 / Lj D jf .x1 / Lj C jL f .x2 /j jf .x1 / L C L f .x2 /j D jf .x1 / f .x2 /j showing 2 > jf .x1 / f .x2 /j which cannot hold. Here is the complete proof (Fig. 3.7). Fig. 3.7 f has no limit at xD2
ε ε
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3 Limits
PROOF: The function
4x 5 if x < 2 10 2x if x 2
has no limit as x ! 2.
4x 5 if x < 2 . 10 2x if x 2 Given any value for L, let D1, and let ı > 0 be given. Let x1 D 2 2ı and x2 D min 2 C 2ı ; 2 C 14 . Note that 0 < jx1 2j < ı and 0 < jx2 2j < ı. Since x1 < 2, it follows that f .x1 / < 3. Since x2 > 2 and x2 < 2 14 , it follows that f .x2 / > 5. As a consequence jf .x1 /f .x2 /j D f .x2 /f .x1 / > 53 D 2. If jf .x1 / Lj < D 1 and jf .x2 / Lj < D 1, it would follow that 2 D 1 C 1 > jf .x1 / Lj C jf .x2 / Lj D jf .x1 / Lj C jL f .x2 /j jf .x1 / L C L f .x2 /j D jf .x1 / f .x2 /j > 2. This shows that 2 > 2 which is a contradiction. Thus, it cannot hold that both jf .x1 / Lj < and jf .x2 / Lj < , and the limit does not exist.
• Let f .x/ D • • • • •
•
It is even easier to show that the function f .x/ D sin 1x has no limit as x approaches 0. This is because for every ı > 0 it is easy to find x1 and x2 between 0 and ı such that f .x1 / D 1 and f .x2 / D 1. This makes it impossible to find an L where jf .x1 / Lj < 1 and jf .x2 / Lj < 1. Thus, the proof follows the given template for proving that a limit does not exist (Fig. 3.8). Fig. 3.8 Graph of sin
1 x
3.6 Proving That a Limit Does Not Exist
73
PROOF: The function sin 1x has no limit as x ! 0. Let f .x/ D sin 1x . Given any value for L, let D 1, and let ı > 0 be given. 1 2 2 Select integer k > 2ı . Let x1 D .4kC1/ and x2 D .4kC3/ . 2 1 Note that both x1 and x2 are positive and less than 4k D 2k < ı,
1 3 f .x1 / D sin .2k C 2 / D 1, and f .x2 / D sin .2k C 2 / D 1. If jf .x1 / Lj < D 1 and jf .x2 / Lj < D 1, it would follow that 2 D 1 C 1 > jf .x1 / Lj C jf .x2 / Lj D jf .x1 / Lj C jL f .x2 /j jf .x1 / L C L f .x2 /j D jf .x1 / f .x2 /j D 2. This shows that 2 > 2 which is a contradiction. • Thus, it cannot hold that both jf .x1 / Lj < and jf .x2 / Lj < , and the limit does not exist.
• • • • • •
If the function f .x/ is unbounded as x approaches a, then there is an even easier template to use for the proof that f .x/ has no limit. The idea is that since f .x/ is unbounded, for any proposed limit L one can find an x close to a such that jf .x/j > jLj C 1. Then the difference jf .x/ Lj will be forced to be greater than 1. Consider, xC3 for example, the function f .x/ D .x5/ 2 as x approaches 5. Given L, you will want an x with 1 .x5/2
xC3 >L .x5/2 1 , so by jx5j
C 1. But with x within 1 of 5, you could claim that 1 jLjC1
xC3 .x5/2
>
> making jx 5j < you will have the inequality that you need. Note that the absolute value function was introduced in jLj C 1 to take care of the embarrassing circumstance that L is negative, and in particular, when L D 1. The proof is as follows. PROOF: The function • • • • • • •
xC3 .x5/2
has no limit as x ! 5.
xC3 Let f .x/ D .x5/ 2. Given any value for L, let D 1, and let ı > 0 be given. 1 . Select a value of x between 5 and 5 C min 1; ı; jLjC1 Note that 0 < jx 5j < ı xC3 1 1 and f .x/ D .x5/ 2 > .x5/2 > x5 > jLj C 1. It follows that jf .x/ Lj > jLj C 1 L 1. Thus, it cannot hold that jf .x/ Lj < , and the limit does not exist.
3.6.4 Exercises Write the negation of each of the following statements. 1. There exists x such that x2 D A. 2. For all x there is a y such that g.x/ D f .y/.
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3 Limits
3. There is an integer k such that f .x/ f .k/ for all x between k and k C 1. 4. For all x > 0 and all y > 0 there exists a z < 0 such that f .z/ xf .y/. Prove that the following limits do not exist. x 5. f .x/ D jxj as x ! 0 1 as x ! 1 6. f .x/ D x sin x1 5x if x < 3 7. f .x/ D as x ! 3 4x if x 3 8. f .x/ D x244 as x ! 2
3.7 Accumulation Points A set A has an accumulation point p if for every > 0 there is an x 2 A with x ¤ p and jx pj < . Informally, p is an accumulation point of A if there are points of A that are arbitrarily close to p. Note that the fact that p is an accumulation point of the set A has˚nothing to do with whether p is actually an element of A. For example, the set A D 1n j n 2 N has one accumulation point, 0, because for every > 0 there is an n 2 N with 1n < . Here the accumulation point 0 is not an element of the set A. The set B D Œ0; 4 (the closed interval from 0 to 4) has infinitely many accumulation points. Indeed, every element of the interval B is an accumulation point of B because for each x 2 Œ0; 4 and each > 0 there are infinitely many points in B within of x. Here all of the accumulation points of B are in B. Each point x 2 Œ0; 4 is also an accumulation point of the set C D .0; 4/ \ Q, the set of rational numbers between 0 and 4. Here, some of the accumulation points are in C, and some are not. The set of natural numbers, N, has no accumulation points. An element a of a set that is not an accumulation point of that set is called an isolated point of the set. For any isolated point a, there is an > 0 such that a is the only element of the set in the interval .a ; a C / (Fig. 3.9). A word of warning is needed here. The term accumulation point is not used the same way by all authors. Many texts, especially those in Topology, will use the terms limit point or cluster point instead of accumulation point. Even more confusing is that some texts use the term accumulation point for something different.
b
Fig. 3.9 Set with accumulation point a and isolated point b
a
3.7 Accumulation Points
75
The first observation to make about accumulation points is that if p is an accumulation point of set A, then for every > 0 there is not only one point of A within of p but infinitely many points of A within of p. The definition of accumulation point guarantees at least one point of A within of p, but once one point, x 2 A, is found to be within of p, the definition also says that there must be another point y 2 A with 0 < jy pj < jx pj. Since for each x 2 A close to p there must be another point y 2 A even closer to p, it follows that there are infinitely many points of A within of p. Perhaps the most used fact about accumulation points is known as the Bolzano– Weierstrass Theorem which states that every infinite bounded set of real numbers has an accumulation point. As pointed out earlier, N has no accumulation points, and it is an infinite set. But N is not a bounded set. Intuitively, one cannot have a bounded infinite set without an accumulation point because one runs out of places to put the infinite number of points. If the points of a set are not allowed to bunch up anywhere, then one will not be able to find room for infinitely many of the points within a bounded interval. There are several good strategies used to prove the Bolzano–Weierstrass Theorem, and two of those strategies are presented here. Of course, one only needs one good strategy to prove a theorem, but these proofs are instructive and use techniques commonly employed in Analysis proofs. One begins each proof with a statement about the set A being an infinite bounded set. Since A is a bounded set, it will have a lower bound, a, and an upper bound, b showing that A Œa; b. The first strategy is to construct the set S D fx a j Œa; x \ A is finiteg, that is, a value x a is in the set S if there are finitely many element of A which fall in the interval Œa; x. First observe that the set S is an interval. This follows because if y 2 S, then Œa; y \ A is finite, so if x is between a and y, then Œa; x \ A Œa; y \ A must also be finite, and x 2 A. The next observation is that S is not empty because the point a, whether or not it is in A, is in S since Œa; a \ A contains at most one point, so it is finite. Since Œa; b \ A D A is an infinite set, the set S is bounded above by b. The Completeness Axiom now shows that S must have a least upper bound, p. It will follow that p is an accumulation point of A because for all > 0, the set A will have only finitely many elements less than p but infinitely many elements less than p C implying that there are infinitely many elements of A within of p. Here is the complete proof.
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PROOF (Bolzano–Weierstrass Theorem): Every infinite bounded set of real numbers has an accumulation point. • Let A be an infinite bounded set of real numbers. • Because A is bounded, it has a lower bound, a, and an upper bound, b, showing that A Œa; b. • Define set S D fx a j Œa; x \ A is finiteg. • Note that a 2 S since Œa; a \ A is finite, so S is nonempty. • Note that if z b, then Œa; z \ A D A is an infinite set, so z … S showing that S is bounded above by b. • By the Completeness Axiom, S has a least upper bound, p. • Given > 0, p < p so p is not an upper bound of S. Hence, there is a y 2 S with y > p . It follows that there are only finitely many elements of A less than or equal to y. • Also, p C > p, so p C … S. It follows that Œa; p C \ A is infinite. • Thus, there must be infinitely many elements of A between p and p C , and there must be an element of A not equal to p within of p. • This shows that p is an accumulation point of A. The second strategy also begins with the interval Œa; b that contains the infinite bounded set, A. One can rename the end points of this interval to be a1 D a and 1 b1 D b. Since Œa1 ; b1 \ A D A is infinite, it follows that either Œa1 ; a1 Cb \ A or 2 a1 Cb1 a1 Cb1 Œ 2 ; b1 \ A is an infinite set. If Œa1 ; 2 \ A is infinite, define a2 D a1 and 1 1 b2 D a1 Cb . Otherwise, define a2 D a1 Cb and b2 D b1 . In either case, Œa2 ; b2 \ A 2 2 is an infinite set. This procedure can be repeated so that for every n 2 N, one gets an interval Œan ; bn where Œan ; bn \ A is infinite, and each interval is half the length of the previous interval. Also, the sequence of left endpoints, , is a monotone increasing sequence bounded above by b, and the sequence of right endpoints, , is a monotone decreasing sequence bounded below by a. Thus, both of these sequences converge. In fact, both of these sequences must converge to the same limit, p. This follows because the distances between the terms of the sequences, bn an , keep getting smaller and converge to 0. Given an > 0, it will follow that there is an n such that an and bn are both within of p. Thus, .p ; p C / \ A contains Œan ; bn \ A which is infinite. Here is the complete proof.
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77
PROOF (Bolzano–Weierstrass Theorem): Every infinite bounded set of real numbers has an accumulation point. • Let A be an infinite bounded set of real numbers. • Because A is bounded, it has a lower bound, a1 , and an upper bound, b1 , showing that A Œa1 ; b1 and Œa1 ; b1 \ A is an infinite set. • Define sequences and recursively as follows. • Suppose, for natural number n, an and bn have been defined so that Œan ; bn \ n A is an infinite set. If Œan ; an Cb \ A is infinite, then define anC1 D an and 2 an Cbn n and bnC1 D bn . In either bnC1 D 2 . Otherwise, define anC1 D an Cb 2 case, ŒanC1 ; bnC1 \ A is an infinite set. • By the way the sequences are constructed, for each n it follows that an anC1 < bnC1 bn showing that is a monotone increasing sequence bounded above by each bi , and is a monotone decreasing sequence bounded below by each ai . a1 • Also, by the way the sequences are constructed, for each n, bn an D b21n1 . • Thus, the bounded monotone sequence must converge to a number pa , and the bounded monotone sequence must converge to a number a1 pb . But pb pa bn an D b21n1 and, therefore, pb pa must be zero. Let p D pa D pb , and note that for each n, p 2 Œan ; bn . a1 • Given > 0, select a natural number n such that b21n1 < . Then p < an p bn < p C . Hence, Œan ; bn \ A .p ; p C / \ A is infinite showing that there is an element of A not equal to p but within of p. • This shows that p is an accumulation point of A. You now have the machinery necessary to prove the result mentioned in Sect. 3.6 that all Cauchy sequences converge. The difficulty in proving this result earlier was that given a Cauchy sequence , it was not clear what real number would play the role of the limit L of the sequence. Now, the Bolzano–Weierstrass Theorem can provide an accumulation point to serve as this limit. There are two cases to consider. If the set of values in the sequence, fan g, is a finite set, then for the sequence to be Cauchy, the sequence will necessarily need to be constant from some point on, and, therefore, the sequence will converge. If the set of values in the sequence is infinite, then since all Cauchy sequences are bounded, the set of values in the sequence will be bounded and will have to have an accumulation point. It is then straightforward to show that the sequence converges to this accumulation point.
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PROOF: All Cauchy sequences converge. • Let be a Cauchy sequence. Let A be the set of terms of the sequence, fan g. • CASE 1: The set A is finite. If A contains only one value, then the sequence is constant and converges to that constant. If A contains more than one value, then, since the range of the sequence is finite, so is the set of differences an am of values in the sequence. Let d be the smallest positive difference between any two values in the sequence, and let D d2 . Because the sequence is Cauchy, there is an N such that whenever m; n > N, the difference jam an j < . But the smallest positive difference between any two terms of the sequence is d > , so it follows that am an D 0. Thus, the sequence is constant for all terms an with n > N, and, again, the sequence must converge. • CASE 2: The set A is infinite. Since all Cauchy sequences are bounded, A is a bounded infinite set, and thus, by the Bolzano–Weierstrass Theorem, A has an accumulation point, p. Because is Cauchy, given > 0 there is an N such that for all m; n > N, jam an j < 2 . Also, since p is an accumulation point of A, there are infinitely many values of A within 2 of p. Surely there is a natural number k > N such that jak pj < 2 . Then, for all n > N, it follows that jan pj D j.an ak /.pak /j jan ak jCjak pj < 2 C 2 D . Thus, the sequence converges to p. • Therefore, all Cauchy sequences must converge. Up to this point, the discussion of the limit lim f .x/ took place only for those x!a
functions defined for all x ¤ a in an open interval containing a. The definition of limit can now be extended. It should not be required that the function f be defined for all x in an open interval containing a but that f be defined at enough points so that it makes sense to allow x to approach a. In other words, a only needs to be an accumulation point of the domain of f . Note that if a is not an accumulation point of the domain of f , then there will be an open interval containing a where f were not defined (except perhaps at a itself). Thus, no sense could be made out of x approaches a. On the other hand, if a is an accumulation point of f , it makes sense to define the limit of f at a to be L or lim f .x/ D L to mean that for all > 0 there is x!a a ı > 0 such that for all x in the domain of f , 0 < jx aj < ı implies jf .x/ Lj < . Similarly, to define lim f .x/ D L one does not need f to be defined in an x!1 entire interval stretching to positive infinity. It is sufficient that f .x/ is defined for arbitrarily large values of x so that x can be allowed to approach infinity. One way of saying this is that the domain of f should be unbounded above. This is what was done, for example, when defining the limit of a sequence which is the limit of a function defined for the natural numbers only. Similarly, lim f .x/ D L can be x!1 defined for f when the domain of f is unbounded below.
3.8 Infinite Limits
79
3.7.1 Exercises 1. Write a definition for lim f .x/ where a is an accumulation point of the domain x!aC
of f . Identify the accumulation points, if any, of the following sets. ˚ n ˇ ˇn 2 N 2. nC2 ˇ ˚ 3. x 2 Q ˇ x2 < 2 ˚1 3 ; 2; 52 ; : : : 4. 2 ; 1; ˚m ˇ 2 5. 2n ˇ m; n 2 N n o n C4 ˇ ˇn 2 N 6. 2n.1/ 3nC5
3.8 Infinite Limits xC3 2 x!5 .x5/
In Sect. 3.6 it was shown that lim
does not exist. But more can be said about
this limit. The reason that the limit does not exist is that the function grows without bound and, therefore, does not approach any real number value. This behavior can be quantified by saying that the limit of the function is infinity. Of course, it does not make sense to say that the function is getting close to infinity, since no real number is very close to infinity. In the definition of lim f .x/ where it had to be made clear x!1 what x approaching infinity meant, it was said that there was a number N such that jf .x/ Lj was small whenever x > N. Similarly, to say that f .x/ approaches infinity, one needs to say that for any real number M, f .x/ can be made larger than M. Thus, if a is an accumulation point of the domain of function f .x/, the following two similar definitions can be given. • The limit of f at a is infinity or lim f .x/ D 1 means that for every M 2 R there x!a is a ı > 0 such that if x is in the domain of f with 0 < jx aj < ı, then f .x/ > M. • The limit of f at a is negative infinity or lim f .x/ D 1 means that for every x!a M 2 R there is a ı > 0 such that if x is in the domain of f with 0 < jx aj < ı, then f .x/ < M. What if f .x/ approaches infinity or negative infinity as x is allowed to approach either infinity or negative infinity? Each of these ideas can be accommodated resulting in four similar definitions. Remember that the limit of f as x approaches infinity makes sense only if the domain of f is unbounded above, and as x approaches negative infinity only if the domain of f is unbounded below. Here are the four definitions. • The limit of f as x approaches infinity is infinity or lim f .x/ D 1 means that x!1 for every M 2 R there is an N 2 R such that if x is in the domain of f with x > N, then f .x/ > M.
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• The limit of f as x approaches infinity is negative infinity or lim f .x/ D 1 x!1 means that for every M 2 R there is an N 2 R such that if x is in the domain of f with x > N, then f .x/ < M. • The limit of f as x approaches negative infinity is infinity or lim f .x/ D 1 x!1 means that for every M 2 R there is an N 2 R such that if x is in the domain of f with x < N, then f .x/ > M. • The limit of f as x approaches negative infinity is negative infinity or lim f .x/ D 1 means that for every M 2 R there is an N 2 R such that x!1 if x is in the domain of f with x < N, then f .x/ < M. xC3 2 x!5 .x5/
For example, how would you prove that lim
show that for every M 2 R there is a ı > 0 such that
D 1? You would need to xC3 .x5/2
> M when x is within
xC3 ı of 5 with x ¤ 5. Working backwards, you would start with .x5/ 2 > M. This is a complicated inequality with which to work, so it would be more convenient to work xC3 with an inequality that is easier to solve. If you want f .x/ D .x5/ 2 to be bigger than M, it would be sufficient to make some fraction smaller than f .x/ bigger than M. 1 For example, for all x > 2 the fraction .x5/ 2 is smaller than f .x/. Moreover, for
x within 1 of 5, 1 jx5j
1 jx5j
is smaller than
1 . .x5/2
Thus, it would be sufficient to make
> M which, under the condition of M > 0, happens when jx 5j < M1 . A proof would need to take care of the embarrassing case of M 0, perhaps by 1 making ı D jMjC1 since jMj C 1 is always bigger than M and is always positive. Another way to handle this is to write a proof that assumes that M is positive. In fact, one could just stipulate that M > 1 by inserting the often used phrase without loss of generality. This phrase means that even though a restriction is being placed on one of the assumptions in the proof, if one can complete the proof using this restriction, then it would be very easy to give a proof without the restriction. In this case, if it is assumed that M > 1, one could just as easily handled cases where M 1 by finding a ı > 0 that ensured f .x/ > 1 M, so being able to produce a proof that works for 1 does provide a proof for M 1. The phrase without loss of generality is used so frequently that many authors abbreviate it as WLOG. These ideas give the following proof. xC3 2 x!5 .x5/
PROOF: lim
D1
xC3 Let f .x/ D .x5/ 2. Let M 2 R be given. Without loss of generality, assume that M > 1. Let ı D M1 > 0. Note that ı < 1. If 0 < jx 5j < ı, then since jx 5j < 1, it follows that jx 5j > .x 5/2 . Also, since x > 4, it follows that x C 3 > 7 > 1. xC3 1 1 1 • Then f .x/ D .x5/ 2 > .x5/2 > jx5j > ı D M.
• • • •
xC3 2 x!5 .x5/
• This shows that lim
D 1.
3.9 The Arithmetic of Limits
81
3.8.1 Exercises Write a proof of each of the following infinite limits. x 2 D 1 x!4 .x4/ 2 lim x 5x D 1 x!1 lim x2 D 1 x!0 jxj
1. lim 2. 3. 4.
lim 10
x!1
x x!2C x2
5. lim
p
4 x D 1
D1
3.9 The Arithmetic of Limits The fact that the limits of some functions are easy to prove hides the fact that there are some limits whose validity is considerably more difficult to prove. Fortunately, the limits of most arithmetic combinations of functions work as expected due to the behavior of the arithmetic operations of addition, subtraction, multiplication, and division. In the words of the next chapter, these operations behave well because they are themselves continuous functions of their arguments. That is, for example, the function of two variables f .x; y/ D x C y is a continuous function of x and y. That continuity allows you to prove the following theorem. THEOREM: Suppose that f and g are functions both defined on a set with accumulation point a. Let lim f .x/ D L and lim g.x/ D H. Then x!a
x!a
1. lim f .x/ C g.x/ D L C H. x!a
2. lim f .x/ g.x/ D L H. x!a
3. lim f .x/g.x/ D LH. x!a
f .x/ x!a g.x/
4. if H ¤ 0, lim
D
L . H
Consider how to prove each part of the above theorem. In each case you will need to prove the validity of a limit, so the proof can follow the usual proof template for establishing a limit. These proofs differ from limit proofs found earlier in the chapter in that you know less about the functions whose limits you are trying to establish. On the other hand, you do know that the limits of the functions f and g exist, and that gives you a lot of tools with which to work.
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3.9.1 Limit of a Sum So what needs to be done to prove that the limit of the sum of two functions is the sum of their respective limits? As with all limit proofs, the proof will begin with a statement about what is being assumed about two functions f and g. In this case, that would essentially be a restatement of the hypothesis of the theorem that says that the limits of f and g at a are L and H, respectively. The second step of the proof would be to say Let > 0 be given which sets the tolerance to be met by the proof. You know that the end of the proof will need to show that the function in question, f .x/ C g.x/, needs to be within of the proposed limit, L C H. In other words, you will need to establish j f .x/ C g.x/ .L C H/j < . Clearly, this inequality will depend on properties of the functions f and g. But you know very little about these functions. Actually, knowing very little about the functions makes your job easier. All you know about these functions is that f has L for a limit, and g has H for a limit. This means that your proof can only use these two facts. Because these two limits exist, you will be able to set up conditions that ensure that jf .x/ Lj and jg.x/ Hj are small. How does this help?It helps because the triangle inequality will allow you to show that the expression j f .x/ C g.x/ .L C H/j is no bigger than the sum of the two small quantities jf .x/ Lj and jg.x/ Hj. That is, j f .x/ C g.x/ .L C H/j D j.f .x/ L/ C .g.x/ H/j jf .x/ Lj C jg.x/ Hj. For example, if both jf .x/ Lj and jg.x/ Hj can be made less than 2 , then their sum will be less than , and the value of j f .x/ C g.x/ .L C H/j will, in turn, be less than , as desired. How can you arrange for jf .x/ Lj and jg.x/ Hj both to be less than 2 ? You are given that the limits of f and g are L and H, respectively, so, by the definition of limit, you can arrange for each of these quantities to be smaller that any given positive value, such as 2 , with appropriate choices of ı > 0. The only subtlety here is that the value of ı > 0 needed to assure that jf .x/ Lj is less than 2 cannot be assumed to be the same value as the ı > 0 needed to assure that jg.x/ Hj is less than 2 . Thus, two different values of ı should be chosen, and then the minimum of those two will be small enough to guarantee both of the needed inequalities. Thus, after the proof proposes a given > 0, it can produce a ı1 > 0 small enough so that if x is in the domain of f and 0 < jx aj < ı1 , then jf .x/ Lj will be less than 2 . The existence of this ı1 comes from the definition of lim f .x/ D L. x!a
Similarly, the proof can produce a ı2 > 0 coming from the definition of lim g.x/ D x!a
H such that if x is in the domain of g and 0 < jx aj < ı2 , then jg.x/ Hj will be less than 2 . The proof then easily follows as described above.
3.9 The Arithmetic of Limits
83
PROOF: Suppose that f and g are functions both defined on a set with accumulation point a. If lim f .x/ D L and lim g.x/ D H, then x!a x!a lim f .x/ C g.x/ D L C H. x!a
• Let f and g be functions both defined on a set with accumulation point a with lim f .x/ D L and lim g.x/ D H. x!a x!a • Let > 0 be given. • By the definition of limit, there is a ı1 > 0 such that if x is in the domain of f and 0 < jx aj < ı1 , then jf .x/ Lj < 2 . • Similarly, there is a ı2 > 0 such that if x is in the domain of g and 0 < jx aj < ı2 , then jg.x/ Hj < 2 . • Let ı D min.ı1 ; ı2 / > 0. • Then if x is in the domain of f C g with 0 < jx aj < ı, • j f .x/Cg.x/ .LCH/j D j.f .x/L/C.g.x/H/j jf .x/LjCjg.x/Hj < C 2 D . 2 • This shows that lim f .x/ C g.x/ D L C H. x!a
A proof that the limit of the difference f .x/ g.x/ equals the difference of the individual limits, L H, is very similar to the above proof and is left as an exercise.
3.9.2 Limit of a Product Proving that the limit of the product f .x/g.x/ equals the product of the individual limits, LH, uses the same techniques as the proof for the limit of a sum but has an added complexity requiring the use of a commonly used trick. The proof of lim f .x/g.x/ D LH follows the usual template for proving the existence of a limit. x!a
Its goal is to establish the inequality jf .x/g.x/ LHj < . Again, you can use the definition of limit to make jf .x/ Lj and jg.x/ Hj as small as you need, but how small these have to be to ensure that jf .x/g.x/LHj is less than is not immediately obvious. The problem is that it is difficult to gauge how close f .x/g.x/ is to LH when you know that f .x/ is close to L, and g.x/ is close to H. The difficulty stems from having to move from f .x/g.x/ to LH, where f .x/ changes to L and g.x/ changes to H at the same time. If only one of these two changes were made, then it might be easier to make the needed estimate. That is, it would be easier to work with an expression like f .x/g.x/ f .x/H than with f .x/g.x/ LH. Of course, f .x/g.x/ LH is not the same as f .x/g.x/ f .x/H, so one cannot just use f .x/g.x/ f .x/H in place of f .x/g.x/ LH. Sometimes, though, it is worth replacing one expression with another expression that is easier to handle,
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and then adjusting the second expression to make it equivalent to the first. In this case, the change can be accomplished by employing one of the oldest tricks used in mathematical proofs, that of adding and subtracting the same quantity. In particular, you can rewrite jf .x/g.x/ LHj as jf .x/g.x/ f .x/H C f .x/H LHj. The advantage of doing this is that now you can see how the difference between f .x/g.x/ and LH depends on the differences between f .x/ and L and g.x/ and H. Indeed, jf .x/g.x/ LHj D jf .x/g.x/ f .x/H C f .x/H LHj D jf .x/.g.x/ H/ C H.f .x/ L/j jf .x/j jg.x/ Hj C jHj jf .x/ Lj. If each of the two terms, jf .x/j jg.x/ Hj and jHj jf .x/ Lj, can be made smaller than 2 , then it will have been shown that jf .x/g.x/ LHj is less than as needed. So how small does jf .x/ Lj need to be to ensure that jHj jf .x/ Lj is less than 2 ? Less than 2jHj appears to be small enough, although one needs to handle the embarrassing situation where H D 0. You could handle H D 0 and H ¤ 0 as two separate cases, or you can take care of both cases at once by making jf .x/ Lj less than since jHj C 1 is larger than jHj and can never be 0. Thus, you can 2 jHjC1
select a ı1 > 0 so that if 0 < jx aj < ı1 , then jf .x/ Lj <
.
2 jHjC1
How small does jg.x/ Hj need to be to ensure that jf .x/j jg.x/ Hj is less than It would be nice to say that jg.x/ Hj < 2jf.x/j suggesting that you set ı small enough to ensure jg.x/ Hj < 2jf.x/j , but there is a problem here. The definition of limit requires that the choice of ı come before the choice of x, so you cannot have the value of ı depending on x. What is needed is an upper bound for jf .x/j because, if jf .x/j M, the value of ı can be found to ensure jg.x/ Hj < 2M which will always be small enough to guarantee jf .x/j jg.x/ Hj < 2 . You can find such an upper bound for jf .x/j because the limit of f .x/ exists as x approaches a, and so jf .x/j can be restricted to being not much larger than jLj. You could, for example, find ı2 > 0 so that if 0 < jx aj < ı2 , then jf .x/ Lj < 1. This would ensure that f .x/ is a distance of no more than 1 from L so that jf .x/j < jLj C 1. Then you would only need jg.x/ Hj < to get jf .x/j jg.x/ Hj < 2 . This gives you all the ? 2
2 jLjC1
pieces necessary to complete the proof as follows.
3.9 The Arithmetic of Limits
85
PROOF: Suppose that f and g are functions both defined on a set with accumulation point a. If lim f .x/ D L and lim g.x/ D H, then x!a x!a lim f .x/g.x/ D LH. x!a
• Let f and g be functions both defined on a set with accumulation point a with lim f .x/ D L and lim g.x/ D H. x!a x!a • Let > 0 be given. • By the definition of limit, there is a ı1 > 0 such that if x is in the domain of f and 0 < jx aj < ı1 , then jf .x/ Lj < . 2 jHjC1
• By the definition of limit, there is a ı2 > 0 such that if x is in the domain of f and 0 < jxaj < ı2 , then jf .x/Lj < 1. Then jLjC1 > jLjCjf .x/Lj jL C .f .x/ L/j D jf .x/j. • By the definition of limit, there is a ı3 > 0 such that if x is in the domain of g and 0 < jx aj < ı3 , then jg.x/ Hj < . 2 jLjC1
• Let ı D min.ı1 ; ı2 ; ı3 / > 0. • Then if x is in the domain of f g with 0 < jx aj < ı, • jf .x/g.x/ LHj D jf .x/g.x/ f .x/H C f .x/H LHj D jf .x/.g.x/ H/ C H.f .x/ L/j jf .x/j jg.x/ Hj C jHj jf .x/ Lj .jLj C 1/ C jHj < 2 C 2 D . 2 jLjC1
2 jHjC1
• This shows that lim f .x/g.x/ D LH. x!a
3.9.3 Limit of a Quotient Finally, the proof that the limit of a quotient is the quotient of the individual limits is much like the proof about the product of limits, although the algebra is more complicated. As in the preceding ˇproof, you can start with the needed ˇ ˇ f .x/ Lˇ inequality which, in this case, is ˇ g.x/ H ˇ < . Using the trick of adding and subtracting the same quantity, the left side of the inequality can be written ˇ as ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ .f .x/L/HCL Hg.x/ ˇ ˇ ˇ f .x/ ˇ ˇ ˇ f .x/HLg.x/ f .x/HLHCLHLg.x/ ˇ ˇ g.x/ HL ˇ D ˇ g.x/H ˇ D ˇ ˇ D ˇˇ ˇ g.x/H g.x/H ˇ ˇ ˇ L.g.x/H/ ˇ jf .x/Lj jf .x/Lj C ˇ g.x/H ˇ. Again, the goal will be to make each of the terms jg.x/j jg.x/j ˇ ˇ ˇ ˇ less than 2 by selecting an appropriate sequence of ı’s. and ˇ L.g.x/H/ g.x/H ˇ Both of these terms have a factor of jg.x/j in the denominator. To make the fractions small, you will need to know that jg.x/j does not get too close to zero. What you do know is that lim g.x/ D H is not zero because the hypothesis of the x!a
theorem will make that assumption. How far away from zero can you require jg.x/j to be? Certainly, this will depend on the value of H. If H is close to zero, then jg.x/j
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3 Limits
will be close to zero as x approaches a. The best you can do is require that jg.x/j be so close to H that it will keep a known distance from zero. For example, you could . That will ensure that jg.x/j is at least jHj require that jg.x/ Hj be less than jHj 2 2 which keeps it a known distance away from zero. So, select a ı1 > 0 such that if x , and jg.x/j will is in the domain of g with 0 < jx aj < ı1 , then jg.x/ Hj < jHj 2 jHj be greater than 2 . .x/Lj 2 < jf .x/ Lj jHj . Thus, it would Now for these values of x you will have jf jg.x/j be sufficient if jf .x/Lj were to be less than
jHj 4
which will ensure that
jf .x/Lj jg.x/j
< 2 .
jHj This can be done ˇby choosing ˇ ı2 > 0 small enough so that jf .x/Lj is less than 4 . ˇ ˇ term less than 2 , you can select a ı3 > 0 so that if To make the ˇ L.g.x/H/ g.x/H ˇ 2
x is within ı3 of a you will have jg.x/ Hj less than H because that will give 4jLj ˇ ˇ 2 H 2 H jLj 4jLj ˇ L.g.x/H/ ˇ < H42 D 2 . Well, OK, did you catch that the preceding does ˇ g.x/H ˇ < jg.x/jjHj 2
not work if L D 0? To avoid this problem it would be better to make jg.x/ Hj less 2 than H . Putting all of these ideas together gives the following proof. 4 jLjC1
PROOF: Suppose that f and g are functions both defined on a set with accumulation point a. If lim f .x/ D L and lim g.x/ D H with H ¤ 0, then f .x/ x!a g.x/
lim
x!a
D
x!a
L . H
• Let f and g be functions both defined on a set with accumulation point a with lim f .x/ D L and lim g.x/ D H ¤ 0. x!a x!a • Let > 0 be given. • By the definition of limit, there is a ı1 > 0 such that if x is in . the domain of g and 0 < jx aj < ı1 , then jg.x/ Hj < jHj 2 jHj For these x it follows that jg.x/j C 2 > jg.x/j C jg.x/ Hj D jg.x/j C jH g.x/j jg.x/ C H g.x/ j D jHj which implies that D jHj . jg.x/j > jHj jHj 2 2 • By the definition of limit, there is a ı2 > 0 such that if x is in the domain of . f and 0 < jx aj < ı2 , then jf .x/ Lj < jHj 4 • By the definition of limit, there is a ı3 > 0 such that if x is in the domain of 2 g and 0 < jx aj < ı3 , then jg.x/ Hj < H . 4 jLjC1
• Let ı D min.ı1 ; ı2 ; ı3 /. • Then if x is in the domain of gf with 0 < jx aj < ı, ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ f .x/HLg.x/ ˇ ˇ f .x/HLHCLHLg.x/ ˇ ˇˇ .f .x/L/HCL Hg.x/ ˇˇ ˇ f .x/ Lˇ • ˇ g.x/ H ˇ D ˇ g.x/H ˇ D ˇ ˇDˇ ˇ g.x/H g.x/H ˇ ˇ 2 ˇ L.g.x/H/ ˇ jHj 2 jf .x/Lj 2jLj H 2 < 2 C 2 D . jg.x/j C ˇ g.x/H ˇ < 4 jHj C H 4 jLjC1
• This shows that
lim f .x/ x!a g.x/
D
L . H
3.9 The Arithmetic of Limits
87
3.9.4 Limit of Rational Functions As a demonstration of the power of these results about the arithmetic of limits, you can now easily prove the following list of results which will allow you to easily calculate limits of polynomials and rational functions of x. • For any constant c in the real numbers, lim c D c. x!a • lim x D a. x!a • For any n in the natural numbers, lim xn D an . x!a
• For any polynomial p.x/, lim p.x/ D p.a/. x!a
p.x/ x!a q.x/
• For any polynomials p.x/ and q.x/ with q.a/ ¤ 0, lim
D
p.a/ . q.a/
The first two results are very easy to prove, and are left as exercises. The next two results can be proved by using mathematical induction which is often the first technique one considers using when trying to prove statements such as these that depends on a natural number. Here, mathematical induction will be employed to prove statements about the limits of polynomials, and the degree of the polynomial provides a natural number to use as the induction variable. To begin with, try using mathematical induction to prove that lim xn D an for x!a any natural number n. In this mathematical induction argument, the base case is lim x D a, that is, when n D b D 1. The proofs of statements similar to this base x!a case were covered earlier. The induction step in the proof will need to show that if lim xk D ak for some natural number k, then lim xkC1 D akC1 . But xkC1 is just the x!a
x!a
product xk x, so this result follows immediately from the theorem about the limits of products. That leads to the following proof that uses the template for proofs by mathematical induction. PROOF: lim xn D an for any natural number n. x!a
• SET THE CONTEXT: The statement will be proved for all natural numbers n by mathematical induction on n. • PROVE S.b/: When n D 1, the statement says that lim x D a which has x!a already been established. • STATE THE INDUCTION HYPOTHESIS: Assume that for some natural number k, lim xk D ak . x!a • PERFORM THE INDUCTION STEP: Then since the limit of a product of two functions is the product of the two individual limits, it follows that lim xkC1 D lim xk x D .lim xk /.lim x/ D ak a D akC1 . So the statement x!a x!a x!a x!a is true for n D k C 1. • STATE THE CONCLUSION: Therefore, by mathematical induction, lim xn D an is true for all natural numbers n. x!a
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3 Limits
Mathematical induction can again be employed to prove that for every polynomial, p.x/, lim p.x/ D p.a/. As a reminder, a polynomial of degree n is a function, x!a
p.x/ D cn xn C cn1 xn1 C cn2 xn2 C C c1 x C c0 where c0 ; c1 ; c2 ; : : : ; cn are constants with cn ¤ 0. Previously it has been proved that lim cj D cj and x!a
lim x j D a j , from which one gets that the limit of a monomial is lim cj x j D cj a j . x!a x!a A polynomial is just a sum of such monomials, so mathematical induction is a convenient tool for showing that this sum of an arbitrary number of monomials has the desired limit. PROOF: For any constants c0 ; c1 ; c2 ; : : : ; cn and a 2 R, the polynomial p.x/ D cn xn C cn1 xn1 C cn2 xn2 C C c1 x C c0 satisfies lim p.x/ D p.a/. x!a
• SET THE CONTEXT: The statement will be proved by mathematical induction on the degree of the polynomial n. • PROVE S.b/: lim c1 xCc0 D .lim c1 /.lim x/C lim c0 D .c1 /.a/Cc0 which x!a x!a x!a x!a shows that the statement is true for n D 1. • STATE THE INDUCTION HYPOTHESIS: Assume that for some natural number k, if p.x/ D ck xk C ck1 xk1 C C c1 x C c0 , then lim p.x/ D p.a/. x!a
• PERFORM THE INDUCTION STEP: If p.x/ D ckC1 xkC1 C ck xk C ck1 xk1 C C c1 x C c0 , lim ckC1 xkC1 C ck xk C ck1 xk1 C C c1 x C c0 D x!a
.lim ckC1 /.lim xkC1 / C lim ck xk C ck1 xk1 C C c1 x C c0 D x!a
x!a
x!a
.ckC1 akC1 / C .cn an C cn1 an1 C cn2 an2 C C c1 a C c0 / D p.a/. This shows that the statement is true for n D k C 1. • STATE THE CONCLUSION: Therefore, by mathematical induction, lim p.x/ D p.a/ is true for all polynomials p.x/. x!a
Recall that a rational function is just a ratio of polynomials, that is, if p.x/ and q.x/ are polynomials, then p.x/ is a rational function. It is only a simple step to get q.x/ the following theorem. PROOF: For any polynomials p and q and a 2 R where q.a/ ¤ 0, it follows p.x/ that lim q.x/ D p.a/ . q.a/ x!a
1. Let p and q be polynomials, and a 2 R such that q.a/ ¤ 0. 2. Because p and q are polynomials, lim p.x/ D p.a/ and lim q.x/ D q.a/. x!a x!a 3. Since the limit of the quotient is equal to the quotient of the individual limits D when the limit of the denominator is not zero, it follows that lim p.x/ q.x/ x!a
lim p.x/
x!a
lim q.x/
x!a
D
p.a/ . q.a/
3.10 Other Limit Theorems
89
3.9.5 Other Types of Limits It is time to note that even though all of these limit theorems concerned limits as x approaches a, most can be extended to cover limits as x approaches a from the left, as x approaches a from the right, as x approaches infinity, and as x approaches negative infinity. In particular, most of the theorems apply to the limits of sequences. Many of these statements can be found in the exercises.
3.9.6 Exercises Write proofs of each of the following statements. 1. If f and g are defined in an open interval containing a, and if lim f .x/ D L and x!a
lim g.x/ D H, then lim f .x/ g.x/ D L H. x!aC
x!a
2. For any constant c in the real numbers, lim c D c. x!a 3. lim x D a. x!a
4. If f and g have a common domain with lim f .x/ D L and lim g.x/ D H, then x!a
lim f .x/ C g.x/ D L C H.
x!a
x!a
5. If f and g have a common domain with lim f .x/ D L and lim g.x/ D H, then x!aC
lim f .x/ g.x/ D L H.
x!aC
x!aC
6. If f and g have a common domain with lim f .x/ D L and lim g.x/ D H, then x!1
lim f .x/g.x/ D LH.
x!1
x!1
7. If and are sequences with lim an D L and lim bn D H ¤ 0, then an n!1 bn
lim
D
n!1
L . H
8. lim f .x/ D L if and only if lim f x!1
x!0C
1 x
n!1
D L. 1 x!a f .x/L lim f .x/CM D 1. x!a f .x/CN
9. If f .x/ > L for all x ¤ a, then lim f .x/ D L if and only if lim x!a
10. If lim f .x/ D 1, then for any constants M and N x!a
D 1.
3.10 Other Limit Theorems This section discusses a few other useful results about limits. They provide an interesting variety of proof strategies to consider.
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3 Limits
3.10.1 The Limit of a Positive Function What can you say about lim f .x/ D L if you know that f .x/ > 0 for all x, or at x!a least for all x in an open interval containing a? Assuming that this limit exists, it is clear that the limit cannot be negative because, from the definition of limit, you know that jf .x/ Lj can be made as small as you like which would not be possible if f .x/ were always positive and L were negative. But how would you prove this? The key lies in the inequality jf .x/ Lj < since, if L were negative, you could choose to be so small that the inequality could not hold. How small would need to be? Well, the only thing you know about f .x/ is that it is positive, or, in other words, cannot be smaller than 0. At the same time, L is negative which means that f .x/ and L must be at least jLj apart, noting that jLj > 0. So set D jLj. Then jLj D > jf .x/ Lj D f .x/ C jLj which implies f .x/ < 0 which is a contradiction. This leads to the following proof. PROOF: Let f be a function such that f .x/ > 0 for all x in the domain of f . If lim f .x/ D L, then L 0. x!a
• Suppose that lim f .x/ D L and that for all x, f .x/ > 0. x!a • Assume that L < 0. • By the definition of limit, there is a ı > 0 such that for all x in the domain of f satisfying 0 < jx aj < ı, it follows that jf .x/ Lj < L. • For these values of x it must be that L > jf .x/ Lj D f .x/ L implying that 0 > f .x/ which contradicts the fact that f .x/ is always positive. • Therefore, it must hold that L 0. Similar statements can be made about the limits of functions f satisfying f .x/ > b or f .x/ < b for all x where b is a constant real number. One can also extend this to limits from the left, limits from the right, and limits to infinity and negative infinity. Several of these possibilities have been left for the exercises.
3.10.2 Uniqueness of Limits There is nothing in the definition of lim f .x/ D L that a priori precludes lim f .x/ D x!a
x!a
M for some M ¤ L. But, in fact, limits are unique, that is, the only way for the limit to be L and the limit to be M is for L and M to be equal. Intuitively, this should make sense. If the values of f .x/ are getting close to L, then they should not also be able to get close to a value distinct from L. So how can you prove this using nothing but the definition of limit as a tool? The result can be proved by contradiction, that is, if you assume that the function f has two distinct limits, L and M, as x approaches a, then this leads to a statement which must be false. Assuming that both limits exist, the definition of limit will
3.10 Other Limit Theorems
91
allow you to force both jf .x/ Lj < and jf .x/ Mj < for any positive that you choose. Why can’t this happen? Well, if it did, you could get C > jf .x/ Lj C jf .x/ Mj D jf .x/ Lj C jM f .x/j jf .x/ L C M f .x/j D jM Lj. If M ¤ L, then jM Lj is a positive number, so if is chosen less than or equal to jMLj , it will be impossible to have jM Lj < 2 as guaranteed by the definition of 2 limit. That gives you the following proof. PROOF: If lim f .x/ D L and lim f .x/ D M, then L D M. x!a
x!a
• Suppose that lim f .x/ D L and lim f .x/ D M. x!a
x!a
• Assume that L ¤ M which implies that jM Lj > 0. • By the definition of limit, there is a ı1 > 0 such that for all x in the domain . of f satisfying 0 < jx aj < ı1 , it follows that jf .x/ Lj < jMLj 2 • By the definition of limit, there is a ı2 > 0 such that for all x in the domain . of f satisfying 0 < jx aj < ı2 , it follows that jf .x/ Mj < jMLj 2 • Let x be in the domain of f with 0 < jx aj < min.ı1 ; ı2 /. • Then jM Lj D jMLj C jMLj > jf .x/ Lj C jf .x/ Mj D 2 2 jf .x/ Lj C jM f .x/j jf .x/ L C M f .x/j D jM Lj showing that jM Lj > jM Lj which is a contradiction. • Thus, L ¤ M must be false, and L D M.
3.10.3 The Squeezing Theorem The Squeezing Theorem, also known as the Sandwich Theorem or the Scrunch Theorem, says that if the values of f .x/ are always between g.x/ and h.x/, then if g and h both have the same limit, L, at x D a, then f must also have limit L at a. The proof of this is not hard once you write down everything that you know about the functions f , g, and h. So what do you know? You can assume that for every x that g.x/ f .x/ h.x/, and you can assume that lim g.x/ D lim h.x/ D L. This means x!a
x!a
that for every > 0 there is a ı1 > 0 such that when x satisfies 0 < jxaj < ı1 , then jg.x/ Lj < . Similarly, for that same , there is a ı2 > 0 such that when x satisfies 0 < jx aj < ı2 , then jh.x/ Lj < . Thus, you can show for values of x near a that g.x/ f .x/ h.x/, < g.x/ L < , and < h.x/ L < . Putting these three sets of inequalities together shows that < g.x/ L f .x/ L h.x/ L < from which jf .x/ Lj < follows. This gives the following proof.
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3 Limits
PROOF (Squeezing Theorem): Let f , g, and h be three functions with the same domain, and let a be an accumulation point of that domain. Assume that for all x in that domain g.x/ f .x/ h.x/, and that lim g.x/ D lim h.x/ D L. Then lim f .x/ D L. x!a
x!a
x!a
• Assume that a is an accumulation point of the domain shared by the three functions f , g, and h. • Also assume that lim g.x/ D lim h.x/ D L. x!a
x!a
• Finally, assume that g.x/ f .x/ h.x/ for all x in the common domain of f , g, and h. • Let > 0 be given. • By the definition of limit, there is a ı1 > 0 such that for all x in the domain of g that satisfy 0 < jx aj < ı1 , it follows that jg.x/ Lj < . • By the definition of limit, there is a ı2 > 0 such that for all x in the domain of h that satisfy 0 < jx aj < ı2 , it follows that jh.x/ Lj < . • Then for all x in the common domain of f , g, and h satisfying 0 < jx aj < min.ı1 ; ı2 /, jg.x/ Lj < and jh.x/ Lj < . • Thus, for those x, < g.x/ L f .x/ L h.x/ L < from which it follows that jf .x/ Lj < . • Therefore, lim f .x/ D L. x!a
3.10.4 Limits of Subsequences If the sequence converges to L, it means that the terms of the sequence are getting close to L. This should mean that the terms of any subsequence should also be getting close to L, and it is not hard to prove that every subsequence of has the same limit. Given the fact that lim an D L and given a subsequence , how do you n!1 use this to prove that the subsequence converges to L? What do you know about this subsequence? Only that there is a strictly increasing sequence of natural numbers, , that tells which terms of are found in the subsequence. A nice property of a strictly increasing sequence of natural numbers, , is that for any natural number j, nj j. This can easily be proved by mathematical induction on j. Certainly, n1 1 since n1 is a natural number, so the claim is true for j D 1. If nk k for some k, then because is strictly increasing, nkC1 nk C 1 k C 1 showing that if the claim is true for k, then it is true for k C 1. This proves the claim. The definition of limit gives you that for any > 0 there is an N such that if j > N, then jaj Lj < . But since nj j, it follows that for all j > N, nj is also greater than N, so janj Lj < as needed.
3.11 Liminf and Limsup
93
PROOF: Let be a sequence with lim an D L L, and let be any subsequence. Then lim anj D L.
n!1
j!1
• Let be a sequence with lim an D L, and let be any n!1 subsequence. • Let > 0 be given. • By the definition of limit, there is an N such that for all n > N, jan Lj < . • By the definition of subsequence, is a strictly increasing sequence of natural numbers and, as such, satisfies nj j for all natural numbers j. • Thus, for all j > N, nj j > N implies janj Lj < . • This proves that lim anj D L. j!1
Of course the converse of this theorem is trivially true. That is, if all subsequences of a given sequence converge, then the original sequence converges. This is trivial since the original sequence is one of its subsequences.
3.10.5 Exercises Write proofs of each of the following statements. 1. If lim f .x/ D L and f .x/ < b for all x, then L b. x!a
2. If f .x/ > b for all x and lim f .x/ D L, then L b. x!1
3. If f .x/ > 0 for all x, then lim f .x/ cannot equal negative infinity. x!a 4. Suppose that sequences , , and satisfy an bn cn for every natural number n. If lim an D lim cn D L, then lim bn D L. n!1
n!1
n!1
3.11 Liminf and Limsup Even when a limit does not exist, there is often something that can be said about the values that the function approaches. Consider, for example, the sequence 1; 1; 0; 1; 1; 0; 1; 1; 0; : : : which just oscillates among the numbers 1, 1, and 0. This sequence does not have a limit, but it has subsequences that do converge. Some of its subsequences converge to 1, some converge to 1, and some converge to 0. 2 sin x 2 Now consider the function f .x/ D 2xx2 C1 . The function x22xC1 has a limit of 2 as x goes to infinity, but f .x/ oscillates without approaching a limit. Some of its
94
3 Limits
values do approach 2, but other values approach 2 and every value in between. More precisely, for each L 2 Œ2; 2, you can find sequences where lim xn D n!1
1 and lim f .xn / D L. n!1 So suppose that the function f is defined for positive real numbers. How might f .x/ behave as x goes to infinity? f might diverge to infinity or minus infinity as 3 2x2 do f .x/ D x2 and f .x/ D x1x 2 C4 . It might have a finite limit as does x2 C1 . It might
sin x . Finally, it oscillate among values within some bounded range such as .3xC100/ xC10 might oscillate and be unbounded like x j sin xj. Even when f oscillates so that it does not have a finite or infinite limit, it is helpful to quantify which values the function f .x/ approaches repeatedly as x grows. This can be done by considering the range of f .x/ when x is restricted to an interval .M; 1/, and then watching what happens to that range as M gets large. sin x For example, consider the function f .x/ D .3xC100/ whose graph is shown in xC10 3xC100 70 Fig. 3.10. The function xC10 D 3 C xC10 is a decreasing function of x for x > 0, so on the interval .M; 1/, the function f oscillates in a range bounded between 3MC100 and 3MC100 . What can be said about the sequence where is MC10 MC10 and 3xxnn C100 , a sequence with lim xn D 1? The values f .xn / are between 3xxnn C100 C10 C10 n!1
so as xn gets large, f .xn / is forced to be inside or very near the interval Œ3; 3. Clearly, for no sequence can f .xn / approach a limit outside of the interval Œ3; 3, but there are sequences for which f .xn / approaches 3 and others for which f .xn / approaches 3 as shown in the figure. Finding the greatest and least values to which f .xn / could converge is the idea behind the limit superior and limit inferior often referred to simply as the lim sup and lim inf, respectively. In the sin x example of f .x/ D .3xC100/ , the values of 3 and 3 came from looking at the xC10 greatest lower bound and least upper bound of the set ff .x/ j x > Mg and then letting M go to infinity. In general, let f be a function whose domain is unbounded above. For each real number M let AM be the range of f for x > M, that is, AM D ff .x/ j x is in the domain of f with x > Mg. Then define lim sup f .x/ to be x!1
lim sup AM . Similarly, define lim inf f .x/ to be lim inf AM . Some books use the
M!1
x!1
M!1
notation lim and lim for lim sup and lim inf, respectively. Fig. 3.10 Sequences approaching the lim sup and .3xC100/ sin x lim inf of f .x/ D xC10
3.11 Liminf and Limsup
95
If f .x/ is unbounded above as x gets large, then sup AM D 1 for each M, so lim sup f .x/ D 1. If lim f .x/ D 1, then sup AM will also approach 1, x!1
x!1
so lim sup f .x/ will be 1. Analogously, if f .x/ is unbounded below as x gets x!1
large, lim inf f .x/ D 1, and if lim f .x/ D 1, then lim inf f .x/ D 1. Note x!1 x!1 x!1 that since sup AM and inf AM are both monotone function of M, their limits always exist, although they might be infinite limits. Thus, unlike lim f .x/, the values of x!1
lim sup f .x/ and lim inf f .x/ always exist. x!1
x!1
If f .x/ remains bounded as x gets large, lim sup f .x/ and lim inf f .x/ are finite x!1
x!1
values. This means lim sup AM is finite. For each natural number n, there must M!1
be an xn > n such that f .xn / is within, say 1n of sup An . Then, is a sequence that diverges to infinity such that for each n, sup An 1n < f .xn / < sup An . By the Squeezing Theorem, lim f .xn / D lim sup An D lim sup f .x/. Similarly, there n!1
n!1
x!1
must be a sequence diverging to infinity with lim f .xn / D lim inf f .x/. This n!1 x!1 means that there is a sequence such that f converges to its limit superior on that sequence and another sequence such that f converges to its limit inferior on it. Consider the three examples: sin x, x j sin xj, and x sin x. None of these functions has a limit as x approaches infinity because each function oscillates and does not approach one particular value. On the other hand, in each case it is easy to see upper and lower bounds to the oscillations. The values ofthe function sin x clearly stay in the interval Œ1; 1. For each integer n, when x D 2n C 12 , the function sin x D 1 D lim sup sin x and when x D 2n 12 , the function sin x D 1 D lim inf sin x. x!1
x!1
The function x j sin xj is unbounded above but is nonnegative for positive x. Again, for integer n, when x D 2n C 12 , the function x j sin xj D x which goes to infinity, the lim sup of x j sin xj, and when x D n, the function x j sin xj D 0 D lim inf x j sin xj. The function x sin x behaves similarly except, now, when x!1 x D 2n 12 , the function x sin x is x which goes to negative infinity, the lim inf of x sin x. The limit superior and limit inferior can be defined for limits at points other than infinity. For example, if a is an accumulation point of the domain of f , one can define the limit superior and limit inferior of f .x/ as x approaches a. Rather than defining AM to be the values of f .x/ for x > M which essentially contains the values of f for x restricted to an interval ending at infinity, one can define Aı for any ı > 0 as Aı D ff .x/ j x is in the domain of f with 0 < jx aj < ıg which contains the values of f for x restricted to an open interval containing a with the point a removed. Then lim sup f .x/ D lim sup Aı and lim inf f .x/ D lim inf Aı . ı!0C
x!a
x!a
ı!0C
These definitions of lim sup and lim inf have properties similar to the definitions of lim sup and lim inf at infinity. That is, sup Aı and inf Aı are both monotone in ı, so their limits as ı goes to 0 always exist. Moreover, there is a sequence where lim xn D a such that lim f .xn / D lim sup f .x/ and another such sequence n!1
n!1
x!a
96
3 Limits
such that lim f .xn / D lim inf f .x/. Similar definitions can be given for lim inf f .x/, n!1
x!aC
x!a
lim sup f .x/, lim inf f .x/, lim sup f .x/, lim inf f .x/, and lim sup f .x/. x!a
x!a
x!aC
x!1
x!a1
The most important theorem concerning lim inf and lim sup is that lim f .x/ D L x!a
if and only if lim inf f .x/ D lim sup f .x/ D L. Notice first that this is a biconditional x!a
x!a
statement; that is, an “if and only if” statement. This requires that its proof have two parts; one that assumes lim f .x/ D L and proves lim inf f .x/ D lim sup f .x/ D L x!a
x!a
x!a
and another that assumes lim inf f .x/ D lim sup f .x/ D L and proves lim f .x/ D L. x!a
x!a
x!a
So, given lim f .x/ D L, how can you conclude that lim inf f .x/ D lim sup f .x/ D x!a
x!a
x!a
L? What you know is that given > 0, there is a ı > 0 such that for all x in the domain of f for which 0 < jx aj < ı, you have jf .x/ Lj < . But this means that for small ı > 0, the supremum sup Aı and infimum inf Aı are both within of L and, therefore, the limits of sup Aı and inf Aı must both approach L as ı decreases to 0. Conversely, suppose that lim inf f .x/ D lim sup f .x/ D L. Note that for any x ¤ a in x!a
x!a
the domain of f , it follows that f .x/ 2 A2jxaj . Thus, inf A 2jxaj f .x/ sup A2jxaj which implies that lim inf A2jxaj lim f .x/ lim sup A2jxaj from which it x!a
follows that lim f .x/ D L.
x!a
x!a
x!a
PROOF: Let a be an accumulation point of the domain of the function f . Then lim f .x/ D L if and only if lim inf f .x/ D lim sup f .x/ D L. x!a
x!a
x!a
• Assume that a is an accumulation point of the function f . • For any ı > 0, define Aı D ff .x/ j x is in the domain of f with 0 < jx aj < ıg. PART I: the limit equals L implies lim inf and lim sup equal L • Assume that lim f .x/ D L. x!a • Let > 0 be given. • Then there is a ı > 0 such that if x is in the domain of f with 0 < jxaj < ı, then jf .x/ Lj < . • This says that inf Aı L and sup Aı L C . • It follows that lim inf f .x/ L and lim sup f .x/ L implying that x!a
x!a
lim inf f .x/ D lim sup f .x/ D L, completing the first part of the proof. x!a
x!a
(continued)
3.11 Liminf and Limsup
97
PART II: lim inf and lim sup equal L implies that the limit equals L • Assume that lim inf f .x/ D lim sup f .x/ D L. x!a
x!a
• For any x in the domain of f with x ¤ a, it follows that inf A2jxaj f .x/ sup A2jxaj . • Because lim inf A2jxaj D lim inf f .x/ D L, and lim sup A2jxaj D x!a
x!a
x!a
lim sup f .x/ D L, the Squeezing Theorem shows that lim f .x/ D L, which x!a
x!a
completes the second part of the proof. As discussed earlier, this theorem holds even when L D 1 or 1. It also holds for limits at ˙1 and for one-sided limits.
3.11.1 Exercises 1. Write definitions for each of the following. (a) lim inf f .x/ x!aC
(b) lim sup f .x/ x!a
(c) lim inf f .x/ x!1
2. Determine each of the following. x x2 x2 x x2 x!2 x2 x x is rational (c) lim inf x!2 5 x is irrational 5 (d) lim sup n n!1 4 C .1/ 1 1C n (e) lim sup 1 C n.1/n n!1
98
3 Limits
3. Prove that if a is any accumulation point of the domain of f , then lim inf f .x/ x!a
lim sup f .x/. x!a
4. Prove that lim f .x/ D 1 if and only if lim inf f .x/ D 1. x!a
x!a
5. Suppose that lim inf f .x/ D L and lim inf g.x/ D M. What can you say about x!a
lim inf.f C g/.x/?
x!a
x!a
6. Suppose that f is a positive-valued function with lim sup f .x/ D L > 0. Prove 1 that lim inf f .x/ D L1 . x!a
x!a
Chapter 4
Continuity
4.1 The Definition of Continuity As with the definition of limit, most Calculus students will develop an intuitive feel for what it means for a function to be continuous. This usually involves knowing that a function is continuous on an interval if the graph of that function over that interval can be drawn without lifting one’s pencil from the page. The important property here is that as the pencil is tracing out the graph of the function, and the pencil is approaching the point where x D a, the points on the graph are getting close to their destination at the point .a; f .a//. In particular, it does not happen that as the points on the graph are getting close to .a; L/ that the graph suddenly jumps to a different point .a; f .a// where f .a/ ¤ L, a situation where the pencil would have to be lifted from the page to get from .a; L/ to .a; f .a//. This intuitive understanding leads directly to the key property of f being continuous at a which is that lim f .x/ D f .a/. x!a How can one state a definition for continuity that embodies this intuitive feel for the function having its own value as its limit? Clearly, the definition of a function f being continuous at a point x D a must be similar to a definition of the limit of f as x approaches a. As a reminder, here is the definition of limit. Suppose that the point a is an accumulation point of the domain of the function f. Then lim f .x/ D L means that for every > 0 there exists a ı > 0 such that for x!a every x in the domain of f satisfying 0 < jx aj < ı, it follows that jf .x/ Lj < . The definition of continuity of f at point a needs to include the fact that the function is defined at the point a, so references to the limit L in the definition of limit can be replaced by references to f .a/. Thus, the definition of continuity will contain the conclusion jf .x/f .a/j < . In the definition of limit, it was not required that the function f be defined at x D a, and if it were defined, f .a/ did not need to be equal to the limit L. For this reason, the definition of limit took care to ensure that even though jf .x/ Lj < was required to hold for x values near x D a, this inequality did not need to hold at x D a. The definition of limit excluded x D a by © Springer International Publishing Switzerland 2016 J.M. Kane, Writing Proofs in Analysis, DOI 10.1007/978-3-319-30967-5_4
99
100
4 Continuity
a
b
c
d
e
f
Fig. 4.1 Continuity of a function
only requiring the inequality to hold for those x values satisfying 0 < jx aj < ı which excludes x D a. This restriction is not necessary in the definition of continuity of a function at a point. Suppose that the point a is in the domain of the function f. Then f is continuous at a means that for every > 0 there exists a ı > 0 such that for every x in the domain of f satisfying jx aj < ı, it follows that jf .x/ f .a/j < . Notice that the requirement that the point a be an accumulation point of the domain of f has been dropped. As a result, if the function f is defined at an isolated point a, then f is continuous at that point. A function that is not continuous at the point a is discontinuous at the point a. A function f is continuous on a set A if it is continuous at each point a 2 A. The function whose graph appears in Fig. 4.1 is discontinuous at x D b because its limit at x D b does not exist. Similarly, it is discontinuous at x D c. It is discontinuous at x D d because it is not defined at that point even though the function has a limit there. The function is continuous on the intervals Œa; b/, .b; c/, and .c; d/, and at the points x D e and x D f . The function is not continuous on the intervals Œa; b or Œc; d. It is a direct consequence of the definition of continuity that if f is continuous at a point a, and if a is an accumulation point of the domain of f , then the limit of f .x/ at a exists and is, in fact, f .a/. To prove this you would just need to show that if f satisfies the definition of continuity at a, then f also satisfies the definition of lim f .x/ D f .a/. Writing down the definition of continuity gives you that for every x!a
> 0 there is a ı > 0 such that jx aj < ı implies jf .x/ f .a/j < . But if this is true, then certainly 0 < jx aj < ı implies jf .x/ f .a/j < , so the definition of limit is satisfied.
4.2 Proving the Continuity of a Function
101
PROOF: If the function f is continuous at a, and a is an accumulation point of the domain of f , then lim f .x/ D f .a/. x!a
• Let f be a function continuous at a where a is an accumulation point of the domain of f . • Given > 0, • the definition of continuity says that there is a ı > 0 such that if x is in the domain of f with jx aj < ı, then jf .x/ f .a/j < . • But then if 0 < jx aj < ı, it follows that jf .x/ f .a/j < satisfying the definition of lim f .x/ D f .a/. x!a
• Therefore, lim f .x/ D f .a/. x!a
Similarly, if f is defined at a and lim f .x/ D f .a/, then f is continuous at a. x!a Again, the proof of this follows directly from the definitions. PROOF: If the function f is defined at a and lim f .x/ D f .a/, then f is x!a continuous at a. • Let f be a function defined at a where lim f .x/ D f .a/. x!a • Given > 0, • the definition of limit says that there is a ı > 0 such that if x is in the domain of f with 0 < jx aj < ı, then jf .x/ f .a/j < . • Certainly, if x D a, then jf .x/ f .a/j D jf .a/ f .a/j D 0 < . • Thus, it follows that jx aj < ı implies jf .x/ f .a/j < satisfying the definition of f being continuous at a. • Therefore, f is continuous at a.
4.2 Proving the Continuity of a Function The template for proofs of lim f .x/ D L followed directly from the definition of x!a limit. Similarly, a template for proofs of the continuity of a function f at a point a will follow directly from the definition of continuity. Indeed, the definition of continuity requires that for every > 0 there exist a ı > 0 which satisfies a particular condition. This suggests that a proof of continuity should select an arbitrary > 0 and proceed to display a value of ı > 0 that causes the needed condition to be satisfied. This is similar to the procedure taken for a limit proof except that the needed condition is slightly different. Thus, here is a template for proofs about the continuity of a function at a point.
102
4 Continuity
TEMPLATE for proving the function f is continuous at the point a • SET THE CONTEXT: Make statements about what is known about the function f and the numbers a and f .a/. • SELECT AN ARBITRARY : Given > 0, • PROPOSE A VALUE FOR ı: let ı D . Here you would insert an appropriate value for ı. • SELECT AN ARBITRARY x: Select x in the domain of f such that jx aj < ı. • LIST IMPLICATIONS: Derive the result jf .x/ f .a/j < . • STATE THE CONCLUSION: Therefore, f is continuous at the point a. As a start, consider how to prove that the function defined for all real numbers x as f .x/ D 5x3 is continuous at x D 4. The proof would begin with “Let f .x/ D 5x3. Given > 0; : : : .” The task is then to find a ı > 0 so that jf .x/ f .4/j < for every x satisfying jx 4j < ı. Working backwards, to get jf .x/ f .4/j < one needs > j.5x 3/ .5 4 3/j D 5jx 4j. Therefore, it seems clear that jx 4j needs to be less than 5 , so letting ı D 5 will work. Note that because > 0, ı is also greater than 0 as required by the definition of continuity. Putting this into the template results in the following proof. PROOF: The function f .x/ D 5x 3 is continuous at x D 4. Let f .x/ D 5x 3. Given > 0, let ı D 5 which is greater than 0 since > 0. Select x such that jx 4j < ı D 5 . Then ı > jx4j implies jf .x/f .4/j D j.5x3/.543/j D j5x20j D 5jx 4j < 5ı D . • Therefore, the function f is continuous at 4.
• • • • •
For a more challenging example, consider proving that the function f .x/ D 2x3 4x C 1 is continuous for all real numbers. This proof not only tackles a more complicated function than the one in the previous example, it is supposed to demonstrate the continuity of the function at the general real number a rather than at a specific value such as a D 4. This requires the proof to select an arbitrary a and prove the continuity of f at the point a. By showing that the function is continuous at any arbitrarily chosen a, it shows that the function is continuous at every point a. Again, the proof will select an arbitrary > 0 and needs to produce a ı > 0 such that jf .x/ f .a/j < for all x satisfying jx aj < ı. The proof needs to select an arbitrary a and an arbitrary > 0. Does it matter which it does first? In this case where the choice of a does not depend on which is chosen, and the choice of does not depend on which a is chosen, the order is not critical. It makes sense to select the a first because you are then challenged to prove that f is continuous at a for which you should choose an > 0. But since both quantifiers are universal quantifiers (for all a 2 R and for all > 0), the order does not matter. If it had been
4.2 Proving the Continuity of a Function
103
a universal quantifier and an existential quantifier such as “for all > 0 there exists a ı > 0,” then the order would matter a great deal. Working backwards from > jf .x/f .a/j you can see that you need > j.2x3 4xC1/.2a3 4aC1/j D j2.x3 a3 /4.xa/j D j2.xa/.x2 CxaCa2 /4.xa/j D jxajj2.x2 CxaCa2 /4j. You should not be surprised and, in fact, be quite pleased to see that this last expression contains a factor of jx aj because this will facilitate making the expression small when jx aj is made small. One only needs to control the size of the other factor j2.x2 C xa C a2 / 4j. Of course, if x is allowed to wonder too far from a, this other factor could get arbitrarily large, so care must be taken to restrict how far x gets from a. This can be done by requiring that ı not be larger than some conveniently selected value such as 1. That means that jx aj < ı 1 would imply, for example, that jxj < jaj C 1. Given this, there are many ways to find an upper bound for the quantity j2.x2 C xa C a2 / 4j where the upper bound does not depend on x. For example, j2.x2 C xa C a2 / 4j 2x2 C 2jxjjaj C 2a2 C 4 2.jaj C 1/2 C 2.jaj C 1/jaj C 2a2 C 4. One can afford to be sloppy here and get a simpler looking upper bound by saying 2.jaj C 1/2 C 2.jaj C 1/jaj C 2a2 C 4 2.jaj C 1/2 C 2.jaj C 1/.jaj C 1/ C 2.jaj C 1/2 C 4.jaj C 1/2 D 10.jaj C 1/2 . All you need is an upper bound that depends only on a. This leads to the following proof. PROOF: The function f .x/ D 2x3 4x C 1 is continuous on the real numbers. • Let f .x/ D 2x3 4x C 1, and let a 2 R. • Given > 0, which is greater than 0 since 1, , and 10.jajC1/2 • let ı D min 1; 10.jajC1/ 2 are all positive. • Select x such that jx aj < ı. Then ı 1 implies jxj < jaj C 1. • Also, ı 10.jajC1/ 2 implies that 3 jf .x/f .a/jDj.2x 4x C 1/ .2a3 4a C 1/j D j2.x3 a3 / 4.x a/jD j2.x a/.x2 C xa C a2 / 4.x a/j D jx aj j2.x2 C xa C a2 / 4j jx aj Œ2.jaj C 1/2 C 2.jaj C 1/jaj C 2a2 C 4 jx aj 2.jaj C 1/2 C 2.jaj C 1/.jaj C 1/ C 2.jaj C 1/2 C 4.jaj C 1/2 D 2 jx aj 10.jaj C 1/2 < 10.jajC1/ 2 10.jaj C 1/ D . • Therefore, the function f is continuous at every real number a. Not all functions can be expressed with nice formulas. Take, for example, the 2x if x is rational function f .x/ D which behaves differently on the rational x C 1 if x is irrational numbers than it does on the irrational numbers. Such functions that are defined one way on the rational numbers and another way on the irrational numbers make interesting examples because both the rational and the irrational numbers are dense in the real numbers; that is, in every nonempty open interval .a; b/, you can find both rational and irrational numbers. For the given function, in every nonempty open interval .a; b/ there are values of x where f .x/ D 2x and other values of x
104
4 Continuity
Fig. 4.2 A function equal to 2x for rational x (blue) and x C 1 for irrational x (red) The blue and red lines are not solid (1,2)
where f .x/ D x C 1. Indeed, for most real numbers a, lim f .x/ does not exist. Only x!a at x D 1, where 2x and x C 1 coincide, does this limit exist, and, in fact, at that point f .x/ is continuous (Fig. 4.2). A proof that f is continuous at x D 1 would be similar to the two preceding proofs, but you need to be careful to handle f .x/ differently depending on whether x is rational or irrational. As in other continuity proofs, given an > 0 you are faced with producing a value for ı > 0 which will ensure that jf .x/ f .1/j < whenever jx aj < ı. If the function in the proof were equal to x C 1 for every value of x, then the value ı D would work because jx 1j < shows that jf .x/ f .1/j D j.x C 1/ .1 C 1/j D jx 1j < . If the function in the proof were equal to 2x for every value of x, then the value ı D 2 would work because jx 1j < 2 shows that jf .x/ f .1/j D j.2x/ .2 1/j D 2jx 1j < . In this proof, then, you can choose ı D min.; 2 / D 2 . After selecting an x with jx 1j < ı, you merely consider two separate cases, one where x is rational, and one where x is irrational. These ideas allow you to produce the following proof. PROOF: The function f .x/ D x D 1. • Let f .x/ D
2x if x is rational x C 1 if x is irrational
2x if x is rational . x C 1 if x is irrational
is continuous at
Given > 0, let ı D 2 which is greater than 0 since > 0. Select x such that jx 1j < ı D 2 . If x is a rational number, then jf .x/ f .1/j D j2x 2j D 2jx 1j < 2ı D . If x is an irrational number, then jf .x/ f .1/j D j.x C 1/ 2j D jx 1j < ı < . • In either case, jf .x/ f .1/j < . • Therefore, the function f is continuous at 1.
• • • • •
4.3 Uniform Continuity
105
4.2.1 Exercises Write proofs of each of the following statements. 1. 2. 3. 4. 5. 6. 7.
f .x/ D 4x C 7 is continuous at x D 2. f .x/ D 5x2 C 3x 2 is continuous at x D 8. f .x/ D 10x3 C 25 is continuous for all real numbers x. f .x/ D jxj is continuous at x D 0. f .x/ D p jx2 9j is continuous for all real numbers x. f .x/ D px is continuous for all x 0. f .x/ D jx2 4j is continuous for all real numbers x.
4.3 Uniform Continuity Continuity of a function is a local property, that is, whether or not a function f is continuous at a point x D a depends only on how f behaves close to a. In fact, f 1 can be continuous at a and yet have very erratic behavior at points just 10 unit from 1 1 a or 100 or even 1;000;000 from a. The last example in the previous section shows a function continuous at x D 1 which is continuous for no other value of x. Even if f is continuous at all points of a set A, it could be that proofs of the continuity of f at two points x D a and x D b might need to be quite different. Certainly, there is no reason to believe that, given an > 0, a value of ı > 0 that works in a proof of the continuity of f at the point a would also work in a proof of the continuity of f at point b. Consider, for example, the function f .x/ D 1x which is continuous for all x ¤ 0. To prove that f is continuous at x D 2, given > 0 one can use ı D min.1; / or even be as generous as to let ı D min.1; 2/. But to prove that f is continuous at x D 12 where the function f changes much more rapidly, for the same > 0, one might need to use ı D min. 41 ; 8 /. You can easily see from the graph of f .x/ D 1x that as a gets closer to 0, the ı > 0 chosen for a particular > 0 will need to get smaller (Fig. 4.3). Suppose that you wanted to prove that a particular function f was continuous at every a in the domain of f . Such a proof was discussed in the previous section using f .x/ D 2x3 4x C 1. In that proof, the formula for the ı > 0 chosen for a given > 0 depended on the point a where f was being shown to be continuous. Clearly, this would have to be the case because f is a cubic function of x which grows increasingly more rapidly as x gets large. But it is not true that every function behaves this way. Some functions change at a constant rate like f .x/ D 6x 13 or change at a rate that does not continue to grow such as f .x/ D x2 1C1 . When writing a proof of the continuity of such functions, it is possible to pick a single value for ı > 0 that depends on > 0 (as it certainly would have to unless f were constant on each interval in its domain), but where the choice of ı > 0 does not depend on
106
4 Continuity
Fig. 4.3 f .x/ D 1x is not uniformly continuous
the point a where the continuity needs to be shown. These functions are special and satisfy the following definition. A function f is uniformly continuous on the set A if for every > 0 there is a ı > 0 such that jf .x/ f .y/j < for every x and y in A satisfying jx yj < ı. You should compare this definition to the definition of continuity at a point. The difference centers on when the value of ı > 0 needs to be determined. For continuity at a single point, given > 0, one must specify the value of ı > 0 after being given the value of a but before being given a value for x. Thus, the value of ı > 0 can depend on the value of a even though it cannot depend on the value of x. On the other hand, for uniform continuity, given > 0, one must specify the value of ı > 0 before learning the values of either x or y, and, therefore, its value cannot depend on either x or y. The definition of uniform continuity suggests a template for how to prove that a given function f is uniformly continuous on a set A. As in the proof for continuity at a point, you would say that a value for > 0 has been given. Then you would present a value for ı > 0. Once these two values have been specified, you would need to show that any x and y in A that satisfy jxyj < ı also satisfy jf .x/f .y/j < . This suggests the following. TEMPLATE for proving the function f is uniformly continuous on the set A • SET THE CONTEXT: Make statements about what is known about the function f . • SELECT AN ARBITRARY : Given > 0, • PROPOSE A VALUE FOR ı: let ı D . Here you would insert an appropriate value for ı. • SELECT ARBITRARY x and y in A with jx yj < ı: Let x and y be in A such that jx yj < ı. (continued)
4.3 Uniform Continuity
107
• LIST IMPLICATIONS: Derive the result jf .x/ f .y/j < . • STATE THE CONCLUSION: Therefore, f is uniformly continuous on the set A. Proving the function f .x/ D 6x 13 is uniformly continuous on the entire real line is straightforward since the function f changes at a constant rate. This allows you to select a value for ı > 0 based on that rate of change, 6. PROOF: The function f .x/ D 6x 13 is uniformly continuous on the real numbers. • • • • • •
Let f .x/ D 6x 13. Given > 0, let ı D 6 which is greater than 0 since > 0. Let x and y be real numbers such that jx yj < ı D 6 . Then jf .x/ f .y/j D j.6x 13/ .6y 13/j D 6jx yj < 6ı D . Therefore, the function f is uniformly continuous on the real numbers.
Less clear is how to choose a value for ı > 0 when proving f .x/ D x2 1C1 is uniformly continuous on the real numbers. To do this, you need to find a way to show ˇjf .x/ f .y/jˇ < . You would try to find an upper bound for 2 2 C1/j ˇ ˇ jxCyj D .x2 C1/.y jf .x/f .y/j D ˇ x2 1C1 y2 1C1 ˇ D j.y.x2C1/.x 2 C1/ jxyj. This expression C1/.y2 C1/ is complicated, so it is convenient to find ways to simplify it. The nice thing about working with inequalities rather than equalities is that you are not prevented from making changes that increase the value of your expression. That is, if you can simplify an expression by substituting an expression that is a little larger, that might not be a problem. The numerator in the previous expression is jx C yj which does not simplify algebraically, but it does suggest a possible application of the triangle inequality, jx C yj jxj C jyj. Changing jx C yj to jxj C jyj allows the fraction to be broken into two simpler fractions. It allows you to continue with jf .x/ f .y/j D jxCyj jyj jxj jx yj. jx yj .x2 C1/.y2 C1/ C .x2 C1/.y2 C1/ jx yj x2jxj C y2jyj .x2 C1/.y2 C1/ C1 C1 When jxj < 1, you can conclude that jxj < 1 x2 C 1. When jxj 1, you can 2 conclude that jxj x2 < x2 C 1. In either case .x2jxj xx2 C1 D 1. This lets you C1/ C1 jxCyj jyj jxj jx yj C state that jf .x/ f .y/j D .x2 C1/.y 2 C1/ jx yj 2 2 2 2 .x C1/.y C1/ .x C1/.y C1/ 2jx yj. This suggests that ı D 2 will work in the proof.
108
PROOF: The function f .x/ D numbers.
4 Continuity 1 x2 C1
is uniformly continuous on the real
Let f .x/ D x2 1C1 . Given > 0, let ı D 2 which is greater than 0 since > 0. Let x and y be real numbers such that ˇ ˇ jx 2 yj < ı2 D 2 . ˇ ˇ C1/j • Then jf .x/ f .y/j D ˇ x2 1C1 y2 1C1 ˇ D j.y.x2C1/.x D C1/.y2 C1/ jxCyj jyj jxj jx yj jx yj .x2 C1/.y 2 2 2 C1/ C .x2 C1/.y2 C1/ .x C1/.y C1/ jxj jx yj C y2jyj x2 C1 C1
• • • •
• Note that if jxj < 1, then jxj < x2 C 1, and if jxj 1, then jxj x2 < x2 C 1. • In either case, jxj < x2 C 1, so x2jxj < 1, and similarly, y2jyj < 1. C1 C1 jyj jxj • It follows that jf .x/ f .y/j x2 C1 C y2 C1 jx yj < 2jx yj < 2ı D . • Therefore, the function f is uniformly continuous on the real numbers. One of the most memorable theorems from Calculus is the Mean Value Theorem which states that if the function f is continuous on the interval Œa; b and differentiable on the interval .a; b/, then there is a c 2 .a; b/ such that .a/ f 0 .c/ D f .b/f . If the function f has a bounded derivative on the interval ba Œa; b, that is, if there is a positive real number M such that jf 0 .x/j M for all values of x 2 Œa; b, then one can easily see that f is uniformly continuous on that interval. Indeed, if x and y are in Œa; b, then there is a c between x and y such that jf .x/ f .y/j D jf 0 .c/j jx yj M jx yj. This implies that given > 0, the value ı D M > 0 can be used in a proof that f is uniformly continuous on Œa; b for then jx yj < ı implies jf .x/ f .y/j D jf 0 .c/j jx yj < M jx yj < Mı D . This is summarized by saying that a function with a bounded derivative on an interval is uniformly continuous there. Whenever you learn of the truth of a conditional statement such as the one at the end of the previous paragraph (bounded derivative implies uniform continuity), it is natural to ask whether the converse of the statement is also true (uniform continuity implies bounded derivative). The answer to this particular question is “no, not all functions uniformly continuous on an interval have bounded derivatives there.” In particular, the function f .x/ D jxj is an example of a function uniformly continuous on the entire real line, yet it fails to be differentiable at x D 0. The function f .x/ D p x is uniformly continuous for x 0, but its derivative is unbounded near x D 0. A more complex example is the function defined by f .x/ D x2 sin x12 when x ¤ 0 and f .0/ D 0. This function is uniformly continuous on the interval Œ10; 10 even though its derivative, which exists on the entire real line, is not bounded as x approaches 0. p Because the function f .x/ D x has an increasingly large rate of change as x approaches 0, proving that the function is uniformly continuous for x 0 provides an interesting challenge. The proof will need to conclude that > jf .x/ f .y/j D
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p p p p p p j x yj. xC y/ p p p . As expected, there is a factor of jx yj in j x yj D D pjxyj xC y xC y this expression, so that you can try to make the expression small by the prestricting p size of jx yj. This is easy if the denominator of the expression, x C y, does not get too small. The problem is if x and y get close to 0, the denominator of the expression will also get close to 0. At first this seems p likepa significant roadblock. But this roadblockppresents its own resolution for if x C y is very small, it must p certainly be that j x yj is even smaller is the conclusion that you want. p which p In other words, there are two cases: either x C y is small which would imply that p p jf .x/ f .y/j is small, or x C y is large which would imply that jf .x/ f .y/j D jxyj p p is small. You only need to decide what to use as the dividing line between xC y p p “large” and “small.” A natural choice would be itself because x C y < p p p p jxyj implies j x yj < . If x C y , then jf .x/ f .y/j D pxCpy jxyj 2
which suggests letting ı D 2 so that jx yj < ı gives jf .x/ f .y/j < D . The complete proof follows. p PROOF: The function f .x/ D x is uniformly continuous on the interval x 0. p • Let f .x/ D x. • Given > 0, • let ı D 2 which is greater than 0 since ¤ 0. • Let x and y be nonnegative real numbers such that jx yj < ı.p p p p • In the case that x C y < , it follows that jf .x/ f .y/j D j x yj p p x C y < . p p p p • In the case that x C y , it follows that jf .x/ f .y/j D j x yj D p p p p j x yj. xC y/ p p xC y
D
jxyj p p xC y
jxyj
<
ı
D
2
D .
• In either case, jx yj < ı implies that jf .x/ f .y/j < , so the function f is uniformly continuous on the interval x 0. There is an important lesson to be learned from this example. When planning how to write a proof, you can pursue one line of thinking which may solve the problem in most but not all cases. Sometimes the special cases where the argument does not work are enough to cause you to abandon your original line of reasoning altogether. But often you can just break your argument into two or more cases and find other techniques to handle the special cases where the original argument does not work.
4.3.1 Exercises Write proofs of each of the following statements. 1. f .x/ D 3x C 11 is uniformly continuous on the set of real numbers. 2. f .x/ D 14x C 5 is uniformly continuous on the set of real numbers. 3. f .x/ D jxj is uniformly continuous on the set of real numbers.
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4. 5. 6. 7. 8.
4 Continuity
f .x/ D 8x2 is uniformly continuous on the interval Œ6; 6. 4 f .x/ D 5xC1 is uniformly continuous for x 0. p f .x/ D 3 x is uniformly continuous on the set of real numbers. f .x/ D x2 is not uniformly continuous on the set of real numbers. f .x/ D x12 is not uniformly continuous on the set .0; 1/.
4.4 Compactness and the Heine–Borel Theorem 4.4.1 Open Covers and Subcovers Let a and b be real numbers with a < b. It turns out that if a function f is continuous on the closed interval Œa; b, then f is uniformly continuous on that interval. How might you prove this result? As a first try, you might say that for each > 0 and for each y 2 Œa; b there is a ı > 0 such that if x 2 Œa; b with jx yj < ı, then jf .x/ f .y/j < . Then, having produced a value for ı for each y 2 Œa; b, you might want to pick the smallest of all of those ı’s and hope that this minimum ı would be sufficiently small to work for every y 2 Œa; b. Unfortunately, you started out with an infinite collection of ı’s, each greater than 0. Such an infinite set might not have a minimum value. The set of such ı’s is certainly nonempty and bounded below, so the collection does have a greatest lower bound, but that greatest lower bound could be 0, too small to use for the ı in the proof. A finite set of positive numbers always has a minimum value that is positive, but an infinite set of positive numbers might have a greatest lower bound of 0. Suppose that T is a collection of open intervals, and A R. If the set A is contained in the union of the open intervals in T, that is, if A [ .s; t/, then .s;t/2T
T is called an open cover of A. A subset T 0 T which is also an open cover of A is called a subcover of A. In the above suggested proof that the continuity of f on Œa; b implies the uniform continuity of f on Œa; b, the definition of continuity at each point of y 2 Œa; b produced a collection of open intervals which form an open cover T of Œa; b. If that open cover had a finite subcover T 0 , then you would be dealing with only a finite number of ı > 0 values, and you could expect to produce a smallest such ı > 0. Whether such a finite subcover exists has nothing to do with the continuous function f that motivated this discussion. A closed bounded interval Œa; b in the real numbers is compact which means that every open open cover of Œa; b contains a finite subcover. The fact that every closed bounded interval in the real numbers is compact is known as the Heine–Borel Theorem, and it is central to proving the above result about continuous functions on closed bounded intervals being uniformly continuous there. In fact, the Heine–Borel Theorem is an important tool for proving many results in analysis. Suppose that for every rational number in Œ0; 1 you represent the rational number in lowest terms as pq . Then for each of these rational numbers you
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associate the open interval . 4p1 ; 4pC1 /. For example, the number 27 would be 4q 4q 7 9 associated with the open interval . 28 ; 28 /. Since the set of rational numbers in Œ0; 1 is infinite, this collection of open intervals is also infinite. The collection forms an open cover of Œ0; 1. One possible finite subcover is the collection of intervals associated with rational numbers 01 ; 14 ; 13 ; 12 ; 23 ; 34 ; and 11 giving the intervals 3 5 3 5 7 9 ; 16 /; . 12 ; 12 /; . 38 ; 58 /; . 12 ; 12 /. 11 ; 13 /; and . 34 ; 54 /. You should verify that . 14 ; 14 /; . 16 16 16 these intervals are in the original open cover and do produce the claimed finite subcover. On the other hand, if you associate with each natural number n > 1 the open interval . n1 ; 1/, you get an open cover of the set .0; 1/, yet no finite subset of this collection of intervals can cover the entire interval .0; 1/. Indeed, any finite collection will only cover the interval . m1 ; 1/ for some natural number m > 1. Since these intervals form an open cover of .0; 1/ which does not have a finite subcover, the set .0; 1/ is not a compact set.
4.4.2 Proofs of the Heine–Borel Theorem Presented next are two quite different proofs of the Heine–Borel Theorem. The techniques used in both proofs are instructive, and it is interesting to see how a single result can be proved using two completely different strategies. Given in each case are real numbers a < b and a set of open intervals T that forms an open cover of the closed bounded interval Œa; b. Both proofs seek to show that there must be a finite subset of T that covers Œa; b. The strategy in the first proof suggests that, whether or not you can cover Œa; b with a finite number of open intervals, you can certainly cover some of the interval starting at a and working at least part of the way toward b. The proof proposes looking at the set S D fx 2 Œa; b j T has a finite subcover that covers the interval Œa; xg: The proof first shows that S is not empty because it contains the point a. The set S is bounded above by b, so S has a least upper bound, r. This is not to say that r 2 S, but if r is not in S, there must be values in S that are arbitrarily close to r. Certainly r is in Œa; b, so there is an open interval from T that covers r. Since there are values of S arbitrarily close to r, there are some inside this open interval containing r. This open interval then extends the finite subcover to values greater than r. One can only conclude that r must be b, and, in fact, b 2 S. Thus, Œa; b has a finite subcover, and the proof is complete (Fig. 4.4).
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PROOF (Heine–Borel Theorem): Let a < b be two real numbers, and let T be an open cover of Œa; b. Then T contains a finite subcover of Œa; b. • Let a < b be two real numbers, and let T be an open cover of Œa; b. • Define set S D fx 2 Œa; b j T has a finite subcover that covers the interval Œa; xg. • The set T is an open cover of Œa; b, and a 2 Œa; b, so T must contain at least one open interval, .p; q/ which contains the point a, that is, p < a < q. Since the interval Œa; a is covered by .p; q/ 2 T, the point a 2 S, and S is not an empty set. • The set S is bounded above by b. • Since S is nonempty and bounded above, it has a least upper bound r. • Since r must be at least a and cannot be greater than b, r 2 Œa; b, so there is an interval .p; q/ in T which contains the point r, that is, p < r < q. • Since p < r and r is the least upper bound of S, p is not an upper bound of S. Thus, there is a point y 2 S with p < y. This means that there is a finite set of intervals in T that covers Œa; y. • Let z D min. rCq ; b/. Since z r and z 2 .p; q/, adding the interval .p; q/ 2 to the finite set of intervals of T that covers Œa; y produces a finite set of intervals in T that covers Œa; z, and z 2 S. • But r is the least upper bound for S, implying that z r. Because z D min. rCq ; b/ and rCq > r, it must be that z D b. 2 2 • Because z 2 S, it follows that b 2 S which completes the proof of the theorem. The second proof of the Heine–Borel Theorem is a proof by contradiction. It begins as the first proof by assuming that a < b are real numbers, and that the interval Œa; b has an open cover T. Then it makes the additional assumption that no finite collection of intervals in T can cover Œa; b. This will lead to a contradiction. This proof is not one that the beginning student is likely to invent on their own unless they have seen the technique before. First, the proof sets a0 D a and b0 D b so that the interval Œa0 ; b0 D Œa; b. Let 0 m0 D a0 Cb be the midpoint of Œa0 ; b0 . It must be the case that at least one of the 2 intervals Œa0 ; m0 or Œm0 ; b0 cannot be covered by a finite number of intervals in T because, if both can be covered by a finite number of intervals, putting those two collections together would give a finite collection of intervals that covered the entire interval Œa0 ; b0 D Œa; b contradicting the assumption that this could not be done. p
[ a
Fig. 4.4 Heine–Borel Theorem first proof
(
q
y
r
z
)
] b
4.4 Compactness and the Heine–Borel Theorem
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So, if it is the case that Œa0 ; m0 cannot be covered by a finite number of intervals in T, let a1 D a0 and b1 D m0 . Otherwise, if Œm0 ; b0 cannot be covered by a finite number of intervals in T, let a1 D m0 and b1 D b0 . In either case, the new interval Œa1 ; b1 Œa; b cannot be covered by a finite collection of intervals in T. Now the proof continues iteratively. If for some j > 0, there is an interval Œaj ; bj contained in Œa; b which cannot be covered by any finite collection of intervals in a Cb T, let mj D j 2 j be the midpoint of the interval. Either Œaj ; mj or Œmj ; bj cannot be covered by a finite collection of intervals from T, so if Œaj ; mj cannot be covered by a finite collection of intervals, let ajC1 D aj and bjC1 D mj . Otherwise, let ajC1 D mj and bjC1 D bj . In either case ŒajC1 ; bjC1 cannot be covered by a finite collection of intervals from T. Notice that this process constructs a sequence of intervals Œa0 ; b0 ; Œa1 ; b1 ; Œa2 ; b2 ; : : : contained in Œa; b, none of which can be covered by a finite collection of intervals in T. Also note that a D a0 a1 a2 : : : while b D b0 b1 b2 : : :, and for each j, the length of the jth interval is bj aj D ba . Since each aj term is less than all of the bk terms, both of the 2j monotone sequences are bounded and, therefore, converge. Moreover, since for each k, lim bj lim aj bk ak D ba , it follows that lim aj D lim bj D r 2 Œa; b. 2k j!1
j!1
j!1
j!1
Note that since the sequence of aj ’s increases to r, and the sequence of bj ’s decrease to r, the limit r 2 Œaj ; bj for each j. Because the limit, r, is in Œa; b, there is an open interval .p; q/ 2 T such that r 2 .p; q/. The distance the limit r is from the boundary of the interval .p; q/ is D min.r p; q r/ > 0. Since lim ba D 0, you can 2j j!1
< . Then it follows that p r < aj r bj r C < q, select a j so that ba 2j and, so, Œaj ; bj .p; q/. But this shows that Œaj ; bj is covered by the single open interval .p; q/ 2 T contradicting the fact that Œaj ; bj could not be covered by a finite collection of intervals in T. Thus, you must conclude that the assumption that Œa; b cannot be covered by a finite number of intervals is false. A formal proof follows (Fig. 4.5).
r Fig. 4.5 Heine–Borel Theorem second proof
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PROOF (Heine–Borel Theorem): Let a < b be two real numbers, and let T be an open cover of Œa; b. Then T contains a finite subcover of Œa; b. • Let a < b be two real numbers, and let T be an open cover of Œa; b. • Assume that T contains no finite subcover of Œa; b. • Let a0 D a and b0 D b so that the interval Œa0 ; b0 D Œa; b, and note that no finite collection of intervals in T will cover Œa0 ; b0 . • Define sequences and inductively. For j 0, let Œaj ; bj Œa; b be an interval which cannot be covered by a finite collection of open intervals in T, and where bj aj D ba . 2j aj Cbj • Let mj D 2 be the midpoint of Œaj ; bj . • It must be the case that at least one of the intervals Œaj ; mj or Œmj ; bj cannot be covered by a finite number of intervals in T because, if both can be covered by a finite number of intervals, putting those two collections together would give a finite collection of intervals that covered the entire interval Œaj ; bj . • If Œaj ; mj cannot be covered by a finite collection of intervals, let ajC1 D aj and bjC1 D mj . Otherwise, let ajC1 D mj and bjC1 D bj . In either case ŒajC1 ; bjC1 cannot be covered by a finite collection of intervals from T, and ba j
bjC1 ajC1 D 22 D 2ba jC1 . • Thus, there are monotone sequences a D a0 a1 a2 : : : and b D b0 b1 b2 : : :, and for each j, the length of the Œaj ; bj interval is bj aj D ba . 2j • Since each aj term is less than all of the bk terms, both of the monotone sequences are bounded and, therefore, converge. The fact that lim aj j!1
lim bj lim .aj C
j!1
j!1
ba /, 2j
shows that lim aj D lim bj D r 2 Œa; b. j!1
j!1
• Because the limit, r, is in Œa; b, there is an open interval .p; q/ 2 T such that r 2 .p; q/. • The distance the limit r is from the boundary of the interval .p; q/ is D min.r p; q r/ > 0. Since lim ba D 0, there is a j such that ba < . 2j 2j j!1
• It follows that p r aj r bj rC < q, and, so, Œaj ; bj .p; q/. • But then Œaj ; bj is covered by the single open interval .p; q/ 2 T contradicting the fact that Œaj ; bj could not be covered by a finite collection of intervals in T. • Thus, the assumption that Œa; b cannot be covered by a finite number of intervals is false, and the theorem is proved. The fact that the interval Œa; b in the Heine–Borel Theorem is both closed and bounded is crucial. The interval Œ1; 1/ is covered by the collection of open intervals .j; j C 2/ for j D 0; 1; 2; 3; : : :, but no finite collection of these open intervals can cover Œ1; 1/. The interval .0; 5/ is covered by the collection . 1j ; 5/ for j D 1; 2; 3; 4; : : :, but, again, no finite collection of these open intervals can cover .0; 5/.
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4.4.3 Uniform Continuity on Closed Bounded Intervals With the Heine–Borel Theorem, it can now be shown that every continuous function on a closed bounded interval is uniformly continuous on that interval. The idea is simple enough: if f is continuous on the closed bounded interval Œa; b, then, given > 0, at each point x 2 Œa; b there is a ı > 0 such that for any y 2 .x ı; x C ı/, it follows that jf .x/ f .y/j < . Thus, there is an open interval around each x 2 Œa; b that has the desired property, and the Heine–Borel Theorem shows that Œa; b can be covered by just a finite number of these open intervals. Since each of these finitely many open intervals is associated with a positive ı, you can select the smallest ı to serve as the ı > 0 needed in your proof of uniform continuity. There are, though, a couple of subtleties that get in the way of this simple argument. First of all, for any y in one of the open intervals .x ı; x C ı/ you can conclude that jf .y/ f .x/j < , but the proof will require that jf .y/ f .z/j < for any y and z that are within the chosen ı of each other, not just for z D x, the middle point of the interval. One can get around this problem by arranging that jf .y/ f .x/j < 2 for all y 2 .x ı; x C ı/. This is a common trick in analysis proofs. The definition of continuity allows you to find a ı > 0 that works for any given > 0, so why not for 2 which is also greater than 0? Then for any y and z in .x ı; x C ı/, you can use the triangle inequality to conclude that jf .y/f .z/j D j f .y/f .x/ f .z/f .x/ j jf .y/f .x/jCjf .z/f .x/j < 2 C 2 D . There is a second problem with the this strategy. If you select y and z within ı of each other, how do you know that they both lie within the same interval .xı; xCı/? The interval Œa; b is covered by a finite number of such intervals, but just because the two numbers y and z are close to each other does not mean that they will both fall within the same interval in your finite collection of open intervals. There are a couple of ways to get around this problem. One method is to consider the endpoints of the intervals in your finite collection of open intervals. Since the number of open intervals is finite, there are only finitely many endpoints to these intervals. You could select the ı in the proof not to be the least of the ı’s used for any of the intervals but to be the least distance between any two distinct elements of the collection of endpoints of these intervals. That ensures that if y and z are closer together than ı, there can be at most one endpoint between y and z. That will guarantee that y and z will both be within one of the finitely many open intervals. This follows from the fact that intervals in an open cover must overlap, so that each endpoint of one of the open intervals must be a member of one of the other open intervals in the open cover as seen in the following diagram (Fig. 4.6).
(
y
(z )
)
Fig. 4.6 y and z straddle one endpoint but remain in an interval of the open cover
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A cleaner way to ensure that any y and z within ı of each other are in one of the finite number of intervals in the open cover of Œa; b is to be more clever about choosing the original open intervals. Suppose that for all y 2 .x ı; x C ı/, it follows that jf .y/ f .x/j < . You can be very conservative and use the open interval .x 2ı ; x C 2ı / as the interval chosen to cover x in the open cover of Œa; b. Then if y and z are very close, and y 2 .x 2ı ; x C 2ı / for some x, it will follow that, since y and z will be closer together than 2ı , guaranteeing that both y and z will be in .x ı; x C ı/, and the result will follow. The following proof uses the first strategy. PROOF: A function continuous on a closed bounded interval is uniformly continuous on that interval. • Let a < b be two real numbers, and let f be continuous on the interval Œa; b. • Let > 0 be given. • By the definition of continuity, for each x 2 Œa; b there is a ıx > 0 such that for all y in Œa; b with jy xj < ıx it follows that jf .y/ f .x/j < 2 . • Because for each x 2 Œa; b, the point x 2 .x ı2x ; x C ı2x /, this collection of open intervals covers Œa; b. • By the Heine–Borel Theorem, there is a finite set fx1 ; x2 ; x3 ; : : : ; xn g Œa; b such that the collection I D f.xj ıxj ; xj C ıxj / j j D 1; 2; 3; : : : ; ng forms an open cover of Œa; b. • The set of endpoints of these intervals, E D fxj ˙ ıxj j j D 1; 2; 3; : : : ; ng, is a finite set, so let ı be the smallest positive difference between two elements of E. • Let y and z be elements of Œa; b with jy zj < ı. • Because the collection of intervals I is an open cover of Œa; b, there are j and k such that y 2 .xj ıxj ; xj C ıxj / and z 2 .xk ıxk ; xk C ıxk /. If j D k, then y and z are in the same interval of I. If j ¤ k, then because jy zj < ı, there is at most one endpoint in E between y and z. Thus, either there is at most one endpoint of .xj ıxj ; xj C ıxj / or .xk ıxk ; xk C ıxk / between y and z. This implies that either y and z are both in .xj ıxj ; xj C ıxj / or both in .xk ıxk ; xk C ıxk /. In either case, there is a single interval .xm ıxm ; xm C ıxm / 2 I such that y; z 2 .xm ıxm ; xm C ıxm /. • Now it follows that jf .y/ f .z/j D j .f .y/ f .xm // .f .z/ f .xm // j jf .y/ f .xm /j C jf .z/ f .xm /j < 2 C 2 D . • This shows that for every > 0 there is a ı > 0 such that if y; z 2 Œa; b with jy zj < ı, then jf .y/ f .z/j < . This completes the proof that f is uniformly continuous on Œa; b. Note that the fact that Œa; b is both closed and bounded is crucial. The function f .x/ D x2 is continuous on the unbounded interval Œ0; 1/, but f is not uniformly continuous on this interval. Similarly, the function f .x/ D 1x is continuous on the open interval .0; 1/, but f is not uniformly continuous on this interval.
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4.4.4 Exercises 1. Determine which of the following sets of real numbers are compact. (a) (b) (c) (d) (e) (f) (g) (h)
Œ0; 12 .2; 2 Œ1; 4 [ Œ8; 15 f1; 3; 5; 7; 9g R ; Œ2; 6/ [ .6; 11 S h i 1 1 1 f0g [ jD1 2jC1 ; 2j
2. For each of the following open covers, find a finite subcover. (a) Œ1; 10 is covered by the collection of open intervals r 14 ; r C 14 where r 2 Q. (b) Œ0; 3 is covered by the collection of open intervals 1j ; 4 for j D 1; 2; 3; : : : 1 1 ; 10 . along with the open interval 10 (c) Œ0; 1 is covered by the collection of open intervals 2j1 ; 2j for j D 1; 2; 3; : : : 1 1 along with 10 ; 10 . Write proofs of each of the following statements. 3. The intersection of two compact sets is a compact set. 4. The union of two compact sets is a compact set. 5. If C is a compact set and .a; b/ is an open interval, then the set difference Cn.a; b/ is a compact set. 6. If the function f is uniformly continuous on the interval Œa; b and uniformly continuous on the interval Œb; c for a < b < c, then f is uniformly continuous on the interval Œa; c. 7. If a set A has a cover consisting of a finite number of open intervals, then A has a subcover such that for each x 2 A, x is an element of at most two of the open intervals in the subcover.
4.5 The Arithmetic of Continuous Functions Chapter 3 discusses several theorems about how one can calculate limits when faced with the addition, subtraction, multiplication, or division of functions whose limits are known. As one might expect, since continuity and limits are closely related, the proofs of the corresponding theorems about functions continuous at a point are, in fact, very similar. Before starting, it is worth pointing out that if f and g are two functions, then you can define the new functions f C g, f g, f g, and gf at all
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points in the intersection of the domain of f and the domain of g and, in the case of gf , only where g is not 0. Generally, one is interested in functions that have a common domain, but sometimes this is not the case. Pathological examples do exist. It could be, for example, that f is only defined for positive real numbers, and g is only defined for negative real numbers as with f .x/ D p1x and g.x/ D p1x . Then f C g has domain and is the empty function, one that contains no ordered an empty pairs, x; f .x/ . Oddly, the definition of continuity says that the empty function is continuous because it satisfies the definition at each point of its empty domain. Suppose that functions f and g have a common domain where the point a is an accumulation point of that domain. Also suppose that lim f .x/ D L and lim g.x/ D x!a x!a H. Recall that when proving that the limit of f C g is L C H, you are given an > 0 and can use the definition of limit to conclude that there are ı1 > 0 and ı2 > 0 such that if x is in the common domain of f and g with 0 < jx aj < ı1 , then jf .x/ Lj < 2 , and if 0 < jx aj < ı2 , then jg.x/ Hj < 2 . Then the triangle inequality allows you to conclude that for all x with 0 < jx aj < min.ı1 ; ı2 / that j .f .x/ C g.x// .L C H/j D j .f .x/ L/ C .g.x/ H/ j jf .x/ Lj C jg.x/ Hj < C 2 D . The same method works for the proof about continuity of f C g at a 2 with minor changes made to match the template for writing proofs about continuity of a function at a point. Of course, the same logic works for proving the continuity of f g, so the two results might as well be combined as follows. PROOF: Suppose that f and g are functions with common domain containing the point a. If both f and g are continuous at the point a, then so are the functions f C g and f g. • Let f and g be functions both defined on a set A containing the point a, and assume that f and g are both continuous at a. • Let > 0 be given. • By the definition of continuity, there is a ı1 > 0 such that if x 2 A and jx aj < ı1 , then jf .x/ f .a/j < 2 . • Similarly, there is a ı2 > 0 such that if x 2 A and jx aj < ı2 , then jg.x/ g.a/j < 2 . • Let ı D min.ı1 ; ı2 /. • Then if x 2 Awith jx aj < ı, • j f .x/ ˙ g.x/ f .a/ ˙ g.a/ j D j f .x/ f .a/ ˙ g.x/ g.a/ j jf .x/ f .a/j C jg.x/ g.a/j < 2 C 2 D . • This shows that f C g and f g are continuous at a. Now suppose that f and g are functions as discussed above with lim f .x/ D L x!a
and lim g.x/ D H. Recall how you prove that lim f .x/g.x/ D LH. Again, as with x!a
x!a
the proof for the sum of the limits, given > 0 you find ı > 0 so that both jf .x/ Lj and jg.x/ Mj are small when 0 < jx aj < ı. How small do these need to be? The idea was to write jf .x/g.x/ LHj as jf .x/g.x/ f .x/H C f .x/H LHj jf .x/j jg.x/ Hj C jHj jf .x/ Lj. Thus, ı1 > 0 can be chosen to ensure that
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jf .x/ Lj is less than 1, ı2 > 0 so that jf .x/ Lj is less than 2.jHjC1/ , and ı3 so that jg.x/ Hj is less than 2.jLjC1/ . Then ı can be set to the least of ı1 , ı2 , and ı3 . The proof for continuity of fg at the point a follows this same strategy.
PROOF: Suppose that f and g are functions with common domain containing the point a. If both f and g are continuous at the point a, then so is the function fg. • Let f and g be functions both defined on a set A containing the point a, and assume that f and g are both continuous at a. • Let > 0 be given. • By the definition of continuity, there is a ı1 > 0 such that if x 2 A and jx aj < ı1 , then jf .x/ f .a/j < 1, and thus, jf .x/j < jf .a/j C 1. • There is a ı2 > 0 such that if x 2 A and jx aj < ı2 , then jf .x/ f .a/j < . 2.jg.a/jC1/ • There is a ı3 > 0 such that if x 2 A and jx aj < ı3 , then jg.x/ g.a/j < . 2.jf .a/jC1/ • Let ı D min.ı1 ; ı2 ; ı3 /. • Then if x 2 A with jx aj < ı, • jf .x/g.x/ f .a/g.a/j D jf .x/g.x/ f .x/g.a/ C f .x/g.a/ f .a/g.a/j C jg.a/j jf .x/ f .a/j < jf .x/j jg.x/ g.a/j jf .a/j C 1 2.jf .a/jC1/ C jg.a/j 2.jg.a/jC1/ 2 C 2 D . • This shows that fg is continuous at a. Finally, suppose that f and g are functions as discussed above with lim f .x/ D L x!a
f .x/ and lim g.x/ D H and H ¤ 0. This time recall how you prove that lim g.x/ D HL . x!a x!a The idea is the same ˇ but the algebra ˇ took ˇ the ˇproof for products, ˇ ˇ as with ˇ f .x/HLg.x/ ˇ ˇ ˇ ˇ f .x/ Lˇ a few more steps. ˇ g.x/ H ˇ D ˇ g.x/H ˇ D ˇ f .x/HLHCLHLg.x/ ˇ D g.x/H ˇ ˇ ˇ .f .x/L/HCL Hg.x/ ˇ ˇ ˇ jf .x/Lj C jLjjg.x/Hj . Then, given an > 0, you can ˇ ˇ g.x/H jg.x/j jg.x/jjHj
choose ı1 > 0 so that jx aj < ı1 would ensure jg.x/ Hj < jHj which, in 2 jHj turn, implies that jg.x/j > 2 . Then you choose a ı2 > 0 so that jx aj < ı2 gives jf .x/ Lj < jHj . Lastly, you choose a ı3 > ˇ0 so that jxˇ aj < ı3 gives 4 ˇ ˇ H 2 jg.x/ Hj < 4.jLjC1/ . This allowed you to conclude ˇ f .x/HLg.x/ ˇ < . Again, the g.x/H proof for continuity can be constructed by changing the limit L to jf .a/j, the limit H to jg.a/j, and making some other minor wording changes.
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PROOF: Suppose that f and g are functions with common domain containing the point a with g.a/ ¤ 0. If both f and g are continuous at the point a, then so is the function gf . • Let f and g be functions both defined on a set A containing the point a, and assume that f and g are both continuous at a with g.a/ ¤ 0. • Let > 0 be given. • Note that jg.a/j > 0. By the definition of continuity, there is a ı1 > 0 such that if x 2 A and jx aj < ı1 , then jg.x/ g.a/j < jg.a/j . For these x it 2 > jg.x/jCjg.x/g.a/j D jg.x/jCjg.a/g.x/j follows that jg.x/jC jg.a/j ˇ ˇ2 ˇg.x/ C g.a/ g.x/ ˇ D jg.a/j which implies that jg.x/j > jg.a/j jg.a/j D 2 jg.a/j . 2 • By the definition of continuity, there is a ı2 > 0 such that if x 2 A and jx aj < ı2 , then jf .x/ f .a/j < jg.a/j . 4 • By the definition of continuity, there is a ı3 > 0 such that if x 2 A and 2 . jx aj < ı3 , then jg.x/ g.a/j < 4.jfg.a/ .a/jC1/ • Let ı D min.ı1 ; ı2 ; ı3 /. • Then if x 2ˇ A with ˇ 0 < jx aj ˇ< ı,ˇ ˇ ˇ ˇ f .x/ f .a/ ˇ ˇ f .x/g.a/f .a/g.x/ ˇ ˇ f .x/g.a/f .a/g.a/Cf .a/g.a/f .a/g.x/ ˇ • ˇ g.x/ g.a/ ˇ D ˇ ˇDˇ ˇD g.x/g.a/ g.x/g.a/ ˇ ˇ ˇ .f .x/f .a//g.a/Cf .a/ g.a/g.x/ ˇ jf .x/f .a/j ˇˇ f .a/.g.x/g.a// ˇˇ ˇ ˇ C ˇ g.x/g.a/ ˇ < ˇ ˇ g.x/g.a/ jg.x/j 2
jg.a/j 4
2 jg.a/j C 4.jfg.a/ 2jf .a/j < 2 C .a/jC1/ jg.a/j2 • This shows that fg is continuous at a.
2
D .
4.5.1 Exercises 1. Suppose that f and g are functions that are both uniformly continuous of a set A. Find an example showing that their product need not be uniformly continuous on A. Write proofs for each of the following statements. 5
2. The function f .x/ D x 2 is continuous for x 0. 3. All polynomials are continuous on R. 4. All rational functions are continuous on R except at points where their denominators are 0. 5. If f and g are uniformly continuous on the set A, then f C g and f g are also uniformly continuous on A.
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6. Suppose f and g have common domain A and f C g is continuous at a 2 A. If f is discontinuous at a, then g is discontinuous at a. 7. Suppose f and g have common domain A and fg is continuous at a 2 A. If f is discontinuous at a, then g is either discontinuous at a or g.a/ D 0.
4.6 Composition, Absolute Value, Maximum, and Minimum Recall that the two functions g W A ! B and f W B ! C can be composed to obtain f ı g W A ! C. An important property of composition is that if the function g is continuous at a 2 A and the function f is continuous at g.a/ 2 B, then the composition f ı g is continuous at a. The proof of this result can follow the template for proofs of continuity at a point. Such a proof would introduce an > 0 and end with concluding that jf ı g.x/ f ı g.a/j D jf g.x/ f g.a/ j < . The continuity of f at g.a/ allows you to claim that jf g.x/ f g.a/ j is small if g.x/ is close to g.a/. But it is easy to ensure that g.x/ is close to g.a/ because g is continuous at a. So, you can choose a ı > 0 to make jg.x/ g.a/j as small as necessary. How small 0 is that? The continuity of f tells you how small. So, given > 0, choose a ı > 0 0 so that jy g.a/j < ı implies jf .y/ f g.a/ j < .ˇThen choose a ı ˇ> 0 so that jx aj < ı implies jg.x/ g.a/j < ı 0 . This gives ˇf g.x/ f g.a/ ˇ < . The complete proof can be written as follows. PROOF: Suppose that function g has domain A with its range contained in the set B, and that function f has domain B. If g is continuous at a 2 A and f is continuous at g.a/, then the composition f ı g is continuous at a. • Let g be a function with domain A with its range contained in the set B, and let f be a function with domain B. Assume g is continuous at a 2 A, and f is continuous at g.a/. • Let > 0 be given. • Because f is continuous at g.a/, there is a ı 0 > that if y is in the 0 such domain of f with jy g.a/j < ı 0 , then jf .y/ f g.a/ j < . • Because g is continuous at a, there is a ı > 0 such that if x 2 A with jx aj < ı, then jg.x/ g.a/j < ı 0 . 0 • If ˇ x 2 A, then jx ˇaj < ı implies jg.x/ g.a/j < ı which, in turn, implies ˇf g.x/ f g.a/ ˇ < . • This shows that f ı g is continuous at a. As an example of how useful this theorem is consider the function jxj. One can prove that this function is continuous fairly easily by following the template for proofsˇ that a function f is continuous at a point a. Indeed, such a proof must end ˇ with ˇjxj jajˇ < , but by considering all ˇthe possible ˇ cases for x and a being negative or nonnegative,ˇ it can beˇ seen that ˇjxj jajˇ jx aj, so if jx aj is made less than , then ˇjxj jajˇ < . This, in fact, shows that jxj is uniformly continuous.
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PROOF: The function jxj is uniformly continuous. Let f .x/ D jxj. Let > 0 be given, and set ı D . Let a be any real number. Note that if x and a areˇ either both ˇ greater than or equal to 0, or both less than or equal to 0, then ˇjxj jajˇˇ D jx ˇaj, but that if x and ˇ opposite ˇ a have ˇ. In either case, ˇjxjjajˇ jxaj. signs, then jxaj D jxjCjaj > ˇjxjjaj ˇ ˇ • So, if jx aj < ı, it follows that ˇjxj jajˇ jx aj < ı D . • Thus, jxj is uniformly continuous.
• • • •
As easy as this proof is, the continuity of jxj can more easily be proved as follows. PROOF: The function jxj is continuous. p • Let g.x/ D x2 , and f .x/ D x. • Let a be any real number. 2 • Then g is continuous at a, and, p since g.a/ D a 0, f is continuous at g.a/. 2 • Because the function jxj D x D .f ıg/.x/, it follows that jxj is continuous at a. In turn, this result can be used to show that if f and g are functions with domain A, and f and g are both continuous at a 2 A, then the functions min.f ; g/ and max.f ; g/ are both continuous at a. This is because the functions min.f ; g/ and max.f ; g/ can be expressed in terms of absolute value. PROOF: If f and g are functions with the same domain A, and both functions are continuous at a 2 A, then the function min.f ; g/ is continuous at a. • Let f and g be functions with the same domain A, and assume that both functions are continuous at a 2 A. • Note that for any two real numbers y and z, if y > z, then y C z jy zj D y C z .y z/ D 2z, but if y z, then y C z jy zj D y C z .z y/ D 2y. In either case, y C z jy zj D 2 min.y; z/. .x/g.x/j • Thus, for any x, min f .x/; g.x/ D f .x/Cg.x/jf . 2 • Since f and g are continuous at a, so is f g. • Since f g is continuous at a, so is jf gj. gj • It then follows that the combination min.f ; g/ D f Cgjf is continuous 2 at a.
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4.6.1 Exercises 1. Find examples of functions f and g defined on R with lim f .x/ D L and lim g.y/ D M such that lim g .f .x// ¤ M.
y!L
x!a
x!a
Write proofs for each of the following statements. 2. If g is uniformly continuous on its domain A, and f is uniformly continuous on the range of g, then f ı g is uniformly continuous on A. 3. If f and g are functions with the same domain A, and both f and g are continuous at a 2 A, then max.f ; g/ is continuous at a. 4. If f and g are functions uniformly continuous on the same domain A, then min.f ; g/ is uniformly continuous on A. 5. If f1 ; f2 ; f3 ; : : : ; fn are all uniformly continuous on the same domain A, then so is max.f1 ; f2 ; f3 ; : : : ; fn /.
4.7 Other Continuity Theorems 4.7.1 Boundedness of Continuous Functions A function that is continuous on a closed bounded interval Œa; b satisfies some important properties. In particular, there are points u and v in Œa; b such that for all x 2 Œa; b, f .u/ f .x/ f .v/. In this case, f .u/ is the minimum value of f on Œa; b and f .v/ is the maximum value of f on Œa; b. This result is often stated as “a function continuous on a closed bounded interval obtains its minimum and its maximum value.” Note that the function f .x/ D x is continuous on the open interval .1; 2/, but it does not take on a minimum or maximum value there. Clearly, f .x/ < 2 for each value of x 2 .1; 2/, and 2 is the least upper bound of all values achieved by f on that interval, but the least upper bound is never achieved. Before showing that the function f continuous on the closed bounded interval Œa; b obtains its minimum and maximum values, it is convenient to first show that such a function must be bounded, that is, there is a real number M such that for all x 2 Œa; b, jf .x/j M. This can be proved by contradiction by assuming that f is continuous on Œa; b, but that no bound exists. How do you quantify the assumption “f is not bounded”? You cannot just assume that f takes on an infinitely large value because f .x/ is a real number for each value of x 2 Œa; b and, hence, f .x/ cannot be infinitely large for any value of x. You need to construct the negation of the statement “there is an M such that for all x 2 Œa; b, jf .x/j M.” This is a statement with two quantifiers: “there is an M” and “for all x 2 Œa; b.” The two quantifies are followed by the proposition (statement) that “jf .x/j M.” The first quantifier is existential
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quantifier, and the second quantifier is a universal quantifier. Thus, the statement “there is an M such that for all x 2 Œa; b, jf .x/j M” has an existential quantifier stating that there exists a number M satisfying a property. This is followed by a universal quantifier stating that all x in the interval Œa; b satisfy a property. Finally, the property is given as “jf .x/j M.” The rule of thumb for constructing the negation of statements with quantifiers is to replace each existential quantifier with a corresponding universal quantifier, replace each universal quantifier with a corresponding existential quantifier, and replace the property with the negation of that property. In this example, the existential quantifier “there exists a number M” would be replaced by the universal quantifier “for all numbers M.” Then the universal quantifier “for all x 2 Œa; b” would be replaced by the existential quantifier “there exists an x 2 Œa; b.” Finally, the property “jf .x/j M” would be replaced by its negation “jf .x/j > M.” The resulting negation is “for all numbers M there is an x 2 Œa; b such that jf .x/j > M.” Your proof of the boundedness of f would begin by introducing f and the interval Œa; b. Then it would assume negation just discussed. The remainder of the proof would be to derive a contradiction, and that would show that the assumption made at the outset of the proof is false, so its negation, the statement you were trying to prove, must be true. Thus, the proof would begin with a statement about f being a continuous function on the closed bounded interval Œa; b which would be followed by the negation of the statement you want to prove. So how do you use this negation to reach a contradiction? Well, just see where this assumption leads you. If for each M you can find an x 2 Œa; b where jf .x/j > M, it means that there is an x1 such that jf .x1 /j > 1. Similarly, there is an x2 such that jf .x2 /j > 2. In this way, you can assert that there is a sequence x1 ; x2 ; x3 ; : : : such that for each n 1, jf .xn /j > n. Note that this gives you an infinite sequence of values in the closed bounded interval Œa; b. The Bolzano–Weierstrass Theorem states that every infinite bounded set has an accumulation point. Does the sequence x1 ; x2 ; x3 ; : : : produce such an infinite bounded set? Well, it is certainly bounded because each xn is in the interval Œa; b. Is it possible that the sequence does not give an infinite collection of points? For that to happen, it would have to be the case that infinitely many of the value in the sequence were equal to each other. Actually, just because you choose x1 so that jf .x1 /j > 1 does not preclude having jf .x1 /j > 100, so the value x1 could appear in the sequence many times. This is awkward. It would be easier if you chose a sequence of distinct values. This is actually not hard to do. Rather than choosing xn so that jf .xn /j > n, why not choose x1 as above, and for each n 1 choose xnC1 so that jf .xnC1 /j > jf .xn /j C 1. This would not only imply that for each n, jf .xn /j > n but also that xn could not equal any of the values that appear earlier in the sequence. So what can you do with the infinite sequence of xn values with its guaranteed accumulation point, y? First note that the accumulation point y is also in Œa; b because all of the xn values satisfy both a xn and xn b, so the accumulation point y must also satisfy a y b. Otherwise, there would be an interval around y
4.7 Other Continuity Theorems
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Fig. 4.7 Proving that a continuous function on Œa; b is bounded
a
x1
x3 x4
x6 x5 x2 b
that did not share any points with Œa; b, so it would not contain any of the xn values. This means that f is defined and continuous at y. That implies that there is a ı > 0 such that for all x 2 Œa; b satisfying jx yj < ı, it follows that jf .x/ f .y/j < 1. But that means jf .x/j < jf .y/j C 1. But y is an accumulation point for the sequence of xn values, so there are infinitely many of the xn within ı of y, and some of them will necessarily have the property that jf .xn /j > jf .y/j C 1. This gives the needed contradiction (Fig. 4.7). PROOF: A function continuous on a closed bounded interval is bounded. • Let a b be real numbers, and let f be a function continuous on the interval Œa; b. • Assume that f on Œa; b is not bounded. That is, for every real number M, there is an x 2 Œa; b such that jf .x/j > M. • Select x1 2 Œa; b so that jf .x1 /j > 1. • Construct a sequence inductively as follows. Assume that for some n 1 the sequence x1 ; x2 ; x3 ; : : : ; xn has been selected. Choose xnC1 2 Œa; b such that jf .xnC1 /j > jf .xn /j C 1. • Note that for each n the xn chosen in this manner must be distinct from all values of xj chosen before it in the sequence, and that jf .xn /j > n. • The terms of the sequence x1 ; x2 ; x3 ; : : : form an infinite set contained in the interval Œa; b, so it is an infinite bounded set. By the Bolzano–Weierstrass Theorem the set has an accumulation point y. • If y lies outside of Œa; b, then there is an open interval containing y that contains no points of Œa; b. Thus, that open interval would not contain any terms of the sequence which cannot happen if y is an accumulation point of the sequence. Therefore, y 2 Œa; b. (continued)
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• The function f is continuous at y, so there is a ı > 0 such that for all x 2 Œa; b with jx yj < ı, it follows that jf .x/ f .y/j < 1, and thus, jf .x/j < jf .y/j C 1. • Since y is an accumulation point of the sequence x1 ; x2 ; x3 ; : : :, there must be infinitely many terms of the sequence within ı of y. Thus, there must be an n such that n > jf .y/j C 1 and xn is within ı of y. This implies that jf .xn /j > n > jf .y/j C 1 which contradicts the fact that jf .xn / f .y/j < 1. • Therefore, the assumption that f is not bounded must be false, and the theorem is proved.
4.7.2 Obtaining Extreme Values Using the fact that continuous functions on closed bounded intervals are bounded, there is a nice trick to show that a function f continuous on the closed bounded interval Œa; b must achieve its extreme values, that is, its minimum and maximum. The fact that the set of values that f takes on is a bounded set implies that the set of values has a least upper bound, M. If f is never equal to M, then the function M f .x/ is positive for all x 2 Œa; b because M is an upper bound, and f .x/ is never equal to M. This implies that the function Mf1 .x/ is also continuous on the interval Œa; b. But then you can again apply the previous theorem to show that there is a number K such that for all x 2 Œa; b, Mf1 .x/ K. Taking reciprocals one more time shows M f .x/ K1 which implies that f .x/ M K1 . This shows that M K1 < M is an upper bound for f on Œa; b when M was assumed to be the least upper bound. This is a contradiction, and you must conclude that f .x/ D M for at least one value of x 2 Œa; b. The formal proof can be written as follows (Fig. 4.8). Fig. 4.8 The maximum and minimum of a function f .x/ on an interval
maximum y = f(x)
minimum
a
b
4.7 Other Continuity Theorems
127
PROOF (Extreme Value Theorem): A function continuous on a closed bounded interval obtains its maximum value and its minimum value at some points in the interval. • Let a b be real numbers, and let f be a function continuous on the interval Œa; b. • The set B D ff .x/ j x 2 Œa; bg is not empty because it contains f .a/, and B is bounded above because all functions continuous on a closed bounded interval are bounded. • Let M be the least upper bound of set B. • Assume that for all x 2 Œa; b, f .x/ ¤ M. • The function M f .x/ is continuous on Œa; b and is never equal to 0. Hence, M f .x/ > 0 on Œa; b. • It follows that the function Mf1 .x/ is continuous on Œa; b. • Because all functions continuous on a closed bounded interval are bounded, there is a real number K > 0 such that Mf1 .x/ K on Œa; b. • But then M f .x/ K1 on Œa; b, so f .x/ M K1 on Œa; b. • Since K > 0, the set B is bounded above by M K1 < M. This means that M K1 is an upper bound for B which contradicts the fact that M was the least upper bound of B. • Therefore, the assumption that f was never equal to M is false, and there must be a value x 2 Œa; b such that f .x/ D M. • Applying the proceeding argument to the function f , which is also continuous on Œa; b, shows that there is an x 2 Œa; b such that f .x/ is equal to the maximum of f on Œa; b. But then f .x/ is the minimum value of f on Œa; b. This completes the proof of the theorem.
4.7.3 The Intermediate Value Property Suppose the function f is defined on an interval containing c and d, and the graph of f passes through the points .c; f .c// and .d; f .d//. It might be that the graph of the function passes through every value of y between f .c/ and f .d/ as it moves between the points .c; f .c// and .d; f .d// as shown in the figure (Fig. 4.9). For example, the function f .x/ D 2x2 3 is defined for all real numbers with f .1/ D 1 and f .2/ D 5. q
lies between 1 and 2 and f .x/ D For each y between 1 and 5, the value x D yC3 2 y. Formally, a function defined on an interval Œa; b is said to have the intermediate value property on that interval if for each choice of c and d with a c d b and each y between f .c/ and f .d/, there is an x 2 Œc; d such that f .x/ D y. The Intermediate Value Theorem states that any function continuous on an interval has the intermediate value property there. If you consider the intuitive notion of continuity where you say that f is continuous on Œa; b if you can draw the graph of
128 Fig. 4.9 f passing through each y between f .c/ and f .d/
4 Continuity
f(c)
y
f(d)
c
x
d
f without lifting your pencil from the paper, then this intermediate value property becomes clear because in going from f .c/ to f .d/, your pencil will necessarily cross over all the y values between f .c/ and f .d/. To prove the Intermediate Value Theorem you would begin by setting the context by introducing a function f continuous on an interval Œa; b and points c and d with a c d b. Then you would select an arbitrary y between f .c/ and f .d/. The proof would have to demonstrate the existence of an x between c and d with f .x/ D y. How is this to be done? As with many other proofs in Analysis, one shows the existence of a real number by constructing a set for which that number is a least upper bound. Consider, for example, the case where f .c/ < y < f .d/. You could construct the set S D fx 2 Œc; d j f .x/ yg. This set is not an empty set because c 2 S, and S is certainly bounded above by d. Thus, the Completeness Axiom says that the set has a least upper bound, s. Now you can refer to the continuity of f which will show that if f .s/ < y, then there is a ı > 0 such that jx sj < ı implies that f .x/ < y showing that there are values greater than s for which f .x/ < y contradicting the fact that s is an upper bound of S. If f .s/ > y, then there is a ı > 0 such that jx sj < ı implies that f .x/ > y showing that s ı < s is an upper bound for S contradicting the fact that s is the least upper bound of S. The only remaining conclusion is that f .s/ D y which provides the needed example, x D s, needed to prove the theorem. Note that the above argument did not cover the general case where f .c/ and f .d/ can be in any order. The argument so far only covers the specific case where f .c/ < f .d/. So is there more proof to write? It is easy to see that the case f .c/ > f .d/ can be proved with an argument virtually identical to the one given above by changing the sense of some of the inequalities. The case of f .c/ D f .d/ is even easier because the only possible y between f .c/ and f .d/ is f .c/, so the value x D c gives the needed f .x/ D y. Thus, giving the argument for f .c/ < f .d/ essentially covers all the needed cases, and it would be very easy for the reader to add the needed arguments to complete the proof for the missing cases. In this situation it is common for the proof to cover only the specific condition f .c/ < f .d/ and introduce it with the phrase without loss of generality. In this case the phrase means that although the following assumption looks like it only covers some of the necessary cases, in order
4.7 Other Continuity Theorems
129
to make the argument completely general, the omitted cases are either very easy or virtually identical to the case being considered. With this in mind, the following is a proof of the Intermediate Value Theorem. PROOF (Intermediate Value Theorem): Let the function f be continuous on the interval Œa; b containing c and d. If y is any value between f .c/ and f .d/, then there exists x between c and d such that f .x/ D y. Let f be a function continuous on Œa; b, and let c and d be in Œa; b. Let y be any value between f .c/ and f .d/. Without loss of generality, assume that c d and f .c/ y f .d/. Let set S D fx 2 Œc; d j f .x/ yg. S is not empty because f .c/ y implying c 2 S. S is bounded above by d. By the Completeness Axiom S has a least upper bound s which will be an element of Œa; b. • If f .s/ < y, then by the continuity of f , there is a ı > 0 such that if x 2 Œa; b with jx sj < ı, then jf .x/ f .s/j < yf2.s/ , and, in particular, f .x/ < y. This shows that there is an x > s with f .x/ < y, so x 2 S contradicting the fact that s is an upper bound of S. • If f .s/ > y, then by the continuity of f , there is a ı > 0 such that if x 2 Œa; b with jx sj < ı, then jf .x/ f .s/j < f .s/y , and, in particular, f .x/ > y. 2 This shows that for all x between s ı and s that f .x/ > y, so s ı is an upper bound of S contradicting the fact that s is the least upper bound of S. • It follows that f .s/ must equal y which completes the proof of the theorem.
• • • • • • •
In the above proof the steps which begin “If f .x/ < y” and “If f .x/ > y” are written in exactly the same style using almost identical words. If you were writing a short story, you would avoid writing in this style because it might sound monotonous to the reader. In creative writing, you would want to be more creative, and you would reach for your thesaurus to find alternate words to enhance your writing. But in a mathematical proof, using such parallel construction of sentences actually makes the proof easier to read. A reader only needs to parse the first of the two steps in order to have a good idea of what is going to be done in the second of the two steps. This gives the reader a head start on processing the second step. What is passed off as boring in creative writing can be applauded in the writing of proofs because of the way it simplifies the understanding. In fact, one often begins the second of two such steps with the word similarly to indicate that the argument to follow looks a lot like the one just completed, again alerting the reader to the parallel construction. The Intermediate Value Theorem says that functions continuous on an interval have the intermediate value property there. But a function need not be continuous for it to have the intermediate value property. Clearly, if a function has a jump discontinuity at a point a, that is, if lim f .x/ and lim f .x/ both exist but are x!a
x!aC
different as shown in Fig. 4.10, then there could well be values of y that the function misses as it passes from .c; f .c// to .d; f .d//.
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4 Continuity
Fig. 4.10 A function with a jump discontinuity
f(c)
y
f(d)
c
Fig. 4.11 Graph of sin
d
1 x
For a discontinuous function to have the intermediate value property, the function must necessarily oscillate wildly (Fig. 4.11). A typical example is the function sin 1x if x > 0 f .x/ D . 0 if x 0
4.7.4 Exercises Write proofs for each of the following statements. Each statement can be proved using one or more of the theorems in this section. 1. Let A R be a bounded set, and let f be a function defined on A. If f is unbounded on A, then for every > 0, there exists a and b in R with b a < such that f is unbounded on A \ .a; b/. 2. If a < b and f is a continuous function on Œa; b with f .a/ D f .b/, then there is a c 2 .a; b/ such that f obtains an extreme value (either a minimum or maximum) at c. 3. Suppose that f is a continuous function defined on R such that lim f .x/ D x!1
lim f .x/ D 1. Then f obtains its minimum value for some x 2 R.
x!1
4.8 Discontinuity
131
4. If p is an odd degree polynomial with real coefficients, then p has at least one real root. 5. Suppose that a plane contains be a polygon G and a line L. Then there is a line L0 in the plane parallel to L such that exactly half the area of G lies on each side of L0 . r 1 2 6. There is a value of x between 0 and 1 such that x equals . 1 C x2
4.8 Discontinuity In Calculus students learn about a great many continuous functions. These include the elementary functions: polynomials, rational functions, algebraic functions, exponential functions, logarithmic functions, and circular and hyperbolic trigonometric functions and their inverses. How badly can a function be discontinuous? A function can 8 be discontinuous 9 at a single point such as the signum or sign function < 1 if x < 0 = sgn.x/ D 0 if x D 0 or at a sequence of points such as the floor or greatest : ; 1 if x > 0 integer function bxc D n if n is the integer satisfying n x < n C 1 (Fig. 4.12). A function ( can be discontinuous at a sequence of points ) that converge such as with 1 1 1 if < x ; for positive integer n nC1 n f .x/ D n . This function is discontin0 otherwise uous at each x D 1n for positive integers n, but it is continuous everywhere else including at x D 0 (Fig. 4.13). A function can be discontinuous at every x such as 0 if x is rational with f .x/ D . 1 if x is irrational But one of the most surprising examples is the following often called Thomae’s function but also known as the popcorn function, the raindrop function,
Fig. 4.12 Graphs of sgn.x/ and bxc
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4 Continuity
Fig. 4.13 Graphs of functions with discontinuities Fig. 4.14 Graph of Thomae’s function
or the modified Dirichlet function. It is defined on the interval .0; 1/ by 1 m if x is rational written in lowest terms as n n . Its graph is shown in f .x/ D 0 if x is irrational Fig. 4.14. It is not hard to see that this function is discontinuous at each rational number mn 2 .0; 1/. Indeed if mn is in lowest terms, then f . mn / D 1n . If is set 1 for every ı > be irrational numbers x 2 .0; 1/ satisfying ˇ 0 there will ˇat 2n ,mthen ˇ ˇ ˇx ˇ < ı for which ˇf .x/ f m ˇ D j0 1 j > . On the other hand, at each n n n irrational number a in .0; 1/, the function is continuous. To see this, given an > 0, notice that there are only finitely many rational numbers r 2 .0; 1/ such that f .r/ . If there are such rational numbers, there is one, r0 , closest to a, so choose ı D jr0 aj. If there are no such rational numbers, you can choose ı D 1. In either case, for all x 2 .0; 1/ with jx aj < ı, it follows that jf .x/ f .a/j < , showing that f is continuous at a.
Chapter 5
Derivatives
5.1 The Definition of Derivative Anybody who was even half paying attention in their first course in Calculus got the strong impression that the differentiation of functions has an enormous number of applications. Not only does it provide a great tool for understanding the behavior of functions, but it also has applications to a very wide range of other fields, most notably Physics, Engineering, Chemistry, Biology, and Economics. In particular, being able to use the derivative to determine where a function is increasing and decreasing in itself justifies this reputation. Merely knowing the average rate of change of a function over an interval is valuable. But the limit concept allows you to refine this idea to get the instantaneous rate of change of the function at a point. This allows for more precise information about the function as well as providing what is often a simpler expression than that of the average rate of change from which it is derived. This chapter will discuss the theorems needed to calculate derivatives efficiently as well as theorems highlighting some of the important properties and applications of the derivative. Let f be a function defined on an open interval containing the point a. Then for values of x near but not equal to a one can calculate the slope of the secant line passing through the two points on the graph of the function a; f .a/ and x; f .x/ . As shown in Fig. 5.1, the slope of this secant .a/ line is given by the difference quotient f .x/f . If f is continuous, as x xa approaches a, the point x; f .x/ approaches the point a; f .a/ , and the secant line may approach a tangent line, the line that passes through a; f .a/ and most closely approximates the graph of the function near a (Fig. 5.2). The derivative of f at a is the slope of this tangent line. More formally, if a is an accumulation point of the domain of the function f , and f is defined at a, then .a/ the derivative of f at a is f 0 .a/ D lim f .x/f . The derivative is said to exist if this xa x!a
limit exists. When the limit exists, f is said to be differentiable at a. Equivalently, .a/ the limit can be written f 0 .a/ D lim f .aCh/f . h h!0
© Springer International Publishing Switzerland 2016 J.M. Kane, Writing Proofs in Analysis, DOI 10.1007/978-3-319-30967-5_5
133
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5 Derivatives
Fig. 5.1 Slope of a Secant Line
(a, f(a))
f(x) - f(a)
x -a
(x, f(x))
Fig. 5.2 Tangent Line (a, f(a))
5.2 Differentiation and Continuity The first important consequence of the definition of the derivative is that if a function f has a derivative at a point, then f is also continuous at that point. As part of the definition of derivative, f needs to be defined at the point a for it to have a derivative .a/ at a. It remains to show that lim f .x/ D f .a/ whenever the limit lim f .x/f exists. xa x!a x!a For this difference quotient to have a finite limit when the denominator is clearly approaching 0, the numerator must also be approaching 0. This last statement is intuitively true, so you would hope that it has an easy justification. Consider what .a/ sort of algebraic operations you could apply to the difference quotient f .x/f in xa order to produce the numerator f .x/ f .a/. It should be clear that if the difference quotient is multiplied by x a, the product will be the desired difference f .x/ f .a/. This suggests the method that works in the following simple proof.
5.3 Calculating Derivatives
135
PROOF: If the function f has a derivative at a point a, then f is continuous at a. • Suppose that f has a derivative at the point a. • It follows from the definition of derivative that f is defined at a, and that a is an accumulation point of the domain of f . .a/ • Also from the definition of derivative it follows that f 0 .a/ D lim f .x/f xa x!a exists. h i .a/ .a/ • Then lim f .x/ f .a/ D lim .x a/ f .x/f D lim x a lim f .x/f D xa xa x!a
x!a
x!a
x!a
0 f 0 .a/ D 0. • Thus, f .x/ is both defined at x D a, and lim f .x/ f .a/ D 0, or lim f .x/ D f .a/. x!a • It follows that f is continuous at x D a.
x!a
So, f differentiable at a implies that f is continuous at a. Is the converse true? You should know several counterexamples that show that the converse is false, that is, there are functions f continuous at a point a that are not differentiable at a. First of all, f can be continuous at a where a is an isolated point of the domain of f , and at such points, the derivative of f is not defined. But even if f is continuous for all real numbers, f need not have a derivative at a particular a. The best known example is the absolute value function, f .x/ D jxj, which is continuous for all real numbers but .0/ fails to have a derivative at x D 0. This is because the difference quotient f .x/f is x0 equal to 1 for all x > 0 and 1 for all x < 0, so the limit of the difference quotient does not exist at x D 0. Of course, the absolute value function has a derivative at all x ¤ 0. There is a well-known example known as the Weierstrauss function that is continuous for real numbers x but does not have a derivative at any point.
5.3 Calculating Derivatives The proof that a function f has a particular derivative at a point a is just a proof about the limit of a difference quotient, and as such, is no different than a proof of any other limit. On the other hand, there are some similarities among the proofs of derivatives, so it is worth working through a few examples. The key observation is that whenever you need to calculate a derivative directly from the definition, you must calculate the limit of a difference quotient which, by design, is a fraction whose numerator and denominator are both approaching zero. In such a case, one would expect to be able to perform some algebraic manipulation that would result in the x a expression in the denominator canceling with an equivalent factor in the numerator. This allows you to use other limit theorems to complete the evaluation.
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5 Derivatives
For example, consider the function f .x/ D 3x2 8x. To calculate the derivative of f at a D 4, one needs to evaluate the limit
2
3x 8x 3 42 8 4 f .x/ f .4/ 3x2 8x 16 lim D lim D lim x!4 x!4 x!4 x4 x4 x4 .3x C 4/.x 4/ D lim D lim 3x C 4 D 16: x!4 x!4 x4 Since each step of this derivation follows either from rules of algebra or from the theorems about calculating the limits of various arithmetic combinations of functions, the calculation given is a complete proof that the derivate of f at x D 4 is 16. In a more general setting, consider proving that the derivative of f .x/ D 5x4 at the point x D a is f 0 .a/ D 20a3 . Here you would calculate 5.x a/ x3 C x2 a C xa2 C a3 f .x/ f .a/ 5x4 5a4 lim D lim D lim x!a x!a x!a xa xa xa D lim 5.x3 C x2 a C xa2 C a3 / D 5.a3 C a2 a C aa2 C a3 / D 20a3 : x!a
Again, finding a factor of x a in the numerator of the difference quotient is the key to evaluating the needed limit.
5.4 The Arithmetic of Derivatives One quickly learns in Calculus that although the derivative is defined as a limit of a difference quotient, there is a small collection of algorithms that reduce the finding of the derivative of any combination of elementary functions to a fairly mechanical exercise. The algorithms show you how to take the derivatives of the sum, difference, product, and quotient of two differentiable functions as well as a constant multiple of a differentiable function, the inverse of a differentiable function, and the composition of two differentiable functions. Those rules along with the knowledge of how to differentiate the elementary functions, xn , ax , loga x, sin x, and cos x give you all the tools necessary to differentiate virtually any function you are likely to see in a lifetime of applications. This and the next sections discusses the proofs of the theorems that provide these needed algorithms. The simplest of these results is the theorem that states that if f is a function differentiable at a and c is any constant, then the function cf is also differentiable at a with .cf /0 .a/ D cf 0 .a/. In the proof of this theorem, you would assume that f 0 .a/ .a/ D f 0 .a/. Since the limit needed exists. That provides for you the limit lim f .x/f xa x!a
to show that .cf /0 .a/ D cf 0 .a/ is just a multiple of a known limit, the needed result follows immediately from the fact that the limit of a constant times a function is the constant times the limit of the function.
5.4 The Arithmetic of Derivatives
137
PROOF: If f 0 .a/ exists, and c is a constant, then .cf /0 .a/ D cf 0 .a/. • Suppose that f has a derivative at the point a. .a/ • From the definition of derivative f 0 .a/ D lim f .x/f . xa cf .x/cf .a/ xa x!a 0
• Then .cf /0 .a/ D lim
• Thus, .cf /0 .a/ D cf .a/.
x!a
.a/ D lim c f .x/f D c lim xa x!a
x!a
f .x/f .a/ xa
D cf 0 .a/.
To show that the derivative of the sum or difference of two differentiable functions is the sum or difference of their derivatives, one is faced with finding the limit of a difference quotient which can easily be written as the sum or difference of two difference quotients whose limits are already known. Thus, if f and g are two functions defined on the same domain and both differentiable at a, calculating the derivative of f C g at a requires the limit f .x/ f .a/ C g.x/ g.a/ .f C g/.x/ .f C g/.a/ D lim lim x!a x!a xa xa f .x/ f .a/ g.x/ g.a/ C lim D f 0 .a/ C g0 .a/ D lim x!a x!a xa xa as needed. PROOF: Let f and g be functions defined on a common domain, and let f and g both be differentiable at a. Then .f C g/0 .a/ D f 0 .a/ C g0 .a/ and .f g/0 .a/ D f 0 .a/ g0 .a/. • Suppose that f and g are functions defined on a common domain, and that f and g are both differentiable at a. .a/ • From the definition of derivative f 0 .a/ D lim f .x/f and xa x!a
g0 .a/ D lim
g.x/g.a/ . xa x!a
•
f .x/f .a/ C g.x/g.a/ Cg/.a/ Then .f C g/ .a/ D lim .f Cg/.x/.f D lim xa xa x!a x!a .a/ g.x/g.a/ 0 0 lim f .x/f C lim D f .a/ C g .a/. xa xa x!a x!a 0 0 0 0
D
• Thus, .f C g/ .a/ D f .a/ C g .a/. • Because .f g/.x/ D f .x/ g.x/ D f .x/ C .1/g.x/, and the derivative the derivative of g.x/, it follows that .f g/0 .a/ D of .1/g.x/is 1 times 0 f C .1/g .a/ D f .a/ C .1/g0 .a/ D f 0 .a/ g0 .a/ completing the proof of the theorem. Why does the first step in this proof make the assumption that f and g are defined on the same domain? This is to avoid the embarrassing situation that the intersection of the domains of f and g isolates the point a. For example, if f is defined for all x 1 and g is defined for all x 1, it could be that both f 0 .1/ and g0 .1/ are defined, but the function f Cg is defined only at 1, so its derivative cannot be defined. Another
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5 Derivatives
example would be for f to p be defined at all rational numbers, and g to be defined at all rational multiples of 2. Each function could be differentiable at each point of its domain, but f C g is only defined at 0, so its derivative cannot be defined. It is certainly worth noting here that the theorems discussed so far show that for functions f and g and constants a and b, the derivative of the linear combination of functions af .x/Cbg.x/ is the linear combination of the derivatives af 0 .x/Cbg0 .x/. In the words of Linear Algebra, this says that the derivative is a linear operator. This fact alone has a long list of ramifications in Differential Equations and other fields. It is important for the beginning Calculus student to learn that even though the derivative behaves in an intuitive way with respect to addition and subtraction, that this intuition ceases when discussing the derivative of a product or quotient. The proof that .fg/0 D fg0 C f 0 g involves one trick reminiscent of the proof that the limit of a product is the product of the limits. That is, one adds and subtracts the same quantity so that rather than making a change in two different factors at the same time, one makes a change in one factor at a time. Indeed, the difference quotient you obtain for the function fg is f .x/g.x/ f .a/g.a/ f .x/g.x/ f .x/g.a/ C f .x/g.a/ f .a/g.a/ D xa xa g.x/ g.a/ f .x/ f .a/ D f .x/ C g.a/: xa xa Taking the limits at each step produces the following proof. PROOF (Product Rule): Let f and g be functions defined on a common domain, and let f and g both be differentiable at a. Then .fg/0 .a/ D f .a/g0 .a/ C f 0 .a/g.a/. • Suppose that f and g are functions defined on a common domain, and that f and g are both differentiable at a. .a/ • From the definition of derivative f 0 .a/ D lim f .x/f and xa x!a
g0 .a/ D lim g.x/g.a/ . xa x!a • Because f is differentiable at a, it is continuous at a. This implies that lim f .x/ D f .a/. x!a
• Then .fg/0 .a/ D lim
.fg/.x/.fg/.a/
xa x!a f .x/g.x/f .x/g.a/Cf .x/g.a/f .a/g.a/ lim xa x!a
lim f .x/
x!a
D lim
x!a
f .x/g.x/f .a/g.a/ xa
D lim f .x/ x!a
.a/ lim g.x/g.a/ C lim f .x/f xa xa x!a x!a 0 0 0
D
g.x/g.a/ xa
C
f .x/f .a/ xa
g.a/ D
lim g.a/ D f .a/g0 .a/ C f 0 .a/g.a/. x!a
• Thus, .fg/ .a/ D f .a/g .a/ C f .a/g.a/. 0 The proof that gf D assumption that g.a/ ¤ 0.
gf 0 fg0 g2
involves the same strategy along with the extra
5.4 The Arithmetic of Derivatives
139
PROOF (Quotient Rule): Let f and g be functions defined on a common domain, and let f and g both be differentiable at a. If g.a/ ¤ 0, then 0 0 0 .a/ f .a/ D g.a/f .a/f.a/g . 2 g g.a/
• Suppose that f and g are functions defined on a common domain, and that f and g are both differentiable at a with g.x/ ¤ 0. .a/ • From the definition of derivative f 0 .a/ D lim f .x/f and xa x!a
. g0 .a/ D lim g.x/g.a/ xa x!a • Because g is differentiable at a, it is continuous at a. This implies that lim g.x/ D g.a/. x!a f f 0 f .x/ f .a/ .x/ .a/ g.a/ g g f g.x/ • Then D lim D .a/ D lim x!a x!a g xa xa f .x/ f .a/ f .a/ f .a/ g.x/ C g.x/ g.a/ g.x/ lim D x!a xa ! 1 1 g.a/ 1 f .x/ f .a/ g.x/ C f .a/ lim D x!a xa g.x/ xa 1 f .a/ g.a/ g.x/ f .x/ f .a/ C D lim x!a xa g.x/ g.x/g.a/ xa lim
x!a
f .x/ f .a/ 1 f .a/ g.a/ g.x/ lim C lim lim D x!a g.x/ x!a g.x/g.a/ x!a xa xa
f .a/ 1 f 0 .a/g.a/ f .a/g0 .a/ 0 . 2 g .a/ D 2 g.a/ g.a/ g.a/ 0 f f 0 .a/g.a/ f .a/g0 .a/ • Thus, .a/ D . 2 g g.a/ f 0 .a/
5.4.1 Exercises Write proofs for each of the following statements. For any constant c, the function f .x/ D c has derivative f 0 .x/ D 0. The function f .x/ D x has derivative f 0 .x/ D 1. For any positive integer n, the function f .x/ D xn has derivative f 0 .x/ D nxn1 . Any polynomial function f .x/ D an xn Can1 xn1 Can2 xn2 C Ca1 xCa0 has derivative f 0 .x/ D nan xn1 C .n 1/an1 xn2 C .n 2/an2 xn3 C C a1 . n 5. For any positive integer n, the function f .x/ D x1n has derivative f 0 .x/ D xnC1 .
1. 2. 3. 4.
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5 Derivatives
6. Given the collection of functions f1 ; f2 ; f3 ; ; fn each defined on the same domain and constants c1 ; c2 ; c3 ; ; cn , the each differentiable at a, and given 0 derivative c1 f1 C c2 f2 C c3 f3 C C cn fn .a/ D c1 f10 .a/ C c2 f20 .a/ C c3 f30 .a/ C C cn fn0 .a/. 7. Given the collection of functions f1 ; f2 ; f3 ; ; fn each defined on the same 0 domain and each differentiable at a, the derivative f1 f2 f3 fn .a/ D f10 .a/f2 .a/f3 .a/ fn .a/ C f1 .a/f20 .a/f3 .a/ fn .a/ C f1 .a/f2 .a/f30 .a/ fn .a/ C C f1 .a/f2 .a/f3 .a/ fn0 .a/.
5.5 Chain Rule and Inverse Functions The Chain Rule shows how to differentiate a function that is a composition of other functions. Since composition is an invaluable tool for constructing functions, the Chain Rule deserves its place among the important algorithms for calculating derivatives. It states that if g is a function differentiable at a, and if f is a function defined on the range of g and differentiable at g.a/, then the function .f ı g/.x/ is differentiable at a, and .f ı g/0 .a/ D f 0 g.a/ g0 .a/. To prove this, f g.x/ f g.a/
. you would need to find the limit of the difference quotient D D xa 0 which has limit g .a/, you might try both Expecting to see the expression g.x/g.a/ xa multiplying and dividing the difference quotient D by the factor g.x/ g.a/ to get f g.x/ f g.a/ g.x/g.a/
g.x/g.a/ . xa
This idea leads to the following almost correct proof.
PROOF ATTEMPT: Let g be a function differentiable at a, and f be a function defined on the range of g and differentiable at g.a/. Then .f ı g/0 .a/ D f 0 g.a/ g0 .a/. • Let g be a function differentiable at a, and f be a function defined on the range of g and differentiable at g.a/. • From the definition of derivative, g0 .a/ D lim g.x/g.a/ and xa x!a f .y/f g.a/ f 0 .g.a// D lim . yg.a/ y!g.a/
• Because g is differentiable at a, it is continuous at a. This implies that lim g.x/ D g.a/. x!a f g.x/ f g.a/ .f ıg/.x/.f ıg/.a/ 0 • Then .f ı g/ .a/ D lim D lim D xa xa x!a x!a f g.x/ f g.a/ f g.x/ f g.a/ lim g.x/g.a/ g.x/g.a/ D lim g.x/g.a/ lim g.x/g.a/ . xa xa x!a x!a x!a f .y/f g.a/ f g.x/ f g.a/ • Therefore, .f ıg/0 .a/ D lim g.x/g.a/ lim g.x/g.a/ D lim xa yg.a/ x!a
g.x/g.a/ xa x!a
lim
D f 0 .g.a// g0 .a/.
x!a
y!g.a/
5.5 Chain Rule and Inverse Functions
141
This proof attempt does include the intuitive reasoning behind why the Chain Rule works, but the proof is not correct. Can you spot the error? The problem is that even though g.x/ is approaching g.a/ as x approaches a, there is no guarantee where g.x/ is different from g.a/. In fact, it is quite easy to construct functions g.x/ which are differentiable at a for which g.x/ is equal to g.a/ for infinitely many values of x as x approaches a. The simplest example is when g is a constant function. A more 2 x sin 1x if x ¤ 0 complicated example is g.x/ D which has a derivative of 0 if x D 0 1 0 at x D 0 and is equal to g.0/ D 0 at n for all nonzero integers n. Clearly, when g.x/ D g.a/, one cannot both multiply and divide the difference quotient by g.x/ g.a/ and expect to get anything except nonsense. This problem does not present anenormous hurdle because, in the cases where g.x/ D g.a/, the difference f g.x/ f g.a/ xa
is itself equal to 0. A typical way around the problem is to 8 9 f .y/f g.a/ ˆ < yg.a/ if y ¤ g.a/ > = introduce the function h.y/ D . This function has the ˆ > : 0 ; f g.a/ if y D g.a/ nice property that it is equal to the desired difference quotient when g.x/ differs from g.a/, and it is continuous at g.a/. Introducing this function into the proof gets around the technical difficulties of the previously attempted proof. quotient
PROOF (Chain Rule): Let g be a function differentiable at a, and f be a function defined on the range of g and differentiable at g.a/. Then .f ı g/0 .a/ D f 0 g.a/ g0 .a/. • Let g be a function differentiable at a, and f be a function defined on the range of g and differentiable at g.a/. • From the definition of derivative, g0 .a/ D lim g.x/g.a/ and f 0 .g.a// D xa x!a f .y/f g.a/ lim . yg.a/ y!g.a/
• Because g is differentiable at a, it is continuous at a. This implies that lim g.x/ D g.a/. x!a 9 8 f .y/f g.a/ ˆ = < yg.a/ if y ¤ g.a/ > , and note that h is continuous • Define h.y/ D > ˆ ; : 0 f g.a/ if y D g.a/ at g.a/. f g.x/ .f ıg/.a/
ıg.a// D lim • Then .f ı g/0 .a/ D lim .f ıg/.x/.f xa xa x!a x!a g.x/g.a/ g.x/g.a/ D lim h g.x/ lim xa D lim h g.x/ xa x!a x!a x!a 0 0 g.a/ g D f .a/. h lim g.x/ lim g.x/g.a/ xa x!a x!a 0 0 0 • Therefore, .f ı g/ .a/ D f g.a/ g .a/.
D
142
5 Derivatives
Recall that a function f W A ! B is called a bijection if it is both surjective and injective; that is, for each y 2 B there is one and only one x 2 A such that f .x/ D y. In this case f is a one-to-one correspondence between the points of A and the points of B. When f is a bijection, one can define the inverse of f to be the function f 1 W B ! A by letting f 1 .y/ be the unique value of x such that f .x/ D y. In other words f 1 is the set of ordered this pairs f.y; x/ j .x; y/ 2 f g. From definition it is clear that for all x 2 A, f 1 f .x/ D x, and for all y 2 B, f f 1 .y/ D y. This says that f 1 ı f is the identity function on A, and f ı f 1 is the identity function on B. Note that if f W A ! B is not a bijection, then one cannot define f 1 as a mapping from B to A. If f is not surjective, then there is a y 2 B which is not in the range of f , so there is no way to define f 1 .y/. If f is not injective, then there is a y 2 B such that f .x/ D y for more than one value of x, and there may be no natural way to select which x should be f 1 .y/. For example, the function f .x/ D x2 maps the real numbers into the real numbers. The function is neither surjective (its range is the nonnegative real numbers) nor injective since f .2/ D f .2/. One can restrict the codomain to the nonnegative real numbers. Then f becomes a surjective function, but it is still not injective. To get an inverse to f you can substitute a different function for f which restricts the domain of f to the nonnegative real numbers. If f is thought of as a function from the nonnegative real numbers to pthe nonnegative real numbers, then f is a bijection, and it has the inverse function x. The same procedure is done to obtain inverses of the trigonometric functions. For example, the function f .x/ D sin x maps the real numbers to the interval Œ1; 1. The function is surjective, but it is not injective. an injective function, the
To obtain domain of sin x is restricted to the interval 2 ; 2 . On this interval sin x is both injective and surjective and has the inverse sin1 y, sometimes written as arcsin y (Fig. 5.3). Now suppose that f is bijective and has inverse function f 1 . If f has a nonzero derivative at the point a, the Chain Rule can be used to find the derivative of f 1 at f .a/. Indeed, one has that .f ı f 1 /.x/ D x, so the Chain Rule implies that 1 0 1 1 0 . Is this conclusion valid? That f f .x/ .f / .x/ D 1, or .f 1 /0 .x/ D 0 1 f
f
.x/
is, can you justify taking the derivative of .f ı f 1 / using the Chain Rule before you know that the derivative of f 1 exists? The answer is yes, the use of the Chain Rule the limit of is justified here. The proof of the Chain Rule includes taking which is broken into the product of the limit of h g.x/ and the h g.x/ g.x/g.a/ xa
Fig. 5.3 Restricting sin.x/ to get sin1 .x/
5.6 Increasing Functions, Decreasing Functions, and Critical Points
143
In its application to the equation .f ı f 1 /.x/ D x one can rewrite xa the difference quotient as g.x/g.a/ D xa . Now there is no a priori assumption xa
limit of
g.x/g.a/ . xa
h g.x/
that the limit of
g.x/g.a/ xa
xa
exists; its limit is just the limit of the quotient xa which h g.x/
exists as the quotient of limits. As an application of differentiating an inverse function, consider finding the p derivative of n x for integer values of n ¤ 0. It is known that, for integer values of n, the derivative of f .x/ D xn is f 0 .x/ D nxn1 . For n ¤ 0 and x 0, the inverse of p 1 1 the function of f is f 1 .x/ D n x, so its derivative must be 0 1 D p . n. n x/n1 f f .x/
5.6 Increasing Functions, Decreasing Functions, and Critical Points Perhaps the most important property of the derivative is its ability to determine where a function is increasing or decreasing. Let f be a function defined on an interval I. If for all x and y in I, x < y implies that f .x/ f .y/, then f is said to be increasing on I, and if x < y implies that f .x/ < f .y/, then f is said to be strictly increasing on I. Similarly, if x < y implies that f .x/ f .y/, then f is said to be decreasing on I, and if x < y implies that f .x/ > f .y/, then f is said to be strictly decreasing on I. So what can be said if it is known that function f has a positive derivative at a? .a/ What is known is that the difference quotient f .x/f has a positive limit, so it is xa positive when x is close to a. How close to a does x have to be? What the limit .a/ definition gives you is that for any > 0, you can find a ı > 0 so that f .x/f is xa within of its limit, f 0 .a/, which is positive. So, if > 0 is chosen to be f 0 .a/, then the difference quotient which has to be within f 0 .a/ of f 0 .a/ will have to be positive. .a/ Thus, for x within ı of a (and not equal to a), the difference quotient f .x/f is xa positive. Then if x > a, it follows that f .x/ > f .a/, and if x < a, it follows that f .x/ < f .a/. Does this mean that f is increasing? The answer is no. There are functions with a positive derivative at a which are not increasing over any open interval containing a. An example of such a function is given in the last section of this chapter. All one can say is the following.
144
5 Derivatives
PROOF: Let f be a function with a positive derivative at a. Then there is a ı > 0 such that for all x in the domain of f with jx aj < ı, if x > a, then f .x/ > f .a/, and if x < a, then f .x/ < f .a/. Similarly, if f has a negative derivative at a, there is a ı > 0 such that for all x in the domain of f with jx aj < ı, if x > a, then f .x/ < f .a/, and if x < a, then f .x/ > f .a/. • Let f be a function with a positive derivative at a. .a/ • Then lim f .x/f D f 0 .a/ > 0. xa x!a
• From the definition of limit, there ˇis a ı > 0 such that ˇ if x is in the domain ˇ ˇ .a/ 0 of f and 0 < jx aj < ı, then ˇ f .x/f f .a/ ˇ < f 0 .a/ implying that xa f .x/f .a/ xa
• • • •
> 0. .a/ > 0, so f .x/f .a/ > If x satisfies a < x < aCı, then xa > 0 and f .x/f xa 0, and f .x/ > f .a/. .a/ If x satisfies a > x > aı, then xa < 0 and f .x/f > 0, so f .x/f .a/ < xa 0, and f .x/ < f .a/. This proves the first part of the theorem. If instead f 0 .a/ < 0, apply the above argument to the function f to obtain the analogous result.
A function f defined at the point a is said to have a relative maximum (sometimes called a local maximum) at a if there is a ı > 0 such that for all x in the domain of f satisfying jx aj < ı the value of f .a/ f .x/. Similarly, one can define relative minimum (or local minimum) where, in this case, f .a/ f .x/. If f has a relative maximum or relative minimum at a, one can say that it has a relative extremum (sometimes called a local extremum) at a. Another very important property of the derivative is its ability to identify points where a function has relative extrema. This ability follows immediately from the previous theorem. PROOF: Let f be a function defined on an open interval containing a, and let f be differentiable at a. Then if f has a relative extremum at a, the value of f 0 .a/ is 0. • Let f be a function defined on an open interval I containing the point a. • Assume that f is differentiable at a and has a relative maximum at a. • If f 0 .a/ is positive, there is a ı > 0 such that for all x 2 I with a < x < aCı, f .x/ > f .a/. This contradicts the fact that f has a relative maximum at a. • If f 0 .a/ is negative, there is a ı > 0 such that for all x 2 I with aı < x < a, f .x/ > f .a/. This contradicts the fact that f has a relative maximum at a. • Thus, f 0 .a/ must be 0. • Applying this argument to the function f shows that if f has a relative minimum at a, it must be that f 0 .a/ D 0.
5.6 Increasing Functions, Decreasing Functions, and Critical Points
145
f(x)
a
b
c
d
e
g
h
Fig. 5.4 This graph of f .x/ on the interval Œa; h shows relative maxima at b, d, and g, relative minima at a; c; e; and h; an absolute maximum at b, and an absolute minimum at h. The derivative f 0 .x/ does not exist at x D d
Any student of Calculus will see applications of this result where one is asked to identify relative extrema for a particular function, and applications to what are fondly called Max/Min problems where one is first asked to construct an appropriate function to fit the application and then find a particular extremum of that function. One defines a critical point of f to be a value a where either f 0 .a/ D 0 or f 0 .a/ does not exist. Not all of these points will end up being relative extrema for some may just be a saddle point of f where f 0 .a/ D 0, but f has no relative extrema at that point. For example, the function f .x/ D x3 has a saddle point at x D 0 where f 0 is 0, but f is a strictly increasing function over the entire real line. A function is said to have an absolute maximum (sometimes called a global maximum) at a if f is defined at a, and for all other x in the domain of f , f .x/ f .a/. The term absolute minimum (sometimes called a global minimum) is defined in the analogous way with f .x/ f .a/, and an absolute extremum (sometimes called a global extremum) is either an absolute maximum or absolute minimum. The theorem about relative extrema shows that if f is defined on any interval I, then the only places f can have relative extrema or absolute extrema are critical points or at endpoints of I. You should be able to identify example functions where each of these criteria give extrema (Fig. 5.4).
5.6.1 Exercises Identify the relative extrema and absolute extrema of the given functions on the given intervals. 1. 2. 3. 4.
f .x/ D x3 8x on the interval Œ3; 2 f .x/ D 3x C 3x on the interval Œ1; 1/ f .x/ D jx2 16j on the interval Œ5; 6 p 2 f .x/ D 3 2 3 x on the interval Œ2; 2
146
5 Derivatives
5.7 The Mean Value Theorem The Mean Value Theorem is one of the better known results about derivatives, and for good reason. It is invoked frequently when one needs to estimate the maximum possible change between the values of a function at two different points. This can be a valuable tool when finding approximations to functions or when it is necessary to know how much variation is exhibited by a particular function. The theorem states that the average rate of change of a function between two points a and b .a/ given by f .b/f is equal to the value of the derivative f 0 .c/ for some c between a ba and b. This allows you to use information about the derivative to make statements about the change f .b/ f .a/. The theorem is usually proved in two steps by first proving Rolle’s Theorem which is a simpler version of the Mean Value Theorem. Rolle’s Theorem states that if a < b, and if f is a function continuous on the interval Œa; b, differentiable on the interval .a; b/, and satisfying f .a/ D f .b/, then there is a c 2 .a; b/ for which f 0 .c/ D 0. What tools do you have to prove this result? Your proof needs to conclude that f 0 .c/ D 0. Think through what you know about derivatives, and see if any of the results conclude that the derivative is equal to 0. The only results that come to mind are the result that the derivative of any constant function is 0, and the result that if f reaches a relative extremum at a point where the function is differentiable, then its derivative at that point must be 0. It is unlikely that the first of these two results will be of much help except in the very special case where f is a constant function. So how can you use the result about extreme values to show that there is a place where the function has a derivative of 0? What you do know is that f is continuous on a closed interval Œa; b, and the Extreme Value Theorem states that such a function obtains its maximum and minimum values on this interval. You also know that these extreme values can only occur at places where the derivative is 0, where the derivative does not exist, or at the endpoints of the interval. OK, there are no places on .a; b/ where the derivative does not exist, but could both the maximum and minimum occur at endpoints of the interval? The hypothesis of Rolle’s Theorem says that f .a/ D f .b/, so the only way that the two endpoints can be both maximum and minimum values of f on the interval is for f to be constant on the interval. In the case of a constant function, the theorem is clearly true. In any other case, it could be that f .a/ and f .b/ are maximum values for f or minimum values for f , but they cannot be both. If f is not constant, its maximum and minimum values must be different. That guarantees that f must have either an absolute maximum or an absolute minimum (possibly both) between a and b. That gives the result (Fig. 5.5).
5.7 The Mean Value Theorem
147
f(x)
a
c
b
Fig. 5.5 The proof of Rolle’s Theorem finds an extreme point c between a and b for which f 0 .c/ D 0
PROOF (Rolle’s Theorem): For a < b, let f be a function continuous on the interval Œa; b and differentiable on the interval .a; b/ satisfying f .a/ D f .b/. Then there is a c 2 .a; b/ for which f 0 .c/ D 0. • For a < b, let f be a function continuous on the interval Œa; b and differentiable on the interval .a; b/ satisfying f .a/ D f .b/. • Because f is continuous on the closed bounded interval Œa; b, it obtains a maximum and a minimum value there. • If both the maximum and minimum values of f occur at endpoints of the interval, then, since f .a/ D f .b/, the maximum and minimum values of f are equal, and f is constant on the interval Œa; b. • In this case, f 0 .c/ D 0 for each c 2 .a; b/, and the conclusion of the theorem holds. • If the maximum and minimum values of f do not both occur at endpoints of the interval, then there must be a c 2 .a; b/ such that f reaches a maximum or a minimum value at c. • In this case, f 0 .c/ D 0, and the conclusion of the theorem holds. • In either case, the conclusion of the theorem holds which completes the proof. Rolle’s Theorem takes care of the case where f .a/ D f .b/. To prove the Mean Value Theorem in the more general case where f .a/ need not equal f .b/, you would want to reduce this general case to the previously proved case where f .a/ and f .b/ are equal. An easy way to do this is to subtract a linear function from f to get a new function h which does satisfy the hypothesis of Rolle’s Theorem. This linear function can be any linear function that takes on a value at b which differs by xa f .b/ f .a/ from the value it takes on at a. One such function is f .b/ f .a/ ba because it takes on the value 0 at a and f .b/ f .a/ at b (Fig. 5.6).
148 Fig. 5.6 Point c between a and b where the tangent line is parallel to the secant line from a to b
5 Derivatives f(x)
a
c
b
PROOF (Mean Value Theorem): For a < b, let f be a function continuous on the interval Œa; b and differentiable on the interval .a; b/. Then there .a/ is a c 2 .a; b/ for which f 0 .c/ D f .b/f . ba • For a < b, let f be a function continuous on the interval Œa; b and differentiable on the interval .a; b/.xa • Let h.x/ D f .x/ f .b/ f .a/ ba . xa • Since both f .x/ and f .b/ f .a/ ba are continuous on Œa; b and differentiable .a; b/, h is also continuous on Œa; b and differentiable on .a; b/. • h.b/ D f .b/ f .b/ f .a/ ba D f .b/ f .b/ f .a/ D f .a/ D h.a/. ba • Thus, h satisfies the hypothesis of Rolle’s Theorem, so there is a c 2 .a; b/ such that h0 .c/ D 0. 1 .a/ • Then 0 D h0 .c/ D f 0 .c/ f .b/ f .a/ ba , so f 0 .c/ D f .b/f . ba f .b/f .a/ 0 • Therefore, there is a c 2 .a; b/ with f .c/ D ba which completes the proof. The following are two instructive applications of the Mean Value Theorem. First, if you know that a function f is differentiable on an interval, and its derivative is nonnegative on that interval, then the function must be increasing on the interval. To show that a function is increasing, you need to show that if x and y are in the interval with x < y, then f .x/ f .y/. This would follow from knowing that if y x 0, .x/ 0. What the Mean Value Theorem gives you then the difference quotient f .y/f yx is that this difference quotient is equal to the derivative of f at some point c between x and y. So, if you know that the derivative on the interval is always nonnegative, then the difference quotient must be nonnegative as needed.
5.7 The Mean Value Theorem
149
PROOF: Let f be a function whose derivative is nonnegative at every point of an interval. Then f is an increasing function on that interval. • Let f be a function whose derivative is nonnegative at each point of the interval I. • Let x and y be in I with x < y. • Then by the Mean Value Theorem, there is a c between x and y such that f .y/f .x/ D f 0 .c/. yx • Since I is an interval and x and y are in I, c is also in I, implying that f 0 .c/ 0. .x/ .x/ • Thus, f .y/f 0 so .y x/ f .y/f D f .y/ f .x/ 0. yx yx • Therefore, f is increasing on I. Clearly, if f 0 is strictly positive on an interval, then you can prove that f is strictly increasing on the interval. This can be done by altering the above proof by changing the greater than or equal signs to greater than signs where needed. Is the converse of the above theorem true? Well, one cannot conclude that a function is differentiable on an interval by just knowing that the function is increasing there. But what if you are given a differentiable function that is increasing? What can you conclude about the derivative? If a function is increasing, it does mean that every difference .x/ quotient f .y/f will be greater than or equal to 0, and, thus, the derivative which yx is the limit of such difference quotients will have to be greater than or equal to 0. If f is strictly increasing, can you conclude that its derivative is positive? In this case you cannot. You can conclude that all difference quotients will be positive, but the limit of positive difference quotients can be 0. For example, f .x/ D x3 is a function differentiable on the entire real line, and it is strictly increasing, but its derivative is 0 at x D 0. Another important consequence of the Mean Value Theorem is that if a function has a derivative equal to 0 at every point of an interval, then f is constant on that interval. Again, this follows directly from what you can say about any difference quotient. PROOF: Let f be a function whose derivative is 0 at every point of an interval. Then f is constant on that interval. • Let f be a function whose derivative is 0 at each point of the interval I. • Let x and y be in I with x < y. • Then by the Mean Value Theorem, there is a c between x and y such that f .y/f .x/ D f 0 .c/. yx • Since I is an interval and x and y are in I, c is also in I, implying that f 0 .c/ D 0. .x/ .x/ • Thus, f .y/f D 0 so .y x/ f .y/f D f .y/ f .x/ D 0. yx yx • Therefore, f .x/ D f .y/ for all x and y in the interval, and, thus, f is constant.
150
5 Derivatives
How important is it that the set where f 0 is 0 is an interval? The fact that the 0 if x < 0 set is an interval is crucial. For example, the function f .x/ D 1 if x > 0 is not defined at 0. The derivative, f 0 , is equal to 0 at each point of the domain of f , but clearly, f is not a constant function, although it is constant on each interval contained in its domain. Looking back to the previous theorem, note that the function f .x/ D 1x has a strictly positive derivative at each point of its domain, but, again, its domain does not include 0. This function is strictly increasing on each interval contained in its domain, but it is not an increasing function because f .1/ > f .1/.
5.7.1 Exercises Write proofs for each of the following statements. 1. If f is a function whose derivative is negative for all points in an interval, then f is a decreasing function on the interval. 2. If f and g are functions differentiable on an interval with f 0 .x/ D g0 .x/ for each x in the interval, then there is a constant C such that f .x/ D g.x/ C C for all x in the interval. 3. If f .0/ D g.0/ and f 0 .x/ g0 .x/ for each x 0, then f .x/ g.x/ for each x > 0.
5.8 L’Hopital’s Rule It seems like most students who take Calculus remember L’Hopital’s Rule. Even those who do not remember what the rule states seem to remember its name. Perhaps this is because it is so much fun to pronounce, but more students remember L’Hopital’s Rule than some far more important results such as the Fundamental Theorem of Calculus. L’Hopital’s Rule states that if f and g are differentiable functions defined on an interval containing the point a with lim f .x/ D lim g.x/ D x!a
0
x!a
f .x/ 0, then lim gf 0.x/ D L implies that lim g.x/ D L. This is very useful because the x!a .x/ x!a theorem stating that the limit of a quotient is the quotient of the limits does not apply in cases when the denominator has a limit of 0. How would you prove L’Hopital’s Rule? You might try to prove it by using the Mean Value Theorem because the quotient you are considering is
f .x/ f .a/ f .x/ D D g.x/ g.x/ g.a/
f .x/f .a/ xa g.x/g.a/ xa
:
5.8 L’Hopital’s Rule
151
This is not exactly correct because, as far as you know, f .a/ and g.a/ might not even be defined, and if they are, they need not be equal to lim f .x/ and lim g.x/. x!a x!a This is not a big stumbling block, because you can always redefine f and g at a to be equal to 0 without changing the result of the theorem. You also would need to know that g.x/ ¤ g.a/ for x near a so that the needed quotient can be calculated. Once the quotient of f and g is rewritten as the quotient of the difference quotient of f and the difference quotient of g, you can apply the Mean Value Theorem to replace the difference quotients with derivatives, and then take the limit. It might look something like the following. PROOF ATTEMPT: Let both f and g be functions differentiable for all x ¤ a in an interval which contains a. Assume that lim f .x/ D x!a f 0 .x/ 0 x!a g .x/
lim g.x/ D 0, and g0 .x/ ¤ 0 for all x ¤ a in the interval. Then lim
x!a
implies
lim f .x/ x!a g.x/
DL
D L.
• Let f and g be functions differentiable for all x ¤ a in an interval which contains a. • Assume that lim f .x/ D lim g.x/ D 0, and g0 .x/ ¤ 0 for all x in the interval x!a
with x ¤ a. • Assume that lim
f 0 .x/ 0 x!a g .x/
x!a
D L.
• Without loss of generality, it can be assumed that f .a/ D g.a/ D 0 because redefining the functions at a does not change the limits at a of f , g, f 0 , g0 , or their ratios. With f .a/ D g.a/ D 0, both f and g are continuous at x D a. • For x in the given interval with x ¤ a, both f and g are continuous on the closed interval with endpoints at x and a, and both f and g are differentiable on the open interval with these endpoints. • Thus, by the Mean Value Theorem, there is a cf .x/ between x and a such .a/ that f 0 cf .x/ D f .x/f , and there is a cg .x/ between x and a such that xa g.x/g.a/ g0 cg .x/ D xa . • Because cf .x/ and cg .x/ are both between x and a, lim cf .x/D lim cg .x/Da. x!a
• Then because f .a/ D g.a/ D 0, 0 f 0 cf .x/ . lim 0 D lim gf 0.x/ .x/ x!a g cg .x/
lim f .x/ x!a g.x/
D
lim f .x/f .a/ x!a g.x/g.a/
x!a
D lim
x!a
f .x/f .a/ xa g.x/g.a/ xa
D
x!a
• This completes the proof. There is a significant problem with this proof. The problem stems from the fact that, although both functions cf .x/ and cg .x/ do approach a as x approaches a, the two functions can approach a at different rates. Why is this a problem? Consider 2 calculating lim xx , a limit which is clearly equal to 0. But what if cf .x/ D x x!0
and cg .x/ D x2 ? Even though it is true that cf .x/ and cg .x/ both approach 0 as
152
5 Derivatives cf .x/2 x!0 cg .x/
x approaches 0, the limit lim
D 1. Just knowing that cf .x/ and cg .x/ are
approaching a does not allow you to use both of these expressions in place of x f .x/ x!0 g.x/
when taking the limit. What the proof attempt does show is that lim
D
lim f 0 .x/
x!0
lim g0 .x/
,
x!0
a result that is not as useful as L’Hopital’s Rule. A second less crucial problem with this proof attempt is that it defines cf .x/ to .a/ be the value of c such that f .x/f D f 0 .c/. But this condition may well be satisfied xa by more than one value of c, so there is a problem with which of the possible values of c is chosen. One can get around this difficulty, but that still does not address the previously stated problem. A common way to correct the problem in the proof attempt is to use a more powerful version of the Mean Value Theorem known as the Extended Mean Value Theorem or sometimes as the Cauchy Mean Value Theorem. It allows you to select 0 f .x/f .a/ one value of c so that gf 0.c/ D g.x/g.a/ , that is, it allows you to select the ratio of .c/ derivatives equal the ratio of the difference quotients at a single value of c rather than selecting one value of c for the numerator and a possibly different value of c for the denominator. One can prove the Extended Mean Value Theorem by 0 f .x/f .a/ manipulating the desired relation gf 0.c/ D g.x/g.a/ . This equation can be rewritten as .c/ f 0 .c/Œg.x/ g.a/ D g0 .c/Œf .x/ f .a/ and then f 0 .c/Œg.x/ g.a/ g0 .c/Œf .x/ f .a/ D 0. This may be confusing because there are three variables involved, x, a, and c, but you can make better sense of it by thinking of x and a as being fixed. That is, if you define the function h.t/ D f .t/Œg.x/ g.a/ g.t/Œf .x/ f .a/, then h0 .c/ D f 0 .c/Œg.x/ g.a/ g0 .c/Œf .x/ f .a/ as needed. How do you know that there is a c such that h0 .c/ D 0? That follows from Rolle’s Theorem because it is easy to verify that h.x/ D h.a/. PROOF (Extended Mean Value Theorem): For a < b, let both f and g be functions continuous on Œa; b and differentiable on .a; b/. Then there is a c 2 .a; b/ such that f 0 .c/Œg.b/ g.a/ D g0 .c/Œf .b/ f .a/. • Let a < b and assume f and g are functions continuous on Œa; b and differentiable on .a; b/. • For x 2 Œa; b define h.x/ D f .x/Œg.b/ g.a/ g.x/Œf .b/ f .a/. • Then h is also continuous on Œa; b and differentiable on .a; b/. • Note that h.a/ D f .a/Œg.b/g.a/g.a/Œf .b/f .a/ D f .a/g.b/g.a/f .b/, and h.b/ D f .b/Œg.b/ g.a/ g.b/Œf .b/ f .a/ D f .a/g.b/ g.a/f .b/ D h.a/. • Thus, h satisfies the hypothesis of Rolle’s Theorem on the interval Œa; b. • It follows that there is a c 2 .a; b/ such that h0 .c/ D 0, so f 0 .c/Œg.b/ g.a/ g0 .c/Œf .b/ f .a/ D 0. • This is equivalent to f 0 .c/Œg.b/ g.a/ D g0 .c/Œf .b/ f .a/ which is the conclusion of the theorem.
5.8 L’Hopital’s Rule
153
Now the Extended Mean Value Theorem can be used to give a correct proof of L’Hopital’s Rule. PROOF (L’Hopital’s Rule, Part 1): Let f and g be functions differentiable for all x ¤ a in an open interval which contains a. Assume that lim f .x/ D x!a f 0 .x/ 0 x!a g .x/
lim g.x/ D 0, and g0 .x/ ¤ 0 for all x ¤ a in the interval. Then lim
x!a
implies
lim f .x/ x!a g.x/
DL
D L.
• Let f and g be functions differentiable for all x ¤ a in an open interval which contains a. • Assume that lim f .x/ D lim g.x/ D 0, and g0 .x/ ¤ 0 for all x in the interval x!a
x!a
with x ¤ a. • Assume that lim
f 0 .x/ 0 x!a g .x/
D L.
• Without loss of generality, it can be assumed that f .a/ D g.a/ D 0 because redefining the functions at a does not change the limits at a of f , g, f 0 , g0 , or their ratios. With f .a/ D g.a/ D 0, both f and g are continuous at x D a. • Let > 0 be given. 0 • Since lim gf 0.x/ D L, there is a ı > 0 such that 0 < jx aj < ı implies that .x/ f 0 .x/ g0 .x/
• •
• •
• •
x!a
is within of L. Fix x in the given interval with 0 < jx aj < ı. Since f and g are continuous on the closed interval from a to x and differentiable on the open interval from a to x, f and g satisfy the hypothesis of the Extended Mean Value Theorem on the interval from a to x. It follows that there is a c between x and a such that f 0 .c/Œg.x/ g.a/ D g0 .c/Œf .x/ f .a/. By assumption g0 is not 0, so g0 .c/ ¤ 0. Also the Mean Value Theorem shows that g.x/ g.a/ D g0 .t/.x a/ for some t between x and a, and this shows g.x/ g.a/ ¤ 0. 0 f .x/f .a/ It follows that g.x/g.a/ D gf 0.c/ . .c/ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ f .x/f .a/ ˇ ˇ 0 ˇ ˇ f .x/ L Thus, ˇ g.x/ Lˇ D ˇ g.x/g.a/ Lˇ D ˇ gf 0.c/ ˇ < . .c/ f .x/ x!a g.x/
• This implies that lim
D L which proves the theorem.
L’Hopital’s Rule also holds in cases where lim g.x/ is infinite rather than x!a zero.
154
5 Derivatives
PROOF (L’Hopital’s Rule, Part 2): Let f and g be functions differentiable in an open interval which contains a. Assume that lim g.x/ is either x!a
positive or negative infinity, and g0 .x/ ¤ 0 for all x in the interval with 0 f .x/ x ¤ a. Then lim gf 0 .x/ D L implies lim g.x/ D L. .x/ x!a
x!a
• Let f and g be a function differentiable in an open interval which contains a. • Assume that lim g.x/ is positive or negative infinity, g0 .x/ ¤ 0 for all x in x!a
0
the interval with x ¤ a, and lim gf 0.x/ D L. x!a .x/ • Let > 0 be given. • Because g0 .x/ is never 0, the Mean Value Theorem shows that if both x and y are in the interval and are both on the same side of a, then g.x/ ¤ g.y/. 0 0 • Since lim gf 0.x/ D L, there is a ı 0 such that 0 < jx aj < ı 0 implies that gf 0.x/ .x/ .x/ x!a
• •
• • • •
•
is within 2 of L. Fix x in the given interval with 0 < jx aj < ı 0 . Since f and g are differentiable between x and a and continuous at x, for any y between x and a it follows from the Extended Mean Value Theorem 0 f .y/f .x/ f .y/f .x/ that there is a c between x and y such that g.y/g.x/ D gf 0.c/ . Thus, g.y/g.x/ is .c/ within 2 of L. f .y/ f .x/ g.y/ f .y/ f .x/ g.y/ D . Note that g.y/ g.x/ 1 g.x/ g.y/ ˇ f .y/ f .x/ ˇ ˇ ˇ ˇ g.y/ g.y/ ˇ f .y/f .x/ f 0 .c/ Because g.y/g.x/ D g0 .c/ is within 2 of L, it follows that ˇ L ˇ< . ˇ 1 g.x/ ˇ 2 g.y/ ˇ ˇ ˇ ˇ ˇ ˇ ˇ f .y/ f .x/ ˇ < 2 ˇ1 g.x/ Then ˇ g.y/ g.y/ L 1 g.x/ g.y/ ˇ g.y/ ˇ. Because g.y/ approaches positive or negative infinity as y approaches a, there is a ı > 0 with ı < ı 0 such that for all y with 0 < jy aj < ı, the fraction jf .x/jCjLg.x/jCjg.x/j < 2 . jg.y/j ˇ ˇ ˇ ˇ f .y/ f .x/ < g.y/ L 1 g.x/ Then for y with 0 < jy aj < ı, ˇ g.y/ g.y/ ˇ ˇ ˇ g.x/ ˇ ˇ ˇ1 g.y/ ˇ implies ˇ2 ˇ ˇ ˇ ˇ f .y/ ˇ ˇ f .x/ g.x/ ˇ jg.x/j jf .x/jCjLg.x/jCjg.x/j < 2 C 2 D . ˇ g.y/ Lˇ < ˇ g.y/ L g.y/ ˇ C 2 C 2jg.y/j < 2 C jg.y/j
• This completes the proof. There are several variations of L’Hopital’s Rule covering the cases of one sided limits and limits at positive or negative infinity. These are covered in the following exercises.
5.9 Intermediate Value Property and Limits of Derivatives
155
5.8.1 Exercises Use L’Hopital’s Rule to calculate the following limits. sin2 .2x3 / 6 x!0 px xx lim pxCx x!0C p
1. lim 2.
3. lim
x!0C
4. lim
x!1
x ln x
ln x p x
5. lim .sin.2x//x x!0C
tan1 x 1 x!0 tan .3x/
6. lim
Write proofs of each of the following statements. 7. If f and g are differentiable functions for all x > 0, lim f .x/ D lim g.x/ D 0, f 0 .x/ 0 g x!1 .x/
and g0 .x/ > 0 for all x > 0, then lim
x!1
f .x/ x!1 g.x/
D L implies lim
x!1
D L.
8. If f and g are differentiable functions for all x > 0, lim g.x/ D 1, and g0 .x/ > 0 for all x > 0, then
0 lim f 0.x/ x!1 g .x/
D L implies
lim f .x/ x!1 g.x/
x!1
D L.
9. If f and g are functions differentiable for all x > a, lim f .x/ D lim g.x/ D 0, and g0 .x/ ¤ 0 for all x with x > a, then lim
x!aC
f 0 .x/ g0 .x/
x!aC
x!aC f .x/ g.x/ x!aC
D L implies lim
D L.
5.9 Intermediate Value Property and Limits of Derivatives The Intermediate Value Theorem says that if a function is continuous on an interval, then it has the intermediate value property on that interval. That is, if f is continuous on the interval I, and a; b 2 I, then for any K between f .a/ and f .b/, there is a c between a and b with f .c/ D K. Suppose that f is differentiable at each point of an interval I. If f 0 is continuous on I, then certainly it obeys the Intermediate Value Theorem and has the intermediate value property on I. But f 0 .x/ can exist 0 for all x 2 I without f being a continuous function. One example is f .x/ D 1 2 x sin x2 if x ¤ 0 . This function is differentiable for all x. When x ¤ 0, the 0 if x D 0 1 2 1 0 derivative is f .x/ D 2x sin 2 cos 2 , and f 0 .0/ D 0. As x approaches x x x 0, f 0 is not even bounded and, in fact, oscillates wildly. In spite of its discontinuity at 0, f 0 does have the intermediate value property. For example, for any x ¤ 0, the function f 0 obtains every value between f 0 .x/ and f 0 .0/ D 0 on the interval between
156
Fig. 5.7 x2 sin
5 Derivatives
1 x2
and its derivative
0 and x. Moreover, it obtains each of those values infinitely often. In fact, between 0 and x, the function f 0 takes on every real number infinitely often (Fig. 5.7). Note that the function f .x/ C x has a derivative of 1 at x D 0. This is an example of a function with a positive derivative at 0 which is not an increasing function over any open interval containing 0. This can easily be seen by the fact that in every open interval containing 0 there are intervals where the derivative of f .x/ C x is negative. So, how can you prove that if a function f has a derivative f 0 on an interval I, that f 0 has the intermediate value property on I? The hypothesis suggests that you start by taking a function f differentiable on an interval I and values a; b 2 I. Then you select a value K between f 0 .a/ and f 0 .b/. Without loss of generality, you can assume that a < b and f 0 .a/ < K < f 0 .b/. The goal would be to show that there is a c between a and b such that f 0 .c/ D K. One simplification is to replace f with the function g.x/ D f .x/ Kx. This function is also differentiable on I, and if f 0 .c/ D K, then g0 .c/ D 0. Which theorems about derivatives allow you to conclude that a derivative is 0 at some point in an interval? First there is a theorem that states that if a differentiable function reaches an extreme value at a point in an interval, then the point is either a critical point of the function or an endpoint of the interval. A second theorem is Rolle’s Theorem which talks about a differentiable function which takes on the same value at the endpoints a and b. Since you do not have any information about the values of g at the endpoints of the interval, the theorem about extreme values may be the more promising choice for this proof. What is known about the function g? You know that g is differentiable at each point of the interval from a to b. Additionally, g0 .a/ D f 0 .a/ K < 0 and g0 .b/ D f 0 .b/ K > 0. Does this mean that the function g is decreasing at a and increasing at b? Well, it would if you knew that g0 were continuous because then g0 would be negative in an interval around a and positive in an interval around b. But, as you now know, g0 need not be continuous. On the other hand, there is a theorem that says that if g0 .a/ is negative, then there is a ı > 0 such that if x satisfies a < x < a C ı, then g.x/ < g.a/. This does not show much, but you can use it to conclude that g does not take on its minimum value on Œa; b at a. A similar argument uses the fact that g0 .b/ > 0 to show that g does not take on its minimum value on Œa; b at b.
5.9 Intermediate Value Property and Limits of Derivatives
157
Is g a continuous function on Œa; b? It is differentiable at each point of Œa; b, so it is continuous. All continuous functions on a close bounded interval take on both their minimum and maximum values on the interval. Thus, you know that g takes on its minimum value on Œa; b at some point c strictly between a and b. Such a point must be a critical point of g, so g0 .c/ D 0. This is the idea behind the following proof. PROOF: A function differentiable on an interval has the intermediate value property on that interval. • • • • • • • • • • • • •
Let f be a function differentiable at each point of an interval I. Let a; b 2 I, and assume that f 0 .a/ ¤ f 0 .b/. Without loss of generality assume that a < b and f 0 .a/ < f 0 .b/. Let K be a value satisfying f 0 .a/ < K < f 0 .b/. Let g.x/ D f .x/ Kx. Then g0 .x/ D f 0 .x/ K for all x 2 Œa; b, and g0 .a/ < 0 < g0 .b/. Since g is differentiable at each point of Œa; b, it is continuous on Œa; b. Since g is continuous on Œa; b, it obtains a minimum value at some point c 2 Œa; b. g0 .a/ < 0 implies that there is a ıa > 0 such that g.x/ < g.a/ for all x satisfying a < x < a C ıa . In particular, g does not obtain its minimum at a. g0 .b/ > 0 implies that there is a ıb > 0 such that g.x/ < g.b/ for all x satisfying b ıb < x < b. In particular, g does not obtain its minimum at b. It follows that g obtains its minimum on Œa; b at a point c strictly between a and b. Since c is not an endpoint of Œa; b, g0 .c/ D 0. Thus, f 0 .c/ D g0 .c/CK D K which shows that f 0 has the intermediate value property on I.
There are simple examples of functions that have discontinuous derivatives that do not have the intermediate value property; functions such as f .x/ D jxj. This function’s derivative is the constant 1 for all x > 0 and 1 for all x < 0. This derivative is not continuous at x D 0 because it is not defined there. Clearly, f 0 does not have the intermediate value property on any interval containing both positive and negative numbers, but then f does not satisfy the hypothesis of the previous theorem on any such interval because f 0 .0/ is not defined. Functions that have discontinuous derivatives that are defined at all points will have to exhibit wild oscillations of those discontinuities similar to the example in the neighborhoods 2 x sin x12 if x ¤ 0 . 0 if x D 0 Suppose f is a function whose derivative is defined at all points of an interval except perhaps at some point c in the interval. What can be said if lim f 0 .x/ exists? x!c Such a derivative does not exhibit wild oscillations near c, and, in fact, it must have a continuous derivative at c. The proof is a consequence of the Mean Value Theorem.
158
5 Derivatives
PROOF: Let f be a function differentiable at all points of an interval .a; b/ except perhaps at some point c 2 .a; b/. Suppose that the limit lim f 0 .x/ x!c exists. Then f 0 is continuous at c. • Let f be a function differentiable on the interval .a; b/ except perhaps at c 2 .a; b/. • Assume lim f 0 .x/ D L. x!c
• From the definition of limit, given > 0, there is a ı > 0 such that if y 2 .a; b/ with 0 < jy cj < ı, then jf 0 .y/ Lj < . • Let x 2 .a; b/ with 0 < jx cj < ı. • By the Mean Value Theorem there is a y between x and c such that f .x/f .c/ D f 0 .y/. xc ˇ ˇ ˇ ˇ .c/ • Then y 2 .a; b/ with 0 < jy cj < ı, so ˇ f .x/f Lˇ D jf 0 .y/ Lj < . xc f .x/f .c/ D L, so f 0 .c/ D L. xc f 0 .c/ D lim f 0 .x/, it follows that f 0 .x/ x!c
• Thus, lim
x!c
• Because • This completes the proof.
is continuous at c.
Chapter 6
Riemann Integrals
6.1 Area The first application one usually sees of the Riemann Integral is that of finding the area of a region in the plane bounded by the graph of a function and the lines x D a, x D b, and the x-axis. Thus, before discussing integration, it makes sense to review what is meant by the area of a region in the plane. Clearly, the measure of area should be a way to assign a size to a region in a way that is compatible with the well-established rules from Geometry for assigning areas to regions such as rectangles, triangles, and circles. But there is a need to go beyond these simple regions so that area can be calculated for far more complicated regions. For example, consider the region in the coordinate plane f.x; y/ j 0 x 1; 0 y 1; at least one of x or y is rationalg. Regions such as these are not typically considered in a Geometry course, but being able to calculate areas for such sets is important in the more general discussion of integration. This chapter, therefore, begins by considering two different measures of the sizes of sets which will aid the understanding of integration.
6.2 Cardinality of Sets What does the set fA; B; C; D; Eg have in common with the set f2; 4; 6; 8; 10g? One thing they have in common is that the two sets have the same number of elements. What does the set of positive integers have in common with the set of positive multiples of 2? These sets are both infinite sets, and the second set is clearly a proper subset of the first, but, here again, the two sets have the same number of elements. To see this consider the function f .n/ D 2n which is a bijection from the set of positive integers one-to-one and onto the set of positive multiples of 2. This function provides a one-to-one matching of the elements of one set
© Springer International Publishing Switzerland 2016 J.M. Kane, Writing Proofs in Analysis, DOI 10.1007/978-3-319-30967-5_6
159
160
6 Riemann Integrals
to the elements of the other set. One says that two sets A and B have the same cardinality if there is a bijection f W A ! B. The bijection demonstrates a oneto-one correspondence between the elements of set A and the elements of set B, so one concludes that A and B are the same size. Some sets are finite, meaning that the set is either empty (has cardinality 0) or, for some positive integer n, is in one-toone correspondence with the set f1; 2; 3; ; ng. A set is called denumerable if it can be put in one-to-one correspondence with the set of positive integers. Thus, the set of positive multiples of 2 is denumerable. So is the set of all integers since the positive can be mapped onto the set of all integers using the bijection integers x if x is even 2 f .x/ D . The verification that this map is a bijection is left if x is odd 1 xC1 2 as an exercises. It shows that the integers and the positive integers have the same cardinality. Sets that are either finite or denumerable are called countable because they can be “counted out” by listing a first, second, third, and so forth. Thus, a good way to think about a countable set is a set whose elements can be written down in a finite or infinite sequence x1 ; x2 ; x3 ; because this listing shows the one-to-one correspondence between the set and the natural numbers or one of its finite subsets. The union of two countable sets is also countable. This can be seen by representing one set by the sequence x1 ; x2 ; x3 ; and the other by y1 ; y2 ; y3 ; . Then the elements of the union of the two sets can be written as x1 ; y1 ; x2 ; y2 ; x3 ; y3 ; . If there are elements that belong to both sets, then one can just leave the second copies of those elements out of the listing. Clearly, this can be extended to the union of any finite collection of countable sets, so the union of a finite number of countable sets is countable. What might seem surprising is that the union of a countable number of countable sets is still countable. That is, if A1 ; A2 ; A3 ; is 1
a sequence of countable sets, then the union [ Ak is also countable. To see this, kD1
suppose that the elements in each Ak can be listed in a sequence ak;1 ; ak;2 ; ak;3 ; . 1
One can now list all the elements of [ Ak by listing the ak;j elements in increasing
kD1 a1;1 ; a2;1 ; a1;2 ; a3;1 ; a2;2 ; a1;3 ; a4;1 ; a3;2 ; a2;3 ; a1;4 ; a5;1 ; .
order of k C j resulting in As above, duplicate elements occurring because they belong to more than one set can be left out of this listing. Figure 6.1 shows the order that the elements enter the list. Note that this result can be used to show that the set of rational numbers is countable. Indeed, the rational numbers can be written as the union R1 [R2 [R3 [ where Rk are the rational numbers that can be written as a fraction with an integer in the numerator and the positive integer k in the denominator. For example, R2 D f 02 ; 12 ; 12 ; 22 ; 22 ; 32 ; 32 ; g. Thus, the rational numbers is a countable union of countable sets showing that it is countable. The cardinality of a denumerable set is often written using the symbol @0 (read “Aleph knot” or “Aleph null”). The symbol represents the size of the natural numbers and the size of any set that can be placed in one-to-one correspondence with the natural numbers. A set which is not a countable set is called uncountable. There is a standard argument that shows that the set of real numbers in the interval .0; 1/ is not a countable set. The method, known as a diagonalization argument, first assumes
6.2 Cardinality of Sets Fig. 6.1 The union of countably many countable sets is countable
Fig. 6.2 Determining y using a diagonalization argument
161
a1,1
a1,2
a1,3
a1,4
a1,5
a1,6
a2,1
a2,2
a2,3
a2,4
a2,5
a2,6
a3,1
a3,2
a3,3
a3,4
a3,5
a3,6
a4,1
a4,2
a4,3
a4,4
a4,5
a4,6
a5,1
a5,2
a5,3
a5,4
a5,5
a5,6
x1 = 0. 4 9 0 3 2 5 5 9 0 9 9 0 x2 = 0. 1 7 7 3 8 8 0 0 0 0 0 0 x3 = 0. 7 4 1 1 8 9 1 8 2 5 4 4 x4 = 0. 1 1 8 8 8 3 7 2 9 0 0 1 x5 = 0. 5 5 2 7 7 7 1 0 6 4 2 3 x6 = 0. 0 0 0 0 0 2 1 0 9 3 7 3 x7 = 0. 8 2 1 7 4 9 0 3 2 8 5 5 x8 = … y = 0. 7 3 7 7 3 7 7 7 7 7 7 3
that the real numbers between 0 and 1 can all be written down in a sequence x1 ; x2 ; x3 ; x4 ; . Then one constructs a real number y between 0 and 1 where the kth digit to the right of the decimal point in y is chosen as follows. If the kth digit to the right of the decimal point of xk is 7, then let the kth digit to the right of the decimal point in y be 3. Otherwise, if the kth digit to the right of the decimal point of xk is not 7, then let the kth digit to the right of the decimal point in y be 7. Figure 6.2 illustrates the process of determining y. The point of this construction is that the number y is a real number in the interval .0; 1/, but it cannot be one of the numbers in the sequence x1 ; x2 ; x3 ; x4 ; . This is because for each k, y cannot equal xk because y and xk differ in their kth digits. This is a contradiction to the assumption that the sequence contained all of the real numbers in .0; 1/ and shows that it is impossible to list all the elements of .0; 1/ in a sequence. Thus, this interval is an uncountable set. If there is a bijection from the set .0; 1/ to a set B, then it follows that B will also be uncountable. You may wonder whether all uncountable sets have the same cardinality. They do not, but that fact will not be needed for the proofs discussed in this book. Refer to a standard text in Set Theory for a far more in-depth look at the cardinality of sets.
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6 Riemann Integrals
6.2.1 Exercises 1. Determine whether each of the following sets is finite, denumberable, or uncountable. (a) the set of points in the coordinate plane where both x and y coordinates are rational numbers (b) the set of points in the coordinate plane where at least one of its x and y coordinates is a rational number (c) the set of polynomials p.x/ with integer coefficients (d) the set of real numbers whose decimal representation does not contain the digit 5 (e) the set of functions f W f0; 1; 2; 3; 4; 5g ! f1; 2; 3; : : : ; 100g (f) the set of functions f W f2; 4; 6; 8; 10; : : : g ! f0; 1g x if x is even 2 2. Show that the function is a bijection from the set of if x is odd 1 xC1 2 natural numbers to the set of integers. 3. Show that .0; 1/ and .1; 5/ have the same cardinality. 4. Show that .0; 1/ and the entire set of real numbers, R, have the same cardinality. 5. Show that .0; 1/ and the interval Œ0; 1 have the same cardinality. (Hint: Find a way to “bury” the endpoints of Œ0; 1 inside of .0; 1/ by mapping a sequence x1 ; x2 ; x3 ; : : : to x3 ; x4 ; x5 ; : : : .) 1
6. Show that if A1 ; A2 ; A3 ; : : : are sets and [ An is an uncountable set, then for at nD1
least one n, the set An must be uncountable. This can be thought of as an infinite form of the Pigeonhole Principle. 7. The interval Œ0; 1 on the real line and the unit square in the plane have the same cardinality. (Hint: for a point in Œ0; 1 split up its decimal digits between the x and y coordinates of a point in the unit square.) 8. Show that the equality of cardinality is an equivalence relation. That is, if A, B, and C are any sets, then • A has the same cardinality as A. • If A has the same cardinality as B, then B has the same cardinality as A. • If A has the same cardinality as B, and B has the same cardinality as C, then A has the same cardinality as C. 9. Suppose that you apply the diagonalization argument to the set of rational numbers in the interval .0; 1/. That is, suppose you list all of the rational numbers in a sequence x1 ; x2 ; x3 ; : : : and use the diagonalization argument to construct a number y in .0; 1/ that differs from each element of the sequence. Why is this not a proof that the rational numbers are uncountable?
6.3 Measure Zero
163
6.3 Measure Zero Cardinality is used to compare the sizes of sets by considering how many elements the sets have. But two sets such as Œ0; 3 and Œ0; 6 can have the same cardinality and yet be quite different in what we traditionally think of as size in the geometric sense. So there is a need to develop a different way to compare the sizes of sets that embodies the notion of the length of a set of real numbers and of the area of a set in the plane. A general theory of measure is not a topic that can be covered in a book at this level, but it is helpful to introduce how one determines which sets should be assigned a length or an area equal to 0. If measure is to mean anything useful, you would want each finite interval Œa; b to have measure equal to its length, ba. How about the measure of the open interval .a; b/? Likely, you would say that its measure should also be b a. This suggests that the set of endpoints fa; bg should be assigned a measure of 0. More generally, a set S R is said to have measure zero if for each > 0 there is a sequence of open intervals .a1 ; b1 /; .a2 ; b2 /; .a3 ; b3 /; : : : such that S is contained in the union of 1
the intervals S [ .aj ; bj / and the total length of the intervals is less than , that jD1
is, for every natural number n,
n P
.bj aj / < . In other words, a set has measure
jD1
zero if you can cover it with a sequence of intervals whose total length is as small as you want. In particular, any finite set consisting of n real numbers has measure zero because for any > 0, each point x in the set can be covered by the interval .x 3n ; x C 3n /, 2 and the total length of these intervals is 3 . Similarly, any countable set of real numbers fx1 ; x2 ; x3 ; : : : g can be covered by intervals .xj 32 j ; xj C 32j /, and the total 1 P 2 D 23 . Thus, the set of rational numbers, which length of these intervals is 32j jD1
is countable, has measure zero. A similar argument shows that if A1 ; A2 ; A3 ; : : : is 1
a sequence of sets all of which have measure zero, then the union [ Aj also has jD1
measure zero. Indeed, given > 0, for each j you can cover Aj with a sequence of open intervals whose total length is less than 2j . Then the sequences of open 1
intervals can be combined into one sequence of intervals which cover [ Aj and has total length less than
1 P jD1
jD1
2j
D .
Since any countable set of real numbers has measure zero, if a set does not have measure zero, it must be an uncountable set. A natural question is whether an uncountable set of real numbers can have measure zero. The answer to this question is yes. The most famous example of this is known as the Cantor set which is constructed as follows. The construction begins with the closed unit interval C0 D Œ0; 1. At the first stage, the open interval of length 13 is removed from the middle of this set leaving two intervals each with length 13 so that C1 D Œ0; 13 [Œ 23 ; 1.
164
6 Riemann Integrals Stage 0 Stage 1 Stage 2 Stage 3 Stage 4 Stage 5
Fig. 6.3 Construction of the Cantor set
At the second stage, open intervals of length 19 are removed from the middle of each of the two remaining intervals leaving four intervals each with length 19 so that C2 D Œ0; 19 [ Œ 29 ; 39 [ Œ 69 ; 79 [ Œ 89 ; 99 . This process is repeated so that at stage n, open intervals of length 31n are removed from each of 2n1 closed intervals of length 1 leaving 2n closed intervals each with length 31n (Fig. 6.3). The Cantor set C 3n1 1
is then defined to be the intersection of all of these Cn sets, that is, C D \ Cn . nD1
The Cantor set is sometimes called the Cantor middle thirds set, because, at each stage, the middle thirds of the remaining intervals are removed. Other similar types of Cantor-like sets can be constructed by removing other portions of each interval. It is clear that the Cantor set has measure zero because it is contained in Cn which is made up of 2n closed intervals each with length 31n . The total length of the closed n intervals in Cn is 23n , a quantity that goes to 0 as n gets large. Cn can be covered by n a finite collection of open intervals whose total length is 10 percent larger than 23n showing that the Cantor set can be covered by open intervals whose total length is as small as you want. So how do you show that the Cantor set is uncountable? To see this, consider writing each number in the unit interval Œ0; 1 in base three. The numbers in the interval Œ0; 13 are the numbers between 0 and 1 whose base-three representation begins with 0.0, and the numbers in the interval Œ 23 ; 1 are the numbers between 0 and 1 whose base-three representation begins with 0.2. The numbers in the middle third of the interval that are removed at the first stage of the construction process are the numbers between 0 and 1 whose base-three representation begins with 0.1. Note that numbers at the endpoints of the removed interval, 13 and 23 each has two different representations. Indeed, in base three 13 D 0:1 D 0:0222 and 2 D 0:2 D 0:1222 . One could say that C1 consists of all the numbers between 0 3 and 1 that can be represented in base three without a 1 in the first place to the right of the decimal point, the one-third place. Similarly, C2 are the numbers between 0 and 1 that have a base-three representation with no 1 in either of the first two places to the right of the decimal point. The Cantor set C is the set of numbers between 0
6.3 Measure Zero
165
and 1 that have a base-three representation that contains no digit equal to 1. Then consider the map that takes each element of the Cantor set and divides it by 2. This is an injective map that maps the numbers in the Cantor set to the set of numbers in the unit interval that have base-three representations that include only the digits of 0 and 1 because it takes numbers with representations that only included the digits of 0 and 2 and divides each of the digits by 2. Now, the numbers between 0 and 1 with base-three representations that include only the digits of 0 and 1 are clearly in one-to-one correspondence with base-two representations of numbers in between 0 and 1. But all the real numbers between 0 and 1 have base-two representations containing only 0 and 1, so the numbers in the Cantor set are as numerous as the real numbers between 0 and 1. Thus, the Cantor set must be uncountable since the set of real numbers between 0 and 1 is an uncountable set. The concept of measure zero can be extended to sets in the plane, although here, rather than being interested in the length of a set, the interest is in the area of the set. Thus, rather than trying to cover a set with intervals whose total length is small, in the plane one would try to cover a set with a sequence of squares whose total area is small. Just as on the real line, it was taken as given that the length of an interval Œa; b was b a, in the plane it will be taken as given that the area of a square with side length x is x2 . Then, a region in the plane is said to have measure zero (or area zero) if for each > 0, the set is contained in the union of a sequence of squares whose total area is less than . As it was with sets of real numbers, any countable set of points in the plane has area zero because, for any < 0, you can cover the sequence x1 ; x2 ; x3 ; : : : with a sequence of squares with total area less than . Moreover, let Y be a line segment with length y > 0. Then Y has area zero. How would you prove this? Certainly, this line segment is contained in a square with side length y which has area y2 , so the square’s area could be rather large and, in particular, the area of the square is not zero. Notice, though, that Y can also be covered by two side-by-side squares each 2 with side length 2y and each with area y4 giving a total area of the two squares equal 2
to y2 . This is the key to covering Y with squares with very small total area. If Y is covered by a sequence of n adjacent squares each with side length ny , then the total 2
2
area of the n squares is yn . Since yn can be made arbitrarily small by choosing n large, it follows that Y has measure zero (Fig. 6.4).
Fig. 6.4 Covering a line segment with smaller and smaller squares
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6 Riemann Integrals
6.3.1 Exercises 1. Rather than constructing the Cantor set only on the interval Œ0; 1, perform the same construction on each interval Œn; n C 1 for every integer n. Show that the resulting set has measure zero. 2. Beginning with the interval Œ0; 1 construct a Cantor-like set, but instead of removing intervals of length 13 at stage 1, 19 at stage 2, and so forth removing intervals of length 31n at stage n, you remove an interval of length 14 at stage 1, 1 at stage 2, and so forth removing intervals of length 41n at intervals of length 16 stage n. Show that the total lengths of the intervals remaining after stage n does not approach zero as n approaches infinity. 3. Which of the following sets of real numbers have measure zero? (a) (b) (c) (d)
the integers the irrational numbers p p fa C b 2 C c 3 j a; b; c are integers g the Cantor-like set where instead of removing the middle 13 of each remaining interval at stage n, you remove the middle 14 of each remaining interval
4. Show that if the set A has measure zero and B A, then the set B has measure zero. 5. Show that a line in the plane has area zero. 6. Show that the set in the plane f.x; y/ j x is rationalg has area zero. 7. Suppose that the set A R has measure zero. Show that the set f.x; y/ j x 2 A; y 2 Œ0; 1g has area zero. 8. Suppose that the set A R has measure zero. Show that the set f.x; y/ j x 2 Ag has area zero. 9. Show that f.x; y/ j 0 x 1; 0 y 1; at least one of x or y is rationalg has area zero. 10. Show that the interval Œ0; 1 does not have measure zero. (Hint: Use the Heine– Borel Theorem to reduce any cover to a finite subcover.)
6.4 Areas in the Plane When discussing area, it is not possible to avoid the limit concept, and this brings a topic usually associated with Geometry into the field of Analysis. One could even make a case for including much of Geometry as a subtopic of Analysis since Geometry involves properties of distance, a distinguishing feature of Analysis. What properties of area can be taken as given? One would hope that whatever axioms are chosen, they would let you prove results about area that you know to be true from Euclidean Geometry. The following axioms accomplish this.
6.4 Areas in the Plane
167
Axioms for Area 1. The area of a set in the plane is a nonnegative real number. 2. A square with side length 1 has area equal to 1. 3. (Similarity) If sets A and B are similar in the geometric sense with lengths in B equal to t times the corresponding lengths in A, then the area of B is t2 times the area of A. 4. (Area Zero) Let A be a set. Suppose that for each > 0 there is a sequence of squares S1 ; S2 ; S3 ; with areas s1 ; s2 ; s3 ; , respectively, such that the 1
set A is contained in the union of the squares [ Sk , and for every natural kD1 P number n, nkD1 sk < . Then A has area 0. 5. (Union) If set A has area a, set B has area b, and their intersection A \ B has area 0, then the union A [ B has area a C b. 6. (Exhaustion) Let B be a set. If for each > 0 there are sets A and C with A B C such that the area of A is greater than b , and the area of C is less than b C , then the area of B is b. Axioms 1, 2, and 3 should agree with what you know about area from Geometry, and they can be used to prove some simple results. For example, since a 1 1 square has area 1, Axiom 3 can be used to show that an s s square has area s2 . The result from the previous section that a line segment has area 0 is particularly useful because of the way it can be used in conjunction with the Union area axiom. In particular, suppose A and B are two squares or other polygons set side-by-side so that they only share an edge. Because the shared edge is a line segment, it has 0 for its area, and the Union area axiom shows that A [ B has an area equal to the sum of the area of A and the area of B. By using mathematical induction, this result can be extended to the union of many polygons that share borders. In particular, consider finding the area of a rectangle with width x and length y. If xy is a rational number equal to pq , where p and q are positive integers, then the x y rectangle is the union of p q squares all with side length px . Indeed, the width of the rectangle which has length x is spanned by p such squares, and the length of the rectangle which has length y is spanned by q such squares showing that the entire rectangle can be tiled 2 by a p q array of squares, each with area px . The Union axiom then shows that 2 the area of the x y rectangle is p q px D x qp x D x y. It will require the last of the area axioms to conclude that the area of any rectangle is equal to its length times its width even when the length of the rectangle is an irrational multiple of its width. The last area axiom is essentially the Method of Exhaustion used some by Euclid and much more extensively by Archimedes to calculate areas and volumes. It is an example of a use of Calculus about 1800 years before the foundation of Calculus was formally established by Newton and Leibniz. This axiom says that if a region in the plane can be closely approximated by sets whose areas you know, then you can figure out the area of the region. Take, for example, a rectangle B with
168
6 Riemann Integrals
width x > 0 and length y > 0 where the ratio xy D ˛ is irrational. It is certainly possible to find other rectangles close to the size of B whose length to width ratios are rational. To prove that B has area xy, the axiom requires that for each > 0 you find a subset A B whose area is greater than xy and a set C containing B whose area is less than xy C . Suppose you choose A to be a rectangle with width x and length just a bit short of y, say rx, where r is a rational number chosen to be less than but suitably close to xy . How close is suitably close? Well, you would need the area of A, which is x rx D rx2 , to be within of xy, that is, xy rx2 < . Solving for r shows that r > xy x2 . Is there such an r which is rational and between xy x2 and xy ? Of course there is. The rational numbers are dense in the real line; there are rational numbers in every interval of positive length. Thus, you can select a rational number r between xy x2 and xy and let A be an x rx rectangle. Then A can be placed inside of B, and the area of A is within of xy. Similarly, you can choose a rectangle C with width x and length sx, where s is a rational number chosen to be greater than but suitably close to xy . You need the area of C to be within of xy, so choose s so that x sx xy < . This happens if xy < s < xy C x2 . Since you have found a rectangle A contained inside B and a rectangle C containing B with the areas of A and C within of xy, the Exhaustion area axiom shows that B has area xy. The familiar formula for the area of a triangle given as one half the base times the height can be derived geometrically, but to prove this formula using the area axioms requires more work. To begin, consider a right triangle with legs with lengths x and y. Place this triangle in a rectangle with side lengths x and y. For any natural number n, the rectangle can be overlaid with an n n grid of rectangles with side lengths x and ny . The hypotenuse of the triangle is the diagonal of the x y rectangle and n spans the diagonals of n of the smaller rectangles as shown in Fig. 6.5 exhibiting the case where n D 8. Because there are n grid rectangles along the hypotenuse of the triangle, it 2 must be that there are n 2n grid rectangles inside the triangle with a total area 2 of n 2n nxy2 D 1 1n xy2 . Similarly, the triangle is enclosed inside the union of n2 Cn grid rectangles with a total area of 1 C 1n xy2 . Clearly, n can be chosen large 2 Fig. 6.5 An 8 8 grid of rectangles overlaying a triangle
6.5 Definition of Riemann Integral
169
enough to make both the total area of grid rectangles inside the triangle and the total area of grid rectangles enclosing the triangle within a particular > 0 of xy2 . Thus, the Exhaustion axiom shows that the area of the triangle is xy2 as expected. Since any triangle can be partitioned into two right triangles, the well-known area formula for the area of a triangle follows. Since any polygon can be partitioned into triangles, the usual formulas from Geometry for the areas of polygons can be derived in the same way they would be in Geometry. You may wonder whether these techniques can be used to find the area of any region in the plane, or at least any bounded region in the plane. This is a really good question with a very complicated answer. The Area Axioms listed in this section are designed to give the reader a feel for proofs about areas that will be useful in the upcoming discussion of proofs about Riemann integrals. The axiom list is not complete enough to allow the calculation of the area of many of the sets that one might encounter. The area of Analysis known as Measure Theory provides a somewhat richer environment for this study, but the complexities of measure theory go beyond the aim of this text. What can be said is that even with the use of measure theory, there are sets in the plane complex enough that one cannot assign an area measure to them.
6.4.1 Exercises 1. Show that a circle with radius r has area r2 . 2. Suppose the polygonal region A in the coordinate plane has area K. Show that the region f.x; y/ j .x; 3y / 2 Ag has area 3K.
6.5 Definition of Riemann Integral The definition of the Riemann Integral is motivated by the Method of Exhaustion which attempts to approximate a planar region with sets, perhaps collections of rectangles or other polygons, whose area can be easily calculated. If the region whose area is to be calculated is not a polygon itself, then one needs to fill the region with a sequence of smaller and smaller polygons until a limit is realized. Such a region might be bounded by the horizontal x-axis, the vertical lines given by x D a and x D b for some real numbers a and b, and finally by the graph of some nonnegative function f . Given this region, one attempts to fill the region with rectangles whose sides are parallel to the axes, have one side along the x-axis, and have a length determined in some way by the graph of the given function.
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6 Riemann Integrals
a
b
a
b
Fig. 6.6 Approximating the area under a curve with narrowing rectangles
If the function is in some sense well behaved, then as the widths of these rectangles are chosen to be smaller and smaller, the total area of the rectangles will approach the area of the region (Fig. 6.6). What is meant by well behaved will be a main focus of the theorems presented in this chapter. To make the definition of Riemann Integral precise, there needs to be a way to talk about the placement of the vertical rectangles used to approximate the area under a curve. This is done by designating the position of the vertical sides of the rectangles with a collection of x values in the interval Œa; b. One defines a partition of the interval Œa; b to be a finite sequence of x values P W a D x0 x1 x2 xn D b for some natural number n. These values of x break the interval Œa; b into n subintervals Œxj1 ; xj . Note that the definition of partition does not say anything about the lengths of the subintervals for the partition. Indeed, it could be that the jth subinterval length xj xj1 could be 0 or could be as large as b a. In particular, there is no requirement that all the interval lengths be the same size. Since the lengths of the subintervals xj xj1 are used frequently in the discussion of Riemann Integrals, one often uses the shorthand notation xj D xj xj1 . Given a partition, P W a D x0 x1 x2 xn D b, one defines the norm of the partition P , jjP jj, to be the maximum length of a subinterval of the partition, that is, jjP jj D max xj . For example, if Œa; b D Œ1; 4 has the partition jn
1; 1 12 ; 1 34 ; 2 12 ; 2 23 ; 2 23 ; 3 14 ; 3 12 ; 3 56 ; 4, then the norm of the partition is 34 D 2 12 1 34 , the largest distance between any two of the adjacent points in the partition. As seen in the previous section, one can get increasingly better approximations to the area of a region by attempting to approximate the region by smaller and smaller polygons. Thus, by requiring the norm of a partition to be smaller, the rectangles used to approximate the area of a region bounded by a curve become smaller in width and can give a better approximation. For the Riemann Integral, a partition will determine the widths of the rectangles used to approximate the area of a region. What will be used as the lengths of those rectangles? Suppose a rectangle rests on the x-axis between xj1 and xj . If the rectangle is going to fit inside the region bounded by the curve y D f .x/, then
6.5 Definition of Riemann Integral
171
the length of the rectangle (its height above the x-axis) cannot exceed
inf
xj1 xxj
f .x/.
If the rectangle is going to enclose the part of the region between xj1 and xj , then the length of the rectangle must be at least sup f .x/. The definition of the Riemann xj1 xxj
Integral uses a value between these two possible extremes. It requires the choice of a sequence of x values 1 ; 2 ; 3 ; : : : ; n with xj1 j xj for each j. Then the rectangle on Œxj1 ; xj is given the length f .j / so that it has area f .j /xj . Clearly, the choice of j 2 Œxj1 ; xj results in the length of the rectangle being f .j / which is between the two extremes inf f .x/ and sup f .x/, so the rectangles that xj1 xxj
xj1 xxj
result might neither be contained in the region bounded by the curve nor cover the region. Instead, the lengths of the rectangles are allowed to be in between these two extremes. The total area of all the rectangles is then given by the Riemann Sum n P f .j /xj . jD1
Now, given a function f defined on the interval Œa; b, one can define the Riemann Rb Integral of f on Œa; b to be I D f .x/dx if for every > 0 there is a ı > 0 a
such that for every partition P W a D x0 x1 x2 xn D b of Œa; b with jjP jj 0 there is a ı > 0 such that a particular inequality holds. But unlike previous kinds of limits, the inequality that must hold for Riemann sums is supposed to be true for every choice of a partition P and every choice of j ’s as long as jjP jj < ı. Thus, it is not just that a region in the plane is being approximated by a sequence of rectangles, but that the region must be closely approximated by every possible
172
6 Riemann Integrals
sequence of rectangles that arise from the Riemann sum
n P
f .j /xj . Also worth
jD1
noting is that the Riemann Integral is not the only way to define integration. Most of the other definitions give the same value as the Riemann Integral for functions where the Riemann Integral exists, but some of the other definitions give values to integrals in situations where the Riemann Integral does not exist. Some examples of other integration definitions include the Riemann–Stieltjes Integral, the Lebesgue Integral, the Darboux Integral, and the Daniell Integral. There are some fairly easy to describe functions that do not have a Riemann 1 if x is rational integral. One simple example is the function f .x/ D whose 0 if x is irrational Riemann integral is not defined on any interval Œa; b with a < b. To see why this is, n P consider any Riemann sum f .j /xj . Because both the rational numbers and the jD1
irrational numbers are dense in the real numbers, in any subinterval of the partition which has positive length, there are values of j in the subinterval where f .j / D 0, and other values of j in the subinterval where f .j / D 1. Thus, for any partition, n P there are choices of the j ’s that make the Riemann sums equal to 0 xj D 0 and other choices that make the Riemann sum equal to
n P
jD1
1 xj D b a > 0. Thus,
jD1
no limit can exist.
6.5.1 Exercises 1. Let f .x/ D x. Partition the interval Œ1; 3 into n subintervals with 1 D x0 and xj D xj1 C 2n for j D 1; 2; 3; : : : ; n. (a) Find the minimum and maximum possible values for an associated Riemann n P sum f .j /xj . jD1
(b) Show that as n gets large, the Riemann sum must approach 4. 2. Let f .x/ D x2 . Partition the interval Œ1; 2 into n subintervals with 1 D x0 and xj D xj1 C 3n for j D 1; 2; 3; : : : ; n. (a) Find the minimum and maximum possible values for an associated Riemann n P sum f .j /xj . jD1
(b) Show that as n gets large, the Riemann sum must approach 3.
6.6 Properties of Integrals
173
6.6 Properties of Integrals There are many theorems about the properties satisfied by Riemann integrals. Some of the proofs of these theorems merely rely on properties of summations since n P the definition of the Riemann Integral is based on the Riemann sum, f .j /xj . jD1
Consider the following results. • If a, b, and c are a constants, then
Rb
c dx D c.b a/.
a
• If f is an integrable function on the interval Œa; b and c is a constant, then Rb Rb c f .x/dx D c f .x/dx. a
a
Rb • If f and g are functions integrable on the interval Œa; b, then .f C g/.x/dx D Rb a
f .x/dx C
Rb
a
g.x/dx.
a
• If f and g are functions integrable on the interval Œa; b, and f .x/ g.x/ for all Rb Rb x 2 Œa; b, then f .x/dx g.x/dx. a
To prove that
Rb
a
cdx D c.b a/, one needs to find a ı > 0 so that if the norm of a
a
partition is less than ı, then the Riemann sum
n P jD1
f .j /xj is within some > 0 of
c.b a/. But in this case f .j / is always equal to the constant c, so the Riemann sum is always equal to the desired integral, c.b a/. This makes the proof particularly easy. Note that the first four steps of this proof merely set up the assumptions required by the definition of the Riemann Integral. That is, one needs to have constants a and b and function f defined on the interval Œa; b. Then one needs to take an arbitrary > 0, find an appropriate ı > 0, and consider an arbitrary Riemann sum which satisfies the needed condition on the norm of the partition. Although straightforward, these steps are necessary in order to show that the definition of Riemann Integral is being satisfied.
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6 Riemann Integrals
PROOF: If a, b, and c are constants, then
Rb a
c dx D c.b a/.
• Let constants a, b, and c be given. • Without loss of generality, assume that a b, and let f .x/ D c for all x in Œa; b. • Let > 0 be given. • Let ı D 1, and let P W a D x0 x1 x2 xn D b be a partition of Œa; b with jjP jj < 1. ˇ ˇ ˇP ˇ ˇ n ˇ • Then for any choices of j 2 Œxj1 ; xj , it follows that ˇ f .j /xj c.ba/ˇ ˇjD1 ˇ ˇ ˇ ˇ ˇ ˇP ˇ ˇ P ˇ n n ˇ ˇ ˇ ˇ D ˇ cxj c.b a/ˇ D ˇc .xj xj1 / c.b a/ˇ D ˇjD1 ˇ ˇ jD1 ˇ jcj j.xn x0 / .b a/j D 0 < . Rb • Thus, cdx D c.b a/. a
Now consider the next theorem which states that
Rb a
c f .x/dx D c
Rb a
f .x/dx.
Rb In the proof of this result you will need to use the fact that f .x/dx D I to a ˇ ˇ ˇP ˇ n ˇ ˇ say something about the size of ˇ c f .j /xj cI ˇ. But this expression equals ˇjD1 ˇ ˇ ˇ ˇ ˇ ˇP ˇP ˇ ˇ n ˇ n ˇ ˇ ˇ jcj ˇ f .j /xj I ˇ suggesting that if you can arrange for ˇ f .j /xj I ˇ to be ˇjD1 ˇjD1 ˇ ˇ ˇ ˇ ˇP ˇ ˇ n ˇ small, then you can arrange for the product jcj ˇ f .j /xj I ˇ to be small. You ˇjD1 ˇ to be less than some given > 0, so it is tempting to require ˇwill need the product ˇ ˇP ˇ ˇ n ˇ . This is fine except for the embarrassing case where c D 0. ˇ f .j /xj I ˇ < jcj ˇjD1 ˇ One could handle this problem by breakingˇ the proof into two ˇ cases: c D 0 and ˇP ˇ n ˇ ˇ c ¤ 0. Easier, though, is to simply ask for ˇ f .j /xj I ˇ to be less than jcjC1 . ˇjD1 ˇ The use of jcj C 1 in the denominator is just a trick that takes care of the case where
6.6 Properties of Integrals
175
jcj is large ˇ and the case ˇwhere jcj is 0 both at the same time. Of course, you can ˇP ˇ Rb ˇ n ˇ arrange ˇ f .j /xj I ˇ < jcjC1 because that follows from f .x/dx D I. ˇjD1 ˇ a PROOF: If f is an integrable function on the interval Œa; b and c is a Rb Rb constant, then c f .x/dx D c f .x/dx. a
a
• Let interval Œa; b and constant c be given. • Let f be a function defined on Œa; b such that
Rb
f .x/dx D I.
a
• Let > 0 be given. • From the definition of Riemann Integral, there is a ı > 0 such that if P W a D x0 x1 x2 ˇxn D b is a partition with jjP jj < ı, then for ˇ ˇP ˇ n ˇ ˇ every choice of j 2 Œxj1 ; xj , ˇ f .j /xj I ˇ < jcjC1 . ˇjD1 ˇ ˇ ˇ ˇ ˇ ˇP ˇ ˇ ˇP ˇ n ˇ ˇ ˇ n < . • Then ˇ c f .j /xj cI ˇ D jcj ˇ f .j /xj I ˇ jcj jcjC1 ˇjD1 ˇ ˇ ˇjD1 Rb Rb • Thus, c f .x/dx D cI D c f .x/dx. a
a
The third theorem in this section can be summarized by saying that the integral of a sum is the sum of the integrals. Its proof is reminiscent of the proof of the theorem stating that the limit of a sum is the sum of the limits, and of the theorem stating that the derivative of a sum is the sum of the derivatives. In this Rb Rb case, you are given that f .x/dx D I and g.x/dx D J and are then faced with a
a
the distance that ˇthe Riemann sum for f C ˇg is from the value of the integral ˇP ˇ ˇ n ˇ I C J given by ˇ .f C g/.j /xj .I C J/ˇ. This easily breaks into the two ˇjD1 ˇ ˇ ! !ˇ ˇ ˇ P n P ˇ ˇ n differences ˇ f .j /xj I C g.j /xj J ˇ. The existence of the two ˇ ˇ jD1 jD1 given integrals then lets you choose a value of ı > 0 that will ensure that the two parts to this sum are both small.
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PROOF: If f and g are integrable functions on the interval Œa; b, then Rb Rb Rb .f C g/.x/dx D f .x/dx C g.x/dx. a
a
a
• Let interval Œa; b be given, and let f and g be integrable functions on Œa; b Rb Rb with f .x/dx D I and g.x/dx D J. a
a
• Let > 0 be given. • From the definition of Riemann Integral, there is a ı1 > 0 such that if P W a D x0 x1 x2 xˇn D b is a partition ˇ with jjP jj < ı1 , then ˇP ˇ ˇ n ˇ for every choice of j 2 Œxj1 ; xj , ˇ f .j /xj I ˇ < 2 . ˇjD1 ˇ • Similarly, there is a ı2 > 0 such that if P W a D x0 x1 x2 xˇ n D b is a partition ˇ with jjP jj < ı2 , then for every choice of j 2 Œxj1 ; xj , ˇP ˇ n ˇ ˇ ˇ g.j /xj J ˇ < 2 . ˇjD1 ˇ • Let ı D min.ı1 ; ı2 /. • Let P W a D x0 x1 x2 xn D b be a partition of Œa; b with jjP jj < ˇ ı, and let j ’s be chosen with ˇ j 2 Œxj1 ; xj . ˇP ˇ n ˇ ˇ • Then ˇ .f C g/.j /xj .I C J/ˇ D ˇjD1 ˇ ˇ ! !ˇ ˇ P ˇ n P ˇ n ˇ f .j /xj I C g.j /xj J ˇ ˇ ˇ jD1 ˇ jD1 ˇ ˇ ˇ ˇ ˇP ˇ ˇP ˇ ˇ n ˇ ˇ n ˇ ˇ f .j /xj I ˇ C ˇ g.j /xj J ˇ < 2 C 2 D . ˇjD1 ˇ ˇjD1 ˇ Rb Rb Rb • Thus, .f C g/.x/dx D I C J D f .x/dx C g.x/dx. a
a
a
The final theorem in this section states that if f .x/ g.x/ for all x 2 Œa; b, then if Rb Rb the functions are integrable, f .x/dx g.x/dx. It is sufficient to prove this result a
a
when f is the identically 0 function, because if h.x/ 0 implies
Rb
h.x/dx 0, this
a
would imply that if f .x/ g.x/, then h.x/ D g.x/ f .x/ 0, so From there
Rb a
.g f /.x/dx D
Rb a
g.x/dx
Rb a
Rb
h.x/dx 0.
a
f .x/dx 0 and the needed result
follows. With the assumption that h.x/ 0 for all x 2 Œa; b, it is not hard to Rb show that h.x/dx 0, because the value of every associated Riemann sum must a
6.6 Properties of Integrals
177
be nonnegative. How do you turn this into a proof? Recall how the proof went when showing that if f .x/ 0, then lim f .x/ cannot be negative. If you assume that x!a
the limit L is negative, then you can choose an D L2 . If f is never negative, it follows that jf .x/ Lj is always greater than giving a contradiction. A very similar argument works here where f is replaced by the Riemann sum. PROOF: If f and g are functions integrable on the interval Œa; b, and Rb Rb f .x/ g.x/ for all x 2 Œa; b, then f .x/dx g.x/dx. a
a
• Let interval Œa; b be given, and let f and g be integrable functions on Œa; b with f .x/ g.x/ for all x 2 Œa; b. • Define h.x/ D g.x/f .x/ which is greater than or equal to 0 for all x 2 Œa; b. Rb Rb Rb Since f and g are integrable, so is h, and h.x/dx D g.x/dx f .x/dx. Thus, it suffices to prove that • Assume instead that
Rb a
Rb
a
a
a
h.x/dx 0.
a
h.x/dx D I < 0, and let D 2I > 0.
• From the definition of Riemann Integral, there is a ı > 0 such that if P W a D x0 x1 x2 xˇn D b is a partition ˇ with jjP jj < ı, then for ˇP ˇ n ˇ ˇ every choice of j 2 Œxj1 ; xj , ˇ h.j /xj I ˇ < . ˇjD1 ˇ ˇ ˇ ˇP ˇ ˇ n ˇ • But for every choice of j , h.j / 0, so ˇ h.j /xj I ˇ D ˇjD1 ˇ n P h.j /xj I I > . jD1
• This contradicts the assumption that I < 0 which completes the proof.
6.6.1 Exercises Write proofs for each of the following statements. 1. If functions f1 ; f2 ; f3 ; : : : ; fn are integrable on interval Œa; b, and c1 ; c2 ; c3 ; : : : ; cn Rb .c1 f1 .x/ C c2 f2 .x/ C c3 f3 .x/ C C cn fn .x// dx are constants, then D c1
Rb a
f1 .x/dx C c2
Rb a
a
f2 .x/dx C c3
Rb a
f3 .x/dx C C cn
Rb
fn .x/dx. (In the words
a
of Linear Algebra, this says that the Riemann integral is a linear operator.)
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6 Riemann Integrals
2. If f is a function integrable on Œa; b with f .x/ c for all x 2 Œa; b, then Rb f .x/dx c.b a/. a ˇ ˇ ˇRb ˇ ˇ ˇ 3. If f is a function such that both f and jf j are integrable on Œa; b, then ˇ f .x/dxˇ ˇa ˇ Rb jf .x/jdx. a
6.7 Integrable Functions It is helpful to have a characterization of those functions which are Riemann integrable. This section will discuss several theorems which establish some properties of functions that guarantee that they are integrable. Then the following three sections present a series of results that give a complete characterization of Riemann integrable functions. Recall that f is called bounded on Œa; b if there is a number M such that jf .x/j M for all x 2 Œa; b. It is important to note that if f is integrable on an interval, then f must be bounded on that interval. The way to prove this result is reminiscent of the way one proves that a function continuous on a closed bounded interval is bounded. That is, one uses an indirect proof assuming that you have an integrable function that is not bounded, and from that, you produce a contradiction. Think about what can be done with a Riemann sum if the function f is not bounded. Given a partition P W a D x0 x1 x2 xn D b, for some choice of the j ’s the Riemann n P sum is f .j /xj . If f is unbounded on Œa; b, then it must be unbounded on at jD1
least one of the subintervals Œxj1 ; xj ; otherwise, if there is a bound for f on each of the n subintervals, one merely needs to select the largest of those n bounds to have a bound for f on the entire interval Œa; b. So what happens if f is not bounded on the kth subinterval Œxk1 ; xk ? It means that k could be changed to be some other value in the subinterval, say , to make the term f . /xk as large as you like. Thus, you can make the entire Riemann ˇ sum as largeˇas you like. So how large do ˇP ˇ ˇ n ˇ you want f . /xk to be? You want ˇ f .j /xj I ˇ to be larger than for some ˇjD1 ˇ preassigned > 0 such as D 1. The proof below does this by selecting a value to replace k in such a way that the kth term of the Riemann sum, f . /xk , is ˇ larger by at least ˇ 1 than the sum of the absolute values of all the other terms of ˇP ˇ n ˇ ˇ ˇ f .j /xj I ˇ guaranteeing that the resulting expression will be bigger than 1. ˇjD1 ˇ This gives the needed contradiction.
6.7 Integrable Functions
179
PROOF: If f is a function integrable on the interval Œa; b, then f is bounded on Œa; b. • Let f be an integrable function on the interval Œa; b. • Assume that f is not bounded on Œa; b. Rb • Let D 1, and f .x/dx D I. a
• Then from the definition of Riemann integral, there is a ı > 0 such that if P W a D x0 x1 x2 xn D b is a partition with jjP ˇ jj < ı and j ’sˇ are chosen with j 2 Œxj1 ; xj , the Riemann sum satisfies ˇP ˇ ˇ n ˇ ˇ f .j /xj I ˇ < D 1. ˇjD1 ˇ • Let a particular partition P with jjP jj < ı and choices for j 2 Œxj1 ; xj be given. • Because f is not bounded on Œa; b, it follows that there is a k between 1 and n such that f is not bounded on the interval Œxk1 ; xk . Otherwise, f is bounded on each of the subintervals of the partition implying that it is bounded on the entire interval Œa; b. Note that xk > 0 because a function cannot be unbounded on an interval of length 0. n P • Let J D jf .j /jxj jf .k /jxk C jIj. jD1
• Because f is unbounded on Œxk1 ; xk , there is exists 2 Œxk1 ; xk such that jf . /j > JC1 . xk • Then the Riemann sum resulting from the partition P with the choices of jˇ ˇ ˇP ˇ ˇ n ˇ f .j /xj C f . /f .k / xk I ˇ ˇjD1 ˇ !
where k is replaced by must satisfy 1 > ˇ jf . /jxk
n P
jf .j /jxj jf .k /jxk C jIj >
jD1
JC1 xk xk
J D 1.
• This is a contradiction, so the assumption that f is unbounded must be false. This completes the proof. Knowing that integrable functions must be bounded is very helpful. If you can claim that jf .x/j P M for some constant M, then you know that any one term of a Riemann sum njD1 f .j /xj can contribute at most M xj to the sum. By forcing the norm of the partition, jjP jj, to be very small, you can control the maximum size of xj and, thus, the maximum size of a term in the Riemann sum. This is the key idea behind the proof of the next theorem which states that if f is integrable Rc Rb Rc on Œa; b and on Œb; c, then f .x/dx D f .x/dx C f .x/dx. To prove this, it is a
a
b
natural to consider finding a ı1 > 0 so that Riemann sums arising from partitions of Rb Œa; b with norm less than ı1 are close to I D f .x/dx and finding a ı2 > 0 so that a
Riemann sums arising from partitions of Œb; c with norm less than ı2 are close to Rc J D f .x/dx. You would consider allowing ı to equal the minimum of the ı1 and ı2 . b
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6 Riemann Integrals
Then you could take a partition of Œa; c with a norm less than ı. Unfortunately, this partition of Œa; c does not separate into a partition of Œa; b and a partition of Œb; c because there is no guarantee that the given partition of Œa; c includes the point b as one of the xj values in the partition. But if you change the Riemann sum by altering the interval of the partition containing the point b by adding b as an extra point to the partition, you are not making a large change in the total sum. More precisely, suppose the partition is P W a D x0 x1 x2 xn D c with the point b in the interval Œxk1 ; xk . A resulting Riemann sum has the term f .k /.xk xk1 /. If this term is replaced by two terms f .b/.b xk1 / C f .b/.xk b/, how much does this change the Riemann sum? The change is exactly f .b/.b xk1 / C f .b/.xk b/ f .k /.xk xk1 / D .f .b/ f .k //.xk xk1 /. Given that f is integrable on Œa; b and on Œb; c, you know that there is a bound M such that jf .x/j M for all x 2 Œa; c. An upper bound for the size of this change is, therefore, 2M.xk xk1 / < 2Mı. This says that by choosing ı small enough, you can control the amount of change made in the Riemann sum by introducing b as a point in the partition of Œa; c. If ı is also Rb Rc chosen small enough so that the resulting Riemann sums for f .x/dx and f .x/dx a
b
are close to the corresponding integral, then the total difference between original Rb Rc Riemann sum and the sum of the integrals f .x/dx C f .x/dx is small enough. a
b
This is the idea behind the following proof. PROOF: If f is a function integrable on the interval Œa; b and on the Rc Rb Rc interval Œb; c, then f .x/dx D f .x/dx C f .x/dx. a
a
b
• Without loss of generality assume that a < b < c, and let f be a function Rb integrable on the interval Œa; b and on the interval Œb; c with I D f .x/dx and J D
Rc
a
f .x/dx.
b
• Because f is integrable on Œa; b, jf j is bounded on that interval by some value M1 . Because f is integrable on Œb; c, jf j is bounded on that interval by some value M2 . It follows that jf j is bounded on the interval Œa; b by M D max.M1 ; M2 /. • Let > 0 be given. • From the definition of Riemann integration, there is a ı1 > 0 such that for every partition P of Œa; b with jjP jj < ı1 and every choice of j 2 Œxj1 ; xj on the intervals of the partition, the associated Riemann sum will be within of the integral I. 3 • Similarly, there is a ı2 > 0 such that for every partition P of Œb; c with jjP jj < ı2 and every choice of j 2 Œxj1 ; xj , the associated Riemann sum will be within 3 of the integral J. (continued)
6.7 Integrable Functions
181
. • Let ı D min ı1 ; ı2 ; 6MC1 • Let P W a D x0 x1 x2 xn D c be a partition of Œa; c with jjP jj < ı. • Let ’s be chosen such that j 2 Œxj1 ; xj . • Since b 2 Œa; c, there is a k such that b 2 Œxk1 ; xk . • Then ˇ ˇ ˇ ˇP ˇ ˇ n ˇ f .j /xj .I C J/ˇ D ˇ ˇjD1 ˇ ! ! ˇ k1 n P ˇ P f . /xj Cf .b/.b xk1 / C f .b/.xk b/ C f .j /xj C ˇ ˇ jD1 j jDkC1 ˇ ˇ ˇ .f .xk / f .b//xk .I C J/ˇ ˇ ˇ ˇ ˇ ˇ ˇk1 ˇ ˇ ˇ n P ˇP ˇ ˇ ˇ f .j /xj J ˇ C ˇ f .j /xj Cf .b/.b xk1 / I ˇ C ˇf .b/.xk b/C ˇjD1 ˇ ˇ ˇ jDkC1 jf .xk / f .b/jxk : • Since the partition a D x0 x1 x2 xk1 b D ˇb is a partition of Œa; b withˇ norm less than ı1 , it follows that ˇk1 ˇ ˇP ˇ ˇ f .j /xj C f .b/.b xk1 / I ˇ < 3 . ˇjD1 ˇ • Similarly, since the partition b D b xk xkC1 norm less than ı2 , it follows that ˇxn D c is a partition of Œb; c with ˇ ˇ ˇ n P ˇ ˇ f .j /xj J ˇ < 3 . ˇf .b/.xk b/ C ˇ ˇ jDkC1 < 3 . • Also, jf .xk / f .b/jxk < 2M 6MC1 ˇ ˇ ˇP ˇ ˇ n ˇ • Therefore, ˇ f .j /xj .I C J/ˇ < 3 C 3 C 3 D . ˇjD1 ˇ Rc • This proves that f .x/dx D I C J and completes the proof of the theorem. a
Note that you can easily show that this theorem also holds if a > b or b > c by simply rearranging the order of the limits on one or more of the integrals. The previous section discusses the theorem stating that if integrable functions Rb Rb satisfy f g on Œa; b, then f .x/dx g.x/dx. Can this statement be made a
a
stronger? That is, if f .x/ g.x/ for x 2 Œa; b, with f .x/ < g.x/ for some x 2 Œa; b, Rb Rb can you conclude that f .x/dx < g.x/dx? The answer is no. For example, if a
a
f and g only differ for a finite number of x values, then f and g will have identical integrals. To prove this, start with two integrable functions, f and g, that are identical for all x 2 Œa; b except for some t 2 Œa; b. How would you prove that f and g have
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6 Riemann Integrals
identical integrals? Again, you should consider the Riemann sums associated with f n P and g, that is, consider a Riemann sum g.j /xj for g with a particular partition jD1
and choice of j ’s, and compare it to the corresponding sum
n P
f .j /xj for f . If
jD1
f .x/ D g.x/ at all points except x D t, how many of the corresponding terms in these two Riemann sum could be different? Well, only those terms for which the chosen j D t and xj ¤ 0. This could happen at most twice (twice in the unusual case of t D xj D j D jC1 ). Thus, the Riemann sum for g is identical to the Riemann sum for f plus at most two terms. By controlling the size of xj which you can do by limiting the norm of the partition, you can control the contribution of those at most two terms in the Riemann sum, thus ensuring that the sums for f and g are close. That is the idea behind the following proof. PROOF: Suppose that f and g are functions integrable on the interval Œa; b, and that f .x/ D g.x/ for all x 2 Œa; b except perhaps at t 2 Œa; b. Rb Rb Then f .x/dx D g.x/dx. a
a
• Let f and g be a functions integrable on the interval Œa; b, and suppose that f .x/ D g.x/ for all x 2 Œa; b except perhaps at t 2 Œa; b. Rb • Let f .x/dx D I. a
• Let M D max.jf .t/j; jg.t/j/ C 1. • Let > 0 be given. • From the definition of Riemann Integration, there is a ı1 > 0 such that for every partition P W a D x0 x1 x2 xn D b with norm less than ıˇ 1 , and every choice ˇ of j 2 Œxj1 ; xj , the associated Riemann sum satisfies ˇP ˇ n ˇ ˇ ˇ f .j /xj I ˇ < 2 . ˇjD1 ˇ • Let ı2 D 8M , and set ı D min.ı1 ; ı2 /. • Select any partition P W a D x0 x1 x2 xn D b with norm less than ı, and select any sequence of j 2 ˇ Œxj1 ; xj . Then ˇ the ˇP ˇ ˇ n ˇ associated Riemann sum for the function g satisfies ˇ g.j /xj I ˇ ˇjD1 ˇ ˇ ˇ ˇP ˇ n ˇ ˇ D . ˇ f .j /xj I ˇ C jg.t/ f .t/j2ı < 2 C 2M 2 8M ˇjD1 ˇ Rb Rb • Thus, g.x/dx D I D f .x/dx which proves the theorem. a
a
It is left as an exercise to extend this theorem to the case where f and g differ at a finite number of points. In fact, this can be extended to f and g which differ on an infinite sequence of points in Œa; b as long as the sequence has a limit.
6.8 Step Functions
183
6.7.1 Exercises Write proofs for each of the following statements. 1. If f and g are functions integrable on the interval Œa; b, and f .x/ D g.x/ for all x 2 Œa; b except perhaps at the finite set of points ft1 ; t2 ; t3 ; : : : ; tk g Œa; b, then Rb Rb f .x/dx D g.x/dx. a
a
2. If f and g are functions integrable on the interval Œa; b, and f .x/ D g.x/ for all x 2 Œa; b except perhaps on a sequence of points ft1 ; t2 ; t3 ; : : : g Œa; b where Rb Rb lim tj D L, then f .x/dx D g.x/dx. j!1
a
a
3. If f is defined on the interval Œ0; 1 by f .0/ D 0 and for each natural number n, R1 1 , then f .x/dx exists and is equal to 13 . f .x/ D 21n for all x with 21n < x 2n1 0
4. If f is defined on the interval Œ0; 1 by f .0/ D 0 and for each natural number n R1 1 , then f .x/dx does not exist but n, f .x/ D 32 for all x with 21n < x 2n1 lim
R1
r!0C r
0
f .x/dx D 3.
5. If f is integrable on Œa; b, then the function defined on Œa; b by F.x/ D is continuous on Œa; b.
Rx
f .t/dt
a
6.8 Step Functions Step functions play an important role in the theory of the Riemann integration. A step function s on the interval Œa; b is associated with a partition P W a D x0 x1 x2 xn D b of Œa; b and has the property that s is constant 9 8 on each interval ˆ 3 0 x < 2> > ˆ > ˆ > ˆ > ˆ = < 1 2Dx of the partition, .xj1 ; xj /. For example, the function s.x/ D 4 2 < x < 4 is > ˆ > ˆ > ˆ > ˆ 0 4Dx > ˆ ; : 1 4 < x 5 a step function defined on the interval Œ0; 5 (Fig. 6.7). It follows easily that a step function on an interval Œa; b is integrable there. Indeed, suppose that P W a D x0 x1 x2 xn D b, and s.x/ D cj for all x satisfying xj1 < x < xj . Clearly, the constant function cj is integrable on the interval Œxj1 ; xj , and the function s.x/ differs from this constant function at at most the two endpoints, xj1 and xj . Thus,
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6 Riemann Integrals
Fig. 6.7 The step function s.x/
by the previous theorem, cj xj D
n P
Rxj
s.x/dx D cj xj . Then,
Rb
xj1
a
s.x/dx D
n Rxj P
s.x/dx D
jD1 xj1
cj xj .
jD1
The importance of step functions comes from the fact that a function f is integrable on Œa; b if and only if f can be closely approximated by step functions. Precisely, f has a Riemann integral on the interval Œa; b if and only if for every > 0, there exist step functions u.x/ and v.x/ on Œa; b with the property that for Rb Rb all x 2 Œa; b, v.x/ f .x/ u.x/, and u.x/dx v.x/dx < . That is, f has a
a
an integral precisely when for every > 0 there is a lower step function v that is always less than or equal to f and an upper step function u that is always greater than or equal to f with the property that the integrals of v and u are within of each other. This squeezes f between two step functions whose integrals are as close as you want. This should remind you of the Exhaustion Area Axiom. The statement of this theorem is a biconditional statement; that is, it is an “if and only if” statement. This means that the proof will have two distinct parts. One proof must show that if a function is integrable, then it can be approximated by very close upper and lower step functions. The other proof must show that if a function can be approximated by very close upper and lower step functions, then it is integrable. Consider how you would approach the proofs of each of these statements. For the first part of the proof, you would consider a function, f , integrable on an interval Œa; b. Given an > 0, somehow you need to show that there are upper and lower step functions, u and v, whose integrals are within of each other. Where do you start? All you know about f is that it has a Riemann integral on Œa; b, thus, all you have to go on is the definition of Riemann integration which makes a statement about the properties of Riemann sums. The key observation here is that a Riemann n P sum f .j /xj is equal to the integral of a step function defined to be equal to the jD1
constant f .j / on the interval .xj1 ; xj /. Since the definition of the integral guarantees
6.8 Step Functions
185
that you can find Riemann sums that are very close to the value of the integral, this suggests how you might choose step functions whose integrals are close to each other. How can you assure that you choose a step function that is less than f .x/ for each x 2 Œa; b? For each interval of the partition .xj1 ; xj / you could consider selecting j so that f .j / is the minimum value of f on that interval. Unfortunately, f might not achieve a minimum value on that interval. Certainly, if f is continuous on Œxj1 ; xj , then it obtains its minimum on that interval, but there is nothing here indicating that f is continuous. On the other hand, you do know that, because f is integrable, it is bounded. Thus, there is a greatest lower bound Mj D inf f .x/. x2.xj1 ;xj /
There may not be any x 2 .xj1 ; xj / with the property that f .x/ D Mj , but you know that there are values of x in the interval such that f .x/ is as close as you like to Mj . Getting specific, now, your goal is to find upper and lower step functions whose integrals are within some given > 0 of each other. It makes sense, therefore, to find upper and lower step functions whose integrals are both within 2 of the value of the integral of f because then the two step functions will be within of each other. From the definition of Riemann integral, you can find a partition of Œa; b such that all associated Riemann sums are within 4 of the integral of f . Then you can define a lower step function, v.x/, that is equal to the infimum of f on each interval of the chosen partition. On each interval of the partition you can find j values so that f .j / is within 4.ba/ of v.j /. Then the integral of the lower step function will be within .b a/ D 4 of a Riemann sum for f which in turn is within 4 of the integral 4.ba/ of f . This produces a lower step function with the properties you want. A similar construction will produce an upper step function whose integral is also within 2 of the integral of f , and that will complete the first part of the proof (Fig. 6.8). For the second part of the proof, you consider a function, f , such that for each > 0 you can find a lower step function, v.x/, and an upper step function, u.x/, whose integrals are within of each other. You must then show that f has an integral. The first task is to figure out what value I will serve as the integral of f . Your proof will need to show that Riemann sums for f approach this value of I, so you first
4(b – a) inf f(x) ξj xj–1 Fig. 6.8 Choosing j on .xj1 ; xj /
xj
186
6 Riemann Integrals
need a target I for that purpose. To do this, consider the collection of all possible lower step functions, v.x/. That is, let L D fv j v is a step function with v.x/ f .x/ for all x 2 Œa; bg. Each v 2 L has an integral, and each integral should be less than or equal to the needed value of I. How about taking the least upper bound of all of those integrals? Does the least upper bound exist? It does if the collection of integrals of elements of L is bounded above. To get that, all you need is one upper step function u. For each v 2 L and for each x 2 Œa; b, you know that Rb Rb v.x/ f .x/ u.x/. This ensures that for each v 2 L, v.x/dx u.x/dx showing a
a
that the set of integrals of elements in L is bounded above. That allows you to set Rb I D sup v.x/dx. This makes sense because I would then be greater than or equal v2L a
to the integral of any lower step function. It would also have to be less than or equal to the integral of any upper step function. Since the assumption is that the integrals of lower step functions and upper step functions can be found arbitrarily close to each other, and each integral of an upper step function must be greater than or equal to any integral of a lower step function, you would expect that the least upper bound of the lower step function integrals would be equal to the greatest lower bound of the upper step function integrals, and this value is what you will choose for I. After determining I, your proof can proceed naturally. You need to show that by restricting the norm of a partition of Œa; b, you can force an associated Riemann sum for f to be close to I. What you have at your disposal is the ability to find upper and lower step functions whose integrals are close to each other. A helpful observation is that if you have a lower step function v and an upper step function u, then for any partition and choice of j in the intervals of the partition, you know that n n n P P P v.j /xj f .j /xj u.j /xj . So you can choose upper and lower step jD1
jD1
jD1
functions, u and v whose integrals are each within, say 2 , of I. Then you can choose a norm of a partition so that any Riemann sum for v is within 2 of the integral of v, and any Riemann sum for u is within 2 of the integral of u. That will force the corresponding Riemann sum for f to be within of I completing the proof. PROOF: The function f is integrable on the interval Œa; b if and only if for every > 0 there are step functions, u and v, such that for each x 2 Œa; b, Rb Rb v.x/ f .x/ u.x/ and u.x/dx v.x/dx < . a
a
• Let the function f and the interval Œa; b be given. • Without loss of generality, assume that a < b, for if a D b, the result follows trivially. (continued)
6.8 Step Functions
187
PART I: Integrability implies close upper and lower step functions • Assume that f is an integrable function with
Rb
f .x/dx D I.
a
• Let > 0 be given. • By the definition of Riemann integration, there is a ı > 0 such that for any partition of Œa; b with norm less than ı and any choice of j ’s in the intervals n P of the partition, the associated Riemann sum f .j /xj is within 4 of I. jD1
• Let P W a D x0 < x1 < x2 < < xn D b be such a partition. • Note that since f is integrable, f is a bounded function on the interval Œa; b. • Because f is bounded, for each j D 1; 2; 3; : : : ; n, the value of inf f .x/ exists. Therefore, there exists j 2 .xj1 ; xj / such that f .j / < xj1 0 be given. Because f is continuous on Œa; b, it is uniformly continuous there. Thus, there is a ı > 0 such that jf .x/ f .y/j < 2.ba/ holds for every x and y in Œa; b with jx yj < ı. • Let n be a positive integer with ba < ı. n . • For each j D 0; 1; 2; 3; : : : ; n let xj D a C j ba n • Define step function v.x/ by v.xj / D f .xj / for each j D 0; 1; 2; : : : ; n and v.x/ D min f .y/ for each j D 1; 2; 3; : : : ; n. Thus, v is a lower step
• • • • •
y2Œxj1 ;xj
function for f . • Similarly, define step function u.x/ by u.xj / D f .xj / for each j D 0; 1; 2; : : : ; n and u.x/ D max f .y/ for each j D 1; 2; 3; : : : ; n. Thus, y2Œxj1 ;xj
u is an upper step function for f . • For each j D 1; 2; 3; : : : ; n, because xj xj1 < ı, it follows that max f .y/ min f .y/ 2.ba/ implying that for all x 2 Œa; b, y2Œxj1 ;xj
y2Œxj1 ;xj
. u.x/ v.x/ < ba Rb Rb Rb Rb u.x/ v.x/ dx < • Thus, u.x/dx v.x/dx D a
a
a
a
dx ba
D .
• Therefore, u and v are upper and lower step functions for f whose integrals on Œa; b differ by less than , so it follows that f is integrable on Œa; b which completes the proof. One thing nice about knowing that a function is integrable on an interval Œa; b is that rather than having to consider all partitions of Œa; b, you can determine the value of the function’s integral by using any collection of partitions of Œa; b whose norms approach zero. Thus, if you know that f is integrable on Œa; b, then for every n P natural number n you could calculate I.n/ D f a C .b a/ nj ba which is the n jD1
Riemann sum for f based on the very specific partition where xj D a C .b a/ nj and with j D xj . This is not the more general Riemann sum required by the definition of the integral, but if you already know that the integral exists, then it must be equal to lim I.n/. n!1
As an example, consider the function f .x/ D x which is continuous on the interval Œ0; 4, so you know that it is integrable there. You can then consider n n n P P n.nC1/ j 4 16 P I.n/ D f a C .b a/ nj ba D .4 / D j D 16 . Then 2 n n n 2 n n2 jD1
jD1
lim I.n/ is easily seen to be 8 which is
n!1
jD1
R4 0
x dx. On the other hand, if you try
6.9 Integrals of Continuous Functions
191
f(c)
a
c
b
Fig. 6.9 The mean value theorem for integration
0 x is rational this with the function f .x/ D on the interval Œ0; 1, you obtain 1 x is irrational n P f nj 1n D 1. So lim I.n/ D 1 which is not the integral of f . That I.n/ D n!1
jD1
integral does not exist. Now that it has been established that continuous functions are integrable, it is appropriate to investigate the properties of the integrals of continuous functions. The first of these properties is known as the Mean Value Theorem for Integration. It states that the integral of a continuous function, f , on an interval, Œa; b, is given by the length of the interval, b a, times one of the values f achieves on the interval. Rb That is, there exists a c 2 Œa; b such that f .x/dx D f .c/ .b a/. This result has a a
nice visual interpretation showing that the area under a continuous curve is equal to the area of a rectangle with length b a and width f .c/ for some c 2 Œa; b as shown in Fig. 6.9. Another way to think about this is that there is a c 2 Œa; b such that f .c/ Rb 1 is the mean value of f which could be defined as ba f .x/dx. a
The proof of this theorem follows easily from three earlier results: (1) the Intermediate Value Theorem, (2) a continuous function on a closed interval takes on its extreme values, and (3) if one integrable function is greater than or equal to a second integrable function, then the integral of the first is greater than or equal to the integral of the second. The proof starts with a function f continuous of an interval Œa; b. That function achieves its minimum value K and its maximum value M on the interval. Thus, for all x 2 Œa; b, it follows that K f .x/ M from which it follows Rb that .b a/K f .x/dx .b a/M. Then, by the Intermediate Value Theorem, a
on the interval Œa; b the function f achieves every value between K and M including Rb 1 f .x/dx. ba a
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6 Riemann Integrals
PROOF (Mean Value Theorem for Integration): Assume the function f is continuous on the interval Œa; b with a < b. Then there is c 2 Œa; b Rb 1 f .x/dx. satisfying f .c/ D ba a
• Let f be a function continuous on the interval Œa; b with a < b. • Because f is continuous on the interval Œa; b there are s and t in Œa; b such that f .s/ D K is the minimum value for f on Œa; b, and f .t/ D M is the maximum value for f on Œa; b. Rb • Since for all x 2 Œa; b, K f .x/ M, it follows that K.b a/ D K dx Rb
f .x/dx
Rb
a
a
M dx D M.b a/.
a 1 ba
• Because f .s/ D K
Rb
f .x/dx M D f .t/, the Intermediate Value
a
Theorem says that there is a c between s and t such that f .c/ D
1 ba
Rb
f .x/dx.
a
• Thus, c 2 Œa; b satisfies the needed requirement and completes the proof. It can be very exciting to take a first course in Calculus. After learning what a limit is, you learn about two very different-looking limit processes: the derivative and the integral. Both differentiation and integration have important applications which justify the amount of attention they receive. But then comes the seemingly amazing revelation that these two processes, although they are defined in extremely different ways, are, in fact, very closely related in that they are essentially inverse operations of each other. This fact is the point of the Fundamental Theorem of Calculus, often presented as the pinnacle of the first course in Calculus. The Fundamental Theorem of Calculus starts with a function f integrable on Œa; b. The result of the theorem is generally stated in two parts. The first part Rx defines a new function F.x/ D f .t/dt and states that if f is continuous at some a
point c 2 .a; b/, then F 0 .c/ D f .c/. The second part states that if f is continuous on Œa; b, and if F is any function satisfying F 0 .x/ D f .x/ for all x 2 Œa; b, then Rb f .x/dx D F.b/ F.a/. It is fairly straightforward to prove the second part using a
the first part. To prove the first part of the theorem, you would assume that a function f is integrable on an interval Œa; b and that f is continuous at c 2 .a; b/. To find Rx the derivative of F.x/ D f .t/dt at c, you would just apply the definition of a
the derivative. That is, you would start with the difference quotient F.x/F.c/ D xc x x c R R R 1 1 f .t/dt f .t/dt . This simplifies to xc f .t/dt. Now if you knew that f xc a
a
c
6.9 Integrals of Continuous Functions
193
were continuous between c and x, you could apply the just completed Mean Value Theorem for Integration to conclude that this difference quotient is equal to f .y/ for some y between c and x. Then by forcing x to be close to c, you could force f .y/ to be close to f .c/ to complete the proof. But you do not know that f is continuous between c and x; only that f is continuous at c. Still this is enough. You can use the continuity of f at c to say that for a given > 0 there is a ı > 0 that ensures that if t satisfies jt cj < ı, then jf .t/ f .c/j < . This shows that for x within Rx Rx Rx 1 1 1 ı of c, xc .f .c/ /dx < xc f .t/dx < xc .f .c/ C /dx which simplifies to c
f .c/ <
1 xc
Rx
c
c
f .t/dx < f .c/ C , and the result follows.
c
PROOF (Fundamental Theorem of Calculus: Part I): Assume the function f is integrable on the interval Œa; b and continuous for some c 2 .a; b/. Rx Then the function F.x/ D f .t/dt is differentiable at c with F0 .c/ D f .c/. a
• Let f be a function integrable on the interval Œa; b and continuous at some c 2 .a; b/. Rx • For x 2 Œa; b, define F.x/ D f .t/dt. a
• Let > 0 be given. • From the definition of continuity, there is a ı > 0 such that if x 2 Œa; b with jx cj < ı, then jf .x/ f .c/j < . • Select any x 2 Œa; b with 0 < jx cj < ı. Rx Rc Rx • Then F.x/ F.c/ D f .t/dt f .t/dt D f .t/dt. a
a
c
• Since f .c/ < f .t/ < f .c/ C for all t between c and x, it follows that Rx Rx Rx 1 1 1 .f .c//dx < xc f .t/dx < xc .f .c/C/dx D f .c/C. f .c/ D xc c c c ˇ ˇ ˇ ˇˇ ˇ Rx ˇ ˇ 1 • Thus, ˇ F.x/F.c/ f .c/ˇ D ˇˇ xc f .t/dx f .c/ˇˇ < . xc c
• This proves that F 0 .c/ D lim F.x/F.c/ D f .c/ completing the proof of the xc x!c theorem. The second part of the Fundamental Theorem of Calculus now follows easily. Rx Indeed, if f is continuous on Œa; b, then the function F.x/ D f .t/dt is an a
antiderivative of f , that is, a function whose derivative is f . If G.x/ is any other antiderivative of f , then G0 .x/ D F 0 .x/ on Œa; b. It follows from the Mean Value Theorem (for derivatives) that G and F differ by a constant because G F has a derivative that is identically 0. Thus, F.x/ F.a/ D G.x/ G.a/ for all x 2 Œa; b Rb showing that f .t/dt D G.b/ G.a/ for any antiderivative G. a
194
6 Riemann Integrals
PROOF (Fundamental Theorem of Calculus: Part II): Assume the function f is continuous on the interval Œa; b and that F is any antiderivative Rb of f . Then f .x/dx D F.b/ F.a/. a
• Let f be a function continuous on the interval Œa; b. • Without loss of generality, a < b. Rx • Define F.x/ D f .t/dt. a
Since f is continuous at each x 2 Œa; b, it follows that F 0 .x/ D f .x/. Let G be any antiderivative of f . Then for all x 2 Œa; b, the derivative of G.x/ F.x/ is f .x/ f .x/ D 0. By is a c 2 .a; b/ such that G.b/ F.b/ the Mean Value Theorem there G.a/ F.a/ D f .c/ f .c/ .b a/ D 0. Rb • Thus, G.b/ G.a/ D F.b/ F.a/ D F.b/ D f .x/dx which completes the
• • • •
a
proof. The importance of the Fundamental Theorem of Calculus cannot be overstated. It turns the complex operation of finding limits of difficult to calculate Riemann sums into the somewhat more routine job of finding antiderivatives of functions.
6.9.1 Exercises 1. If F.x/ D
Rx3 x2
t dt, 1Ct2
find F 0 .x/.
2. Suppose f has a jump discontinuity at c 2 Œa; b (that is, lim f .x/ and lim f .x/ x!cC
x!c
both exist and are unequal). If f is integrable on Œa; b, what is the behavior of Rx F.x/ D f .t/dt at c? a
3. Suppose f is integrable on Œa; b. If the derivative of F.x/ D c 2 Œa; b, what can you say about f at c?
Rx
f .t/dt exists at
a
6.10 Characterization of Integrable Functions A function continuous on the closed interval Œa; b is integrable there. Some functions which are not continuous are still integrable, so the question is, how badly can a function behave and still be integrable? If a continuous function is changed
6.10 Characterization of Integrable Functions
195
at one point, it is no longer continuous, but changing a function at a single point does not affect whether or not it is integrable. If a function has a jump discontinuity at a point (that is, it has a right limit and a left limit at the point, but those two limits are not equal) but is continuous elsewhere, then the function is still integrable. This is because if a function is integrable on Œa; b and integrable on Œb; c, then it is integrable on Œa; c whether or not the function is continuous at b. It follows that bounded piecewise continuous functions are integrable. Let the function f be defined on an interval Œa; b. Define the set of discontinuities of f , Df , to be the subset of Œa; b where f fails to be continuous. For example, 0 x D 15 ; 25 ; 35 ; 45 , then Df D f 15 ; 25 ; 35 ; 45 g. If on the interval Œ0; 1 if f .x/ D x otherwise 0 x is rational f .x/ D , then Df is the entire set Œ0; 1 because f is discontinuous 1 x is irrational 0 x is in the Cantor set everywhere. Finally, if f .x/ D , then Df is equal to 1 x is not in the Cantor set the Cantor set because for any x not in the Cantor set, there is an open interval containing x such that f is identically 0 on that open interval. These examples suggest that a function defined on Œa; b is integrable as long as its set of discontinuities does not get too large. In fact, a function defined on Œa; b is Riemann integrable if and only if it is bounded and its associated set of discontinuities, Df , has measure zero. Thus, any bounded function which is discontinuous only on a countable set of points must be Riemann integrable. The first function in the preceding paragraph with Df D f 15 ; 25 ; 35 ; 45 g is, therefore, integrable. The second function which has Df D Œ0; 1 is not integrable as seen earlier in this chapter. The third function which has Df equal to the Cantor set is interesting because its set of discontinuities is not countable, yet the function is integrable because the Cantor set does have measure zero. The statement that the Riemann integrable functions on Œa; b are exactly those whose set of discontinuities has measure zero is a biconditional statement. It says both that if a function is Riemann integrable, then it is bounded with a set of discontinuities that has measure zero, and that if a function is bounded with a set of discontinuities that has measure zero, then the function is Riemann integrable. Thus, a proof of this statement will have two parts, one for each conditional. The proof of the theorem is somewhat longer than others seen in this book, but it requires only one new concept not yet discussed. Assume first that the function f is defined on the interval Œa; b, is bounded, and its set of discontinuities, Df , has measure zero. You can prove that f is integrable on Œa; b if you can show that for every > 0 the function f has upper and lower step Rb functions, u and v, such that u.x/ v.x/dx < . The key point here is that f is well a
behaved near points where it is continuous, and the set where it is not well behaved is very small (has measure zero). The strategy, then, is to construct step functions, u and v, so that u.x/ v.x/ is very small near points where f is continuous, and
196
6 Riemann Integrals
to limit the size of the intervals where u.x/ v.x/ is large. Suppose, for example, that near points where f is continuous, you could limit u.x/ v.x/ to be less than . Then the total contribution to the integral of u v over those sections of the 2.ba/ .b a/ D 2 . The function f is bounded, so step functions would be at most 2.ba/ there is an M such that jf .x/j < M for all x 2 Œa; b. It is possible, therefore, to define upper and lower step functions that differ by at most 2M at points of Df . If you can limit the regions where u.x/ v.x/ is large to intervals whose total length is at most 4M , then the total contribution to the integral of u v over those sections of the step functions would be at most 2M 4M D 2 . Accomplishing both of these goals would then show that the integral of u v is less than 2 C 2 D . Can this be accomplished? By the definition of continuity, for each point x where f is continuous there is a ı > 0 such that if y is in Œa; b with jy xj < ı, then jf .y/ f .x/j < 4.ba/ . That would ensure that for any two values y1 and y2 in the interval .x ı; x C ı/, the difference jf .y1 /f .y2 /j jf .y1 /f .x/jCjf .x/f .y2 /j < 4.ba/ C 4.ba/ D 2.ba/ . By the definition of measure zero, the set of discontinuities of f can be covered by a collection of open intervals whose total length is less than the needed 4M . Thus, each point of Œa; b can be covered by one of the open intervals covering Df or by one of these .x ı; x C ı/ intervals constructed at each point of continuity. The Heine– Borel Theorem then lets you reduce this covering of Œa; b with open intervals to a finite subcovering, and from that subcovering, the appropriate upper and lower step functions can be constructed. That completes the strategy for the first part of the proof. Assume, conversely, that the function f defined on the interval Œa; b is integrable. You already know that this implies that jf j is bounded by some constant M, so all you need to prove is that the set of discontinuities of f , Df , has measure zero. This can be done with a proof by contradiction. That is, by assuming that Df does not have measure zero, you can show that for any upper and lower step functions, u and v, the integral of u v is bounded away from 0. To do this it is helpful to consider how much f can vary near a particular value x. For a point x 2 Œa; b and a ı > 0, you would like to know how much f can change over the interval .x ı; x C ı/. So define Wı .x/ D sup f .y/ inf f .y/ where the supremum and infimum are calculated for y varying over the interval .x ı; x C ı/ \ Œa; b. Note that if f had upper and lower step functions that were both constant on the interval .x ı; x C ı/, then the two step functions would have to differ by at least Wı .x/ on that interval. Now define the variation of a function f at a point x to be W.x/ D lim Wı .x/. Since ı!0C
0 Wı .x/ 2M is nonincreasing as ı ! 0C , the limit W.x/ always exists and is equal to inf Wı .x/. The following lemma gives an important property of W.
6.10 Characterization of Integrable Functions
197
PROOF: Let f be any bounded function defined on the interval Œa; b. Then for any x 2 Œa; b, the variation of f at x is 0 if an only if f is continuous at x. • Let f be a bounded function defined on the interval Œa; b. PART I: Continuity implies W D 0 • Assume that for some x 2 Œa; b the function f is continuous at x. • Then for every > 0 there is a ı > 0 such that if y 2 .x ı; x C ı/ \ Œa; b, then jf .y/ f .x/j < 2 . Thus, Wı .x/ < .jf .x/j C 2 / .jf .x/j 2 / D . • Thus, there are ı for which Wı .x/ is within of 0 implying that W.x/ D inf Wı .x/ D 0. PART II: W D 0 implies continuity • Assume that for some x 2 Œa; b the variation of f at x is W.x/ D 0. • Since lim Wı .x/ D 0, for every > 0, there is a ı > 0 such that for ı!0C
0 < ı < ı , Wı .x/ < . • Select ı > 0 with ı < ı . • Then for any y 2 .x ı; x C ı/ \ Œa; b, it follows that jf .y/ f .x/j sup f .z/ inf f .z/ D Wı .x/ < . jzxj 1n g to be the points of Œa; b where the variation of f at x is greater than 1n . If the variation of f at x is positive, then it must be greater than 1n for some n. Thus, the set of all discontinuities of f must be the union of these Dnf 1
sets, that is, Df D [ Dnf . The key observation here is that if for each n the Dnf set nD1
has measure zero, then the entire set of discontinuities, Df , must have measure zero because it is just a countable union of sets with measure zero. So, if you assume that Df does not have measure zero, it requires that there is a natural number n such that Dnf also does not have measure zero. What does it mean for Dnf not to have measure zero? It means that there is an > 0 such that no collection of open intervals with total length less than can cover all of Dnf . This will be the key to showing that upper and lower step functions for f cannot have integrals that are arbitrarily close to each other, and thus, f cannot be integrable. The result is known as Lebesgue’s Theorem.
198
6 Riemann Integrals
PROOF (Lebesgue’s Theorem): The function f defined on the interval Œa; b is Riemann integrable if and only if f is bounded and the set of points in Œa; b where f is discontinuous has measure zero. • Let f be a function defined on the interval Œa; b with a < b. PART I: Boundedness and discontinuities with measure zero imply integrable • Assume that there is a real number M such that jf .x/j < M for all x 2 Œa; b, and assume the set Df , the set of x 2 Œa; b such that f is discontinuous at x, has measure zero. • Let > 0 be given. • By the definition of measure zero, there is a sequence of open intervals I1 ; I2 ; I3 ; : : : with total length less than 4M such that Df is contained in the union of those intervals. • By the definition of continuity, for each x 2 Œa; b where f is continuous, there is a ıx > 0 such that jf .y/ f .x/j < 4.ba/ for all y 2 Œa; b with jy xj < ıx . Let Jx be the interval .x ıx ; x C ıx /. • Since each x 2 Œa; b is either a point of continuity of f or a member of Df , each x 2 Œa; b is either a member of one of the intervals Ij that covers Df or in the interval Jx . Thus, the collection of open intervals consisting of I1 ; I2 ; I3 ; : : : together with the Jx intervals forms an open covering of Œa; b. • By the Heine–Borel Theorem, there exists a finite collection of these open intervals than covers Œa; b. Let E D fx1 ; x2 ; x3 ; : : : ; xn g be the set of distinct endpoints for the intervals in this finite cover of Œa; b where x1 < x2 < x3 < < xn . • Define step functions u.x/ and v.x/ as follows. If x D xj for one of the endpoints xj 2 E, then define u.x/ D v.x/ D f .x/. • For each j the open interval .xj1 ; xj / must be a subset of one of the finite number of intervals that cover Œa; b. If .xj1 ; xj / is contained in one of the Ik intervals that covers Df , define u.x/ D M and v.x/ D M for each x 2 .xj1 ; xj /. Since jf j is bounded by M, v.x/ f .x/ u.x/ for each x 2 .xj1 ; xj /. • Otherwise, .xj1 ; xj / is contained in one of the Jx intervals. In this case, define u.y/ D f .x/ C 4.ba/ and v.y/ D f .x/ 4.ba/ for each y 2 .xj1 ; xj /. Since jf .y/ f .x/j < 4.ba/ for all y 2 Jx , it follows that v.y/ < f .y/ < u.y/ for each y 2 .xj1 ; xj /. • It follows that v is a lower step function of f , and u is an upper step function of f . n Rb Rxj P • u.x/ v.x/dx D u.x/ v.x/dx. a
jD2 xj1
(continued)
6.10 Characterization of Integrable Functions
199
• Over the intervals that were subsets of the Ij intervals, u.x/ v.x/ D 2M. The total length of such intervals cannot exceed 4M . As a result, the integral of u.x/ v.x/ over these intervals cannot exceed 2M 4M D 2 . • Over the intervals that were subsets of the Jx intervals, u.x/ v.x/ < 2.ba/ . As a result, the integral of u.x/ v.x/ over these intervals cannot exceed Rb D 2 . 2.ba/ a
• Thus, f has upper and lower step functions, u and v, with the property that Rb u.x/ v.x/dx < 2 C 2 D . a
• Therefore, f is Riemann integrable on Œa; b. PART II: Integrable implies bounded and discontinuities with measure zero • Let f be Riemann integrable on Œa; b. • Since all integrable functions are bounded, there is an M such that jf .x/j < M for all x 2 Œa; b. • For each x 2 Œa; b, let Wı .x/ D sup f .y/ inf f .y/ where the supremum and infimum are calculated for y varying over the interval .x ı; x C ı/ \ Œa; b. • For each x 2 Œa; b define the variation of f at x to be W.x/ D lim Wı .x/. ı!0C
By the preceding lemma, W.x/ is 0 if and only if f is continuous at x 2 Œa; b. • For natural number n define Dnf D fx 2 Œa; b j W.x/ > 1n g to be the points of Œa; b where the variation of f at x is greater than 1n . 1
• The set of discontinuities of f is then Df D [ Dnf . nD1
• Assume that Df does not have measure zero. • A countable union of sets with measure zero is itself a set with measure zero. Since Df is the union of the Dnf , there must exist a natural number n such that Dnf does not have measure zero. • Since Dnf does not have measure zero, there is an > 0 such that if Dnf is covered by a sequence of open intervals, the total length of those intervals must exceed . • Let u be an upper step function for f and v be a lower step functions for f . From the definition of step function, there is a sequence a D x0 < x1 < x2 < < xk D b such that both u and v are constant on the open intervals Ij D .xj1 ; xj / for each j D 1; 2; 3; : : : ; k. • For each x 2 Ij , u.x/ cannot be less than sup f .z/, and v.x/ cannot be greater z2Ij
than inf f .z/. As a consequence, u.x/ v.x/ sup f .z/ inf f .z/. Thus, if z2Ij
Dnf
\ Ij is not empty, then u.x/ v.x/
z2Ij 1 n
z2Ij
for all x 2 Ij . (continued)
200
6 Riemann Integrals
• Dnf cannot be covered by open intervals whose total length is less than . Thus, it follows that the total length of the intervals Ij that contain points of Df must be at least . Rb • It follows that u.x/ v.x/dx 1n . a
• Thus, f cannot have upper and lower step functions whose integrals differ by less than n . This implies that f is not integrable which is a contradiction. Therefore, the assumption that Df does not have measure zero is false, which completes the proof. The last section of Chap. 4 introduced Thomae’s function, a function defined on Œ0; 1 which is discontinuous at each rational number but is continuous at each irrational number. Since the rational numbers is a countable set, it has measure zero. Thus, Thomae’s function is bounded, and its set of discontinuities has measure zero, so Thomae’s function is Riemann integrable. Compare this to the function that is equal to 1 for all rational numbers and equal to 0 for all irrational numbers. That function is discontinuous everywhere, so its set of discontinuities does not have measure zero, and it is not integrable as seen earlier.
6.10.1 Exercises 1. Suppose f W Œ0; 2 ! Œ5; 9 is integrable and g W Œ5; 9 ! Œ0; 2 is continuous. Show that g ı f is integrable. Write proofs for each of the following statements. 2. If f .x/ is a function integrable on the interval Œ0; 10, then so is the function f .x/f .10 x/. 3. If f .x/ is a function integrable on the interval Œa; b, and p.x/ is a polynomial, then p f .x/ is also integrable on Œa; b. 4. If f .x/ and g.x/ are integrable functions on the interval Œa; b, then so is f .x/g.x/.
Chapter 7
Infinite Series
7.1 Convergence of Infinite Series The axioms for the real numbers define addition as a binary operation and establish the rules for adding two real numbers together. One can use mathematical induction to extend axioms and theorems about addition to get theorems about the addition of any finite number of terms. But there is nothing in the axioms that suggests how to add an infinite number of terms together or what such a sum would mean. You need to make a separate definition in order to make sense out of adding infinitely 1 P many terms together. An infinite series a1 C a2 C a3 C D an has a sequence nD1
of terms a1 ; a2 ; a3 ; : : : which are written with plus signs or minus signs between the terms of the sequence. In this chapter, most series will begin with a first term a1 , although there is no problem with beginning the series at other subscript values 1 P such as the commonly seen an . Also in this chapter the terms of the series will nD0
be real numbers, although it is possible to extend the definition to series of other kinds of terms such as complex numbers or matrices. This explains what an infinite series looks like, but it does not prescribe any meaning to the symbols. In Abstract Algebra one can study formal power series, a study that looks at one type of infinite series and considers how to manipulate the series without regard to whether these series can be assigned any meaningful numerical values. But in Analysis, one is interested in the cases where it makes sense to assign a numerical value to the series. The difference in the two studies is in the interpretation of a series like 1 2 C 3 4 C . If you ask what happens if you multiply this series by 2, a purely algebraic answer would be that you just use the Distributive Law and multiply each term of the series by 2 to get 2 4 C 6 8 C . But an analytical answer to the question is that it makes little sense to assign a numerical value to the series, so multiplying the series by 2 cannot yield a meaningful result.
© Springer International Publishing Switzerland 2016 J.M. Kane, Writing Proofs in Analysis, DOI 10.1007/978-3-319-30967-5_7
201
202
7 Infinite Series
To each series a1 C a2 C a3 C , one associates the sequence of partial sums s1 D a1 s2 D a1 C a2 s3 D a1 C a2 C a3 ::: sk D a1 C a2 C a3 C C ak : Since each of these sums is just the sum of a finite number of terms, they are easily defined. The series is said to converge to real number L if the sequence of partial 1 P sums converges to L, that is, if lim sk D L. In this case one writes an D L and k!1
nD1
says that the series has limit L or even that the series has value L. If the sequence of partial sums does not converge, then the series is said to diverge. If the limit of partial sums converges to infinity or negative infinity, the series is said to diverge to 1 P infinity or negative infinity, respectively. In that case one could write an D 1 or nD1
1 P
an D 1.
nD1
The definition of convergence suggests that for each series derive a simple expression for its partial sums sk D
k P
1 P
an one should
nD1
an and then calculate the
nD1
limit of the partial sums lim sk . Unfortunately, there are relatively few series that k!1
admit simple closed-form expressions for their partial sums, and this technique for finding the value of a series has limited use. Still, it is important to know about some of the cases when this technique does work. Perhaps the best known examples of series whose partial sums can be explicitly calculated are the geometric series. These are the series whose sequence of terms can be written in the form an D arn1 , where a and r are given real numbers. Then the first term of the series is a and the n1 an common ratio of adjacent terms is an1 D ar D r, at least in the interesting cases arn2 when ar ¤ 0. When r is not equal to 1, there is a simple algebraic trick that gives the expression for the partial sums. sk D
k X
arn1
nD1
r sk D
k X nD1
arn
7.1 Convergence of Infinite Series
203
sk r sk D
k X
arn1
nD1
k X
arn D a ark
nD1
sk .1 r/ D a.1 r / k
sk D a
1 rk : 1r
(Of course, there is an even simpler trick for the case when r D 1.) Thus, except in the trivial case where a D 0, the limit of the partial sum diverges if jrj 1. On a which can easily be the other hand, when jrj < 1, lim rk D 0 so lim sk D 1r k!1
k!1
remembered as “the first term divided by 1 minus the common ratio.” The geometric series is particularly important because one can often compare other series to a geometric series to determine if the other series converges. It also gives a nice example showing that series that make a lot of sense when they converge can lead you to very strange and very incorrect conclusions when they do not converge. In 1 P r particular, rn D 1r whenever jrj < 1. But when you take a limit as r approaches nD1
1, you get lim series
1 P
1 P
r!1 nD1
rn D lim
r r!1 1r
D
1 2
which is not the same as the nonsensical
lim rn D .1/ C 1 C .1/ C 1 C .1/ C 1 C .
nD1 r!1
Another class of series whose partial sums can be calculated are the telescoping series. This is a class of series where each term an can be written as a difference of two terms an D bn bnC1 . Then sk D .b1 b2 / C .b2 b3 / C .b3 b4 / C C .bk bkC1 / D b1 C.b2 Cb2 /C.b3 Cb3 /C.b4 Cb4 /C .bk C bk / bkC1 D b1 bkC1 . Hence, if lim bkC1 exists, the series converges. The best known example k!1 1 1 P P 1 1 1 1 D 1 lim nC1 D nC1 D 1. of this type is the series 2 n n Cn nD1
nD1
n!1
Fortunately, even though it is often difficult to determine the exact values for the partial sums of a series, one can very often determine whether or not the series converges and sometimes the value to which it converges even without knowing an explicit formula for its partial sums. There are many tools that can be used to do this. These tools consist of a large collection of convergence tests which can be applied to determine if a particular series converges. Calculus students often get a great deal of practice selecting appropriate convergence tests for series. This chapter will be more interested in proving the theorems that provide these tests. The simplest and possibly most important convergence test is the Limit of Terms Test which says that a series can converge only if its sequence of terms has a limit 1 P of 0. That is, if an converges, then lim an D 0. This is a direct consequence of nD1
the fact that if
1 P
nD1
n!1
an converges, then its sequence of partial sums converges.
204
7 Infinite Series
The point is, if lim sk exists, then is a Cauchy sequence whose term must get k!1
close to each other, and so sk sk1 D ak must approach 0. PROOF (Limit of Terms Test): The series
1 P
an converges only if
nD1
lim an D 0.
n!1
• Assume that the series
1 P
an converges to the limit L.
nD1
• Then the sequence of partial sums sk D
k P
an converges to L.
nD1
• This implies lim an D lim .sn sn1 / D lim sn lim sn1 D L L D 0 n!1 n!1 n!1 n!1 which completes the proof. The convergence of one series can often be inferred from the convergence of a similar series. For example, inserting extra terms equal to 0 into a series does not affect whether the series converges, nor can inserting extra 0 terms affect the value to which the series converges. This is because the insertion of terms equal to 0 into a series does not change the sequence of partial sums for that series except to allow some of the partial sums to be repeated, and that does not change the limit of the sequence of partial sums. Another useful observation is that if two series differ in only a finite number of terms, then either both series converge or both series diverge. Suppose, for example, 1 1 P P that an and bn are two series such that for some positive integer N, the terms nD1
nD1
an D bn for all n > N. Why would the convergence of one of the series imply the convergence of the other? It must depend on the convergence of their partial k k P P sums, so let sk D an and tk D bn be the sequences of partial sums for nD1
nD1
the two series. The agreement of an and bn for all n > N shows that for k > N, k k P P tk D t N C bn D tN C an D tN C sk sN . Thus, lim sk exists if and only nDNC1
nDNC1
k!1
if lim sk C tN sN D lim tk exists. k!1
k!1
7.1.1 Exercises Find limits for the following series or show that the limit does not exist. 1. 2.
1 P nD1 1 P nD1
5 3n 4 22nC1
7.2 Absolute and Conditional Convergence
3. 4. 5.
1 P nD1 1 P nD1 1 P nD1
p
23
205
n
7 n2 C5n 1 n2 C9nC14
1 1 1 1 1 1 1 1 C 2 C 3 3 4 C 4 C 2 3 22 3 2 3 2 3 1 1 1 1 7. C 0 C C 0 C 0 C C 0 C 0 C 0 C C 0 C 0 C 0 C 0 2 4 8 16 1 1 1 1 1 5 C C C C 8. 11 C 3 C C 9 12 23 34 45 56
6.
7.2 Absolute and Conditional Convergence Before going on, it is necessary to distinguish between two types of convergent 1 P series. The series an is said to be absolutely convergent if the series of the nD1
absolute value of its terms, convergent. That is, if
1 P
1 P
jan j, converges. All absolutely convergent series are
nD1
jan j converges, then so does
nD1
1 P
an . This can be proved
nD1
by using the fact that a series converges if and only if its sequence of partial sums is 1 1 P P Cauchy. The proof can begin with the absolutely convergent series an , so jan j converges. Then, knowing that sequence of partial sums Sk D
nD1
1 P
nD1
jan j converges, the proof can conclude that its
nD1 k P
jan j converges. From this you need to reach the
nD1
conclusion that the original series
1 P nD1
that the sequence of partial sums sk D
an converges, that is, you need to conclude k P nD1
an converges. But if converges, it
must be Cauchy, which means that for each > 0 there is an N such that for any m > k > N, jSm Sk j < . What do you need to know for to be Cauchy? You need to know that for each ˇ> 0 there ˇ is an mN such that for any m > k > N, m ˇ ˇ P P ˇ jsm sk j < . But jsm sk j D ˇ an ˇˇ jan j D jSm Sk j which you nDkC1 nDkC1 already know can be made small. It follows that the sequence is Cauchy, so it converges.
206
7 Infinite Series
PROOF: If the series
1 P
an is absolutely convergent, then it converges.
nD1
• Assume that the series series
1 P
1 P
an is absolutely convergent which means that the
nD1
jan j converges.
nD1
• For each k > 0 let Sk D
k P
jan j and sk D
nD1
k P
an .
nD1
• Then, since the series is absolutely convergent, the sequence converges implying that is a Cauchy sequence. • Given > 0 there is an N such that for all m and k greater than N, jSm Sk j < . ˇ ˇ m m ˇ ˇ P P • Let m > k > N. Then jsm sk j D ˇˇ an ˇˇ jan j D jSm Sk j < . nDkC1 nDkC1 • Thus, the sequence is a Cauchy sequence and is, therefore, a convergent sequence. 1 P • This shows that an is convergent which proves the theorem. nD1
If the series
1 P
an converges, but it is not absolutely convergent, then it is called
nD1
conditionally convergent. An absolutely convergent series converge because its terms get small fast enough that its partial sums must rapidly get close to each other and to a limit. A conditionally convergent series converges because its negative terms balance the growth of its positive terms. For example, the series 1 1 C 12 1 C 13 13 C 14 14 C clearly converges to 0 due to this type of cancelation. 2 Thus, every series can be categorized as either absolutely convergent, conditionally convergent, or divergent.
7.3 The Arithmetic of Series Because the definition of the convergence of a series involves the limit of partial sums, many results that are true for finite sums are easily proved for infinite sums. 1 1 1 P P P For example, if an converges, and c is any constant, then can D c an . To nD1
prove this you would have to consider the partial sums Law works for finite sums, so c
1 P nD1
k P nD1
an .
can D c
k P nD1
k P
nD1
nD1
c an . But the Distributive
nD1
an , and the limit of this is the needed
7.3 The Arithmetic of Series
207 1 P
PROOF: If the series 1 P
c an D c
nD1
1 P
an converges, and c is any real number, then
nD1
an .
nD1
• Assume that • Then
1 P
an converges, and that c is a real number.
nD1
1 P
can D lim
k P
k!1 nD1
nD1
k P
can D lim c k!1
an D c lim
k P
k!1 nD1
nD1
an D c
1 P
an
nD1
proving the result. Another easy result is that if 1 P
.an C bn / D
nD1
1 P
1 P
an C
nD1
1 P
an and
nD1
1 P
bn are both convergent series, then
nD1
bn . Again, this is easy because the result follows
nD1
immediately from properties of finite sums. PROOF: If the series 1 P
an C
nD1
1 P
an and
nD1
1 P
1 P
bn both converge, then
nD1
1 P
.an C bn /D
nD1
bn .
nD1
• Assume that the series • Then lim
1 P nD1 k P
k!1 nD1
1 P
an and
nD1
.an C bn / D lim
an C lim
k P
k!1 nD1
k P
1 P
bn both converge.
nD1
.an C bn / D lim
k!1 nD1
bn D
1 P
nD1
an C
k!1
1 P
k P
an C
nD1
k P
bn
D
nD1
bn proving the result.
nD1
With these theorems you can often start with a series whose value you know and derive the values of other series. For example, what is the value of the 1 1 1 1 1 series 1 C 12 14 C 18 C 16 32 C 64 C 128 256 ? This series looks something like the geometric series with first term 1 and common ratio 12 which is 1 C 12 C 14 C 18 C . That series has limit 11 1 D 2. But the new series 2 is clearly not a geometric series because the terms are not all the same sign, which would be the case for a geometric series with a positive common ratio, nor are the terms alternating in sign, which would be the case for a geometric series with a negativecommon ratio. The new series can be written, though, as 2 2 1 C 12 C 14 C 18 C 0 C 0 C 24 C 0 C 0 C 32 C 0 C 0 C 256 C . This is the difference of two series: the geometric series with first term 1 and common ratio 1 , and a series whose value is the same as a geometric series with first term 24 D 12 2 1 1 10 2 . and common ratio 18 . Thus, the new series has value D 1 1 7 1 2 1 8
208
7 Infinite Series
The series 1 12 C 13 14 C 15 16 C is a conditionally convergent series. In the next chapter it will be shown that this series converges to ln 2. So how about the 1 series 1 C 13 12 C 15 C 17 14 C 19 C 11 16 C ? At first it appears that this series is the same as the previous series because it includes the same terms rearranged in a 1 1 different order. Indeed, both series include the terms 2n1 and 2n for each positive integer n. But one can write 1C
1 1 1 1 1 1 1 1 1 1 1 C C C C C D 1 C 0 C C 3 2 5 7 4 9 11 6 3 2 5 1 1 1 1 1 C D C0C C C0C 7 4 9 11 6 1 1 1 1 1 1 C C C 2 3 4 5 6 1 1 1 1 C 0 C C 0 C 0 C C 0 C 0 C D 2 4 6 8 1 1 1 1 1 1 C C C 2 3 4 5 6 1 1 1 1 1 1 C 1 C C C D 2 2 3 4 5 6 ln 2 C
1 3 ln 2 D ln 2: 2 2
It is not unusual that rearranging the order of terms in a series results in the series converging to a different quantity. This is, in fact, a characteristic of all conditionally convergent series as will be shown later in this chapter.
7.3.1 Exercises 1. Prove that if the series
1 P
an and
nD1
1 P
bn both converge, and c and d are real
nD1
numbers, then 1 1 1 P P P .c an C d bn / D c an C d bn . nD1
2. Prove that if the series
nD1
1 P
nD1
converges to 0.
nD1
an converges, then its sequence of tails tn D
1 P mDn
am
7.4 Tests for Absolute Convergence
209
3. Find the value of each of the following series. 1 1 1 1 1 1 1 (a) 1 C C 2 3 C 4 C 5 C 6 7 C 3 3 3 3 3 3 3 1 1 1 1 1 1 C C C (b) 23 34 45 56 67 78 1 1 1 1 1 1 1 1 1 C C C C C C (c) 2 4 3 6 8 5 10 12 7 1 1 1 1 1 1 1 1 1 (d) 1 C C C C C C C C 3 5 7 2 9 11 13 15 4
7.4 Tests for Absolute Convergence There are many tests for the convergence of series. Presented here are four very useful tests that apply to series whose terms are all positive real numbers. Of course, 1 1 P P since the convergence of jan j implies the convergence of an , these tests can nD1
nD1
be thought of as tests for the absolute convergence of series.
7.4.1 Comparison Test After the Limit of Terms Test, the Comparison Test is likely the most important convergence test because it is used to prove most of the other convergence tests. It states that if the terms of one series are less than or equal to the corresponding terms of a second series, then the convergence of the second series implies the convergence 1 1 P P of the first series. Specifically, suppose there are two series an and bn , and for each n, the terms satisfy 0 an bn . Then if
1 P
nD1
nD1
bn converges, it follows that
nD1
1 P
an
nD1
must converge. The contrapositive of this statement is then also true and states that 1 1 P P if an diverges, then bn must also diverge. nD1
nD1
Consider how you would prove that this test is valid. The proof would assume that 1 P 0 an bn for each n, and assume that bn converges. Then it must show that 1 P
nD1
an converges. One shows that a series converges by showing that its sequence
nD1
of partial sums converges. You do know that the sequence of partial sums for
1 P
bn
nD1
converges, so how can you use that to make a conclusion about the partial sums of 1 P an ? One idea is to use the technique from the proof that absolutely convergent nD1
210
7 Infinite Series
series are convergent; that is, a series converges if and only if its sequence of partial 1 m P P sums is Cauchy. If the partial sums of bn form a Cauchy sequence, then bn nDk
nD1
gets small whenever k m are large. Now, the given fact that an bn lets you m m 1 P P P conclude that an bn which implies that the partial sums of an are nDk 1 P
Cauchy. Thus,
nDk
nD1
an must converge.
nD1
PROOF (Comparison Test): Suppose that
P
P
an and
nD1
bn are series
nD1
with nonnegative terms and N is a real number such that for every 1 P integer n > N, the terms satisfy 0 an bn . Then if bn converges, so does
1 P
nD1
an .
nD1
• Assume that
P nD1
an and
P
bn are series with nonnegative terms.
nD1
• Assume that there is an N such that for every n > N, the terms of the series satisfy 0 an bn . 1 P • Assume that the series bn converges. nD1
• This means that the sequence of partial sums
k P
bn converges and is,
nD1
therefore, a Cauchy sequence. • Thus,ˇ given > 0 ˇthere is an M N such that if M < k m, then m k m ˇP ˇ P P > ˇˇ bn bn ˇˇ D bn . nD1 nD1 nDkC1 m m P P • But then whenever M < k m, it follows that > bn an D m P nD1
an
k P
nDkC1
nDkC1
an .
nD1
• This implies that the sequence of partial sums of • Therefore, the sequence of partial sums of
1 P
1 P
an is Cauchy.
nD1
an converges, so the series
nD1
converges. • This proves that the Comparison Test is valid.
The Comparison Test can be used in many cases when you are faced with a series which is similar to a series that you know converges. For example, you already know 1 P 1 that the series converges because it forms a telescoping series. Can this fact n2 Cn nD1
7.4 Tests for Absolute Convergence
be used to show that the series
211
1 P nD1
1 n2
converges? Well, the Comparison Test cannot
be applied directly because for each n you have n2 1Cn < n12 which is not what you need. You need to find a convergent series whose terms are greater than or equal to 1 or a divergent series whose terms are less than or equal to them. You have neither. n2 2 2 On the other hand for each positive integer n, it is true that n12 D n2 Cn 2 n2 Cn . 1 P 2 is twice a convergent series, so it is also convergent. Thus, the The series n2 Cn nD1
Comparison Test shows that
1 P nD1
1 n2
converges.
In this way the Comparison Test can be used to simplify the task of testing the convergence of many complicated looking series. As another example, consider the 1 P 2nC7 series . Note that the first two terms of this series are negative. Because n3 5nC1 nD1
the convergence of a series does not depend on the value of any finite set of its terms, it is sufficient to test the series by considering the terms where n 3. In the terms n32nC7 the degree of the polynomial in the denominator is 3 while the 5nC1 degree of the polynomial in the numerator is 1. This suggests that the terms could be compared to the terms n12 of a known convergent series. The strategy is to compare 2nC7 to a fraction that is greater but look more like n12 . If the series with greater n3 5nC1 fractions converges, the Comparison Test shows that the original series converges. This can be done by attempting to eliminate lower degree terms of the numerator and denominator polynomials, thus, ending up with a simpler fraction greater than the original. Clearly, when considering the numerator 2n C 7, the constant term, 7, will be dwarfed by the size of the linear term 2n suggesting that you replace 2n C 7 by the larger quantity 2n C 7n D 9n. This replacement will result in a larger fraction, but it should not affect whether or not the series converges. Similarly, it would be good to replace the denominator n3 5n C 1 with a smaller polynomial of the same degree which will result in obtaining a fraction larger than n32nC7 . One 5nC1 can drop the constant term altogether, but one cannot drop the 5n term without making the denominator polynomial larger. This can be handled by writing n3 as 1 n3 C 12 n3 . For large enough values of n, the value of 12 n3 will exceed 5n making 2 1 n3 5n a positive quantity which could be removed from the polynomial to make 2 the polynomial smaller. Indeed, you need 12 n3 5n 0 implying n2 10. Thus, if n 4, you can conclude that n3 5n C 1 > 12 n3 . This shows that for n 4, the 1 1 P P 1 1 18 < 19n D 18 . fraction n32nC7 2 . Since the series 2 converges, so does 3 5nC1 n n n2 n 2
Thus, by the Comparison Test, the series
1 P nD1
nD1
2nC7 n3 5nC1
nD1
converges.
As a final example of using the Comparison Test, consider the series 12 C 14 C 1 1 1 1 1 1 1 1 C 18 C 18 C 18 C 18 C 16 C 16 C 16 C 16 C 16 C 16 C 16 C 16 C . For each 1 k k k 0 this series has 2 terms equal to 2kC1 , and these 2 terms add to 12 . Thus, the 1 4
212
7 Infinite Series
sequence of partial sums for this series contains the subsequence 12 ; 1; 32 ; 2; 52 ; 3; which clearly diverges. Thus, the series diverges. Now, compare this series to the 1 P 1 harmonic series and note that each term of the harmonic series is greater than n nD1
or equal to the corresponding term of the first series. Thus, by the Comparison Test, the harmonic series must diverge.
7.4.2 Ratio Test The geometric series
1 P
a rn1 converges whenever jrj < 1. The Ratio Test
nD1
essentially is an application of the Comparison Test where one compares terms of 1 P a series to the terms of an appropriate geometric series. Suppose an is a series nD1
a
with positive terms, and suppose that the sequence of ratios of adjacent terms, nC1 an has limit L as n approaches infinity. If L < 1, then the series can be compared to a convergent geometric series with a common ratio between L and 1. If L > 1, then the terms of the series increase in value and do not approach 0, so the series diverges. When L D 1, the ratio test fails because there are series for which L D 1 that converge and other series for which L D 1 that diverge. 1 P To prove that the Ratio Test is valid, you would start by assuming that an nD1
a
is a series with positive terms such that lim nC1 D L < 1. Then you would n!1 an compare this series to a well-chosen geometric series known to converge so that 1 P the Comparison Test can be used to conclude that an converges. To compare the nD1
given series to the geometric series with nth term arn1 , you would need to know that for all n greater than some N, the terms an are less than arn1 . If you know that anC1 is always less than r, then the an terms will grow more slowly than the arn1 an terms, and the Comparison Test can be used. In general, r cannot be set equal to L a because knowing that nC1 approaches L in the limit does not ensure that the ratio an is ever actually smaller than L and certainly not that it is always less than L. But if a the limit of nC1 is L, then by the definition of limit, there is an N such that for all an , which is half n N, the ratio is less than some value greater than L such as LC1 2 way between L and 1. Then, if the value of a is chosen so that aN a, you will have aNCk ark for each k 0, and the result follows.
7.4 Tests for Absolute Convergence
213
PROOF (Ratio Test): Suppose that
1 P
an is a series of positive terms such
nD1
that lim anC1 D L. Then if L < 1, the series converges, if L > 1, the series n!1 an diverges, and if L D 1 the test fails. 1 P
• Assume that
an is a series of positive terms such that lim
n!1
nD1
anC1 an
D L.
CASE 1: L < 1 a
• If L < 1, there is an integer N such that nC1 < LC1 for all n N. an 2 LC1 • Let a D aN , and r D 2 < 1. a • Assume that for some k 0, aNCk ark . Then NCkC1 < r, so aNCkC1 < aNCk k kC1 aNCk r ar r D ar . • Therefore, by mathematical induction it follows that aNCk ark for all k 0. 1 P • Since arn is a convergent geometric series, it follows from the ComparnD0
1 P
ison Test that
an is also convergent.
nD1
CASE 2: L > 1 a
• If L > 1, there is an integer N such that nC1 > 1 for all n N. an • Then for all n N, anC1 > an > 0, so the sequence of terms increases from aN and cannot have a limit of 0. • Therefore, the series diverges by the Limit of Terms Test. CASE 3: L D 1 • Note that the constant series • The series
1 P nD1
1 P
1 diverges, and lim
n!1
nD1 1 n2
converges, and lim
n!1
anC1 an
anC1 an
n2 2 n!1 .nC1/
D lim
1 n!1 1
D lim
D 1.
D 1.
• Therefore, no conclusion can be drawn when L D 1, and the Ratio Test fails. The ratio test is not helpful for series where the nth term is a rational function of n a because the limit of nC1 will always be 1, and the test is inconclusive. The ratio test an is particularly useful for series whose nth terms involve powers or factorials. For 5nC1 1 n P .nC1/Š 5 , you get lim D example, when you apply the ratio test to the series 5n nŠ lim 5 n!1 nC1
nD1
D 0 < 1, so the series converges.
n!1
nŠ
214
7 Infinite Series
Note that rather than requiring lim that lim sup n!1
anC1 an
n!1
anC1 an
to have a limit, it is enough to assume
< 1 to assure that the series converges and lim inf n!1
anC1 an
> 1 to
assure that the series diverges. The proofs of these facts are left as exercises, but they are important refinements of the Ratio Test since the lim inf and lim sup always exist even if the limit does not. For example, consider the series 1 C 23 C 13 C 322 C a 1 C 323 C 313 C . For this series, the ratio nC1 oscillates between 23 and 12 , so the an 32 limit of the ratio does not exist. But the lim sup of the ratio is 23 < 1 implying that the series converges. The Ratio Test will play a major role in the discussion of power series in the next chapter.
7.4.3 Root Test The Root Test is similar to the Ratio Test and can often be used for the same series for which the Ratio Test can be used. This is because, like the Ratio Test, it compares a series to a geometric series. For some series where the general term an involves the nth powers of expressions, the Root Test can be easier to apply than the Ratio Test. To test a series with positive terms an with the Root Test, you calculate p the limit lim n an D L. Then, as with the Ratio Test, if L < 1, the series converges, n!1 if L > 1, the series diverges, and if L D 1, the test fails. Proving that the Root Test is valid is very straightforward. Given that p p lim n an D L < 1, there is an integer N such that for all n N the root n an is n!1 n < 1. Then, for n N, the terms an are less than LC1 , the terms less than LC1 2 2 of a convergent geometric series. Thus, the series converges by the Comparison Test. 1 P PROOF (Root Test): Suppose that an is a series of positive terms such nD1 p that lim n an D L. Then if L < 1, the series converges, if L > 1, the series n!1 diverges, and if L D 1 the test fails. 1 P
• Assume that
nD1
an is a series of positive terms such that lim
n!1
p n a D L. n
CASE 1: L < 1 p • If L < 1, there is an integer N such that n an < LC1 for all n N. 2 LC1 n • Then, for n N, each term an < , the corresponding term of a 2 < 1. geometric series with common ratio LC1 2 • Therefore, since the geometric series converges, the Comparison Test shows 1 P that an converges. nD1
(continued)
7.4 Tests for Absolute Convergence
215
CASE 2: L > 1 p • If L > 1, there is an integer N such that n an > LC1 > 1 for all n N. 2 LC1 n • Then, for all n N, an > 2 which diverges to infinity. • Therefore, the series diverges by the Limit of Terms Test. CASE 3: L D 1 • Note that the constant series • The series
1 P nD1
1 P
1 diverges, and lim
nD1 1 n2
p n
n!1
converges, and lim
n!1
p n a D lim n
n!1
an D lim 1 D 1. n!1
1 p n 2. n
• Since natural function is continuous at 1, it follows that logarithmp thep ln lim n an D lim ln n an D lim 2 lnn n . Then by L’Hopital’s Rule, n!1 n!1 n!1 2 p this limit is lim 1n D 0, from which it follows that lim n an D 1. n!1 n!1 • Therefore, no conclusion can be drawn when L D 1, and the Root Test fails. p As with the Ratio Test, it is sufficient to know that lim sup n an < 1 to conclude n!1 p that the series converges, and that lim inf n an > 1 to conclude that the series n!1
diverges. For example, the series 12 C 13 C 212 C 312 C 213 C 313 C has general term p p a2n D 31n and a2n1 D 21n . Thus, lim n an does not exist, but lim sup n an D p1 n!1
n!1
2
which is less than 1, so the series converges.
7.4.4 Integral Test The definition of the Riemann integral considers the integrals of functions over closed bounded intervals, Œa; b. This definition can be extended to integrals on an infinite interval. An improper Riemann integral of the first kind defines integrals over intervals where one or both of the endpoints of the interval are infinite. R1 Rb Rb Rb One defines f .x/dx as lim f .x/dx. Similarly, f .x/dx D lim f .x/dx and
R1
a
f .x/dx D
1 lim 1x jb1 b!1
D 1.
lim
b!1 a Rb
lim
a!1 b!1 a
1 R1
f .x/dx. For example,
1
1 x2
dx D
a!1 a Rb lim x12 b!1 1
dx D
After seeing a definition of the improper Riemann integral of the first kind, the reader may be curious whether there is also an improper Riemann integral of the second kind. Although this text will not need to deal with improper Riemann integrals of the second kind, the definition is given here for completeness. Recall that Riemann integrals over an interval Œa; b exist only if the integrand is bounded. So, an improper integral of the second kind is an integral where the integrand is unbounded in every neighborhood of a point c 2 Œa; b. In this case, the Riemann
216
7 Infinite Series
integral on Œa; b can be calculated on a region that excludes c and then the limit can R4 be taken as the region expands toward c. For example, one would define p1x dx as lim
R4
a!0C a
1 p
0
p dx D lim 2 xj4a D 4. x a!0C
The Integral Test for the convergence of a series of positive terms involves the comparison of an infinite series with an improper Riemann integral. It applies to series whose terms are equal to a monotonically decreasing function f defined on an interval Œa; 1/ such that for all n a, the nth term of the series an is equal to the function at the point n, that is, an D f .n/. The following figure makes this comparison clear. Let k be an integer greater than or equal to a. If f is a monotonically decreasing function, then whenever n x > k, the function Rn Rn f .x/ f .n/ D an showing that f .x/dx f .n/dx D f .n/ D an . Thus, by the Comparison Test, the series
1 P
n1
n1
1 nC1 R P
an converges if the series
nD1
Then, because f is a positive function,
1 nC1 R P
f .x/dx D
nDk n
f .x/dx converges.
nDk n R1
f .x/dx. Alternatively, if
k
f is a monotonically decreasing function, then whenever x n > k, the function nC1 nC1 R R f .x/dx f .n/dx D f .n/ D an . Thus, f .x/ f .n/ D an showing that n
n
again by the Comparison Test, the improper integral converges if the series
1 P
f .n/ converges. Therefore, the series
if and only if the improper integral R1 kC1
f .x/dx
1 P nDkC1
an
R1
R1
1 nC1 R P
f .x/dx D
k
nDk
k a,
R1
1 P
f .x/dx
nDk n
an converges
nD1
f .x/dx converges. Moreover, for any integer
k
f .x/dx giving a fairly narrow range for the value
k
of the infinite series and a good way to obtain an approximation to the value of the series. This is helpful because it is often easier to evaluate the integral than the corresponding infinite series (Fig. 7.1).
Fig. 7.1 Comparing the series with the integral in the Integral Test
7.4 Tests for Absolute Convergence
217
PROOF (Integral Test): Suppose f is a positive monotonically decreasing 1 P an is a series such that function on the interval Œa; 1/. Suppose nD1
an D f .n/ for all n greater than or equal to an integer k a. Then the 1 R1 P series an converges if and only if the improper integral f .x/dx nD1
k
converges.
• Suppose f is a positive monotonically decreasing function on the interval Œa; 1/. 1 P • Suppose an is a series such that an D f .n/ for all n greater than or equal nD1
to an integer k a. • Because f is monotonically decreasing, for any n > k it follows that f .x/ f .n/ for all x 2 Œn 1; n. Rn Rn • Thus, for any n k it follows that an D f .n/ D f .n/dx f .x/dx. • By the Comparison Test the series 1 P
Rn
f .x/dx D
nDkC1 n1
R1
1 P
n1
n1
an converges if the series
nD1
f .x/dx converges.
k
• Because f is monotonically decreasing, for any n k it follows that f .n/ f .x/ for all x 2 Œn; n C 1. nC1 nC1 R R • Thus, for any n > k it follows that f .x/dx f .n/dx D f .n/ D an . • By the Comparison Test the series if the series
1 P
nC1 R
nDkC1 n
n
f .x/dx D
R1
f .x/dx converges
kC1
an converges.
nD1
• Thus, the series R1
n 1 P
1 P
an converges if and only if the improper integral
nD1
f .x/dx converges, proving that the Integral Test is valid.
k
As an example, consider the collection of p-series which are the series
1 P nD1
1 np
where p is some constant greater than 0. For which p does the p-series converge? You have already seen that it converges when p D 2 and diverges when p D 1, the harmonic series. All the p-series can be handled at once using the Integral Test. Indeed, since the function f .x/ D x1p is monotonically decreasing in x for each p > 0, R1 the p-series converges exactly when the integral x1p dx converges. But the integral 1
218
7 Infinite Series
is easy to calculate. When p ¤ 1,
R1 1
1 dx xp
1 1 D 1p xp1 j1 . This is infinite when p < 1
but converges to 1p when p > 1. When p D 1, the integral is ln xj1 1 which is infinite. Thus, by the Integral Test, the integral and the series converge exactly when p > 1. Consider the p-series when p D 2. The value of this series can be estimated using the integral estimate associated with the Integral Test. The estimate would be 1 1 1 R1 1 R1 P P P 1 1 1 dx > > x12 dx or 1 C 1 > a1 C > 1 C 12 , so is between x2 n2 n2 n2 1
2
nD2
nD2
nD1
1.5 and 2. This is not very precise, but one can apply this technique a few terms 1 1 R1 R1 P P 1 1 farther down the series to get x12 dx > > x12 dx which shows that n2 n2 10
nD11
11
is between 1.6406 and 1.649. In fact, the limit of the series is
2 6
nD1
1:6449.
7.4.5 Exercises 1. Suppose that
1 P
bn is a convergent series,
nD1
1 P
an is a series, and there are
nD1
1 P
constants N and K such that 0 an Kbn for all n > N. Prove that
an
nD1
converges. 2. Suppose that
1 P
bn is a convergent series with positive terms,
nD1
1 P
an is a series
nD1 an n!1 bn
with positive terms, and lim
1 P
D L for some real number L. Prove that
an
nD1
converges. This is sometimes called the Limit Comparison Test. 1 P a 3. Assume that an is a series of positive terms that satisfies lim sup nC1 an nD1
D L < 1. Prove that 4. Assume that
1 P
lim sup n!1
anC1 an
an converges.
an is a series of positive terms that satisfies lim inf n!1
D L > 1. Prove that 1 P
n!1
nD1
nD1
5. For the series
1 P
1 P
anC1 an
an diverges.
nD1 1 C 221 C 212 C 222 C 213 C 223 C 214 C 224 21 a lim inf nC1 . What can you conclude about the an n!1
an D
C calculate
nD1
and
the series? 6. Assume that
1 P
convergence of
an is a series of positive terms that satisfies lim sup
nD1
D L < 1. Prove that
1 P nD1
n!1
an converges.
p n a n
7.5 Alternating Series Test
7. Assume that
1 P
219
an is a series of positive terms that satisfies lim sup
nD1
D L > 1. Prove that
1 P
p n a n
n!1
an diverges.
nD1
1 P 8. For the series an D 211 C 312 C 213 C 314 C 215 C 316 C 217 C 318 C calculate nD1 p p lim sup n an and lim inf n an . What can you conclude about the convergence of n!1
n!1
the series? 9. Use the integral estimate from the Integral Test to estimate the size of the series 1 P 1 . n3 nD1
10. Determine which of the following series are absolutely convergent by applying an appropriate convergence test. (a) (b) (c) (d) (e) (f) (g)
1 P nD1 1 P nD1 1 P
7n6 18n4 C12n2 183 n10 5n5 19 n2 C5 n3 5 3n 2n C5n
nD1 1 P nD1 1 P nD1 1 P nD1 1 P nD1
n 3
nŠ .2n/Š 5n nŠ nn .nŠ/2
7.5 Alternating Series Test So what can you do with a series which is not absolutely convergent? There are fewer tools to handle conditionally convergent series. One tool that does help is the Alternating Series Test which considers series whose terms alternate in sign. Specifically, if the absolute values of the terms of the series are monotonically decreasing to 0, and the signs of the term alternate, then the series converges. For example, the series seen earlier 1 12 C 13 14 C 15 16 C satisfies these conditions. The series formed by the absolute values of these terms 1 C 12 C 13 C 14 C 15 C 16 C is the harmonic series which does not converge, so the given series is not absolutely convergent. Seeing how the partial sums of this series behave will give you an idea how to prove that the Alternating Series Test is valid. In particular, the first few partial sums of this series are
220
7 Infinite Series
1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
Fig. 7.2 Converging to ln 2 with an alternating series
s1 D 1 D 1 1 2 1 1 s3 D 1 C 2 3 1 1 1 s4 D 1 C 2 3 4 1 1 1 1 s5 D 1 C C 2 3 4 5 s2 D 1
1 2 5 D 6 7 D 12 47 D : 60 D
The progression can be seen graphically in Fig. 7.2. Notice that the partial sums of an odd number of terms are all greater than the limit, ln 2, while the partial sums of an even number of terms are all less 1than the 1 1 limit. Also, if n is odd, then snC2 D sn C nC1 D sn .nC1/.nC2/ < C nC2 sn , showing that the partial sums of an odd number of terms forms a decreasing 1 1 1 D sn C .nC1/.nC2/ sequence. Similarly, if n is even, then snC2 D sn C nC1 nC2 > sn , showing that the partial sums of an even number of terms forms an increasing nC1 sequence. Because the terms of the series .1/n approach 0, the odd partial sums and the even partial sums approach each other. They both form bounded monotonic sequences which both converge to the common limit. This behavior is typical of all series satisfying the hypothesis of the Alternating Series Test.
7.5 Alternating Series Test
221
PROOF (Alternating Series Test): Suppose
1 P
an is a series such that
nD1
lim an D 0, and for each n 1, an and anC1 have opposite signs, and jan j janC1 j. Then the series converges.
n!1
• Assume that
1 P nD1
an is a series such that lim an D 0, and for each n 1, an n!1
and anC1 have opposite signs, and jan j janC1 j. • Without loss of generality, assume that a1 > 0. k P • Let the series have partial sums sk D an . nD1
• Note that if n 1 is odd, then anC1 is negative and anC2 is positive with janC1 j janC2 j implying that snC2 D sn C .anC1 C anC2 / sn . • Similarly, if n 1 is even, then anC1 is positive and anC2 is negative with janC1 j janC2 j implying that snC2 D sn C .anC1 C anC2 / sn . • Thus, the subsequence of odd numbered partial sums forms a monotonically decreasing sequence while the subsequence of even numbered partial sums forms a monotonically increasing sequence. • Because the subsequence of even numbered partial sums is increasing, when n is an odd positive integer it follows that sn > snC1 s2 showing the subsequence of odd numbered partial sums is bounded below by s2 implying that that sequence converges to a limit L1 . • Similarly, the subsequence of even number partial sums is an increasing sequence that is bounded above by s1 implying that that sequence converges to a limit L2 . • Then L1 L2 D lim s2nC1 s2n D lim a2nC1 D 0 showing that L1 D L2 n!1 n!1 and that the odd numbered partial sums and the even numbered partial sums both converge to the same limit. • Therefore, the sequence of partial sums converges and the series converges. This proof not only says that the given alternating series converges; it gives a way to estimate the limit of the series. For any series that satisfies the hypothesis of the theorem, any two adjacent partial sums, sn and snC1 , are on opposite sides of the limit L of the series. Thus, the distance that sn is from the limit of the series is less than the distance sn is from snC1 , and that distance is just janC1 j. Therefore, it is easy to remember that for these series, the distance that a partial sum is from the limit of the series is no more than the first term that is not part of the sum, janC1 j. Note that the Alternating Series Test for convergence and this limit estimate apply to series without regard to whether the series is absolutely convergent or conditionally convergent. For example, the number 1e D 0Š1 1Š1 C 2Š1 3Š1 C . This is an absolutely convergent series as seen by the ratio test. But it is also a series whose terms alternate in sign, and the absolute values of the terms decrease monotonically to 0. Thus, 1 the partial sum of the series 0Š1 1Š1 C 2Š1 3Š1 C 4Š1 is already within 100 of 1e because 1 the first neglected term is 5Š1 D 120 . This technique gives an easy proof that the
222
7 Infinite Series
number e is irrational. It goes like this: If e were rational, then it could be expressed as pq , where p and q are positive integers. Then 1e D qp D 0Š1 1Š1 C 2Š1 3Š1 C . Multiplying both sides of this equation by pŠ yields q.p 1/Š D pŠ pŠ C pŠ2 pŠ3 C 1 1 ˙ 1 pC1 C .pC1/.pC2/ . Thus, the integer q.p 1/Š would be an integer
1 1 .pC1/.pC2/ C . But this infinite series is an alternating series plus (or minus) pC1 1 where the absolute value of the terms decrease to 0, so its value is between pC1 and 1 1 .pC1/.pC2/ . Thus, there would have to be an integer between those two values, pC1 something clearly not possible. This is a contradiction, so the assumption that e is rational must be false.
7.5.1 Exercises Determine which of the following series are conditionally convergent, absolutely convergent, or divergent. 1. 1 2. 1 3. 1 4. 1 C 5. 1 C
1 C ln13 ln14 C ln 2 1 C 3 ln1 3 4 ln1 4 C 2 ln 2 1 C 3.ln13/2 4.ln14/2 C 2.ln 2/2 p1 p1 C p1 C p1 p1 C 3 2 5 7 4 p1 p1 C p1 C p1 p1 C 3 4 5 7 8
p1 9 p1 9
C C
p1 11 p1 11
p1 C 6 p1 C 12
7.6 The Smallest Divergent Series Recall that the p-series
1 P nD1
1 np
converges when p > 1 and diverges otherwise. This
raises a natural question about whether there is, in some sense, a largest series that converges, or, perhaps a smallest series that diverges. If there were, that might provide a good series to use in the Comparison Test because all series smaller would converge, and all series larger would diverge. This turns out not to be the case. For 1 P every series of positive terms, an , that diverges, there is a sequence of positive nD1
numbers that converges to 0 such that the series
1 P
an bn also diverges. In
nD1
fact, one can take bn D
1 sn
where sn is the nth partial sum of
series diverges, then sn goes to infinity, so bn D
1 sn
1 P nD1
goes to 0.
an . Clearly, if the
7.6 The Smallest Divergent Series
223
To prove this result you would begin with a divergent the series with positive 1 P terms, an . Because the series is divergent, you know that the sequence of partial nD1
sums must diverge to infinity. The strategy is to show that the partial sums of the new 1 P an series are not Cauchy. In particular, for every integer m, there is an integer k sn nD1
such that
k P
nDmC1
1 . 2
1 2
>
an sn
showing that the mth and kth partial sums differ by at least
Suppose you are given a positive integer m. Since the original series diverges, there is a positive integer k such that sk > 2sm . Then the difference between the kth and the mth partial sums of the new series is sk sm sk
>1
1 2
nDmC1
D 12 . 1 P
PROOF: Let k P
sums sk D
an . Then the series
nD1
sk D
k P
an sn
k P
>
nDmC1
k P
an sk
D
an
nDmC1
sk
D
an be a divergent series with positive terms and partial
nD1
1 P nD1
• Assume that
k P
1 P
an sn
also diverges.
an is a divergent series with positive terms and partial sums
nD1
an .
nD1
• Let m be any positive integer. • Since the partial sums sn diverge to infinity, there is a positive integer k such that sk > 2sm . 1 P an • Then the difference between the mth and kth partial sums of the series sn nD1
is
k P nDmC1
an sn
>
k P
k P
an sk
nDmC1
D
nDmC1
sk
an
D
sk sm sk
>1
1 2
D 12 .
• This shows that the sequence of partial sums of the series
1 P nD1
an sn
is not a
Cauchy sequence, so it cannot converge. 1 P an • Therefore, the series diverges. sn nD1
For example, consider the harmonic series 1 C 12 C 13 C 14 C 15 C which diverges. The Integral Test suggests that the kth partial sum of this series is close to ln k. If for all n > 1, the nth term, 1n , of the harmonic series is divided by ln n, 1 P 1 . The Integral Test shows that this series also diverges the resulting series is nln n since the integral
R1 2
nD2
1 dx xln x
D ln.ln x/j1 2 D 1 diverges.
224
1 P
7 Infinite Series
It is interesting to note that even though for positive termed divergent series 1 1 P P an an , the series also diverges, for positive termed series an , the series sn
nD1 1 P nD1
nD1
nD1
always converges. To see this, note that for n > 1 the term
an s2n
an s2n
D
sn sn1 s2n
<
1 P an Thus, the terms of the series are less than the terms of a sn1 s2 nD1 n 1 P 1 convergent telescoping series s1n D s11 lim s1n . Whether the original sn1 sn sn1 sn sn1
D
1
1 . sn
n!1
nD2
series diverges so that sn goes to infinity, or it converges to a finite value L so that sn goes to L, the telescoping series converges.
7.7 Rearrangement of Terms 7.7.1 Addition of Parentheses The series 11C11C11C does not converge. Yet, if you insert parentheses to group some of the terms together, it can result in a convergent series such as .11/C .11/C.11/C which converges to 0 or 1C.1C1/C.1C1/C.1C1/C which converges to 1. So, inserting parentheses can turn a divergent series into a convergent series. Equivalently, removing parentheses from a convergent series can 1 P turn it into a divergent series. What if the series an converges? Can inserting nD1
parentheses change whether or not it converges or change the limit to which the 1 P series converges? The answer to this is no. The point is, if an converges, it means nD1
that its sequence of partial sums converges. By inserting parentheses into the series, you are just removing some of the terms in the sequence of partial sums. You end up with a new series whose sequence of partial sums is a subsequence of the sequence 1 P of partial sums of an , and any such subsequence will converge to the same limit nD1
as the original series. Slightly more can be said. Suppose
1 P
an is a series whose terms approach 0. If
nD1
parentheses are inserted in such a way that the number of terms contained within each set of parentheses is bounded, then the insertion of parentheses cannot affect whether the series converges or the limit to which the series converges. To see this 1 P assume that each set of parentheses encloses at most K terms. If an converges nD1
to L, then, as suggested above, no insertion of parentheses can affect the limit of 1 P the series. So suppose that the series an diverges, and that its partial sums are sk D
k P nD1
nD1
an . Because lim an D 0, for each > 0, there is an N such that for all n!1
7.7 Rearrangement of Terms
225
n > N, the size of the terms jan j must be less than K . Suppose that for some m > N one term of the series with parentheses added is .amC1 CamC2 CamC3 C CamCk /. Then sm and smCk are both partial sums for the series with parentheses added. For any j D 1; 2; 3; : : : ; k, jamC1 CamC2 CamC3 C CamCj j K j , showing that for any of those j, jsmCj sm j < . The sequence of partial sums for the original series does not converge either because the sequence is unbounded or because its lim sup and lim inf approach distinct values. Because the subsequence of partial sums for the original series remains within of the subsequence corresponding to the series with parentheses added, the subsequence must also either be unbounded or have distinct lim sup and lim inf values. Thus, the series with parentheses added cannot converge. This observation can be very helpful. Consider again the series 1 C 13 12 C 15 C 1 1 14 C 19 C 11 16 C . This series is not absolutely convergent, and it does not 7 satisfy the hypothesis of the Alternating Series Yet, if parentheses are inserted Test. 1 to group each set of three terms: 1 C 13 12 C 15 C 17 14 C 19 C 11 16 C , 1 1 4n3 C 2n1 1n D n.2n1/.2n3/ . The series with one gets a general term equal to 2n3 4n3 terms n.2n1/.2n3/ converges absolutely as can be seen by comparing it to the p4n3 4n series with p D 2 since, for n 3, one has n.2n1/.2n3/ < n.2nn/.2nn/ D n42 . So, the series with parentheses added converges, and since each set of parentheses contains a maximum of three terms, and the terms of the original series approach 0, this means that the original series converges. Of course, if the number of terms enclosed by sets of parentheses is not bounded, one cannot draw the same type of conclusions. The series .1/ C . 12 C 12 12 12 / C . 13 C 13 C 13 13 13 13 / C . 41 C 14 C 14 C 14 14 14 14 14 / C converges, but if parentheses are removed, the series diverges even though its terms do approach 0. The partial sums oscillate between 1 and 2.
7.7.2 Order of Terms It has already been shown that the terms of the series 1 12 C 13 14 C 15 16 C 1 16 C to get a series that can be rearranged as 1 C 13 12 C 15 C 17 14 C 19 C 11 converges to a different limit. This is typical for a conditionally convergent series. 1 P In fact, if an is a conditionally convergent series, and L is any real number, then nD1
there is a rearrangement of the terms of this series that converges to L. To prove this, first note that a conditionally convergent series must have both positive and negative terms. Define two new series so that for each n, bn D an if an 0 and bn D 0 if an < 0, and cn D bn an . Then, for each n, both bn 0 and cn 0, 1 1 1 1 1 P P P P P and an D .bn cn /. Since jan j D .bn C cn /, it must be that both bn and
nD1 1 P nD1
nD1
cn diverge to infinity.
nD1
nD1
nD1
226
7 Infinite Series
Suppose you are given a target limit L. You have isolated the positive terms of the series, the bn terms, and the negative terms of the series, the cn terms, so you can play a cute game by taking a few bn terms such that the sum of those terms exceeds L, and then subtract off a few cn terms until the sum decreases below L. You can then add on more bn terms to make the sum again exceed L, and subtract of a few cn terms until the sum decreases below L. Thus, by alternating between adding on bn terms and subtracting off cn terms, you can arrange for the resulting series to have limit L. More precisely, construct a new series inductively as follows: select u1 so that u1 P bn > L. This is always possible because the series with bn terms diverges to nD1
infinity. Then select v1 to be the least positive integer such that
v2 to be the least positive integer such that
u2 P
bn
nD1
v2 P
bn
nD1
nD1
u2 P
Then select u2 to be the least positive integer such that
u1 P
bn v1 P
v1 P
cn < L.
nD1
cn > L, and
nD1
cn < L. For k 2, having
nD1
selected uk and vk , select ukC1 and vkC1 to be the least positive integers such that uP vP vk kC1 kC1 P bn cn > L, and bn cn < L. It is then the case that the series
uP kC1 nD1
nD1
nD1
nD1
b1 C b2 C b3 C C bu1 c1 c2 c3 cv1 C bu1 C1 C bu1 C2 C bu1 C3 C C bu2 cv1 C1 cv1 C2 cv1 C3 cv2 C is a rearrangement of the terms of the original series with some extra 0 terms added. Since the terms of the series approach 0, the partial sums of the series approach L. This provides the desired rearrangement (Fig. 7.3).
b1
b2
b3
b4
b5
c3
c2
b6
b7
c1
b8
b9
c6
c5 c4
b8
c8
0
Fig. 7.3 Rearranging terms to converge to L
c7
L
7.7 Rearrangement of Terms
PROOF: Let
1 P
227
an be a conditionally convergent series, and let L be any
nD1
real number. Then there is a rearrangement of the terms of the series which converges to L. • Let
1 P
an be a conditionally convergent series, and let L be any real number.
nD1
• For each n, define bn D an if an 0, and bn D 0 if an < 0, and define cn D bn an . • Thus, for each n, bn 0, cn 0, and an D bn cn . 1 1 1 1 P P P P • Because an is conditionally convergent, jan j D bn C cn nD1
diverges. Thus, at least one of and because
1 P
nD1
1 P
bn and
nD1 1 P
.bn cn / D
nD1
1 P
nD1
nD1
cn must diverge to infinity,
nD1
an converges, both series must diverge to
nD1
infinity. • The Limit of Terms Test shows that lim an D 0 and, thus, lim bn D n!1 n!1 lim cn D 0. n!1 1 P • Because bn is unbounded, there is a least positive integer, u1 , such that u1 P
nD1
bn > L.
nD1
• Because u1 P nD1
bn
1 P nD1 v1 P
cn is unbounded, there is a least positive integer, v1 , such that cn < L.
nD1
• Having selected uk and vk for some k 1, let ukC1 > uk be the least uP v1 kC1 P positive integer such that bn cn > L. Then let vkC1 > vk be the nD1
least positive integer such that
uP kC1
nD1
bn
nD1
vP kC1
cn < L. Thus, by mathematical nD1 can be constructed so that for vk P
induction, the sequences and uP uk vk kC1 P P each k, bn cn < L and bn nD1
nD1
nD1 u1 P
cn > L.
nD1
dn be given by nD1 c1 ; c2 ; c3 ; : : : ; cv1 ,
• Let the terms of the new series
terms b1 ; b2 ; b3 ; : : : ; bu1
followed by the terms followed by the terms bu1 C1 ; bu1 C2 ; bu1 C3 ; : : : ; bu2 follows by the terms cv1 C1 ; cv1 C2 ; cv1 C3 ; : : : ; cv2 , and so forth, alternating between the sequence of bn terms for uk < n ukC1 and the sequence of cn terms for vk < n vkC1 . (continued)
228
7 Infinite Series
• The resulting series 1 P
1 P
dn is a rearrangement of the terms in the series
nD1
an with some terms equal to 0 inserted.
nD1
• Given > 0 there is an N1 u1 such that if n > N1 , then bn < , and there is an N2 such that if n > N2 , then cn < . • Then there is a k1 such that uk1 > N1 and a k2 such that vk2 > N2 . • Let k D max.k1 ; k2 /, and let N D uk C vk . • Then for all m > N, either there is an r such that dm D bp for some p with N1 < ur < p urC1 or is anˇ s such that dm D cq for some q with N2 < ˇ there m ˇP ˇ vs < q vsC1 . Thus, ˇˇ dn Lˇˇ is bounded by either max cvr ; burC1 < nD1 or by max .bus ; cvs / < . 1 P • This shows that dn converges to L implying that a rearrangement of the series
1 P
nD1
an converges to L as claimed.
nD1
This theorem takes care of the case of conditionally convergent series, but what happens when terms of an absolutely convergent series are rearranged? The answer is that nothing happens; that is, every rearrangement of an absolutely convergent 1 P series converges to the same limit. Suppose, for example, the series an is absolutely convergent with rearrangement
1 P
bn . Because
nD1
> 0 there is an integer N such that for all k N, limit. Alternatively,
1 P nDNC1
jan j < . Because
1 P
k P
1 P
nD1
jan j converges, given
nD1
jan j is within > 0 of its
nD1
bn is a rearrangement of
nD1
1 P
an ,
nD1
there is an integer K such that all the terms a1 ; a2 ; a3 ; : : : ; aN are among the terms k k P P b1 ; b2 ; b3 ; : : : ; bK . So, if k K, by how much can an and bn differ? Both nD1
nD1
sums contain the terms a1 ; a2 ; a3 ; : : : ; aN , so the two sums differ only by a finite 1 P number of the terms aNC1 ; aNC2 ; aNC3 ; : : : which add to at most jan j < . nDNC1
This shows that the series and its rearrangement have partial sums within of each other and completes the argument.
7.7 Rearrangement of Terms
PROOF: Let
1 P
229
an be a series that converges absolutely to L. Then every
nD1
rearrangement of the series also converges to L. • Let • Let
1 P
an be a series that converges absolutely to L.
nD1 1 P
bn be any rearrangement of the series
nD1 1 P
1 P
an converges absolutely, given > 0 there is an integer N such
• Since
nD1
that if k N,
k P
jan j is within of its limit,
nD1
1 P
an .
nD1
1 P
jan j. This means that
nD1
jan j < .
nDNC1
• Since
1 P
bn is a rearrangement of
nD1
1 P
an , there is an integer K such that all
nD1
of the terms a1 ; a2 ; a3 ; : : : ; aN are among the terms b1 ; b2 ; b3 ; : : : ; bK . • For k K, the difference between the kth partial sums of the two series k k k P P P is an bn . This difference is a sum of the terms in an that are nD1
not in
k P
nD1
bn minus the sum of the terms in
nD1
k P
nD1
bn that are not in
nD1 a1 ; a2 ; a3 ; : : : ; aN ,
k P
an .
nD1
Neither sum contains any of the terms nor are there any terms that appear in both sums. It follows that the difference of partial sums equals a sum minus another sum where ˇ k each ksum ˇcontains distinct terms ˇP P ˇ from aNC1 ; aNC2 ; aNC3 ; : : : . Thus, ˇˇ an bn ˇˇ is bounded above by nD1 nD1 1 P jan j < . nDNC1
• Thus, given > 0, there is a K such that for all k K, the k partial sum of 1 1 P P an and the kth partial sum of bn are within of each other. nD1
1 P
nD1
an converges to L, given > 0, there is and N1 such that if ˇnD1 ˇ k ˇP ˇ ˇ k N1 , ˇ an Lˇˇ < 2 .
• Because
nD1
(continued)
230
7 Infinite Series
ˇ ˇ k k ˇ ˇP P • Also, there is an N2 such that if k N2 , ˇˇ an bn ˇˇ < 2 . ˇ ˇ k nD1 ˇ nD1 ˇ k k ˇP ˇ ˇP ˇ P ˇ ˇ ˇ • Then for all k max.N1 ; N2 /, ˇ bn Lˇ ˇ bn an ˇˇ C nD1 nD1 nD1 ˇ k ˇ ˇP ˇ ˇ an Lˇˇ < 2 C 2 D . ˇ nD1
• Thus, the series
1 P
an and its rearrangement
nD1
1 P
bn must both converge L.
nD1
7.7.3 Exercises 1. In which of the following series can the parentheses be removed without affecting the convergence of the series? 1 1 (a) 1 12 C C 12 13 13 C 13 C 13 14 14 14 2 3 1 1 1 1 1 1 1 1 C C C C 4 4 4 5 5 5 5 4 1 1 C 12 (b) 12 14 C 16 18 C 10 1 1 1 1 1 1 (c) 2 2 C 2 2 C 2 2 C 1 1 1 1 1 1 C C 9 C 10 C 11 12 13 (d) 12 13 C 14 C 15 16 17 8 1 1 1 1 1 1 C 16 C 17 18 19 C 14 15 1 C (e) .1/C 1 12 C 1 12 14 C 1 12 14 18 C 1 12 14 18 16 2. Write a proof to show that if
1 P
an is a conditionally convergent series, then
nD1
there is a rearrangement of the terms of the series that diverges to infinity and a rearrangement that diverges to negative infinity. 3. Write a proof to show that if a1 , a2 , and a3 are real numbers, the series a11 C a22 C a3 C a41 C a52 C a63 C converges if and only if a1 C a2 C a3 D 0. 3 1 1 P P an is an absolutely convergent series, and bn 4. Write a proof to show that if nD1
is a convergent series, then
1 P
nD1
an bn converges.
nD1
5. Give an example of convergent series
1 P nD1
an and
1 P nD1
bn where
1 P
an bn diverges.
nD1
6. Using the method described in this section find the first 20 terms of the rearrangement of the series 1 1 C 12 12 C 13 13 C 14 14 C that converges to 1.
7.8 Cauchy Products
231
7.8 Cauchy Products Earlier it was shown that if
1 P
an converges to L and
nD1
1 P
1 P
bn converges to M, then
nD1
.an C bn /, converges to L C M. What can be said about 1 1 P P an bn ? First of all, can this product even be the product of the series the sum of the series,
nD1
nD1
nD1
written as an infinite series? One could, of course, write
1 P 1 P
an bp , and some
nD1 pD1
sense can be made out of this expression. The notation suggests that for each n, 1 P one would calculate a limit of an bp , and then one would consider the series of pD1
those limits. This raises interesting questions about whether that limit, if it should 1 P 1 P exist, has anything to do with the similar looking an bp . In fact, as seen in pD1 nD1
the exercises, there are examples where interchanging the order of summation in a double summation can result in a different limit. 1 1 P P A simpler approach is to group the terms of the product an bn in a nD1
nD1
way that might allow you to calculate the sum. One strategy is to group the terms an bp where n C p is a given constant. For example, when the constant is 2, there is only one term a1 b1 . When the constant is 3, there are two terms a1 b2 C a2 b1 . n1 P ap bnp . This In general, the grouping of the terms whose subscripts add to n is pD1 ! 1 n1 P P gives what is known as the Cauchy product of the two series ap bnp . nD2
pD1
Note that this definition is symmetric in a and b, so the Cauchy product of and
1 P
bn is the same as the Cauchy product of
nD1 1 P
1 P
bn and
nD1
1 P
1 P
an
nD1
an .
nD1
For example, what is the Cauchy product for the square of the geometric series n1 P 1 ? Here you have two identical series where an D bn D 21n , so ap bnp D 2n
nD1 n1 P pD1
pD1
1 2p
1 2np
D
n1 P pD1
1 2n
D
n1 . 2n
Thus, the Cauchy product is
1 P nD2
shows that this series converges to some value S. So, SD
1 X n1 nD2
2n
n1 . 2n
The Ratio Test
232
7 Infinite Series
2S D
1 X n1 nD2
2n1
1 X n n 2 nD1
D
1
1 X 1 2S S D C D1 2 nD2 2n SD1 This Cauchy product converges to 1 which is the expected limit since
1 P nD1
1 2n
D 1.
But Cauchy products do not always behave so nicely. For example, find the Cauchy 1 1 P P .1/n .1/n p p product of the two series and . The Alternating Series Test shows n nC4 nD1
nD1
that both of these series converge, but the Integral Test shows that neither converges absolutely. The nth term of the Cauchy product of these two series is n1 n1 P .1/p .1/np P p 1 p p D .1/n . For even values of n, this is a sum of n1 p npC4 p.npC4/
pD1
pD1
positive terms of the form
p
when p D
nC4 , 2
1 . p.npC4/
Since the product p.n p C 4/ is maximum
each term of the sum is greater than or equal to
p2 nC4
2 pnC4 D
4 . nC4
which approaches 4 as This means that the sum is greater than or equal to 4.n1/ nC4 n gets large. Thus, the terms of the Cauchy product do not approach 0 as n goes to infinity, and the Limit of Terms Test shows that the Cauchy product does not converge. This last example shows what can go wrong with the Cauchy product of two conditionally convergent series, but the results are better when at least one of the series is absolutely convergent. For example, if both series are absolutely convergent, then the Cauchy product is absolutely convergent to the product of the series. To see why this is, just consider the difference between a partial sum of the Cauchy product of the two series and the product of two partial sums of the individual series. That is, let k1 and k2 be positive integers, and find the difference ! k1P Ck2 n1 P between the .k1 C k2 /th partial sum of the Cauchy product, ap bnp , nD2
pD1
and the product of the k1 th and k2 th partial sums, respectively, of the two series, k1 k2 P P am bn . These are both just finite sums where the Cauchy product partial mD1
nD1
sum includes all the terms am bn where the sum of the subscripts of m C n add to something less than or equal to k1 C k2 and the other sum includes all the terms am bn where M k1 and n k2 . Thus, the is the difference sum of the remaining k1 Ck k1 Ck k1 Ck k1 Ck P2 1 P2 m P2 1 P2 n am bn terms bn C am . So by choosing k1 and mDk1 C1
nD1
k2 large, you can ensure that both necessary convergence.
nDk2 C1 k1 Ck P2 1 mDk1 C1
mD1 k1 Ck P2 1
am and
nDk2 C1
bn are small showing the
7.8 Cauchy Products
233
PROOF: The Cauchy product of the two absolutely convergent series converges absolutely to the product of the two series. 1 P
• Let
1 P
am and
mD1
bn be absolutely convergent series.
nD1
• Let > 0 be given. 1 P • Because am converges absolutely, there exists an integer N1 such that mD1
1 P
jam j <
mDN1
2 1C
!.
1 P
jbn j
nD1
1 P
• Similarly, because
bn converges absolutely, there exists an integer N2
nD1
such that 1 P jbn j < nDN2
!.
1 P
2 1C
jam j
mD1
• Then for all k1 N1 and k2 N2 , the difference between the .k1 C k2 /th partial sum of the Cauchy product of the two series and the product of the ˇ ! k1 th and k2 th ˇpartial sums of the two series is ˇ ˇk1P Ck2 n1 k1 k2 P P P ˇ ˇ ap bnp am bn ˇ D ˇ ˇ ˇ nD2 pD1 mD1 nD1 ˇ ˇˇ ˇk1 Ck k1 Ck k1 Ck k1 Ck P2 m P2 1 P2 n ˇ P2 1 ˇ a bn bn C am ˇ ˇ ˇmDk1 C1 m ˇ nD1 nDk2 C1 mD1 k1 Ck P2 1 mDk1 C1 1 P mDk1 C1 1 P
jam j jam j
k1 Ck P2 m nD1 1 P
jbn j C
nD1
jbn j
nD1
jbn j C
2 1C
1 P
! jbn j
k1 Ck P2 1 nDk2 C1
1 P
nDk2 C1 1 P
C
jbn j 1 P
jbn j
k1 Ck P2 n
jam j
mD1
jam j
mD1
jam j
mD1
2 1C
nD1
1 P mD1
! jam j
<
2
C
2
D .
• Therefore, since the .k1 C k2 /th partial sum of the Cauchy product of the two series and the product of the k1 th and k2 th partial sums of the two series are within of each other when k1 and k2 are large, both expressions must converge to the same quantity when k1 and k2 approach infinity, and that limit is the product of the two series. ˇ ˇ ˇ n1 ˇ 1 P ˇP ˇ • Also note that for k1 N1 and k2 N2 , the sum ap bnp ˇ ˇ ˇ nDk1 Ck2 C1 ˇpD1 1 1 1 1 P P P P jam j jbn j C jbn j jam j < showing that the Cauchy mD1
nDk2 C1
nD1
product converges absolutely.
mDk1 C1
234
1 P
7 Infinite Series
Another way! to think about this theorem is that the Cauchy product n1 1 1 P P P ap bnp and the product of the two series am bn are rearrangements
nD2
pD1
mD1
nD1
of each other. Thus, if either converges absolutely, both converge absolutely to the same limit. Of course, to make this rigorous, one would need to find at least one rearrangement of the terms into a form c1 C c2 C c3 C and then show that that series converges absolutely. If one series converges absolutely and the other only converges conditionally, then the Cauchy product of the two series still converges to the product of the two series, but absolute convergence is not guaranteed. The proof is similar to the proof of the previous theorem in that it carefully considers the difference between the partial sum of the Cauchy product and product of the two series. This difference can be broken into three differences each of which can be bounded. Specifically, assume 1 1 P P that am is absolutely convergent and bn is convergent. Then consider the mD1
nD1
difference between the Nth partial sum of the Cauchy product and the product of the N n1 1 1 N1 1 Nm P P P P P P P ap bnp am bn D am bn am series. That difference is nD2 pD1 mD1 nD1 mD1 nD1 mD1 Nm N1 1 1 N1 1 1 P P P P P P P bn D am bn bn C am am bn . In the second term nD1 mD1 nD1 nD1 mD1 mD1 nD1 N1 1 1 P P P bn is fixed, and the factor am am can be made of this sum, the factor nD1
mD1
mD1
as small as necessary by choosing The Nm N large. first term of the sum is a little trickier 1 P P to handle. In the terms am bn bn the am factor can be made small by making m large, and the
Nm P nD1
nD1
bn
1 P
nD1
bn factor can be made small by making N m
nD1
large, or by keeping m small. Both of these can be done, but not at the same time. The technique one would use here would be to break the sumfrom m D 1 to m D N1 Nm 1 P P P am bn bn D N 1 at some intermediate value K < N 1 writing mD1 nD1 nD1 K Nm 1 N1 Nm 1 P P P P P P am bn bn C am bn bn . You can now choose K mD1
nD1
nD1
mDKC1
nD1
nD1
so that for m > K the value of am is small, and when m K, the N m is large so Nm 1 P P that bn bn will be small. This gives the following proof. nD1
nD1
7.8 Cauchy Products
235
1 P
PROOF: If
am is an absolutely convergent series and
mD1
1 P
bn is a
nD1
convergent series, then the Cauchy product of the two series converges to the product of the two series. 1 P
• Let
am be an absolutely convergent series, and
mD1
1 P
bn be a convergent
nD1
series. • Then for integers N and K with 1 < K < N 1, the difference between the Nth partial sum of the Cauchy product of the two series and the product of the two series is N n1 1 1 P P P P ap bnp am bn D nD2 pD1 N1 Nm P P
mD1 1 P
nD1 1 P
bn am bn D nD1 mD1 nD1 N1 1 Nm 1 1 P P P P P am bn bn C am am bn D mD1 nD1 nD1 mD1 mD1 nD1 K Nm 1 N1 Nm 1 P P P P P P am bn bn C am bn bn C mD1 nD1 nD1 nD1 nD1 mDKC1 N1 1 1 P P P am am bn : am
mD1 N1 P
mD1
mD1
• Because
nD1
T P
bn converges as T goes to infinity, it remains bounded. Thus, ˇ T ˇ 1 ˇP ˇ P there is a number M such for all T, ˇˇ bn bn ˇˇ < M. nD1 nD1 • Let > 0 be given. 1 P • Because am converges absolutely, there is an integer K such that nD1
1 P
mD1
jam j <
mDKC1
. 3M
1 P bn converges to bn , there is a positive integer N1 such that nD1 nD1 ˇ ˇ N 1 ˇ ˇP P !. for all N N1 , ˇˇ bn bn ˇˇ < 1
• Because
N P
nD1
nD1
3 1C
P
jam j
mD1
1 P am converges to am , there is a positive integer N2 such that mD1 mD1ˇ ˇ N 1 ˇ ˇP P for all N N2 , ˇˇ am am ˇˇ < ˇ 1 ˇ! . ˇP ˇ
• Because
N1 P
mD1
mD1
• Let N max.N1 C K; N2 C 1/.
3 1Cˇˇ
nD1
bn ˇˇ
(continued)
236
•
7 Infinite Series ˇ ˇ ˇ ˇ ˇP ˇ ˇ N1 Nm 1 1 1 1 ˇ P P P P P P P ˇ N n1 ˇ Then ˇˇ ap bnp am bn ˇ D ˇˇ am bn am bn ˇˇ D ˇ nD2 pD1 mD1 nD1 mD1 nD1 mD1 nD1 ˇ Nm Nm N1 1 ˇˇ ˇP 1 N1 1 1 P P P P P P P P ˇ ˇ K a b b C am bn bn C am am bn ˇ ˇ ˇmD1 m nD1 n nD1 n ˇ nD1 nD1 mD1 mD1 nD1 mDKC1 ˇ ˇ ˇ N1 ˇˇ 1 ˇ ˇNm ˇNm K 1 N1 1 1 ˇ ˇ ˇP ˇˇP ˇ ˇP ˇP P P P P P jam j ˇˇ bn bn ˇˇ C jam j ˇˇ bn bn ˇˇ C ˇˇ am am ˇˇ ˇˇ bn ˇˇ < mD1 nD1 nD1 nD1 nD1 mD1 mD1 nD1 mDKC1 ˇ ˇ K 1 ˇP ˇ P ! C MC ˇ 1 ˇ! ˇˇ jam j bn ˇˇ < 3 C 3 C 3 D . 1 3M ˇP ˇ P mD1
3 1C
3 1Cˇˇ
jam j
mD1
• Therefore, the Cauchy product 1 P
the series
1 P
am
mD1
nD1
bn ˇˇ
nD1
N Nn P P
ap bnp converges to the product of
nD2 pD1
bn .
nD1
Cauchy products play a particularly useful role in the study of power series, a topic covered in the next chapter.
7.8.1 Exercises 1. Let am;n be the nth number in the mth row of the following table where m and n both range from 1 to infinity. 1 1
0
0
0
0
0
0
0
0
0
0
0
0
0
12 12
0
0
0
0
0
0
0
0
0
0
0
0
14 14 14 14
0
0
0
0
0
0
0
0
0
1 2
1 2
1 4
1 4
1 4
1 4
1 8
1 8
1 8
1 8
Show that
1 P 1 P mD1 nD1
1 8
1 8
1 8
1 8
18 18 18 18 18 18 18 18
am;n is not equal to
1 P 1 P
am;n .
nD1 mD1
2. Show that the Cauchy product for the square of the conditionally convergent 1 P .1/n series converges. n nD1
3. Show that the Cauchy product for the square of the series
1 P nD1
.1/n p n
diverges.
4. Suppose you have two series whose indices begin with 0 rather than 1 as in 1 1 P P an and bn . Show that the Cauchy product of these two series is then nD0 1 P n P nD0 pD0
nD0
ap bnp .
7.8 Cauchy Products
237
5. In the next chapter it will be shown that for all real values of x, the exponential 1 n P x function has the series representation ex D . Use the Cauchy product of nŠ series to show that ea eb D eaCb .
nD0
Chapter 8
Sequences of Functions
8.1 Pointwise Convergence Chapter 3 introduces the idea of a sequence of real numbers and discusses theorems related to the limit lim an , limit superior lim sup an , limit inferior n!1
n!1
lim inf an , and subsequences of such a sequence. If instead of requiring n!1 the terms of the sequence an to be constants, the an were allowed to depend on the value of a variable as in fn .x/, then the sequence is a sequence of functions. Thus, for each value of x, if all the functions fn .x/ are defined at x, then there is a sequence of real numbers, . This sequence changes as x changes, and, indeed, there is a different sequence of real numbers for each choice of x. The limit of the sequence, if it exists, could be different for each x, and, therefore, the limit would also be a function, f .x/. The first question that arises is, what is meant by the convergence of such a sequence? In fact, there are many different definitions for the convergence of a sequence of functions, each with its own applications and properties. The next question is, what can one say about the properties of the limit of the sequence? For example, under what conditions can you know that the limit function is continuous, differentiable, or integrable? In particular, if the sequence of integrable functions converges to an integrable function f .x/, when can Rb Rb you conclude that lim fn .x/dx D f .x/dx? n!1 a
a
The simplest form of convergence of a sequence of functions is to say that the sequence of functions converges pointwise to the function f .x/ on a set A if for each x 2 A, lim fn .x/ D f .x/. This type of convergence is referred n!1
to as pointwise convergence. For example, the sequence of functions fn .x/ D nx converges pointwise to the function f .x/ D 0 on the entire real line because for each x 2 R, lim nx D 0. A more interesting example is the sequence fn .x/ D xn which n!1
converges pointwise on the interval .1; 1. When jxj < 1, the powers xn get small as n gets large so lim xn D 0. But when x D 1, the powers xn D 1, so the limit n!1
© Springer International Publishing Switzerland 2016 J.M. Kane, Writing Proofs in Analysis, DOI 10.1007/978-3-319-30967-5_8
239
240
8 Sequences of Functions
Fig. 8.1 The sequence of functions xn converging to a discontinuous function
-1
1
Fig. 8.2 The sequence of nC1 functions jxj n converging to the function f .x/ D jxj
0 if 1 < x < 1 of the sequence is 1. Thus, the limit function is f .x/ D : Note 1 if x D 1 that this is a sequence of continuous functions that converges to a function that is not continuous. The sequence does not converge at x D 1 because the terms oscillate between 1 and 1 (Fig. 8.1). Continuity is not the only property not preserved by functions converging nC1 pointwise. The terms of the sequence fn .x/ D jxj n are differentiable functions for all real numbers, but the limit of the sequence is the function f .x/ D jxj which is not differentiable9at x D 0 (Fig. 8.2). The terms of the sequence f .x/ D 8 2 ˆ n x if 0 x 1n > > ˆ > ˆ > ˆ > ˆ = < R2 2 1 2 all have integral fn .x/dx D 1, yet the sequence n2 . n x/ if n < x < n > ˆ > ˆ 0 > ˆ > ˆ > ˆ ; : 2 0 if n x 2 converges pointwise on the interval Œ0; 1 to the function f .x/ D 0 which has integral equal to 0 (Fig. 8.3).
8.2 Uniform Convergence
241
Fig. 8.3 A sequence of functions with integral 1 converging to the function f .x/ D 0 f3
f2 f1
8.1.1 Exercises Determine the pointwise limits of the following sequences of functions. For which sequences is the limit continuous? For which sequences is the limit of the integrals of the terms equal to the integral of the limit? p 1. fn .x/ D n x for x 2 Œ0; 16. 2. Let r1 ; r2 ; r3 ; : : : be a sequence consisting of all the rational numbers in the 1 if x D rk for some k n interval Œ0; 1. Let fn .x/ D for x 2 Œ0; 1. 0 otherwise n nx for x 2 .0; 1/. 3. fn .x/ D ( ) n nC4 n if 2nC4 < x < 2nC4 4. fn .x/ D for x 2 Œ0; 1. 0 otherwise 8 9 1 .2 C .1/n / n2 x if 0 < x < 2n < = 1 5. fn .x/ D .2 C .1/n / n2 1n x if 2n x < 1n for x 2 Œ0; 1. : ; 0 otherwise
8.2 Uniform Convergence The sequence converges pointwise to f on the set A if given > 0, for each x 2 A there is an integer N such that jfn .x/ f .x/j < for all n N. So, for each x there is an integer N that ensures the inequality. The value of N can depend on the choice of x. If this dependence is dropped, and you are able to specify a value of N that does not depend on the choice of x, then the speed of convergence becomes similar for all x 2 A; that is, the rate of convergence is uniform for all x 2 A. The sequence converges uniformly to f on the set A if, given > 0, there is an integer N such that for each x 2 A, jfn .x/ f .x/j < for all n N. The difference between a sequence of functions converging uniformly and converging pointwise is that with uniform convergence there can be no points of
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8 Sequences of Functions
Fig. 8.4 A sequence of functions converging uniformly
the set A where convergence “lags behind.” For any region of width > 0 around the limit function, all of the terms suitably far down the sequence enter that region. Compare, for example, the uniformly convergent sequence depicted in Fig. 8.4 with the pointwise convergent sequences depicted in Figs. 8.1 and 8.3. In Fig. 8.4 the functions of the sequence get close to the limit function for all the values of x, whereas for each function in Figs. 8.1 and 8.3 there is an x for which the function is far from its limit. Clearly, if a sequence of functions converges uniformly, then it also converges pointwise. Thus, to converge uniformly is a stronger condition than to converge pointwise. As seen in the previous section, the terms of the sequence can have many properties that are not automatically inherited by the limit of the sequence, f , when the convergence is pointwise. Under uniform convergence, more of the properties of the terms of the sequence are retained by the limit. This is because under uniform convergence there are no points x 2 A for which the values of lag behind as n gets large. For all values of x 2 A, the sequence of values get close to the corresponding f .x/ at a rate at least as fast as some fixed rate. For example, if is a sequence of functions continuous on the set A which converges uniformly to f on A, then the limit is guaranteed to be continuous. Actually, a stronger statement can be made. If all the terms fn are continuous at some point a 2 A, then the limit, f , will also be continuous at a. To prove that f is continuous at a, you will need to show that for each > 0 there is a ı > 0 such that if x is in A with jx aj < ı, then jf .x/ f .a/j < . How can you arrange for f .x/ to be close to f .a/? What you know is that the functions fn get uniformly close to f , and that the fn functions are continuous at a. Since, for any particular n, the term fn is continuous at a, you can arrange for fn .x/ to be close to fn .a/. The uniform convergence allows you to choose an integer n so that for every x 2 A, fn .x/ is close
8.2 Uniform Convergence
243 f(x)
3
fn(x)
(
) a
x
Fig. 8.5 If continuous function fn is close to f , then f .x/ is close to f .a/
to f .x/. That is, jf .x/ f .a/j D jf .x/ fn .x/ C fn .x/ fn .a/ C fn .a/ f .a/j jf .x/ fn .x/j C jfn .x/ fn .a/j C jfn .a/ f .a/j. Each of these three terms can be made small, say less than 3 , so that the sum is less than . The key point here is that only one value of n needs to be chosen so that jf .x/ fn .x/j can be made less than 3 no matter which x is chosen (Fig. 8.5). PROOF: If the sequence converges uniformly to the limit f on the set A and if for each n, fn is continuous at a 2 A, then f is continuous at a 2 A. In particular, if each fn is continuous on A, then f is continuous on A. • Let be a sequences of functions that converge uniformly to the function f on a set A. • Assume that each fn is continuous at point a 2 A. • Let > 0 be given. • Because the sequence converges uniformly, there is an integer N such that jfn .x/ f .x/j < 3 for all x 2 A and all n N. • Because fN is continuous at a, there is a ı > 0 such that jfN .x/ fN .a/j < 3 for all x 2 A satisfying jx aj < ı . • Then, for all x 2 A satisfying jx aj < ı, it follows that jf .x/ f .a/j D jf .x/ fN .x/ C fN .x/ fN .a/ C fN .a/ f .a/j jf .x/ fN .x/j C jfN .x/ fN .a/j C jfN .a/ f .a/j < 3 C 3 C 3 D . • Therefore, the function f is continuous at a 2 A. • Moreover, if each function fn is continuous at each a 2 A, then f is continuous at each x 2 A, so f is continuous on A. It is worth considering where this proof breaks down if all you assume is that the sequence converges pointwise to f . The problem comes in the fact that although jfN .a/ f .a/j and jfN .x/ fN .a/j can be made smaller than 3 , there could
244
8 Sequences of Functions
be values of x very close to a for which jfN .x/ f .x/j is no longer small. Thus, the needed inequality jf .x/ fN .x/j C jfN .x/ fN .a/j C jf .a/ f .a/j < might not hold. Also consider the function f defined on the interval Œ0; 2 by f .x/ D x if x ¤ 1 and f .1/ D 3. If for each positive integer n you let fn .x/ D f .x/ C 1n , then it is clear that the sequence converges uniformly to f . At the points where each fn is continuous, that is, for x ¤ 1, the limit function f is also continuous. Suppose that is a sequence of functions Riemann integrable on an interval Œa; b and that this sequence converges to a limit f . Examples in the last section show Rb that if the convergence is pointwise, the limit lim fn .x/dx does not necessarily equal
Rb
n!1 a
f .x/dx. Moreover, the function f does not even need to be Riemann
a
integrable, and the limit of the integrals of the fn might not exist. On the other hand, if the convergence is uniform, then the limit function f will be Riemann integrable and the limit of the integrals of the fn will equal the integral of f . Showing that the uniform limit of Riemann integrable functions is Riemann integrable is not difficult and is based on the characterization of Riemann integrable functions given by Lebesgue’s Theorem. Recall that a function is Riemann integrable on an interval if and only if it is bounded and the set of points where the function is discontinuous has measure zero. If each term of the sequence, fn , has these properties, then the limit function, f , must also have them. By the definition of uniform convergence, there is an integer N such that jfN .x/ f .x/j < 1 for all x 2 Œa; b. So, if the function fN is bounded by some constant M, then the function f must be bounded by M C 1 since for all x 2 Œa; b it follows that M 1 fN .x/ 1 < f .x/ < fN .x/ C 1 M C 1. As for points of discontinuity of f , for each positive integer n, let Dn be the set of points in Œa; b where the function fn is discontinuous. Because each fn is Riemann 1
integrable, each Dn has measure zero. But then D D [ Dn is a countable union nD1
of sets of measure zero, so it also has measure zero. The sequence converges uniformly on the set A D Œa; bnD, and each term of the sequence is continuous at each point of A, so, by the preceding theorem, the limit f must be continuous on A. Thus, the set of discontinuities of f must be contained in D, so the set of discontinuities of f has measure zero. Therefore, f is Riemann integrable. Rb Rb So why does it follow that lim fn .x/dx D f .x/dx? From the definition of n!1 a
a
uniform convergence, for every > 0 there is an integer N such that n N implies that jfn .x/ f .x/j < for every x 2 Œa; b. This means that for every n N and Rb every x 2 Œa; b it follows that f .x/ < fn .x/ < f .x/ C , so f .x/dx .b a/ D Rb a
.f .x//dx
Rb a
Rb
Rb
a
a
fn .x/dx .f .x/C/dx D
a
f .x/dxC.ba/. Thus, by selecting
a more appropriate value for , this shows that lim
Rb
n!1 a
fn .x/dx D
Rb a
f .x/dx.
8.2 Uniform Convergence
245
PROOF: Assume that is a sequence of functions that are Riemann integrable on the interval Œa; b. If the sequence converges uniformly to f , Rb Rb then f is also Riemann integrable on Œa; b, and lim fn .x/dx D f .x/dx. n!1 a
a
• Assume that is a sequence of functions Riemann integrable on the interval Œa; b which converges uniformly to the function f . • Because the sequence converges uniformly, there is an integer N such that for all n N and all x 2 Œa; b it follows that jfn .x/ f .x/j < 1. • Because fN is Riemann integrable on Œa; b, fN is bounded, so there exists an M such that jfN .x/j < M for all x 2 Œa; b. • Then for each x 2 Œa; b it follows that M 1 fN .x/ 1 < f .x/ < fN .x/ C 1 M C 1. Thus, jf .x/j is bounded by M C 1 and f is a bounded function. • For each positive integer n let Dn be the set of x 2 Œa; b where fn fails to be continuous. Because each fn is Riemann integrable, Lebesgue’s Theorem shows that Dn has measure zero. • Since the countable union of sets of measure zero is a set with measure zero, 1
the set D D [ Dn has measure zero. nD1
• For each n, the function fn is continuous on the set A D Œa; bnD. • Because converges uniformly to f on Œa; b, the limit function, f , is continuous at each point in A. • Thus, the set of points where f is discontinuous is a subset of D, and, hence, the set of discontinuities of f has measure zero. • It follows from Lebesgue’s Theorem that f is integrable on the interval Œa; b. • Let > 0 be given. • Because converges uniformly to f on Œa; b, there is an integer N such that for all n N and all x 2 Œa; b, jfn .x/ f .x/j < baC1 . • Thus, for each x 2 Œa; b, f .x/ baC1 < fn .x/ < f .x/ C baC1 . b b Rb Rb R R • Then, f .x/dx < f .x/ baC1 dx fn .x/dx f .x/C baC1 dx <
Rb
a
a
a
a
f .x/dx C .
a
• This proves that lim
Rb
n!1 a
fn .x/dx D
Rb
f .x/dx and completes the proof of the
a
theorem. . The Note the use of b a C 1 rather than just b a in the denominator of baC1 extra C1 avoids the embarrassing case of a D b. The addition of C1 allows the proof to handle the easy special case without having to provide a separate argument for it.
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8 Sequences of Functions
8.2.1 Exercises 1. Show that lim xn converges uniformly to 0 on any interval Œa; a where 0 < n!1 a < 1. 1 converges uniformly to 0 on any interval Œa; 1/ for a > 0 but 2. Show that lim nx n!1
not on the interval .0; 1/. 3. Another way to show that the uniform limit of Riemann integrable functions is Riemann integrable is to show that the limit function has upper and lower step functions, u and v, such that the integrals of u and v are within > 0 of each other. Write a proof that uses this strategy. 4. Suppose that the sequence of functions converges uniformly to the function f and that g is a uniformly continuous function defined on the range of f and the ranges of each of the f functions. Prove that the functions g f .x/ converge n n uniformly to g f .x/ .
8.3 Monotone Convergence Chapter 3 introduces monotonically increasing sequences of numbers where an anC1 for each n 1 and monotonically decreasing sequences of numbers where an anC1 for each n 1. One can similarly define a to be a monotonically increasing sequence of functions or a monotonically decreasing sequence of functions on a set A if for each x 2 A the sequence of numbers is always monotonically increasing or always monotonically decreasing, respectively. Such a sequence is said to converge monotonically to the limit function f on A if the monotone sequence converges to f . That convergence could be a pointwise convergence or a uniform convergence. Even if convergence is pointwise, sometimes knowing that the convergence is also monotone gives results similar to knowing that the convergence is uniform. In the previous section it was shown that if a sequence of continuous functions converges uniformly to f , then f is continuous. If the convergence is actually monotone, then the converse holds, that is, if the terms of the sequence are continuous on an interval Œa; b and the sequence converges monotonically to a limit function f that is also continuous on Œa; b, then the convergence is actually uniform. To prove this you would need to take an > 0 and show there was an integer N such that for all n N and all x 2 Œa; b it was true that jfn .x/ f .x/j < . What do you have working for you? What you have is that the limit function f is continuous at each point x 2 Œa; b, each of the terms of the sequence fn is continuous, and the values fn .x/ are approaching f .x/ monotonically. Let x 2 Œa; b be given. Using the continuity of f you can find an interval around x such that for any y in that interval, f .y/ is close to f .x/. Using the fact that the fn are converging to f pointwise, you can find an integer N such that for all n N the value of fn .x/ is close to the value of f .x/. Finally, using the continuity of fN you can find an interval around x such that for any y in that interval fN .y/ is close to fN .x/.
8.3 Monotone Convergence
247
Combining these you can show that for any y in some interval around x, the value of fN .y/ is close to the value of f .y/. The crucial observation here is that once you know that fN .y/ and f .y/ are close, the monotonicity of the convergence gives you that fn .y/ is between fN .y/ and f .y/ for all n N, and, thus, fn .y/ will be close to f .y/ for all n N. Note, though, that the value of N can vary with the value of x. Well, this means that for each x 2 Œa; b there is an interval around x where fn .y/ is close to f .y/ for all y in the interval and all n N. Now you can use the compactness of the interval Œa; b, that is, you can use the Heine–Borel Theorem to show that there is a finite collection of these x values, say x1 ; x2 ; x3 ; : : : ; xk , such that the entire interval Œa; b is covered by these intervals you constructed around each of the xj s. Each of the xj s was associated with an Nj , and now one can select the maximum of these Nj values to get a single function fN which is uniformly close to f . Again, because the convergence is monotone, once you know that fN is close to f , you know that fn is close to f for all n N. This will complete the proof. PROOF: Assume that is a sequence of functions continuous on the interval Œa; b that converges monotonically to the function f that is also continuous on Œa; b. Then the sequence converges uniformly to f on Œa; b. • Assume that is a sequence of functions continuous on the interval Œa; b that converges monotonically to the function f that is also continuous on Œa; b. • Let > 0 be given. • Let x 2 Œa; b. • Because the function f is continuous at x, there is a ı1 > 0 such that if y 2 Œa; b with jy xj < ı1 , then jf .y/ f .x/j < 3 . • Because lim fn .x/ D f .x/, there is an integer Nx such that if n Nx , then n!1
• • •
• •
• • • • •
jfn .x/ f .x/j < 3 . Because fNx is continuous at x, there is a ı2 > 0 such that if y 2 Œa; b with jy xj < ı2 , then jfNx .y/ fNx .x/j < 3 . Let ıx D min.ı1 ; ı2 /. Then, if y 2 Œa; b with jy xj < ıx , it follows that jfNx .y/ f .y/j D jfNx .y/ fNx .x/ C fNx .x/ f .x/ C f .x/ f .y/j jfNx .y/ fNx .x/j C jfNx .x/ f .x/j C jf .x/ f .y/j < 3 C 3 C 3 D . The interval Œa; b is covered by the collection of open intervals .x ıx ; x C ıx / for x 2 Œa; b. By the Heine–Borel Theorem, there is a finite collection of these x values, x1 ; x2 ; x3 ; : : : ; xk , such that the intervals .xj ıxj ; xj C ıxj / for j D 1; 2; 3; : : : ; k covers the interval Œa; b. Let N D max.Nx1 ; Nx2 ; Nx3 ; : : : ; Nxk /. Let y 2 Œa; b. There is a value of j between 1 and k such that y 2 .xj ıxj ; xj C ıxj /. Because the sequence converges monotonically to f , for all n N Nxj , fn .y/ is between fNxj .y/ and f .y/, and jfNxj .y/ f .y/j < . This shows that the sequence converges uniformly to f .
248
8 Sequences of Functions p
x
The sequence of functions fn .x/ D 2xC n converges monotonically to f .x/ D 2x on the interval Œ0; 4. Since each fn and f is continuous on Œ0; 4, you can conclude that the convergence of the sequence is uniform. On the other hand, the sequence of continuous functions fn .x/ D xn converges monotonically on the interval Œ0; 1 to a function discontinuous at 1. Clearly, then, the sequence does not converge uniformly. Another important theorem about monotone convergence is that if is a sequence of functions Riemann integrable on the interval Œa; b that converge monoRb Rb tonically to the Riemann integrable function f , then lim fn .x/dx D f .x/dx. The n!1 a
a
result is called the Monotone Convergence Theorem for Riemann Integrals. It is generally not proved in a book of this type because it is an easy consequence of the Monotone Convergence Theorem of Lebesgue which is covered in any beginning course in measure theory, but that study requires the development of Lebesgue measure, a topic which is beyond the scope of this book. It does need to be pointed out that even if all the terms of a sequence are Riemann integrable functions, and the sequence converges monotonically to a function f , it may be that the limit, f , is not itself Riemann integrable. For example, let r1 ; r2 ; r3 ; : : : be a sequence consisting of all the rational numbers in the interval Œ0; 1. Let fn .x/ be the function equal to 1 for x D r1 ; r2 ; r3 ; : : : ; rn and equal to 0 elsewhere. Then each fn has finitely many points of discontinuity so each fn has a Riemann integral on Œ0; 1 equal to 0. Yet the sequence converges monotonically to the function f equal to 1 for rational values of x and 0 for irrational values of x, so f is discontinuous everywhere, and, as a result, it is not Riemann integrable. So, suppose that is a sequence of functions Riemann integrable on the interval Œa; b that converge monotonically to a limit function f that is also Riemann integrable on Œa; b. Without loss of generality one can assume that the sequence is monotonically decreasing to f because if the sequence were increasing, the same argument could just be applied to the sequence . Also, it can be assumed that the function f is identically 0 on Œa; b because if that is not the case, the argument could be applied to the sequence which does decrease monotonically to 0, Rb Rb Rb and lim Œfn .x/ f .x/dx D 0 is equivalent to lim fn .x/dx D f .x/dx. 1 n!1 a
n!1 a
a
A proof of the Monotone Convergence Theorem for Riemann Integrals would start with an > 0, and the goal of the proof would be to show that there is an Rb integer N such that for all n N, it follows that fn .x/dx < . The proof presented a
here is based on the fact that for any Riemann integrable function, fn , you can find upper and lower step functions, un .x/ and vn .x/, satisfying vn .x/ fn .x/ un .x/ Rb for every x 2 Œa; b so that a .un .x/ vn .x//dx is as small as you like. Suppose Rb you select un and vn so that a .un .x/ vn .x//dx < 2n . That is, find upper and lower step functions for each fn such that they give increasingly better and better 1
This proof is based on ideas from the article Monotone Convergence Theorem for the Riemann Integral by Brian S. Thomson from the American Mathematical Monthly, June–July 2010.
8.3 Monotone Convergence
249
approximations to the integrals of fn as n gets large. In particular, with the stated 1 R P b precision, you would be able to know that the entire sum a .un .x/ vn .x//dx would be less than
1 P
nD1
n
2
D . Actually, the bound is not small enough for this
nD1
proof, but later this value can be adjusted when you see just how small the bound needs to be in order to make the proof work. For each n these un and vn functions are step functions on the interval Œa; b, so there must be a positive integer k and a partition of Œa; b given by a D x0 < x1 < x2 < < xk D b such that both the un and the vn functions are constant on each of the open intervals .xj1 ; xj / for j D 1; 2; 3; : : : ; k. Well, for this proof, there will be a different k and a different partition for each fn function, so it would be better to name the positive integer kn and the partition a D xn;0 < xn;1 < xn;2 < < xn;kn D b. For the purposes of this proof, it is important that the endpoints of the partition associated with the un and vn , that is, xn;1 ; xn;2 ; xn;3 ; ; xn;kn 1 , do not match any of the endpoints of the partition associated with the next case unC1 and vnC1 . This is easy to arrange because an upper or lower step function for fn that is constant on the two intervals .xj1 ; xj / and .xj ; xjC1 / can be altered to be constant on the three intervals .xj1 ; xj ı/, .xj ı; xj C ı/, and .xj C ı; xjC1 / for some suitably small ı > 0 without significantly changing the value of the integral of the step function and without destroying whether the step function is an upper or lower step function of fn . Indeed, if, for example, the upper step function un were constant on the function un by redefining un on the .xj1 ; xj / and .xj ; xjC1 /, you could define interval .xj ı; xj C ı/ to equal max un .xj ı/; un .xj /; un .xj C ı/ . Then un would Rb be slightly larger than un on a small interval so that un .x/ un .x/ dx is less than a
2ıun .xj / which can be made arbitrarily small by selecting ı small. Because un is greater than or equal to un , it is also an upper step function of fn . For each y 2 Œa; b, consider the sequence of numbers f1 .y/; f2 .y/; f3 .y/; : : : which decreases monotonically to 0. For each y, select a positive integer n.y/ such that fn.y/ .y/ < . Thus, n.y/ associates a term of the sequence fn.y/ with y. That term is also associated with un and vn and the partition a D xn;0 < xn;1 < xn;2 < < xn;kn D b. As stated in the previous paragraph, it can be assumed that y is not equal to any of the endpoints xn;1 ; xn;2 ; xn;3 ; : : : ; xn;kn 1 , so y can be associated with an open interval .xn;j1 ; xn;j / containing y unless y is a or b in which case y will be associated with the open interval .a 1; xn;1 / or .xn;kn 1 ; b C 1/, respectively. Each point y 2 Œa; b has been associated with an open interval that contains y. Thus, these open intervals provide an open cover of the interval Œa; b. The Heine– Borel Theorem says that there exists a finite subcover of Œa; b. That is, there is a sequence of y 2 Œa; b, say y1 ; y2 ; y3 ; : : : ; ym , such that the intervals associated with these y values cover Œa; b. Something stronger can be said. In this finite subcover of open intervals, you can assume that there are no values of y 2 Œa; b that belong to more than two of the open intervals in that subcover. Indeed, suppose y is an element of the three intervals of the subcover .a1 ; b1 /, .a2 ; b2 /, and .a3 ; b3 /. Suppose a1 is the least of a1 , a2 ; and a3 , and that b2 is the greatest of b1 ; b2 ; and b3 . Then a1 < a3 < y < b3 < b2 , so .a3 ; b3 / .a1 ; b1 /[.a2 ; b2 /, and the interval .a3 ; b3 / can
250
8 Sequences of Functions
be dropped from the subcover. Because the subcover contains only a finite number of open intervals, all of these superfluous intervals can be dropped from the subcover. Consider the intervals associated with each of the yj values. For simplicity, let the interval associated with yj be renamed .aj ; bj /. Note that the endpoints a and b will be among the yj values because for every n each of these endpoints was covered by only one possible open interval. At this point the left endpoint associated with a can be set to a and the right endpoint of the interval associated with b can be set to b. It is important to note that if n.yi / D n.yj / for some distinct i and j, then the intervals associated with yi and yj do not overlap. This is because the intervals associated with yi and yj are distinct intervals from .a1; x1 /; .x1 ; x2 /; .x2 ; x3 /; : : : ; .xk1 ; bC1/. Let N be the maximum of the finitely many n.yj / values for j D 1; 2; 3; : : : ; m. Because no value of y 2 Œa; b appears in more than two of the open intervals associated with the yj , it can be concluded that Zb a
m Z X
bj
fN .x/dx
jD1 a
m Z X
bj
fN .x/dx
jD1 a
j
fn.yj / .x/dx D
j
2 2 3 3 Zbj Zbj m m X X 6 6 7 7 fn.yj / .x/ fn.yj / .yj / dx C fn.yj / .yj /.bj aj /5 un.yj / .x/ vn.yj / .yj / dx C .bj aj /5 4 4 jD1
jD1
aj
aj
N Z N X X 2p C 2.b a/ < .2b 2a C 1/: up .x/ vp .x/ dx C 2.b a/ b
pD1 a
pD1
There were two places in the above argument where quantities were forced to be less than the given value . It can now be seen that those quantities should have been Rb made smaller than 2b2aC1 so that the final inequality would show fN .x/dx < as a
needed. It is also worth noting that there were two places in the argument that use the fact that the sequence converges monotonically. The first was to conclude that when, for a particular y 2 Œa; b, the value of fn .y/ is small, then the values of fm .y/ are also small for all m n. The second important use of monotonicity takes Rb Rb the final result that fN .x/dx < and concludes that fm .x/dx < for all m N. a
a
PROOF: Assume that is a sequence of functions Riemann integrable on the interval Œa; b that converges monotonically to the function f that is Rb Rb also Riemann integrable on Œa; b. Then lim fn .x/dx D f .x/dx. n!1 a
a
• Assume that is a sequence of functions Riemann integrable on the interval Œa; b that converges monotonically to the function f that is also Riemann integrable on Œa; b. • Without loss of generality assume that decreases monotonically to f .x/ 0 on Œa; b. If this were not the case, the argument could be applied to the sequence of functions . (continued)
8.3 Monotone Convergence
251
• Let > 0 be given. • It is left to show that there is an integer N such that for all n N, Rb fn .x/dx < . a
• For each positive integer n the function fn is Riemann integrable on Œa; b so there exists upper and lower step functions, un and vn , satisfying for each Rb x 2 Œa; b, vn .x/ fn .x/ un .x/ and un .x/ vn .x/ dx < 2b2aC1 2n . a
• Because un and vn are step functions, there exist a positive integer kn and a partition of Œa; b given by a D xn;0 < xn;1 < xn;3 < < xn;kn D b such that for each j D 1; 2; 3; : : : ; kn , the functions un and vn are constant on each open interval .xn;j1 ; xn;j /. • Because there is flexibility in selecting the upper and lower step functions, it can be assumed that for each positive integer n, except for a and b, the endpoints of the partition associated with the upper and lower step functions for fn are distinct from the endpoints of the partition associated with the upper and lower step functions for fnC1 . • For each y 2 Œa; b the sequence f1 .y/; f2 .y/; f3 .y/; : : : decreases monotonically to 0, so for each y there is a positive integer n.y/ such that fm .y/ < 2b2aC1 for all m n.y/. In particular, it can be assumed that, unless y is a or b, y is not an endpoint of the partition of Œa; b associated with the upper and lower step functions of fn.y/ . • Associate with each y 2 Œa; b an open interval as follows. If y D a, then let the open interval be .a 1; xn.a/;1 /. If y D b, then let the open interval be .xn.b/;kn.b/ 1 ; b C 1/. Otherwise, associate y with the open interval .xn.y/;j1 ; xn.y/;j / that contains y. • Thus, each y 2 Œa; b is associated with an open interval that contains y, so this collection of open intervals provides an open cover of Œa; b. • By the Heine–Borel Theorem, there exists a finite subcovering of Œa; b consisting of m open intervals associated with m values y1 ; y2 ; y3 ; : : : ; ym in Œa; b. Note that since a and b are each covered by at most one of the open intervals in the covering of Œa; b, both a and b appear in the list of y1 through ym . • Let the interval associated with yj be called .aj ; bj /. Reset the interval associated with a so that its aj value is equal to a rather than a 1, and reset the interval associated with b so that its bj value is equal to b rather than b C 1. It can be assumed that no value of y 2 Œa; b belongs to more than two of the open intervals of the subcovering. (continued)
252
8 Sequences of Functions
• Let N D max n.y1 /; n.y2 /; n.y3 /; : : : ; n.ym / . • Then Zb
m Z X
bj
fN .x/dx
jD1 a
a
m Z X
bj
fN .x/dx
jD1 a
j
fn.yj / .x/dx D
j
3 2 Zbj m X 7 6 fn.yj / .x/ fn.yj / .yj / dx C fn.yj / .yj /.bj aj /5 4 jD1
aj
2 3 Zbj m N Zb X 6 7 X un.yj / .x/ vn.yj / .yj / dx C .bj aj /5 up .x/ vp .x/ dx C 4 jD1
pD1 a
aj N X pD1
2.b a/ 2b 2a C 1
2p C 2.b a/ < .2b 2a C 1/ D : 2b 2a C 1 2b 2a C 1 2b 2a C 1
• Because the sequence decreases monotonically to 0, it follows that Rb Rb for all n N that fn .x/dx fN .x/dx < which completes the proof. a
a
Pointwise convergence and uniform convergence are not the only methods of convergence of sequences of functions. Another method suggested by the above theorem is called convergence in mean or convergence in L1 . A sequence of Riemann integrable functions is said to converge in mean to the Riemann Rb integrable function f on the interval Œa; b if lim jfn .x/ f .x/jdx D 0. For n!1 a
example, consider the following sequence of functions defined on the interval Œ0; 1. Define f .xI a; b/ be the function that is 1 for x in the interval Œa; b and 0 for all other x. Then for positive integer n and for integer k with 2n1 k < 2n , let n1 kC12n1 n1 kC12n1 fk .x/ D f .xI k2 ; 2n1 /. The integral of f .xI k2 ; 2n1 / from 0 to 1 is 2n1 2n1 1 , so the integrals of f .x/ approach 0 as k gets large. Thus, fk converges in mean k 2n1 to the zero function. Yet this sequence of functions does not converge pointwise for any single value of x.
8.4 Series of Functions The infinite series f .x/ D
1 P
an .x/ is a function of x. For each value of x where the
nD1
terms of the series are defined and the series converges, f .x/ is just an infinite series 1 P 1 is defined for each x of real numbers given by an .x/. For example, f .x/ D n2 Cx nD1
that is not the negative of a perfect square. If x is the negative of a perfect square, then there is a term of the series that is not defined. Otherwise, the series converges
8.4 Series of Functions
253
by the Comparison Test for if x 0, then with n2 > 2jxj it follows that 1 P nD1
1 n2
1 n2 Cx
D
1 n2 Cx
1 , n2
and if x < 0, then for any n 1 P 1 n22 . Then converges since n2 Cx
2 2n2 2jxj
nD1
converges.
Another example is f .x/ D
1 P
xn which is just a geometric series which
nD1
x converges to f .x/ D 1x for all x satisfying jxj < 1. Note here that the function x f .x/ D 1x is defined for all x ¤ 1, but the infinite series is only defined for jxj < 1. This is an example of a power series dealt with in considerably more detail in the next section. The results concerning the convergence of sequences of functions discussed earlier in this chapter apply to the study of infinite series of functions because an infinite series is just defined to be the sequence of its partial sums. Still other questions arise such as, can one find the derivative or the integral of an infinite series by simply differentiating or integrating the terms of the series and then finding the limit of the resulting partial sums? The answer to this question is that sometimes one gets a correct answer by differentiating or integrating a series term by term, but other times this process results in nonsense. For example, consider again the function 1 1 1 R R R P P P 1 1 1 f .x/ D . Here, the statement that f .x/ dx D dx D dx 2 2 n Cx n Cx n2 Cx nD1
is not valid since
1 R P nD1
1 n2 Cx
dx D
1 P
nD1
nD1
ln n2 C x C C which does not converge
nD1
for any value of x. Alternatively, for this particular series it is valid to use the 1 1 Ry Ry Ry P P 1 1 definite integral from 0 to y and write f .x/ dx D dx D dx D 2 n Cx n2 Cx nD1 0 0 0 nD1 1 P 2 which does converge for each y > 1. The integral and derivative of ln n nCy 2
nD1
the series f .x/ D
1 P
xn make perfectly good sense in the range jxj < 1.
nD1
One simple observation about series series of positive numbers
1 P
1 P
an .x/ is that if there is a convergent
nD1
Mn such that for each n, the term an is bounded by
nD1
Mn for all x in some set A, then the series
1 P
an .x/ converges uniformly on A. This
nD1
in known as the Weierstrass M-Test. Consider how the proof of this result would 1 P go. First, of course, you would assume that you had a series of functions, an .x/, and a convergent series of positive numbers,
1 P nD1
nD1
Mn , such that for each positive
integer n, jan .x/j Mn for every x 2 A. You should note that for each x 2 A, the
254
8 Sequences of Functions
series
1 P
an .x/ converges by the Comparison Test and, thus,
nD1
1 P
an .x/ converges
nD1
pointwise. You are to prove that the sequence of function converges uniformly, so you would need to take an > 0 and show that there is an integer N such that m P whenever m N and x 2 A, the partial sum an .x/ is within of the limit nD1
1 P
1 P an .x/. The difference between the mth partial sum of an .x/ and its limit nD1ˇ nD1 ˇ 1 1 1 ˇ P ˇ P P is ˇˇ an .x/ˇˇ jan .x/j Mn which can be made less than by nDmC1
nDmC1
nDmC1
selecting m large. The value of m does not depend on x showing that the convergence is uniform. This gives the following proof. PROOF (Weierstrass M-Test): Let defined on the set A, and let
1 P
1 P
an .x/ be a series of functions
nD1
Mn be a convergent series of positive
nD1
numbers. If for each n and each x 2 A it holds that jan .x/j Mn , then 1 P an .x/ converges uniformly on A. nD1
• Let
1 P
1 P
an .x/ be a series of functions defined on the set A, and let
nD1
Mn be
nD1
a convergent series of positive numbers. • Assume that for each n and each x 2 A it holds that jan .x/j Mn . • Then for each x 2 A it follows from the Comparison Test that
1 P
an .x/
nD1
converges absolutely and, thus, the series converges. • Let > 0 be given. 1 1 P P • Because Mn converges, there is an integer N such that Mn < for nDm
nD1
all m N. • But, then, for each x 2 A and each m N, the difference between the mth ˇ 1 ˇ 1 1 ˇ P ˇ P P partial sum of an .x/ and its limit is ˇˇ an .x/ˇˇ jan .x/j 1 P nDmC1
• Thus,
nD1
nDmC1
nDmC1
Mn < . 1 P
an .x/ converges uniformly on A.
nD1
For example, the series
1 P nD1
because for all x 2 Œ0; 1/,
1 n2 Cx 1
n2 Cx
converges uniformly on the interval Œ0; 1/
1 , n2
and the series
1 P nD1
1 n2
converges. Since
all the partial sums of the series are continuous functions, it follows from this
8.5 Power Series
255
uniform convergence that the limit function is continuous on Œ0; 1/. Similarly, the ˇ ˇ 1 P ˇ sin.n2 x/ ˇ sin.n2 x/ n12 series converges uniformly on the entire real line because ˇ n2 n2 ˇ nD1
for every positive integer n. Again, you can conclude that the limit function is continuous because all the partial sums are continuous functions. Notice, though, 1 P that if you differentiate each term of this series, you get cos.n2 x/ which does not nD1
converge for any value of x because the terms do not approach 0.
8.5 Power Series Power series form a class of infinite series of functions that stands out because of the particularly nice properties they satisfy, the ease in which pthey can be produced, the many well-known elementary functions they represent ( x; ex ; sin x; cos x; ln x), and the enormous number of applications they have. A power series is a series of 1 P the form an .x c/n , where the real number an is the nth coefficient and c is the nD0
center of the power series. This book will consider such series where the variable, coefficients, and center are real numbers, although most of what is said here holds when these quantities are allowed to be complex numbers. In fact, such series play a central role in Complex Analysis.
8.5.1 Absolute Convergence The first important result about power series is that they converge in an interval .c R; c C R/ where c is the center of the power series and R, called the radius of convergence, is a nonnegative real number or possibly even infinity. In fact, if the 1 P power series an .xc/n converges for a particular real number y, then it converges nD0
absolutely for any x satisfying jx cj < jy cj, that is, for any x closer to c than y. The proof is based on the Weierstrass M-Test where the power series ˇ ˇ at the point x ˇ ˇ which is less is compared to a convergent geometric series with common ratio ˇ xc yc ˇ than 1.
256
8 Sequences of Functions
PROOF: If the power series
1 P
an .x c/n converges when x D y, then the
nD0
series converges absolutely for all x satisfying jx cj < jy cj. • Let
1 P
an .x nD0 1 P
c/n be a power series that converges at x D y.
an .y c/n converges, its terms must approach 0 by the Limit of
• Since
nD0
• • • •
•
Terms Test. Thus, the terms must be bounded, and there exists a real number M such that jan .y c/n j M for every nonnegative integer n. Let x be any real number satisfyingˇ jx ˇn cj < jyˇ cj. ˇ ˇ ˇn n n ˇ xc ˇ Then jan .x c/ j D jan .y c/ j ˇ yc ˇ M ˇ xc . yc ˇ ˇ ˇn 1 P ˇ ˇ The series M ˇ xc is a convergent geometric series with common ratio yc ˇ nD0 ˇ ˇ ˇ xc ˇ ˇ yc ˇ < 1. 1 P Thus, jan .y c/n j converges absolutely by the Weierstrass M-Test. nD0
8.5.2 Interval of Convergence It follows immediately from the previous theorem that the radius of convergence for ˇ a power series is R D supfjy cj ˇ the series converges at yg, and that the power series converges absolutely for all x 2 .c R; c C R/. This does not say anything about how the power series behaves at the end points c R and c C R. There are examples of power series that converge at both endpoints, that converge at one of the two endpoints, or converge at neither endpoint. It also follows from the above proof, that if the power series converges absolutely at y, then it converges uniformly for all x satisfying jx cj jy cj. In particular, since all the partial sums of the series are continuous functions, if the power series has radius of convergence R > 0 and ı is any positive number less than R, then the series converges absolutely at x D cCRı, so the series converges absolutely and uniformly on Œc R C ı; c C R ı. As a 1 P result, the function f .x/ D an .x c/n is continuous on Œc R C ı; c C R ı for nD0
all small ı > 0, so it is continuous on the open interval .c R; c C R/. If the series converges absolutely for x D c C R, then f .x/ is continuous on the closed interval 1 P Œc R; c C R. What if the series an .x c/n converges conditionally at x D c C R nD0
or x D c R? Does this mean that the function is continuous at that endpoint? The answer is yes, but this takes some proof and is known as Abel’s Theorem.
8.5 Power Series
257
PROOF (Abel’s Theorem): Suppose the power series
1 P
an .x c/n has
nD0
positive radius of convergence R < 1, and that the series converges at one of the endpoints c R or c C R. Then the series is continuous on an interval from c R to c C R containing that endpoint. • Let
1 P
an .x c/n be a power series with positive radius of convergence
nD0
R < 1. • Assume that the series converges at one of the endpoints of the interval of convergence, c R or c C R. • Without loss of generality c D 0 and R D 1 because the argument can be applied to the series where x is replaced by Rx C c. Thus, assume that the 1 P series is an xn with radius of convergence 1. nD0
• Also, it can be assumed that the series converges at 1, because if it converges 1 P at 1, the argument can be applied to the series an .1/n xn which nD0
converges at 1. • Finally, by subtracting a constant from the constant term of the series, a0 , it 1 P can be assumed that an D 0. nD0
• Let this series have partial sums sk D
k P
an .
nD0
• Because lim sk D 0, the sk are bounded, and, in particular, k!1
1 P
sn x n
nD1
converges for all x with jxj < 1. 1 1 1 1 P P P P • Then an xn D a0 C .sn sn1 /xn D s0 C sn x n sn1 xn D nD0
s0 C
1 P
nD1
sn x .1 x/ s0 x D .1 x/ n
nD1
1 P
nD1
nD1
n
sn x .
nD0
• Let > 0 be given. • Because lim sn D 0, there is an integer N such that for all n N, jsn j < 2 . ˇ ˇ ˇ ˇ ˇ ˇ 1n!1 ˇ ˇ 1 N 1 ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇP P P P nˇ nˇ nˇ nˇ ˇ ˇ ˇ ˇ • Then ˇ an x ˇ D ˇ.1x/ sn x ˇ ˇ.1x/ sn x ˇ C ˇ.1x/ sn x ˇ nD0 nDNC1 ˇ nD0 N ˇ ˇ ˇ nD0 N 1 ˇ ˇ ˇ ˇ P P P NC1 ˇˇ.1x/ sn xn ˇˇ C.1x/ xn D ˇˇ.1 x/ sn xn ˇˇC 2 .1x/ x1x D 2 nD0 nDNC1 nD0 ˇ ˇ N ˇ NC1 ˇ P n ˇ.1 x/ sn x ˇˇ C 2 x . ˇ nD0
• Because the limit of this quantity as x approaches 1 from the left is 2 , there exists ı > 0 such that for all x between 1 ı and 1, this expression is less than . 1 P an xn D 0 which completes the proof. • This shows that lim x!1
nD0
258
8 Sequences of Functions
The Root Test can be used to determine the radius of convergence of a 1 P power series. The Root Test says that the series an .x c/n must converge if nD0 p . Conversely, the lim sup n jan j jx cjn < 1. Equivalently, jx cj < lim sup1 p n ja j n!1
series diverges if jx cj > must be R D
1p . lim sup n jan j
1p . lim sup n jan j
n!1
n
Thus, the radius of convergence of the series
n!1
n!1
1 P an .x c/n , you see that the series will converge If you apply the Ratio Test to nD0 ˇ ˇ ˇ a .xc/nC1 ˇ nj < 1. Equivalently, jx cj < lim jajanC1 . The series diverges if if lim ˇ nC1 ˇ n an .xc/ j n!1
n!1
nj nj showing that R D lim jajanC1 . This expression is fine as long jx cj > lim jajanC1 j j n!1 n!1 as the limit of the ratio of terms exists, but it is less helpful when it does not. It is worth considering a few examples.
•
1 P
nxn
nD0
The center is c D 0, and the radius of convergence is R D lim
n!1
1 p n
n
D
D 1. Clearly, the series diverges at both endpoints x D 1 and x D 1 by the Limit of Terms Test. 1 P n • .1/n .x1/ n lim nC1 n!1 n
nD1
1 The center is c D 1, and the radius of convergence is R D lim p D n 1 n!1
•
n
lim n D 1. At the right endpoint x D 0 the series is the harmonic series which n!1 nC1 diverges to infinity, but at the right endpoint x D 2 the series is the alternating harmonic series which converges conditionally. 1 P .xC4/n n2 5n
nD0
The center is c D 4, and the radius of convergence is R D lim .nC1/2 5nC1
n!1
1 p n 2 n n 5
D
D 5. At the right endpoint x D 1 the series converges absolutely lim n2 5n which means that the series will also converge at the left endpoint x D 9. 1 P xn
n!1
•
nD0
.2n/Š
The center is c D 0, and the radius of convergence is R D lim
n!1
lim
n!1
.2nC2/Š .2n/Š
D 1, so this series converges for all real numbers x.
q1 n
1 .2n/Š
D
8.5 Power Series
•
1 P
259
nn xn
nD0
The center is c D 0, and the radius of convergence is R D lim lim
•
nn
n!1
4
6
C 234 x4 C 215 x5 C 236 x6 C The center is c D 0. This is an example of a series where lim n!1
D
D 0. This series only converges at its center.
nC1 n!1 .nC1/ 1 22 2 x C 32 x C 213 x3 21
but lim inf
1 p n n n
1 p n a n
D
3 2
n!1
1 p n a n
does not exist, n
n D R. Note that lim inf anC1 D 0 and lim sup .nC1/ nC1 D 1, an
n!1
n!1
neither of which shed any light on the value of R. This series diverges at both endpoints by the Limit of Terms Test.
8.5.3 Differentiability A function represented by a power series in an interval with positive length is said to be analytic in that interval. Perhaps the most unusual property of analytic functions is that they are differentiable, and the derivative of a power series can be found by differentiating the series term by term to get a new series which converges with the 1 P same radius of convergence as the original series. That is, if f .x/ D an .x c/n 0
for all x satisfying jx cj < R, then f .x/ D
1 P
nD0
n an .x c/
n1
, where the new
nD1
series converges for these same values of x. It is easy to check that the radius of 1 P convergence of the derivative series n an .x c/n1 is the same as the original nD1 p p p series. This follows from the fact that lim sup n n an D lim n nlim sup n an D R1 , n!1
n!1
n!1
where R is the radius of convergence of the original series. 1 P It is, therefore, the case that g.x/ D n an .x c/n1 is a power series with the nD1
same radius of convergence as the power series f .x/ D
1 P
an .xc/n . The question is
nD0
whether this new power series is, in fact, the derivative of the original series. That is, does g.x/ D f 0 .x/ hold for all x in the open interval where the two series converge? .x/ This needs to be proved. The proof needs to show that lim f .xCh/f D g.x/ for h h!0
each x within the interval of convergence ˇ To construct a proof ˇ of the power series. ˇ ˇ .x/ g.x/ of this, you might express the difference ˇ f .xCh/f ˇ in terms of power series h and see if this simplifies to an expression that has a limit of 0 as h approaches 0. The calculation is simpler if you assume that c D 0. Then,
260
8 Sequences of Functions
ˇP ˇ 1 P ˇ1 ˇ ˇ ˇ ˇ ˇ an .x C h/n an xn X 1 ˇ ˇ f .x C h/ f .x/ ˇ ˇ nD0 nD0 n1 ˇ ˇ ˇDˇ g.x/ n a x n ˇ ˇD ˇ ˇ h h ˇ ˇ nD1 ˇ ˇ ˇ1 ˇ 1 1 n ˇP ˇ P P P n p np n ˇ h a x a x hn an xn1 ˇˇ n n ˇ nD0 pD0 p nD0 nD1 ˇ ˇ ˇ ˇ: ˇ ˇ h ˇ ˇ ˇ ˇ A careful accounting of the terms in the numerator shows that all the terms of 1 1 P P an xn and all of the terms of hn an xn1 cancel leaving nD0
nD1
ˇ1 ˇ n ˇP P ˇ ˇ n p np ˇˇ ˇ ! h a x ˇ1 ˇ n ˇ nD2 n pD2 p ˇ X X ˇ ˇ n ˇ ˇ p2 np ˇ ˇ an h x ˇ: ˇ ˇ D jhj ˇ ˇ ˇ h ˇ nD2 pD2 p ˇ ˇ ˇ ˇ ˇ The factor jhj clearly goes to 0 as h goes to 0, but there is a question about what happens to the other factor. This infinite sum will not be a problem if it remains bounded as h gets small. Here is where you can use the fact that power series with radius of convergence R converge absolutely at points less than a distance R from the center of the series. Assume that jhj is smaller than some fixed value s > 0. Then the second factor can be estimated as follows. ˇ ˇ ! ! ! ˇ1 ˇ 1 n 1 n n X X X ˇX X n p2 np ˇˇ X n n p2 np p2 np ˇ h jhj s an x jan j jxj jan j jxj ˇ ˇ p p ˇnD2 pD2 p ˇ nD2 pD2 nD2 pD2 ! n 1 1 X jan j X n p np 1 X s jxj D 2 jan j.jxj C s/n : 2 s s p nD2 pD0 nD2
This last expression converges as long as jxjCs is a point where the power series for f converges absolutely. But if x were chosen so that jxj < R, then for any positive s with s < R jxj, this will happen. Because you are freeˇ to choose any s ˇ> 0, you ˇ ˇ .x/ can choose one less than R jxj which will ensure that ˇ f .xCh/f g.x/ˇ is small h whenever 0 < jhj < s, so the proof can be completed.
8.5 Power Series
261
PROOF: Suppose the function f is defined by the power series 1 P an .x c/n which has a positive radius of convergence R 1. f .x/ D nD0
Then for all x satisfying jx cj < R, the derivative of f at x is given by 1 P n an .x c/n1 . f 0 .x/ D nD1
• Let
1 P
an .x c/n be a power series with positive radius of convergence
nD0
R 1. • The power series for f and its derivative depend on x c and not on c, so there is no loss of generality to assume p that c D 0. p p • Note that lim sup n n an D lim n n lim sup n an , so the two series 1 P
n!1
an .x c/n and
nD0
1 P
n!1
n!1
n an .x c/n1 have the same radius of convergence,
nD1
so each is absolutely convergent for all x with jxj < R. • Let x be chosen with jxj < R. • Let > 0 be given. • If R < 1, let s D Rjxj , and if R D 1, let s D 1. 2 1 P an .jxj C s/n converges absolutely. • Because jxj C s < R, the series nD0 0 1 • Let ı D min @s;
1C
1 P
s2 jan j.jxjCs/n
A > 0.
nD2
• Let h be chosen with 0 < jhj < ı. • Then ˇ ˇ 1 ˇ 1 ˇ ˇ ˇ ˇ ˇ P an .x C h/n P an xn 1 1 ˇ ˇ f .x C h/ f .x/ X ˇ ˇ X ˇ ˇ ˇ nD0 nD0 n1 n1 ˇ nan x n an x ˇD ˇ ˇDˇ ˇ ˇ ˇ ˇ h h nD0 nD1 ˇ ˇ ˇ ˇ ˇ 1 ˇ ˇ 1 ˇ 1 1 n n P P P P ˇ P ˇ ˇ P n p np n p np ˇˇ ˇ an x n hn an xn1 ˇˇ ˇˇ an p h x ˇ nD0 an pD0 p h x ˇ nD0 nD1 ˇ ˇ ˇ nD2 pD2 ˇ ˇ ˇDˇ ˇD ˇ ˇ ˇ ˇ h h ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ! ! ˇ1 ˇ 1 n n X X ˇX X n p2 np ˇˇ n ˇ jhj ˇ jhj h jhjp2 jxjnp an x ja j n ˇ p ˇnD2 pD2 p ˇ nD2 pD2
(continued)
262
8 Sequences of Functions
jhj
1 X nD2
jan j
! ! n 1 n 1 X X jan j X n p np n p2 np 1 X s jxj s jxj jhj D jhj 2 jan j.jxj C s/n < : 2 s s p p pD2 nD2 pD0 nD2
Thus, the derivative of f at x is
1 P
nan xn1 .
nD1
An immediate consequence of this theorem is that not only can you obtain the 1 P first derivative of f .x/ D an .x c/n by differentiating term by term, but you nD0
can also get all the higher derivatives of f by repeating the process. This follows by induction because, if the mth derivative of f is equal to the series formed by the mth derivatives of the terms of the series for f , and if that series has the same radius of convergence as the series for f , then the theorem says that the mC1st derivative of f can be obtained by differentiating the terms of the series for the mth derivative of f , and the radius of convergence of that series will remain the same. Moreover, one can find an antiderivative for f by integrating each term of the series for f . That is, if 1 1 P P an f .x/ D an .x c/n for all x with jx cj < R, then the series .x c/nC1 will nC1 nD0
nD0
have the same radius of convergence as the series for f , and the theorem says that the derivative of the new series is equal to f . It is important to note that if a function is analytic by virtue of having a power series representation in an open interval of radius R around c, then that function is infinitely differentiable in that interval. These results make it very simple to derive new series from previously known 1 P 1 series. For example, you already know that 1x D xn for all x with jxj < 1. nD0
From this one can get • by substituting x for x in the series for • by substituting x2 for x in the series for • by differentiating the series for
1 , 1 1x 1Cx
D
1 , 1 1Cx 1Cx2
D
1 , 1 1x .1x/2
D
1 P
1 P
.1/n xn .
nD0 1 P
.1/n x2n .
nD0
nxn1 .
nD1
1 and noting that ln 1 D 0, ln.1 C x/ D • by integrating the series for 1Cx 1 1 P P nC1 n .1/n xnC1 D .1/n1 xn . In particular, by Abel’s theorem, nD0
nD1
ln 2 D 1 12 C 13 14 C . 1 1 0 D 0, tan1 x D • by integrating the series for 1Cx 2 and noting that tan 1 P 2nC1 .1/n x2nC1 . In particular, by Abel’s Theorem, 4 D 1 13 C 15 17 C . nD0
8.5 Power Series
263
8.5.4 Taylor’s Theorem If f .x/ D
1 P
an .x c/n for all x with jx cj < R, then f .c/ D a0 , the
nD0
constant term of the series for f . Finding the mth derivative of the series for f and evaluating it at the center of the series, c, gives that f .m/ .c/ D mŠam . So, for all .m/ integers m 0, am D f mŠ.c/ . This gives a straightforward way to generate the power series representing any analytic function. Moreover, even if f is not infinitely differentiable, if it is m times differentiable, one can generate the mth degree Taylor m .n/ P f .c/ polynomial for f centered at c given by g.x/ D .x c/n . Then g is an mth nŠ nD0
degree polynomial that is equal to f at c, and all of its derivatives up to order m agree with the corresponding derivatives of f at c. In particular, the first degree Taylor polynomial is just the familiar linear approximation to f given by the line tangent to the graph of f at c. If f is m-times differentiable at c, one can generate the mth degree Taylor polynomial, g.x/, for f centered at c, but this does not say whether the value of g.x/ is even remotely related to the value of f .x/ when x is different from c. This issue is what is addressed by Taylor’s Theorem which states that f .x/ D g.x/ C Rm .x/ for some remainder function Rm .x/. Depending on various characteristics of f , one can show that Rm .x/ is suitably small so that g.x/ is a good approximation for f .x/. There are many forms of Taylor’s Theorem that express the remainder term, Rm .x/, in different ways. The one discussed here is sometimes called Lagrange’s form. It says that if f is m C 1 times differentiable on the interval between c and x, then the difference between f .x/ and the mth degree Taylor polynomial for f centered at c can be expressed in terms of f .mC1/ ./ for some strictly between c and x. Its proof follows easily from the following generalization of Rolle’s Theorem. PROOF (Higher Order Rolle’s Theorem): Let f be an m C 1 times differentiable function on the open interval from a to b with a ¤ b, let f be continuous on the closed interval from a to b, and suppose that 0 D f .a/ D f 0 .a/ D f 00 .a/ D D f .m/ .a/ D f .b/. Then there is an x strictly between a and b where f .mC1/ .x/ D 0. • Let f be an m C 1 times differentiable function on the open interval from a to b with a ¤ b, let f be continuous on the closed interval from a to b, and suppose that 0 D f .a/ D f 0 .a/ D f 00 .a/ D D f .m/ .a/ D f .b/. • Since f .a/ D f .b/, f is continuous on the closed interval between a and b, and f is differentiable between a and b, then by Rolle’s Theorem there is an x1 strictly between a and b such that f 0 .x1 / D 0. • Assume for some k with 1 k m, that f .k/ .a/ D f .k/ .xk / D 0, f .k/ is continuous on the closed interval between a and xk , and f .k/ is differentiable between a and xk . Then by Rolle’s Theorem there is an xkC1 strictly between a and xk such that f .kC1/ .xkC1 / D 0. • Thus, by mathematical induction, there is an x D xmC1 strictly between a and b such that f .mC1/ .x/ D 0 completing the proof.
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8 Sequences of Functions
This Higher Order Rolle’s Theorem can now be used to prove Taylor’s Theorem. If the function f is m C 1 times differentiable between c and x, then f has an m .n/ P f .c/ mth degree Taylor polynomial g.y/ D .y c/n . Notice that the difference nŠ nD0
f .y/ g.y/ has the property that this function and its first m derivatives are all equal to 0 at c. The remainder term RmC1 .x/ will include a factor of f .mC1/ evaluated at some between c and x, and that value of will come from an application of Rolle’s Theorem. Of course, to apply Rolle’s Theorem, f .y/g.y/ would need to be 0 at y D x. One needs to add a term to f .y/ g.y/ which will not affect the function and its derivatives at c but will make the function equal to 0 at x. The term that accomplishes .yc/mC1 this is f .x/g.x/ .xc/ mC1 since this term equals f .x/g.x/ at x, and it and its first m derivatives are equal to 0 at c. But now Rolle’s Theorem can be applied to the .yc/mC1 function h.y/ D f .y/g.y/ f .x/g.x/ .xc/ mC1 to find a value of between c and x .mC1/Š such that h.mC1/ ./ D 0, or 0 D f .mC1/ ./g.mC1/ ./.f .x/g.x// .xc/ mC1 . Noting
that the mC1st derivative of g at c is equal to 0 gives f .x/ D g.x/Cf .mC1/ ./ .xc/ .mC1/Š as desired.
mC1
PROOF (Taylor’s Theorem): Let f be an m C 1 times differentiable function on the open interval from c to x with c ¤ x, and let f be continuous on the closed interval from c to x. Then there is an between m .n/ P mC1 f .c/ c and x such that f .x/ D .x c/n C f .mC1/ ./ .xc/ . nŠ .mC1/Š nD0
• Let f be an m C 1 times differentiable function on the open interval from c to x with c ¤ x, and let f be continuous on the closed interval from c to x. m .n/ P f .c/ • Let g.x/ D .x c/n , and define the function nŠ nD0 .yc/mC1 h.y/ D f .y/ g.y/ f .x/ g.x/ .xc/ mC1 . • Then 0 D h.c/ D h0 .c/ D h00 .c/ D D h.m/ .c/ D h.x/. • Thus, by the Higher Order Rolle’s Theorem, there exists between c and x such that h.mC1/ ./ D 0. m .n/ P mC1 f .c/ • This implies that f .x/ D .x c/n C f .mC1/ ./ .xc/ which nŠ .mC1/Š nD0
completes the proof. For example, the cosine function is analytic, and its power series which converges 2 4 6 8 for all real numbers is cos x D 1 x2Š C x4Š x6Š C x8Š . So, how accurate 2 4 of an approximation is 1 x2Š C x4Š at x D 2? It is clear that the given Taylor polynomial includes the terms for n D 0, 1, 2, 3, and 4, but it is beneficial to note that it also includes the term for n D 5 which is 0. Therefore, Taylor’s Theorem 6 26 D cos./ 720 . Since the cosine says that the remainder at x D 2 is f .6/ ./ .20/ 6Š function is bounded by 1, the error introduced by using the Taylor polynomial as an 26 approximation to cos 2 is at most 720
0:09. In fact, at x D 2 the polynomial is 13 while cos 2 0:416146 with a difference of 0:08281.
8.5 Power Series
265
8.5.5 Arithmetic of Power Series Given two analytic functions each represented by power series with common center c and positive radii of convergence, it is straightforward to find the power series representing the sum, difference, product, and quotients of these series. Suppose 1 1 P P two functions have power series f .x/ D an .x c/n and g.x/ D bn .x c/n nD0
nD0
which both converge when jx cj < R for some R > 0. Then theorems about the sum and difference of series of real numbers ensure that the sum and difference, 1 1 P P .an Cbn /.xc/n and .f g/.x/ D .an bn /.xc/n , both converge .f Cg/.x/ D nD0
nD0
when jx cj < R. Of course, it is possible that the new series converges in an even larger interval. For example, the series 1 C x C x2 C x3 C and 2 x x2 x3 both have radius of convergence equal to 1, but the sum of the two series is the constant function 3, and its power series converges for all x. The product of two power series can be found by using the Cauchy product of 1 P the two series. If f .x/ D an .x c/n has radius of convergence R1 > 0 and g.x/ D
1 P
nD0
bn .x c/ has radius of convergence R2 > 0, then both series converge n
nD0
absolutely when jxcj < min.R1 ; R!2 / implying that their Cauchy product, .fg/.x/ D ! 1 n 1 n P P P P ap .x c/p bnp .x c/np D ap bnp .x c/n , converges for nD0
pD0
nD0
pD0
jx cj < min.R1 ; R2 /. Again, the radius of convergence can be larger as is with 1 the product of 1x D 1 C x C x2 C x3 C and 1 x which converges for all x. 1 1 P P an .xc/n has radius of convergence R1 > 0 and g.x/D bn .xc/n If f .x/ D nD0
nD0
has radius of convergence R2 > 0, and g.c/ is not zero, then one can find the power f .x/ series for the quotient h.x/ D g.x/ centered at c by working backwards from the 1 P Cauchy product of h and g. That is, if you assume that h.x/ D qn .x c/n , then nD0 ! 1 1 n P P P f .x/ D an .x c/n D h.x/g.x/ D bp qnp .x c/n . Because of the nD0
nD0
pD0
assumption that g.c/ ¤ 0, it follows that b0 ¤ 0. Then equating like terms in the product gives the sequence of equations a0 Db0 q0 a1 Db0 q1 C b1 q0 a2 Db0 q2 C b1 q1 C b2 q0 a3 Db0 q3 C b1 q2 C b3 q1 C b4 q0
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8 Sequences of Functions
and so forth. The first equation can be solved to give q0 . Then the second equation can be solved to give q1 , and so forth. The fact that g.c/ ¤ 0 says that the coefficient b0 ¤ 0 which allows the equation for am to be solved for qm for each m 0. Often this results in a recursive formula for qn . For example, it is known that ln.1 C x/ D 2 3 4 x x2 C x3 C x4 , so you can find the series centered at 0 for the quotient 1 1 P P ln.1Cx/ n n D qn x by writing .1 C x/ qn x giving 0 D 1 q0 so q0 D 0. Then 1Cx nD0
for each n > 0, with qn D
nD0
.1/n1 n
.1/n1 n
D qn C qn1 , so q1 D 12 , q2 D 56 , q3 D 13 , and so forth 12
qn1 .
8.5.6 Exercises Determine for which x the following power series converge. 1. 2. 3. 4.
1 P nD0 1 P nD0 1 P nD0 1 P nD0
4n .xC4/n 3n C5n n5 .x2/n 8n nŠxn .2n/Š nŠxn nn
Determine power series representations for the following functions centered at c D 0. 5. 6. 7. 8. 9. 10.
ex e2x sin.3x/ sin x 1Cx
ln.cos x/ 3 5 2 4 Using the fact that sin x D x x3Š C x5Š and cos x D 1 x2Š C x4Š , show that the powers series satisfy the identity sin2 x C cos2 x D 1. 11. Find the first four nonzero terms of the series for tan x centered at 0 by finding the quotient of the series for sin x and the series for cos x. Then check your work by generating those terms using ( ) Taylor’s Theorem. 1
e x2 if x > 0 . Prove that for each positive integer n, the 0 if x 0 derivative f .n/ .0/ D 0. Then show that the mth degree Taylor polynomial for f centered at 0 is p.x/ D 0, and the remainder term is Rm .x/ D f .x/.
12. Let f .x/ D
8.6 Fundamental Question of Analysis
267
8.6 Fundamental Question of Analysis In a sense Analysis can be thought about as the study of limiting processes. So far this book has discussed limits of functions and sequences, the continuity of functions, differentiation of functions, integration of functions, the convergence of infinite series, and now the convergence of sequences and series of functions. In each of these studies one fundamental question recurs: when is it valid to interchange the order of limiting processes. For example, the question of continuity is a question of whether lim f .x/ is the same as f lim x . The Fundamental Theorem of Calculus x!a x!a establishes when the derivative of an integral is equal to the integral of a derivative. The discussion of convergence of sequences of functions included questions about Rb Rb when lim fn .x/ dx is the same as lim fn .x/ dx. Power series give an example
a n!1 1 d P where dx an .x nD0
n!1 a 1 P
c/n is the same as
whether it is valid to write lim x!R
1 P nD0
nD0
an xn D
d a .x dx n 1 P
c/n . Abel’s Theorem discusses
lim an xn . Thus, the fundamental
nD0 x!R
question of Analysis asks “when can you interchange the order of two limiting processes?” It is instructive to watch for other occurrences of this question as your study of Analysis continues.
Chapter 9
Topology of the Real Line
9.1 Interior, Exterior, and Boundary In the field of Analysis the concepts of the limit and the continuity of a function f at a point x D a are defined in terms of open intervals. For example, the condition jf .x/ Lj < says that f .x/ is in an open interval centered at L, and the condition jx aj < ı says that x is in an open interval centered at a. These intervals are specified in terms of the distance between x and y given by jx yj. Topology is a branch of Mathematics where these concepts are extended to spaces where one can discuss intervals without having to rely on a distance formula. As a result the concepts of limit and continuity can be extended to such spaces, and it can be shown that many of the properties associated with continuous functions defined on the real line are shared by continuous functions defined on these more general spaces. Although the theorems discussed in this chapter are presented in the context of sets on the real line, virtually all of the theorems are true in the more general context of any topological space. Many of the techniques used to prove these theorems are the same techniques one would use for a general topological space, and, therefore, this chapter can be thought of as an introduction to the field of Topology even though general topological spaces are not discussed here. A good way to begin is by taking a set S R and identifying the points s 2 S that are not only inside of S but are, in a sense, completely surrounded by points in S. The point s is said to be in int.S/, called the interior of S, if there is an > 0 such that all x within of s are in S, that is, jx sj < implies x 2 S. You can think of the interior of S as those points which are a positive distance from the complement of S, Sc D RnS. For example, if S is the closed interval Œ0; 4, then the open interval .0; 4/ is the interior of S. This is because if s 2 .0; 4/ and D min.s; 4 s/, then all x satisfying jx sj < are elements of S. The two endpoints of the interval Œ0; 4, 0 and 4, do not have this property. No open interval containing either 0 or 4 is completely contained inside of S. Clearly, if x > 4 or x < 0, then x … S, so x … int.S/. Thus, int.S/ D .0; 4/. The interior of the set Q of rational numbers is © Springer International Publishing Switzerland 2016 J.M. Kane, Writing Proofs in Analysis, DOI 10.1007/978-3-319-30967-5_9
269
270
9 Topology of the Real Line
Fig. 9.1 The point x is in the interior of S: The point y is on the boundary of S: The point z is in the exterior of S
z
y x S SC
the empty set because all nonempty open intervals contain irrational numbers, so no nonempty open interval is contained in Q. One sometimes says that Q has no interior even though it does have an interior; it is just that its interior is the empty set. Then ext.S/, called the exterior of S, is just defined to be the interior of Sc , that is, s 2 ext.S/ if there is an > 0 such that all x satisfying jx sj < are in Sc D RnS. The exterior of S is the set of points that are completely surrounded by points outside of S. You can think of the exterior of S as the collection of points bounded away from S, that is, the points that are a positive distance from S. The exterior of the set Œ0; 4 is the union of two open intervals, .1; 0/ [ .4; 1/. The exterior of the set Q is the empty set. If a point s is in neither int.S/ nor ext.S/, then it must be that no open interval containing s is completely inside of S and no open interval containing s is completely outside of S. Thus, for every > 0, the interval .s ; s C / contains at least one element of S and at least one element of Sc . Such points are said to be in @S, called the boundary of S (Fig. 9.1). Note that the symbol used for boundary is @ which is the same symbol use for partial derivatives in Calculus. There are connections between derivatives and boundaries that justify the use of the same symbol for both concepts. The boundary of Œ0; 4 is the set f0; 4g. The boundary of Q is the entire real line, R. It is important to note that for any set S R, the three sets int.S/, ext.S/, and @S partition R, that is, each real number x is in exactly one of these three sets. A proof of this fact must show two things about a set S: that R D int.S/ [ ext.S/ [ @S, and that no point x belongs to more than one of these sets. To show that R is a union of the three sets, you would take an arbitrary x 2 R and show that it is in at least one of these sets. One way to show that a point must be one of three things is to assume that it not one of the first two, and then prove that it must be the third. In this case, you can assume that a point x 2 R is not in int.S/ or in ext.S/. If x is not in int.S/, then
9.1 Interior, Exterior, and Boundary
271
for every > 0, the open interval .x ; x C / is not contained in S, and if x is not in ext.S/, then for every > 0, the open interval .x ; x C / is not contained in Sc . The only alternative is that if x is in neither int.S/ nor ext.S/, then for every > 0, the open interval .x ; x C / contains points in both S and its complement. This means that x is in @S implying that x must be in at least one of int.S/, ext.S/, or @S. To show that the three sets are disjoint, show that if x belongs to one of the three sets, it cannot belong to either of the other two sets. These inferences follow directly from the definitions of the sets. PROOF: For every set S R, R D int.S/ [ ext.S/ [ @S and the three sets int.S/, ext.S/, and @S are mutually disjoint. • Let S R. • Assume that x is a real number that is not a member of int.S/ or ext.S/. • Then, because x … int.S/, for every > 0, the open interval .x ; x C / is not contained in S, so it contains points of Sc . • And because x … ext.S/, for every > 0, the open interval .x ; x C / is not contained in Sc , so it contains points of S. • It follows that for every > 0, the open interval .x ; x C / contains points in S and points in Sc . • Thus, by the definition of boundary, x 2 @S, and this shows that x must be in at least one of the three sets, int.S/, ext.S/, or @S. • If x 2 int.S/, then there is an > 0 such that the open interval .x ; x C / S. • But then x 2 S, so x … ext.S/, and .x ; x C / S shows that x … @S. • Similarly, if x 2 ext.S/, then it cannot be in int.S/ or @S. • Thus, no x 2 R is a member of more than one of the three sets which completes the proof. There are many results that follow directly from the definitions of interior, exterior, and boundary. For example, if S and T are any subsets of R, then • • • • • • • • • • • • • •
int.int.S// D int.S/. int.ext.S// D ext.S/. int.S/ ext.ext.S//. ext.S/ ext.int.S//. @
[email protected]// @S. @.int.S// @S. @.ext.S// @S. @.S/ D @.Sc /. int.S/ [ int.T/ int.S [ T/. ext.S [ T/ ext.S/ \ ext.T/. int.S \ T/ D int.S/ \ int.T/. @.S [ T/ @S [ @T. if S T, then int.S/ int.T/. if S T, then ext.T/ ext.S/.
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9 Topology of the Real Line
Each of these results is a statement about either two sets being equal to each other or one set being a subset of another. Thus, one would prove these results using the techniques discussed in Chap. 2 for proving subset and set equality statements. For example, how would you write a proof that for any set S, int.int.S// D int.S/? This would be a proof that two sets are equal, so the proof would consist of two parts: showing int.int.S// int.S/ and showing int.S/ int.int.S//. The fact that int.int.S// int.S/ is just a consequence of the definition of interior. For any set T, int.T/ T, so certainly int.int.S// int.S/. Showing that int.S/ int.int.S// is showing that one set is a subset of another. So, you would let x be an element of int.S/, and then show that x is also an element of int.int.S//. By the definition of interior, there is an > 0 such that the open interval .x ; x C / S. Thus, you need to show that .x ; x C / is contained in int.S/. That is, each y 2 .x ; x C / must be in the interior of S. But it is easy to find an open interval centered at y that is contained in .x ; x C /. Just let ı D min.y .x /; x C y/ > 0 because then .y ı; y C ı/ .x ; x C /. This shows each point of .x ; x C / is in int.S/ which completes the proof. PROOF: For every set S R, int.int.S// D int.S/. • • • • • • • •
Let S R. For any set T, int.T/ T, so int.int.S// int.S/. So let x 2 int.S/. By the definition of interior, there is an > 0 such that the open interval .x ; x C / is contained in S. Let y 2 .x ; x C /, and let ı D min.y .x /; x C y/ > 0. Then .y ı; y C ı/ .x ; x C / S. This shows that .x ; x C / int.S/ implying that x is in int.int.S//. This proves that int.S/ int.int.S// and completes the proof of the theorem.
For a more difficult challenge, consider writing a proof that for any set S, @.@S/ @S which, in words, says that the boundary of the boundary of a set is contained in the boundary of the set. For example, let S be the set of rational numbers in the interval Œ0; 4. You should prove to yourself that the boundary of this set is the entire interval Œ0; 4. The boundary of that interval is just f0; 4g which indeed is contained in @S D Œ0; 4. To show that @.@S/ is a subset of @S, you would take an arbitrary point x 2 @.@S/ and show that it is in @S. So what do you know if x 2 @.@S/? The only tool you have at your disposal here is the definition of the boundary of a set, so you would proceed to use that definition. It says that for every > 0 the open interval .x ; x C / contains elements of @S and elements of the complement of @S. You want to show that x is in @S, so you would need to show that the open interval .x ; x C / contains elements of S and elements of Sc . Well, what is the consequence of saying that the open interval .x; xC/ contains elements of @S? It must mean that there is a y 2 .x ; x C / such that y 2 @S. What does it mean for y to be in @S? It means that for every ı > 0, the interval .yı; yCı/ contains elements of S and elements of Sc . But this is sufficient if .y ı; y C ı/ .x ; x C / because
9.1 Interior, Exterior, and Boundary
273
Fig. 9.2 x in @.@S/, y in @S
((S
) c
y S
x
)
that would put elements of both S and Sc in .x ; x C /. This can be arranged by selecting ı small enough (Fig. 9.2). PROOF: For every set S R, @.@S/ @S. • Let S R. • Let x 2 @.@S/. • Then by the definition of boundary, for every > 0, the open interval .x ; x C / contains points of @S and points of the complement of @S. • Let > 0 be given, and let y 2 .x ; x C / such that y 2 @S. • Let ı D min.y.x/; xCy/ > 0 so that the open interval .yı; yCı/ .x ; x C /. • By the definition of boundary, the interval .y ı; y C ı/ contains an element of S and an element of Sc . • But .y ı; y C ı/ .x ; x C / shows that .x ; x C / contains an element of S and an element of Sc , so x is in @S which completes the proof. As a third example, consider proving that for any two sets S and T, that int.S/ [ int.T/ int.S [ T/. Again, this is proving that one set is a subset of a second set, so your proof would start by selecting an arbitrary element of the first set and then proceed to show that that element belongs to the second set. Here the first set is int.S/ [ int.T/. If you select an x from this set, all you know about x is that it is in the union of the two sets int.S/ and int.T/. So, the only tool you can use is the definition of union to say that x must be either a member of int.S/ or a member of int.T/. In the case that x 2 int.S/, you can then apply the definition of interior to say that there is an > 0 such that the interval .x ; x C / S. But this is all you need since S S [ T showing .x ; x C / S [ T proving that x 2 int.S [ T/. The case where x 2 int.T/ is analogous, completing the proof. PROOF: For any sets of real numbers S and T, int.S/ [ int.T/ int.S [ T/. • Let S and T be sets of real numbers. • Let x 2 int.S/ [ int.T/. • Then by the definition of the union of two sets, either x 2 int.S/ or x 2 int.T/. • Without loss of generality, assume that x 2 int.S/. • Then there is an > 0 such that the open interval .x ; x C / S. • But since S S [ T, it follows that .x ; x C / S [ T showing that x 2 int.S [ T/, completing the proof. Can it be that int.S/ [ int.T/ is not equal to int.S [ T/? The answer is yes. See if you can think of an example.
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9 Topology of the Real Line
9.1.1 Exercises For each of the following sets, find the interior, exterior, and boundary of the set. 1. Œ0; 3/ [ .3; 6 1 1 1 2. [ 2n . ; 2n1 nD1
3. Œ0; 4 \ Q Write proofs for each of the following statements. For exercises involving the subset relation rather than the equality relation, give examples showing that the subset relation in the statement cannot be replaced by an equality. If S T, then int.S/ int.T/. If S T, then ext.T/ ext.S/. int.ext.S// D ext.S/. ext.S/ ext.int.S//. @.int.S// @S. @.S/ D @.Sc /. @.ext.S// @S. int.S \ T/ D int.S/ \ int.T/. ext.S [ T/ D ext.S/ \ ext.T/. @.S [ T/ @S [ @T. int.S/ ext.ext.S//
4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
9.2 Open and Closed Sets A set S of real numbers is called open if for every s 2 S there is an > 0 such that the open interval .x; xC/ S. A set S of real numbers is called closed if @S S. The intervals that are called open intervals are, in fact, open sets. In particular, .1; 7/, .2; 1/, and ; are all open sets as well as .3; 3/ [ .5; 9/ [ .10; 41/ and 1
[ .2n; 2n C 1/. The intervals that are called closed intervals are, in fact, closed sets.
nD1
In particular, Œ5; 3, Œ4; 1/, and ; are all closed sets. There are actually many equivalent ways to define open and closed sets, so one usually begins this discussion by proving that all the different definitions are equivalent. In particular, if S R, then the following are equivalent: 1. 2. 3. 4.
S is an open set. S D int.S/. S \ @S D ;. Sc is a closed set.
Many theorems in mathematics are statements of the form p , q, and the proof of these statements is often broken into two steps: p ) q and q ) p. Theorems of that type state that two conditions are equivalent. But it is not uncommon to have
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a theorem that states that several statements are equivalent, that is, p1 , p2 , p3 , , pk . One way to prove theorems of this form is to show in a sequence of steps that p1 ) p2 , p2 ) p3 , p3 ) p4 , . . . , pk1 ) pk , and then pk ) p1 . This is the technique you can use to prove the list of statements about open sets. You would begin by assuming condition 1, that a set S is open and then prove condition 2, that S D int.S/. This can be done by noting that for any set, elements of the set are either in the interior of the set or on the boundary of the set. But if the set S is open, it means that for each x 2 S there is an > 0 such that the open interval .x ; x C / S. Thus, .x ; x C / contains no elements of Sc showing that x cannot be in @S, so it must be that x 2 int.S/ which proves that S D int.S/. Now, assuming condition 2 that S D int.S/ it follows immediately that S \ @S D int.S/ \ @S D ;, which is condition 3. If you assume condition 3 that S \ @S D ;, how can you conclude that Sc is closed? Well, if S contains no elements of @S, it must be that all the elements of @S (if there are any) must belong to Sc . But as seen in the exercises of the previous section, the boundary of S and the boundary of Sc are always the same. This follows from the fact that the definition of boundary is symmetric in its references to S and Sc . Therefore, Sc contains its boundary proving that Sc is a closed set, which is condition 4. Finally, assuming condition 4 that Sc is a closed set, you know that Sc contains its boundary, so Sc contains the boundary of S. You must show that for each x 2 S, there is an open interval centered at x such that the entire interval is contained in S. But if for every > 0 the open interval .x ; x C / contains elements in Sc , then x would be in the boundary of S which is false. Thus, there is an > 0 such that the open interval .x ; x C / is contained in S. This proves that S is an open set, which is condition 1 (Fig. 9.3). Fig. 9.3 An open set S, its boundary, and its complement Sc
S
∂S SC
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PROOF: Let S R. Then the following statements are equivalent. 1. 2. 3. 4.
S is an open set. S D int.S/. S \ @S D ;. Sc is a closed set.
• Let S R. Condition 1 ) Condition 2 • Assume that S is an open set. • If x 2 S, then by the definition of open set, there exists an > 0 such that .x ; x C / S. • But .x ; x C / S shows that x is not an element of @S. • Since S int.S/ [ @S, it can be concluded that S int.S/. • Because the interior of any set is contained in the set, it follows that int.S/ S implying that S D int.S/, which is condition 2. Condition 2 ) Condition 3 • Assume that S D int.S/. • Then S \ @S D int.S/ \ @S D ; because the interior and the boundary of any set are disjoint. • Thus, S \ @S D ;, which is condition 3. Condition 3 ) Condition 4 • Assume that S \ @S D ;. • Then @S must be contained in Sc . • Because @S D @.Sc /, it follows that @.Sc / Sc implying that Sc is a closed set, which is condition 4. Condition 4 ) Condition 1 • Assume that Sc is a closed set, which means that Sc contains @.Sc /. • Let x 2 S. • If for every > 0, the open interval .x ; x C / contains elements of Sc , then x would be an element of @S D @.Sc /. • But all elements of @.Sc / are contained in Sc , so there must be an > 0 such that the interval .x ; x C / contains no elements of Sc implying that .x ; x C / S. • This shows that S is an open set, which is condition 1. A similar theorem can be proved concerning closed sets.
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PROOF: Let S R. Then the following statements are equivalent. 1. 2. 3. 4.
S is a closed set. S D int.S/ [ @S. Every accumulation point of S is an element of S. Sc is an open set.
• Let S R. Condition 1 ) Condition 2 • Assume that S is a closed set. • Because no point in the exterior of S is a member of S, it is clear that S int.S/ [ @S. • Because S is closed, it contains @S, and because all sets contain their interior, S contains int.S/. • Thus, S contains int.S/ [ @S proving that S D int.S/ [ @S, which is condition 2. Condition 2 ) Condition 3 • Assume that S D int.S/ [ @S. • Let x be an accumulation point of S. • If x … S, then for all > 0, the open interval .x ; x C / contains elements of S (since x is an accumulation point of S) and elements of Sc (in particular, x). • Thus, x 2 @S implying that x 2 S, a contradiction. • Therefore, all accumulation points of S must be elements of S, which is condition 3. Condition 3 ) Condition 4 • Assume that every accumulation point of S is an element of S. • Let x 2 Sc . • Because S contains all of its accumulation points, x is not an accumulation point of S. • Thus, there is an > 0 such that the open interval .x ; x C / contains no elements of S, and is, therefore, contained in Sc . • This shows that Sc is an open set, which is condition 4. Condition 4 ) Condition 1 • Assume that Sc is an open set. • Then Sc \ @.Sc / D ; implying that @.Sc / S, so @S S. • Thus, S contains @S, so S is a closed set, which is condition 1. Of course, a set need not be either open or closed as is seen by the interval .0; 5 which contains one but not both of its boundary points (Fig. 9.4), so it is neither open (because it contains a boundary point) nor closed (because it does not contain all of its boundary points).
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Fig. 9.4 Boundaries are drawn dotted to indicate open sets. Boundaries are drawn solid to indicate closed sets
Open
Closed
9.2.1 Exercises Determine which of the following sets of real numbers are open and which are closed. 1. 2. 3. 4.
.2; 2/ [ .2; 6/ [ .6; 10/ R the irrational numbers the real numbers that are not integers
Write proofs for each of the following statements. 5. 6. 7. 8. 9. 10.
If S is an open set, and T is a closed set, then SnT is an open set. If S is an open set, and T is a closed set, then TnS is a closed set. ; is both an open set and a closed set. If x is an accumulation point of a set S, and x … S, then x 2 @S. If x 2 @S and x … S, then x is an accumulation point of S. Let be any sequence. Let A be the set of all values y such that there exists a subsequence of that converges to y. Then A is a closed set.
9.3 Unions and Intersections Perhaps the most important properties open sets have are that the union of any collection of open sets is itself an open set and that the intersection of a finite collection of open sets is itself an open set. In fact, these two properties of open sets are the defining conditions required to hold in the more general setting of topological spaces (Fig. 9.5). In the context of the real numbers, it is not hard to show that the union of any collection of open sets is itself an open set. But before this proof can be started, there needs to be a convenient way to discuss an arbitrary collection of open sets.
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Fig. 9.5 The union of open sets is an open set
If the open sets are listed as A1 ; A2 ; A3 ; : : : ; Ak , then there is an implication that this is a collection of a finite number of sets because the indices used to describe the collection, f1; 2; 3; : : : ; kg, is a finite set. If the collection is listed as a sequence, A1 ; A2 ; A3 ; : : : , then there is an implication that the collection of sets is denumerable, that is, there is one set for each natural number. But to allow the collection of open sets to be any size, even to be an uncountable collection of sets, one generally wants to represent the collection as a collection of sets Ai where the index i is allowed to range over a particular index set, I. That is, the collection is given by fAi j i 2 Ig. Thus, since there is no restriction on the size of the index set I, there is no restriction on the size of the collection of open sets. So assume that fAi j i 2 Ig is a collection of open sets. How would you prove that its union, A D [ Ai , is an open set? Given the theorem about open sets in the i2I
previous section, you could prove that the set A is open by using the definition of open set, by showing that A D int.A/, by showing that A \ @A D ;, or by showing that Ac is a closed set. In this case, it is simple enough to use the definition of open set. Thus, for each x 2 A, you would need to show that there is an open interval centered at x such that this open interval is contained in A. All you know about A is that it is a union of a collection of open sets, so the first thing you should try is invoking the definition of union. That is, if x 2 A, then there must be a j 2 I such that x 2 Aj . What do you know about Aj ? Only that it is an open set. That means that there is an > 0 such that the open interval .x ; x C / is contained in Aj . But by the definition of union, Aj A implying that .x ; x C / A which is what you needed to prove.
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PROOF: Assume that for each i in the index set I, Ai is an open set. Then [ Ai is an open set. i2I
• Assume that for each i in the index set I, Ai is an open set. • Let x 2 [ Ai . i2I
• By the definition of set union, there is a j 2 I such that x is an element of the open set Aj . • By the definition of open set, there is an > 0 such that the open interval .x ; x C / Aj . • But by the definition of set union, Aj [ Ai showing that .x ; x C / [ Ai , which proves the theorem.
i2I
i2I
Now consider proving the result that the intersection of a finite collection of open sets is itself an open set. This time there is no need to consider an arbitrarily large collection of open sets; you can just use the finite collection of open sets A1 ; A2 ; A3 ; : : : ; Ak . Again you would take an arbitrary x 2 A1 \ A2 \ A3 \ \ Ak . You know from the definition of intersection that for each j D 1; 2; 3; : : : ; k, this element x must be in Aj . And you know that since Aj is an open set, there must be an j > 0 such that the interval .x j ; x C j / Aj . Now you have a collection of k open intervals each centered at x. By selecting D min.1 ; 2 ; 3 ; : : : ; k /, you will have the least of these j values which is a positive number. This is crucial. The fact that you have a finite collection of open sets ensures that you can find a finite number of open intervals centered at x and can find the shortest of these intervals. If the collection of open sets were infinite, there would be no guarantee that there would be a minimum j . The fact that there is a minimum value that is greater than 0 allows you to claim that the interval .x ; x C / is contained in each of the Aj sets, and thus, .x ; x C / is contained in the intersection of the Aj ’s. PROOF: Assume that A1 ; A2 ; A3 ; : : : ; Ak A1 \ A2 \ A3 \ \ Ak is an open set.
are
open
sets.
Then
• Assume that A1 ; A2 ; A3 ; : : : ; Ak are open sets. • Let x 2 A1 \ A2 \ A3 \ \ Ak . • Then for each j D 1; 2; 3; : : : ; k, x is an element of the open set Aj , and because Aj is an open set, there exists an j > 0 such that the open interval .x j ; x C j / Aj . • Let D min.1 ; 2 ; 3 ; : : : ; k / > 0. • Then the open interval .x ; x C / is contained in Aj for each j D 1; 2; 3; : : : ; k. • This shows that .x ; x C / A1 \ A2 \ A3 \ \ Ak proving that this intersection is an open set. There are analogous results about the union and intersections of closed sets. In particular, the intersection of an arbitrary collection of closed sets is itself a closed set, and the union of a finite number of closed sets is itself a closed set. One can prove these results by relying on the definition of a closed set, but it is much easier
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to use the results from the previous section that show that a set is a closed set if and only if it is the complement is an open set. For example, to show that the union of a finite number of closed sets is closed, let A1 ; A2 ; A3 ; : : : ; Ak be a finite collection of closed sets. Then for each j, Acj is the complement of a closed set, so it is open. By the previous theorem, the intersection of a finite number of open sets is an open set, so Ac1 \ Ac2 \ Ac3 \ \ Ack is an open set. But DeMorgan’s Law says that Ac1 \ Ac2 \ Ac3 \ \ Ack D .A1 [ A2 [ A3 [ [ Ak /c which is an open set, so its complement, A1 [ A2 [ A3 [ [ Ak is a closed set as desired. Although not needed in this textbook about writing proofs in Analysis, for completeness, it makes sense at this point to introduce the definition of a topological space to be a set S together with a collection T of subsets of S satisfying the conditions • Both ; and S are in T . • The union of any collection of sets in T is also a set in T . • The intersection of any finite collection of sets in T is also a set in T . If these conditions are satisfied, then the set T is said to be a topology for the topological space S. From the definitions and theorems presented so far in this chapter it follows that the real numbers R along with its collection of open sets forms a topological space. The advantage of introducing the more general concept of a topological space is that many theorems about the real numbers extend to all topological spaces, so once you justify the fact that you are dealing with a topological space, you then know many theorems about your new space. As an example of another topological space consider the set of integers, Z, along with the collection T of subsets of Z consisting of the empty set, ;, and the sets A Z with the property that Ac D ZnA is a finite set. It is easy to see that both ; and Z are elements of T . To show that T is closed under unions, suppose you have a collection of sets in T . There are two cases to consider: (1) all the sets in the collection are the empty set, and (2) at least one of the sets in the collection is not empty. In the first case, the union of all the sets in the collection is the empty set which is in T . In the second case, if the collection includes a set A, then the union of the sets in the collection contains A, and because the complement of the union lies inside the complement of A which is finite, the union will have to have a finite complement and be a set in T . To show that T is closed under finite intersections, suppose you have a finite collection of sets in T . Again, there are two cases to consider: (1) at least one set in the collection is the empty set, and (2) none of the sets in the collection is the empty set. In the first case, the intersection of the collection of sets is the empty set which is in T . In the second case, the complement of the intersection of the finite collection of sets is the union of the complements of the sets. If all the complements are finite, then the union of the finite number of complements is also finite, so the intersection is in T . This verifies that T is a topology for Z. This is known as the finite complement topology for Z. It is clearly not the usual topology associated with the integers which is just the usual topology of R restricted to Z. Generally, a set can have many different topologies, each giving rise to a different topological space. Most of these topologies are uninteresting and have few if any applications.
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9.3.1 Exercises 1. Find a sequence of open sets whose intersection is neither an open nor a closed set. 2. Find a sequence of closed sets whose union is neither an open nor a closed set. 3. Prove that the intersection of a collection of closed sets is a closed set. 4. Prove that an open set of real numbers is the union of all the open intervals contained in the set. 5. Verify that if S is any set, then the power set of S, P .S/, consisting of all the subsets of S, is a topology for S. This is called the discrete topology for S. 6. Let S be the interval Œ0; 5, and let T include the empty set, the set S, and any interval of the form Œ0; x/ where x 2 .0; 5/. Verify that T is a topology for S.
9.4 Continuous Functions Applied to Sets Sometimes rather than focusing your attention on the entire real line, you are interested in the open sets within a particular subset of the real numbers. For example, if the real valued function f has domain A D Œ4; 4, you might be interested in the open sets contained in A. Moreover, you might want to consider some new sets to be open which were not considered to be open sets in R. For example, within A the interval Œ4; 0/ should be considered open in the topological space consisting just of the set A. This is because, within A, each point of Œ4; 0/ is an interior point. The only controversial point here is 4, but it makes sense to claim that 4 is in the interior of A if your entire universe of interest is A. Certainly, all the points of A that are within a distance of 12 of 4 are elements of Œ4; 0/. Generalizing this idea leads to the definition of the inherited topology in the set A R. In the inherited topology, a set B A is said to be open in A if B is the intersection of A with some set that is open in R. For example, if A D Œ4; 4 as above, then the set Œ4; 0/ is open in A because Œ4; 0/ D .5; 0/ \ A, and .5; 0/ is an open set in R. With this same reasoning, within A the set Œ4; 3 [ Œ2; 0/ [ f1; 2g [ Œ3; 4/ has interior Œ4; 3/ [ .2; 0/ [ .3; 4/ and boundary f3; 2; 0; 1; 2; 3; 4g. Similarly, a set B A is said to be closed in A if B is the intersection of A with some set that is closed in R. Note that all of the properties proved earlier in this chapter pertaining to open or closed sets in R hold equally well for sets open or closed in A. In particular, the union of any collection of sets open in A is itself a set that is open in A. The motivation for developing the properties of open and closed sets and for defining topological spaces is that one can now generalize the idea of a continuous function. One defines P .X/, the power set of a set X, to be the collection of all subsets of the set X. For example, if X is a finite set with n elements, then P .X/ contains the 2n subsets of X. If f W A ! B is a function which maps elements of the
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set A to elements of the set B, the function f can be extended to f W P .A/ ! P .B/ which maps subsets of the set A to subsets of the set B. If C A, then define f .C/ to be the set fy 2 B j y D f .a/ for some a 2 Cg. Then f .C/ is called the image of C under f . The notation f .C/ could be confusing because f was originally defined for elements of A, not subsets of A. The application of f to subsets of A is really defining a new function f W P .A/ ! P .A/ whose domain is the power set of A and codomain is the power set of B. The confusion arises because the same name, f , is given to both functions. The confusion is cleared up by recognizing the distinction that if the argument of f is an element a 2 A, then f .a/ refers to an element of the codomain, B, while if the argument of f is a subset C A, then f .C/ is a subset of B, f .C/ B. For example, the function f .x/ D x2 is defined to be a function with domain R and codomain R. It is then easily understood that f .3/ D 9 and f .2/ D 4. But taking C to be the interval .3; 2/, the expression f .C/ now refers to the function f W P .R/ ! P .R/, and f .C/ is the set of all elements of R that are images under f of elements of C. That is, f .C/ D Œ0; 9/. If the function f W A ! B is not a bijection mapping A one-to-one and onto B, then it is not possible to define the inverse function f 1 W B ! A. One problem is that if f is not surjective (mapping A onto B), there might be an element of b 2 B for which there is no corresponding element a satisfying f .a/ D b, so f 1 .b/ cannot be defined. Another problem is that if f is not injective (mapping A one-to-one to B), then there will be an element b 2 B such that f .x/ D b is satisfied by more than one value of x, so f 1 .b/ would not be unique. On the other hand, if D B, it is always possible to define the function f 1 W P .B/ ! P .A/ mapping the power set of B to the power set of A. Indeed, one defines f 1 .D/ D fx 2 A j f .x/ 2 Dg. In this case f 1 .D/ is called the preimage of D under f . For example, returning to f .x/ D x2 , 1 1 .1; 16/ D .4; 4/. it follows that f .Œ4; 9/ D .3; 2/ [ .2; 3/ and f 2 Note that when the continuous function f .x/ D x was applied to an open set as in f .3; 2/ D Œ0; 9/, the image did not end up being an open set. But when f 1 was 1 .1; 16/ D .4; 4/, the preimage was also an open applied to an open set as in f set. This is an important distinction. A continuous function need not map open sets to open sets; functions that do map all open sets to open sets are called open functions. But all continuous functions have the property that their inverses always map open sets to open sets. Conversely, a function whose inverse always maps open sets to open sets must be a continuous function. Of course, these statements require proof, but the proofs follow directly from the definition of continuity and definition of open set. Assume, for example, that f W A ! B is a continuous function and D B is an open set in B. You are challenged to show that f 1 .D/ is an open set in A. To show that f 1 .D/ is open, you would need to show for every a 2 f 1 .D/ there is a ı > 0 such that .a ı; a C ı/ \ A f 1 .D/. From the definition of f 1 .D/, you know 1 that if a 2 f .D/, then f .a/ 2 D. Because D is open, there is an > 0 such that f .a/ ; f .a/ C \ B D. This means that if y 2 B such that jy f .a/j < , then y 2 D. But now, by the definition of continuity, there is a ı > 0 such that if x 2 A
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Fig. 9.6 Mapping sets
f
C A
b
a f-1
D
f(a) = b f-1(D) = C B
with jxaj < ı, then jf .x/f .a/j < implying that f .x/ is in f .a/; f .a/C \B, and thus, f .x/ is in D. This shows that x 2 f 1 .D/ proving that .a ı; a C ı/ f 1 .D/, so f 1 .D/ is open. Conversely, suppose that f has the property that f 1 .D/ is an open set in A whenever D is an open set in B. Then let a 2 A. This time you are challenged to show that for every > 0, there is a ı > 0 such that if x 2 A with jx aj < ı, then jf .x/ f .a/j < . But the set D of all y 2 B satisfying jy f .a/j < is an open set in B implying that f 1 .D/ is an open set in A containing the point a. This means that there is a ı > 0 such that .a ı; a C ı/ \ A is contained in f 1 .D/. In other words, if x 2 A with jx aj < ı, then x is in f 1 .D/, so f .x/ is in D and jf .x/ f .a/j < , completing the proof that f is continuous (Fig. 9.6). PROOF: Let A and B be sets of real numbers, and let f W A ! B be a function from A to B. Then f is continuous on A if and only if for every open set D B, its preimage under f , f 1 .D/, is an open set in A. • Let A and B be sets of real numbers, and let f W A ! B be a function from A to B. Continuity implies that the preimages of open sets are open • Assume that f W A ! B is a continuous function. • Let D be an open set in B, and let a 2 f 1 .D/. • Because D is open in B, there is an > 0 such that .f .a/ ; f .a/ C / \ B D. • Thus, if y 2 B with jy f .a/j < , then y 2 D. • Because f is a continuous function, there is a ı > 0 such that for all x 2 A with jx aj < ı it follows that jf .x/ f .a/j < . • Thus, if x 2 .a ı; a C ı/ \ A, then jf .x/ f .a/j < , so f .x/ 2 D and x 2 f 1 .D/. • Therefore, .a ı; a C ı/ f 1 .D/ and f 1 .D/ is an open set in A. This proves that the preimage under f of any open set is open. (continued)
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The preimages of open sets are open implies continuity • Assume that the preimage under f of any set D open in B is an open set in A. • Let a 2 A, and let > 0 be given. • The setˇ f .a/ ; f .a/ C \ B is an open set in B, so its preimage, C D fx 2 A ˇ jf .x/ f .a/j < g is an open set in A. • Because C is an open set containing a, there is a ı > 0 such that .a ı; a C ı/ \ A C. • Thus, if x 2 A with jx aj < ı, then x 2 .a ı; a C ı/ \ A C, so f .x/ 2 f .C/ implying that jf .x/ f .a/j < . • Therefore, f is continuous which completes the proof of the theorem. There is a similar theorem that states that a function f W A ! B is continuous if and only if the preimage of every set closed in B is a closed set in A. The proof is left as an exercise. As it is with open sets, continuous functions do not always map closed set onto closed sets. Functions that do map all closed sets onto closed sets are called closed functions. In general, then, one can define what continuity means for any function from one topological space into another topological space. If A and B are topological spaces, the function f W A ! B is continuous if the preimage under f of every set open in B is a set open in A. Note that this definition makes sense even in topological spaces where there is no distance measure, and the definition does not involve the selection of a ı > 0 given an > 0.
9.4.1 Exercises Write proofs for each of the following statements. 1. 2. 3. 4. 5.
The union of any collection of sets open in A is itself a set open in A. The intersection of any finite collection of sets open in A is itself a set open in A. If f W A ! B and C A, then C f 1 f .C/ . If f W A ! B and D B, then f f 1 .D/ D. If f is a function from set A into set B, then f is continuous on A if and only if the preimage under f of every set closed in B is a set closed in A.
9.5 Closure Recall that if S is any subset of R, then a is an accumulation point of S if for every > 0 the open interval .a ; a C / contains at least one point of S other than a itself. An important property of closed sets is that if a closed set, S, has an accumulation point, a, then a 2 S. You should be able to construct a
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short proof of this fact that relies only on the definitions of accumulation point, closed set, and boundary. Such a proof would start with the assumption that a is an accumulation point of the closed set S. One way to continue is to construct a proof by contradiction, that is, to assume that a … S and hope that this will lead to a contradiction. Interestingly, you can proceed in more than one way. You could use the fact that S is a closed set which implies that, since a … S, then a 2 ext.S/. This means that there is an > 0 such that the open interval .a ; a C / is contained in Sc . But the definition of accumulation point says that every open interval containing a also contains points of S, so this contradicts the fact that a is an accumulation point of S. Alternatively, you could use the fact that a is an accumulation point of S. This means that for every > 0, the open interval .a ; a C / contains points in S. All of these open intervals also contain a … S implying that each of these open intervals contains points in S and points in Sc . Thus, a satisfies the definition of being an element of @S. From the definition of closed set, @S S. Thus, a 2 S. PROOF: If S is a closed set, then S contains all of its accumulation points. • Let S be a closed set, and let a be an accumulation point of S. • Assume that a … S. • From the definition of accumulation point, for every > 0 it follows that the open interval .a ; a C / contains elements in S. • Because a … S, it follows that for every > 0 the open interval .a; aC/ contains elements of S and elements of Sc , so a 2 @S. • From the definition of closed set, @S S, so a 2 S which contradicts the assumption that a … S. • Thus, every accumulation point of S must be contained in S. The collection of all the accumulation points of S is called the derived set of S which is written S0 . The previous theorem shows that if S is closed, then S0 S. The converse is also true, that is, if S0 S, then S must be closed. This follows from the fact that if a is in the boundary of S but a is not an element of S, then a must be an accumulation point of S. This should make sense to you. A boundary point is a point close both to S and to Sc . An accumulation point is close to S, and if it is not in S, it is close to Sc . PROOF: If set S contains all of its accumulation points, then S is a closed set. • Let S be a set that contains all of its accumulation points. • Assume that a 2 @SnS. • Because a 2 @S, for every > 0, the open interval .a ; a C / contains elements of S and elements of Sc . • Thus, because a itself is not a member of S, .a; aC/ contains an element of S not equal to a. • It follows that a 2 S0 S which contradicts the assumption that a … S. • Therefore, @S S which proves that S is a closed set.
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287
You can conclude from this result that for any set S, if a 2 @S \ Sc , it is an accumulation point of S, and, by symmetry, if a 2 @S \ S, then it is an accumulation point of Sc . The set S is closed if it contains its boundary, @S. But for any set S, the elements of @S that are not in S are accumulation points of S, so S is closed if and only if it contains all of its accumulation points. It is important to recognize, though, that the derived set S0 need not be contained in @S since points in the interior of S are accumulation points of S, and @S need not be contained in S0 since isolated points of S are in the boundary of S without being accumulation points of S. On the other hand, S [ @S D S [ S0 . For any set S, define the closure of S or cl.S/ to be S [ S0 D S [ @S. Some books use the notation S or S for the closure of S. Intuitively, the closure of a set S takes the elements of the boundary of S and adds them to the set so that you now have S along with its boundary (Fig. 9.7). The closure also has the following properties. • • • •
For any set S, the closure cl.S/ is a closed set. The set S is closed if and only if S D cl.S/. cl.S/ is the intersection of every closed set that contains S. cl.S/ is the smallest closed set that contains S.
All of these results have short proofs. For example, to get the first result, recall that if x is in the boundary of the union of two sets, S [ T, then x is either in the boundary of S or the boundary of T. Thus, if x 2 @ cl.S/, it means that x 2 @.S [ @S/ and, therefore, x 2 @S or x 2 @.@S/. It was shown in the first section of this chapter that @.@S/ @S implying that x 2 @S proving that x is in cl.S/. Thus, cl.S/ contains its boundary, so it is closed. For the second result, note that if S is closed, it contains its boundary so cl.S/ D S [ @S D S. Conversely, if S D cl.S/, then S is closed because cl.S/ is always a closed set. The third and fourth results follow quickly after noticing that any closed set containing S must also contain the boundary of S.
S
(
](
Fig. 9.7 The closure of a set
cl(S)
)
[
][
]
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9.5.1 Exercises For each of the following sets S, determine @S, S0 , and cl.S/. 1. 2. 3. 4. 5.
The real numbers, R. The integers, Z. f 1n j n 2 Zg .0; 3/ [ .3; 5/ [ .5; 7/ .1; 3/ \ Q [ f0; 4g
Write proofs of each of the following statements. 4. 5. 6. 7. 8.
For any set S, the closure cl.S/ is a closed set. The set S is closed if and only if S D cl.S/. cl.S/ is the intersection of every closed set that contains S. cl.S/ is the smallest closed set that contains S. For any set S, its derived set, S0 , is a closed set.
9.6 Compactness The topics of open cover, finite subcover, compactness, and the Heine–Borel Theorem were introduced in Chap. 4 because of their usefulness in proving that a function continuous on a closed bounded interval is uniformly continuous on that interval. Compactness also played an important role in showing that a continuous function on a closed bounded interval is bounded, a continuous function on a closed bounded interval obtains its extreme values (maximum and minimum), and a continuous function on a closed bounded interval has a Riemann integral. Recall that an open cover of a set S was defined to be a collection open intervals T where for each x 2 S there is an open interval .p; q/ 2 T such that x 2 .p; q/. After the introduction of the topological ideas in this chapter, that definition can be generalized to allow T to be a collection of open sets rather than just open intervals, that is, a collection of open sets, T, is called an open cover of S if for each x 2 S there is an open set U 2 T such that x 2 U. Moreover, the Heine–Borel Theorem can now be extended in two ways: the concept of an open cover by intervals can be generalized to an open cover by open sets, and the concept of closed bounded interval can be generalized to closed bounded set. PROOF (Heine–Borel Theorem): Let S be any closed bounded set of real numbers, and let T be a cover of S by open sets. Then T contains a finite subcover of S. • Let S be a closed bounded set and T be a cover of S by open sets. • Because S is bounded, there are real numbers a and b with a < b such that S Œa; b. (continued)
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289
• Let U D .a 1; b C 1/nS which is the intersection of the open interval .a 1; b C 1/ and the open set Sc , so U is an open set. • Then T 0 D T [ fUg is an open cover of Œa; b. • For each x 2 Œa; b there is an open set Vx 2 T 0 that contains x. • Because Vx is open, there is an open interval .px ; qx / Vx that contains x. • Thus, the collection T 00 D f.px ; qx / j x 2 Œa; bg is a cover of Œa; b by open intervals. • Now, by the previously proved version of the Heine–Borel Theorem, T 00 has a finite subcover of Œa; b, say f.p1 ; q1 /; .p2 ; q2 /; .p3 ; q3 /; : : : ; .pk ; qk /g for some natural number k. • For each j D 1; 2; 3; : : : ; k, the open interval .pj ; qj / in the subcover is contained in an open set Vj 2 T 0 , so it is clear that the subcover V1 ; V2 ; V3 ; : : : ; Vk covers Œa; b and, therefore, covers S. • If one of the open sets, Vj , happens to be the set U added to T, this set can be discarded from the subcover of S because it contains no elements of S. • This gives a finite subcover of S which completes the proof. So, this shows that all closed bounded sets of real numbers are compact. The converse is also true, that is, all compact subsets of real numbers are both closed and bounded. These two results together, then, completely characterize the compact sets of real numbers. PROOF: A subset of R is compact if and only if it is closed and bounded. • The Heine–Borel Theorem shows that closed bounded sets of real numbers are compact. • Conversely, assume that S is a compact subset of R. • The collection of open intervals .j; j/ where j ranges over the natural numbers is a collection of open sets that covers all of R, so it certainly covers S. • Because S is compact, S can be covered by a finite collection of the .j; j/ intervals. • It follows that there exists a natural number k such that S .k; k/, and S is a bounded set. • Suppose that there is a real number x in the boundary of S that is not an element of S. • For each > 0, let U D .1; x / [ .x C ; 1/ which is an open set. • The collection of all such U covers all of Rnfxg, and since x is not an element of S, the collection is an open cover of S. • Because S is compact, it is covered by a finite collection of the U sets. • It follows that there is an ı > 0 such that S Uı . (continued)
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• Thus, the interval .x ı; x C ı/ contains no elements of S which contradicts the assumption that x is in the boundary of S. • Therefore, there are no elements x in the boundary of S that are not elements of S implying @S S and S is closed. • This shows that all compact sets are closed and bounded completing the proof of the theorem. Continuous functions need not map bounded sets onto bounded sets as is seen by f .x/ D 1x which maps the bounded interval .0; 1/ continuously onto the interval .1; 1/ which is not bounded. Continuous functions need not map closed sets onto closed sets as seen by f .x/ D 1x which maps the closed interval Œ1; 1/ onto Œ1; 0/ which is not closed. But continuous functions always map compact sets onto compact sets. This is a result that is true in any topological space, so its proof need not use any more than the properties of open sets, compact sets, and continuous functions. To write the proof you would start by assuming that the function f W A ! B is continuous on A, and that C A is a compact set. You must then show that the image of C, f .C/ B, is compact. How would you show this set is compact? The definition of compact set suggests that you would take an open cover of the set and proceed to show that that cover has a finite subcover. So let I be an index set and assume that fUi j i 2 Ig is an open cover of f .C/. Somehow you must show that this cover has a finite subcover. All you know is that f is a continuous function and that the set C is compact. Since C is compact, you know that open covers of C have finite subcovers, but you have an open cover of f .C/, not an open cover of C. You need to use the fact that f is a continuous function which means that for each i 2 I, the preimage of the open set Ui , f 1 .Ui /, is an open set in A. Does the collection of f 1 .Ui / sets form a cover of C? Follow what happens: if x 2 C, then f .x/ 2 f .C/. Thus, there is at least one i 2 I such that f .x/ 2 Ui . Therefore, x 2 f 1 .Ui /. So, indeed, the collection of f 1 .Ui / sets forms an open cover of C. Hence, there is a finite subcover of C given (by renaming subscripts) as f 1 .U1 /, f 1 .U1 /; f 1 .U1 /; : : : ; f 1 .Uk /, for some natural number k. For each x 2 C, there is a j between 1 and k such that x 2 f 1 .Uj /, so f .x/ 2 f f 1 .Uj / Uj . Because each element of f .C/ is the image of at least one x 2 C, and each x 2 C is an element of at least one of the finite number of f 1 .Uj /, it follows that the finite collection of open sets, U1 ; U2 ; U3 ; : : : ; Uk , covers f .C/ proving that f .C/ is compact. PROOF: If f W A ! B is continuous on A, and if C A is a compact set, then f .C/ is a compact set in B. • Assume that f W A ! B is continuous on A, and C A is a compact set. • Let I be an index set, and fUi j i 2 Ig be a collection of open sets that cover f .C/. • For each x 2 C there is an i 2 I such that f .x/ 2 Ui . • Since f is continuous, and, for each i 2 I, Ui is an open set in B, f 1 .Ui / is an open set in A. • Thus, ff 1 .Ui / j i 2 Ig is an open cover of C. (continued)
9.7 Connectedness
291
• Since C is compact, this open cover has a finite subcover. • By renaming subscripts, the subcover is given as f 1 .U1 /; f 1 .U2 /; f 1 .U3 /; : : : ; f 1 .Uk / for some natural number k. • Let y be any element of f .C/. Then y D f .x/ for some x 2 C. • Since x 2 f 1 .Uj / for one of the j D 1; 2; 3; : : : ; k, it follows that y D f .x/ 2 Uj showing that the finite collection U1 ; U2 ; U3 ; : : : ; Uk covers f .C/. • Therefore, every open cover of f .C/ has a finite subcover, and f .C/ is compact. Notice that it is an immediate consequence of this theorem that a real valued continuous function on a closed bounded interval on the real line is bounded and obtains its maximum and minimum values. This is because every closed bounded interval on the real line is a compact set, so its image under a continuous function is compact which means the image is closed and bounded. The image being bounded is just another way of saying that the function is bounded. The image being closed shows that the image contains its boundary which includes the maximum and minimum values of the function. The Heine–Borel Theorem can be extended to n-dimensional Euclidean space Rn . That is, the compact sets in Rn are the sets that are both closed and bounded. One can use mathematical induction to show that a rectangular box that is a cross product of n closed intervals is compact, and then, that can be extended to any closed bounded set.
9.6.1 Exercises 1. Find an example of a function f and a set C such that f 1 f .C/ is notequal to C. 2. Find an example of a continuous function f and a set D such that f f 1 .D/ is not equal to D. 3. Find an example of a continuous function f W A ! B and a compact set D B such that f 1 .D/ is not compact. 4. Suppose that the continuous function f has domain Œ0; 10 and codomain .4; 4/. Show that the function is not surjective.
9.7 Connectedness The intervals on the real line were discussed in Chap. 2. A set of real numbers is an interval if whenever x and y are elements of the interval, then all the real numbers between x and y are also elements of the interval. The intervals are the connected sets on the real line, but the concept of connectedness can be extended to any topological space. In a general topological space, two nonempty sets A and B are
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[
]
|
0
1
2
|
(
)
3
4
5
Fig. 9.8 The sets Œ0; 1 and .4; 5/ are disconnected
disconnected if there are disjoint open sets U and V with A U and B V. For example, the sets Œ0; 1 and .4; 5/ are disconnected because Œ0; 1 .1; 2/ and .4; 5/ .4; 5/ where .1; 2/ and .4; 5/ are disjoint open sets (Fig. 9.8). The sets Œ0; 3 and .3; 5/ are disjoint nonempty sets, but they are not disconnected because any open set that contains Œ0; 3 will necessarily share points with any open set containing .3; 5/, and, in particular, both open sets will contain the element 3. A set is called connected if it is not the union of two disconnected nonempty sets. Even though the connected sets of real numbers are just the intervals, the concept of connectedness gets far more interesting in more general topological spaces. If f W A ! B is continuous, then it always maps connected sets to connected sets, that is, if C A is a connected set, then so is f .C/. This is easy to see since, if f .C/ is disconnected, then there are two disjoint open sets U and V in B and two nonempty sets S and T in B such that f .C/ D S [ T and S U and T V. But then C f 1 .U/ [ f 1 .V/ where f 1 .U/ and f 1 .V/ are disjoint open sets in A. Because S and T are nonempty, C \ f 1 .U/ and C \ f 1 .V/ are nonempty implying that C is a disconnected set. Thus, if C is connected, f .C/ must also be connected. PROOF: If f W A ! B is continuous on A, and if C A is a connected set, then f .C/ is a connected set in B. • Let f W A ! B be a continuous function on A, and assume that C A such that f .C/ is disconnected. • This means that there are disjoint open sets U and V in B, and nonempty sets S and T in B with S U and T V such that B D S [ T. • Since f is continuous, f 1 .U/ and f 1 .V/ are open sets in A. • Since S and T are nonempty sets whose union is f .C/, both C \ f 1 .U/ and C \ f 1 .V/ are nonempty. • This shows that C is a disconnected set. • Therefore, if C is a connected set, f .C/ must also be connected. When this theorem is applied to functions from the real numbers to the real numbers, the result is the Intermediate Value Theorem which states that if f is a real valued function on the interval Œa; b, then for every c between f .a/ and f .b/ there is an x 2 Œa; b such that f .x/ D c. This is because f must map the connected set Œa; b into a connected set which must include all the elements c between f .a/ and f .b/. Note that f 1 need not bring connected sets to connected sets. In n-dimensional Euclidean space the concept of connectedness gets considerably richer as the connected sets are not merely the cross products of intervals (Fig. 9.9). In R2 one introduces what it means for a set to be path-connected which makes precise the intuitive notion that a set is connected if you can draw a path
9.7 Connectedness
293 N
C
Fig. 9.9 The set C is a connected set. The set N is not a connected set Fig. 9.10 Graph of sin with the y-axis
1 x
between any two of its points where the path stays inside the set. On the real line, this just means that for any two points in the set, the interval between the two points stays in the set. But in R2 where paths need not be straight lines, the examples are far more varied. In fact, in R2 there are examples of connected sets that are not path connected, a phenomenon that cannot occur on the real line. A famous example is the set consisting of the graph of the equation y D sin 1x along with the y-axis. This is a connected set because any open set that contains the y-axis must intersect parts of the graph of y D sin 1x both to the left and to the right of the y-axis. On the other hand, this set is not path-connected because there is no way to construct a path that stays inside the set and connects the points . 1 ; 0/ and . 1 ; 0/ (Fig. 9.10).
9.7.1 Exercises 1. Find an example of a continuous function f W A ! B and connected set D B such that f 1 .D/ is not connected. 2. Show that in any topological space A, if S and T are connected sets with A D S[T and S \ T ¤ ;, then A is connected.
Chapter 10
Metric Spaces
10.1 Definition of Metric Space This book has discussed at length how one writes proofs about the limits and continuity of functions whose domains and ranges are subsets of the real numbers, R. Although the real numbers is a far simpler set to study than many other naturally arising sets in Analysis, the techniques learned while dealing with realvalued functions of a real variable can be applied almost exactly to prove similar theorems about functions defined on other domains with other types of ranges. It is instructive to take note of the properties of the real numbers that play important roles in these proofs. In particular, most of the proofs about limits and continuity involve measuring the distance between two real numbers x and y. This is done by calculating the absolute value of the difference between the numbers, jx yj. This distance measure has important properties that allow the proofs about limits and continuity to proceed. Among the useful properties of this distance measure is that if jx yj < for every > 0, then it follows that x D y, and if jx yj > 0, then x is surely different from y. Another property use repeatedly in these proofs is the triangle inequality. For example, if f and g are two functions, and x and y are both elements in the domains of these functions, then knowing that g.y/j < 2 allows the proofs to conclude that ˇ jf .x/ f.y/j < 2 and jg.x/ ˇ ˇ ˇ ˇ f .x/ C g.x/ f .y/ C g.y/ ˇ D ˇ f .x/ f .y/ C g.x/ g.y/ ˇ jf .x/f .y/jC jg.x/ g.y/j < 2 C 2 D . The fact that the triangle inequality holds true for this chosen measure of distance is crucial in the argument. The conclusion is, then, that if there were a set, X, and a distance measure that assigned to each x and y in X a real number, d.x; y/, that had many of the same properties that the jx yj distance measure does in the real numbers, then it might be possible to prove limit and continuity theorems for functions defined on X by just adopting the same proof techniques used for the theorems about functions of
© Springer International Publishing Switzerland 2016 J.M. Kane, Writing Proofs in Analysis, DOI 10.1007/978-3-319-30967-5_10
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Fig. 10.1 Metric distances in the plane
x d(x,y)
d(x,z)
y d(y,z)
z
real numbers. With this in mind a nonempty set X together with distance function d is defined to be a metric space if, for all x, y, and z in X, this distance function satisfies the following properties: • • • •
d.x; y/ 2 R with d.x; y/ 0 (the distance function is a nonnegative real number). d.x; y/ D 0 if and only if x D y (the distance function separates points). d.x; y/ D d.y; x/ (the distance function is symmetric). d.x; y/ C d.y; z/ d.x; z/ (the distance function satisfies the triangle inequality).
The distance function defines a metric for the metric space, and the metric space is designated as (Fig. 10.1). This definition is a generalization of the distance function defined on the real numbers, d.x; y/ D jx yj. Clearly, for all real numbers x, y, and z, • • • •
d.x; y/ D jx yj 0 0 D d.x; y/ D jx yj if and only if x D y d.x; y/ D jx yj D jy xj D d.y; x/ d.x; y/ C d.y; z/ D jx yj C jy zj j.x y/ C .y z/j D jx zj D d.x; z/
showing that where d.x; y/ D jx yj is a metric space. In general, it is a fairly straightforward process to construct a proof that is a metric space. Most proofs would follow this template: TEMPLATE for proving is a metric space • SET THE CONTEXT: Give the definitions of X and d. • METRIC DEFINITION: Show that d maps each x; y 2 S to a nonnegative real number. • SEPARATION OF POINTS: Show that d.x; y/ D 0 implies x D y. • ZERO DISTANCE: Show that for all x 2 X that d.x; x/ D 0. • SYMMETRY: Show that for all x; y 2 X that d.x; y/ D d.y; x/. • TRIANGLE INEQUALITY: Show that for all x; y; z 2 S that d.x; y/ C d.y; z/ d.x; z/. Given a metric space , an element a 2 X, and a positive real number r, define the neighborhood of a with radius r to be N.a; r/ D fx 2 X j d.a; x/ < rg. Sometimes, as in the definition of a limit at point a, one needs to exclude the point a from the neighborhood of a. In this case, one can define the deleted neighborhood of a with radius r to be N ı .a; r/ D fx 2 X j 0 < d.a; x/ < rg. These neighborhoods play a central role in defining limits and continuity of functions defined on X and in establishing a topology for the space X. It is not uncommon for there to be
10.2 Inequalities
297
several different distance functions defined on a particular set X that make X into a metric space. Each new distance function results in different shaped neighborhoods. Some give rise to the same topology of X while others may result in quite different topologies. Many examples of these different distance functions will be explored in the sections that follow.
10.2 Inequalities Most proofs in Analysis involve establishing one or more inequalities. Some inequalities seem to keep reappearing in different guises throughout Analysis, so they provide great tools for writing proofs. This section presents two very common inequalities that will be used later in the chapter to justify the triangle inequality for some examples of metric spaces.
10.2.1 Cauchy–Schwarz Inequality For natural number n let a D .a1 ; a2 ; a3 ; : : : ; an / and b D .b1 ; b2 ; b3 ; : : : ; bn / be any two points in n-dimensional Euclidean space. The Cauchy–Schwarz Inequality states that ja1 b1 C a2 b2 C a3 b3 C C an bn j
q
q a21 C a22 C a33 C C a2n b21 C b22 C b33 C C b2n :
The student familiar with vectors and the dot q product of two vectors will find this inequality easy to remember. If jaj D a21 C a22 C a23 C C a2n refers to the magnitude of vector a and the dot product a b D .a1 ; a2 ; a3 ; : : : an / .b1 ; b2 ; b3 ; : : : bn / D a1 b1 C a2 b2 C a3 b3 C C an bn , then a b D jaj jbj cos where is the angle between the two vectors. Then the Cauchy–Schwarz Inequality is just the statement that jaj jbj ja bj which follows because j cos j 1. To prove the Cauchy–Schwarz Inequality note that for given a; b 2 Rn and every n P real number x the quantity .aj C xbj /2 is a sum of squares of real numbers, so jD1
it must be nonnegative. By expanding the squares one gets x2
n P jD1
n P jD1
a2j C 2x
n P
aj bj C
jD1
b2j 0. Thus, this is a quadratic polynomial in x that is nonnegative for every
real number x. Any quadratic polynomial Ax2 CBxCC with A > 0 is nonnegative for every x if and only if its discriminant B2 4AC is not positive. But the discriminant
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2 of the previous polynomial is 4 4
n P
!2 aj bj
jD1
n P jD1
! a2j
n P jD1
!3 b2j 5. The statement
that this discriminant is less than or equal to 0 is exactly the statement of the Cauchy–Schwarz Inequality. An even stronger statement can now be made. Equality occurs in the Cauchy–Schwarz Inequality if and only if the given discriminant is 0 so that the underlying quadratic polynomial has exactly one root, meaning that the n P sum .aj C xbj /2 is 0 for exactly one value of x. This happens if and only if a is jD1
a multiple of b. Thus, the Cauchy–Schwarz Inequality always holds, and equality holds exactly when one of the points .a1 ; a2 ; a3 ; : : : ; an / and .b1 ; b2 ; b3 ; : : : ; bn / is a scalar multiple of the other.
10.2.2 Minkowski Inequality Starting with the Cauchy–Schwarz Inequality a1 b1 C a2 b2 C a3 b3 C C an bn
q q a21 C a22 C a33 C C a2n b21 C b22 C b33 C C b2n
doubling it and adding a21 C a22 C a23 C C a2n C b21 C b22 C b23 C C b2n to both sides yields ˚ .a1 C b1 /2 C .a2 C b2 /2 C .a3 C b3 /2 C C .an C bn /2 q q 2 a1 C a22 C a23 C C a2n C 2 a21 C a22 C a33 C C a2n b21 C b22 C b33 C C b2n C .b21 C b22 C b23 C C b2n /
or by taking square roots q q q .a1 C b1 /2 C .a2 C b2 /2 C .a3 C b3 /2 C C .an C bn /2 a21 C a22 C a23 C C a2n C b21 C b22 C b23 C C b2n
which is a special case of the Minkowski Inequality which can be restated jaCbj jaj C jbj. Again, equality occurs only when one of the points is a scalar multiple of the other.
10.2.3 Exercises 1. Show that the Cauchy–Schwarz Inequality extends to infinite s series. That s is, if ˇ ˇ1 1 1 1 1 ˇ ˇ P 2 P 2 P P 2 P an and bn are both convergent series, then ˇˇ an bn ˇˇ an b2n . nD1
nD1
nD1
nD1
nD1
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1 P 2. Show that the Minkowski Inequality extends to infinite series. That is, if a2n nD1 s s 1 1 1 P P P 2 2 and bn are both convergent series, then .an C bn / a2n C nD1 nD1 nD1 s 1 P b2n . nD1
3. Show that for any real numbers a1 ; a2 ; a3 ; : : : ; an and positive real numbers a2 a2 a2 a2 b1 ; b2 ; b3 ; : : : ; bn , the following inequality holds: 1 C 2 C 3 C C n b1 b2 b3 bn .a1 C a2 C a3 C C an /2 . This can be shown by mathematical induction on n, b1 C b2 C b3 C C bn but can also be shown using the Cauchy–Schwarz Inequality.
10.3 Examples of Metric Spaces For any natural number n one can define n-dimensional Euclidean space, Rn , with R1 being the real numbers, R2 being the Euclidean plane, R3 being 3-dimensional Euclidean space, and so forth. Elements of Rn can be represented as ordered n-tuples of real numbers, .x1 ; x2 ; x3 ; : : : ; xn /. You should be familiar with the Euclidean distance between two points in n-dimensional Euclidean space, x D .x1 ; x2 ; x3 ; : : : ; xn / and y D .y1 ; y2 ; y3 ; : : : ; yn /, given by the generalization of the Pythagorean Theorem as d.x; y/ D
p
.x1 y1 /2 C .x2 y2 /2 C .x3 y3 /2 C C .xn yn /2 :
For any x; y 2 Rn , the distance d.x; y/ is a nonnegative real number since it is a square root of the sum of squares of real numbers. Moreover, the distance is 0 exactly when the sum of squares is 0 which happens only when x D y. The fact that d is symmetric follows from the fact that for all real numbers a and b, .a b/2 D .b a/2 . The fact that the Euclidean distance satisfies the triangle inequality is just the statement of the Minkowski Inequality with aD.x1 y1 ; x2 y2 ; x3 y3 ; : : : ; xn yn / and b D .y1 z1 ; y2 z2 ; y3 z3 ; : : : ; yn zn /. Then p d.x; y/ C d.y; z/ D .x1 y1 /2 C .x2 y2 /2 C .x3 y3 /2 C C .xn yn /2 C p .y1 z1 /2 C .y2 z2 /2 C .y3 z3 /2 C C .yn zn /2 p .x1 z1 /2 C .x2 z2 /2 C .x3 z3 /2 C C .xn zn /2 D d.x; z/: Together these facts show that Rn with Euclidean distance is a metric space (Fig. 10.2).
300 Fig. 10.2 Euclidean distance is R2
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(x1, y1) d(x,y) =
√
(x1– x2)2+(y1– y2)2
|y2– y1|
|x2– x1|
(x2, y2)
PROOF: For natural number n, n-dimensional Euclidean space with Euclidean distance function p the d.x; y/ D .x1 y1 /2 C .x2 y2 /2 C .x3 y3 /2 C C .xn yn /2 is a metric space. • SET THE CONTEXT: For natural number n, let x D .x1 ; x2 ; x3 ; : : : ; xn /, y D .y1 ; y2 ; y3 ; : : : ; yn /, and z D .z1 ; z2 ; z3 ; : : : ; zn / be elements of Rn . • METRIC DEFINITION: Define d.x; y/ p D .x1 y1 /2 C .x2 y2 /2 C .x3 y3 /2 C C .xn yn /2 which is the square root of a sum of squares of real numbers, so it is a nonnegative real number. • SEPARATION OF POINTS: If x ¤ y, then for some j between 1 and n, .xj yj /2 must be positive implying that d.x; y/ > 0. • ZERO DISTANCE: For each j between 1 and n, .xj xj /2 D 0, so d.x; x/ D 0. • SYMMETRY: Since for each j between 1 and n, .xj yj /2 D .yj xj /2 , it follows that d.x; y/ D d.y; x/. • TRIANGLE INEQUALITY: The fact that d.x; y/ C d.y; z/ d.x; z/ is just a restatement of the Minkowski Inequality with aj D xj yj and bj D yj zj for each j between 1 and n. • This shows that Rn with the Euclidean distance is a metric space. The Euclidean distance, sometimes called the Euclidean metric, may be the most commonly seen distance function used for Euclidean space, but there are many other distance functions which can make Rn into a metric space. One example is d.a; b/ D ja1 b1 jCja2 b2 jCja3 b3 jC Cjan bn j. This is sometimes called the taxicab metric because the distance d.a; b/ is the distance you would travel between the two points a and b if you could only travel in directions parallel to one of the coordinate axes as a taxicab would do on a rectangular grid of streets. Proving that this distance function makes Rn into a metric space is quite easy.
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PROOF: For natural number n, n-dimensional Euclidean space with the Euclidean distance function d.x; y/ D j.x1 y1 j C jx2 y2 j C jx3 y3 jC C jxn yn j is a metric space. • SET THE CONTEXT: For natural number n, let x D .x1 ; x2 ; x3 ; : : : ; xn /, y D .y1 ; y2 ; y3 ; : : : ; yn /, and z D .z1 ; z2 ; z3 ; : : : ; zn / be elements of Rn . • METRIC DEFINITION: Define d.x; y/ D jx1 y1 j C jx2 y2 j C jx3 y3 j C C jxn yn j which is the sum of nonnegative absolute values so it is a nonnegative real number. • SEPARATION OF POINTS: If x ¤ y, then for some j between 1 and n, jxj yj j must be positive implying that d.x; y/ > 0. • ZERO DISTANCE: For each j between 1 and n, jxj xj j D 0, so d.x; x/ D 0. • SYMMETRY: Since for each j between 1 and n, jxj yj j D jyj xj j, it follows that d.x; y/ D d.y; x/. • TRIANGLE INEQUALITY: Since for each j between 1 and n, jxj yj j C jyj zj j jxj zj j, it follows that d.x; y/ C d.y; z/ d.x; z/. • This shows that Rn with the d distance function is a metric space. Still another distance function that can be used for Euclidean space is called the supremum metric given by d.a; b/ D max.ja1 b1 j; ja2 b2 j; ja3 b3 j; : : : ; jan bn j/. It is constructive to compare the shapes of the neighborhoods that you get using the Euclidean metric, the taxicab metric, and the supremum metric as shown in Fig. 10.3. Since the Euclidean distance is the familiar distance from Euclidean Geometry, it is easy to see that if a 2 Rn and r > 0, then N.a; r/ is an open ball with center a and radius r. On the other hand, using the taxicab metric, N.a; r/ is a union of 2n n-dimensional triangular pyramids. That is, when n D 2, N.a; r/ is a diamond made up of four isosceles right triangles, and when n D 3, N.a; r/ is a union of 8 tetrahedra, one in each octant, forming a regular octahedron. For the supremum metric, N.a; r/ is an n-dimensional cube. Note that in the Euclidean metric, if the coordinate axes are rotated (performing an orthogonal change of coordinates), there is no change in the neighborhood whereas with the other two metrics, rotating the axes changes the orientation of the neighborhoods. It turns out that all three of these
Fig. 10.3 N.0; 1/ in the Euclidean, taxicab, and supremum metrics in 2 and 3 dimensions
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metrics give rise to the same topology on Rn because each metric gives the same open sets even though the open neighborhoods are different in shape. But the three metrics have different algebraic properties, and sometimes it is easier to prove a particular theorem using one of these metrics rather than the others. Distance measures in metric spaces need not be complicated. For any set X you can define d.x; x/ D 0 for all x 2 X and d.x; y/ D 1 for all x and y in X with x ¤ y. It is very easy to see that d.x; y/ is nonnegative, symmetric, and equal to 0 if and only if x D y. Also, for any x; y; z 2 S, if d.x; z/ D 1, then x ¤ z, so at least one of d.x; y/ and d.y; z/ must be 1 which implies the triangle inequality d.x; y/Cd.y; z/ d.x; z/. Thus, any set X is a metric space with this metric sometimes called the discrete metric, and is called a discrete metric space. Note that for this metric, each neighborhood, N.a; r/ is either all of X or just the single point fag depending on whether or not r is greater than 1. Next, consider a space that looks much different than Euclidean space. Let CŒ0; 1 be the set of all real-valued functions continuous on the interval Œ0; 1. Certainly, this set contains all the polynomials with real coefficients, but it also includes the rational functions that are defined on Œ0; 1, exponential functions, many elementary functions, and a much larger class of functions continuous but not differentiable on Œ0; 1. This set is truly very large as compared, say, to the set of real numbers. There are many ways you might try to measure the distance between two functions in this set. For example, you could evaluate the function at one or more points and measure how much the functions differ at those points. ˇ That is,ˇif f and g are in CŒ0; 1, you could define d.f ; g/ D jf .0/ g.0/j C ˇf 12 g 12 ˇ C jf .1/ g.1/j. The only problem with this definition is that there are continuous functions f and g which are equal at 0, 12 , and 1 but not equal at other points such as f .x/ D x.2x 1/.x 1/ and g.x/ D 2x.2x 1/.x 1/. Because the given distance function gives a distance of 0 between two unequal functions, it cannot serve as a metric for the space of continuous functions on Œ0; 1 (Fig. 10.4). As a result, a distance function that makes CŒ0; 1 into a metric space really needs to take into account the values of the functions at all the points (or at least a dense set of points) in Œ0; 1. One distance measure that does this is called the supremum metric or sup metric for short. It is defined for all f and g in CŒ0; 1 as d.f ; g/ D sup jf .x/ g.x/j. It is clear that if f ¤ g, then there are x2Œ0;1
values of x 2 Œ0; 1 where f .x/ ¤ g.x/, so d.f ; g/ will be positive, yet when f g, then d.f ; g/ D 0 as needed. It is necessary to check that this distance Fig. 10.4 Some functions in CŒ0; 1
0
1
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303
function has a valid definition, that is, for every f and g in CŒ0; 1 the distance function gives a nonnegative real number. But if f and g are continuous functions on Œ0; 1, then so is jf .x/ g.x/j. Since all functions continuous on Œ0; 1 are bounded and jf .x/g.x/j is a continuous function, the needed supremum is defined. The triangle inequality follows from the fact that the triangle inequality works for real numbers. Since for any three continuous functions f , g, and h and for each x 2 Œ0; 1it is true that jf .x/ g.x/j C jg.x/ h.x/j jf .x/ h.x/j, it follows that sup jf .x/ g.x/j C jg.x/ h.x/j sup jf .x/ h.x/j. Then, the inequality x2Œ0;1
x2Œ0;1
sup.A C B/ sup A C sup B shows that sup jf .x/ g.x/j C sup jg.x/ h.x/j x2Œ0;1 x2Œ0;1 sup jf .x/ g.x/j C jg.x/ h.x/j sup jf .x/ h.x/j, and d.f ; g/ C d.g; h/ x2Œ0;1
x2Œ0;1
d.f ; h/.
PROOF: The set CŒ01 with distance function d.f ; g/ D sup jf .x/ g.x/j x2Œ0;1
is a metric space. • SET THE CONTEXT: Let CŒ0; 1 be the set of real-valued functions continuous on the interval Œ0; 1. • METRIC DEFINITION: For any f and g in CŒ0; 1, the function jf .x/ g.x/j is also in CŒ0; 1. Define d.x; y/ D sup jf .x/ g.x/j which is x2Œ0;1
the supremum of a nonnegative continuous function, so it is a nonnegative real number. • SEPARATION OF POINTS: For f ; g 2 CŒ0; 1 if f ¤ g, then for some x 2 Œ0; 1, jf .x/ g.x/j must be positive implying that d.f ; g/ > 0. • ZERO DISTANCE: For all x 2 Œ0; 1 and f 2 CŒ0; 1, jf .x/ f .x/j D 0, so sup jf .x/ f .x/j D 0 and d.f ; f / D 0. x2Œ0;1
• SYMMETRY: Since for all x 2 Œ0; 1 and all f ; g 2 CŒ0; 1, jf .x/ g.x/j D jg.x/ f .x/j, it follows that d.f ; g/ D d.g; f /. • TRIANGLE INEQUALITY: Since for all x 2 Œ0; 1 and all f ; g; h 2 CŒ0; 1, it holds that jf .x/ g.x/j C jg.x/ h.x/j jf .x/ h.x/j, it follows that sup jf .x/ g.x/j C sup jg.x/ h.x/j sup jf .x/ g.x/j C x2Œ0;1 x2Œ0;1 x2Œ0;1 jg.x/ h.x/j sup jf .x/ h.x/j, and d.f ; g/ C d.g; h/ d.f ; h/. x2Œ0;1
• This shows that CŒ0; 1 with the supremum distance function is a metric space. The supremum metric provides only one of many possible distance functions for the space CŒ0; 1. Another example is called the L1 metric and is defined by R1 d.f ; g/ D jf .x/ g.x/jdx. Since all functions continuous on a closed interval are 0
integrable there, this distance function is defined. Moreover, since jf .x/ g.x/j 0 for all x 2 Œ0; 1, its integral is also nonnegative. If f ¤ g, then there is an a 2 Œ0; 1 where f .a/ ¤ g.a/. Because jf .x/ g.x/j is continuous and positive at x D a, there is a ı > 0 such that for all x 2 CŒ0; 1 with jxaj < ı, jf .x/g.x/j > 12 jf .a/g.a/j.
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This implies that d.f ; g/ D
R1 0
jf .x/ g.x/jdx >
aCı R
jf .x/ g.x/jdx > 0. Of course,
aı
a rigorous proof will take care that the limits of integration in the previous sentence are chosen in a way that the integral is guaranteed to be defined. The symmetry of d follows from its definition. For all f ; g; h 2 CŒ0; 1 and each x 2 Œ0; 1, the triangle inequality gives jf .x/ g.x/j C jg.x/ h.x/j jf .x/ h.x/j. Thus, R1 R1 R1 jf .x/ g.x/jdx C jg.x/ h.x/jdx D jf .x/ g.x/j C jg.x/ h.x/jdx 0
R1 0
0
0
jf .x/ h.x/jdx, so d.f ; g/ C d.g; h/ d.f ; h/, and the needed triangle inequality
holds. PROOF: The set CŒ01 with distance function d.f ; g/ D is a metric space.
R1 0
jf .x/ g.x/jdx
• SET THE CONTEXT: Let CŒ0; 1 be the set of real-valued functions continuous on the interval Œ0; 1. R1 • METRIC DEFINITION: Define d.x; y/ D jf .x/ g.x/jdx which is the 0
• • • •
integral of a nonnegative continuous function, so it is a nonnegative real number. SEPARATION OF POINTS: For f ; g 2 CŒ0; 1 if f ¤ g, then for some a 2 Œ0; 1, jf .a/ g.a/j must be positive. Because jf .x/ g.x/j is a continuous function, there is a ı > 0 such that jf .x/ g.x/j > 12 jf .a/ g.a/j for all x 2 Œ0; 1 satisfying jx aj < ı. In particular, there are ˛ and ˇ in Œ0; 1 with ˛ < ˇ such that jf .x/ g.x/j > 1 jf .a/ g.a/j for all x satisfying ˛ < x < ˇ. 2 R1 Rˇ Then d.f ; g/ D jf .x/ g.x/jdx jf .x/ g.x/jdx > 1 jf .a/ 2
˛
0
g.a/j.ˇ ˛/ > 0, so d.f ; g/ > 0 whenever f ¤ g. • ZERO DISTANCE: For all x 2 Œ0; 1 and f 2 CŒ0; 1, jf .x/ f .x/j D 0, so R1 R1 jf .x/ f .x/jdx D 0 dx D 0 and d.f ; f / D 0. 0
0
• SYMMETRY: Since for all x 2 Œ0; 1 and all f ; g 2 CŒ0; 1, jf .x/ g.x/j D jg.x/ f .x/j, it follows that d.x; y/ D d.y; x/. • TRIANGLE INEQUALITY: Since for all x 2 Œ0; 1 and all f ; g; h 2 CŒ0; 1, it holds that jf .x/ g.x/j C jg.x/ h.x/j jf .x/ h.x/j, it follows that R1 R1 R1 jf .x/ g.x/jdx C jg.x/ h.x/jdx D jf .x/ g.x/j C jg.x/ h.x/jdx 0
R1 0
0
0
jf .x/ h.x/jdx, and d.f ; g/ C d.g; h/ d.f ; h/.
• This shows that CŒ0; 1 with the d.f ; g/ distance function is a metric space.
10.3 Examples of Metric Spaces
305
It is important to note that the supremum metric and the L1 metric are distinctly different. In particular, 8 9 consider the sequence of functions fn .x/ D 1 ˆ > 0 if 0 ˆ > nC1 ˆ > ˆ > ˆ > < = 1 1 for all natural numbers n. In the L1 metric, n.n C 1/x n if nC1 < x n ˆ > ˆ > ˆ > ˆ > ˆ > : ; 1 1 if n < x 1 these functions converge to the function which is identically 1 on Œ0; 1. On the other hand, this sequence is not even a Cauchy sequence in the supremum metric since d.fn ; fm / D 1 for all n ¤ m. All metrics for CŒ0; 1 need to measure the distance between two continuous functions. The supremum metric measures the maximum distance between two functions whereas the L1 metric measures a mean distance between two functions.
10.3.1 Exercises Write proofs for each of the following statements. 1. Let C be a circle. For x and y in C, define d.x; y/ to be the number in Œ0; equal to the measure of the central angle in C of the arc bounded by x and y. Show that C with this distance function is a metric space. 2. If d is defined for points .x1 ; y1 / and .x2 ; y2 / in R2 by 2jx1 x2 j C 3jy1 y2 j, then R2 with distance function d is a metric space. 3. Let X be the set consisting of all integers plus one extra point M. For each 1 x 2 X, let d.x; x/ be 0. For integers x ¤ y, let d.x; y/ D min.jxj;jyj/C1 , and for 1 each integer x, let d.x; M/ D d.M; x/ D jxjC1 . Then is a metric space. 4. Let X be the collection of all sequences of real numbers a1 ; a2 ; a3 ; : : : for which there exists a natural number n such that aj D ak for all j and k greater than or equal to n. In other words, X is the collection of all sequences which are constant from some point on, such as 1; 2; 3; 4; 3; 3; 3; 3; : : : or 12 ; 23 ; 23 ; 23 ; 23 ; : : : . Define the distance between two sequences and to be 0 if the sequences are identical, and to be the least natural number n for which the difference between the two sequences is constant for all terms j n. Then X is a metric space with this metric. 5. Let p be any prime number. Then for any two rational numbersn r and s, define d.r; s/ D 0 if r D s. Otherwise, if r ¤ s, then jr sj D pba where a and b are relatively prime natural numbers, n is an integer, and neither a nor b is divisible by p. Define d.r; s/ D pn . Then the rational numbers with distance function d is a metric space. 6. If is a metric space, then for any c > 0, is also a metric space. 7. If and are both metric spaces, then is also a metric space.
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8. If and are both metric spaces, then X Y D f.x; y/ j x 2 X and y 2 Yg with distance function d .x1 ; y1 /; .x2 ; y2 / D dX .x1 ; x2 / C dY .y1 ; y2 / is a metric space. s 2 R1 f .x/ g.x/ dx is a metric space. 9. CŒ0; 1 with distance function d.f ; g/ D 0
This distance function is called the L2 metric. R1 10. The L1 metric d.f ; g/ D jf .x/ g.x/j dx is not a metric for the space of all 0
Riemann integrable functions defined on the interval Œ0; 1.
10.4 Topology of Metric Spaces Recall that in the real numbers, R, the interior of a set S is defined to be the set of points x 2 S such that there is an > 0 for which the entire interval .x ; x C / is contained in S. The exterior of a set is defined to be the set of points x … S such that there is an > 0 for which the entire interval .x ; x C / is contained in the complement of S. The boundary of a set S is defined to be the set of points neither in the interior nor the exterior of the set, or the points x such that for all > 0 the set .x; xC/ contains elements of S and elements of Sc . All three of these definitions generalize in a natural way to all metric spaces. Indeed, one just has to replace the role of the open interval .x ; x C / with the neighborhood N.x; /. That is, if is a metric space, and S X, the interior of S, int.S/, is the set of x 2 S such that there is an > 0 for which N.x; / S, the exterior of S, ext.S/, is the set of x 2 Sc such that there is an > 0 for which N.x; / Sc , and the boundary of S, @S, is the set of x 2 X such that for every > 0, the set N.x; / contains points in S and points in Sc . The definitions of interior, exterior, and boundary, in turn, allow one to define open and closed sets, accumulation point, derived set, and closure in ways analogous to how they are defined for the set of real numbers. That is, if S is a subset of a metric space X, S is an open set if S D int.S/, S is a closed set if @S S, S has accumulation point a if, for every > 0, N ı .a; /\S ¤ ;, the derived set of S, S0 , is the set of accumulation points of S, and the closure of S; cl.S/; is S[S0 . It is worth noting that for every x 2 X and every > 0 that N.x; / is an open set. This is easy to show by thinking about how you prove that an open interval in the real numbers is an open set. In the real numbers, if a < b, then .a; b/ is open because if y 2 .a; b/, the interval .y ı; y C ı/ .a; b/ when ı D min.y a; b y/. Similarly, then, in metric space , if a 2 X and > 0 are given, let y 2 N.a; /. It follows from the definition of N.a; / that ı D d.a; y/ > 0. Then, if x 2 N.y; ı/, d.x; y/ < ı D d.a; y/, so by the triangle inequality d.a; x/ d.a; y/ C d.y; x/ < , and x 2 N.a; /. Thus, you can conclude that N.y; ı/ N.a; / when ı D d.a; y/ which proves that N.a; / is open.
10.4 Topology of Metric Spaces
307
Fig. 10.5 Proving that the union of open sets is open
x
Many of the theorems pertaining to the topology of the real numbers proved in the preceding chapter can now be reproved in the context of metric spaces by merely changing references to open intervals .x ; x C / with the new neighborhood notation, N.x; /. For example, consider the proof that the union of open sets is also an open set (Fig. 10.5). PROOF: In metric space assume that for each i in the index set I, Ai is an open set. Then [ Ai is an open set. i2I
• In metric space assume that for each i in the index set I, Ai is an open set. • Let x 2 [ Ai . i2I
• By the definition of set union, there is an j 2 I such that x is an element of the open set Aj . • By the definition of open set, there is an > 0 such that the N.x; / Aj . • But by the definition of set union, Aj [ Ai showing that N.x; / [ Ai , i2I
i2I
which proves the theorem. Several other examples are left for the exercises. Note that any metric space with the given definition of open set is a topological space as defined in Chap. 9.
10.4.1 Exercises Write proofs for each of the following statements.
1. For every subset, S, of metric space , int int.S/ D int.S/. 2. For every subset, S, of metric space , @.@S/ @S. 3. For subsets S and T of metric space , ext.S [ T/ ext.S/ \ ext.T/.
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4. A subset S of a metric space is open if and only if its complement is closed. 5. In any metric space, the intersection of a finite number of open sets is an open set. 6. The subset S of a metric space is closed if and only if S D cl.S/. 7. For any subset S of a metric space, its derived set, S0 , is a closed set. 8. A set U is an open set in Rn using the taxicab metric if and only if U is open in Rn using the Euclidean metric.
10.5 Limits in Metric Spaces Let A and B be subsets of the real numbers and f W A ! B. Recall that if a is an accumulation point of the set A, the limit lim f .x/ D L means that for every > 0 x!a
there is a ı > 0 such that whenever x 2 A with 0 < jx aj < ı, then jf .x/ Lj < . This definition can now be extended to general metric spaces. Thus, if A and B are metric spaces and f W A ! B, then if a 2 A and L 2 B, the limit lim f .x/ D L x!a
means that for every > 0 there is a ı > 0 such that whenever x 2 N ı .a; ı/, then f .x/ 2 N.L; / (Fig. 10.6). For example, consider f W R2 ! R3 given by f .x; y/ D .x y; x C y; x2 C xy/ where R2 and R3 are the metric spaces using their respective Euclidean metrics. Then lim f .x; y/ D .0; 2; 2/. Consider how you would prove this. The proof .x;y/!.1;1/
would need to end with showing that the distance from .x y; x C y; x2 C xy/ to .0; 2; 2/ is less p than some > 0 whenever .x; y/ is within some ı > 0 of .1; 1/. That is, .x y 0/2 C .x C y 2/2 C .x2 C xy 2/2 < . At first this inequality may appear quite intimidating, but keep in mind that when you write proofs about limits, you have a great deal of flexibility because you do not have to find an optimal value for ı > 0, just one that works. In particular, your task would be done if you could simultaneously make jx yj small, jx C y 2j small,
f a
X Fig. 10.6 Limit of f W X ! Y as x approaches a is L
L
Y
10.5 Limits in Metric Spaces
309
and jx2 C xy 2j small. In fact, if these three quantities were each less than 2 , q p 2 then .x y 0/2 C .x C y 2/2 C .x2 C xy 2/2 would be less than 3 4 < . So how can you arrange for each of p jx yj, jx C y 2j, and jx2 C y2 2j to be less than 2 when you know that .x 1/2 C .y 1/2 < ı? One thing that p .x 1/2 C .y 1/2 < ı tells you is that each of jx 1j and jy 1j must be less than ı because, if either of them exceeded ı, the square root of the sum of their squares would also exceed ı. So look at each inequality separately. To make jx yj < 2 , it would be enough to have jx 1j and jy 1j both less than 4 because jxyj D j.x1/.y1/j jx1jCjy1j. Similarly, if jx1j and jy1j were both less than 4 , then jxCy2j D j.x1/C.y1/j jx1jCjy1j would be less than . As for jx2 Cxy2j you again want to write the expression using terms that include 2 factors of x1 or y1 so you can make those terms small. One way to do this would be to write jx2 Cxy2j D j.x2 1/C.xy1/j D j.x1/.xC1/C.x1/yC.y1/j. As with other limits of quadratic expressions with which you have dealt, it is convenient to limit the size of ı so that x and y cannot grow too large. So, if ı is less than 1, both jxj and jyj will be bounded by 2, and jx C 1j and jy C 1j will each be bounded by 3. Thus, it would be good enough to know that jx 1j and jy 1j do not exceed 12 because then, jx2 C y2 2j D j.x 1/.x C 1/ C .x 1/y C .y 1/j jx 1j jx C 1j C jx 1j jyj C jy 1j 12 3 C 12 2 C 12 D 2 . This results in the following proof. PROOF: If f .x; y/D.x y; x C y; x2 C xy/, then • • • • • • • • • • •
lim
.x;y/!.1;1/
f .x; y/D.0; 2 ; 2/.
Let f .x; y/ D .x y; x C y; x2 C xy/ be a function from R2 into R3 . For > 0, let ı D min.1; 12 / > 0. 2 Let .x;py/ 2 R such that .x; y/ is within ı of .1; 1/. Then .x 1/2 C .y 1/2 < ı from which it follows that jx 1j < ı and jy 1j < ı. Because ı 1, 0 < x < 2 so jxj < 2 and jx C 1j < 3. Similarly, jyj < 2 and jy C 1j < 3. Also, since jx 1j < ı 12 and jy 1j < ı 12 , it follows that jx yj D j.x 1/ .y 1/j jx 1j C jy 1j < 2 12 < 2 . Similarly, jx C y 2j D j.x 1/ C .y 1/j jx 1j C jy 1j < 2 . Finally, jx2 C xy 2j D j.x 1/.x C 1/ C .x 1/y C .y 1/j jx 1j jx C 1j C jx 1j jyj C jy 1j < 12 3 C 12 2 C 12 D 2 . 2 Thus, the distance from .x y; x C y; x C xy/ to 2; 2/ is q q .0; 2 2 .x y/2 C .x C y 2/2 C .x2 C xy 2/ < 3 2 < . Therefore, lim f .x; y/ D .0; 2; 2/. .x;y/!.1;1/
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Fig. 10.7 Limit of a sequence in a metric space
a3
a2 L a1
In Chap. 8 it was seen that the sequence of functions fn .x/ D xn as a sequence in CŒ0; 1 does not converge in the supremum metric. On the other hand, if one uses the L1 metric instead, then this sequence converges to the function f .x/ D 0. To see this, just calculate the distance between fn and f and find its limit as n approaches R1 infinity. The required distance is jxn 0jdx D 1n which indeed does converge to 0 0
(Fig. 10.7). This raises an interesting question. Suppose and are both metric spaces. Can you have a sequence in X which converges to one limit in metric d1 and converges to a different limit in metric d2 ? This differs from the example in the previous paragraph where a sequence converges in one metric and diverges in the other. The answer to this question is yes, you can have lim an D L n!1
in the d1 metric, and lim an D M in the d2 metric with L ¤ M. For example, let n!1
X be the real numbers and d1 be the familiar jx yj (Euclidean) distance function. Let d2 be exactly the same as d1 except that it “confuses” the numbers 0 and 1. That is, let d2 .x; y/ D d1 .x; y/ whenever neither x nor y is either 0 or 1, but if either x or y is either 0 or 1, then exchange 0 for 1 and 1 for 0 when calculating the distance. Then d2 .3; 5/ D j3 5j D 2; d2 .0; 6/ D j1 6j D 5; d2 .1; 1/ D j 1 0j D 1, and d2 .0; 1/ D j1 0j D 1. Essentially, d2 acts as if the real line were relabeled with the names for the points 0 and 1 interchanged. Then, of course, any sequence of non-zero numbers that converges to 0 in the d1 metric will converge to 1 in the d2 metric.
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10.5.1 Exercises Write proofs for each of the following statements. 1. Let a1 ; a2 ; a3 ; : : : be a sequence in the metric space . Then lim an D a n!1
if and only if lim d.an ; a/ D 0. n!1 2. If a sequence in a metric space converges, then the sequence is Cauchy. 3. In any metric space if lim f .x/ D L and lim f .x/ D M, then L D M. x!a x!a 4. Suppose that real-valued functions f and g are defined on metric space X. Then lim f .x/ D L and lim g.x/ D M imply that lim .f C g/.x/ D L C M. x!a
x!a
x!a
5. A sequence in Rn converges when using Euclidean metric if and only if it converges using the supremum metric.
10.6 Continuous Functions on Metric Spaces Recall that if f is a real-valued function defined on a subset A of the real numbers, then f is continuous at a point a 2 A if for every > 0 there is a ı > 0 such that jf .x/ f .a/j < for all x 2 A satisfying jx aj < ı. Then f is said to be continuous on A if it is continuous at each a 2 A. This definition is easily extended to functions f W X ! Y where and are metric spaces. The function f W X ! Y is continuous at a 2 X if for all > 0 there is a ı > 0 such that dY f .x/; f .a/ < for all x 2 X satisfying dX .x; a/ < ı. Such a function is said to be continuous on X if it is continuous at each point a 2 X. The function is uniformly continuous on X if for each > 0 there is a ı > 0 such that dY f .x/; f .y/ < whenever x; y 2 X with dX .x; y/ < ı. Many theorems about continuous functions from the real numbers into the real numbers can now be extended to continuous functions from one metric space into another metric space, or at least to continuous functions from a metric into the real numbers. For example, in the previous chapter it was shown that if A and B are subsets of the real numbers, then a function f W A ! B is continuous on A if and only if f 1 .D/ is open in A whenever D is open in B. This can now be extended to the case where A and B are any two metric spaces. The following proof is identical to the proof in the previous chapter obtained by replacing occurrences of jx yj in the proof with distance functions from the metric spaces.
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PROOF: Let and be metric spaces, and let f W A ! B be a function from A to B. Then f is continuous on A if and only if for every open set D B, its preimage under f , f 1 .D/, is an open set in A. • Let and be metric spaces, and let f W A ! B be a function from A to B. Continuity implies that the preimages of open sets are open Assume that f W A ! B is a continuous function. Let D be an open set in B, and let a 2 f 1 .D/. Because D is open in B, thereis an > 0 such that N.f .a/; / D. Thus, if y 2 B with dB f .a/; y < , then y 2 D. Because f is a continuous function, there is aı > 0 such that for all x 2 A with dA .a; x/ < ı it follows that dB f .a/; f .x/ < . • Thus, if x 2 N.a; ı/, then f .x/ 2 N f .a/; , so f .x/ 2 D and x 2 f 1 .D/. • Therefore, f 1 .D/ is an open set in A proving that in preimage under f of any open set is open.
• • • • •
The preimages of open sets are open implies continuity • Assume that the preimage under f of any set D open in B is an open set in A. • Let a 2 A, and let > 0 be given. ˇ ˇ • The set N f .a/; is an open set in B, so its preimage, C D fx 2 A f .x/ 2 N f .a/; g is an open set in A. • Because C is an open set containing a, there is a ı > 0 such that N.a; ı/ C. • Thus, if x 2 A with dA.a; x/ < ı, then x 2 N.a; ı/ C, so f .x/ 2 f .C/ implying that f .x/ 2 N f .a/; . • Therefore, f is continuous which completes the proof of the theorem. Another theorem about continuous functions from the real numbers to the real numbers is that the composition of two continuous functions is a continuous function. This theorem generalized to metric spaces, but due to the previous result, it now has a very simple proof which is not only valid for metric spaces, it is valid for continuous functions between topological spaces. PROOF: Suppose , , and are metric spaces and functions f W X ! Y and g W Y ! Z are both continuous functions. Then g ı f W X ! Z is a continuous function. • Let , , and be metric spaces, and let functions f W X ! Y and g W Y ! Z both be continuous functions. • Let U Z be an open set. • Then, because g is a continuous function, the set g1 .U/ is an open set in Y. (continued)
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• Then, because f is a continuous function, the set f 1 g1 .U/ is an open set in X. • If x 2 f 1 g1 .U/ , then f .x/ 2 g1 .U/ and g f .x/ 2 U implying that x 2 .g ı f /1 .U/. • Conversely, if x 2 .g ı f /1 .U/, then .g ı f /.x/ D g f .x/ 2 U implying 1 that x 2 f 1 g .U/ . • Thus, f 1 g1 .U/ D .g ı f /1 .U/ is an open set, so g ı f is a continuous function. For some natural number n, consider functions that map a metric space into the metric space Rn using the Euclidean metric. In particular, if f W X ! Rn , then for each x 2 X, f .x/ 2 Rn . As a point in Rn , f .x/ has n coordinates that each depends on the value of x, that is, f .x/ D f1 .x/; f2 .x/; f3 .x/; : : : ; fn .x/ . It is important that f is a continuous function if and only if fj is continuous for each j D 1; 2; 3; : : : ; n. What will be the format of a proof of this result? First of all, since the statement of the result is the biconditional “f is continuous if and only if fj is continuous for each j,” the proof will have two parts. If it is assumed that f is a continuous function, then you must show that fj is continuous for each j. But this will not be hard since the Euclidean distance between two points f .x/ and f .a/ will be greater than or equal to the distance between their jth coordinates, fj .x/ and fj .a/. Thus, a ı > 0 that ensures that f .x/ is within > 0 of f .a/ will also insure that fj .x/ is within of fj .x/. Conversely, if fj is continuous for each j, then a ıj > 0 can be found that will ensure that fj .x/ is very close to fj .a/. But by making each jfj .x/ fj .a/j small, say smaller than n , the Euclidean distance from f .x/ to f .a/ will be less than . This will be what is needed to show that f is continuous. PROOF: Let n be a natural number, and suppose is a metric space. Let f be a function from X to n-dimensional Euclidean space, Rn , using the Euclidean metric. Then f is a continuous function if and only if for each j D 1; 2 ; 3; : : : ; n the jth coordinate function for f , fj W X ! R, is continuous. • Let n be a natural number, and suppose is a metric space. • Let f be a function from X to n-dimensional Euclidean space, Rn , using the Euclidean metric. Continuity of f implies continuity of fj • Assume that f is a continuous function. • Let a 2 X, and let > 0 be given. • Because f is continuous, there is a ı > 0 such that for all x 2 X with d.x; a/ < ı, the distance in Rn from f .x/ to f .a/ is less than . (continued)
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But, for each j D 1; 2; 3; : : : ; n and each x 2 X with d.x; a/ < ı, jfj .x/ fj .a/j q 2 2 2 2 f1 .x/f1 .a/ C f2 .x/f2 .a/ C f3 .x/f3 .a/ C C fn .x/fn .a/ < . Thus, fj is continuous at each a 2 X, so fj is continuous on X. Continuity of fj implies continuity of f • Assume that for each j D 1; 2; 3; : : : ; n the coordinate function fj is continuous. • Let a 2 X, and let > 0 be given. • For each j D 1; 2; 3; : : : ; n, because fj is continuous at a, there is a ıj > 0 such that whenever x 2 X with d.x; a/ < ıj , it follows that jfj .x/fj .a/j < n . • Let ı D min.ı1 ; ı2 ; ı3 ; : : : ; ın / > 0, and let x 2 X with d.x; a/ < ı. • q Then the Euclidean distance in Rn from f .x/ to f .a/ is 2 2 2 2 f1 .x/f1 .a/ C f2 .x/f2 .a/ C f3 .x/ f3 .a/ C C fn .x/ fn .a/ q 2 < n n D pn . • This shows that f is continuous at each a 2 X, so f is a continuous function completing the proof of the theorem. Recall that is a discrete metric space when d.x; y/ D 1 whenever x ¤ y. In this case N.x; 1/ is the singleton fxg for every x 2 X, so each singleton is an open set. Since arbitrary unions of open sets are open sets, all subsets of X are open. As a consequence, all subsets of X are also closed. Suppose that is any metric space, and the function f maps X into Y. Then f is automatically continuous. Indeed, for any open set U Y, the set f 1 .U/ X is open in X because all subsets of X are open.
10.6.1 Exercises Write proofs for each of the following statements. 1. Let be a metric space and a 2 X. Then f .x/ D d.a; x/ is a continuous function from X to R. 2. Let be a metric space, and suppose f and g are both continuous functions from X to R. Then the product f g is a continuous function from X to R. 3. Let d1 be the taxicab metric on R2 and d2 be the supremum metric on R2 . Then f W ! given by f .x/ D x is a continuous function, and so is its inverse. 4. Let d1 be the supremum metric in CŒ0; 1 and d2 be the L1 metric in CŒ0; 1. Then f W ! given by f .x/ D x is a continuous function.
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315
5. Suppose and are both metric spaces, and that there are positive real numbers c and C such that for all x and y in X, the distance function satisfies cd1 .x; y/ d2 .x; y/ Cd1 .x; y/. Then a function is continuous on X with metric d1 if and only if it is a function continuous on X with metric d2 . 6. The three metrics for Euclidean n-space: the Euclidean metric, the taxicab metric, and the supremum metric are related in pairs as described in the previous problem. 7. If is a metric space and D X, then if f W D ! R is uniformly continuous on D, there is a continuous function g W cl.D/ ! R such that g.x/ D f .x/ for all x 2 D.
10.7 Homeomorphism The interval Œ0; 1 and the interval Œ5; 30 certainly are different in length and 5 have different arithmetic properties such as 01 ¤ 30 . On the other hand, the two intervals have identical topological properties in the sense that there is a one-to-one correspondence between the points of these two intervals such that a set is open in the first interval if and only if the corresponding set is open in the second interval. It can easily be seen that the function f .x/ D 25x C 5 is a bijection from Œ0; 1 to Œ5; 30 and a set U is open in Œ0; 1 if and only if f .U/ is open in Œ5; 30. This follows from the fact that both f and its inverse function f 1 .x/ D x5 are continuous 25 and the fact that inverse images of open sets under continuous maps are open. Two topological spaces X and Y are called homeomorphic if there is a continuous bijection f W X ! Y whose inverse f 1 W Y ! X is also continuous. In such a case, the bijection f is called a homeomorphism. If you know that two topological spaces are homeomorphic, then you know that all of the topological properties of the two spaces are the same. In particular, if X and Y are homeomorphic spaces, then if you can prove that X has a particular property, it may follow immediately that Y also has that property. Since metric spaces are topological spaces, one has that two metric spaces and are homeomorphic if there is a continuous bijection from X to Y whose inverse is continuous. Thus, Œ0; 1 and Œ5; 30 are homeomorphic with homeomorphism f .x/ D 25x C 5. Note that if f W X ! Y is a continuous bijection, then the function f 1 W Y ! X will be defined, but it need not be continuous. For example, the function f . / D .cos ; sin / is a continuous bijection that maps the interval Œ0; 2/ on the real line onto the circle radius 1 centered at the origin in R2 . The inverse is not a continuous function because it maps every neighborhood of the point .1; 0/ 2 R2 to points that are close to 2 and other points that are close to 0. This argument does not prove that the interval is not homeomorphic to the circle; just that this particular function is not a homeomorphism. A proof that the two spaces are not homeomorphic will be discussed in the next section.
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10.7.1 Exercises Write proofs for each of the following statements. 1. The interval .0; 1/ in R is homeomorphic to the entire real line. 2. The open interval .0; 1/ is not homeomorphic to the closed interval Œ0; 1.
10.8 Connected Metric Spaces Recall that a set S is connected if whenever it is contained in the union of two disjoint open sets, U and V, then either S \ U or S \ V is empty. In the previous chapter it was shown that the continuous image of a connected set must be connected. Again, since metric spaces are topological spaces, this result pertains to metric spaces as well. Also seen in the previous chapter is a simple proof of the Intermediate Value Theorem which is an immediate consequence of the fact that the continuous image of an interval on the real line must be a connected set. A metric space is disconnected if it can be written as the union of two disjoint nonempty open sets. Of course, if X D U [ V for open sets U and V, then U is the complement of V in X, so U is the complement of an open set implying that U is a closed set. Thus, another way to say that X is connected is to say that the only subsets of X that are both open and closed are the empty set and X itself. Also, if X is the union of the two disjoint open sets, U and V, then the function f W X ! R defined by f .x/ D 1 if x 2 U and f .x/ D 0 if x 2 V is a continuous function since the preimages of open sets in R are open in X. Thus, X is connected if and only if every continuous function from X to f0; 1g is a constant function. The concept of connectedness can be used to prove results about what may or may not be the image of a particular connected set. For example, using a connectedness argument, you can show that it is impossible to have a continuous function map a closed interval on the real line injectively (one-to-one) onto a circle in R2 . That is, if f W Œa; b ! R2 is a continuous injective function, then the image of f cannot be a circle. In particular, a closed interval is not homeomorphic to a circle. To see why, suppose that f does map Œa; b continuously and injectively into a circle C in R2 . Consider the points f .a/ and f .b/ which must be elements of the circle C, and since f is injective, f .a/ ¤ f .b/. Thus, f .a/ and f .b/ are the end points of two disconnected open arcs, C1 and C2 , of circle C. The open interval .a; b/ is a connected set which f maps injectively into C, and since f is injective, f cannot map any point of .a; b/ to f .a/ or f .b/. Hence, f maps .a; b/ continuously into C1 [ C2 . But the image of the connected set .a; b/ must be a connected set, so it cannot be the entire union C1 [ C2 . Of course, if f .a/ is allowed to equal f .b/, it is easy to construct a continuous map of Œa; b onto a circle that is injective on .a; b/. Similarly, you can find a continuous injective map of the half-open interval Œa; b/ onto a circle.
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Generalizing the argument used in the previous paragraph leads to a very powerful tool for showing two specific spaces are not homeomorphic. Suppose that for metric spaces and , the function f W X ! Y is a homeomorphism, that is, a one-to-one map from X onto Y that is continuous with a continuous inverse f 1 W Y ! X. Then for any C X with f .C/ D D, the function f W C ! D is a homeomorphism of C to D. Indeed, the function f restricted to C is a one-to-one continuous function that maps C onto D whose inverse is the restriction of f 1 to D, so it is also continuous. As a result C is a connected set in X if and only if D is a connected set in Y. With this observation it can easily be shown that the interval Œ0; 2/ in R is not homeomorphic to a circle in R2 . Suppose that there were a homeomorphism f mapping the interval Œ0; 2/ onto a circle. Then f also maps Œ0; 1/ [ .1; 2/ onto the circle with one point missing. But this cannot be since Œ0; 1/ [ .1; 2/ is not a connected set and its image, a circle with one point missing, is a connected set. This is a contradiction implying that there is no homeomorphism from Œ0; 2/ to the circle.
10.8.1 Exercises Write proofs for each of the following statements. 1. If f is a continuous function from R to R with the property that f .x/ is a rational number for each irrational x 2 R, then f is a constant function. (Hint: This is a question of cardinality.) 2. Let Y be a set in R2 shaped like Y, that is, three line segments that share a common endpoint. Then there is no continuous injective function that maps the interval Œa; b in R onto Y. 3. The real line R is not homeomorphic to the plane R2 .
10.9 Compact Metric Spaces Recall that a set S in any topological space is compact if every cover of S by a collection of open sets has a finite subcover, that is, there are finitely many open sets in the collection which together form a cover of S. There is an equivalent way to state the property of compactness by considering the complements of the open sets in the open cover of S. In particular, if the sets in the open cover are open sets, their complements are closed sets. If the union of the sets in the open cover contains all of S, then the intersection of the complements contains no points of S. Thus, the compactness of S is equivalent to the property that if the intersection of a collections of sets closed in S is the empty set, then there is a finite subcollection of those closed sets whose intersection is also the empty set.
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Since metric spaces are topological spaces, the concept of compactness can be applied to metric spaces. That is, is said to be a compact metric space if every cover of X by open sets contains a finite subcover. It follows easily that if X is compact, then every close subset of X is also compact. The proof of this is left as an exercise. Any set S in a metric space is said to be bounded if there is a positive real number r and a point a 2 X such that S N.a; r/. Equivalently, if S is not an empty set, one can define the diameter of set S to be sup d.x; y/. Then S is bounded x;y2S
if its diameter is finite. As it is in the real numbers, any compact set S in a metric space is both closed and bounded. To prove that a compact set S is closed, suppose instead that a is an accumulation point S that is not contained in S. For each real number r > 0 let Ur D fx j d.x; a/ > rg. Then S is contained in the union of the open sets Ur sets, yet, because a is an accumulation point of S, no finite collection of the Ur sets will cover S. Thus, if S is compact, it must contain all of its accumulation points, so S is closed. To prove that a compact set S is bounded, let a be any point in X. Then S is certainly contained in the collection of sets N.a; r/ where r ranges over the positive real numbers. Thus, if S is compact, it is contained in a finite number of the N.a; r/. If t is the largest of the r values associated with the finite number of neighborhoods, then S is contained in N.a; t/ showing that S is bounded (Fig. 10.8).
a a S
S X
Fig. 10.8 A compact set of a metric space is closed and bounded
X
10.9 Compact Metric Spaces
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PROOF: A compact set in a metric space is both closed and bounded. • Let S be a compact subset of metric space . S is closed • Assume that a … S is an accumulation point of S. • Because a … S, the set S is contained in the collection of open sets Ur D fx j d.x; a/ > rg where r ranges over the positive real numbers. • Since S is compact and contained in the union of the open sets fUr g, it must be contained in a finite subcollection of fUr g. • Let t be the least of the r values associated with this finite subcollection of Ur sets. • Then S Ut . • But a is an accumulation point of S, so N.a; t/ \ S is not empty, so S cannot be contained in Ut . • This is a contradiction, so there must not be any accumulation points of S contained in the complement of S implying that S is a closed set. S is bounded • Let a be any point in X. • S is contained in the union of the open neighborhoods N.a; r/ where r ranges over the positive real numbers. • Since S is compact and contained in the union of the N.a; r/ sets, it must be contained in a finite subcollection of the N.a; r/ sets. • Let t be the greatest of the r values associated with this finite subcollection of N.a; r/ sets. • Then S N.a; t/ implying that S is a bounded set. • Thus, any compact set S in a metric space must be both closed and bounded. The Heine–Borel Theorem says that for the real numbers, the converse of this theorem is also true; that is every closed and bounded set of real numbers is compact. The fact that it is also true for Rn using the Euclidean metric is left as an exercise. But there are metric spaces in which some closed and bounded sets are not compact. Consider, for example, the metric space CŒ0; 1 using the of functions in CŒ0; 1 given by fn .x/ D 9 8 supremum metric. The sequence 1 > ˆ 0 if x nC1 = < 1 n.n C 1/x n if nC1 < x < 1n for natural numbers n. Let S D ffn j n > 0g. > ˆ ; : 1 if 1n x Then if m and n are distinct natural numbers, the distance between fm and fn is 1, so the set S can have no accumulation point. Thus, S is a closed set. Each fn 2 S is a distance 1 from all the other functions in S, so S has diameter 1, and, thus, S is a bounded set. Note that for each natural number n the open neighborhood N.fn ; 12 /
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contains fn but no other function in S. Thus, S is contained in the union of an infinite collection of open sets, but is not contained in any finite subcollection of those open sets. Therefore, S is a closed bounded set that is not a compact set. In Chap. 9 it was shown that if f is a continuous function and C is a compact set contained in the domain of f , then f .C/ is also compact. This proof used only properties of open sets and continuous functions, so the proof applied to functions between any topological spaces. As a consequence, since all metric spaces are topological spaces, this result applies to functions between metric spaces. In Chap. 4 it was shown that a continuous real-valued function on a closed bounded interval Œa; b is a bounded function, and such a function obtains its minimum and maximum values at points in Œa; b. These results can now be generalized to continuous functions from any compact metric space into the real numbers. They are, in fact, simple consequences of the fact that the continuous image of a compact set is compact (Fig. 10.9). PROOF: If f is a continuous function from a compact metric space into the real numbers, then f is a bounded function. • Let be a compact metric space, and let f be a continuous real-valued function with domain X. • Because the continuous image of a compact set is compact, the image of f , f .X/, is a compact subset of R. • Since all compact subsets of R are bounded, it follows that f is a bounded function.
Fig. 10.9 A continuous real-valued function on a compact set in R2 obtains its maximum and its minimum
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PROOF: If f is a continuous function from a compact metric space into the real numbers, then f obtains its minimum and its maximum value on X. • Let be a compact metric space, and let f be a continuous real-valued function with domain X. • Because the continuous image of a compact set is compact, the image of X, f .X/, is a compact subset of R. • Since all compact subsets of R are closed, it follows that the infimum and supremum of f .X/ are elements of f .X/. • Thus, f .X/ contains its minimum and maximum values implying that there are a; b 2 X such that f .a/ and f .b/ are the minimum and maximum values of f .C/, respectively. • Therefore, f obtains its minimum and maximum values on X. You might reflect for a minute on the fact that the preceding two proofs are considerably simpler than analogous theorems proved in Chap. 4 about continuous real-valued functions defined on closed bounded intervals. Of course, at that time, you had not been introduced to open and closed sets, did not have at your disposal the theory of compact sets, had not proved the Heine–Borel Theorem that established that closed bounded sets are compact, and had not proved that the continuous images of compact sets are compact. It is actually comforting to realize that the effort to learn about all of these newer concepts does give you many powerful tools that simplify such proofs. Earlier in the chapter in the section on homeomorphisms there was an example of a continuous bijection between two metric spaces that failed to be a homeomorphism because the function did not have a continuous inverse. When compact metric spaces are involved, this does not happen, that is, any continuous bijection of a compact metric space to another (necessarily compact) metric space has a continuous inverse, and thus, is a homeomorphism. To prove this you would want to assume that f W X ! Y was a continuous bijection from compact metric space to metric space . Because f is injective, it does have an inverse function f 1 W Y ! X. How would you prove that such a function is continuous? One way would be to show that the inverse of f 1 , that is f , maps open sets to open sets. But note that f maps closed sets to closed sets because every closed subset of the compact metric space, X, is compact, so its image under the continuous function f must be compact and thus closed. So if f maps closed sets onto closed sets, then since f is a bijection, it must map the complement of a closed set onto the complement of a closed set. This means f maps open sets onto open sets (Fig. 10.10).
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Fig. 10.10 A continuous bijective function between compact metric spaces is a homeomorphism
f f(U)
U f-1 X
Y
PROOF: If f is a continuous bijective function from a compact metric space onto another metric space, then f is a homeomorphism. • Let be a compact metric space and f be a continuous bijection from X onto the metric space . • Because f is injective, the function f 1 W Y ! X is defined. • Let U be an open set in X. • Then the complement of U in X, U c , is a closed set. • Since all closed subsets of a compact set are compact, U c is a compact set. • Since the continuous image of a compact set is compact, f .U c / is a compact set in Y. • Since f .U c / is compact, it is closed, and its complement Ynf .U c / is open. • Because f is a bijection, Ynf .U c / D Ynf .U/c D f .U/ which is open. • Thus, f maps open sets in X onto open sets in Y. • Since the inverse of f 1 is f which maps open sets onto open sets, f 1 is a continuous function, and, therefore, f is a homeomorphism.
10.9.1 Exercises Write proofs for each of the following statements. 1. A closed subset of a compact metric space is compact. 2. A closed bounded subset of Rn using the Euclidean metric is a compact set. (Hint: Use the fact that each n-dimensional cube is the union of 2n cubes that each has side lengths half that of the original cube. Then if an open cover of a cube has no finite subcover, then at least one of the 2n smaller cubes has no finite subcover.) 3. A nonempty set S in a metric space has a finite diameter if and only if there is an a 2 X and a positive real number r such that S N.a; r/. 4. The Bolzano–Weierstrass Theorem holds in a compact metric space, that is, if is a compact metric space and S is an infinite subset of X, then S has an accumulation point.
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5. Suppose set S in metric space has diameter r. Show that cl.S/ also has diameter r.
10.10 Complete Metric Spaces The Completeness Axiom discussed in Chap. 2 says that every nonempty subset of the real numbers that is bounded above has a least upper bound. As seen in Chap. 3, this has many important consequences such as • The Intermediate Value Theorem • Every bounded monotone sequence of real numbers converges. • Every Cauchy sequence of real numbers converges. The concept of completeness can be generalized to metric spaces. is a complete metric space if every sequence in X that is Cauchy is a convergent sequence. This is not a concept that can apply to topological spaces in general because the concept of a Cauchy sequence only makes sense when there is some sort of distance measure that can be used to determine if the terms of the sequence are getting close to each other. Any compact metric space is a complete metric space. To prove this you would assume that is a compact metric space and that is a Cauchy sequence in X. Your goal is to prove that the sequence has a limit in X. First you would need to identify a point as the limit and then prove that that point is the limit of the sequence. The property of compact metric spaces that works here is the property that if the intersection of a collection of closed sets is the empty set, then a finite subcollection of those closed sets also has empty intersection. Can you use the property of Cauchy sequences to find a sequence of closed sets decreasing in size whose intersection is not empty? Here is one technique that works. Because the sequence is Cauchy, there is a natural number k1 such that for all m and n greater than or equal to k1 , the distance d.am ; an / < 1. That means that for all n k1 , an 2 N.ak1 ; 1/. Of course, N.ak1 ; 1/ is not a closed set, but its closure N1 D cl N.ak1 ; 1/ is. Then N1 is a closed set in X that contains the entire sequence from the k1 term on. Similarly, for each natural number p, find a natural number kp such that for all m and n greater than or equal to kp , the distance d.am ; an / < 1p . Then let Np D cl N.akp ; 1p / , which is a closed set such that for all n akp , an 2 Np . Now it is easy to verify that the intersection of any finite number of the Np sets contains infinitely many terms of the sequence, so the intersection of any finite collection of these closed sets is not empty. Thus, since X is a compact metric space, the intersection of all the Np sets is not empty. But since the diameter of the set Np is at most 2p which goes to 0 as p gets large, there can be no more than one point in the intersection of all of the Np sets. Call this intersection point L. By selecting a p large enough so that 2 is smaller than a given > 0, you can assure that for all n > kp the term an is in p Np and, thus, within of L. This shows that the sequence converges to L.
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PROOF: Every compact metric space is complete. • Let be a compact metric space and be a Cauchy sequence in X. • Because is Cauchy, for each natural number p there is a natural number kp such that for all m; n kp , d.am ; an / < 1p . • Thus, for every m kp , the term am 2 N.akp ; 1p /. • Define Np D cl N.akp ; 1p / which is a closed set because it is the closure of a set, and it has diameter at most 2p . • Each set Np contains the tail end of the sequence from the kp term onward. • Thus, it follows that the intersection of a finite number of the Np sets also contains the tail end of the sequence, and, therefore, is not empty. • Because X is a compact metric space, it follows that the intersection of all of the Np sets is not empty, that is, \Np ¤ ;. • If this intersection contains points L and M, then L; M 2 Np for each p so d.L; M/ 2p . • It follows that d.L; M/ D 0, so L D M, and there is only one point L in the intersection of the Np sets. • Given > 0 select p so that 2p < . • Then for all n kp , an 2 Np so jan Lj < 2p < . • This proves that lim an D L, proving that all Cauchy sequences in X n!1 converge, so is a complete metric space. In Chap. 4 there are two proofs for the Heine–Borel Theorem, which states that closed bounded sets of real numbers are compact. The second of these proofs begins with a closed interval which is assumed to have an open covering. If the interval cannot be covered by a finite subcover, then at least one of the two halves of the interval cannot be covered by a finite subcover. That half interval can, in turn, be broken into two halves, and at least one of those intervals does not have a finite subcover. This process can be continued until the collection of ever smaller intervals converge to an individual point. The properties of the real numbers that makes this argument work is that every interval can be written as the union of two intervals each with half the length of the original, and the fact that the real numbers are complete which guarantees that there will be a point in the intersection of these nested intervals that decrease to 0 in length. This suggests a way to prove an analogous result for other complete metric spaces that have the property that bounded closed sets can be written as the union of a finite number of closed sets each with a diameter that is at most half that of the original. By following the argument from the Heine–Borel Theorem, it is possible to prove that the compact sets in such a metric space are exactly the closed and bounded sets.
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PROOF: Let be a complete metric space with the property that every closed bounded subset of X can be written as the union of a finite number of closed sets each with a diameter that is at most half the diameter of the original closed bounded subset. Then a subset of X is compact if and only if it is both closed and bounded. • Let be a complete metric space, and assume that every closed and bounded subset of X can be written as the union of a finite number of closed sets each with a diameter that is at most half the diameter of the original closed bounded subset. • It has already been shown that every compact set in X is both closed and bounded. • Assume that S0 X is both closed and bounded with diameter r. • Let T be a collection of open sets that covers S0 , and assume that no finite collection of open sets in T covers S0 . • By assumption S0 can be written as a union of a finite number of closed subsets each diameter at most 2r . • At least one of these finitely many subsets cannot be covered by finitely many open sets in T. Call one of those sets S1 . • Clearly, S1 is not the empty set, or it could be covered by a single open set in T. • Assume, inductively, that for some k > 0 the set Sk Sk1 has been chosen with diameter at most 2rk such that Sk cannot be covered by a finite number of open sets in T. • Sk can be written as the union of a finite number of closed subsets each with r diameter at most 2kC1 . • Because Sk has no finite subcover in T, at least one of these finitely many subsets of Sk has no finite subcover in T. Call one of these sets SkC1 . • Thus, by mathematical induction, there is a nested sequence of sets S0 S1 S2 where each Sk cannot be covered by a finite number of open sets in T and each Sk has diameter at most 2rk . • Because none of the Sk are the empty set and the sets are nested, the intersection of any finite collection of Sk sets is nonempty. • From each Sk set select one element and call it ak . • Because the Sk sets are nested, for each natural number k and each n k, an 2 Sk . • Let > 0 be given. • Select k such that 2rk < . • Then for all m; n k, am and an are in Sk , so d.am ; an / 2rk < . • This shows that the sequence is Cauchy, and because is a complete metric space, the sequence has a limit L. • Because Sk is closed for each natural number k, and an 2 Sk for each n k, it follows that the limit of the sequence, L, is also an element of Sk . (continued)
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• Since T is an open cover of S0 , there must be an open set U 2 T such that L 2 U. • Because U is an open set, there is a ı > 0 such that N.L; ı/ U. • Select k so that 2rk < ı. • Then Sk , which contains L and has diameter of at most 2rk < ı, satisfies Sk N.L; ı/ U implying that Sk has a finite subcover of open sets in T. • This is a contradiction implying that S0 must have a finite subcover of open sets in T, and S0 is compact. • Therefore, all closed bounded sets in X are compact. For example, suppose that S is a closed bounded set in the metric space R2 using the Euclidean metric. Let S have diameter r > 0. One could be very careful and show that S can be enclosed in a square with side length r, or one could be sloppy and more easily show that S can be enclosed in a square with side length 2r. In the latter case, just select any point s 2 S and draw two parallel lines a distance 2r apart where s is half way between the two lines. Since the diameter of S is r, S will be bounded by these two lines. Then draw two lines a distance 2r apart where s is half way between the two lines and the two lines are perpendicular to the original two lines. The four lines now determine a square of side length 2r that contains S. Drawing a 6 6 grid in thispsquare partitions the set S into at most 36 sections each with diameter at most 2r 6 2 < 3r6 D 2r . Thus, R2 satisfies the condition of the theorem, so the compact sets in R2 are exactly those that are both closed and bounded (Fig. 10.11). Earlier of functions from CŒ0; 1 given by 8 in this chapter you saw the sequence 9 1 ˆ > 0 if x < = nC1 1 fn .x/ D n.n C 1/x n if nC1 < x < 1n . This is an example of a closed bounded ˆ > : ; 1 if 1n x set with diameter 1, but it is not possible to cover this set with a finite number of sets with diameter 12 since each such set could contain at most one term of the sequence. As seen before, CŒ0; 1 is a metric space which contains closed bounded sets that are not compact. It is left as an exercise to show that CŒ0; 1 with the supremum metric is complete. Fig. 10.11 Enclosing a closed bounded set in a grid
s
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10.10.1 Exercises Write proofs for each of the following statements. 1. The integers, Z, with the usual distance metric, d.x; y/ D jx yj, is a complete metric space. 2. The space CŒ0; 1 with the supremum metric is a complete metric space. 3. The space CŒ0; 1 with the L1 metric is not a complete metric space. (Hint: Find a Cauchy sequence of continuous functions that converge in the L1 metric to a discontinuous function.) 4. In the metric space Rn with the Euclidean metric, a set is compact if and only if it is both closed and bounded. 5. A closed set in R2 with diameter r > 0 can be enclosed in a square with side length r. 6. If is a sequence in a complete metric space such that 1 P d.an ; anC1 / is a convergent series, then the sequence converges. Show nD1
that the converse does not hold.
10.11 Contraction Mappings 10.11.1 Contraction Mapping Theorem A very powerful tool for showing that a particular equation has a solution comes from the properties of complete metric spaces. Let be a complete metric space, and suppose that there is a function f W X ! X. The function f is called a contraction mapping if there is a positive real number r < 1 such that for every x; y 2 X, the distance from f .x/ tof .y/ is smaller by at least a factor of r than the distance from x to y. That is, d f .x/; f .y/ r d.x; y/. In other words, the mapping contracts the space X by making all points move closer to each other by a factor of at least r. Note that all contraction mappings are continuous functions, in fact, uniformly continuous functions, because given any > 0, you can choose ı D and have that d f .x/; f .y/ r d.x; y/ < r < for all x; y 2 X with d.x; y/ < ı. The important theorem about contraction mappings states that every contraction mapping on a complete metric space has a unique fixed point, that is, exactly one point L such that f .L/ D L so f maps L to itself (Fig. 10.12). To prove this theorem about contraction mappings, you first have to identify the point L that is a fixed point, and then you must show that it is unique. Finding L is easy. Just start at any point x0 2 X and follow its orbit, that is, follow the sequence x0 ; x1 D f .x0 /; x2 D f .x1 /; : : : . By the property of contraction mappings, the distance between the terms of this sequence keeps shrinking by a factor of at least r. It follows that for any natural number k, the distance between xk and xkC1 is less than rk d.x0 ; x1 /, and, thus, the distances between successive terms of the orbit decrease at least as fast as a convergent geometric series. This is sufficient to prove
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Fig. 10.12 Contraction mapping
f that the orbit is a Cauchy sequence. Since X is complete, the sequence must converge to some point L 2 X. Note that if you begin the sequence at some other point y0 2 X, its orbit also converges, and since d.x0 ; y0 /; d.x1 ; y1 /; d.x2 ; y2 /; : : : must converge to 0, it follows that the orbit of y0 must also converge to L. In particular, the orbit of L must converge to L. Thefact that from f .L/ D L follows the triangle inequality which shows that d f .L/; L d f .L/; f .xn/ Cd f .x /; L n for any natural number n, and, in particular, for n large enough that d f .xn /; L and d f .xn /; f .L/ r d.xn ; L/ are small. The uniqueness of the fixed point comes from the simple fact that if L and M are both fixed points, then d.L; M/ D d f .L/; f .M/ r d.L; M/ which can only happen if d.L; M/ D 0 and L D M. PROOF (Contraction Mapping Theorem): Every contraction mapping on a complete metric space has a unique fixed point. • Let be a complete metric space, f be a function mapping X to X, and r be a real number in Œ0; 1/ such that for all x; y 2 X, d f .x/; f .y/ r d.x; y/. • Select any x0 2 X and for each natural number k define xk D f .xk1 /. • Note that for any natural numbers m > n, the triangle inequality shows that d.xm ; xn / d.xn ; xnC1 / C d.xnC1 ; xnC2 / C C d.xm1 ; xm / rn d.x0 ; x1 /Œrn C rnC1 C rnC2 C C rm1 d.x0 ; x1 / 1r . n r • Since 1r goes to 0 as n gets large, d.xm ; xn / can be made as small as desired by requiring m and n to be large which shows that the sequence is a Cauchy sequence. • Because is a complete metric space, the Cauchy sequence has a limit L 2 X. • Since lim xn D L, given any > 0, there is a natural number N, such that n!1
• • • •
d.xn ; L/ < 2 for all n N. Then d f .L/; L d f .L/; f .xN / Cd f .xN /; L/ rd.L; xN /Cd.xNC1 ; L/ < .r C 1/ 2 < . Therefore, f .L/ must equal L, and L is a fixed point of the contraction mapping. If M 2 X with f .M/ D M, then d.L; M/ D d f .L/; f .M/ r d.L; M/ which implies d.L; M/ D 0 and, thus, L D M. Therefore, L is a unique fixed point of f completing the proof of the theorem.
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The power of contraction mappings is that one can find the fixed point of f W X ! X by selecting any x 2 X and following its orbit which will converge to the fixed point at least as fast as a geometric series. The reader may well be familiar, for example, with a method of calculating the square root of a positive real number a by starting with a positive number x and iterating the function 2 Ca f .x/ D x 2x , a formula which comes from applying Newton’s Method to find 2 Ca a root of the function x2 a.pThe fixed point of f is an x satisfying x D x 2x which is satisfied by x D a.pIf a D 2, for example, and you begin with x D 100 as a first guess for 2, then you generate the orbit of 100 to be 100; 50:01; 25:024996; 12:55245805; 6:355894695; 3:335281609; 1:967465562; 1:49200089; 1:416241332; 1:414215014; 1:414213562; : : : with each successive p term being about half as far from 2 as the previous term. Why does this work? First note that f maps the interval Œ1; 1/ to itself. On that interval the derivative 2 of f is f 0 .x/ D 2x which has its maximum value at x D 1 where f 0 .1/ D 12 . 2x2 Thus, byˇ the Mean Value Theorem, when x > y 1, there is a c > 1 such that ˇ ˇ f .x/f .y/ ˇ ˇ xy ˇ D jf 0 .c/j, so jf .x/ f .y/j < 12 jx yj showing that f is a contraction mapping on Œ1; 1/. Another familiar example, known to any student who has played around with a calculator which can calculate the cosine function at the touch of a button, is that the orbit of 0 under the cosine function (using radian measure) is 0; 1; 0:540302306; 0:857553216; 0:65428979; 0:793480359; 0:701368774; 0:763959683; 0:722102425; 0:750417762; 0:731404042; : : : ; 0:739085132; : : : . Again, this is because the function cos x is a contraction mapping on the interval Œ 12 ; 1 because the derivative of cos x is sin x, and the Mean Value Theorem guarantees that there is a c 2 Œ 12 ; 1 such that j cos x cos yj D jx yj sin c < jx yj sin 1 where sin 1 < 1.
10.11.2 Picard’s Theorem The preceding examples suggest why contraction mappings are an important tool in Numerical Analysis, a field that seeks out efficient algorithms for solving numerical problems, often using computers. These examples find fixed points of functions from R to R. But contraction mappings can be defined on other more complex complete metric spaces. For example, the space of continuous functions on a closed interval is a complete metric space. In particular, it can be shown that there is a contraction mapping on this space of continuous functions whose fixed point is a solution to particular differential equation. A differential equation of the first order is an equation relating the variables x, y, and y0 . The equation is called first order because it only involves first order derivatives. A solution to such an equation is a function y.x/ that satisfies this 3 equation. For example, the equation y0 D 9x2 y has solutions y.x/ D Ce3x where C is
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any real number. By differentiating y.x/ D Ce3x , you can verify that it does satisfy the differential equation. It is typical that the solution of a first order equation will contain an arbitrary constant such as the C that appears in this solution. This comes from the fact that knowing a function’s derivative only determines the function up to an additive constant of integration. For this reason, a first order differential equation is often stated by giving an initial condition y.a/ D b for some constants a and b, because then, the arbitrary constant in the solution can be determined. For example, for the equation given above, if it were required that y.1/ D 5, then the solution 3 would be y.x/ D e53 e3x . A differential equation along with an initial condition is called an initial value problem. In a first course in Differential Equations you learn a large number of techniques that can be used to solve various differential equations. For example, the above equation y0 D 9x2 y can be solved by first dividing both sides of the equation 0 by y to get yy D 9x2 and then integrating both sides to find ln jyj D 3x3 C K. Exponentiating both sides of this equation and letting C equal either eK or eK gives the previously stated solution. It comes as a surprise to the student of Differential Equations that even though the first course in the field contains many techniques for solving differential equations, subsequent courses contain very little about how to solve equations and concentrate instead on finding numerical approximations to solutions and on theorems telling when solutions are expected to exist. One very powerful theorem which shows there exist unique solutions to a fairly large class of initial value problems is the Picard Existence Theorem which applies to equations of the form y0 D F.x; y/ with initial condition y.a/ D b. The theorem requires that there exists a compact set R R2 surrounding the initial point .a; b/ such that F is continuous on that compact set R, and there is a constant M such that for any points .x; y1 / and .x; y2 / in R, the function F satisfies jF.x; y1 / F.x; y2 /j Mjy1 y2 j. This condition on the second variable of F is known as a Lipschitz condition. The theorem concludes that there is a ı > 0 and a function y.x/ defined on the interval Œa ı; a C ı such that y.x/ is the unique solution to the initial value problem y0 D F.x; y/ and y.a/ D b. The theorem can be proved by noticing that y is a solution to the given initial Rx value problem if and only if y satisfies the equation y.x/ D b C F t; y.t/ dt. a
Rx But the function G.y/ D b C F t; y.t/ dt turns out to be a contraction mapping a
on a space of continuous functions on an interval of the form Œa ı; a C ı, and the Contraction Mapping Theorem guarantees a unique fixed point to this equation y D G.y/ which solves the initial value problem.
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PROOF (Picard’s Existence Theorem): Let the point .a; b/ be contained in the interior of a compact set R R2 . Let F be a function continuous on R, and assume that there is a constant M such that whenever .x; y1 / and .x; y2 / are in R, then jF.x; y1 / F.x; y2 /j Mjy1 y2 j. Then there is a ı > 0 and a unique function y.x/ such that y.a/ D b and y0 D F.x; y/ for all x in Œx ı; x C ı. • Let the point .a; b/ be contained in the interior of the compact set R R2 . • Let F be a function continuous on R, and assume that there is a constant M such that whenever .x; y1 / and .x; y2 / are in R, then jF.x; y1 / F.x; y2 /j Mjy1 y2 j. • Because F is continuous on the compact set R, it is bounded. Thus, there is a constant K such that jF.x; y/j K for all points .x; y/ 2 R. • Because .a; b/ is in the interior of R, there is a ı > 0 such that Mı < 1 and all the points .x; y/ satisfying jx aj ı and jy bj Kı lie within the compact set R. • Let C be the metric space of all functions y.x/ continuous on the interval Œa ı; a C ı satisfying jy.x/ bj Kı using the supremum metric. • Note that C is a complete metric space because it is just a closed subset of the complete metric space of all the continuous functions on Œa ı; a C ı. Rx • Define the mapping G on C by G.y/ D b C F t; y.t/ dt. a
• For each y 2 C, G.y/ is an integral of the continuous function F.t; y.t//, so G.y/ is a continuous function of x. • ˇIf y 2 C, thenˇfor any x 2 Œa ı; a C ı, it follows that jG.y/.x/ bj D ˇRx ˇ Rx Rx ˇ F t; y.t/ dtˇ jF t; y.t/ j dt K dt D Kjx aj Kı showing that ˇ ˇ a
a
a
for all y 2 C, G.y/ is also in C. • Moreover, if y1 and y2 are bothˇ in C, then for all x with jxaj ˇı, it follows ˇRx ˇ Rx that jG.y1 /.x/ G.y2 /.x/j D ˇˇ F t; y1 .t/ dt F t; y2 .t/ dtˇˇ Rx
a
a
Rx jF t; y1 .t/ F t; y2 .t/ j dt Mjy1 .t/ y2 .t/j dt
a
a
Mı sup jy1 .x/ y2 .x/j. Because Mı < 1, this shows that G is a jxajı
contraction mapping on C. • Thus, by the Contraction Mapping Theorem, there is a unique y 2 C such Rx that y D G.y/ D b C F t; y1 .t/ dt. a
• Clearly, G.y/ is equal to b at x D a, and The Fundamental Theorem of Calculus implies that y0 D F x; y.x/ . • Because y satisfies the given initial value problem if and only if it satisfies y D G.y/, this completes the proof.
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Of course, this theorem only guarantees that there is a function y.x/ satisfying the Œa ı; a C ı. But if the points differential equation in a small neighborhood a ı; y.a ı/ and a C ı; y.a C ı/ are in the interior of the compact set R, the theorem can be applied again to extend the function. As a result, a function y.x/ can be constructed whose graph extends toward the boundary of R. Consider the problem of finding a solution to the first order differential equation p y0 D xy2 C y with the initial condition y.0/ D 2. Common techniques for solving differential equations do not yield a solution to this equation. Yet Picard’s Existence Theorem shows that the equation has a unique solution. In particular, let R be the rectangle of points satisfying jxjp 2 and jy 2j 1. On that rectangle the function p 1 xy2 C y is bounded by 2 32 C 3 < 20 D K. Thus, you can choose ı D 20 so that p p 2 2 ıˇ 2 and Kı D 1 1.ˇThen jF.x; y1 / F.x; y2 /j D jx.y1 y2 / C . y1 y2 /j D ˇ ˇ ˇx.y1 C y2 / C py1 C1 py2 ˇ jy1 y2 j .2 6 C 12 /jy1 y2 j < 13jy1 y2 j, so you can < 1. The theorem then shows that there is a unique choose M D 13. Then Mı D 13 20 1 1 ; 20 . solution to the equation on the interval Œ 20 It is instructive to apply the method of Picard’s Theorem in a situation with a known outcome. For example, the initial value problem y0 D y with initial condition y.0/ D 2 has the solution y.x/ D 2ex . In this case the contraction mapping is Rx G.y/ D 2 C y.t/ dt. Suppose you begin with the function y0 .x/ D 2 and iteratively 0
apply the contraction mapping to see the orbit of y0 . Zx y1 .x/ D G.y0 / D 2 C
2 dt D 2 C 2x 0
Zx y2 .x/ D G.y1 / D 2 C
2 C 2t dt D 2 C 2x C 2 0
Zx y3 .x/ D G.y2 / D 2 C
2 C 2t C 2 0
Zx y4 .x/ D G.y3 / D 2 C
2 C 2t C 0
x2 2Š
x2 x3 x2 dt D 2 C 2x C 2 C 2 2Š 2Š 3Š
x3 x2 x3 x4 x2 C 2 dt D 2 C 2x C 2 C 2 C 2 : 2 3Š 2Š 3Š 4Š
This shows that the orbit of y0 is just the partial sums of the power series for 2ex centered at 0, a satisfying result even if it is not surprising.
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10.11.3 Fractals There is a beautiful application of contraction mappings in the study of fractals. It is difficult to define what a fractal is, but you are invited to look at the large number of pictures of fractals available in books and on the Internet to get a feel for what they are. Loosely, they are sets that show repeated structure at every level of magnification so that if you look at a very small piece of the set, it includes a structure that is either geometrically similar or nearly similar to the entire set. Famous examples are the Mandelbrot set and the Sierpinski triangle as well as the Cantor set already discussed in Chap. 6. Benoit Mandelbrot invented the word fractal in 1975 to describe the phenomenon he observed in the set that now bares his name (Figs. 10.13 and 10.14). One method of generating fractals is based on contraction mappings. In this method one sets up a complete metric space whose elements are compact sets in R2 . Then by selecting an appropriate contraction mapping f W H ! H and any nonempty compact set A 2 H, the orbit of A under the mapping f will converge to the unique fixed point of the contraction mapping which is the desired fractal. To do this let H be the collection of all nonempty compact subsets of R2 . To place a metric on H, consider first the problem of finding the distance from an individual point x 2 R2 to a set A 2 H. If R2 has Euclidean metric d, then since for every fixed y 2 B the distance d.x; y/ is a continuous function of x, it follows that d.x; y/ obtains Fig. 10.13 The Mandelbrot set (black)
Fig. 10.14 Stages in the forming of the Sierpinski triangle
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a minimum value on the compact set A, which is min d.x; y/. Then, if A and B are x2A
both elements of H, define h .A; B/ D max min d.x; y/ which is the largest distance y2B x2A
any point of B is from the set A. Again, since the Euclidean distance function is continuous, and B is compact, this min d.x; y/ obtains a maximum value on B, and x2A
there are points x 2 A and y 2 B such that d.x; y/ D h.A; B/. Because h.A; B/ need not be the same as h .B; A/, define h.A; B/ D max h .A; B/; h .B; A/ . This distance function is known as the Hausdorff metric. For example, let A be the closed disk consisting of all the points in R2 within 1 of the origin, and let B be the closed disk consisting of all the points in R2 within 2 of the origin. Clearly, A B. As a result h .B; A/ D 0. But h .A; B/ D d .2; 0/; .1; 0/ D 1, so h.A; B/ D max h .A; B/; h .B; A/ D max.1; 0/ D 1. Intuitively, there are points in B that are a distance 1 from the set A, and this is the largest distance from points in B to the set A. As a second example, let A be the square with vertices at .1; 1/; .1; 1/; .1; 1/; and .1; 1/, and let B be the line segment from .1; 2/ to .1; 2/. Since each point in B is a distance 1 from A, h .A; B/ D 1. Since the largest distance from a point in A to the set B is d .0; 1/; .0; 2/ D 3, h .B; A/ D 3. Thus, h.A; B/ D 3. What does it take to show that a metric space? Because H is clearly a nonempty set, and h.A; B/ is defined for all A and B in H, one merely has to verify the conditions for h to be a metric. The fact that h is always a nonnegative real number follows immediately from its definition, as does the fact that h.A; B/ D h.B; A/. Is it true that for all A 2 H that h.A; A/ D 0? Yes, this follows from the fact that if y 2 A, then min d.x; y/ D 0, so h.A; A/ D h .A; A/ D max min d.x; y/ D 0. x2A
y2A x2A
If A ¤ B, then there is either a point x 2 AnB or an x 2 BnA. If a set is compact, then any point a distance 0 from the set must be in the boundary of the set and is, therefore, an element of the set. From this it follows that either min d.x; y/ > 0 or y2A
min d.x; y/ > 0, so either h .A; B/ > 0 or h .B; A/ > 0, and h.A; B/ > 0. y2B
Showing that h satisfies the triangle inequality is a little trickier and relies on a careful look at the definitions of h and h. Let A, B, and C be elements of H. To show that h.A; C/ h.A; B/ C h.B; C/, begin by showing that for every a 2 A, min d.a; c/ h.A; B/ C h.B; C/. So if a 2 A, then it is true for all b 2 B c2C that min d.a; c/ min d.a; b/ C d.b; c/ D d.a; b/ C min d.b; c/. In particular, c2C
c2C
c2C
this is true for b 2 B which minimizes d.a; b/, so min d.a; c/ d.a; b / C c2C
min d.b ; c/ min d.a; b/ C max min d.b; c/ h .B; A/ C max min d.b; c/ c2C
b2B
b2B c2C
b2B c2C
h.A; B/ C h .C; B/ h.A; B/ C h.B; C/. Since all the distances min d.a; c/ are c2C
bounded by h.A; B/Ch.B; C/, its maximum also has the same bound, so h .A; C/ h.A; B/ C h.B; C/. The same argument shows that h .C; A/ h.A; B/ C h.B; C/, so h.A; C/ h.A; B/ C h.B; C/ which is the desired triangle inequality.
10.11 Contraction Mappings
335
PROOF: Let H be the set of all nonempty compact subsets of R2 and let d.x; y/ be the Euclidean distance function in R2 . For each A and B in H let h .A; B/ D max min d.a; b/, and define b2B a2A h.A; B/ D max h .A; B/; h .B; A/ . Then is a metric space. • Let H be the space of all nonempty compact subsets of R2 . Clearly, H is not an empty set. • Let d be the Euclidean distance function in R2 . • Let A and B be elements of H. • For fixed b 2 B the distance function d.a; b/ is a continuous function of a, so it obtains a minimum value on the compact set A. • Then since min d.a; b/ is a continuous function of b, it obtains a maximum a2A
value on the compact set B. • Therefore, for each A and B in H the function h .A; B/ D max min d.a; b/ b2B a2A
is defined and is a nonnegative real number. • This shows that h.A; B/ D max h .A; B/; h .B; A/ is also defined, is a nonnegative real number, and satisfies h.A; B/ D h.B; A/. • For any A 2 H, let b 2 A. Then min d.a; b/ D d.b; b/ D 0, so h.A; A/ D a2A
h .A; A/ D max 0 D 0. a2A
• If A and B are in H with A ¤ B, then it cannot be the case that both AnB and BnA are empty, so there is an x in at least one of the two sets. If x 2 AnB, then d.x; b/ > 0 for all b 2 B, and since d.x; b/ is a continuous function of b and B is compact, min d.x; b/ > 0 and h .B; A/ > 0. b2B
Similarly, if x 2 BnA, then h .A; B/ > 0. In either case, h.A; B/ > 0. This shows that h separates points of H. • It remains to show that h satisfies the triangle inequality, so let A, B, and C be elements of H. • Let a 2 A. • It is true for all b 2 B that min d.a; c/ min d.a; b/ C d.b; c/ D c2C
d.a; b/ C min d.b; c/.
c2C
c2C
• In particular, if b 2 B minimizes d.a; b/, it follows that min d.a; c/ d.a; b / C min d.b ; c/ min d.a; b/ C max min d.b; c/ c2C
b2B
c2C
b2B c2C
h .B; A/ C h .C; B/ h.A; B/ C h.B; C/. • Since for each a 2 A, min d.a; c/ is bounded by h.A; B/ C h.B; C/, c2C
it follows that max min d.a; c/ also has the same bound, so a2A c2C
h .A; C/ h.A; B/ C h.B; C/. • The same argument shows that h .C; A/ h.A; B/ C h.B; C/, so h.A; C/ h.A; B/ C h.B; C/ which is the desired triangle inequality. • Thus, h is a metric for the space H.
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10 Metric Spaces
Moreover, is a complete metric space. A natural way to prove this completeness is to take a Cauchy sequence of sets in H and show that this sequence converges in the h metric to some set L 2 H. So let be a Cauchy sequence in H. The strategy for this proof is to construct another sequence of sets, , so that each Tm is close to one of the terms of the sequence. Then it is shown that the sequence converges to a set L, and since the terms of the sequence are close to the terms of the sequence, the sequence will also converge to L. Because is a Cauchy sequence, for each natural number n, there is an Nn such that for every k and m greater than or equal to Nn , h.Ak ; Am / < 21n . Then define Tn D fx 2 R2 j min d.a; x/ 21n g. Tn can be thought of as a cloud or halo around a2An
the set ANn . In particular, Tn contains ANn and all of the points within 21n of An . Since h.Am ; ANn / < 21n for all m Nn , it follows that Am Tn for all m Nn . It is also clear that since Tn contains ANnC1 , that Tn contains TnC1 , and is a sequence of nested sets. It is also easy to show that Tn is nonempty, bounded, and closed so that Tn 2 H. 1
Now you can define L D \ Tn . This set is not empty because if you intersect a nD1
nested sequence of nonempty compact sets, you get a nonempty set. L is bounded because it is contained in the bounded set T1 , and it is closed because it is the intersection of closed sets. Thus, L 2 H. To show that lim Tn D L, you would n!1
want to estimate h.Tn ; L/. Since L Tn , it follows that h .Tn ; L/ D 0. To estimate h .L; Tn /, for each xn 2 Tn , the proof will construct a sequence of points 1 . This is xn ; xnC1 ; xnC2 ; : : : where, for each m n, xm 2 Tm and d.xm ; xmC1 / 2m2 enough to show that the sequence xn ; xnC1 ; xnC2 ; : : : is Cauchy, so it will converge to a point x which must end up being a member the set L. From this, it will follow 1 1 1 that d.xn ; x/ < 2n3 , so h .L; Tn / D max min d.a; b/ < 2n3 . Thus, h.Tn ; L/ < 2n3 . b2Tn a2L
But this is enough to show that L D lim Tn . It is then a simple argument to show n!1 that lim An D L which proves that all Cauchy sequences in H converge, so H is n!1 complete. PROOF: The metric space is complete. • Let R2 be the metric space with Euclidean distance function d.x; y/. 2 • Let H be the space of all nonempty compactsubsets of R with the metric h.A; B/ D max h .A; b/; h .B; A/ where h .A; B/ D max min d.a; b/. b2B a2A
• Let A1 ; A2 ; A3 ; : : : be a Cauchy sequence in H. • Because the sequence is Cauchy, for every natural number n there is an Nn such that for every k and m greater than or equal to Nn , h.Ak ; Am / < 21n . • Define Tn D fx 2 R2 j min d.a; x/ 21n g. a2An
• Tn is nonempty because it contains the nonempty set An . (continued)
10.11 Contraction Mappings
337
• Tn is bounded because no point in Tn is more than 1 away from the bounded set An . • If x is in the boundary of Tn , then forevery > 0 thereis a point y 2 Tn with d.x; y/ < , so min d.a; x/ min d.a; y/ C d.y; x/ < 21n C . From this a2ANn
it follows that min d.a; x/ a2ANn
a2ANn 1 , so 2n
x 2 Tn . Thus, Tn contains its boundary,
so Tn is closed. • Since Tn is a nonempty, bounded, closed set in R2 , it is a nonempty compact set and Tn 2 H. • Note that for all natural numbers n, Tn contains Ak for all k Nn , and Tn contains Tk for all k n. 1
• Let L D \ Tn . nD1
• The intersection of any finite collection of Tn sets contains the smallest set in the collection, so the intersection of any finite collection of Tn sets is nonempty. Thus, the intersection of all the Tn sets cannot be empty, and L is nonempty. • L is the intersection of closed sets, so it is closed. L is contained in the bounded set T1 , so L is bounded. Thus, L 2 H. • Let natural number n be given. • L Tn , so h .Tn ; L/ D 0. • To estimate h .L; Tn /, note that for every natural number m the triangle inequality shows that h.Tm ; TmC1 / h.Tm ; ANm / C h.ANm ; ANmC1 / C 1 1 < 2m2 . h.ANmC1 ; TmC1 / 21m C 21m C 2mC1 1 • Thus, for any x 2 Tm there is a y 2 TmC1 such that d.x; y/ < 2m2 . 1 • Let xn 2 Tn . Then there is an xnC1 2 TnC1 with d.xn ; xnC1 / < 2n2 . By mathematical induction there is a sequence xn ; xnC1 ; xnC2 ; : : : such that for 1 each m n, xm 2 Tm and d.xm ; xmC1 / < 2m2 . • If k and m are natural numbers with m > k, then d.xk ; xm / d.xk ; xkC1 / C d.xkC1 ; xkC2 / C d.xkC2 ; xkC3 / C C d.xm1 ; xm / < 1 1 1 1 C 2k1 C 21k C C 2m3 < 2k3 . 2k2 1 • Because lim 2k3 D 0, the sequence is a Cauchy sequence in R2 , so k!1
it converges to a point x in R2 . • For each k n and each m k, xm 2 Tk , and since Tk is a closed set, it 1
follows that x 2 Tk . This shows that x 2 \ Tn D L. nD1
• Since lim xm D x, for every > 0 there is an m > n such that d.xm ; x/ < . m!1
This shows that d.xn ; x/ d.xn ; xnC1 / C d.xnC1 ; xnC2 / C d.xnC2 ; xnC3 / C 1 1 C d.xm1 ; xm / C d.xm ; x/ < 2n3 C . This shows that d.xn ; x/ 2n3 , 1 1 so h .L; Tn / D max min d.a; b/ 2n3 . Thus, h.Tn ; L/ 2n3 . b2Tn a2L
(continued)
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10 Metric Spaces
• Since lim h.Tn ; L/ D 0, it follows that L D lim Tn . n!1
n!1
1 • Given > 0, let k be a natural number such that 2k5 < . • Then for every m Nk , h.Am ; L/ h.Am ; ANk / C h.ANk ; Tk / C h.Tk ; L/ < 1 1 1 C 21k C 2k3 < 2k5 < . 2k • Therefore, lim An D L which proves that all Cauchy sequences in H n!1 converge, so H is complete.
Knowing that is a complete metric space shows that every contraction mapping on H has a unique fixed point. Some fractals can be generated as the result of being a fixed point of such a contraction mapping f W H ! H. In the words of the study of fractals, these fixed points are called attractors. What it means is that if you begin with any nonempty compact set A, the orbit of A under the contraction mapping, A; f .A/; f f .A/ ; f f f .A/ ; : : : , will converge to this attractor. Where can you find a contraction mapping f W H ! H with an interesting fractal as its attractor? The first thing to notice is that if f W R2 ! R2 is a contraction mapping on R2 , then the usual extension of f to subsets of R2 given by f .A/ D fy 2 R2 j y D f .x/ for some y 2 R2 g gives a function f W H ! H that is a contraction mapping on H. To see this, let d be the usual Euclidean distance metric on R2 , let k be a positive real number less than 1, and let f W R2 ! R2 be a contraction mapping satisfying d f .x/; f .y/ k d.x;y/ for all x and y in R2 . Then, if A and B are in H, it follows that h f .A/; f .B/ D max min d.a; b/ D b2f .B/ a2f .A/ max min d f .a/; f .b/ max min k d.a; b/ D k h .A; B/, and it follows that b2B b2B a2A a2A h f .A/; f .B/ k h.A; B/. Next, suppose that f1 ; f2 ; f3 ; : : : ; fs is a finite set of contraction mappings on R2 (therefore on H) with associated contraction constants k1 ; k2 ; k3 ; : : : ; ks , respecs
tively. Then the mapping F W H ! H given by F.A/ D [ fj .A/ is also a jD1
contraction mapping on H with contraction constant k D max.k1 ; k2 ; k3 ; : : : ; ks /. This follows from the easily established fact that if A, B, C, and D are in H, then h.A [ B; C [ D/ h.A; C/ C h.B; D/. So what functions should one choose for f1 ; f2 ; f3 ; : : : ; fs ? Any contraction mappings of the plane will give rise to some attractor. Linear transformations are easy to construct and easy to understand, but they are limited because they map points near the origin to points near the origin, so the origin is always the fixed point of such transformations. Almost as easy are the affine transformations which are functions that perform a linear transformation followed by a simple translation. In other words, for every affine function on R2 there are six constants, a1 ; a2 ; b1 ; b2 ; c1 ; and c2 , such that the affine function maps a point .x; y/ 2 R2 to .a1 x C b1 y C c1 ; a2 x C b2 y C c2 /. The Sierpinski triangle can be generated using such a collection of transformations where the constants are given by
10.11 Contraction Mappings Function f1 f2 f3
339 a1 0:5 0:5 0:5
a2 0 0 0
b1 0 0 0
b2 0:5 0:5 0:5
c1 0 0:5 0:25
c2 0 0 p 0:25 3
A second famous example is that of the Barnsley fern. It can be generated by a set of four affine transformations given by Function f1 f2 f3 f4
a1
a2
b1
0 0:85 0:2 0:15
0 0:04 0:26 0:28
0 0:04 0:23 0:26
b2 0:16 0:85 0:22 0:24
c1 0 0 0 0
c2 0 1:6 1:6 0:44
Figure 10.15 shows what happens when this transformation is applied to a simple equilateral triangle in the plane as shown by the first picture in the figure. Applying the function once results in the second picture showing that the triangle is mapped into four separate skewed copies by the four pieces of the transformation. The third picture shows what happens to this after a second iteration of the transformation. By the time the transformation has been iterated three times, the image set already appears to be a recognizable object quite different from the original triangle. The figure also shows what the set looks like after 5, 10, 15, and 20 iterations. After 20 iterations, the image appears to be quite close to the attractor for this transformation and does look a great deal like a fern.
Fig. 10.15 Generation of a fractal fern by applying a contraction mapping to a triangle. Shown is the original triangle and its image after 1, 2, 3, 5, 10, 15, and 20 iterations of the transformation
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10 Metric Spaces
There are many fractal generating programs available either as web applications or as stand-alone programs. The interested reader may well find it useful to investigate the convergence properties of the transformations discussed here and other similar transformations obtainable by either small or perhaps large changes in the given affine maps.
10.11.4 Exercises Write proofs for each of the following statements. 1. If f W X ! X and g W X ! X are both contraction mappings on X, then the composition g ı f is also a contraction mapping on X. 2. The function f .x/ D xC2 is a contraction mapping on R with fixed point L D 2. 2 3. Applying Newton’s Method to solve the equation ex D x C 2 yields the function x f .x/ D x e ex2 x 1 . This function has a unique fixed point on the interval Œ0; 2. 4. Let f be a differentiable function from Œa; b into Œa; b such that for some M < 1 the derivative of f satisfies jf 0 .x/j M for each x 2 Œa; b. Then f has a unique fixed point. 5. Show that if the function f W R ! R satisfies the Lipschitz condition jf .x/ f .y/j 5jx yj for x and y in R, then f is uniformly continuous on R. 6. Show that if function f W R ! R is differentiable, and its derivative is a bounded function, then f satisfies a Lipschitz condition on R. p xCy 0 7. The differential equation y D xC2 with initial condition y.1/ D 3 has a unique solution on the interval Œ0; 2. 8. If A; B; C, and D are in H, then h.A [ B; C [ D/ max h.A; C/; h.B; D/ . 9. If f1 and f2 are contraction mappings on H with contraction constants k1 and k2 , respectively, then F.A/ D f1 .A/ [ f2 .A/ is a contraction mapping on H with contraction constant k D max.k1 ; k2 /.
Books for Further Reading
Classical Analysis and Advanced Calculus Abbott, S.: Understanding Analysis. Springer, New York (2015) Apostol, T.M.: Mathematical Analysis, 2nd edn. Pearson, New York (1974) Bartle, R.G., Sherbert, D.R.: Introduction to Real Analysis, 4th edn. Wiley, Hoboken (2011) Berberian, S.K.: Measure and Integration. American Mathematical Society, Providence (2013) Boas, R.P.: A Primer of Real Functions. Mathematical Association of America, Washington, DC (1961) Browder, A.: Mathematical Analysis: An Introduction. Springer, New York (2001) Gelbaum, B.R.: Counterexamples in Analysis. Dover, Mineola (2003) Goldberg, R.R.: Methods of Real Analysis, 2nd edn. Wiley, New York (1976) Kaplan, W.: Advanced Calculus. Pearson, New York (2002) Knapp, A.W.: Basic Real Analysis. Birkhäuser, Boston (2005) Krantz, S.G.: Real Analysis And Foundations, 3rd edn. Chapman and Hall/CRC, Boca Raton (2013) Marsden, J.E., Hoffman, M.J.: Elementary Classical Analysis, 2nd edn. W.H. Freeman, New York (1993) Rosenlicht, M.: Introduction to Analysis. Dover, Mineola (1986) Ross, K.A.: Elementary Analysis, 2nd edn. Springer, New York (2013) Rudin, W.: Principles of Mathematical Analysis. McGraw Hill, New York (1976) Spivak, M.: Calculus, 3rd edn. Cambridge University Press, Cambridge (2006)
Measure and Integration Bartle, R.G.: The Elements of Integration and Lebesgue Measure. Wiley, Hoboken (1995) Johnston, W.: The Lebesgue Integral for Undergraduates. Mathematical Association of America, Washington, DC (2015) © Springer International Publishing Switzerland 2016 J.M. Kane, Writing Proofs in Analysis, DOI 10.1007/978-3-319-30967-5
341
342
Books for Further Reading
Rosentrater, C.R.: Varieties of Integration. Mathematical Association of America, Washington, DC (2015) Royden, H.I., Fitzpatrick, P.M.: Real Analysis, 4th edn. Pearson, New York (2010)
Infinite Series Bonar, D.D., Khoury, M.J.: Real Infinite Series. Mathematical Association of America, Washington, DC (2006) Knopp, K.: Theory and Application of Infinite Series. Dover, Mineola (1990)
Writing Proofs Lay, S.R.: Analysis with an Introduction to Proof, 5th edn. Pearson, New York (2013) Solow, D.: How to Read and do Proofs: An Introduction to Mathematical Thought Processes, 6th edn. Wiley, New York (2014)
Differential Equations Birkhoff, G., Rota, G.-C.: Ordinary Differential Equations, 4th edn. Wiley, New York (1989) Kelley, W.G., Peterson, A.C.: The Theory of Differential Equations. Springer, New York (2010)
Fractals Barnsley, M.F.: Fractals Everywhere. Dover, Mineola (2012) Devaney, R.L.: Chaos Fractals, and Dynamics: Computer Experiments in Modern Mathematics. Addison-Wesley, Boston (1991)
Index
A Abel’s theorem, 256 absolute extremum, 145 absolute maximum, 145 absolute minimum, 145 absolute value, 38, 121 absolutely convergent series, 205 accumulation point, 74, 286, 306, 308, 318, 319 addition property of less than, 32 affine transformation, 338 Algebra, 4 alternating series, 219 alternating series test, 219 Analysis, 4 analytic function, 259 antecedent, 9 antiderivative, 193 Archimedian principle, 36 area, 159 area axioms, 169 area zero, 165 assert the hypothesis, 14 associative law of intersection, 19 associative law of union, 19 associative laws, 32 attractors, 338 axiom, 2, 7, 35–37, 63, 75, 128, 168, 169, 201, 323 axioms for a field, 32 axioms for an ordered field, 32 axioms for area, 166 axioms for the real numbers, 36, 201
B Barnsley fern, 339 biconditional, 11, 96, 184, 195, 313 bijection, 42, 142, 159, 161, 283, 315, 321, 322 bijective, 42 Bolzano–Weierstrass theorem, 75, 76, 78, 124, 125, 322 bound, 35 boundary, 270, 286, 306 bounded, 35 bounded above, 35, 62 bounded below, 35, 62 bounded derivative, 108 bounded function, 62, 123 bounded metric space, 318 bounded sequence, 63
C C[0,1], 302 Cantor set, 163, 166, 333 cardinality, 159 Cauchy product, 231, 265 Cauchy sequence, 66, 67, 77, 78, 204–206, 210, 223, 323, 324, 328, 336 Cauchy sequences converge, 77 Cauchy–Schwarz inequality, 297 chain rule, 140 closed function, 285 closed set, 274, 306 closure of a set, 287, 306 closure properties, 32 codomain, 40, 42
© Springer International Publishing Switzerland 2016 J.M. Kane, Writing Proofs in Analysis, DOI 10.1007/978-3-319-30967-5
343
344 commutative law of intersection, 19 commutative law of union, 19 commutative laws, 32 compact metric space, 318 compact set, 288 compactness, 288, 318 comparison test, 209, 212, 214, 216 complement of a set, 18 complete metric space, 323 completeness axiom, 35–37, 63, 75, 128, 323 composition of continuous functions, 312 composition of functions, 43 conclusion, 9 conditional, 9 conditionally convergent series, 206 conjecture, 2 connected metric space, 316 connected set, 39, 291 connectedness, 291, 316 consequent, 9 continuity of a composition, 121 continuity of a difference, 118 continuity of a product, 119 continuity of a quotient, 119 continuity of a sum, 118 continuous at a point, 100, 134 continuous function on a metric space, 311 continuous function on a topological space, 285 continuous functions are integrable, 189 continuous functions on a set, 302 continuous on a set, 100 contraction mapping, 327, 329–331, 333, 338 contradiction, 10, 36, 70, 90, 112, 123, 126, 161, 177, 178, 196, 222, 286, 317, 318, 324 contrapositive, 10 converge pointwise, 239 converge uniformly, 241 convergence in L1 , 252 convergence in mean, 252 convergent sequence, 62 convergent series, 202 converse, 9 corollary, 2 countable, 160, 163, 165, 195, 197, 200 counterexample, 11 cowgirl, 242 critical point, 145
D decreasing on an interval, 143 decreasing sequence, 60
Index definition, 1 deleted neighborhood in a metric space, 296 DeMorgan’s Laws, 19, 25 denumerable, 160 derivative, 133 derivative of a difference, 137 derivative of a product, 138 derivative of a quotient, 138 derivative of a sum, 137 derivative of power series, 259 derived set, 286, 306 diagonalization argument, 160 diameter of a metric space, 318 difference of power series, 265 difference of sets, 18 difference quotient, 133–143, 148–152, 192 differentiable, 133 differential equation, 329–332 disconnected metric space, 316 disconnected set, 291 discontinuities, 197 discontinuous, 100, 131 discrete metric, 302 discrete metric space, 302 discrete topology, 282 distance, 38 distance function, 296 distributive law of intersection over union, 19 distributive law of multiplication over addition, 32 distributive law of union over intersection, 19 divergent sequence, 62 divergent series, 202 division algorithm, 28 domain, 40 dot product, 297
E element, 17 empty set, 18 Euclidean distance, 299 Euclidean metric, 300 Euclidean plane, 299 Euclidean space, 299 even integers, 27 existence of square roots, 36 existential quantifier, 69 extended mean value theorem, 152 exterior, 270, 306 extreme value, 126
Index F fern, 339 field, 32 field axioms, 32 finite complement topology, 281 finite set, 160 finite subcover, 110–114, 196, 249, 251, 288, 290, 291, 317, 324, 326 first order differential equation, 329 fixed point, 327 fractal, 333–340 function, 40, 41 function applied to sets, 283 fundamental question of Analysis, 267 fundamental theorem of Calculus, 192, 193
G geometric series, 202, 203, 207, 212, 214, 231, 253, 255, 327, 329 greatest lower bound, 35, 94, 110, 185, 186
H harmonic series, 212, 223 Hausdorff metric, 334 Heine–Borel theorem, 110, 115, 196, 247, 288, 319, 324 higher order Rolle’s theorem, 263 homeomorphic metric spaces, 315 homeomorphism, 315, 321 hypothesis, 9
I identity properties, 32 image of a set, 283 improper integral of the first kind, 215 improper integral of the second kind, 215 inconsistent axioms, 3 increasing on an interval, 143 increasing sequence, 60 independence of the axioms, 3 index set, 279 infimum, 36 infinite limit, 79 infinite series, 201–237, 252, 255 inherited topology, 282 initial value problem, 330 injection, 42, 44, 142, 283, 316, 317, 321 injective, 42 integers, 27 integrable function, 171–200, 239, 244, 248, 252
345 integral, 169, 171 integral of continuous function, 189 integral test, 216 interior, 269, 306 intermediate value property, 127–130, 155–157 intermediate value property for derivatives, 155–157 intermediate value theorem, 128 intersection of closed sets, 280 intersection of open sets, 278 intersection of sets, 18 interval, 39 interval of convergence, 256 inverse function, 136, 142, 143, 283, 315, 317, 321 inverse properties, 32
L L1 metric, 303, 305, 310 L’Hopital’s rule, 150 Lagrange’s form of Taylor’s theorem, 263 law, 2 least upper bound, 35–37, 63, 75, 94, 111, 123, 126, 128, 129, 186, 323 least upper bound principle, 36 Lebesgue measure, 248 Lebesgue’s theorem, 197 lemma, 2 less than, 32 limit at a point, 49 limit at infinity, 57 limit at negative infinity, 58 limit comparison test, 218 limit from the left, 54 limit from the right, 54 limit inferior, 94 limit of a difference, 81, 82 limit of a polynomial, 88 limit of a positive function, 90 limit of a product, 81, 84 limit of a quotient, 81, 86 limit of a rational function, 88 limit of a sequence, 62 limit of a sum, 81, 82 limit of terms test, 203 limit superior, 94 limit, infinite, 79 limits are unique, 91 limits in a metric space, 308 Lipschitz condition, 330 list implications, 15 local property, 105
346 lower bound, 35 lower step function, 184 lowest terms, 31
M M-test, 253 Mandelbrot set, 333 mathematical induction, 35, 64, 87, 92, 167, 212, 262, 291, 299 maximum, 122 maximum value, 123, 126 mean value theorem, 108, 146, 147, 152 mean value theorem for integration, 191 measure zero, 163, 195, 200, 244 method of exhaustion, 167 metric, 296 metric space, 296–340 minimum, 122 minimum value, 123, 126 Minkowski inequality, 298, 299 monotone convergence theorem, 248 monotone decreasing sequence, 60 monotone increasing sequence, 60 monotone sequence, 60 monotonically decreasing sequence of functions, 246 monotonically increasing sequence of functions, 246 multiplication property of less than, 32 multiset, 18
N n-dimensional Euclidean space, 299 natural numbers, 27 negation of a statement, 10 negation of statements with quantifiers, 69 negation symbol, 10 neighborhood in a metric space, 296 norm of a partition, 170
O odd integers, 27 one-sided limits, 54 one-to-one, 42 onto, 41 open cover, 110, 288, 317 open function, 283 open set, 274, 306 orbit, 327 ordered field, 32
Index P p-series, 217 paradox, 3 parentheses in series, 224 partial sums, 202 partition, 28 partition of an interval, 170 path-connected set, 292 Picard existence theorem, 330 pointwise convergence, 239 positive integers, 27 postulate, 2 power series, 201, 215, 253, 255–267, 332 power series antiderivative, 262 power series convergence, 255 power series difference, 265 power series differentiability, 259 power series interval of convergence, 256 power series product, 265 power series quotient, 265 power series sum, 265 preimage of a set, 283 principle, 2 product of power series, 265 product rule, 138 proof, 2 proof by contradiction, 10, 36, 70, 90, 112, 123, 126, 161, 177, 178, 196, 222, 286, 317, 318, 324 proof of conditional statements, 11 proof template, 5, 12, 20, 23, 41, 42, 49, 70, 100, 106, 296 proofs about even and odd, 27 proofs about set equality, 22 proofs about subsets, 19 proposition, 2 proving a derivative, 135 proving a limit, 49 proving a metric space, 296 proving continuity, 100 proving no limit exists, 70 proving uniform continuity, 106
Q quantifier, 69 quotient, 28 quotient of power series, 265 quotient rule, 138
R radius of convergence, 255 range, 40
Index ratio test, 212–215, 221, 231, 258 rational numbers, 31 rearrangement of terms of a series, 225 relative extremum, 144 relative maximum, 144 relative minimum, 144 remainder, 28 Riemann integral, 169, 171 Riemann sum, 171, 173, 175, 177, 178, 182, 184, 186, 190, 194 Rolle’s theorem, 146, 263 root test, 214, 215, 258 Russell’s paradox, 3 S saddle point, 145 secant line, 133 sequence, 60 sequence of continuous functions, 243 sequence of functions, 239 sequence of integrable functions, 244 sequence of partial sums, 202 sequence of terms, 201 series, 201–237, 252, 255 series diverges, 202 series of functions, 252 series rearrangement, 225 series with parentheses, 224 set, 17 set builder notation, 18 set containment, 17 set difference, 18 set equality, 22 set notation, 17 set of discontinuities, 197 set the context, 13 Set Theory, 4 shorthand symbols for proofs, 11 Sierpinski triangle, 333 slope, 133 squeezing theorem, 91 state the conclusion, 14 statement, 1 step function, 183 strictly decreasing on an interval, 143 strictly decreasing sequence, 60 strictly increasing on an interval, 143 strictly increasing sequence, 60 subcover, 110–114, 196, 249, 251, 288, 290, 291, 317, 324, 326 subsequence, 61, 92 subset, 17 sum of power series, 265
347 sup metric, 302 supremum, 36 supremum metric, 301, 302 surjection, 41–43, 142, 283 surjective, 41 symbols for proofs, 11
T tangent line, 133 taxicab metric, 300 Taylor polynomial, 263 Taylor series, 263 Taylor’s theorem, 263 telescoping series, 203 template, 5, 12, 20, 23, 41, 42, 49, 70, 100, 106, 296 term of a sequence, 60 theorem, 2 Thomae’s function, 131, 200 topological space, 281 Topology, 4 topology, 269, 281 transitive property, 32 triangle inequality, 38, 39, 66, 71, 82, 107, 115, 118, 295–297, 299, 300, 302–304, 306, 328, 334, 337 trichotomy property, 32, 37, 38
U uncountable, 160, 163, 279 uniform convergence, 241 uniformly continuous, 106, 115, 311, 327 union of closes sets, 280 union of open sets, 278, 307 union of sets, 18 universal quantifier, 69 universal set, 18 upper bound, 35 upper step function, 184
W Weierstrass M-test, 253 without loss of generality, 80, 128, 151, 153, 156, 174, 186, 190, 194, 221, 248, 250, 257, 273 WLOG, 80, 128 working definition, 28
Z Zermelo–Fraenkel axioms, 3
