Conway, John B - Functions of One Complex Variables I (1).pdf

Graduate Texts in Mathematics

11

Editorial Board S. Axler F.W. Gehring K.A. Ribet

Graduate Texts in Mathematics 2 3 4 5 6 7 8 9 10 11

12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

TAKEUTIIZARING. Introduction to Axiomatic Set Theory. 2nd ed. OXTOBY. Measure and Category. 2nd cd. SCHAEFER. Topological Vector Spaces. 2nd ed. HlLTON/STAMMBACH. A Course in Homological Algebra. 2nd ed. MAC LANE. Categories for the Working Mathematician. 2nd ed. HUGHESIPIPER. Projective Planes. J.-P. SERRE. A Course in Arithmetic. TAKEUTIIZARING. Axiomatic Set Theory. HUMPHREYS. Intro.duction to Lie Algebras and Representation Theory. COHEN. A Course in Simple Homotopy Theory. CONWAY. Functions of One Complex Variable I. 2nd ed. BEALS. Advanced Mathematical Analysis. ANDERSON/FuLLER. Rings and Categories of Modules. 2nd ed. GOLUBITSKy/GUILLEMIN. Stable Mappings and Their Singularities. BERBERIAN. Lectures in Functional Analysis and Operator Theory. WINTER. The Structure of Fields. ROSENBLATT. Random Processes. 2nd ed. HALMos. Measure Theory. HALMOS. A Hilbert Space Problem Book. 2nd ed. HUSEMOLLER. Fibre Bundles. 3rd ed. HUMPHREYS. Linear Algebraic Groups. BARNES/MACK. An Algebraic Introduction to Mathematical Logic. GREUB. Linear Algebra. 4th ed. HOLMES. Geometric Functional Analysis and Its Applications. HEWITT/STROMBERG. Real and Abstract Analysis. MANES. Algebraic Theories. KELLEY. General Topology. ZARlSKIlSAMUEL. Commutative Algebra. Vol. I. ZARISKIISAMUEL. Commutative Algebra. Vol.lI. JACOBSON. Lectures in Abstract Algebra I. Basic Concepts. JACOBSON. Lectures in Abstract Algebra II. Linear Algebra. JACOBSON. Lectures in Abstract Algebra III. Theory of Fields and Galois Theory. HIRSCH. Differential Topology.

34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 5! 52 53 54 55 56 57 58 59 60 61 62 63

SPITZER. Principles of Random Walk. 2nd ed. ALEXANDERIWERMER. Several Complex Variables and Banach Algebras. 3rd ed. KELLEy/NAMIOKA et al. Linear Topological Spaces. MONK. Mathematical Logic. GRAUERTIFRITZSCHE. Several Complex Variables. ARVESON. An Invitation to C*-Algebras. KEMENY/SNEWKNAPP. Denumerable Markov Chains. 2nd ed. ApOSTOL. Modular Functions and Dirichlet Series in Number Theory. 2nd ed. J.-P. SERRE. Linear Representations of Finite Groups. GILLMAN/JERISON. Rings of Continuous Functions. KENDIG. Elementary Algebraic Geometry. LoEVE. Probability Theory I. 4th ed. LoEVE. Probability Theory II. 4th ed. MorSE. Geometric Topology in Dimensions 2 and 3. SACHSlWu. General Relativity for Mathematicians. GRUENBERGIWEIR. Linear Geometry. 2nd ed. EDWARDS. Fermat's Last Theorem. "KLINGENBERG. A Course in Differential Gcometry. HARTSHORNE. Algebraic Geometry. MANIN. A Course in Mathematical Logic. GRAVERlWATKlNS. Combinatorics with Emphasis on the Theory of Graphs. BROWNIPEARCY. Introduction to Operator Theory I: Elements of Functional Analysis. MASSEY. Algebraic Topology: An Introduction. CROWELLIFox. Introduction to Knot Theory. KOBLITZ. p-adic Numbers, p-adic Analysis, and Zeta-Functions. 2nd ed. LANG. Cyclotomic Fields. ARNOLD. Mathematical Methods in Classical Mechanics. 2nd ed. WHITEHEAD. Elements of Homotopy Theory. KARGAPOLOV/MERLZJAKOV. Fundamentals of the Theory of Groups. BOLLOBAS. Graph Theory. (continued after index)

John B. Conway

Functions of One Cotnplex Variable I Second Edition With 30 Illustrations

~ Springer

John B. Conway Mathematics Department University of Tennessee Knoxville, Tennessee 37996-1301 USA

Editorial Board S. Axler Mathematics Department San Francisco State University Sall Francisco, CA 94132 USA [email protected]

F.W. Gehring Mathematics Department East Hall University of Michigan Ann Arbor, MI 48109 USA [email protected]

K.A. Ribet Mathematics Department University of California, Berkeley Berkeley, CA 94720-3840 USA [email protected]

Mathematics Subject Classification (2000): 30-01 Library of Congress Cataloging-in-Publication Data Conway, John B. Functions of one complex variable I I John B. Conway.-2nd ed. p. cm. - (Graduate texts in mathematics; 11) Rev. corr. ed. of: Functions of one complex variable. 2nd ed. c1978. "Seventh corrected printing"-T.;. verso. Includes bibliographical references (p. - ) and index.

ISBN- 13: 978-0-387-94234-6 DOl: 10.1007/978-1-4612-6313-5

e_ISBN-13: 978-1-4612-6313-5

1. Functions of complex variables. I. Conway. John B. of one complex variable. II. Title. III. Series. QA331.7C68 1978 515'.93---{jc20 Printed on acid-free paper.

Functions 95-5931

© 1973, 1978 Springer Science+Business Media. Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer Science+Business Media, Inc., 233 Spring Street, New York. NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use in this publication of trade names, trademarks, service marks. and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights.

Solkover reprint of the hardcover 15t Edition 1978 15 14 13 springeronline.com

To Ann

PREFACE FOR THE SECOND EDITION I have been very pleased with the success of my book. When it was apparent that the second printing was nearly sold out, Springer-Verlag asked me to prepare a list of corrections for a third printing. When I mentioned that r had some ideas for more substantial revisions, they reacted with characteristic enthusiasm. There are four major differences between the present edition and its predecessor. First, John Dixon's treatment of Cauchy's Theorem has been included. This has the advantage of providing a quick proof of the theorem in its full generality. Nevertheless, I have a strong attachment to the homotopic version that appeared in the first edition and have proved this form of Cauchy's Theorem as it was done there. This version is very geometric and quite easy to apply. Moreover, the notion of homotopy is needed for the later treatment of the monodromy theorem; hence, inclusion of this version yields benefits far in excess of the time needed to discuss it. Second. the proof of Runge's Theorem is new. The present proof is due to Sandy Grabiner and does not use "pole pushing". In a sense the "pole pushing" is buried in the concept of uniform approximation and some ideas from Banach algebras. Nevertheless, it should be emphasized that the proof is entirely elementary in that it relies only on the material presented in this text. Next, an Appendix B has been added. This appendix contains some bibliographical material and a guide for further reading. Finally, several additional exercises have been added. There are also minor changes that have been made. Several colleagues in the mathematical community have helped me greatly by providing constructive criticism and pointing out typographical errors. I wish to thank publicly Earl Berkson, Louis Brickman, James Deddens, Gerard Keough, G. K. Kristiansen, Andrew Lenard, John Mairhuber, Donald C. Meyers, Jeffrey Nunemacher, Robert Olin, Donald Perlis, John Plaster, Hans Sagan, Glenn Schober, David Stegenga, Richard Varga, James P. Williams, and Max Zorn. Finally, I wish to thank the staff at Springer-Verlag New York not only for their treatment of my book, but also for the publication of so many fine books on mathematics. In the present time of shrinking graduate enrollments and the consequent reluctance of so many publishers to print advanced texts and monographs, Springer-Verlag is making a contribution to our discipline by increasing its efforts to disseminate the recent developments in mathematics. John B. Conway VI

PREFACE This book is intended as a textbook for a first course in the theory of functions of one complex variable for students who are mathematically mature enough to understand and execute E - 8 arguments. The actual prerequisites for reading this book are quite minimal; not much more than a stiff course in basic calculus and a few facts about partial derivatives. The topics from advanced calculus that are used (e.g., Leibniz's rule for differentiating under the integral sign) are proved in detail. Complex Variables is a subject which has something for all mathematicians. In addition to having applications to other parts of analysis, it can rightly claim to be an ancestor of many areas of mathematics (e.g., homotopy theory, manifolds). This view of Complex Analysis as "An Introduction to Mathematics" has influenced the writing and selection of subject matter for this book. The other guiding principle followed is that all definitions, theorems, etc. should be clearly and precisely stated. Proofs are given with the student in mind. Most are presented in detail and when this is not the case the reader is told precisely what is missing and asked to fill in the gap as an exercise. The exercises are varied in their degree of difficulty. Some are meant to fix the ideas of the section in the reader's mind and some extend the theory or give applications to other parts of mathematics. (Occasionally, terminology is used in an exercise which is not defined-e.g., group, integral domain.) Chapters I through V and Sections VI.l and VI.2 are basic. It is possible to cover this material in a single semester only if a number of proofs are omitted. Except for the material at the beginning of Section VI.3 on convex functions, the rest of the book is independent of VI. 3 and VI.4. Chapter VII initiates the student in the consideration of functions as points in a metric space. The results of the first three sections of this chapter are used repeatedly in the remainder of the book. Sections four and five need no defense; moreover, the Weierstrass Factorization Theorem is necessary for Chapter XI. Section six is an application of the factorization theorem. The last two sections of Chapter VII are not needed in the rest of the book although they are a part of classical mathematics which no one should completely disregard. The remaining chapters are independent topics and may be covered in any order desired. Runge's Theorem is the inspiration for much of the theory of Function Algebras. The proof presented in section VIII. I is, however, the classical one involving "pole pushing". Section two applies Runge's Theorem to obtain a more general form of Cauchy's Theorem. The main results of sections three and four should be read by everyone, even if the proofs are not. Chapter IX studies analytic continuation and introduces the reader to analytic manifolds and covering spaces. Sections one through three can be considered as a unit and will give the reader a knowledge of analytic vii

viii

Preface continuation without necessitating his going through all of Chapter IX. Chapter X studies harmonic functions including a solution of the Dirichlet Problem and the introduction of Green's Function. If this can be called applied mathematics it is part of applied mathematics that everyone should know. Although they are independent, the last two chapters could have been combined into one entitled "Entire Functions". However, it is felt that Hadamard's Factorization Theorem and the Great Theorem of Picard are sufficiently different that each merits its own chapter. Also, neither result depends upon the other. With regard to Picard's Theorem it should be mentioned that another proof is available. The proof presented here uses only elementary arguments while the proof found in most other books uses the modular function. There are other topics that could have been covered. Some consideration was given to including chapters on some or all of the following: conformal mapping, functions on the disk, elliptic functions, applications of Hilbert space methods to complex functions. But the line had to be drawn somewhere and these topics were the victims. For those readers who would like to explore this material or to further investigate the topics covered in this book, the bibliography contains a number of appropriate entries. Most of the notation used is standard. The word "iff" is used in place of the phrase "if and only if", and the symbol. is used to indicate the end of a proof. When a function (other than a path) is being discussed, Latin letters are used for the domain and Greek letters are used for the range. This book evolved from classes taught at Indiana University. I would like to thank the Department of Mathematics for making its resources available to me during its preparation. I would especially like to thank the students in my classes; it was actually their reaction to my course in Complex Variables that made me decide to take the plunge and write a book. Particular thanks sho~ld go to Marsha Meredith for pointing out several mistakes in an early draft, to Stephen Berman for gathering the material for several exercises on algebra, and to Larry Curnutt for assisting me with the final corrections of the manuscript. I must also thank Ceil Sheehan for typing the final draft of the manuscript under unusual circumstances. Finally, I must thank my wife to whom this book is dedicated. Her encouragement was the most valuable assistance I received. John B. Conway

TABLE OF CONTENTS Preface for the Second Edition

VI

Preface

viii

Contents, Volume II

1.

The Complex Number System §l. §2. §3. §4. §5. §6.

n.

3 4 6

8

Definition and examples of metric spaces Connectedness. Sequences and completeness Compactness Continuity Uniform convergence

11 14

17

20

24 28

Power series Analytic functions Analytic functions as mapping, Mobius transformations

30 33 44

Complex Integration §1. §2. §3. §4. §5. §6.

§7. §8.

V.

1 1

Elementary Properties and Examples of Analytic Functions §1. §2. §3.

IV.

The real numbers The field of complex numbers The complex plane Polar representation and roots of complex numbers Lines and half planes in the complex plane The extended plane and its spherical representation

Metric Spaces and the Topology of IC §1. §2. §3. §4. §5. §6.

m.

xi

Riemann-Stieltjes integrals Power series representation of analytic functions Zeros of an analytic function The index of a closed curve Cauchy's Theorem and Integral Formula The homotopic version of Cauchy's Theorem and simple connectivity . Counting zeros; the Open Mapping Theorem Goursat's Theorem .

58

68 76 80 83 87 97 100

Singularities §l.

Classification of singularities ix

103

x

Table of Contents

§2. §3.

VI.

Residues The Argument Principle

The Maximum Modulus Theorem The Maximum Principle §2. Schwarz's Lemma §3. Convex functions and Hadamard's Three Circles Theorem §4. Phragmen-LindelOf Theorem

128 130 133 138

Compactness and Convergence in the Space of Analytic Functions §1. The space of continuous functions C(G,Q) §2. Spaces of analytic functions §3. Spaces of merom orphic functions . §4. The Riemann Mapping Theorem . §5. Weierstrass Factorization Theorem §6. Factorization of the sine function §7. The gamma function §8. The Riemann zeta function .

142 151 155 160 164 174 176 187

Runge's Theorem Runge's Theorem §2. Simple connectedness §3. Mittag-Leffler's Theorem

195 202 204

Analytic Continuation and Riemann Surfaces Schwarz Reflection Principle. §2. Analytic Continuation Along A Path §3. Monodromy Theorem §4. Topological Spaces and Neighborhood Systems §5. The Sheaf of Germs of Analytic Functions on an Open Set §6. Analytic Manifolds §7. Covering spaces

210 213 217 221 227 233 245

Harmonic Functions Basic Properties of harmonic functions §2. Harmonic functions on a disk §3. Subharmonic and superharmonic functions §4. The Dirichlet Problem §5. Green's Functions

252 256 263 269 275

§1.

VII.

VIII.

§1.

IX.

§1.

X.

112 123

§1.

Table of Contents

XI.

xi

Entire Functions Jensen's Formula The genus and order of an entire function Hadamard Factorization Theorem

§l. §2. §3.

XII.

280 282 287

The Range of an Analytic Function Bloch's Theorem The Little Picard Theorem Schottky's Theorem The Great Picard Theorem

§l. §2. §3. §4.

292

296 297 300

Appendix A: Calculus for Complex Valued Functions on an Interval

303

Appendix B: Suggestions for Further Study and Bibliographical Notes

307

References

311

Index

313

List of Symbols

317

TABLE OF CONTENTS, VOLUME II Preface 13.

Return to Basics §l. §2. §3. §4.

§S.

14.

Regions and Curves Derivatives and other recollections Harmonic conjugates and primitives Analytic arcs and the reflection principle Boundary values for bounded analytic functions

Conformal Equivalence For Simply Connected Regions §1.

§2. §3. §4.

§S.

Elementary properties and examples Crosscuts Prime Ends Impressions of a prime end Boundary values of Riemann maps

xii

Table of Contents

§6. §7.

15.

Conformal Equivalence For Finitely Connected Regions §I. §2. §3. §4. §5. §6. §7.

16.

The Area Theorem Disk mappings: the class S

Analysis on a finitely connected region Conformal equivalence with an analytic Jordan region Boundary values for a conformal equivalence between finitely connected Jordan regions Convergence of univalent functions Conformal equivalence with a circularly slit annulus Conformal equivalence with a circularly slit disk Conformal equivalence with a circular region

Analytic Covering Maps Results for abstract covering spaces §2. Analytic covering spaces §3. The modular function §4. Applications of the modular function §5. The existence of the universal analytic covering map §1.

17.

De Branges's Proof of the Bieberbach Conjecture §I. §2. §3. §4. §5. §6.

18.

Some Fundamental Concepts From Analysis § I.

§2. §3. §4.

§5. §6. §7.

19.

Subordination Loewner chains Loewner's differential equation The Milin Conjecture Some special functions The proof of de Branges's Theorem

Bergman spaces of analytic and harmonic functions Partitions of unity Convolution in Euclidean space Distributions The Cauchy transform An application: rational approximation Fourier series and Cesaro sums

Harmonic Functions Redux §1. Harmonic functions on the disk §2. Fatou's Theorem §3. Semicontinuous functions §4. Subharmonic functions §5. The logarithmic potential §6. An application: approximation by harmonic functions §7. The Dirichlet problem §8. Harmonic majorants §9. The Green function

Table of Contents

§1O. §ll.

Regular points for the Dirichlet problem The Dirichlet principle and Sobolev spaces

20.

Hardy Spaces on the Disk §l. Definitions and elementary properties §2. The Nevanlinna Class §3. Factorization of functions in the Nevanlinna class §4. The disk algebra §5. The invariant subspaces of HP §6. Szeg6's Theorem

21.

Potential Theory in the Plane §I. Harmonic measure §2. The sweep of a measure §3. The Robin constant §4. The Green potential §5. Polar sets §6. More on regular points §7. Logarithmic capacity: part 1 §8. Some applications and examples of logarithmic capacity §9. Removable singularities for functions in the Bergman space §1O. Logarithmic capacity: part 2 §ll. The transfinite diameter and logarithmic capacity §12. The refinement of a subharmonic function §13. The fine topology §14. Wiener's criterion for regular points

References

List of Symbols Index

xiii

Chapter 1 The Complex Number System §1. The real numbers We denote the set of all real numbers by IR. It is assumed that each reader is acquainted with the real number system and all its properties. In particular we assume a knowledge of the ordering of IR, the definitions and properties of the supremum and infimum (sup and inf), and the completeness of IR (every non-empty set in IR which is bounded above has a supremum). It is also assumed that every reader is familiar with sequential convergence in IR and with infinite series. Finally, no one should undertake a study of Complex Variables unless he has a thorough grounding in functions of one real variable. Although it has been traditional to study functions of several real variables before studying analytic function theory, this is not an essential prerequisite for this book. There will not be any occasion when the deep results of this an:a are needed.

§2. The field of complex numbers We define C, the complex numbers, to be the set of an ordered pairs (a, b) where a and b are real numbers and where addition and multiplication are defined by:

(a, b)+(c, d) = (a+c, b+d) (a, b) (c, d) = (ac-bd, bc+ad) It is easily checked that with these definitions C satisfies all the axioms for a field. That is, C satisfies the associative, commutative and distributive laws for addition and multiplication; (0,0) and (1,0) are identities for addition and multiplication respectively, and there are additive and multiplicative inverses for each nonzero element in C. We will write a for the complex number (a, 0). This mapping a --* (a, 0) defines a field isomorphism of IR into C so we may consider IR as a subset of C. If we put i = (0, 1) then (a, b) = a+bi. From this point on we abandon the ordered pair notation for complex numbers. Note that i 2 = - 1, so that the equation Z2 + 1 = 0 has a root in IC. In fact, for each z in C, Z2+ I = (z+i) (z-i). More generally, if z and ware complex numbers we obtain Z2+ W2

=

(z+iw) (z-iw) 1

2

The Complex Number System

By letting z and w be real numbers a and b we can obtain (with both a and b "# 0)

a~ib = a~~~2 = a2:b2 -

{a 2!b 2)

so that we have a formula for the reciprocal of a complex number. When we write z = a + ib (a, b E IR) we call a and b the real and imaginary parts of z and denote this by a = Re z, b = Im z. We conclude this section by introducing two operations on IC which are not field operations. If z = x + iy(x, y E IR) then we define Izl = (x 2+ y2)!- to be the absolute value of z and z = x - iy is the conjugate of z. Note that 2.1 In particular, if z "# 0 then

Izl2

Z

The following are basic properties of absolute values and conjugates whose verifications are left to the reader. 2.2

Rez = !(z+z)

and

(z+w) = z+w

2.3

Imz =

1

2i (z-z).

zw = zw.

and

Izwl = Izllwl· Izjwl = Izlllwl· Izl = Izl·

2.4 2.5 2.6

The reader should try to avoid expanding z and w into their real and imaginary parts when he tries to prove these last three. Rather, use (2.1), (2.2), and (2.3). Exercises 1. Find the real and imaginary parts of each of the following:

!.

z-~(aEIR)' z3. 3+5i. (-I+iJ3)3;

z ' z+a

"

7i+ 1 '

(-I- J 3)6 . (1J2+ i)n 2

i

;

In;

2

for

2::; n ::; 8.

2. Find the absolute value and conjugate of each of the following:

3-i

i

-2+i; -3; (2+i)(4+3i); J2+3i; i+3 ; (I + i)6; i 17.

The complex plane

3

3. Show that z is a real number if and only if Z = Z. 4. If z and ware complex numbers, prove the following equations: Iz+wl2

= Iz1 2 +2Re zw+ Iwlz.

IZ-WJ2 = Iz1 2 -2Re zw+ Jwi 2 • Iz+ wlz + Iz- wlz = 2(lz12 + IwI Z). 5. Use induction to prove that for z = ZI + .. . +zn; W = WIW2 ••• Wn: Iwl = IWII·· ·Iwnl; 2 = 21 +., .+2n; W = WI ... wn ' 6. Let R(z) be a rational function of z. Show that R(z) = R(i) if all the coefficients in R(z) are real.

§3. The complex plane From the definition of complex numbers it is dear that each z in C can be identified with the unique point (Re z, 1m z) in the plane [R2. The addition of complex numbers is exactly the addition law of the vector space [R2. If z and ware in C then draw the straight lines from z and w to = (0, 0». These form two sides of a parallelogram with 0, z and w as three vertices. The fourth vertex turns out to be z + w. Note also that Iz-wl is exactly the distance between z and w. With this in mind the last equation of Exercise 4 in the preceding section states the parallelogram law: The sum of the squares of the lengths of the sides of a parallelogram equals the sum of the squares of the lengths of its diagonals. A fundamental property of a distance function is that it satisfies the triangle inequality (see the next chapter). In this case this inequality becomes

°(

IZ1 - z 21 s IZI - Z31 + for complex numbers ZI, Z2, Z3' By using easy to see that we need only show 3.1

Iz+wl

s

IZ3-z21

ZI-Z2

Izl + Iwl (z, WE

=

(ZI-Z3)+(Z3-Z2),

it is

q.

To show this first observe that for any z in C,

3.2 Hence, Re (zw)

-IZI

s

s s

Rez

-izi Imz izwi = Jzllwl. Thus,

s s

Izl Izi

Iz+wl2 = Iz1 2 +2Re (zw) + Iwl 2

s

IzI2+2Izllwl+lwI2

= (izi + Iwl)2, from which (3.1) follows. (This is called the triangle inequality because, if we represent z and w in the plane, (3.1) says that the length of one side of the triangle [0, z, z -+ w] is less than the sum of the lengths of the other two sides. Or, the shortest distance between two points is a straight line.) On encounter-

4

The Complex Number System

ing an inequality one should always ask for necessary and sufficient conditions that equality obtains. From looking at a triangle and considering the geometrical significal'lce of (3.1) we are led to consider the condition z = tw for some t E IR, t ;:: O. (or IV = tz if IV = 0). It is clear that equality will occur when the two points are colinear with the origin. In fact, if we look at the proof of (3.1) we see that a necessary and sufficient condition for Iz + wi = Izi + Iw, is that Izwl = Re (zw). Equivalently, this is zw ;:: 0 (i.e., zw is a real number and is non negative). MultiplyinK this by w/w we get IwI 2 (z/w);:: Oifw # O. If t

1 )' 2 = z/w . = (~ 1\V12 I'wi (z/w)

then t ;:: 0 and z = two By induction we also get

3.3 Also useful is the inequality 3.4

Izl-Iwll

s:;

Iz-wl

Now that we have given a geometric interpretation of the absolute value let us see what taking a complex conjugate does to a point in the plane. This is also easy; in fact, Z is the point obtained by reflecting z across the x-axis (i.e., the real axis). Exercises 1. Prove (3.4) and give necessary and sufficient conditions for equality. 2. Show that equality occurs in (3.3) if and only if zdz/ ;:: 0 for any integers k and I, 1 s:; k, I s:; n, for which Zl # O. 3. Let a E rr~ and c > 0 be fixed. Describe the set of points z satisfying

Iz-al-Iz+al =

2c

for every possible choice of a and c. Now let a be any complex number and, using a rotation of the plane, describe the locus of points satisfying the above equation.

§4. Polar representation and roots of complex numbers Consider the point z = x + iy in the complex plane C. This point has polar coordinates Cr, 8): x = r cos 8, y = r sin e. Clearly r = Izi and () is the angle between the positive real axis and the line segment from 0 to Z. Notice that 8 plus any multiple of 211" can be substituted for 8 in the above equations. The angle () is called the argument ofz and is denoted by e = arg z. Because of the ambiguity of e, "arg" is not a function. We introduce the notation 4.1

cis 8 = cos 8+i sin 8.

Polar representation and roots of complex numbers

5

Let ZI = 1'1 cis 8 1 , Z2 = 1'2 cis 82 , Then Z1 Z2 = 1'11'2 cis 01 cis O2 = 1'11'2 [(cos 0 1 cos Oz-sin Ol sin (2)+i (sin 0 1 cos 02+sin 82 cos OJ)]. By the formulas for the sine and cosine of the sum of two angles we get

4.2 Alternately, arg(ZtZ2) = arg ZI + arg Z1' (What function of a real variable takes products into sums?) By induction we get for Z k = r k cis (j h' 1 ~ k ~ 11,

4.3 In particular, 4.4

Z" =

I'"

cis (ne),

for every integer n ;::: O. Moreover if z #- 0, z· [r - 1 cis (- 0)] = 1; so that (4.4) also holds for all integers n, positive, negative, and zero, if z #- O. As a special case of (4.4) we get de M oivre's formula: (cos O+i sin 8t = cos n8+i sin nO. We are now in a position to consider the following problem: For a given complex number a#-O and an integer n ;::: 2, can you find a number z satisfying zn = a? How many such z can you find? In light of (4.4) the solution is easy. Let a = lal cis ex; by (4.4), Z = lal ' /II cis (aln) fills the bill. ]

However this is not the only solution because z' = la1 1/11 cis - (cc+ 27T) also n satisfies (zT = a. In fact each of the numbers 1 (a + 27Tk), 0 ~ k ~ n-l, n

lal 1 / n cis -

4.5

is an nth root of a. By means of (4.4) we arrive at the following: for each non zero number a in IC there are n distinct nth roots of a; they are given by formula (4.5). Example Calculate the nth roots of unity. Since 1 = cis 0, (4.5) gives these roots as

27T

417

217

1, cis - , cis ~-, ... , cis - (n-l). n n n In particular, the cube roots of unity are

Exercises I. Find the sixth roots of unity.

6

The Complex Number System

2. Calculate the following: (a) the square roots of i (b) the cube roots of i (c) the square roots of ..)3 +3i 3. A primilive nth rool oj unity is a complex number a such that l,a,a 2, ••• ,a n - 1 are distinct nth roots of unity. Show that if a and bare primitive nth and mth roots of unity, respectively, then ab is a kth root of unity for some integer k. What is the smallest value of k? What can be said if a and bare nonprimitive roots of unity? 4. Use the binomial equation

(a+b)" = where

f (n)k a"-kbk,

k=O

(z) k!(nn~k)! ' =

and compare the real and imaginary parts of each side of de Moivre's formula to obtain the formulas: cos n(J

= cos" (J- (;)

sin

= (~)

n(J

cos·- 2 sin 2 (J+

COS"-l (J

sin

(~) COS"-4 (J sin4 (J- • ••

(J- (;) COS,,-3 (J

sin 3

(J+ ••.

. 21T t ' . h 1 lor an Integer n ~ 2. S ow that 1 +z+ ... +z"- = O. n 6. Show that «pCt) = cis t is a group homomorphism of the additive group IR onto the multiplicative group T = {z: Izl = I}. 7. If Z E C and Re(zn)~ 0 for every positive integer n, show that z is a non-negative real number.

5. Let z =

CIS -

§5. Lines and half planes in the complex plane Let L denote a straight line in C. From elementary analytic geometry, L is determined by a point in L and a direction vector. Thus if a is any point in Land b is its direction vector then L

Since b

=1=

= {z = a+tb: -

00

< t < oo}.

0 this gives, for z in L,

1m ( z-a) b = O. In fact if z is such that O=lm ( z-a) b

then 1=

-( z-a)

b

7

Lines and half planes in the complex plane

implies that z = a + tb, -

00

< t <

00.

That is

5.1 What is the locus of each of the sets

{z: Im(z~a) > o}, {z: (z~a) < o} 1m

?

As a first step in answering this question, observe that since b is a direction we may assume Ibl = 1. For the moment, let us consider the case where a = 0, and put Ho = {z: 1m (z/b) > O}, b = cis f3. If z = r cis () then z/b = r cis «()-f3). Thus, z is in Ho if and only if sin «()-f3) > 0; that is, when f3 < 8 < 7T+f3. Hence Ho is the half plane lying to the left of the line L if

we are "walking along L in the direction of b." If we put

then it is easy to see that Ha = a+Ho == {a+w: WE Ho}; that is, Ha is the translation of Ho by a. Hence, Ha is the half plane lying to the left of L. Similarly,

is the half plane on the right of L. Exercise 1. Let C be the circle {z:

Iz-cl

=

r}, r > 0; let a = c+r cis

(X

and put

8

Lp

=

{z:

The Complex Number System

1m

(z~a) =

o}

where b = cis {3. Find necessary and sufficient conditions in terms of {3 that Lp be tangent to C at a.

§6. The extended plane and its spherical representation Often in complex analysis we will be concerned with functions that become infinite as the variable approaches a given point. To discuss this situation we introduce the extended plane which is IC U {oo} == 1C00. We also wish to introduce a distance function on 1C00 in order to discuss continuity properties of functions assuming the value infinity. To accomplish this and to give a concrete picture of 1C00 we represent 1C00 as the unit sphere in 1R 3 ,

s=

{(XI,X2,X3)EIR3:xf+x~+x~

= I}.

Let N = (0,0, 1); that is, N is the north pole on S. Also, identify IC with {(Xl' X2' 0): Xl' X 2 E IR} so that IC cuts S along the equator. Now for each point z in IC consider the straight line in 1R3 through z and N. This intersects .'v

the sphere in exactly one point Z #- N. If /z/ > 1 then Z is in the northern hemisphere and if /z/ < I then Z is in the southern hemisphere; also, for /z/ = I, Z = z. What happens to Z as /z/ -+ oo? Clearly Z approaches N; hence, we identify N and the point 00 in 1C00. Thus 1C00 is represented as the sphere S. Let us explore this representation. Put z = X+ iy and let Z = (Xl' X2' X3) be the corresponding point on S. We will find equations expressing Xl' x 2, and X3 in terms of X and y. The line in 1R3 through z and N is given by {tN+(I-t)z: -00 < t < oo}, or by

6.1

{«I-t)x,(I-t)y,t):

-00

< t < oo}.

Hence, we can find the coordinates of Z if we can find the value of t a't

The extended plane and its spherical representation

which this line intersects S. If

1

9

is this value then

1 = (l_t)2x2+(1_t)2y2+t 2 = (l_t)2IzI2+t 2

From which we get 1-12

= (1-1)2IzI2.

Since t :/= 1 (z :/= 00) we arrive at

Thus

6.2 But this gives

1

- i(z - i)

z+i x =-I Iz12+ I

6.3

Iz12+ I

Iz 12 -1

x =---

IzI 2+1·

3

If the point Z is given (Z :/= N) and we wish to find z then by setting X3 and using (6.1), we arrive at

=

Xl +iX2 Z=---

6.4

I-X 3

Now let us define a distance function between points in the extended plane in the following manner: for z, z' in Coo define the distance from z to z', d(z, z'), to be the distance between the corresponding points Z and Z' in 1R3. If Z = (Xl' X2' X3) and Z' = (x;, X2' Xl) then

6.5 Using the fact that Z and Z' are on S, (6.5) gives

6.6 By using equation (6.3) we get

6.7

d(') 2Iz-z'l z, z = [(1 + Iz12) (1 + Iz'12W·'

In a similar manner we get for

Z

(' Z, Z

E

C)

in C

6.8 This correspondence between points of Sand C", is called the stereographic projection.

The Complex Number System

10

Exercises 1. Give the details in the derivation of (6.7) and (6.8). 2. For each of the following points in C, give the corresponding point of S: 0, 1 +i, 3+2i. 3. Which subsets of S correspond to the real and imaginary axes in C? 4. Let A be a circle lying in S. Then there is a unique plane P in 1R3 such that P (\ S = A. Recall from analytic geometry that

P

=

{(X t :X2,X3): XtfJl+X2fJ2+X3fJ3

= /}

where (/31' /32' /3 3) is a vector orthogonal to P and I is some real number. It can be assumed that + fii + fif = 1. Use this information to show that if A contains the point N then its stereo graphic projection on C is a straight line. Otherwise, A projects onto a circle in C. 5. Let Z and Z' be points on S corresponding to z and z' respectively. Let W be the point on S corresponding to z+z'. Find the coordinates of Win terms of the coordinates of Z and Z'.

fir

Chapter II Metric Spaces and the Topology of

C

§1. Definition and examples of metric spaces A metric space is a pair (X, d) where X is a set and d IS a function from X x X into IR, called a distance function or metric, which satisfies the following conditions for x, y, and z in X: 0

d(x,y)

~

d(x, y)

= 0 if and only if x = y

d(x, y)

= dey, x) (symmetry)

d(x, z) ::::; d(x, y)+d(y, z) (triangle inequality)

If x and r > 0 are fixed then define B(x;r) = {yeX: d(x,y) < r} H(x;r) = {yeX: d(x,y)::::; r}. B(x; r) and H(x; r) are called the open and closed balls, respectively, with center x and radius r.

Examples

1.1 Let X = IR or C and define d(z, w) = Jz-wJ. This makes both (IR, d) and (C, d) metric spaces. In fact, (C, d) will be the example of principal interest to us. If the reader has never encountered the concept of a metric space before this, he should continually keep (C, d) in mind during the study of this chapter. 1.2 Let (X, d) be a metric space and let Y eX; then (Y, d) is also a metric space. 1.3 Let X = C and define d(x+iy, a+ib) = Jx-aJ+Jy-bJ. Then (C, d) is a metric space. 1.4 Let X = C and define d(x+iy, a+ib) = max {Jx-aJ, Jy-bJ}. 1.5 Let X be any set and define d(x, y) = 0 if x = y and d(x, y) = I if x of. y. To show that the function d satisfies the triangle inequality one merely considers all possibilities of equality among x, y, and z. Notice here that B(x; E) consists only of the point x if E ::::; 1 and B(x; E) = X if E > 1. This metric space does not appear in the study of analytic function theory. 1.6 Let X = IRn and for x = (Xl> ••• , x n), y = (Yl' ... ,Yn) in IR n define d(x,y)

=

Lt 11

(X j - y )2r

12

Metric Spaces and the Topology of C

1.7 Let S be any set and denote by B(S) the set of all functions f: S such that

IlflL" == sup {If(s)l: s E S} <

~

C

00.

That is, B(S) consists of all complex valued functions whose range is contained inside some disk of finite radius. For f and g in B(S) define d(f; g) = Ilf-g;lw' We will show that d satisfies the triangle inequality. In fact if J, g, and h are in B(S) and s is any point in S then If(s)-g(s)1 = If(s)-h(s)+ h(s)-g(s)1 ::; If(s)-h(s)1 + Ih(s)-g(s)[ ::; Ilf-hil oo + Ilh-giloo" Thus, when the supremum is taken over all s in S, Ilf-gilro ::; IIf-h 1100 +IIh-gll oo , which is the triangle inequality for d.

1.8 Definition. For a metric space (X, d) a set G c X is open if for each x in G there is an E > 0 such that B(x; E) c G. Thus, a set in C is open if it has no "edge." For example, G = {z E C: a < Re z < b} is open; but {z: Re z < O} u {O} is not because B(O; E) is not contained in this set no matter how small we choose E. We denote the empty set, the set consisting of no elements, by D. 1.9 Proposition. Let (X, d) be a metric space; then: (a) The sets X and D are open; (b) If G 1 , . . • , Gn are open sets in X then so is (c)

If

n n

Gk ;

k~l

{G/ j E J} is a collection of open sets in X, J any indexing set, then G = U {G j : j E J} is also open.

n n

Proof The proof of (a) is a triviality. To prove (b) let x

E

G

=

Gk ; then

k~1

x E Gk for k = 1, ... , n. Thus, by the definition, for each k there is an Ek > 0 such that B(x; Ek) C Gk • But if E = min {El' E2, ••• , En} then for 1 ::; k ::; n B(x; E) c B(x; Ek) C Gk • Thus B(x; E) c G and G is open. The proof of (c) is left as an exercise for the reader. • There is another class of subsets of a metric space which are distinguished. These are the sets which contain all their "edge"; alternately, the sets whose complements have no "edge." 1.10 Definition. A set F c X is closed if its complement, X - F, is open. The following proposition is the complement of Proposition 1.9. The proof, whose execution is left to the reader, is accomplished by applying de Morgan's laws to the preceding proposition. 1.11 Proposition. Let (X, d) be a metric space. Then: (a) The sets X and D are closed; n (b) If F 1 , ••• , Fn are closed sets in X then so is U Fk ; k~l

(c) If {Fj : j E J} is any collection of closed sets in X, J any indexing set, then F = n {Fj : j E J} is also closed. The most common error made upon learning of open and closed sets is to interpret the definition of closed set to mean that if a set is not open it is

Definition and examples of metric spaces

13

closed. This, of course, is false as can be seen by looking at {z E C: Re z > O} u {O}; it is neither open nor closed.

n

1.12 Definition. Let A be a subset of X. Then the interior of A, int A, is the {F: F set U {G: G is open and G c A}. The closure of A, A-, is the set is closed and F :::> A}. Notice that int A may be empty and A - may be X. If A = {a+bi: a and b are rational numbers} then simultaneously A- = C and int A = O. By Propositions 1.9 and 1.11 we have that A - is closed and int A is open. The boundary of A is denoted by OA and defined by OA = An (X-A)-. 1.13 Proposition. Let A and B be subsets of a metric space (X, d). Then: (a) A is open if and only if A = int A; (b) A is closed if and only if A = A -; (c) int A = X-(X-A)-; A- -= X-int (X-A); OA = A--int A; (d) (A u B)- = A- u B-; (e) Xo E int A if and only if there is an € > 0 such that B(xo; €) C A; (f) Xo E A - if and only if for every € > 0, B(xo; €) n A :f. O.

Proof. The proofs of (a)-(e) are left to the reader. To prove (f) assume xoEA- = X-int (X-A); thus, xo¢int (X-A). By part (e), for every € > 0 B(xo; €) is not contained in X -A. That is, there is a point y E B(xo; €) which is not in X-A. Hence, yEB(xo; €)nA. Now suppose xo¢A- = X -int (X -A). Then Xo E int (X -A) and, by (e), there is an € > 0 such that B(xo; €) eX-A. That is, B(xo; €) n A = 0 so that Xo does not satisfy the condition . • Finally, one last definition of a distinguished type of set. 1.14 Definition. A subset A of a metric space X is dense if A - = X. The set of rational numbers 0 is dense in IR and {x+iy: x, yEO} is dense in Co

Exercises 1. Show that each of the examples of metric spaces given in (1.2)-(1.6) is, indeed, a metric space. Example (1.6) is the only one likely to give any difficulty. Also, describe B (x; r) for each of these examples. 2. Which of the following subsets of C are open and which are closed: (a) {z:lzl 0 such that B(b; t:) c G. But if Z E B(b; t:) then [b, z] c B(b; t:) c G. Hence the polygon Q = P U [b, z] is inside G and goes from a to z. This shows that B(b; t:) c A, and so A is open. To show that A is closed suppose there is a point Z in G-A and let E > 0 be such that B(z; E) c G. If there is a point b in A n B(z; E) then, as above, we can construct a polygon from a to z. Thus we must have that B(z; E) n A = D, or B(z; E) c G-A. That is, G-A is open so that A is closed. • a E A and G is connected this will give that A

2.4 Corollary. If Gee is open and connected and a and b are points in G then there is a polygon P in G from a to b which is made up of line segments parallel to either the real or imaginary axis. Proof There are two ways of proving this corollary. One could obtain a

16

Metric Spaces and the Topology of C

polygon in G from a to b and then modify each of its line segments so that a new polygon is obtained with the desired properties. However, this proof is more easily executed using compactness (see Exercise 5.7). Another proof can be obtained by modifying the proof of Theorem 2.3. Define the set A as in the proof of (2.3) but add the restriction that the polygon's segments are all parallel to one of the axes. The remainder of the proof will be valid with one exception. If z EO B(b; €) then [b, z] may not be parallel to an axis. But it is easy to see that if z = x+iy, b = p+iq then the polygon [b, p+iy] U [p+iy, z] c B(b; €) and has segments parallel to an axis. III It will now be shown that any set S in a metric space can be expressed, in a canonical way, as the union of connected pieces. 2.5 Definition. A subset D of a metric space X is a component of ,\:' if it is a maximal connected subset of X. That is, D is connected and there is no connected subset of X that properly contains D. If the reader examines the example at the beginning of this section he will notice that both A and B are components and, furthermore, these are the only components of X. For another example let X = {D, I, .1, t, ... }. Then clearly every component of X is a point and each point is a component. Notice that while the components

{M

are all open in X, the component {O}

is not. 2.6 Lemma. Let x 0 E X and let {D j: j E J} be a collection of connected subsets of X such that x 0 EO D j for each j in J. Then D = U {D / j E J} is connected.

Proof Let A be a subset of the metric space CD, d) which is both open and closed and suppose that A i= O. Then A n D j is open in (D j, d) for each j and it is also closed (Exercises l.8 and 1.9). Since D j is connected we get that either A n D j = or An D j = D j • Since A i= 0 there is at least one k such that A n Dk i= 0; hence, A n Dk = D k • In particular x 0 E A so that Xo EA n D j for every j. Thus An D j = D j , or D j C A, for each index). This gives that D = A, so that D is connected. III 2.7 Theorem. Let (X, d) be a metric space. Then: (a) Each Xo in X is contained in a component of X. (b) Distinct components of X are disjoint. Note that part (a) says that X is the union of its components.

Proof (a) Let!» be the collection of connected subsets of X which contain the given point Xo. Notice that {xo} E 9 so that!» i= O. Also notice that the hypotheses of the preceding lemma apply to the collection g. Hence C = U {D: D EO g} is connected and Xo E C. But C must be a component. In fact, if D is connected and C C D then x 0 EO D so that DE!»; but then DeC, so that C = D. Thus C is maximal and part (a) is proved. (b) Suppose C 1 and C z are components, C 1 i= C z , and suppose there is a point Xo in C 1 n C z . Again the lemma says that C 1 U C z is connected.

17

Sequences and completeness

Since both C 1 and C 2 are components, this gives C 1 = C 1 U C 2 = C z, a contradiction. II 2.8 Proposition. (a) If A c X is connected and A c Be A -, then B is connected. (b) If C is a component of X then C is closed. The proof is left as an exercise.

2.9 Theorem. Let G be open in IC; then the components (!f G are open and there are only a countable number of them. Proof. Let C be a component of G and let Xo c C. Since G is open there is an ( > 0 with B(xo; f) c G. By Lemma 2.6, B(xo; €) U C is connected and so it must be C. That is B( xo; () c C and C is, therefore. open. To see that the number of components is countable let S = {a+ ib: a and b are rational and a + bi E G}. Then S is countable and each component of G contains a point of S, so that the number of components is countable. II Exercises

1. The purpose of this exercise is to show that a connected subset of IR is an interval. (a) Show that a set A c IR is an interval iff for any two points a and b in A with a < b, the interval [a, b] c A. (b) Use part (a) to show that if a set A c IR is connected then it is an interval. 2. Show that the sets Sand T in the proof of Theorem 2.3 are open. 3. Which of the following su bsets X of IC are connected; if X is not connected, what are its components: (a) X = {z: Izl :s; I} U {z: IZ-21 < I}. (b) X =

[0,1)

U

{1+~: n ~

I}. (c) X = IC-(A

°

U

B) where A = [0, (0) and B =

{z = r cis 0: r = 0, :s; e:s; oo}? 4. Prove the following generalization of Lemma 2.6. If {D j: j E J} is a collection of connected subsets of X and if for each j and k in J we have D j n Dk -# D then D = U {D j : j E J} is connected. 5. Show that if F c X is closed and connected then for every pair of points a, b in F and each to > there are points Zo, ZI, . . • ,Zn in F with Zo = a, Zn = band d(zk _ 1, Zk) < to for 1 :s; k :s; n. I s the hypothesis that F be closed needed? If F is a set which satisfies this property then F is not necessarily connected, even if F is closed. Give an example to illustrate this.

°

§3. Sequences and completeness One of the most useful concepts in a metric space is that of a convergent sequence. Their central role in calculus is duplicated in the study of metric spaces and complex analysis. 3.1 Definition. If {Xl' x 2 ,

••• }

is a sequence in a metric space (X, d) then

18

Metric Spaces and the Topology of C

{xn} converges to x-in symbols x = lim Xn or Xn -+ x-if for every" > 0 there is an integer N such that d(x, xn) < " whenever n ;:::: N. Alternately, x = lim XII if 0= lim d(x, XII)' If X = C then z = lim z" means that for each" > 0 there is an N such that Iz-z,,1 < Eo" when n ;:::: N. Many concepts in the theory of metric spaces can be phrased in terms of sequences. The following is an example.

3.2 Proposition. A set F c X is closed x = lim xn we have x E F.

iff for

each sequence {xn} in F with

Proof. Suppose F is closed and x = lim X Il where each X Il is in F. So for every E > 0, there is a point x ll in B(x; E); that is B(x; £) n F", D, so that x EO F~ = F by Proposition l.13(f). Now suppose F is not closed; so there is a point Xo in F~ which is not in F. By Proposition 1.13(f), for every Eo" > 0 we have B(x 0; €) n F # D-

In particular for every integer n there is a point d(xo,

X,I)

0 if and only if IZn-zl -3> O. In spite of this, any sequence {zn} with lim IZnl = 00 is Cauchy in Coo, but, of course, is not Cauchy in C. If A c X we define the diameter of A by diam A = sup {d(x, y): x and yare in A}. 3.7 Cantor's Theorem. A metric space (X, d) is complete

iff for

any sequence

{Fn} of non-empty closed sets with F) ~ F2 ~ ... and diam Fn consists of a single point.

-3>

0,

n Fn 00

n~1

Proof Suppose (X, d) is complete and let {Fn} be a sequence of closed sets having the properties: (i) F) ~ F2 ~ ... and (ii) lim diam Fn = O. For each n, let Xn be an arbitrary point in Fn; if n, m ;::: N then Xn, Xm are in FN so that, by definition, d(xm xm) : ••• } : 0; but which has F = F) (J F2 (J ••• either empty or consisting of more than one point. Everyone should get examples satisfying the possible combinations. 3.8 Proposition. Let (X,d) be a complete metric space and let Y eX. Then (Y, d) is a complete metric space iff Y is closed in X. Proof. It is left as an exercise to show that (Y,d) is complete whenever Y is a closed subset. Now assume (Y,d) to be complete; let Xo be a limit point of Y. Then there is a sequence {y n} of points in Y such that Xo = limyn' Hence {y n} is a Cauchy sequence (Exercise 5) and must converge to a point Yo in Y, since (Y,d) is complete. It follows that Yo=xo and so Y contains all its limit points. Hence Y is closed by Proposition 3.4. •

20

Metric Spaces and the Topology of C

Exercises 1. Prove Proposition 3.4. 2. Furnish the details of the proof of Proposition 3.8. 3. Show that diam A = diam A - . 4. Let Zn' Zbe points in IC and let d be the metric on 1C00- Show that IZn - zi -+ 0 if and only if d(zn' z) -+ O. Also show that if IZnl -+ 00 then {zn} is Cauchy in IC",. (Must {zn} converge in ICC()?) 5. Show that every convergent sequence in (X, d) is a Cauchy sequence. 6. Give three examples of non complete metric spaces. 7. Put a metric don IR such that Ixn-xi -+ 0 if and only if d(xn' x) -+ 0, but that {xn} is a Cauchy sequence in (IR, d) when IXnl -+ 00. (Hint: Take inspiration from 1C 8. Suppose {xn} is a Cauchy sequence and {xnJ is a subsequence that is convergent. Show that {xn} must be convergent. 00 . )

§4. Compactness The concept of compactness is an extension of the benefits of finiteness to infinite sets. Most properties of compact sets are analogues of properties of finite sets which are quite trivial. For example, every sequence in a finite set has a convergent subsequence. This is quite trivial since there must be at least one point which is repeated an infinite number of times. However the same statement remains true if "finite" is replaced by "compact." 4.1 Definition. A subset K of a metric space X is compact if for every collection ~ of open sets in X with the property

4.2

Kc

U {G:GE~},

there is a finite number of sets G 1 , ••• , Gn in ~ such that K c G 1 U G 2 U ... U Gn • A collection of sets ~ satisfying (4.2) is called a cover of K; if each member of ~ is an open set it is called an open cover of K. Clearly the empty set and all finite sets are compact. An example of a non compact set is D = 2, 3, ... , then {G 2 , G 3 , finite subcover.

{z E IC: Izi ••• }

< I}. If G. =

{Z: Izi

< 1-

~} for n =

is an open cover of D for which there is no

4.3 Proposition. Let K be a compact subset of X; then: (a) K is closed; (b) ifF is closed and Fe K then F is compact. Proof To prove part (a) we will show that K

= K-. Let

Xo E

K-; by Pro-

position 1.13(0, B(xo; €) n K =f. D for each € > O. Let G. = X and suppose that

Xo

!# K. Then each Gn is open and K c

-B(xo;~)

U G. (because n 00

00

.-1

.-1

21

Compactness

B(XO;~) =

Kc UG n~1

n•

{xo}). Since

But G1 c G2

gives that B( x 0;

~) n

C

Kis compact there is an integer m such that so that K G = X-B(XO;~)' But this m ...

C

m

K = D, a contradiction. Thus K = K - .

To prove part (b) let f§ be an open cover of F. Then, since F is closed, {X-F} is an open cover of K. Let G 1 , ••. , Gn be sets in f§ such that K c G l U ... U Gn U (X -F). Clearly, Fe G 1 U ... U Gn and so F is compact. • If .'F is a collection of subsets of X we say that .'F has the finite intersection property (f.i.p.) if whenever {Fl' F 2, ... , Fn} c .'F, Fl n F2 n ... n Fn =1= D. An example of such a collection is {D - G2 , D - G 3 , ••• } where the sets Gn are as in the example preceding Proposition 4.3. f§ U

4.4 Proposition. A set K c X is compact iff every collection .'F of closed subsets of K with the fi.p. has {F: FE.'F} =1= D·

n

Proof Suppose K is compact and .'F is a collection of closed subsets of K having the f.i.p. Assume that {F: FE.'F} = D and let f§ = {X - F: FE.'F}. Then, U {X-F: FE.'F} = X {F: FE.'F} = X by the assumption; in particular, f§ is an open cover of K. Thus, there are F I , •.• ,

n

Fn E.'F such that K

n

c

U(X -

Fk) = X -

k~1

C

n

n Fk. n

But this gives that

nFk n

k~1

X - K, and since each Fk is a subset of K it must be that

contradicts the f.i.p. The proof of the converse is left as an exercise . •

k

n Fk n

k~1

=

D. This

~I

4.5 Corollary. Every compact metric space is complete.

Proof This follows easily by applying the above proposition and Theorem 3.7. •

4.6. Corollary.

If X is compact then every infinite set has a limit point in

X.

Proof Let S be an infinite subset of X and suppose S has no limit points. Let {ai' a 2 , ••• } be a sequence of distinct points in S; then Fn = {an, an + I, ..• } also has no limit points. But if a set has no limit points it contains all its limits points and must be closed! Thus, each Fn is closed and {Fn:

n Fn 00

n ~ I} has the f.i.p. However, since the points ai' a2 , = D, contradicting the above proposition. •

•••

are distinct,

n~ I

4.7 Definition. A metric space (X, d) is sequentially compact if every sequence in X has a convergent subsequence. It will be shown that compact and sequentially compact metric spaces are the same. To do this the following is needed. 4.8 Lebesgue's Covering Lemma.

If (X, d) is sequentially compact and

22

Metric Spaces and the Topology of C

'1J is an open cover of X thell there is an set G in 'lJ with Bex; IE) c G.

E

> 0 such that if x is in X, there is a

Proof The proof is by contradiction; suppose that 'lJ is an open cover of X and no such f' > 0 can be found. In particular, for every integer n there is a

point x" in X such that B(t"n; \

1)' is not contained in any set Gin '1J. Since X

11

is sequentially compact there is a point Xo in X and a subsequence {xIIJ such that Xo = lim x"". Let Go E 'lJ such that Xo E' Go and choose f' > 0 such that B(xo; E) C Go. Now let N be such that d(xo, xnJ < E/2 for all 11k :.?:: N. Let 11, be any integer larger than both Nand 2/E, and let y E' Bexnk ; link)· Then d(x o, y) ~ d(xo, xnJ+d(Xllk' Y) < E/2+ I < E. That is, B(xn,; link) C B(xo; E) C GIl> contradicting the choice of X llk • II There are two ,common misinterpretations of Lebesgue's Covering Lemma; one implies that it says nothing and the other that it says too much. Since 'lJ is an open covering of X it follows that each x in X is contained in some G in 'lJ. Thus there is an E > 0 such that B(x; E) C G since G is open. The lemma. however, gives one E > 0 such that for any x, B(x; E) is contained in some member of 'lJ. The other misinterpretation is to believe that for the E > 0 obtained in the lemma. B(x; fc) is contained in each G in '!J such that xc G.

4.9. Theorem. Let eX, d) be a metric space; then the following are equimlent statements: (a) X is compact; (b) Every infinite set ill X has a limit point; (c) X is sequentially compact; (d) X is complete and for every E > 0 there are a finite number of points x I, • . . , xn in X such that

U B(x II

X =

k;

fc).

k= 1

(The property mentioned in Cd) is called total boundedness.) Proof That (a) implies (b) is the statement of Corollary 4.6. (b) implies (c): Let {XII} be a sequence in X and suppose, without loss of generality, that the points Xj, X2 • . . . are all distinct. By (b), the set {Xl' x 2 , ••• } has a limit point Xo' Thus there is a point x n , (B(xo; 1); similarly, there is an integer 112 > 111 with x n , E' B(x 0: 1/2). Continuing we get integers n! < 112 < ... , with X"k E B(xo; 11k). Thus, -'0 = lim x nk and X is sequentially compact. (c) implies (d): To see that X is complete let {xn} be a Cauchy sequence, apply the definition of sequential compactness, and appeal to Exercise 3.8. Now let € > 0 and fix Xl E' X. If X = B(x!; E) then we are done; otherwise choose X 2 EX -B(x!; E). Again, if X = B(x!; E) U B(x 2 ; E) we are done;

Compactness

23

if not, let x 3 E X - [B(x 1 ; E) sequence {xn} such that

e)]. If this process never stops we find a

U B(x 2;

X.+ I EX -

n

U B(Xk; 1:-).

k= I

But this implies that for n # m, d(xn, xm) ;;.:: e > O. Thus {xn} can have no convergent subsequence, contradicting (c). (d) implies (c): This part of the proof will use a variation of the "pigeon hole principle." This principle states that if you have more objects than you have receptacles then at least one receptacle must hold more than one object. Moreover, if you have an infinite number of points contained in a finite number of balls then one ball contains infinitely many points. So part (d) says that for every E > 0 and any infinite set in X, there is a point Y E X such that B(y; e) contains infinitely many points of this set. Let {x,,} be a sequence of distinct points. There is a point YI in X and a subsequence {x~t)} of {xn} such that {x~l)} c B(YI; I). Also, there is a point Yz in X and a subsequence {X~2)} of {x~l)} such that {x~)} c B(yz; t). Continuing, for each integer k ;;.:: 2 there is a point Yk in X and a subsequence {~k)} of {x~k-l)}suchthat {X~k)} c B(Yk; Ilk). LetFk = {x~k)}-;thendiamFk:5 21k

n Fk o:J

and Fl ~ F2 ~ .... By Theorem 3.7,

=

{x o }. We claim that X~k)--)oo

k~1

Xo (and {X~k)} is a subsequence of {x.}). In fact, Xo E Fk so that d(xo, 4k » :5 diam Fk :5 21k, and Xo = lim X~k). (c) implies (a): Let '!f be an open cover of X. The preceding lemma gives an E > 0 such that for every x E X there is a G in ~ with R(x; e) c G. Now

(c) also implies (d); hence there are points X =

n

U B(x

k;

k~1

Hence X

=

n

€). Now for 1 :5 k :5 n there is a set Gk

UG

k;

that is, {G 1 ,

••• ,

• , Xn

XI>" E

in X such that

'!f with B(xk ; e) c Gk •

Gn } is a finite subcover of '!f . •

k~1

4.10 Heine-Borel Theorem. A subset K oflR" (n ;;.:: 1) is compact

iff K is closed

and bounded. Proof. If K is compact then K is totally bounded by part (d) of the preceding theorem. It follows that K must be closed (Proposition 4.3); also, it is easy to show that a totally bounded set is also bounded. Now suppose that K is closed and bounded. Hence there are real numbers al, ... ,an and bl, ... ,bn such that KcF=[al,b;lX ... X [a",bnl. If it can be shown that F is compact then, because K is closed, it follows that K is compact (Proposition 4.3(b». Since Rn is complete and F is dosed it follows that F is complete. Hence, again using part (d) of the preceding theorem we need only show that F is totally bounded. This is easy although somewhat "messy" to write down. Let (>0; we now will write F as the union of n-dimensional rectangles each of diameter less than f. m

After doing this we will have FeU B (x,,; £) where each 1 btl x ... x [an, bnl c

[t

~n

k=l

(qk_Pk)2]t.

and let

€

> 0; use Exercise 2 to

show that there are rectangles R l , ... , Rm such that F

=

m

U Rk and

k-I

diam

Rk < € for each k. If Xk E Rk then it follows that Rk c B(xk; f). 4. Show that the union of a finite number of compact sets is compact. 5. Let X be the set of all bounded sequences of complex numbers. That is, {xn } E X iff sup {lx.l: n ~ I} < 00. If x = {x.} and Y = {Yn}, define d(x, y) = sup {Ix. - Y.I : n ~ I}. Show that for each x in X and € > 0, B(x; €) is not totally bounded although it is complete. (Hint: you might have an easier time of it if you first show that you can assume x = (0, 0, ... ).) 6. Show that the closure of a totally bounded set is totally bounded.

§5. Continuity One of the most elementary properties of a function is continuity. The presence of continuity guarantees a certain degree of regularity and smoothness without which it is difficult to obtain any theory of functions on a metric space. Since the main subject of this book is the theory of functions of a complex variable which possess derivatives (and so are continuous), the study of continuity is basic.

° °

5.1 Definition. Let eX, d) and (0, p) be metric spaces and let f: X -+ be a function. If a E X and W EO, then limf(x) = w if for every € > there is a

°such that p(f(x), w) < whenever °< d(x, a) < S. The function f is continuous at the point a if limf(x) = f(a). Iffis continuous at each point of X-wl

S>

€

X-WI

X then f is a continuous function from X to 0.

5.2 Proposition. Let f: (X, d) -+ (0, p) be a function and a E X, Ct = f(a). The following are equivalent statements: (a) f is continuous at a; (b) For every € > O,f-l(B(Ct; €» contains a ball with center at a; (c) Ct = limf(x.) whenever a = lim x •.

The proof will be left as an exercise for the reader. That was the last proposition concerning continuity of a function at a

Continuity

25

point. From now on we will concern ourselves only with functions continuous on all of X. 5.3 Proposition. Let f: (X, d) ~ (0, p) be a function. The following are equivalent statements: (a) fis continuous; (b) If Ll is open in 0 then f-l(Ll) is open in X; (c) -If r is closed in 0 then f- 1 (r) is closed in X. Proof (a) implies (b): Let Ll be open in 0 and let x Ef-l(Ll). If w = f(x) then w is in Ll; by definition, there is an E > 0 with B(w ; E) C Ll. Since f is continuous, part (b) of the preceding proposition gives a 0 > 0 with B(x; 0) c f-l(B(w; E» c f-l(Ll). Hence, f-l(Ll) is open. (b) implies (c): If reO is closed then let Ll = o-r. By (b),J-l(Ll) = X-f-l(r) is open, so thatf-l(r) is closed. (c) implies (a): Suppose there is a point x in X at whichfis not continuous. Then there is an E > 0 and a sequence {xn } such that p(f(xn ), f(x» ~ E for every n while x = lim x n • Let r = O-B(f(x); E); then r is closed and each Xn is inf-l(f). Since (by (c»f-l(r) is closed we have x Ef-l(r). But this implies p(f(x), f(x» ~ E > 0, a contradiction . • The following type of result IS probably well understood by the reader and so the proof is left as an exercise.

5.4 Proposition. Let f and g be continuous functions from X into C and let

IX, (3 E IC. Then IXf+ (3g and fg are both continuous. Also, fIg is continuous provided g(x) of- 0 for every x in X. 5.5 Proposition. Letf: X ~ Yandg: Y ~Z be continuousfunctions. Then gof (where go f(x) = g(f(x») is a continuous function from X into Z. Proof If U is open in Z then g-l(U) is open in Y; hence,J-l(g-l(U» (gaf)-I(U) is open in X . •

=

5.6 Definition. A function f: (X, d) ~ (0, p) is uniformly continuous if for every E > 0 there is a 0 > 0 (depending only on E) such that p(f(x),J(y» < E whenever d(x, y) < o. We say thatfis a Lipschitzfunction if there is a constant M > 0 such that p(f(x),J(y» :s; Md(x, y) for all x and y in X. It is easy to see that every Lipschitz function is uniformly continuous. In fact, if E is given, take 0 = ElM. It is even easier to see that every uniformly continuous function is continuous. What are some examples of such functions? If X = 0 = IR then f(x) = x 2 is continuous but not uniformly continuous. If X = 0 = [0, 1] then f(x) = xt is uniformly continuous but is not a Lipschitz function. The following provides a wealthy supply of Lipschitz functions. Let A be a non-empty subset of X and x E X; define the distance from x to the set A, d(x, A), by d(x, A) = inf {d(x, a): a E A}. 5.7 Proposition. Let A be a non-empty subset of X; then: (a) d(x, A) = d(x, A -);

26

Metric Spaces and the Topology of C

(b) d(x, A) = 0 iff x E A - ; (c) Id(x, A) - dey, A)I ::::; d(x, y) for all x, y in X. Proof (a) If A c B then it is clear from the definition that d(x, B) ::::; d(x, A). Hence, d(x, A -) ::::; d(x, A). On the other hand, if E > 0 there is a point y in A - such that d(x, A -) ~ d(x, y) - E/2. Also, there is a point a in A with dey, a) < E/2. But Id(x, y) - d(x, a)1 ::::; dey, a) < E/2 by the triangle inequality. In particular, d(x, y) > d(x, a)-E/2. This gives, d(x, A-) ~ d(x, a)-E ~ d(x, A) -E. Since E was arbitrary d(x, A -) ~ d(x, A), so that (a) is proved. (b) If x E A - then 0 = d(x, A -) = d(x, A). Now for any x in X there is a minimizing sequence {an} in A such that d(x, A) = lim d(x, an). So if d(x, A) = 0, lim d(x, an) = 0; that is, x = lim an and so x E A -. (c) For a in A d(x, a) ::::; d(x, y)+d(y, a). Hence, d(x, A) = inf {d(x, a): a E A} ::::; inf {d(x, y) + dey, a): a E A} = d(x, y) + dey, A). This gives d(x, A)dey, A) ::::; d(x, y). Similarly dey, A) - d(x, A) ::::; d(x, y) so the desired inequality follows . • Notice that part (c) of the proposition says that f: X --+ IR defined by f(x) = d(x, A) is a Lipschitz function. If we vary the set A we get a large supply of these functions. It is not true that the product of two uniformly continuous (Lipschitz) functions is again uniformly continuous (Lipschitz). For example, f(x) = x is Lipschitz butfIis not even uniformly continuous. However ifbothfand g are bounded then the conclusion is valid (see Exercise 3). Two of the most important properties of continuous functions are contained in the following result.

5.8 Theorem. Let f: (X, d) --+ (0, p) be a continuous function. (a) If X is compact then f(X) is a compact subset of O. (b) If X is connected then f(X) is a connected subset of O. Proof To prove (a) and (b) it may be supposed, without loss of generality, that f(X) = O. (a) Let {w n } be a sequence in 0; then there is, for each n ~ 1, a point Xn in X with Wn = f(x n ). Since X is compact there is a point x in X and a subsequence {xnJ such that x = lim x nk ' But if W = f(x), then the continuity off gives that W = lim w nk ; hence 0 is compact by Theorem 4.9. (b) Suppose ~ c 0 is both open and closed in 0 and that ~ # D. Then, because/eX) = 0, D # /-1(~); also,f-l(~) is both open and closed because / is continuous. By connectivity, /-1(~) = X and this gives 0 = ~. Thus, 0 is connected . •

5.9 Corollary. If f: X --+ 0 is continuous and K c X is either compact or connected in X then f(K) is compact or connected, respectively, in O.

5.10 Corollary. Iff: X

~ R is continuous and X is connected then f( X) is an interval. This foIIows from the characterization of connected subsets of IR as intervals.

Continuity

27

5.11 Intermediate Value Theorem. Iff: [a, b] --+ IR is continuous andf(a) :::; :::; feb) then there is a point x, a :::; x :::; b, with f(x) = g.

g

5.12 Corollary. Iff: X --+ IR is continuous and K c X is compact then there are points Xo and Yo in K withf(x o) = sup {f(x): x E K} andf(yo) = inf {f(x): XEK}. Proof If IX = sup {f(x): x E K} then IX is in f(K) because f(K) is closed and bounded in IR. Similarly f3 = inf {f(x): x E K} is in f(K) . • 5.13 Corollary. If K c X is compact and f: X are points x 0 and Yo in K with If(xo) I = sup {If(x)l: x

E

K} and If(Yo)1

--+

C is continuous then there

= inf {If(x) I: x E K}.

Proof This corollary follows from the preceding one because g(x) = If(x) I defines a continuous function from X into IR. 5.14 Corollary. If K is a non-empty compact subset of X and x is in X then there is a point y in K with d(x, y) = d(x, K). Proof Define f: X --+ IR by fey) = d(x, y). Then f is continuous and, by Corollary 5.12, assumes a minimum value on K. That is, there is a point y in K with fey) :::; fez) for every Z E K. This gives d(x, y) = d(x, K) . • The next two theorems are extremely important and will be used repeatedly throughout this book with no specific reference to the theorem numbers. 5.15. Theorem. Suppose f: X uniformly continuous.

--+ Q

is continuous and X is compact; then f is

Proof Let E > 0; we wish to find a 8 > 0 such that d(x, y) < 8 implies p(f(x), fey)) < E. Suppose there is no such 8; in particular, each 8 = lin will fail to work. Then for every n ~ I there are points Xn and Yn in X with d(xn> Yn) < lin but p(f(xn), f(Yn)) ~ E. Since X is compact there is a subsequence {xnJ and a point x E X with x = lim x nk . Claim. x = lim Ynk. In fact, d(x, Ynk) :::; d(x, xnJ+ link and this tends to zero as k goes to 00. But if w = f(x), w = lim f(xnJ = lim f(YnJ so that E :::;

p(f(xnJ,J(Ynk))

:::; p(f(xnJ, w)+p(w,J(ynJ)

and the right hand side of this inequality goes to zero. This is a contradiction and completes the proof. • 5.16. Definition. If A and B are non-empty subsets of X then define the distance from A to B, dCA, B), by dCA, B)

= inf {dCa, b): a E A, b E B}.

Notice that if B is the single-point set {x} then dCA, {x}) = d(x, A). If

28

Metric Spaces and the Topology of C

A = {y} and B = {x} then d( {x}, {y}) = d(x, y). Also, if A n B of. [: then dCA, B) = 0, but we can have dCA, B) = 0 with A and B disjoint. The most popular type of example is to take A = {(x, 0): x E SIl} C IFR2 and B = {ex, eX): x E SIl}. Notice that A and B are both closed and disjoint and still dCA, B) = O.

5.17 Theorem. If A and B are non-empty di!Jjoint sets in X with B closed and A compact then dCA, B) > O. Proof DefineJ: ) ( + SIl byJ(x) = d(x, B). Since A n B = ::::J and B is closed, J(a) > 0 for each a in A. But since A is compact there is a point a in A such that 0 < J(a) = inf {f(x): x E A} = dCA, B). III

Exercises 1. 2. X 3.

Prove Proposition 5.2. Show that if J and g are uniformly continuous (Lipschitz) functions from into C then so is J+ g. We say that f: X --+ C is bounded if there is a constant M > 0 with IJ(x)! ::; M for all x in X. Show that if J and g are bounded uniformly continuous (Lipschitz) functions from X into C then so is f'5. 4. Is the composition of two uniformly continuous (Lipschitz) functions again uniformly continuous (Lipschitz)? 5. Supposef: X --+ n is uniformly continuolls; show that if {x n } is a Cauchy sequence in X then {l(xn)} is a Cauchy sequence in £2. Is this still true if we only assume that J is continuous? (Prove or give a counterexample.) 6. Recall the definition of a dense set (1.l4). Suppose that n is a complete metric space and that J: (D, d) .~)- en; p) is uniformly continuous, where D is dense in (X, d). Use Exercise 5 to show that there is a uniformly continuous function g: X -+ 0 with g(x) = J(x) for every x in D. 7. Let G be an open subset of C and let P be a polygon in G from a to b. Use Theorems 5.15 and 5.17 to show that there is a polygon Q c G from a to b which is composed of line segments which are parallel to either the real or imaginary axes. 8. Use Lebesgue's Covering Lemma (4.8) to give another proof of Theorem 5.15. 9. Prove the following converse to Exercise 2.5. Suppose (,\:', d) is a compact metric space having the property that for every E > 0 and for any points a, b in X, there are points Zo, ZI' ••. , Zn in X with Zo = a, z" = b, and d(zk-J' Zk) < E for 1 ~ k ::; n. Then (X, d) is connected. (Hint: Use Theorem 5.17.) 10. Let J and g be continuous functions from (X, d) to (D, p) and let D be a dense subset of X. Prove that if J(x) = g(x) for x in D then J = g. Use this to show that the function g obtained in Exercise 6 is unique. §6. Uniform convergence Let X be a set and (n, p) a metric space and suppose f j~, J2' ... are functions from X into n. The sequence {in} converges uniformly to J-written

29

Uniform convergence

f = u-lim fn-if for every € > 0 there is an integer N (depending on alone) such that p(f(x), fn(x» < € for all x in X, whenever n ~ N. Hence,

€

sup {p(f(x),fn(x»: x E X} ::; € whenever n ~ N. The first problem is this: If X is not just a set but a metric space and each j~ is continuous does it follow thatfis continuous? The answer is yes. 6.1 Theorem. Suppose f,,: (X, d) ---+ (0, p) is continuous for each n and that f = u -lim fn; then f is continuous. Proof Fix Xo in X and E > 0; we wish to find a 0 > 0 such that p(f(xo), f(x» < € when d(xo, x) < O. Sincef= u-limfm there is a functionf" with p(f(x), j~(x» < €/3 for all x in X. Since fn is continuous there is a 0 > 0 such that p(f,,(xo),f,,(x» < €/3 when d(xo, x) < O. Therefore, if d(xo, x) 0 there is an integer N such that I Lan - zl < E n=O whenever m ~ N. The series L an converges absolutely if L lanl converges.

L an

1.1 Proposition. If L an converges absolutely then

Proof Let

E

converges.

> 0 and put Zn = aO+a 1 + ... + an" Since 00

there is an integer N such that L lanl < n=N

IZm-zkl =

m

m

E.

lanl converges

Thus, if m > k ~ N, 00

L ani::::; L lanl::::; L n=k+l n=k+l n=N

I

L

lanl

<

E.

That is, {zn} is a Cauchy sequence and so there is a Z in C with Z = lim Zn. Hence Lan = z . • Also recall the definitions of limit inferior and superior of a sequence in IR. If {an} is a sequence in IR then define lim inf an = lim [inf {an> an+1 ,

••• }]

n~oo

lim sup an = lim [sup {an> an+1 ,

••• }]

n~oo

An alternate notation for lim inf an and lim sup an is lim an and lim an" If = inf {an> an + 1 , ••• } then {b n } is an increasing sequence of real numbers or {- oo}. Hence, lim inf an always exists although it may be ± 00. Similarly lim sup an always exists although it may be ± 00. A number of properties of lim inf and lim sup are included in the exercises of this section. 00 A power series about a is an infinite series of the form L a,,(z-a)n. One n=O of the easiest examples of a power series (and one of the most useful) is the

bn

geometric series

L zn. 00

n=O

For which values of z does this series converge and 30

31

Power series

when does it diverge? It is easy to see that I-z so that

n+ 1

1.2

l+z+ ... +zn

If

Izl

< 1 then 0

= (I-z) (I +z+ . .. +zn),

l-zn+ 1

= --l-z

= lim zn and so the geometric series is convergent with 1 = l-z·

00

L zn o

Izl > I then lim Izln = ro and the series diverges. Not only is this result an archetype for what happens to a general power series, but it can be used to explore the convergence properties of power series.

If

1.3 Theorem. For a given power series L anCz-a)n define the number R, o ::; R ::; 00, by n= 0 R, the terms of the series become unbounded and so the

series diverges; (c) if 0 < r < R then the series converges uniformly on {z: Iz - al ::; r}. Moreover, the number R is the only number having properties (a) and (b). Proof We may suppose that a

= O. If Izl < R there is an r with Izl < r < R.

Thus, there is an integer N such that

~}

But then

the tail end

lanl

<

lanl l !.

<

!r for all n ~

N (because

~ and so Ianz· I < (';IJ for all n ~ N.

~r

>

This says that

I a.znis dominated by the series L (~)n, and since ~ < 1 r r

n=N

the power series converges absolutely for each Izl < R. Now suppose r < R and choose p such that r < p < R. As above, let N be such that

lanl

<

~ for

all n

~

N. Then if

Izl ::;

r, lanznl ::;

GJ

and

(~) < 1. Hence the Weierstrass M-test gives that the power series converges Izl ::; To prove (b), let Izi

uniformly on {z:

r}. This proves parts (a) and (c). .

1

Izl

1

> r > R. Hence - < - ; r R from the definition of lim sup, this gives infinitely many integers n with

~<

> R and choose r with

la.ll!n. It follows that lanz·1 >

(';IJ

and, since

(';1)

> 1, these terms

become unbounded . • The number R is called the radius of convergence of the power series.

32

1.4 Proposition. If vergence R, then

L

Elementary Properties and Examples of Analytic Functions

an(z -a)" is a given power series with radius of con-

if this limit exists. Proof Again assume that a = 0 and let ex = lim la./a.+ 11, which we suppose to exist. Suppose that Izi < r < ex and find an integer N such that r < la./an+ti for all n::::: N. Let B= laNlr N ; then la,'HllrN+1 = laN+drr N < laNlr N = B; laN+2Ir N+2 = laN+zlrr N+1 < la N+1 !r N+1 < B;continuingweget

~ B for all n ::::: N. But then lanz" I = lanr"1 EJ: .~ BEJ: for all n ::::: N. ~ ~ Since Izi < r we get that L lanz"1 is dominated by a convergent series and ianr"!

00

n=1

hence converges. Since r < ex was arbitrary this gives that ex ~ R. On the other hand if Izl > r > ex, then lanl < rlan + 11 for all n larger than some integer N. As before, we get jan," 1 ::::: B = jaNrNI for n ::::: N. This

1 gives lanz"1 ::::: B 1 1;1 : which approaches and so R ~ ex. Thus R = ex. III Consider the series

f

n~O

00

as n does. Hence,

L a.z" diverges

z· ; by Proposition 1.4 we have that this series

n!

has radius of convergence 00. Hence it converges at every complex number and the convergence is uniform on each compact subset of C. Maintaining a parallel with calculus, we designate this series by

_

e" = exp z =

~

z"

n= 0

n.

L !'

the exponential series or junction. Recall the following proposition from the theory of infinite series (the proof will not be given).

1.5 Proposition. Let L an and L bn be two absolutely convergent series and put n

Cn

Then

Lc

n

=

L0 akbn-k'

k=

is absolutely convergent with sum

1.6 Proposition. Let L an(z-a)n and I bll(z-at be power series with radius oj convergence::::: r > O. Put n

Cn

=

I

k~'

()

akbn-k;

then both power series I (an+b n ) (z-a)" and I cn(z-a)" have radius of convergence ::::: r, and

Analytic functions

I

for

Iz-al

(an + bn) (z-ay

I

=

cn(z-aY =

33

[I an(z-ay+ I bn(z-ay] [I an(z-ay] [I bn(z-aY]

< r.

Proof We only give an outline of the proof. If 0 < s < r then for lan+bnl Izln ~ lanlsn+ Ibnls n < 00; 1e,,1 Izln ~ (I Ibnls n) < 00. From here the proof can easily be completed . •

we get

I

I

I

I

Izl

~ s,

(I lanls")

Exercises l. Prove Proposition 1.5. 2. Give the details of the proof of Proposition 1.6. 3. Prove that lim sup (a" + b,,) ~ lim sup a" + lim sup b" and lim inf (a" + b,,) 2. lim inf a" + lim inf b" for bounded sequences of real numbers {a,,} and {b ll } . 4. Show that lim inf a n ~ lim sup an for any sequence in ~. 5. If {an} is a convergent sequence in ~ and a = lim an' show that a = lim inf an = lim sup an. 6. Find the radius of convergence for each of the following power series:

I

00

(a)

a"zn, a

E

C; (b)

n~O

co

I

an2 z", a

I

00

E

n~O

C; (c)

I

00

knz", k an integer #-0; (d)

,,~O

zn!.

,,~O

7. Show that the radius of convergence of the power series

i

(-nI)n zn(n+l)

n=1

is 1, and discuss convergence for z = 1, -1, and i. (Hint: The nth coefficient of this series is not (-IY/n.) §2. Analytic functions In this section analytic functions are defined and some examples are given. It is also shown that the Cauchy-Riemann equations hold for the real and imaginary parts of an analytic function. 2.1 Definition. If G is an open set in C and f: at a point a in G if

G~C

then f is differentiable

a _+---,h)'-----=-P---'-(a--,-) lim f'---.:(ch~O

h

exists; the value of this limit is denoted by f'(a) and is called the derivative of . f at a. Iff is differentiable at each point of G we say that f is differentiable on G. Notice that if f is differentiable on G then f'(a) defines a function f': G -+ Co Iff' is continuous then we say thatfis continuously differentiable. Iff' is differentiable then f is twice differentiable; continuing, a differentiable function such that each successive derivative is again differentiable is called infinitely differentiable.

(Henceforward, all functions will be assumed to take their values in C unless it is stated to the contrary.)

Elementary Properties and Examples of Analytic Functions

34

The following was surely predicted by the reader. 2.2 Proposition. continuous at a.

If f: G -? IC is differentiable at a point a in G then f is

Proof In fact,

I.iml/(z) -

:~a

.

l(a)1 =

llim I/(z) - 1(~}J.l· [limlz - all Iz - al ::~a Z---"'U

=1/'(a)I·O=O.1III 2.3 Definition. A function I: G--.IC is analytic if I is continuously differentiable on G. It follows readily, as in calculus. that sums and products of functions analytic on G are analytic. Also, if I and g are analytic on G and G I is the set of points in G where g doesn't vanish, thenl/g is analytic on G I . Since constant functions and the function z are clearly analytic it follows that all rational functions are analytic on the complement of the set of zeros of the denominator. Moreover, the usual laws for differentiating sums, products, and quotients remain valid. 2.4 Chain Rule. Let I and g be analytic on G and Q respective(v and suppose I(G)c Q. Then gal is analytic on G and (gof)'(z) = g'(f(z) )/'(z) for all z in G.

Proof. Fix Zo in G and choose a positive number r such that B (zo; r) c G. We must show that if O 0 be arbitrary except

for the restriction that Rew; 8) c B(O; r). (We will further restrict 8 later in the proof.) Let z E B( w; S); then

2.8

jJ~)~f(w) z-w

_ g(w) =

[S.(Z) -sn(w) z-w

S~(W)J + [s~(w) -

g(w)]

+ [!!.~_~Z) - R_~\~!J z-)t!

Now

But

Hence, Rn(Z)-Rn(W)I:::; I z-w Since r < R,

L'" laklkr

k- 1

I

iaklkr k -

1

k=II+1

converges and so for any

E'

> 0 there is an integer

k~l

N J such that for n 2:: N J

< ~ IRn(Z)-Rn(W)1 z-w 3

(z

E

B(w; 8».

Also, lim s~(w) = g(w) so there is an integer N z such that Is~(w)-g(w)1 <

j

whenever n 2:: N 2 • Let n = the maximum of the two integers Nl and N z .

37

Analytic functions

Then we can choose 8 > 0 such that

_ ISn(Z)-Sn(W) z-w

S~(W)I <

-<

3

whenever 0 < [z - wi < 8. Putting these inequalities together with equation (2.8) we have that

-=[(~! If(Z)z-w

g(W)\ <

IE

for 0 < [z-w[ < 8. That is,!,(w) = g(w). (c) By a straightforward evaluation we get f(O) = f(O)(O) = aQ. Using (2.6) (for a = 0), we get f(k)(O) = k !ak and this gives formula (2.7). II

If the series L anCz -- a)n has radius of convergence R > 00

2.9 Corollary.

=

fez)

0 then

n~O

00

L an(z -

a)" is analytic in B(a; R).

n =0

Hence. expz =

L zn / n! is analytic in C. Before further examining the oc

n=O

exponential function and defining cosz and sinz. the following result must be proved. 2.10 Proposition. If G is open and connected and f:G -» C is differentiable with !'(z) = 0 for all z in G, then f is constant.

Proof Fix z 0 in G and let Wo = fez 0)' Put A = {z E G: fez) = wo}; we will show that A = G by showing that A is both open and closed in G. Let z E G and let {zn} c A be such that z = lim Zn' Since f(z.) = Wo for each n ~ 1 andfis continuous we getf(z) = Wo, or z E A. Thus, A is closed in G. Now fix a in A, and let IE > 0 be such that BCa; E) c G. If z E B(a; £), set g(t) = f(tz+(l-t)a), 0 ::; t ::; 1. Then g(t)-g(s) t-s

2.11 Thus. if we let

(-H

.

lIm

g(t)-g(s) (t-s)z+{s-t)a (t-s)z+(s-t)a t-s

we get (A.4(b), Appendix A)

g(t) - g(s)

1-5

t-s

,

--- = f (sz+(l-s)a)'(z-a) = O.

That is, g'(s) = 0 for 0 ::; s ::; 1, implying that g is a constant. Hence, fez) = g(l)=g(O) = f(a) = woo That is, BCa; IE) c A and A is also open. II Now differentiate fez) = eZ; we do this by Proposition 2.5. This gives that

!'(z)

=

~ ~ zn-l

n~ln!

=

i

_1_ zn-I

n~l(n-l)!

=

i

~ z" =f(z).

.~on!

Thus the complex exponential function has the same property as its real counterpart. That is d

2.12

--eZ=e" dz

38

Elementary Properties and Examples of Analytic Functions

Putg(z) = eZea-Zfor some fixed ain C; theng'(z) = eZea-z+eZ(_ea-z) = O. Hence g(z) = w for all z in C and some constant w. In particular, using eO = 1 we get w = g(O) = ea. Then eZe a- z = ea for all z. Thus ea +b = eae b for all a and b in IC. This also gives 1 = eZe- Z which implies that eZ # 0 for any z and e- = lje z • Returning to the power series expansion of et , since all the coefficients of this series are real we have exp z = exp z. In particular, for (J a real number we get le wl2 = eiOe- iO = eO = l. More generally, le z I2 = eZe' = ezH = exp (2 Re z). Thus, Z

2.13

iexp z! = exp (Re z).

We see, therefore, that e has the same properties that the real function eX has. Again by analogy with the real power series we define the functions cos z and sin z by the power series Z

cos

Z

= 1-

Z2

2!

Z4

z2n

4!

(2n)!

+- + ... + ( - 1 ) " - + _..

_3

sm z = z

-

3!

z5

+

5!

7

2n- 1

+ ... +(-1)"( -

) + ...

2n - 1 !

Each of the series has infinite radius of convergence and so cos z and sin z are analytic in IC. By using Proposition 2.5 we find that (cos z)' = - sin z and (sin z)' = cos z. By manipulating power series (which is justified since these series converge absolutely) 2.14

- z = 2i 1 (iz sm e _... e - iz)

cos z = J( e iz + e - iz)

This gives for z in C, cos 2 z + sin 2 z 2.15 In particular if we let z Hence, for z in IC

e iz =

=

=

1 and

cos z+ i sin z.

a real number

(J

in (2.15) we get e iO = cis

(J.

2.16 where B=argz. Since eX+IY=exe'Y we have lezl=exp(Rez) and arge z = Imz.

A functionfisperiodic with period c iff(z+c)=f(z) for all z in IC. If c is a period of e Z then eZ=ez+c=ezec implies that e C = 1. Since I =lecl= expRe(c), Re(c)=O. Thus c=i(} for some B in IR. But !=ec=eil1=cosB+ i sin () gives that the periods of e Z are the multiples of 2wi. Thus, if we divide the plane into infinitely many horizontal strips by the lines Imz = w(2k - I), k any integer, the exponential function behaves the same in each of these strips. This property of periodicity is one which is not present in the real exponential function. Notice that by examining complex functions we have demonstrated a relationship (2.15) between the exponential function and the trigonometric functions which was not expected from our knowledge of the real case.

Analytic functions

39

Now let us define log z. We could adopt the same procedure as before and let log z be the power series expansion of the real logarithm about some point. But this only gives log z in some disk. The method of defining the logarithm as the integral of t- 1 from 1 to x, x > 0, is a possibility, but proves to be risky and unsatisfying in the complex case. Also, since e" is not a one-one map as in the real case, log z cannot be defined as the inverse of e". We can, however, do something similar. We want to define log w so that it satisfies w = e" when z = log w. Now since eZ =f. 0 for any z we cannot define log O. Therefore, suppose e" = w and w =f. 0; if z = x+iy then Iwl = e" and y = arg w+211'k, for some k. Hence 2.17

{log Iwl +i(arg w+211'k): k is any integer}

is the solution set for e"

= w. (Note that log Iwl is the usual real logarithm.)

2.18 Definition. If G is an open connected set in C and /: G -+- C is a continuous function such that z = exp /(z) for all z in G then / is a branch 0/ the logarithm. Notice that 0, G. Suppose / is a given branch of the logarithm on the connected set G and suppose k is an integer. Let g(z) = /(z) +211'ki. Then exp g(z) = exp/(z) = z, so g is also a branch of the logarithm. Conversely. if/and g are both branches of log z then for each z in G,/(z) = g(z) +211'ki for some integer k, where k depends on z. Does the same k work for each z in G? The answer is

yes. In fact, if h(z)

1 .[/(z)-g(z)] = -2 11'1

then h is continuous on G and h(G)

C I, the integers. Since G is connected, h(G) must also be connected (Theorem II. 5.8). Hence there is a k in I with/(z) + 211'ki = g(z) for all z in G. This gives

2.19 Proposition. If G c C is open and connected and f is a branch of log z on G then the totality 0/ branches 0/ log z are the functions /(z) + 211'ki, k e Z. Now let us manufacture at least one branch of log z on some open connected set. Let G=C-{ZEIR:Z:S:O}; that is, "slit" the plane along the negative real axis. Clearly G is connected and each z in G can be uniquely represented by z = Izle" where -11' < 8 < 11'. For 8 in this range, define f(re " ) = log r+i8. We leave the proof of continuity to the reader (Exercise 9). It follows that/is a branch ofthe logarithm onG. Is/analytic? To answer this we first prove a general fact. 2.20 Proposition. Let G and n be open subsets of C. Suppose that /: G -+- C and g: n -+- C are continuous functions such that /(G) c nand g(f(z» = z for all z in G. If g is differentiable and g'(z) =f. 0, / is differentiable and

I'(z) =

g'~(Z»

40

Elementary Properties and Examples of Analytic Functions

If g is analytic,

f

is analytic.

C such that II # 0 and a+h g(f(a» and a+1I = g(f(a+h» impliesf(a) # f(a+h). Also Proof. Fix a in G and let

hE

g(f(a+

E

G. Hence a

=

h» - g(f(a»

------""-------------_.. _-

h

g(j{a+h»-g(f(a» ---. __

.----------------- - ---

f(a +h) -./(a)

---

. f(a+h)-f(a) --------h

Now the limit of the left hand side as h --+ 0 is, of course, 1; so the limit of the right hand side exists. Since lim [/(a+I1)-f(a)J = 0, h~O

. g(f(a + 11» - g(f(a» = g ,(j(a». . Itm ............................. f(a +h) - ./(a)

h·. 0

Hence we get that . l(a+17)-/(a)

I!m--·--··-

h~O h exists since g'(f(a» # 0, and I = g'(j(a»f'(a). Thus,f'(z) = [g'(f(z))J -1. If g is analytic then g' is continuous and this gives that f is analytic. III

2.21 Corollary. A branch of the logarithm function is analytic and its derivative is

Z-I.

We designate the particular branch of the logarithm defined above on

C - {z E ~: z< O} to be the principal branch of the logarithm. If we write log z as a function we will always take it to be the principal branch of the logarithm unless otherwise stated. Iff is a branch of the logarithm on an open connected set G and if b in C is fixed then deline g: G --+ C by g(z) = exp (bICz». If b is an integer, then g(z) = z/'. In this manner we define a branch of Zb, b in C, for an open connected set on which there is a branch of log z. If we write g(z) = Zb as a function we will always understand that Zb = exp (b log z) where log z is the principal branch of the logarithm; Zb is analytic since log z is. As is evident from the considerations just concluded, connectedness plays an important role in analytic function theory. For example, Proposition 2.W is false unless G is connected. This is analogous to the role played by intervals in calculus. Because of this it is convenient to introduce the term "region." A region is an open connected subset of the plane. This section concludes with a discussion of the Cauchy-Riemann equations. Let f: G --+ C be analytic and let u(x, y) = Re f(x + iy), vex, y) = 1m ./(x+iy) for x+iy in G. Let us evaluate the limit

f

, . I(z+h) - fez) (z) = hm -------h-O h

in two different ways. First let h and h real we get

--+

0 through real values of h. For h # 0

41

Analytic functions

f(z+h)-f(z) h

f(x+h+iy) -f(x+iy) h

= Letting h

-+

u(x+h, y)-u(x, y) h

. v(x+h, y)-v(x, y) h

+1

0 gives ou F(z) = ox (x, y)

2.22

Now let h h real,

-+

ov

+ i ox (x, y)

0 through purely imaginary values; that is, for h t= 0 and

f(z+ ih) - fez) ih

. u(x, y+h)-u(x, y) h

-l

+

vex, y+h)-v(x, y) h

Thus, 2.23

F(z)

=

ou (x, y) oy

-i -

ov (x, y) oy

+-

Equating the real and imaginary parts of (2.22) and (2.23) we get the Cauchy-Riemann equations ou ox

2.24

ov oy

-=-

and

ou oy

ov ox

-=--

Suppose that u and v have continuous second partial derivatives (we will eventually show that they are infinitely differentiable). Differentiating the Cauchy-Riemann equations again we get

Hence, 02U

2.25

o2U

-+--0 8x 2 oy2 - .

Any real-valued function with continuous second derivatives satisfying (2.25) is said to be harmonic. In a similar fashion, v is also harmonic. We will study harmonic functions in Chapter X. Let G be a region in the plane and let u and v be functions defined on G with continuous partial derivatives. Furthermore, suppose that u and v satisfy the Cauchy-Riemann equations. Ifj(z)=u(z)+iv(z) thenjcan be shown to be analytic in G. To see this, let z = x + ry E G and let B (z; r) c G. If h=s+it E B(O;r) then u(x+s, y+t)-u(x, y) = [u(x+s, y+t)-u(x, y+t)]+[u(x, y+t)-u(x, y)]

Applying the mean value theorem for the derivative of a function of one variable to each of these bracketed expressions, yields for each s+it in B(O; r) numbers SI and 11 such that ISll < lsi and Itt! < It I and 2.26

{

u(x+s, y+t) -u(x, y+t) = u..,(x+s 1 , y+t)s u(x, y+t)-u(x, y) = u,(x, y+( 1 )t

Elementary Properties and Examples of Analytic Functions

42

Letting G for some interval [a, bJ in IR. If y'(t) exists for each t in [a, b] and y': [a, b] -> C is continuous then y is a smooth path. Also y is piecewise smooth if there is a partition of [a, b], a = to < t[ < ... < t" = b, such that y is smooth on each subinterval [f}_I' 1 ~j ~ n. To say that a function y: la, b] -> C has a derivative y'(t) for each point t in [a, b] means that

tJ,

.

y{l+h)-y(t)

l~ _·······)-1- _... =Y'(l)

exists for a < t < b and that the right and left sided limits exist for t = a and t = b, respectively. This is, of course, equivalent to saying that Rey and 1m y have a derivative (see Appendix A). Suppose y:[a,b]-,G is a smooth path and that for some to in (a,b), y'(to) 7'" O. Then y has a tangent line at the point zo= yUo). This line goes through the point Zo in the direction of (the vector) y'(lo); or, the slope of the line is tan(argy'(to». If y, and Y2 are two smooth paths with y,«(,)= Y2(t2)=zO and y;(I,)7"'O, y;(l2)7"'0, then define the angle between the paths y, and Y2 at Zo to be

Elementary Properties and Examples of Analytic Functions

46

Suppose y is a smooth path in G andJ: G-C is analytic. Then (1= Joy is also a smooth path and (1'(t)=f'(y(t»y'(t). Let zo=y(to), and suppose that y'(to)*O and f'(zo)*O; then (1'(to)*O and arg(1'(to)=argJ'(zo)+ argy'(to). That is,

3.2

arg(1'(to) - arg y'( to) = argJ'(zo)·

Now let YI and Y2 be smooth paths with YI(t I)= yi t 2) = Zo and y;(tI)* 0* y;(t2); let (11 = JOYI and (12 = J O Y2. Also, suppose that the paths YI and Y2 are not tangent to each other at zo; that is, suppose y;UI)*y;(t2). Equation (3.2) gives

3.3 This says that given any two paths through z 0, f maps these paths onto two paths through Wo = f(zo) and, whenJ'(zo) 'f 0, the angles between the curves are preserved both in magnitude and direction. This summarizes as follows. 3.4 Theorem. Iff: G - IC is analytic then f preserves angles at each point Zo of G where J'(zo) 'f O. A functionf: G - IC which has the angle preserving property and also has . If(z) - f(a) I I1m z-a Iz-al

existing is called a conformal map. If f is analytic and J'(z) 'f 0 for any z then f is conformal. The converse of this statement is also true. If fez) = e Z then f is conformal throughout IC; let us look at the exponential function more closely. If z = c+iy where c is fixed thenf(z) = re iy for r = eC • That is, f maps the line x = c onto the circle with center at the origin and of radius eC • Also, f maps the line y = d onto the infinite ray {reid: 0 < r < co} .

--- ---.

."i

di

e'

-------

~-.-....:

-Tr'

----

We have already seen that e Z is one-one on any horizontal strip of width 0 z-a - Iz-al 2 '

-.~

so that z* lies on the ray {a+t(z-a): 0 < t < oJ} from a through z. Using the fact that Iz - al lz* - al = R2 we can obtain z* from z (if z lies inside r) as in the figure below. That is: Let L be the ray from a through z. Construct

a--~----~------------~~z* z

a line P perpendicular to L at z and at the point where P intersects r construct the tangent to r. The point of intersection of this tangent with L is the point z*. Thus, the points a and Cf) are symmetric with respect to r. 3.19 Symmetry Principle. If a Mijbius transformation T takes a circle 1'1 onto the circle 1'2 then any pair of points symmetric with respect to 1'1 are mapped by T onto a pair of points symmetric with respect to r 2'

52

Elementary Properties and Examples of Analytic Functions

Proof Let =2, Z 3, respect to r 1 then

Z4 E

r 1;

it follows that if z and z* are symmetric with

(Tz*, Tz 2 , Tz 3 , Tz 4 ) = (z*,

=

Z2, Z3, Z4)

(Tz, Tz 2 , Tz 3 , Tz 4 )

by Proposition 3.8. Hence Tz* and Tz are symmetric with respect to r 2 • • Now we will discuss orientation for circles in Coo; this will enable us to distinguish between the "inside" and "outside" of a circle in Coo. Notice that on Coo (the sphere) there is no obvious choice for the inside and outside of a circle. 3.20 Definition. If r is a circle then an orientation for r is an ordered triple of points (ZI' Z2, Z3) such that each Zj is in r. Intuitively, these three points give a direction to r. That is we "go" from ZI to Z2 to Z3. If only two points were given, this would, of course, be ambiguous. az+b Let r = ~ and let z\> Z2' Z3 E~; also, put Tz = (z, ZI' Z2, Z3) = - - . ez+d Since T(~oo) = ~oo it follows that a, b, e, d can be chosen to be real numbers (see Exercise 8). Hence,

az+b Tz=-ez+d az+b lez+dl 2

•

(ez+d)

1

lez+dl 2 [aelzI 2 +bd+bez+adz]

Hence, 1m (z,

(ad-be)

ZI' Z2' Z3)

= lez+dl 2 1m z.

Thus, {z: 1m (z, z l' Z 2, z 3) < o} is either the upper or lower half plane depending on whether (ad-be) > 0 or (ad-be) < O. (Note that ad-be is the "determinant" of T.) Now let r be arbitrary, and suppose that z[, Z2' Z3 are on r; for any Mobius transformation S we have (by Proposition 3.8) {z: 1m (z,

Z[, Z2' Z3)

> O} = {z: 1m (Sz,

=

S-1

SZI' SZ2, SZ3)

{z: 1m (z,

> O}

SZI' SZ2, SZ3)

> O}

In particular, if S is chosen so that S maps r onto ~('" then {z: 1m (z, ZI' z 2, z 3) > O} is equal to S - 1 of either the upper or lower half plane. If (ZI' Z2' Z3) is an orientation of r then we define the right side of r (with respect to (ZI' Z2' Z3» to be {z: 1m (z,

ZI' Z2' Z3)

> O}.

Analytic functions as mappings. Mobius transformations

53

Similarly, we define the left side of l' to be

{z: 1m (z,

ZI' Z2, Z3)

< O}.

The proof of the following theorem is left as an exercise.

3.21 Orientation Principle. Let 1'1 and 1'2 be two circles in C oc and let T be a Mobius tran.~formation such that T(1' 1) = 1'2' Let (Zl, Z2, Z3) be an orientation for r l ' Then T takes the right side and the left side of 1\ onto the right side and left side of r 2 Ivith respect to the orientation (Tz 1, Tz 2 , Tz 3)' Consider the orientation (l, 0, co) of R By the definition of the cross ratio, (z, 1, 0, (0) = z. Hence, the right side of IR with respect to (1, 0, (0) is the upper half plane. This fits our intuition that the right side lies on our right as we walk along IR from 1 to 0 to co. As an example consider the following problem: Find an analytic function f: G -+ IC, where G = {z: Re z > O}, such thatf(G) = D = {z: Izl < I}. We solve this problem by finding a Mobius transformation which takes the imaginary axis onto the unit circle and, by the Orientation Principle, takes G onto D (that is, we must choose this map carefully in order that it does not send G onto {z: Izl > 1 If we give the imaginary axis the orientation ( - i, 0, i) then {z: Re z > O} is on the right of this axis. In fact,

n.

2z

(z, -I, 0, i) = -~; Z-l

2z z+i

z-iz+i 2

= -- .

Iz-W

(!zI2+iz) I

Hence, {z: 1m (z, -I, 0, i) > O} = {z: 1m (iz) > O} = {z: Re z > O}. Giving r the orientation (-i, -1, i) we have that D lies on the right of r. Also, 2i z+ I . (z - i -1 i) = - ..~...~- . , , , i-I z-i If 2z. Sz = Z-l

and

Rz

( ;2i- ) =

1-1

(zz-z+

1) --.

then T = R - 1 S maps G onto D (and the imaginary axis onto f). By algebraic manipUlations we have

z-l z+1

Tz=-····

e"-I

Combining this with previous results we have that g(z) = ---1 maps

e-+

the infinite strip {z: 11m zl < 71"/2} onto the open unit disk D. (It is worth e"-1 mentioning that eX+! = tanh (z/2).)

S4

Elementary Properties and Examples of Analytic Functions

Let G 1, G 2 be open connected sets; to try to find an analytic functionJ such thatJ(G 1) = Gz we try to map both G 1 and G z onto the open unit disk. If this can be done, J can be obtained by taking the composition of one function with the inverse of the other. /L

f,

/

/

/

/

/

/

/

As an example, let G be the open set inside two circles r 1 and r 2, intersecting at points a and b (a t= b). Let L be the line passing through a and b and give L the orientation (00, a, b). Then Tz = (z, 00, a, b) =

(z -

a) z-b maps L onto the real axis (Too = 1, Ta = 0, Tb = 00). Since T must map circles onto circles, T maps fl and f2 onto circles through and 00. That is, T(f l ) and T(f2) are straight lines. By the use of orientation we have that T( G) = {w: - a < arg w < a} for some a > 0, or the complement of some such closed sector. By the use of an appropriate power of z and possibly a rotation we can map this wedge onto the right half plane. Now, composing with the map (z - l)(z + 1)-1 gives a map of G onto D = {z:lzl < I}.

°

Exercises 1. Find the image of {z: Re z < 0, 11m zl < 7T} under the exponential function. 2. Do exercise 1 for the set {z: 11m zl < 7T/2}. 3. Discuss the mapping properties of cos z and sin z. 4. Discuss the mapping properties of z· and zl/n for n ~ 2. (Hint: use polar coordinates. ) 5. Find the fixed points of a dilation, a translation and the inversion on Coo. 6. Evaluate the following cross ratios: (a) (7 + i, I, 0, 00) (b) (2, 1- i, 1, 1 + i) (c)(O, 1, i, -I)(d)(i-l, 00, l+i, 0). az+b . 7. If Tz = - - find Z2' Z3' Z4 (m terms of a, b, c, d) such that Tz = (z, cz+d

Analytic functions as mappings. Mobius transformations

az+b B. If Tz = - - show that T(IR",) cz+d real numbers.

55

.

= IR", Iff we can choose a, b, c, d to be

az+b fi nd necessary an d su ffi' . 9. If Tz = --d' Clent cond'Itl0ns that T(r) = r cz+ where r is the unit circle {z: Izl = I}. 10. Let D = {z: Izl < I} and find all Mobius transformations T such that T(D) = D.

11. Show that the definition of symmetry (3.17) does not depend on the choice of points Z2' Z3' Z4' That is, show that if W2, W3, W4 are also in r then equation (3.1B) is satisfied iff (z*, W2, W3, w4) = (z, W2, W3, w4)' (Hint: Use Exercise B.) 12. Prove Theorem 3.4. 13. Give a discussion of the mappingf(z) = -!(z+ liz). 14. Suppose that one circle is contained inside another and that they are tangent at the point a. Let G be the region between the two circles and map G conformally onto the open unit disk. (Hint: first try (z - a)-I.) 15. Can you map the open unit disk conformally onto {z :0< Izl < l}? 16. Map G = C- {z: -1 ::; z ::; I} onto the open unit disk by an analytic function f Can f be one-one? 17. Let G be a region and suppose that f: G - C is analytic such that f( G) is a subset of a circle. Show that f is constant. z-ia lB. Let - 00 < a < b < 00 and put Mz = - . . Define the lines Ll = z-lb {z: 1m z = b}, L2 = {z: 1m z = a} and L3 = {z: Re z = O}. Determine which of the regions A, B, C, D, E, F in Figure 1, are mapped by M onto the regions U, V, W, X, Y, Z in Figure 2.

A

ib

D

u

E

B

w

o

z

x ia

C

F

Figure 1

Figure 2

19. Let a, b, and M be as in Exercise 18 and let log be the principal branch of the logarithm.

56

Elementary Properties and Examples of Analytic Functions

(a) Show that log (Mz) is defined for all z except z = ie, a :S: c :S: b; and if h(z) = 1m [log Mz] then 0 < h(z) < 71' for Re z > O. (b) Show that log (z-ie) is defined for Re z > 0 and any real number e; also prove that 11m log (z-ie)1 <

71'

2 if Re z

(c) Let h be as in (a) and prove that h(z) (d) Show that

f dt

=

> O. 1m [log (z-ia)-Iog (z-ib)J.

b

--. = z-/l

i[log (z-ib)-log (z-ia)]

a

(Hint: Use the Fundamental Theorem of Calculus.) (e) Combine (c) and Cd) to get that b

h(x+iy) =

fx

2

+(:_t)2

dt = arctan

a

e:

a ) -arctan

(Y:b)

(f) Interpret part (e) geometrically and show that for Rez > O. h(z) is the

angle depicted in the figure.

ih

ia

az+h

Otz+f3

and Tz = ---- ; show that S = T iff there is a non zero cz+d yz+" complex number A such that or: = Aa, f3 = Ab, y = Ae, " = Ad. 21. Let T be a Mobius transformation with fixed points ZI and Z2' If S is a Mobius transformation show that S -1 TS has fixed points S - I Z I and S - I Z 2' 22. (a) Show that a Mobius transformation has 0 and 00 as its only fixed points iff it is a dilation. but not the identity. 20. Let Sz

= --

Analytic functions as mappings. Mobius transformations

57

(b) Show that a Mobius transformation has 00 as its only fixed point iff it is a translation, but not the identity. 23. Show that a Mobius transformation T satisfies T(O) = 00 and T( 00) = 0 iff Tz = az- 1 for some a in C. 24. Let T be a Mobius transformation, T #- the identity. Show that a Mobius transformation S commutes with T if Sand T have the same fixed points. (Hint: Use Exercises 21 and 22.) 25. Find all the abelian subgroups of the group of Mobius transformations. 26. (a) Let GL 2 (C) = all invertible 2 x 2 matrices with entries in C and let vIt be the group of Mobius transformations. Define 'P: GL 2 (C) - vIt by

'P(: :) = : : : : . Show that 'P is a group homomorphism of GL 2(C) onto vIt. Find the kernel of 'P. (b) Let SL 2 (C) be the subgroup of GL 2 (C) consisting of all matrices of determinant 1. Show that the image of SL 2 (C) under 'P is all of vIt. What part of the kernel of'P is in SL 2 (C)? 27. If f§ is a group and.JV is a subgroup then.IV is said to be a normal subgroup of f§ if S -1 TS E.JV whenever T E.IV and S E f§. f§ is a simple group if the only normal subgroups of f§ are {J} (/ = the identity of f§) and f§ itself. Prove that the group vIt of Mobius transformations is a simple group. 28. Discuss the mapping properties of (1- z)'. 29. For complex numbers a and fJ with laI 2+lfJI2=1

U",.p(z)

=

az -13 fJz+ii

and let

U=

{u",.p:l aI2 +lfJI 2 =1}.

(a) Show that U is a group under composition. (b) If SU2 is the set of all unitary matrices with determinant I, show that SU2 is a group under matrix multiplication and that for each A in SU2 there are unique complex numbers a and fJ with lal 2+ I fJ 12 = I and

A=( - fJ~ ~). a (c) Show that ( _;

:

}-~Ua.p is a homomorphism of the group SU2onto

U. What is its kernel? (d) If I E{ 0, I, ~, ... } let HI = all the polynomials of degree $. 2/. For ua.p=u in U define TJ/):Hr-~HI by (TJ I>j)(z)=(fJz+ii)2'J(U(Z». Show that Tu(l) is an invertible linear transformation on HI and U f--+ Tu(l) is an injective homomorphism of U into the group of invertible linear transformations of HI onto HI' 30. For Izl < I define fez) b\

t,

f(z)=ex p { -ilog[i(

:~;)r2}.

(a) Show that f maps D = {z : Izl < I} conformally onto an annulus G. (b) Find all Mobius transformations S(z) that map D onto D and such that f(S (z» = fez) when Izl < I.

Chapter IV Complex Integration In this chapter results are derived which are fundamental in the study of analytic functions. The theorems presented here constitute one of the pillars of Mathematics and have far ranging applications. §1. Riemann-Stieltjes integrals We will begin by defining the Riemann-Stieltjes integral in order to define the integral of a function along a path in C. The discussion of this integral is by no means complete, but is limited to those results essential to a cogent exposition of line integrals. 1.1 Definition. A function y: [a, b] -+ C, for [a, b] c IR, is of bounded variation if there is a constant M > 0 such that for any partition P = {a = to < t 1 < ... < tm = b} of [a, b]

v{y; P)

=

m

L k=l

ly(tk)-y{tk - 1)1 ~ M.

The total variation of y, V{y), is defined by V{y)

= sup {v(y;

P): P a partition of[a, b]}.

Clearly V{y) ~ M < 00. It is easily shown that y is of bounded variation if and only if Re y and 1m yare of bounded variation. If y is real valued and is non-decreasing then .y is of bounded variation and V(y) = y(b) -y(a). (Exercise I) Other examples will be given, but first let us give some easily deduced properties of these functions. 1.2 Proposition. Let y: [a, b] -+ C be of bounded variation. Then: (a) If P and Q are partitions of[a, b] and pc Q then v{y; P)

~ v(y; Q); (b) If a: [a, b] -+ C is also of bounded variation and ~

J ly'(t)1 dt

::;

E

+

tkl. Hence

m

k1;l ly'h)1 Uk-fk-I)

a

f ly'(t)1 dt ::; e+£(b-a) + .1;lly(tk)-y(tk-l)1 m

b

a

=

£[1 +(b-a)J+v(y; P)

::; E[l + (b-a)]+ V(y).

Complex Integration

60

Letting

€

-+ 0

+,

gives

Jly'(t)J dt ::; VCy), b

a

which yields equality . • 1.4 Theorem. Let y: fa, b] -+ C be oj bounded variation and suppose that J: [a, b] -+ C is continuous. Then there is a complex number I slich that Jar every € > 0 there is a 0 > 0 such that when P = {to < tl < .. . 82 > 03 > ... such that if Is-tl < Om'

~. For each m :2: 1 let & m = the collection of all partitions P m of [a, bJ with [[PII < 8m ; so &1 ::> 9 2 ::> Finally define Fm to be the IJ(s) - J(t) I <

closure of the set:

1.5

Ltl J

(Tk)[y(tk)-y(t k-

1)]:PE&m and

tk- J

::;

Tk::; tk}'

The following are claimed to hold:

1.6

(

F!

::>

F2

::> •••

diam Fm ::;

~

m

and V(y)

If this is done then, by Cantor's Theorem (II. 3.7), there is exactly one complex number I such that / E Fm for every m :2: L Let us show that this will complete the proof. If E > 0 let m > (2/E) V(y); then E > (21m) V(y) :2: diam Fm. Since / E Fm , Fm c B(/; E). Thus, if 0 = Om the theorem is proved. Now to prove (1.6). The fact that FI ::> F2 ::> ••• follows trivially from 2 the fact that &1 ::> &2 ::> •••• To show that diam Fm ::; -- V(y) it suffices

m

2 to show that the diameter of the set in (1.5) is s: ---- V(y). This is done in two m stages, each of which is easy although the first is tedious. If P = {to < ... < In} is a partition we will denote by S (P) a sum of the form L!(Tk)[y(tk)-y(lk-l)] where Tk is any point with lk-l ::;Tk ::;lk'

Riemann-Stieltjes integrals

61

the first step will be to show that if P c Q (and 1 hence Q E 9 m ) then IS(P) - S(Q)I < - V(y). We only give the proof for m the case where Q is obtained by adding one extra partition point to P. Let 1 ~ P ~ In and let f p- 1 < t* < tp; suppose that P U {t*} = Q. If tp 1 ~ a ~ t*, t* ~ a' ~ (p' and Fix

:2c 1 and let P E 9

In

S(Q)=

m;

L j(ak)[y(tk)-y(tk .. l)]+](O)[y(t*)-y(tp_l)] k#'p + ](0')[ y(tp } - y(t*)

J,

then, using the fact that 1](1")-}(0)1<

IS(P)- S(Q)I=

-.l m

for 11"-al- C is a path then the set {y(t): a :::; t :::; b} is called the trace of y and is denoted by {y}. Notice that the trace of a path is always a compact set. y is a rectifiable path if y is a function of bounded variation. If P is a partition of [a, b] then vCy; P) is exactly the sum of lengths of line segments connecting points on the trace of y. To say that Y is rectifiable is to say that Y has finite length and its length is V(y). In particular, if y is piecewise smooth then Y is rectifiable and its length is J~ Iy'l dt. If y: [a, b] ----,.. C is a rectifiable path with {y} c E c C and f: E --+ C is a continuous function then f 0 y is a continuous function on [a, b]. With this in mind the following definition makes sense.

Riemann·Stieltjes integrals

63

1.12 Definition. If ,,: [a, b) ~ C is a rectifiable path and I is a function defined and continuous on the trace of" then the (line) integral 01I along" is

f1(,,(/» d,,(/). b

"

LI

This line integral is also denoted by = L/(z) dz. As an example let us take ,,: [0, 21T) ~ C to be ,,(t)

f(z) =

!

z

= e it

and define

for z "i' O. Now " is differentiable so, by Theorem 1.9 we have

Jy!z dz = J5

K

e-it(ie it ) dl

= 21Ti.

Using the same definition of 'I and letting m be any integer 2!: 0 gives exp (i(m+l)t) dl=iH'fI cos (m+l)t dtf~'fIsin{m+ 1)ldl=O. Now let a, bEe and put '1(1) = tb + (1 - t)a for 0:::; t:::; 1. Then Y'{t) = b - a, and using the Fundamental Theorem of Calculus we get that for

J zmdz=f5'f1eiml(ieil)dl=ig'fl

n 2!: 0,

Jy

zndz

= (b-a)

J~

[tb+(l-t)a]n dt

=-

1

(bn+l_an+ I ).

n+l There are more examples in the exercises, but now we will prove a certain "invariance" result which, besides being useful in computations, forms the basis for our definition of a curve. If,,: [a, b] -~ C is a rectifiable path and 'P: [c, d) ~ [a, b) is a continuous non· decreasing function whose image is all of [a, b) (i.e., 'P(c) = a and !P(d) = b) then" 0 'P: [c, d) ~ C is a path with the same trace as". Moreover, "0'P is rectifiable because if c = So < SI < ... < Sn = d then a = 'P(so) :::; 'P(SI) :::; ... :::; 'P(sn) = b is a partition of [a, b). Hence n

L

k=1

1"('P(Sk»-"('P(Sk-I»I:::; V(,,)

so that V(" 0 'P) :::; V(,,) < 00. So if I is continuous on {,,} JyO tp f is well defined.

= {"

0

'P} then

1.13 Proposition. II,,: [a, b) ~ C is a rectifiable path and 'P: [c, d) ~ [a, bj is a continuous non·decreasing lunction with !p(c) = a, 'P(d) = b; then lor any lunction I continuous on {y}

f1= f f

y

yo tp

Proof Let £ > 0 and choose 8 1 > 0 such that for {so < SI < ... < sn}, a partition of [c, d) with (SAl-SAl-I) < 81 , and Sk-I ~ (1k ~ S1 we have

Similarly choose 82 > 0 such that if {to < t I < ... < tn} is a partition of [a, b) with (t" - t"-I) < 82 and tit. -1 ~ 'Tit. ~ tAl, then

64

Complex Integration

1.15 But fP is uniformly continuous on [c, d]; hence there is a.5 > 0, which can be chosen with 8 < 8 b such that IfP(s)-fP(s')1 < 82 whenever Is-s'l < 8. So if {so < SI < ... < sn} is a partition of [c, dJ with (Sk-Sk-l) < .5 < 01 and tk = fP(Sk), then {to :$ tl :$"':$ In} is a partition of [a, b] with (tk-f.-l) < 82 , If Sk-l :$ Uk :$ Sk and 'Tk = fP(u k) then both (1.14) and (1.15) hold. Moreover, the right hand parts of these two differences are equal! It follows that

Since € > 0 was arbitrary, equality is proved . • We wish to define an equivalence relation on the collection of rectifiable paths so that each member of an equivalence class has the same trace and so that the line integral of a function continuous on this trace is the same for each path in the class. It would seem that we should define a and y to be equivalent if a = y fP for some function fP as above. However, this is not an equivalence relation! 0

1.16 Definition. Let u: [c, d]-+ C and y: [a, b] -+ C be rectifiable paths. The path u is equivalent to y if there is a function fP: [c, d] -+ [a, b] which is continuous, strictly increasing, and with fP(c) = a, fP(d) = b; such that a = yo fP. We call the function fP a change 0/ parameter. A curve is an equivalence class of paths. The trace of a curve is the trace of anyone of its members. If/is continuous on the trace of the curve then the integral of/ over the curve is the integral of/over any member of the curve. A curve is smooth (piecewise smooth) if and only if some one of its representatives is smooth (piecewise smooth). Henceforward, we will not make this distinction between a curve and its representative. In fact, expressions such as "let y be the unit circle traversed once in the counter-clockwise direction" will be used to indicate a curve. The reader is asked to trust that a result for curves which is, in fact, a result only about paths will not be stated. Let y: [a, b] -+ C be a rectifiable path and for a :$ t :$ b, let Iyl (t) be V(y; [a, tJ). That is,

Iyl (t) =

sup

{f

iy(tk)-y(tk-1)1: {to,··., t n } is a partition of [a,

k=1

{l}.

Clearly Iii (t) is increasing and so Iii: [a, bj-J> !R is of bounded variation. If / is continuous on {y} define

J/Idzl = J/(y(t)) diyl (t). b

y

a

If y is a rectifiable curve then denote by

-y

the curve defined by (-y) (t) =

Riemann-Stieltjes integrals

65

y( - t) for - b ~ t ~ - a. Another notation for this is y - I. Also if c E IC let y+c denote the curve defined by (y+c) (t) = y(t)+c. The following

proposition gives many basic properties of the line integral. 1.17 Proposition. Let y be a rectifiable curve and suppose that I is a lunction continuous on {y}. Then:

L

I = - S- y I; (a) (b) IL/I ~ L I/lldzl ~ V(y) sup [1/(z)l: Z E {y}]; (c) If c E IC then JJ(z) dz = L+cf'(z-c) dz. The proof is left as an exercise. The next result is the analogue of the Fundamental Theorem of Calculus for line integrals. 1.18 Theorem. Let G be open in IC and let y be a rectifiable path in G with initial and end points a and f3 respectively. f: G --+ IC is a continuous lunction with a primitive F: G --+ IC, then

fr

f1= F(8) - F(a) y

(Recall that F is a primitive of I when F'

=

f)

The following useful fact will be needed in the proof of this theorem. 1.19 Lemma.IJ G is an open set in C, y : [a,bj--+G is a rectifiable path, and J: G --+ C is continuous Then Jor every £ > 0 there is a polygonal path f in G such that f(a)=y(a), f(b)=y(b), and IfyJ-IrJI 0 such that IJ(z)-J(w)I O. Denote - by qJ2; it follows

ot

that qJ2 must be uniformly continuous on [a, b] x [c, d]. Thus, there is a I) > 0 such that IqJzCs " t')-qJ2(S, t)1 < € whenever (S-S')2+(t-t')2 < 1)2. In particular

2.4 whenever It-tol < a ~ S ~ b,

I)

and a ~ S ~ b. This gives that for It-tol <

I)

and

2.5 But for a fixed S in [a, b] (t) = qJ(s, t) - tqJ2(S, to) is a primitive of qJz{s, t)qJz{s, to). By combining the Fundamental Theorem of Calculus with inequality (2.5), it follows that IqJ(s, t)-qJ(s, to)-(t-tO)qJ2(S, to)1 ~

for any S when It-tol <

I).

€

It-tol

But from the definition of g this gives

g(t)-g(to) - Ib qJz{s, to) ds I ~ I~-.:::....:..-= t-t o a

E(b-a)

when 0 < It-tol < I)• • This result can be used to prove that 2"

f -.-ds = 21T e"S-z eis

if

Izl

< 1.

o

Actually, we will need this formula in the proof of the next proposition. e is Let qJ(s, t) = -.-s- for 0 ~ t ~ 1, 0 ~ s ~ 21T; (Note that qJ is continuously e" -tz differentiable because Izl < 1.) Hence get) = J~" qJ(S, t) ds is continuously differentiable. Also, g(O) = 21T; so if it can be shown that g is a constant, then 21T = g(1) and the desired result is obtained. Now

but for t fixed, (s) = zi(eiS-tz)-l has '(s) = -zi(e is -tz)-2(ie is)= ze is(e i'_tz)-2. Hence g'(t) = (21T)-(O) = 0, so g must be a constant.

70

Complex Integration

The next result, although very important, is transitory. We will see a much more general result than this-Cauchy's Integral Formula; a formula which is one of the essential facts of the theory. 2.6 Proposition. Let f: G ---* C be analytic and suppose B(a; r) c G(r > 0). If yet) = a + re il , 0 ::; t ::; 2rr, then 1 ff(W) fez) = ----; dw 2rrl w-z y

for

Iz-al

< r.

H

Proof By considering G1 =

(z-a): z E G} and the function g(z) = f(a+

rz) we see that, without loss of generality, it may be assumed that a = 0 and r = 1. That is we may assume that B(O; 1) c G.

Fix z,

Izi

< 1; it must be shown that fez) = -1. ff(W) - - dw

2m

w-z

1

2K

=LJ"f(~iS)eiS ds; 2rr elS-z o

that is, we want to show that 2"

f(eis)eis 0= f - . - - ds-2rrf(z) eiS-z

o

2.

=

iS f[ f(eiS)e e"-z

- .------ -

] fez) ds

o

We will apply Leibniz's rule by letting rp( s, t ) =

fez -+. tee;s - z))e is .

e"-z

-

f()

z,

for 0 :S t :s 1 and 0 :s s ::; 2'7T. Since Iz + (e z)1 = Iz(1 - t) + te"1 ::; 1, rp is well defined and is continuously differentiable. Let g( t) = (o2wrp(s, t) ds; so g has a continuous derivative. The proposition will be proved if it can be shown that g(l) = 0; this is done by showing that g(O) = 0 and that g is constant. To see that g(O) = 0 IS

----

71

Power series representation of analytic functions

compute:

fo cp(S, 0) ds 2"

g(O) =

2"

=

f [f(z)eiS eis_z -

fez)

]

ds

o 2"

f

ei.

= fez) - .•o

since we showed that

f

2"

=

e' -z

ds-2.".f(z)

0,

ei.

-.-- ds = 2.". prior to the statement of this pro-

o elS-z

position. To show that g is constant compute g'. By Leibniz's rule, g'(t) = J~" cpis, t) ds where

However, for 0 < t :;:; 1 we have that (s) = -U-1f(z+t(eiS-z» is a primitive of cpis, t). So g'(t) = (27T) -(0) = 0 for 0 < t :;:; 1. Since g' is continuous we have g' = 0 and g must be a constant. • How is this result used to get the power series expansion? The answer is that we use a geometric series. Let Jz-aJ < r and suppose that w is on the circle Jw-aJ = r. Then

- 1- = - _ . w-z

w-a

1

1

z-aJ = (w-a) 1- [ - w-a

L (z-a)" w-a }

w-z

dw

y

is analytic on C - {y} and g(n)(z) = n!

f-. _ _~(~L (W-Z)"+I

dw.

y

4. (a) Prove Abel's Theorem: Let Ian (z - a)" have radius of convergence 1 and suppose that I an converges to A. Prove that

Jim La/'

=

r-+ 1-

A.

(Hint: Find a summation formula which is the analogue of integration by parts.) (b) Use Abel's Theorem to prove that log 2 = 1-1+t-... 5. Give the power series expansion of log z about z = i and find its radius of convergence. 6. Give the power series expansion of about z = I and find its radius of convergence. 7. Use the results of this section to evaluate the following integrals:

.Jz

(a)

(b)

(c)

f;:%

dz,

f z-a dz ,

f

Sin z 7dz,

o :::;

t :::;

211' and n 2: O.

Power series representation of analytic functions

7S

8. Use a Mobius transformation to show that Proposition 2.15 holds if the disk B(a; R) is replaced by a half plane. 9. Evaluate the following integrals: (a)

f · -. f f2 ~ z"

dz where n is a positive integer and yet)

= eit,O

SIS 21T;

7

(b)

dz 1

(z - 2)

y

(c)

n

where n is a positive integer and yet)

dz where yet) z +1

= 2e i', 0

=!- +e it , 0 S ts2'IT;

S t S 271. (Hint: expand

(Z2 + 1)-1 by

y

means of partial fractions);

(d)

f f

Sin z . -z- dz where yet) = e"', 0 S t S 271;

y

(e)

zltm

- - ) dz where yet)

(z-I

7

10. Evaluate

m

= I +te i', 0 s

f

z2+1 2 ) dz where yet) z(z +4

s

t

= re ll , 0

271.

S t S 271, for all possible

y

values of r, 0 < r < 2 and 2 < r < 00. II. Find the domain of analyticity of

fez) = also, show that tan fez)

= z (i.e., f

fez) =

(I

~.Iog I-Iz +~z) ;

21

is a branch of arctan z). Show that

2: (_l)k 2Z2k+ -k- for Izl < I +I co

1

k-O

(Hint: see Exercise III. 3.19.) 12. Show that

sec z

=1+

E L: ~ Z2k (2k)! co

k-l

for some constants E l , E 4 ,' ••• These numbers are called Euler's numbers. What is the radius of convergence of this series? Use the fact that I = cos z sec z to show that

E2n-Cn2:2)E2n-2 + Cn2:4) E2n- + ... +( 4

_1)n-l

C2n)E2 +( -It=O.

Evaluate E 2 , E4 , E6 , E 8 • (E10 = 50521 and Ell = 2702765).

Complex Integration

76

eZ-l 13. Find the series expansion of - - about zero and determine its radius z

z

= -:;:-

of convergence. Consider fez)

e"-l

fez) =

and let co

ak k z ~ ....... k~O

k!

be its power series expansion about zero. What is the radius of convergence? Show that

o=

ao

+

(

/ 11

+ 1)· 1

a 1 + ...

+

(n+ 1)

n / an"

Using the fact that f(z) +1Z is an even function show that ak = 0 for k odd and k > 1. The numbers B Zn = (-1)n- 1a 2" are called the Bernoulli numbers for 11 ;::0: 1. Calculate B 2 , B4 ,' .• BI o. 14. Find the power series expansion of tan z about z = 0, expressing the coefficients in terms of Bernoulli numbers. (Hint: use Exercise 13 and the formula cot 2z = 1 cot Z-1· tan z.) §3. Zeros of an analytic function If p(z) and q(z) are two polynomials then it is well known that p(z) = s (z)q(z) + r(z) where s(z) and r(z) are also polynomials and the degree of r(z) is less than the degree of q(z). In particular, if a is such that pea) = 0 then choose (z-a) for q(z). Hence, p(z) = (z-a)s(z)+r(z) and r(z) must be a constant polynomiaL But letting z = a gives 0 = pea) ~~ rea). Thus, p(z) = (z-a)s(z). If we also have that sea) = 0 we can factor (z-a) from s(z). Continuing we get p(z) = (z-a)m fez) where 1 ::: m ::: degree of p(z), and t(z) is a polynomial such that tea) # O. Also, degree t(z) = degree p(z)-m. 3.1 Definition. If f: G --7> C is analytic and a in G satisfies f(a) = 0 then a is a zero off of multiplicity m ;::0: 1 if there is an analytic function g: G --7> C such thatf(z) = (z-a)"'g(z) where g(a) # O. Returning to the discussion of polynomials, we have that the multiplicity of a zero of a polynomial must be less than the degree of the polynomial. If n "-7. the degree of the polynomial p(z) and ai' ... , ak are all the distinct zeros of p(z) thenp{z) = (z_al)m t •• • (z-ak)mks(z) where s(z) is a polynomial with no zeros. Now the Fundamental Theorem of Algebra says that a polynomial with no zeros is constant. Hence, jf we can prove this result we will have succeeded in completely factoring p(z) into the product of first degree polynomials. The reader might be pleasantly surprised to know that after many years of studying Mathematics he is right now on the threshold of proving the Fundamental Theorem of Algebra. But first it is necessary to prove a famous result about analytic functions. It is also convenient to introduce some new terminology.

Zeros of an analytic function 77 3.2 Definition. An entire function is a function which is defined and analytic in the whole complex plane Co (The term "integral function" is also used.) The following result follows from Theorem 2.8 and the fact that C contains BCO; R) for arbitrarily large R. 3.3 Proposition.

If f is

an entire function then f has a power series expansion 00

fez)

=

L

"= 0

anz"

with infinite radius of convergence.

In light of the preceding proposition, entire functions can be considered as polynomials of "infinite degree". So the question arises: can the theory of polynomials be generalized to entire functions? For example, can an entire function be factored? The answer to this is difficult and is postponed to Section VII. 5. Another property of polynomials is that no non constant polynomial is bounded. Indeed, if p(z) = z"+an_Iz n- I + ... +ao then limp(z) = lim z" [1+an_IZ-l+"'+aoz-n] = 00. The fact that this also z-+OO

z-+oo

holds for entire functions is an extremely useful result.

3.4 Liouville's Theorem. If I is a bounded entire lunction then I is constant. Proof Suppose I/(z) :0; M for all z in C. We will show that I'{z) = 0 for all z in Co To do this use Cauchy's Estimate (Corollary 2.14). Since I is analytic in any disk B(z; R) we have that il'(z)l :0; MfR. Since R was arbitrary, it follows that I'(z) = 0 for each z in Co • The reader should not be deceived into thinking that this theorem is insignificant because it has such a short proof. We have expended a great deal of effort building up machinery and increasing our knowledge of analytic functions. We have plowed, planted, and fertilized; we shouldn't be surprised if, occasionally, something is available for easy picking. Liouville's Theorem will be better appreciated in the following application. 1

II p(z) is a non constant polynomial then there is a complex number a with pea) = O.

3.5 Fundamental Theorem of Algebra.

Proof Supposep(z) 1= 0 for all z and let/(z) = [p(Z)J-l; thenlis an entire function. If p is not constant then, as was shown above, lim p(z) = 00;

so lim I(z) = O. In particular, there is a number R > 0 such that I/(z) 1 < 1 z-+oo

if Izl > R. Butlis continuous on BCO; R) so there is a constant M such that I/(z) 1:0; M for Izl :0; R. Hencelis bounded and, therefore, must be constant by Liouville's theorem. It follows that p must be constant, contradicting our assumption. III

If p(z) is a polynomial and aI' ... ,am are its zeros with a j having multiplicity k j thenp(z) = c(z-alll.· ·(z-am)kmlor some constant c and kl + ... +km is the degree 01 p.

3.6 Corollary.

Returning to the analogy between entire functions and polynomials, the

78

Complex Integration

reader should be warned that this cannot be taken too far. For example, if p is a polynomial and a E C then there is a number z with p(z) = a. In fact, this follows from the Fundamental Theorem of Algebra by considering the polynomial p(z)-a. However the exponential function fails to have this property since it does not assume the value zero. (Nevertheless, we are able to show that this is the worst that can happen. That is, a function analytic in C omits at most one value. This is known as Picard's Little Theorem and will be proved later.) Moreover, no one should begin to make an analogy between analytic functions in an open set G and a polynomial p defined on C; rather, you should only think of the polynomials as defined on G. F or example, let

+z) ,Izl < 1.

1 I(z) = cos ( ----

1-z

Notice that

1+ z

1- z

=

{z: Izl < I} onto G = {z: Re z > O}. The

! n1T -

2

}

are the points \ ------------: n is odd and positive ; so f has in• n1T --r- 2 finitely many zeros. However, as n ---> Xi the zeros approach 1 which is not in the domain of analyticity D. This is the story for the most general case. zeros of

f

maps D

3.7 Theorem. Let G be a connected open set and let f: G - IC be an analytic Junction. Then. the Jollowing are equivalent statements: (a) J=- 0; (b) there is a point a in G such that I(O)(a) = 0 Jor each n 2:: 0; (c) {z E G: J(z) = O} has a limit point in G.

Proof Clearly (a) implies both (b) and (c). (c) implies (b): Let a E G and a limit point of Z = {z E G: J(z) = O}, and let R > 0 be such that B(a; R) c G. Since a is a limit point of Z and J is continuous it follows that J(a) = O. Suppose there is an integer n 2:: 1 such that J(a) = /,(a) = ... = p.-Il(a) = 0 and J(o)(a) # O. Expanding J in power series about a gives that

L'" ak(z-a)k

J(z) =

k=n

for

Iz-al

< R. If g(z) =

ro

L ak(z-al-

n

k==n

then g is analytic in B(a; R), J(z) = (z-a)"g(z), and g(a) = an # O. Since g is analytic (and therefore continuous) in B(a; R) we can find an r, 0 < r < R, such that g(z) # 0 for Iz-al < r. But since a is a limit point of Z there is a point b with I(b) = 0 and 0 < Ib-aJ < r. This gives 0 = (b-a)"g(b) and so g(b) = 0, a contradiction. Hence no such integer n can be found; this proves part (b). (b) implies (a): Let A = {z E G: pn)(z) = 0 for all n 2:: O}. From the hypothesis of (b) we have that A # O. We will show that A is both open

Zeros of an analytic function

79

and closed in G; by the connectedness of G it will follow that A must be G and so f == O. To see that A is closed let Z E A - and let Zk be a sequence in A such that Z = lim Zk' Since each pHl is continuous it follows that f(n)(z) = Iimi'n l (zk) = O. So Z E A and A is closed. To see that A is open, let a E A and let R > 0 be such that B(a; R) c::: G. 1 Then fez) = ~ a.(z-a)" for Iz-al < R where a. = - pN)(a) = 0 for each

n!

n ~ O. Hencef(z) = 0 for all z in B(a; R) and, consequently, B(a; R) c::: A. Thus A is open and this completes the proof of the theorem. II 3.8 Corollary. Iff and g are analytic on a region G then f == g iff {z E G: fez) = g(z)} has a limit point in G. This follows by applying the preceding theorem to the analytic function f-g· 3.9 Corollary. Iff is analytic on an open connected set G and f is not identically zero then for each a in G with f(a) = 0 there is an integer n ~ 1 and an analytic function g: G -l> C such that g(a) f= 0 and fez) = (z -a)"g(z) for all z in G. That is, each zero off has finite multiplicity.

Proof. Let n be the largest integer (2: 1) such that /(k)( a) and define g(z)

=

(z - a)-"/(z) for z

*-

a and g(a)

=

=

1

0 for 0 ~ k < n

-/(n)(a). Then g

n!

is clearly analytic in G - {a}; to see that g is analytic in G it need only be shown to be analytic in a neighborhood of a. This is accomplished by using the method of the proof that (c) implies (b) in the theorem. II 3.10 Corollary. Iff: G -l> C is analytic and not constant, a E G, and f(a) = 0 then there is an R > 0 such that 8(a; R) c G andf(z) f= 0 for 0 < Iz - al < R.

Proof By the above theorem the zeros of f are isolated . • There is one instance where the analogy between polynomials and analytic functions works in reverse. That is, there is a property of analytic functions which is not so transparent for polynomials.

3.n Maximum Modulus Theorem.lfG is a region andf: G...,.. C is an analytic function. such that there is a point a in G with If(a) 1 ~ If(z) 1 for all z in G, then f is constant. Proof Let 8(a; r) c::: G, y(t) = a+re it for 0 ~ t ~ 217'; according to Proposition 2.6 f(a)

= ~ f f(w) dw w-a

2m

y

2x

= ~ff(a+re'l) dt 2?T

o

80

Complex Integration

Hence

If· 2.

!/(a)1

:s:

I/(a+relt)1 dt :::~ I/(a) I

27T

o

since I/(a+rei')1

:s:

I/(a)1 for all t. This gives that

°= J 2n

[1/(a)! -1/(a + re it )!] dt;

o

but since the integrand is non-negative it follows that I/(a)1 = I/(a + reit)1 for all t. Moreover, since r was arbitrary, we have that I maps any disk B(a; R) c G into the circle Izl = 10(1 where 0( = I(a). But this implies that I is constant on B(a; R) (Exercise III. 3.17). In particular I(z) = 01. for Iz-al < R. According to Corollary 3.8, 1== 01.. II According to the Maximum Modulus Theorem, a non-constant analytic function on a region cannot assume its maximum modulus; this fact is far from obvious even in the case of polynomials. The consequences of this theorem are far reaching; some of these, along with a closer examination of the Maximum Modulus Theorem, are presented in Chapter VI. (Actually, the reader at this point can proceed to Sections VI. I and VI. 2.)

Exercises L Let I be an entire function and suppose there is a constant M, an R > 0, and an integer n ~ I such that I/(z)1 :s: M Izl" for Izi > R. Show that I is a polynomial of degree :s: n. 2. Give an example to show that G must be assumed to be connected in Theorem 3.7. 3. Find all entire functions I such that I(x) = eX for x in JR. 4. Prove that ez + a = eZe" by applying Corollary 3.8. 5. Prove that cos (a + b) = cos a cos b - sin a sin b by applying Corollary 3.8. 6. Let G be a region and suppose that I: G ->- C is analytic and a E G such that I/(a) 1 :s: I/(z) I for all z in G. Show that either 1(0) = 0 or I is constant. 7. Give an elementary proof of the Maximum Modulus Theorem for polynomials. 8. Let G be a region and let f and g be analytic functions on G such that f(z)g(z) = 0 for all z in G. Show that either f == 0 or g == O. 9. Let U: C ~ R be a harmonic function such that U( z) ~ for all z in C; prove that U is constant. 10. Show that if f and g are analytic functions on a region G such that f g is analytic then either f is constant or g == O.

°

§ 4. The index of a closed curve

It is easy to see that L(z - a)-1 dz = 2nin if i'(t) = a + eZ"int, 0 The following result shows that this is not peculiar to the path y.

:s: t:s:

l.

Cauchy's Theorem

81

4.1 Proposition. If y: [0, I]_C is a closed rectifiable curve and a • {y} then

2~i f /'~a y

is an integer. Proof. An easy argument using Lemma 1.19 shows that it suffices to prove this proposition under the additional assumption that y is smooth. In this case define g: [0, 1] ~ C by 1 y'(s) g(t)= () ds

£ o

s -a

y

Hence, g(O) = 0 and g( I) = f y(z - a)-I dz. We also have that g'(t)=

for

y'(t) y(t)-a

O:5t:51.

But this gives

.!!... e - g ( y dt

a) = e - gy' - g' e - g ( y - a) =e- g [ y'-y'(y-a)-I(y-a)J

=0 So e-g(y-a) is the constant function e-g(O)(y(O)-a)=y(O)-a= e -8(1)( y( I) - a). Since y(O) = y(l) we have that e -g(1) = I or that g(I)= 2'ITik for some integer k . • 4.2 Definition. If y is a closed rectifiable curve in C then for a 1/ {y}

1 .j(z-a)-I dz n(y;a)= -2 'IT!

y

is called the index of y with respect to the point a. It is also sometimes called the winding number of y around a. Recall that if y: [0, I]_C is a curve, - y or y -\ is the curve defined by (-y)(I)=y(I-/) (this is actually a reparametrization of the original definition). Also if y and 0 are curves defined on [0, I] with y(l) = 0(0) then y+o is the curve (y+ 0 )(1)= y(21) if 0:5 1 :5 ~ and (y+ o)(t) =0(2t-l) if ~:5 t:5 1. The proof of the following proposition is left to the reader. 4.3 Proposition. If y and 0 are closed rectifiable curves having the same initial points then (a) n(y;a)= -n(-y;a) for every a* {y}; (b) n(y+o;a)=n(y;a)+n(o;a) for every a. {y}u{o}. Why is n(y; a) called the winding number of y about a? As was said before if y(t)=a+e 2 ,..;", for 0:5 1 S 1 then n(y; a)=n. In fact if Ib-al < 1 then n(y; b)=n and if Ib-al> 1 then n(y; b)=O. This can be shown directly or one can invoke Theorem 4.4 below. So at least in this case n(y; b) measures the number of times y wraps around b- with the minus sign indicating that the curve goes in the clockwise direction.

82

Complex Integration

The following discussion is intuitive and mathematically imprecise. Actually, with a little more sophistication this discussion can be corrected and gives insight into the Argument Principle (V.3). If y is smooth then -I y'(t) (z-a) dz= ( ) dt. 0 y t -a y Taking inspiration from calculus one is tempted to write f/ z - a) -1 dz = log[y(t) - a1l::6. Since y(l) = y(O), this would always give zero. The difficulty lies in the fact that y( t) - a is complex valued and unless y( t) - a lies in a region on which a branch of the logarithm can be defined, the above inspiration turns out to be only so much hot air. In fact if y wraps around the point a then we cannot define log( y( t) - a) since there is no analytic branch of the logarithm defined on C - {a}. Nevertheless there is a correct interpretation of the preceding discussion. If we think of log z = logl z I+ i arg z as defined then

11

f

f (z - a) - dz = log[ y(l) - a] -Iog[ y(O) - a] = 1

y

{iogi y ( I ) - a 1- log/ y (0) - a I} + i { arg [ y ( I ) - a ] - arg [ y (0) - a] }.

Since the difficulty in defining logz is in choosing the correct value of argz, we can think of the real part of the last expression as equal to zero. Since y(l)= y(O) it must be that even with the ambiguity in defining argz, arg[y(l)-a]-arg[y(O)-a] must equal an integral multiple of 2w, and furthermore this integer counts the number of times y wraps around a. Let y be a closed rectifiable curve and consider the open set G = C {y}. Since {y} is compact (z:/zl>R}cG for some sufficiently large R. This says that G has one, and only one, unbounded component. 4.4 Theorem. Let y be a closed rectifiable curve in C. Then n(y;a) is constant for a belonging to a component of G = C - { y }. A/so, n (y; a) = 0 for a belonging to the unbounded component of G.

Proof. Define f: G~C by f(a) = n( y; a). It will be shown that f is continuous. If this is done then it follows that f(D) is connected for each component D of G. But since f(G) is contained in the set of integers it follows that f(D) reduces to a single point. That is, f is constant on D. To show that f is continuous recall that the components of G are open (Theorem II. 2.9). Fix a in G and let r = d (a, (y». If Ia - bl < tr. It follows that If(a)-f(b)I0 is given 1fr then, by choosing 5 to be smaller than ~ rand ('IT r 2() /2 V( y), we see that j must be continuous. (Also, see Exercise 2.3.) Now let U be the unbounded component of G. As was mentioned before the theorem there is an R > 0 such that U::> {z: Izi > R}. If { > 0 choose a with lal > Rand Iz - al > (21f£)-1V(y) uniformly for z on {y}; then In(y; a)1 < (. That is, n(y; a) ~ 0 as a --+ 00. Since n(y; a) is constant on U, it must be zero . •

Exercises 1. Prove Proposition 4.3. 2. Give an example of a closed rectifiable curve y in C such that for any integer k there is a point a $. {y} with n(y; a) = k. 3. Let p(z) be a polynomial of degree n and let R > 0 be sufficiently large so that p never vanishes in {z: Izl 2 R}. If yet) = Reir, 0 S t s 2n, show that

PI(z) f -.-dz y

p(z)

=

.

21f1n.

4. Fix w = re lll *0 and let y be a rectifiable path in C- {O} from I to w. Show that there is an integer k such that iyz - 1 dz = logr + iii + 21fik. § 5. Cauchy's Theorem and Integral Formula We have already proved Cauchy's Theorem for functions analytic in a disk: if G is an open disk then /J=O for any analytic functionj on G and any closed rectifiable curve y in G (Proposition 2.15). For which regions G does this result remalll valid? There are regions for which the result is false. For example, if G=C,-{O} andf(z)=z-I then y(t)=e il for OS IS 21f gives that 1 = 21fi. The difficulty with C- {O} is the presence of a hole (namely {O». In the next section it will be shown that 1/=0 for every analytic function f and every closed rectifiable curve '( in regions G that have no "holes." In the present section we adopt a different approach. Fix a region G and an analytic function f on G. Is there a condition on a closed rectifiable curve y such that I J = O? The answer is furnished by the index of y with respect to points outside G. Before presenting this result we need the following lemma. (This has already been seen in Exercise 2.3.)

J

5.1 Lemma. Let '( be a rectifiable curve and suppose cp is a junction defined and continuous on {y}. For each In 2: I let Fm(z) = 1yCP(w)(w - z)-m dw for z C an analytic Junction. IJ y is a closed rectifiable curve in G such that n(y; w)=O Jor all w in C - G, then for a in G- {y} I

n(y;a)J(a)= -2' 7rI

1 y

J(z) --dz. z-a

Proof. Define qJ: G X G--->C by qJ(z. w) = [fez) - J(w)l/(z - w) if z 1= wand qJ(z ,z) =F(z). It follows that qJ is continuous; and for each w in G, z--->qJ(z.w) is analytic (Exercise I). Let H={wEC:n(y;w)=O}. Since n(y; w) is a continuous integer-valued function of w, H is open. Moreover HuG = C by the hypothesis. Define g: C--->C by g(z) = f yqJ(Z, w)dw if z z)-~(w)dw if z E H. If z E GnH then

f qJ(z, w) dw

=

'Y

f

W -

Z

J(w) - - dw- J(z)n(y;z)27Ti yW-Z

=fJ(w)dW. yW-Z Hence g is a well-defined function.

G and g(z) = f y(w-

fJ(w) - J{z) dw 'Y

=

E

Cauchy's Theorem

8S

By Lemma 5.1 g is analytic on H and by an analogue of Leibniz's rule (for example, see Exercise 2.2) g is analytic on G; that is, g is an entire function. But Theorem 4.4 implies that H contains a neighborhood of 00 in Coo- Since f is bounded on {y} and lim (W-Z)-I=O uniformly for w in {y}, Z --+ 00

5.5

}~~ g(z)= }~~

f ~,~'~

dw=O.

y

In particular (5.5) implies lhere is an R >0 such that Ig(z)l:O such that! is defined and analytic in B(a;R)-{a} but not in B (a; R). The point a is called a removable singularity if there is an analytic function g: B(a; R)-,;C such that g(z)= !(z) for 0< Iz - al < R. . sin z 1 1 . . . The functIons - - , - , and exp - all have Isolated smgularitles at z = O.

z

z

z

sin-z h as a removab ' I ' (see ExerCIse .) However, only Ie smgu anty 1 . It is left to z the reader to see that the other two functions do not have removable singularities. How can we determine when a singularity is removable? Since the function has an analytic extension to B(a; R), J = 0 for any closed curve in the punctured disk; but this may be difficult to apply. Also it must happen that lim J(z) exists. This is easier to verify, but a much weaker criterion is

J

available.

1.2 Theorem. IJ J has an isolated singularity at a then the point z = a is a removable singularity iff lim (z-a)J(z) = 0 Proof. Suppose J is analytic in {z: 0 < Iz-al < R}, and define g(z) = (z-a)J(z) for z #- a and g(a) = O. Suppose lim (z-a)J(z) = 0; then g is z->-a

clearly a continuous function. If we can show that g is analytic then it follows that a is a removable singularity. In fact, if g is analytic we have g(z) = (z-a)h(z) for some analytic function defined on B(a; R) because g(a) = 0 (IV. 3.9). But then h(z) andJ(z) must agree for 0 < Iz-al < R, so that a is, by definition, a removable singularity. 103

Singularities To show that g is analytic we apply Morera's Theorem. Let T be a triangle in B(a; R) and let Ll be the inside of T together with T. If art Ll then T"" 0 in {z: 0 < Jz-aJ < R} and so, hg = 0 by Cauchy's Theorem. If a is a vertex of T then we have T = [a, b, e, aJ. Let x E [a, bJ and y E [e, aJ and 104

c

aL-------~------------------------~b

form the triangle TI = [a, x, y, aJ. If P is the polygon [x, b, e, y, x] then h g = h, g+ g = h, g since P "" 0 in the punctured disk. Since g is continuous and g(a) = 0, for any E > 0 x and y can be chosen such that Jg(z)J ~ Elt for any z on T I , where t is the length of T. Hence Jh gl = Ih I gl ~ E; since E was arbitrary we have h g = O. If a E Ll and T = [x, y, z, x] then consider the triangles TI = [x, y, a, x], T2 = [y, z, a, y], T3 = [z, x, a, z]. From the preceding paragraph hj g = 0

Jp

x'OC:;;"--------------------------------..:::=..\y for j = 1,2,3 and so, hg = h,g+ hzg+ h3g = O. Since this exhausts all possibilities, g must be analytic by Morera's Theorem. Since the converse is obvious, the proof of the theorem is complete. • The preceding theorem points out another stark difference between functions of a real variable and functions of a complex variable. The function f(x) = IxI. x E JR, is not differentiable because it has a "corner" at x = O. Such a situation does not occur in the complex case. For a function to have an honest singularity (i.e., a non-removable one) the function must behave badly in the vicinity of the point. That is, either If(z) I becomes infinite as z nears the point (and does so at least as quickly as (z-a)-I), or If(z) I doesn't have any limit as z -+ a.

Classification of singularities

105

1.3 Definition. If z = a is an isolated singularity of f then a is a pole of f if lim If(z)1 = 00. That is, for any M > 0 there is a number € > 0 such that z--+a

If(z) 1 ?: M whenever 0 < Iz-al < E. If an isolated singularity is neither a pole nor a removable singularity it is called an essential singularity. It is easy to see that (z - a) -m has a pole at z = a for m ?: 1. Also, it is not difficult to see that although z = 0 is an isolated singularity of exp (Z-1), it is neither a pole nor a removable singularity; hence it is an essential singularity. Suppose thatfhas a pole at z = a; it follows that [f(Z)]-l has a removable singularity at z = a. Hence, h(z) = [f(zW 1 for z of. a and h(a) = 0 is analytic in B(a; R) for some R > O. However, since h(a) = 0 it follows by Corollary IV. 3.9 that h(z) = (z-a)mhj(z) for some analytic function hI with 111(a) of. Oandsomeintegerm ?: 1. But this gives that (z-a)"'!(z) = [h 1 (z)rI has a removable singularity at z = a. This is summarized as follows. 1.4 Proposition. If G is a region with a in G and iff is analytic on G - {a} with a pole at z = a then there is a positive integer m and an analytic function g: G -)- C such that

fez)

1.5

=

g(z) . (z-a)m

1.6 Definition. Iff has a pole at z = a and m is the smallest positive integer such that fez) (z - a)'" has a removable singularity at z = a then f has a pole of order m at z = a. Notice that if m is the order of the pole at z = a and g is chosen to satisfy (1.5) then g(a) of. O. (Why?) Letfhave a pole of order m at z = a and putf(z) = g(z) (z-a)-m. Since g is analytic in a disk B(a; R) it has a power series expansion about a. Let

L 00

g(z)

=

A m +A m - 1 (z-a)+'" +A 1 (z-a)m-l+(z-ar

ak(z-at·

k= 0

Hence

1.7 where g 1 is analytic in B(a; R) and Am of. O.

1.8 Definition. Iff has a pole of order m at z = a and f satisfies (1.7) then Am(z - a) -m + ... + A j (z - a) -1 is called the singular part of fat z = a. As an example consider a rational function r(z) = p(z)/q(z), where p(z) and q(z) are polynomials without common factors. That is, they have no common zeros; and consequently the poles of r(z) are exactly the zeros of q(z). The order of each pole of r(z) is the order of the zero of q(z). Suppose q(a) = 0 and let S(z) be the singular part of r(z) at a. Then r(z) - S(z) = r 1 (z) and r 1 (z) is a rational function whose poles are also poles of r(z). More-

106

Singularities

over, it is not difficult to see that the singular part of r I (z) at any of its poles is also the ~ingular part of r(z) at that pole. Using induction we arrive at the following: if a J , " , an are the poles of r(z) and S/z) is the singular part of r(z)atz = ajthen

r(z) =

1.9

n

I

S/z)+P(z)

j '" 1

where P(z) is a rational function without poles. But, by the Fundamental Theorem of Algebra, a rational function without poles is a polynomial! So P(z) is a polynomial and (1.9) is nothing else but the expansion of a rational function by partial fractions. Is this expansion by partial fractions (1.9) peculiar only to rational functions'? Certainly it is if we require P(z) in (1.9) to be a polynomial. But if we allow P(z) to be any analytic function in a region G, then (1.9) is valid for any function r(z) analytic in G except for a finite number of poles. Suppose we have a function f analytic in G except for infinitely many poles (e.g., f( z) = (cos z) -1); can we get an analogue of (1.9) where we replace the finite sum by an infinite sum? The answer to this is yes and is contained in Mittag-Leffler's Theorem which will be proved in Chapter VIII. There is an analogue of the singular part which is valid for essential singularities. Actually we will do more than this as we will investigate functions which are analytic in an annulus. But first, a few definitions. 1.10 Definition. If

{Zn:

complex numbers,

I

n

= 0, ± 1, ± 2, ... } is a doubly infinite sequence of

I

00

n= -

00

Zn

is absolutely convergent if both

00

00

are absolutely convergent. In this case

I

n= -

fun,ction on a set S for n = 0, ± 1, ... and

Zn 00

=

n=O

00

00

E

Z-n

n=l

n=O

I=-" + I

z". If

Un

is a

n=t

C(J

I

un{s) is absolutely convergent

-00

for each s

I

00

z" and

cc

:x;.

S, then the convergence is uniform over S if both

I

Un

and I

n~O

n~l

converge uniformly on S. The reason we are limiting ourselves to absolute convergence is that this is the type of convergence we will be most concerned with. One can define U- n

m

00

convergence of

I

-co

Zm

but the definition is not that the partial sums

I

Zn

n= -m

I -1 satisfies this criterion but it is clearly not a "",0 n series we wish to have convergent. On the other hand, if L z" is absolutely

converge. In fact, the series

00

- 0, {f[ann (a; 0, 8)]} - = c.

=a

Proof Suppose thatfis analytic in ann (a; 0, R); it must be shown that if c and E > 0 are given then for each 8 > 0 we can find a z with Iz-.al < 8 and If(z)-cl < E. Assume this to be false; that is, assume there is a c in C and E > 0 such that If(z)-cl ~ E for all z in G = ann (a; 0, 8). Thus lim Z-+(J

Iz-al-11f(z)-cl=oo, which implies that (z-a)-I(f(z)-c) has a pole at z=a. If m is the order of this pole then lim Iz-alm+1If(z)-cl=O. Z-+(J

no

Singularities

Hence Iz - al m + 11](z)1 $ Iz - al m + 11](z} - cl + Iz - al m + llel gives that lim Iz-alm+'I](z)I=O since m ~l. But, according to Theorem 1.2, this gives that ](z)(z - a)'" has a removable singularity at z = a. This contradicts the hypothesis and completes the proof of the theorem. • z~a

Exercises 1. Each of the following functions I has an isolated singularity at z = O. Determine its nature; if it is a removable singularity define 1(0) so that I is analytic at z = 0; if it is a pole find the singular part; if it is an essential singularity determine/({z: 0 < Izl < S}) for arbitrarily small values of 3. (a) fez)

sin z

= -; z

cos z-1 (e) fez) = - - - - ; z

(e) fez)

= log (~+ 1) z

;

= cos z ;

(d)f(z)

=

z

exp(z-l);

(f) fez) = cos

~1-1)

;

Z

Z2+ 1 (g) fez) = z(z-1) ; (i) fez)

(b) fez)

= z sin! ; z

(h) fez) = (l_e z )-l; (j) fez)

2. Give the partial fraction expansion of r(z)

1

= z" sin - . z

=

2

+1-------------2

(z +z+ l)(z-l) 3. Give the details of the derivation of (1.1 7) from (1.l6).

4. Letf(z)

•

=

1 ; give the Laurent Expansion of fez) in each of z(z-l) (z-2) the following annuli: (a) ann (0; 0, 1); (b) ann (0; 1,2); (c) ann (0; 2, co). 5. Show that fez) = tan z is analytic in C except for simple poles at

Z

=

:2 + n7T, for each integer n.

1T

Determine the singular part of I at each of

these poles. 6. If f: G --» C is analytic except for poles show that the poles of f cannot have a limit point in G. 7. Let f have an isolated singularity at z = a and suppose fez) ¢= o. Show that if either (1.19) or (1.20) holds for some sin IR then there is an integer m such that (1.19) holds if s > m and (1.20) holds if s < m. 8. Let/, a, and m be as in Exercise 7. Show: (a) m = 0 iff z = a is a removable singularity and/(a) # 0; (b) m < 0 iff z = a is a removable singularity andfhas a zero at z = a of order -m; (c) m > 0 iff z = a is a pole off of order m. 9. A function f has an essential singularity at z = a iff neither (U9) nor (1.20) holds for any real number s.

111

Classification of singularities

10. Suppose that f has an essential singularity at z = a. Prove the following strengthened version of the Casorati-Weierstrass Theorem. If C E C and E > 0 are given then for each 1) > 0 there is a number o£, Ie - (XI < E, such that fez) = IX has infinitely many solutions in B(a; 8). 11. Give the Laurent series development of fez) = exp

(D.

Can you

generalize this result? 12. (a) Let A E C and show that exp{!.>.(z for 0 < Izl <

00,

+D}

where for n an =

=

ao

+~an(z' +~)

0

~

"

~ f e~ cos ( cos nt dt o

(b) Similarly, show

{ ( I)}

expp z - ; for 0 < Izl <

00,

where bn

=

~

=

bo

+ ~ bn 00

(

z"

1t)

+ ---;n( _

f "

cos (nt-.>. sin t) dt.

o

13. Let R > 0 and G = {z: Izl > R}; a functionf: G ---')- C has a removable singularity, a pole, or an essential singularity at infinity if f(Z-I) has, respectively, a removable singularity, a pole, or an essential singularity at z = O. If f has a pole at 00 then the order of the pole is the order of the pole of f(Z-I) at z = O. (a) Prove that an entire function has a removable singularity at infinity iff it is a constant. (b) Prove that an entire function has a pole at infinity of order m iff it is a polynomial of degree m. (c) Characterize those rational functions which have a removable singularity at infinity. (d) Characterize those rational functions which have a pole of order m at infinity. 14. Let G = {z: 0 < lzl < I} and let f: G ---')- C be analytic. Suppose that y is a closed rectifiable curve in G such that nCy; a) = 0 for all a in C - G. What is J1f? Why? 15. Let f be analytic in G = {z: 0 < Iz-ai < r} except that there is a sequence of poles {an} in G with an ---')- a. Show that for any w in C there is a sequence {zn} in G with a = lim z" and w = lim f(zn).

112

Singularities

16. Determine the regions

III

which the functions

f (z) = (sin 1. )-, z

and

g(z)= Ia'(t-z)-Jdt are analytic. Do they have any isolated singularities? Do they have any singularities that are not isolated? 17. Letfbe analytic in the region G=ann(a;O,R). Show that if ff)f(x+ iy )/2 dx qy < 00 then f has a removable singularity at z = a. Suppose that p > 0 and f f)f(x+ iy)/P dxqy < 00; what can be said about the nature of the singularity at z = a? § 2. Residues

The inspiration behind this section is the desire for an answer to the following question: If f has an isolated singularity at z = a what are the possible values for f J when y is a closed curve homologous to zero and not passing through a? If the singularity is removable then clearly the integral will be zero. If z = a is a pole or an essential singularity the answer is not always zero but can be found with little difficulty. In fact, for some curves y, the answer is given by equation (1.12) with n=-1. 2.1 Definition. Let f have an isolated singularity at z

fez) =

= a and let

00

L

n= -

a.(z-a)" 00

be its Laurent Expansion about z = a. Then the residue of fat z = a is the coefficient a _ l ' Denote this by Res (/; a) = a _ l ' The following is a generalization of formula (l.l2) for n = -I. 2.2 Residue Theorem. Let f be analytic in the region G except for the isolated singularities ai' a 2 , ••. ,am' If y is a closed rectifiable curve in G which does not pass Ihrough any of the points a k and if y=:O in G then 1 m -2. f = n ( y; a k ) Res (j; a k ) • 71"1

J Y

L

k=l

Proof. Let mk = n( y; ak) for 1~ k :::; m, and choose positive numbers rJ, ... ,rm such that no two disks B(ak;rd intersect, none of them intersects {y}, and each disk is contained in G. (This can be done by induction and by using the fact that y does not pass through any of the singularities.) Let Yk (I) = ak + rk exp( - 271"imkt) for 0:::; t :::; I. Then for I '$ j :::; m m

n(y;a)+ ~ n(Yk;a)=O. k= J

m

n(y;a)+ ~ n(Yk;a)=O k= J

Residues

for all a not in G-{al, ... ,am }. Since Theorem IV.5.7 gives

J

113

is analytic

In

G-{al, ... ,am }

m

2.3

0=

fJ+ L f J. k= I

y

If J(z) =

I:

y,

bn(z - ad n is the Laurent expansion about z

-oc

series converges uniformly on aB(a k; rk )· Hence f

f (z b- f (z -

But

=

ak

then this

oc

a k )" = 0 if

11

"* -}

J = I: bnf (z

y,

- oc

- ak)"·

Yk

since (z - ad" has a primitive. Also

Yk

1

y,

a k )-1

= 2'ITin(Yk; ak)Res(f; ak)' Hence (2.3) implies the de-

sired result. I11III Remark. The condition in the Residue Theorem that J have only a finite number of isolated singularities was made to simplify the statement of the theorem and not because the theorem is invalid when J has infinitely many isolated singularities. In fact, if J has infinitely many singularities they can only accumulate on aGo (Why?) If r=d({y},JG) then the fact that y~O gives that n(y; a) =0 whenever d(a; aG) < tr. (See Exercise IV.7.2.) The Residue Theorem is a two edged sword; if you can calculate the residues of a function you can calculate certain line integrals and vice versa. Most often, however, it is used as a means to calculate line integrals. To use it in this way we will need a method of computing the residue of a function at a pole. Suppose 1 has a pole of order m ::::: I at z = a. Then g(z) = (z-ar/(z) has a removable singularity atz = aandg(a) # O.Letg(z) = bo+b1(z-a)+ ... be the power series expansion of g about z = a. It follows that for z near but not equal to a, bo bm - I ~ 'k I(z) = - ( ) .. + ... + (--) + 1... bm+k(z-a). z-a z-a k=O This equation gives the Laurent Expansion of 1 in a punctured disk about z = a. But then Res (f; a) = bm - 1 ; in particular, if z = a is a simple pole Res (f; a) = g(a) = lim (z-a)/(z). This is summarized as follows. z-+a

2.4 Proposition. Suppose 1 has a pole (z-a)m/(z); then

Res (f; a)

01 order m at z 1

= a and put g(z) =

g(m - I )(a). (m-l)! The remainder of this section will be devoted to calculating certain integrals by means of the Residue Theorem 2.5 Example. Show

=

114

Singularities

2

z If fez) = - 14 then! has as its poles the fourth roots of -l. These are

+z

exactly the numbers e iO where (J

rr 3rr 5rr

=-

4' 4 ' 4 '

7rr

and 4 .

Let

an =

exp({~ +(n-l)~J)

for n = 1, 2, 3, 4; then it is easily seen that each an is a simple pole off. Consequently, Res (f; al) = lim (z-al)!(z) = ai(al-a2)-1(al-a3)-1(al-a4)-1

(rri) 4 .

l-i

= 4~2 = ! exp Similarly Res (f; a 2 )

=

~~~i = t exp (-!rri).

Now let R > 1 and let y be the closed path which is the boundary of the upper half of the disk of radius R with center zero, traversed in the counter-clockwise direction. The Residue Theorem gives

-R -I

0

1 R

~f! = Res (f; a1)+Res (f; a2)

2m

1

-i

2~2

But, applying the definition of line integral,

This gives 2.6

Residues

115 4il For 0 S t S'TT, 1+ R 4e lies on the circle centered at 1 of radius R4; hence II + R 4e4i' l ~ R4-1. Therefore

2

and since ~ ~ 0 for all x in IR, it follows from (2.6) that l+x

f~ co

-

l+x

fR

dx = lim R .... oo

-R

00

x 24 dx l+x

7T

= ../2 2.7 Example. Show

f 00

Sin

xd _

--;- x-

o

7T

2·

ei % The functionJ(z) = - has a simple pole at z = O. If 0 < r < R let Y be the

z

closed curve that is depicted in the adjoining figure. It follows from Cauchy's

Theorem that 0

2.8

Jy f

=

o~ t

Breaking y into its pieces,

dx+ f'>+ T'>+ f'> -R

YR

y,

where YR and Yr are the semicircles from R to - Rand - r to r respectively. But

f

x I feiX-e-'iX dx --dx=-

R

R

Sin

,

x

2i

x

r

R

1 =2i

-r

Ie-dx+1 I -dx ex x 2; i

,

ix

"

-R

116

Singularities

Also

I

I n

e iZ ~dz = i

exp (iRe i9 ) de

o

YR

: ; oJlexp (i R eiB)1 de 1[

Jexp ( - R sin e) de 1t

o

By the methods of calculus we see that, for 0 > 0 sufficiently small, the largest possible value of exp (- R sin 8), with 0 ::; e ::; 7r - 0, is exp (- R sin 0). (Note that 0 does not depend on R if R is larger than 1.) This gives that

I~~:

J

,,- 8

+

dz ::; 20

exp ( - R sin 8) de

YR

5

< 20+11" exp (-R sin 0). If

E

t

> 0 is given then, choosing 0 <

there is an Ro such that exp ( - R

E,

E

sin 0) < - for ali R > Ro. Hence 3'IT lim R .... co

J

e iZ

-

dz

=

O.

Z

YR

eiz_l Since - - has a removable singularity at z = 0, there is a constant

z

iZ

M > 0 such that /e z

11::;

M for

f

eiZ -

Izl ::;

1. Hence,

I

- z - dz ::::; 7rrM;

Yr

that is,

But

I~

°

= lim

dz

= -'ITt

' .... 0

I

eiZ-1

- - - dz.

for each r so that -7ri = lim , .... 0

z

I

e iZ - dz z

Yr

So, if we let r -+ 0 and R -+

00

in (2.8)

117

Residues

Notice that this example did not use the Residue Theorem. In fact, it could have been presented after Cauchy's Theorem. It was saved until now because the methods used to evaluate this integral are the same as the methods used in applying the Residue Theorem. 2.9 Example. Show that for a > I,

f '"

o "0

If z = el then

z=

dO a + cos 0 =

'TT

.ja2 - I .

I

- and so

z

( + -I)

a+cos B = a+t(z+z} = a+-!- z Hence

f '"

o

z

=

z2+2az+1 . 2z

tf f 2",

dB a+cos B -

dB a+cos B

0

dz z2+2a z+ I

= -i

y

where y is the circle Izl = 1. But z2+2a z+ I = (z-or:) (z-{3) where or: = -a+(a 2-I)t, fJ = -a-(a 2 -I)t. Since a > 1 it follows that 1Or:1 < I and IfJl > 1. By the Residue Theorem

f

dz

Z2

'TTi

+ 2a z + I = .ja2 -

I ;

y

by combining this with the above equation we arrive at

f

"

o

dB a + cos B =

'TT

.ja2 - I .

2.10 Example. Show that

f 00

o

log x dx 1+x2

=

O.

To solve this problem we do not use the principal branch of the logarithm. Instead define log z for z belonging to the region G = {Z

E

C: z "# 0 and

-2'TT

3'TT} ; < arg z W

L

7T

cot 'ITa = -1 a

+

2: -2a00

11=1

a 2 _n 2

for a =f. an integer. 9. Use methods similar to those of Exercises 6 and 8 to show that

for a =f. an integer. 10. Let y be the circle Izl Show that

and let m and n be non-negative integers.

(± 1)p(n+2p)! p! (n+p)! '

o

if m

= 2p+n,

p ?:: 0

otherwise

II. In Exercise 1.12, consider all and bn as functions of the parameter A and use Exercise 10 to compute power series expansions for anCA) and bnCA). (bnCA) is called a Bessel function.) 12. Let f be analytic in the plane except for isolated singularities at ai' a2, ... , am' Show that Res (f; 00) =

m

2: Res (f; a k=l

k )·

The Argument Principle

123

(Res(j; (0) is defined as the residue of -Z-1(Z-I) at z=O. Equivalently, Res(f; (0) = - - 12. If when y(t) = Re", 0::; I ::; 21)", for sufficiently large R.) 1)"1 .

What can you say if j has infinitely many isolated singularities? 13. LetJbe an entire function and let a, b E C such that lai < Rand Ibl < R. Ify(t) = Reit,O::; t:::; 27T,eva!uateL[(z-a)(z-b)]-IJ(z)dz. Use this result to give another proof of Liouville's Theorem.

§3 The Argument Principle Suppose thatJis analytic and has a zero of order mat z = a. SoJ(z) (z-a)mg(z) where g(a) of O. Hence

=

I'(z) m g'(z) -=-+-J(z) z-a g(z)

3.1

and g'lg is analytic near z = a since g(a) of O. Now suppose that J has a pole of order m at z = a; that is, J(z) = (z - a) -mg(z) where g is analytic and g(a) of O. This gives I'(z) .............. =_-m ........... J(z) Z-Q

3.2

g'(z) + ....... g(z)

and again g'!g is analytic near z = a. Also, to avoid the phrase "analytic except for poles" which may have already been used too frequently, we make the following standard definition.

3.3 Definition. If G is open andJis a function defined and analytic in G except for poles, then J is a meromorphic Junction on G. Suppose that J is a meromorphic function on G and define f: G --)- Coo by setting J(z) = co whenever z is a pole of I It easily follows that J is continuous from G into Coo (Exercise 4). This fact allows us to think of meromorphic functions as analytic functions with singularities for which we can remove the discontinuity of f, although we cannot remove the nondifferentiability of I 3.4 Argument Principle. Let J be meromorphic in G with poles PI' P2' ... ,Pm and zeros ZI' Z2, . . . ,Zn counted according to multiplicity. IJ y is a closed rectifiable curve in G with y:::::; 0 and not passing through PI' ... ,Pm ; ZI' ••• , Zn; then

3.5

I fI'CZ)

-.

J(z)

27T1

dz =

L n(y; Zk) k=1 n

-

L n(y;pJ. j=1 m

y

Prool By a repeated application of (3.1) and (3.2) I'(z)

J(z)

f

f.

_1__ _1_ k=lz-zk r 1Z - Pj

+ g'(z).. g(z)

124

Singularities

where g is analytic and never vanishes in G. Since this gives that g'lg is analytic, Cauchy's Theorem gives the result. III Why is' this called the "Argument Principle"? The answer to this is not completely obvious, but it is suggested by the fact that if we could define logf(z) then it would be a primitive for I'll Thus Theorem 3.4 would give that as z goes around y, logf(z) would change by 27riK where K is the integer on the right hand side of (3.5). Since 27riK is purely imaginary this would give that 1m log fez) = arg fez) changes by 27TK. Of course we can't define log fez) (indeed, if we could then L f'lf = 0 since f'lf has a primitive). However, we can put the discussion in the above paragraph on a solid logical basis. Since no zero or pole off lies on y there is a disk B(a; r), for each a in {y}, such that a branch of logf(z) can be defined on Bea; r) (simply select r sufficiently small that fez) #- or 00 in B(a; r». The balls form an open cover of {y}; and so, by Lebesgue's Covering Lemma, there is a positive number E > 0 such that for each a in {y} we can define a branch of log fez) on B(a; E). Using the uniform continuity of y (suppose that y is defined on [0, 1]), there is a partition 0 = to < t 1 < ... < tk = I such that y(t)EB(y(tj_l); E) for lj-l.:s; t.:s; tj and I .:s;j.:s; k. Let I j be a branch of log f defined on B(y(tj-I); E) for 1 .:s;j.:s; k. Also, since thej-th and U+ 1)-5t sphere both contain y(t) we can choose II, ... , I~ so that II (y(tl» = 12(y(tj»; t 2(y(t2) = (3(yt 2»;··· ; (k-I(y(tk-l» = tk(y(t k - 1 ) · If I' j is the path y restricted to [t j _ I, t J then, since tj = f'lf,

°

for 1 .:s; j .:s; k. Summing both sides of this equation the right hand side "telescopes" and we arrive at

where a = 1'(0) = 1'(1). That is, tk(a)-tl(a) = 27TiK. Because 2niK is purely imaginary we get 1m tkCa)-lm IICa) = 27TK. This makes precise our contention that as z traces out y, arg f{z) changes by 27T K. The proof of the following generalization is left to the reader (Exercise 1). 3.6 Theorem. Letfbe meromorphic in the region G with zeros Z1, Z2 •••• ' Zn and poles PI' ...• Pm counted according to multiplicity. If g is analytic in G and y is a closed rectifiable curve in G with y ~ 0 and not passing through any z i or Pj then 1 j~ f' . g - = 27T1 f

L g(zi)n(y; Zi) n

;=1

-

L m

j=1

g(p)n(y;p).

y

We already know that a onc-one analytic function f has an analytic inverse (IV. 7.6). It is a remarkable fact that Theorem 3.6 can be used to give

The Argument Principle

125

a formula for calculating this inverse. Suppose R > 0 and that I is analytic and one-one 011 B(a; R); let n = I[B(a; R)]. If Iz-al < Rand g = I(z) E n then I( w) - g has one, and only one, zero in B(a; R). If we choose g( w) == w, Theorem 3.6 gives

f

z = _1 wf'(w) dw 2'T1'i I(w)-g y

where y is the circle Iw-al = R. But z =I-I(g); this gives the following 3.7 Proposition. Let I be analytic on an open set containing B(a; R) and suppose that I is one-one on B(a; R). II n = I[B(a; R)] and y is the circle Iz-al = R thenl-l(w) is definedlor each w in n by thelormula

f

I-I(w) = _1 zf'(z) dz. 2'T1'i I(z)-w y

This section closes with Rouche's Theorem.

3.8 _Rouche's Theorem. Suppose J and g are meromorphic in a neighborhood

oj B (a; R) with no zeros or poles on the circle y = {z: Iz - al = R}. IJ Zf' Zg (Pf , Pg ) are the number oj zeros (poles) oj J and g inside y counted according to their multiplicities and if IJ(z) + g(z)1 < IJ(z)1 + 1 g(z)1 on y, then Proof. From the hypothesis

I IJ(z) I g(z) +1 < g(z) +1 IJ(z) on y. If A = J(z)/g(z) and if A is a positive real number then this inequality becomes A + 1 < A + 1, a contradiction. Hence the meromorphic function f / g maps y into n = C - [0, (0). If I is a branch of the logarithm on n then l(f(z)jg(z» is a well-defined primitive for (f/g)'(f/g)-l in a neighborhood of y. Thus

£ = 2~i f[ j - ~]

0=

2~i

y

(f/g)'(f/g)-l

y

=(Zf- Pf) -(Zg- Pg). •

This statement of Roucbe's Theorem was discovered by Irving Glicksberg (A mer. Math. Monthly, 83 (1976), 186-187). In the more classical statements of the theorem, f and g are assumed to satisfy the inequality

126

Singularities

1/+ gl < Igl on y. This weaker version often suffices in the applications as can be seen in the next paragraph. Rouche's Theorem can be used to give another proof of the Fundamental Theorem of Algebra. If p(z) = z"+a 1 z,,-1 + ... +a" then

p(z) a1 a" -=1+-+"'+z" z z" and this approaches 1 as z goes to infinity. So there is a sufficiently largt. number R with

/p;:) - 11 < 1

for Izl = R; that is, Ip(z)-z"1 < Izl" for Izl = R. Rouchc's Theorem says that p(z) must have n zeros inside Izl = R. We also mention that the use of a circle in Rouchc's Theorem was a convenience and not a necessity. Any closed rectifiable curve" with" """" 0 in G could have been used, although the conclusion would have been modified by the introduction of winding numbers.

Exercises I. Prove Theorem 3.6. 2. Suppose / is analytic on .8(0; I) and satisfies 1/(z)1 < I for Izl = I. Find the number of solutions (counting multiplicities) of the equation/(z)=zn where n is an integer lar~r than or equal to I. 3. Let / be analytic in B(O; R) with /(0)=0, /,(0)*0 and /(z)*O for 0< Izl S R. Put p=min{if(z)I:lzl = R} >0. Define g: B(O; p)_C by

g(w)= _I 2'1Ti

1 y

z/,(z) dz /(z)-w

where y is the circle Izl = R. Show that g is analytic and discuss the properties of g. 4. If / is meromorphic on G andj: G-C"" is defined by j(z) = 00 when z is a pole of / and j(z) = fez) otherwise. show that j is continuous. 5. Let f be meromorphic on the region G and not constant; show that neither the poles nor the zeros of f have a limit point in G. 6. Let G be a region and let B(G) denote the set of all analytic functions on G. (The letter" B" stands for holomorphic. Some authors call a differentiable function holomorphic and call functions analytic if they have a power series expansion about each point of their domain. Others reserve the term "analytic" for what many call the complete analytic function, which we will not describe here.) Show that B(G) is an integral domain; that is, B(G) is a commutative ring with no zero divisors. Show that M(G), the meromorphic functions on G, is a field. We have said that analytic functions are like polynomials; similarly, meromorphic functions are analogues of rational functions. The question

The Argument Principle

127

arises, is every meromorphic function on G the quotient of two analytic functions on G? Alternately, is M(G) the quotient field of H(G)? The answer is yes but some additional theory will be required before this answer can be proved. 7. State and prove a more general version of Rouche's Theorem for curves other than circles in G. 8. Is a non-constant meromorphic function on a region G an open mapping of G into C? Is it an open mapping of G into Coo ? 9. Let A > I and show that the equation A-z-e- z = 0 has exactly one solution in the half plane {z: Re z > O}. Show that this solution must be real. What happens to the solution as A---+ I ? 10. Let f be analytic in a neighborhood of D= ii(O; I). If If(z)1 < I for Izl = I, show that there is a unique z with Izi < I and f(z) = z. If If(z)1 ~ I for Izl = I, what can you say?

Chapter VI The Maximum Modulus Theorem This chapter continues the study of a property of analytic functions first seen in Theorem IV. 3.1 I. In the first section this theorem is presented again with a second proof, and other versions of it are also given. The remainder of the chapter is devoted to various extensions and applications of this maximum principle.

§l. The Maximum Principle Let n be any subset of IC and suppose a is in the interior of n. We can, therefore, choose a positive number p such that B(a; p) C [2; it readily follows that there is a point f in n with If I > 14 To state this another way, if a is a point in n with lal 2': I~i for each ~ in the set n then ex belongs to on.

1.1 Maximum Modulus Theorem---First Version. Iff is analytic in a region G and a is a point in G with If(a)[ 2': If(z)[ for all z in G then f must be a constant function. Proof Let 0 = f(G) and put a = f(a). From the hypothesis we have that IIXI 2': If I for each gin il; as in the discussion preceding the theorem IX is in 00 n n. In particular, the set .0 cannot be open (because then .0 n 8.0 = [J). Hence the Open Mapping Theorem (IV. 7.5) says thatfmust be constant. II

1.2 Maximum Modulus Theorem-Second Version. Let G be a bounded ope/! set in IC and suppose f is a continuous function on G- which is analytic in G. Then max {If(z)l: z E G-} = max {If(z)l: z E oG}. Proof Since G is bounded there is a point a E G-- such that I/(a)1 2': I/(z) I for all z in G-. Iflis a constant function the conclusion is trivial; iflis not constant then the result follows from Theorem 1.1. II Note that in Theorem 1.2 we did not assume that G is connected as in Theorem 1.1. Do you understand how Theorem 1.1 puts the finishing touches on the proof of 1.2? Or, could the assumption of connectedness in Theorem 1.1 be dropped? Let G = {z = x+iy: -t1T < y < 11T} and put/(z) = exp [exp z]. Then I is continuous on G- and analytic on G. If z E oG then z = x ± 11Ti so I/(z) I = lexp (± ieX)1 = 1. However, as x goes to infinity through the real numbers, I(x) -+ 00. This does not contradict the Maximum Modulus Theorem because G is not bounded. In light of the above example it is impossible to drop the assumption of the boundedness of G in Theorem 1.2; however, it can be replaced. The 128

The Maximum Principle

129

substitute is a growth condition on [f(z) [ as z approaches infinity. In fact, it is also possible to omit the condition that f be defined and continuous on G-. To do this, the following definitions are needed. 1.3 Definition. Iff: G -+ IR and a E G- or a = 00, then the limit superior of fez) as z approaches a, denoted by lim sup fez), is defined by

lim supf(z) = lim sup {fez): z r-+O+

z ...... a

E

G n B(a; r)}

(If a = 00, B(a; r) is the ball in the metric of COO') Similarly, the limit. inferior offez) as z approaches a, denoted by lim inf fez), is defined by z-+a

lim inff(z) = lim inf {fez): z r-+O+

z-+a

E

G n B(a; r)}.

It is easy to see that lim fez) exists and equals

lim inf fez).

IX

z-+a

iff

0:

= lim sup fez) =

If G c C then let 000G denote the boundary of G in Coo and call it the extended boundary of G. Clearly 000G = oG if G is bounded and 000G = oG U {oo} if G is unbounded. After these preliminaries the final version of the Maximum Modulus Theorem can be stated. 1.4 Maximum Modulus Theorem-Third Version. Let G be a region in C and f an analytic function on G. Suppose there is a constant M such that lim sup Ifez) [ :::; M for all a in 0ooG. Then Ifez) [ :::; M for all z in G. z-+a

Proof Let S > 0 be arbitrary and put H = {z E G: If(z) [ > M+S}. The theorem will be demonstrated if H is proved to be empty. Since If I is continuous, H is open. Since lim sup [f(z) [ :::; M for each z->a

a in 0ooG, there is a ball B(a; r) such that If(z) I < M+S for all z in G n B(a; r). Hence H- c G. Since this condition also holds if G is unbounded and a = 00, H must be bounded. Thus, H - is compact. So the second version of the Maximum Modulus Theorem applies. But for z in oH, If(z) [ = M+8 since H- c {z: If(z) I ~ M+S}; therefore, H = 0 or fis a constant. But the hypothesis implies that H = 0 if f is a constant. • Notice that in the example G = {z: [1m z[ < 1-1T}, fez) = exp (e f satisfies the condition lim sup [f(z) [ :::; 1 for all a in oG but not for a = 00. Z

),

z-+a

Exercises 1. Prove the following Minimum Principle. If f is a non-constant analytic function on a bounded open set G and is continuous on G-, then either f has a zero in G or If I assumes its minimum value on oG. (See Exercise IV. 3.6.) 2. Let G be a bounded region and suppose f is continuous on G- and

130

The Maximum Modulus Theorem

analytic on G. Show that if there is a constant c ~ 0 such that If(z) 1 = c for all z on the boundary of G then either f is a constant function or f has a zero in G. 3. (a) Letfbe entire and non-constant. For any positive real number c show that the closure of {z: /f(z)/ < c} is the set {z: If(z)1 :s; c}. (b) Let p be a non-constant polynomial and show that each component of {z:lp(z)1 < c} contains a zero of p. (Hint: Use Exercise 2.) (c) If p is a non-constant polynomial and c> 0 show that {z: Ip(z)1 = c} is the union of a finite number of closed paths. Discuss the behavior of these paths as c -> XJ. 4. Let 0 < r < R and put A = {z: r :s; Izl :s; R}. Show that there is a positive number £ > 0 such that for each polynomial p, sup {/p(Z)-Z-l/:

ZE

A} ~

E

This says that Z-1 is not the uniform limit of polynomials on A. 5. Let f be analytic on B(O; R) with If(z)1 :s; M for Izi :s; Rand If(O) I = a > O. Show that the number of zeros off in B(O; lR) is less than or equal to

IO~ 2 log ( ; ) .

Hint: If Z I'

... ,

z. are the zeros off in B(O; 1R), consider

the function

and note that g(O)

= f(O). (Notation:

{I ak = a 1a2 ... an.)

k~1

6. Suppose that bothfand g are analytic on B(O; R) with If(z)1 = /g(z)/ for Izi = R. Show that if neither f nor g vanishes in B(O; R) then there is a constant '\, 1,\1 = I, such thatf = '\g. 7. Let f be analytic in the disk BCO; R) and for 0 :s; r < R define A(r) = max {Re/(z): Izl = r}. Show that unless 1 is a constant, A(r) is a strictly increasing function of r. 8. Suppose G is a region,!: G -)- C is analytic, and M is a constant such that whenever z is on GroG and {zn} is a sequence in G with z = lim Zn we have lim sup /f(zn) I :s; M. Show that I/(z)1 :s; M, for each z in G. §2. Schwarz's Lemma

2.1 Schwarz's Lemma. Let D = {z: Izl < I} and suppose f is analytic on D with (a) If(z) 1:s; 1 for z in D, (b) f(O) = O.

Then 11'(0)1 :s; I and If(z)1 :s; Izlfor all z in the disk D. Moreover ifl1'(O») = 1 or if I/(z)1 = /zlJor some z #- 0 then there is a constant c, lei = 1, such that f(w) = cw for all w in D.

Schwarz's Lemma Proof. Define

g:

D~C by g(z)= f(z) z

for Z'FO and g(O) =1'(0); then

g is

analytic in D. Using the Maximum Modulus Theorem, Ig(z)l:s; r- for Izl $ rand 0 < r < l. Letting r approach 1 gives Ig(z)1 $ 1 for all z in D. That is, If(z)l:s; Izl and 11'(0)1 = Ig(O)1 :s; 1. If If(z)1 = Izi for some z in D, Z9"O, or 11'(0)1 = 1 then Igl assumes its maximum value inside D. Thus, again applying the Maximum Modulus Theorem, g(z)= e for some constant e with lei = l. This yields f(z)= ez and completes the proof of the theorem. II We will apply Schwarz's Lemma to characterize the conformal maps of the open unit disk onto itself. First we introduce a class of such maps. If lal < 1 define the Mobius transformation: 1

z-a fP,,(z) = - - _

l-az

Notice that fPa is analytic for Jzl < lal- 1 so that it is analytic in an open disk containing the closure of D = {z: Izl < 1}. Also, it is an easy matter to check that for

Izi < 1. Hence qJa maps D into itself in a one-one fashion. Let 8 be a real number; then

eiO -

I = e

iO -

a

ii

I

I

= 1 This says that qJa(i3D) = aD, and, from the preceding material, qJ(D) = D. These facts, and other pertinent information which can be easily checked, are summarized as follows. 2.2 Proposition. 1/ lal < 1 then fPa is a one-one map 0/ D = {z: Izi < I} onto itself; the inverse of fPa is fP-a' Furthermore, fP" maps 8D onto 8D, fPia) = 0, fP~(O) = l-laI 2 , andfP~(a) = (1-laI 2)-1. Let us see how these functions fP" can be used in applying Schwarz's Lemma. Suppose / is analytic on D with I/(z) I $ 1. Also, suppose lal < 1 and/(a) = IX (so IIXI < 1 unless/is constant). Among all functions/having these properties what is the maximum possible value of If'(a) I? To solve this problem let g = fP" 0 / 0 fP-.' Then g maps D into D and also satisfies g(O) = rp,,(/(a)) = fP,,(Q() = O. Thus we can apply Schwarz's Lemma to obtain that Ig'(O)1 $ 1. Now obtain an explicit formula for g'(O). Applying the chain rule

132

The Maximum Modulus Theorem

g'(O) = (qJl> f)'( qJ_,,(O))qJ'_JO) 0

= ('1'..

0

1)'(0)(1-101 2 )

= 'P~(a.)f'(0)

(1-101 2)

= 1- 10 12 1'(0). l-11X12

Thus,

11'(0)1

2.3

:5

l-11X12 .

l-lal 2

Moreover equality will occur exactly when Ig'(O)1 = I, or, by virtue of Schwarz's Lemma, when there is a constant c with lei = I and

2.4

f (z) = q; _ a (cCPa (z»)

Izl < 1. We are now ready to state and prove one of the main consequences of Schwarz's Lemma. Note that if lei = 1 and lal < 1 then f = eCPa defines a one-one analytic map of the open unit disk D onto itself. The next result says that the converse is also true. for

2.5 Theorem. Let f: D --'> D be a one-one analytic map of D onto itself and suppose f( a) = O. Then there is a complex number c with lei = 1 such that

f

=

CqJ".

Proof Since I is one-one and onto there is an analytic function g: D -+ D such that g{ I(z» = z for Izl < I. Applying inequality (2.3) to both I and g gives II'(a)1 :5 (I -Ial l ) - I and Ig'(O)! :5 I -lal 2 (since g(O) = a). But since

1= g'(O)I'(a), II'(a)[ = (1-laI 2)-I. Applying formula (2.4) we have that f = CqJ(1 for some e, lei = I. II

Exercises

1. Suppose If(z)1 :5 1 for Izl < 1 and f is a non-constant analytic function. By considering the function g: D --'> D defined by g(z)

I(z)-a

= I-a/(z)

where a = I(O}, prove that

I/(O)I-I~I :5 I/(z)1 :5 1/(0)1 +fl

1+ I/(O)llzl

1-

1/(0)1 Izi

for Izi < 1. 2. Does there exist an analytic function/: D---)o D with/(t) =

,?

i

andl'(t) =

133

Convex functions and Hadamard's Three Circles Theorem

3. Suppose f: D-+C satisfies Re f( z) ~ 0 for all z in D and suppose that f is analytic and not constant. (a) Show that Ref(z»O for all z in D. (b) By using an appropriate Mobius transformation, apply Schwarz's Lemma to prove that if f(O) = I then

I/(z) I ~ I + Izl

l-lzl

for Izi < 1. What can be said if1(0) =F I? (c) Show that if f(O) = I, f also satisfies I/(z)1

~ 1-izi 1+ Izi

.

(Hint: Use part (a». 4. Prove Caratheodory's Inequality whose statement is as follows: Letf be analytic on li(O; R) and let M(r)=max{lf(z)I:lzl=r}, A(r)= max{ Ref(z): Izl = r}; then for 0< r < R. if A (r) ~ 0,

M(r) ~ R+r [A(R) + 1/(0)1] R-r (Hint: First consider the case where 1(0) = 0 and examine the function

g(z) =/(Rz) [2A(R)+/(Rz>r 1 for Izl < 1.) 5. Letl be analytic in D = {z: Izl < I} and suppose that I/(z) I all z in D. (a) If I(zk) = 0 for 1 ~ k ~ n show that

~

M for

°

for Izl < 1. (b) If !(zk) = for 1 ~ k.:5; n, each zk"* 0, and 1(0) = Me ia (z l z2'···' zn)' find a formula for f. 6. Suppose I is analytic in some region containing B(O; 1) and II(z) I = 1 where Izl = 1. Find a formula for f (Hint: First consider the case where I has no zeros in B(O; 1).) 7. Suppose I is analytic in a region containing B(O; 1) and I/(z)1 = 1 when Izl = 1. Suppose that I has a zero at z = i(1 + i) and a double zero at z = 1· Can 1(0) = 1? 8. Is there an analytic function/ on B(O; 1) such that I/(z) I < 1 for Izl < 1, 1(0) = t, and 1'(0) = !? If so, find such an f Is it unique? §3. Conpex!unctiollS lind HtUlIImllrd'. T""ee Circle. Tlteorem In this section we will study convex functions and logarithmically convex functions and show that such functions appear in connection with the study of analytic functions.

134

The Maximum Modulus Theorem

3.1 Definition. If [a, b] is an interval in the real line, a functionf: [a, b] is convex if for any two points Xl and X2 in [a, b]

-+ ~

whenever 0 ::; t ::; 1. A subset A c IC is convex if whenever z and IV are in A, tz + (I - t)w is in A for 0 ::; t ::; 1; that is, A is convex when for any two points in A the line segment joining the two points is also in A. (See IV. 4.3 and IV. 4.4.) What is the r(,lation between convex functions and convex sets? The answer is that a function is convex if and only if the portion of the plane lying above the graph of the function is a convex set. 3.2 Proposition. A function f: [a, b]

-+ ~

is convex

iff the

set

A = {(x,y):a::; x::; b andf(x)::; y} is convex. Proof. Suppose f: [a, b] -+ ~ is a convex function and let (Xl' y,) and (Xz, Y2) be points in A. If 0 ::; t ::; 1 then, by the definition of convex function, f(tx 2 +(I-t)x,)::; (f(x 2)+(l-t)f(x,) ::; tYz+(l-t)y,. Thus t(X2' Yz)+ (I-t) (x" y,) = (tx 2 +(1-t)x" tYz+(l-t)y,) is in A; so A is convex. Suppose A is a convex set and let x" X 2 be two points in [a, b]. Then (tx 2 +(l-t)x" tf(x 2)+(I-t)f(x,» is in A if 0::; t::; 1 by virtue of its convexity. But the definition of A gives that f(tx2+(I-t)x,)::; (f(X2)+ (1- t)f(x ,); that is,f is convex . • The proof of the next proposition is left to the reader.

3.3 Proposition. (a) A function f: [a, b]

-+ ~

iff for

is convex

n

x" ... ,x" in [a, b] and real numbers t" ... , t" 2: 0 with

L tk

k='

any points

= 1,

fet, tkXk) ::; kt, td(Xk)'

(b) A set A

c

IC is convex

numbers t" ... , t" 2: 0 with

n

iff for

L tk

k~'

= 1,

any points z" ... ,z" in A and real n

L tkzk belongs to

k~'

A.

What are the virtues of convex functions and sets? We have already seen the convex sets used in connection with complex integration. Also, the fact that disks are convex sets has played a definite role, although this may not have been apparent since this fact is taken for granted. The use of convex functions may not be so familiar to the reader; however it should be. In the first course of calculus the fact (proved below) that f is convex when f" is non-negative is used to obtain a local minimum at a point to whenever !'(to) = O. Moreover, convex functions (and concave functions) are used to obtain inequalities. Iff: [a, b] -+ IR is convex then it follows from Proposition 3.2 that f(x) ::; max {f(a),f(b)} for all x in [a, b]. We now give a necessary condition for the convexity of a function.

Convex functions and Hadamard's Three Circles Theorem

135

3.4 Proposition. A differentiablefunctionf on [a, b] is convex ifff' is increasing. Proof. First assume that f is convex; to show that I' is increasing let a S x < y s b and suppose that 0 < t < 1. Since 0 < (l-t)x+ty-x = t(y-x), the definition of convexity gives that f«l-t)x+ty)-f(x) f(y)-f(x) S . t(y-x) y-x

Letting t --+ 0 gives that

3.5

f'(x) S fey) -f(x) . y-x

Similarly, using the fact that 0 > (l-t)x+ ty- y = (I-t) (x- y) and letting t --+ I gives that

3.6

f'(y) 2: fey) - f(x) . y-x

So, combining (3.5) and (3.6), we have thatf' is increasing. Now supposing thatf' is increasing and that x < u < y, apply the Mean Value Theorem for differentiation to find rand s· with x < r < u < s < y such that f'(r) = feu) - f(x) u-x

and f'(s)

= fCy) - feu)

.

y-u

Since f'(r) S f'(s) this gives that feu) - f(x) u-x

s

fey) - feu) y-u

whenever x < u < y. In particular by letting u = (1- t)x + ty where t < I, f(u)-f(x) < f(y)-f(u) . t(y-x) - (I-t)(y-x) , and hence (I - t) [feu) - f(x)] s t[f(y) - feu)].

o<

This shows that f must be convex . • In actuality we will mostly be concerned with functions which are not only convex, but which are logarithmically convex; that is, logf(x) is convex. Of course this assumes that f(x) > 0 for each x. It is easy to see that a logarithmically convex function is convex, but not conversely.

3.7 Theorem. Let a < b and let G be the vertical strip {x + iy: a < x < b}.

\36

The Maximum Modulus Theorem

Suppose f: G-[a, b] -'>- IR! by

-0;

C is continuous and f is analytic in G.

If

we define M:

M(x) = sup {If(x+ iy)!: - w < y < w}, and If(z)1 < B for all z in G, then log M(x) is a convex fimction. Before proving this theorem, note that to say that 10gM(x) is convex means (Exercise 3) that for a ::s; x < u < y ::s; b, (y-x) log M(u) ::s; (y-u) log Al(x)+(u-x) log M(y) Taking the exponential of both sides gives

3.8 whenever a ::s; x < u < y ::s; b. Also, since log M(x) is convex we have that log M(x) is bounded by max {log M(a), log M(b)}. That is, for a ::s; x ::s; b

M(x) ::s; max {Mea), M(b)}. This gives the following.

3.9 Corollary. If f and G are as in Theorem 3.7 and f is not constant then !f(z)! < sup {If(w)!: WE cG} for all z in G. To prove Theorem 3.7 the following lemma is used. 3.10 Lemma. Let f and G be as in Theorem 3.7 alld further suppose that If(z)1 ::s; 1 for z on oG. Then If(z)1 ::s; I for all z in G.

Proof For each E > 0 let g,(z) = [l+E(z-a)]-l for each z in G-. Then for z = x+iy in GIg,(z)!::s; !Re[l+E(z-a)ll-l =

[I +E(x-a)r 1

::s; 1. So for z in cG if(z)g,(z) I ::s; I. Also, since f is bounded by B in G,

!f(z)g.(z) I ::s; BII +E(z-a)!-)

3.n

::s; BrE 11m zlr 1

So if R = {x+iy: a::s; x::s; b, Iyl ::s; Bid, inequality (3.11) gives If(z)g,(z) I ::s; 1 for z in oR. It follows from the Maximum Modulus Theorem that IfCz)g.(z) I ::s; 1 for z in R. But if 11m zl > BIE then (3.11) gives that IfCz)gcCz) I ::s; 1. Thus for all z in G.

If(z) I ::s; 11 +E(z-a)l· Letting

E

approach zero the desired result follows.

II

Proof of Theorem 3.7. First observe that to prove the theorem we need only establish

Convex functions and Hadamard's Three Circles Theorem

137

for a < u < b (see (3.8). To do this recall that for a constant A > 0, A" = exp (z log A) is an entire function of z with no zeros. So g(z) defined by

g(z) = M(a)(b-zl!(b-a)M(b)(Z-a)!(b-a) is entire, never vanishes, and (because IAxl

= ARe.) for z =

x+iy

Ig(z) I = M(a)(b-x)!(b-a)M(b)(X-al!(b-a).

3.12

(It is assumed here that Mea) and M(b) -# 0. However, if either M(a) or

M(b) is zero then f == 0.) Since the expression on the right hand side of (3.12) is continuous for x in [a, b] and never vanishes, Igl- 1 must be bounded in G-. Also, Ig(a+iy)1 = M(a) and Ig(b+iy)1 = M(b) so that If(z)/g(z) 1 S; 1 for z in 8G; and fig satisfies the hypothesis of Lemma 3.10. Thus If(z) 1

S;

Ig(z)l, z

E

G.

Using (3.12) this gives for a < u < b

M(u) s; M(a)(b-u){(b-alM(b)(u-a)/(&-a) which is the desired conclusion. III Hadamard's Three Circles Theorem is an analogue of the preceding theorem for an annulus. Consider ann (0; R I , R 2 ) = A where 0 < R\ '< R z < 00. If G is the strip {x+iy: log RI < X < log R 2 } then the exponential function maps G onto A and 8G onto 8A. Using this fact one can prove the following from Theorem 3.7 (the details are left to the reader).

3.13 Hadamard's Three Circles Theorem. Let 0 < R 1 < R 2 < 00 and suppose f is analytic and not identically zero on ann (0; R j , R 2)' If R j < r < R 2' define M(r) = max {If(reIB)I: 0 s; () s; 2'lT}. Then for R j < '1 s; , s; '2 < R2 and,[ '2. iogr 2 -logr logr-Iogr, log M(r) S; log M(r t ) + ------------------- log M(r 2 ). iogr 2 -Iogr 1 log r 2 -10g r 1

*'

Another way of expressing Hadamard's Theorem is to say that log M(r) is a convex function of log r. Exercises 1. Let f: [a, b] -0- ~ and suppose that f(x) > 0 for all x and that f has a continuous second derivative. Show that f is logarithmically convex iff r(x)f(x) - [f'(x)] 2 ~ 0 for all x. 2. Show that iff: (a, b) -0- ~ is convex thenfis continuous. Does this remain true iffis defined on the closed interval [a, b]? 3. Show that a function f: Ca, b] -0- ~ is convex iff any of the following equivalent conditions is satisfied: (a) a

S;

x < u < y

S;

1)

f(U) U b gives det ( f(x) x 1 fey) y 1

~

0;

138

The Maximum Modulus Theorem

(b) a S x < u < y S b give/(u)-f(x) S f(y)-f(x) ; u-x y-x (c) a S x < u < y S b givesf(u)-f(x) S f(y)-f(u) . u-x y-u Interpret these conditions geometrically. 4. Supply the details in the proof of Hadamard's Three Circles Theorem. 5. Give necessary and sufficient conditions on the function f such that equality occurs in the conclusion of Hadamard's Three Circles Theorem. 6. Prove Hardy's Theorem : Iff is analytic on B(O; R) and not constant then I(r)

=~ 211'

f

2"

If(re i8 )ld8

o

is strictly increasing and log I(r) is a convex function of log r. Hint: If r 1 < r < r 2 find a continuous function tp: [0, 211'] --* C such that tp(8)f(re i8 ) = If(re i8 ) I and consider the function F(z) = f(ze i8)tp(8)d8. (Note that r is fixed, so rp may depend on r.) 7. Let f be analytic in ann (0; R I , R 2 ) and not identically zero; define

o<

g"

Iz(r) =

~ 211'

f

2"

If(re i8 )1 2 d8.

o

Show that log Iir) is a convex function of log r, Rl < r < R z .

§4. The Phragmen-LindeliJ! Theorem This section presents some results of E. Phragmlm and E. LindelOf (published in 1908) which extend the Maximum Principle by easing the requirement of bounded ness on the boundary. The Phragmen-LindelOf Theorem bears a relation to the Maximum Modulus Theorem which is analogous to the relationship of the following result to Liouville's theorem. If f is entire and If(z) I S 1+ Izlt then f is a constant function. (Prove it!) So it is not necessary to assume that an entire function is bounded in order to prove that it is constant; it is sufficient to assume that its growth as z --* 00 is restricted by 1+ Izlt. The PhragmenLindelof Theorem places a growth restriction on an analytic function f: G --* C as z nears a point on the extended boundary. Nevertheless, the conclusion, like that of the Maximum Modulus Theorem, is thatfis bounded. 4.1 Phragmen-LindelOf Theorem. Let G be a simply connected region and let f be an analytic function on G. Suppose there is an analytic function tp: G --* C which never vanishes and is bounded on G. If M is a constant and 000G = A u B such that: (a) for every a in A, lim suplf(z)1 z_a

s

M;

139

Phragmen-Linde1of Theorem

(b) for every b in B, and TJ > 0, lim sup If(z)1 l!p(z)I'I z"'b

then If(z)l:s; MforallzinG.

~

M;

Proof. Let 1 0 and z = re i9 is sufficiently large,

If(z)II'P(z)l1J :s; P exp (r b-TJr c cos c/J) :s; P exp (r b -7]r c p) But rb_TJrcp=rC(rb-c_TJp). Since b 00. This shows that the growth condition (4.5) is very delicate and can't be improved. 1

1

Exercises I. In the statement of the Phragmen-Lindelof Theorem, the requirement that G be simply connected is not necessary. Extend Theorem 4.1 to regions G with the property that for each z in a'X)G there is a sphere V in Coo centered at z such that V Ii G is simply connected. Give some examples of regions that are not simply connected but have this property and some which don't.

141

The Phragmen-Lindelof Theorem

2. In Theorem 4.1 suppose there are bounded analytic functions 9),92"'" 9n on C that never vanish and doe C = A u B) u ... U En such that condition (a) is satisfied and condition (b) is also satisfied for each 9" and Ek • Prove that If(z)1 ~ M for all z in C. 3. Let C = {z: 1m zl < 11'} and suppose I: G -71[: is analytic and lim sup %->w

I/(z)] '.S M for win cG. Also, suppose A < 00 and a < 1 can be found such that I/(z)1 < exp [A exp (a IRe zl)] for all z in G. Show that I/(z)1 ~ M for all z in C. Examine exp (exp z) to see that this is the best possible growth condition. Can we take a = 1 above? 4. Let f: C -7 I[: be analytic and suppose M is a constant such that lim sup I/(z,,)I ~ M for each sequence {zn} in G which converges to a point in aooG. Show that I/(z)1 ~ M. (See Exercise 1.8). 5. Let f: G -7 I[: be analytic and suppose that G is bounded. Fix Z 0 in 8G and suppose that lim sup If(z)] ~ M for w in cC, w oF zoo Show that if z-~w

lim Iz-zol' I/(z)1 = 0 for every

€

:>

0 then I/(z)1

~

M for every z in aGo

z -+Zo

(Hint: If a¢. G, consider O} and let I: G -7 I[: be an analytic function with lim sup II(z) I :::; M for w in aG, and also suppose that for every E > 0, z-w

lim sup {exp( -E/r) Il(re i8 ) I : 181 < 1n} = O.

r--+oo

Show that I/(z) ~ Ai for all z in G. 7. Let G = {z: Re z > O} and let I: G-·;..I[: be analytic such that/(l) = 0 1

and such that lim sup

I/(z)1

~

M for

IV

in cG. Also, suppose that for every

z~w

8, 0 < 8 < I, there is a constant P such that Prove that

( Hint: Consider I(z)

G~;).)

Chapter VII Compactness and Convergence in the Space of Analytic Functions In this chapter a metric is put on the set of all fixed region G, and compactness and convergence discussed. Among the applications obtained is a Mapping Theorem. Actually some more general results are obtained study spaces of meromorphic functions.

analytic functions on a in this metric space is proof of the Riemann which enable us to also

§I. The space of continuous functiolls C(G,Q) In this chapter (n, d) will always denote a complete metric space. Although much of what is said does not need the completeness of fl, those results which hold the most interest are not true if (n, d) is not assumed to be complete. 1.1 Definition. If G is an open set in C and (l..!, d) is a complete metric space then designate by C(G, n) the set of all continuous functions from G to n. The set C(G, n) is never empty since it always contains the constant functions. However, it is possible that C(G, n) contains only the constant functions. For example, suppose that G is connected and f2 = N = {l, 2, ... }. Iff is in C(G, £1) then f(G) must be connected in n and, hence, must reduce to a point. However, our principal concern will be when n is either C or C". For these two choices of n, C(G, £1) has many non constant elements. In fact, each analytic function on G is in C(G, q and each meromorphic function on G is in C(G, C",) (see Exercise V. 3.4). To put a metric on C(G, [1) we must first prove a fact about open subsets of IC. The third part of the next proposition will not be used until Chapter

VIII. 1.2 Proposition. If G is open in C then there is a sequence {Kn} of compact

subsets of G such that G

=

U Kn. CD

Moreover, the sets Kn can be chosen to

n~l

satisfy the following conditions: (a) Kn c: int Kn+ 1; (b) K c: G and K compact implies K C Knfor some n; (c) Erery component of Ceo - Kn contains a component of C" - G. 142

The space of continuous functions C(G,O)

143

Proof. For each positive integer n let

K. = {z:

Izl ~

n}

n {z: d(z, C-G) ~ ~}

;

since Kn is clearly bounded and it is the intersection of two closed subsets of C, K. is compact. Also, the set {z:

Izl

< n+ I}

n {z: d(z, C-G)

>

_l_} n+l

is open, contains K., and is contained in K.+ l ' This gives that (a) is satisfied. Since it easily follows that G =

00

00

n~.

n~.

U K. we also get that G = U int K.;

so if

K is a compact subset of G the sets {int K.} form an open cover of K. This

gives part (b). To see part (c) note that the unbounded component of Coo -K. (:::> Coo - G) must contain 00 and must therefore contain the component of Coo - G which contains 00. Also the unbounded component contains {z: Izl > n}. So if D is a bounded component of C;, - K. it contains a point z with d(z, C - G) <

Iw - z I <

~.

But by definition this gives a point w in C - G with

n

~. But then z

E

B( w; ~) c

Coo - Kn; since disks are connected and

z is in the component D of Coo - K", B( w;

~)

c D. If D. is the component

of Ceo - G that contains w it follows that D) cD . • If G =

U Kn

where each Kn is compact and Kn c int K.+ 1 , define

n~1

Pn(/' g) = sup {d(f(z), g(z»: z

1.3

E

Kn}

for all functions f and g in C(G, Q). Also define p(/, g)

1.4

=

f

(t)"

n; 1

P.(/' ~L 1+ Pn(/' g)

L

since t(l + t) - 1 ~ 1 for all t ~ 0, the series in (1.4) is dominated by (t). and must converge. It will be shown that P is a metric for C(G, Q). To do this the following lemma, whose proof is left as an exercise, is needed. 1.5 Lemma.

If (S,

d) is a metric space then

Ji- (s, t)

= _ -:d(,--"-s,-..::t)_

1 +d(s, t)

is also a metric on S. A set is open in (S, d) iff it is open in (S, Ji-); a sequence is a Cauchy sequence in (S, d) iff it is a Cauchy sequence in (S, Ji-).

1.6 Proposition. (C(G, Q), p) is a metric space.

144

Compactness and Convergence

Proof It is clear that p(f, g) = p(g,f). Also, since each Pn satisfies the triangle inequality, the preceding lemma can be used to show that p satisfies the 00

triangle inequality. Finally, the fact that G = U Kn gives thatf = g whenever p(f, g) = O. • n=l The next lemma concerns subsets of C(G, Q) x C(G, Q) and is very useful because it gives insight into the behavior of the metric p. Those who know the appropriate definitions will recognize that this lemma says that two uniformities are equivalent. 1.7 Lemma. Let the metric p be defined as in (1.4). If € > 0 is given then there is a 8 > 0 and a compact set K c: G such that for f and gin C(G, Q), sup {d(f(z), g(z»: z

1.8

E

K} < 8 => p(f, g) <

€.

Conversely, if 8 > 0 and a compact set K are given, there is an that for f and gin C(G, Q), p(f, g) <

1.9

€

=> sup {d(f(z), g(z»: z

E

K} < 8.

€

ao

> 0 such

nt

0 is fixed let p be a positive integer such that L < 1€ n=p+1 . t and put K = Kp. Choose 0 > 0 such that 0 :s; t < 0 gIves 1 + t < -rE. Suppose Proof If

€>

fand g are functions in C(G, Q) that satisfy sup {d(f(z), g(z»: z E K} < o. Since Kn c: Kp = K for 1 :s n :s; p, Pn(f, g) < 0 for 1 :s; n :s; p. This gives Pn(f, g)

<

1E

1 +pif, g)

for 1 :s; n :s; p. Therefore p

p(f, g) <

00

L (W(!-") + n=p+l L (W n=l

< " That is, (1.8) is satisfied. Now suppose K and 0 are given. Since G = compact there is an integer p

~

1 such that K

00

U Kn

n=1

c:

€

.

.

U int Kn and K is

n

=

I

Kp; this gives

pif, g) ~ sup {d(f(z), g(z»: z

Let

00

=

E

K}

s I-s

1

> 0 be chosen so that 0 :s; s < 2P " Impltes - - < 0; then -1-

+1

< 2P €

implies t < O. So if p(f, g) < " then- pif, g) < 2P " and this gives pif, g) 1 +pif, g) < O. But this is exactly the statement contained in (1.9) . • 1.10 Proposition. (a) A set (!) c: (C(G, Q), p) is open iff for eachf in is a compact set K and a 0 > 0 such that (!) :::>

{g: d(f(z), g(z» < 0, z

E

K}

(!)

there

The space of continuous functions C(G,O)

145

(b) A sequence {fn} in (C( G, Q), p) converges to f iff {/,,} converges to f uniformly on all compact subsets of G. € > 0, (2 :::> {g: p(J, g) < €}. But now the first part of the preceding lemma says that there is as> 0 and a compact set K with the desired properties. Conversely, if (2 has the stated property and f E (2 then the second part of the lemma gives an € > 0 such that (2 :::> {g: p(J, g) < €}; this means that (2 is open. The proof of part (b) will be left to the reader. •

Proof If (2 is open and fE (2 then for some

1.11 Corollary. The collection of open sets is independent of the choice of the sets {Kn}. That is, ifG

=

co

U K~ where each K~ is compact and K~ c

int K~+l

n~1

and if I'- is the metric defined by the sets iff it is open in (C( G, Q), p).

{K~}

then a set is open in (C(G, Q), 1'-)

Proof This is a direct consequence of part (a) of the preceding proposition since the characterization of open sets does not depend on the choice of the sets {Kn } • • Henceforward, whenever we consider C(G, Q) as a metric space it will be assumed that the metric p is given by formula (1.4) for some sequence {Kn} of compact sets such that Kn c int Kn+l and G

=

co

U Kn. Actually, the

n~1

requirement that Kn c int Kn+ 1 can be dropped and the above results will remain valid. However, to show this requires some extra effort (e.g., the Baire Category Theorem) which, though interesting, would be a detour. Nothing done so far has used the assumption that Q is complete. However, if Q is not complete then C(G, Q) is not complete. In fact, if {wn } is a non-convergent Cauchy sequence in Q and /,,(z) = Wn for all z in G, then {fn} is a non-convergent Cauchy sequence in C(G, Q). However, we are assuming that Q is complete and this gives the following. 1.12 Proposition. C(G, Q) is a complete metric space. Proof Again utilize Lemma I.? Suppose{/,,} is a Cauchy sequence inC(G, Q). Then for each compact set KeG the restrictions of the functions In to K gives a Cauchy sequence in C(K, Q). That is, for every S > 0 there is an integer N such that

1.13

sup {d(/,,(z),Jm(z)): z

E

K} < S

for n, m ~ N. In particular {fn(z)} is a Cauchy sequence in Q; so there is a point f(z) in n such that f(z) = lim fn(z). This gives a function f: G --+ Q; it must be shown that f is continuous and p(/", f) --+ O. Let K be compact and fix 13 > 0; choose N so that (1.13) holds for n, m ~ N. If z is arbitrary in K but fixed then there is an integer m ~ N so that d(f(z), fm(z)) < 13. But then d(f(z),/,,(z)) < 213

146

Compactness and Convergence

for all n :2: N. Since N does not depend on z this gives sup {d(f(z),fn(z»:

Z E

K}f 0

as n -+ 00. That is, {fn} converges to f uniformly on every compact set in G. In particular, the convergence is uniform on all closed balls contained in G. This gives (Theorem I r. 6.1) that f is continuous at each point of G. Also, Proposition 1.10 (b) gives that p(j~. f) -,. O. • The next definition is derived from the classical origins of this subject. Actually it could have been omitted without interfering with the development of the chapter. However, even though there is virtue in maintaining a low ratio of definitions to theorems, the classical term is widely used and should be known by the reader.

1.14 Definition. A set :F c C(G, il) is normal if each sequence in .'j: has a subsequence which converges to a function f in C(G, £2). This of course looks like the definition of sequentially compact subsets, but the limit of the subsequence is not required to be in the set !IF. The next proof is left to the reader. 1.15 Proposition. A set :::F c C(G, £2) is lIormal if! its closure is compact. 1.16 Proposition. A set !IF c C(G, il) is normal iff for aery compact set KeG alld 8 > 0 there are functions fl, ....f,. in .::F such that for f in .ry; there is at least one k, 1 0 such that (1.9) holds. But since !IF- is compact, .'!i' IS totally bounded (actually there are a few details to fill in here). So there are fl, ... , j~ in :::F such that .ry; c

n

U

k '" ]

But from the choice of

.'F

E n

c

this gives

U {f: d(f(z),J;..(z))

< ii,:: E K};

k=1

that is, ,'F satisfies the condition of the proposition. For the converse, suppose ,ry; has the stated property. Since it readily follows that !IF - also satisfies this condition, assume that ,ry; is closed. But since C(G, ~2) is complete !IF must be complete. And, again using Lemma 1.7, it readily follows that ~ is totally bounded. From Theorem II.4.9, ~ is compact and therefore normal. • This section concludes by presenting the Arzela-Ascoli Theorem. Although its proof is not overly complicated it is a deep result which has proved extremely useful in many areas of analysis. Before stating the theorem a few results of a more general nature are needed.

147

The space of continuous functions C(G,fl)

Let (X., d.) be a metric space for each n ;::: 1 and let X cartesian product. That is, X = {g = {xn}: Xn g = {xn} and '7 = {Yo} in X define d(g, 7])

1.17 1.18 Proposition.

CD

X. for each n ;::: I}. For

l+d.(xn,Yn)

n=1

Xm d), where d is defined by (1.17), is a metric space.

e = {X:}:'_I is in X TI Xn then e =

-+

• -1

Also,

TI Xn be their

n-l

= ~ H)'--- d.(xn' Yn) .

00

If

E

=

00

g = {xn} iff x~ ----;.. x.for each n .

if each (Xm dn) is compact then X is compact,

Proof The proof that d is a metric is left to the reader. Suppose d(gk, g)

--+

0;

SInce

we have that

, I1m k-oo

d"(x~, x.) - 0 k -. I +d.(xm x.)

This gives that x~ -+ x. for each n ;::: I. The proof of the converse is left to the reader. Now suppose that each (X., do) is compact. To show that (X, d) is compact it suffices to show that every sequence in X has a convergent subsequence; this is accomplished by the Cantor diagonalization process. Let gk = {~} E X for each k ;::: 1 and consider the sequence of the first entries of the that is, consider {xnt'_1 C XI' Since Xl is compact there is a point XI in Xl and a subsequence of {x~} which converges to it. We are now faced with a problem in notation. If this subsequence of {Xnt'-l is denoted by {x~j}j": 1 there is little confusion at this stage. However, the next step in the proof is to consider the corresponding subsequence of second entries {x~j} /':~! and take a subsequence of this. Furthermore, it is necessary to continue this process for all the entries. It is easy to see that this is opening up a notational Pandora's Box. However, there is an alternative. Denote the convergent subsequence of {x~} by {x~: kEN I }, where N 1 is an infinite subset of the positive integers N. Consider the sequence of second entries of {e: kEN! }. Then there is a point x 2 in X 2 and an infinite subset N 2 C N 1 such that lim {X~: kEN 2} = X2' (Notice that we still have lim {xt: kEN 2} = XI') Continuing this process gives a decreasing sequence of infinite subsets of N, N 1 ::J N 2::J ... ; and points x n in Xn such that

e;

1.19

lim {~: k

E

Nn }

= xn

{e

J }; we claim that Let kj be the }th integer in Nj and consider {xn} as} -+ 00. To show this it suffices to show that

1.20

tkj -+ g =

148

Compactness and Convergence

for each n ;::: 1. But since!\lJ c !\In for};::: n, {X~J:};::: n} is a subsequence of {X~: k E !\In}. So (1.20) follws from (1.19) . • The following definition plays a central role in the Arzela-Ascoli Theorem.

1.21 Definition. A set §" c C(G, Q) is equicontinuous at a point for every E > 0 there is as> 0 such that for Iz-zol < S, d(f(z),J(zo» <

Zo

in G iff

E

for every fin §". §" is equicontinuous over a set E c G iffor every E > 0 there is as> 0 such that for z and z' in E and Iz-z'l < S,

d(f(z),J(z'» <

E

for all / in §". Notice that if §" consists of a single function/then the statement that §" is equicontinuous at Zo is only the statement that/is continuous at zoo The important thing about equicontinuity is that the same S will work for all the functions in §". Also, for §" = {f} to be equicontinuous over E is to require that / is uniformly continuous on E. For a larger family §" to be equicontinuous there must be uniform uniform continuity. Because of this analogy with continuity and uniform continuity the following proposition should not come as a surprise. 1.22 Proposition. Suppose §" c C( G, Q) is equicontinuous at each point then §" is equicontinuous over each compact subset 0/ G.

Proof Let KeG be compact and fix a Sw > 0 such that

E

0/ G;

> O. Then for each w in K there is

d(f(w'),J(w» <

!E

for allfin§"whenever Iw-w'l 0 such that for each z in K, B(z; 8) is contained in one of the sets of this cover. So if z and z' are in K and Iz-z'l 0; by Lemma 1.10(b) it suffices to find an integer J such that for k, j ;:::: J, 1.26

sup {d(J..(z)Jj(z): z

E

K} <

-1.)

+t

2. Find the sets K. obtained in Proposition 1.2 for each of the following choices of G: (a) G is an open disk; (b) G is an open annulus; (c) G is the plane with n pairwise disjoint closed disks removed; (d) G is an infinite strip; (e) G = C-I. 3. Supply the omitted details in the proof of Proposition 1.18. 4. Let F be a subset of a metric space (X, d) such that F- is compact. Show that F is totally bounded. 5. Suppose {I.} is a sequence in C(G, Q) which converges tofand {zn} is a sequence in G which converges to a point z in G. Show lim I.(zn) = fez). 6. (Dini's Theorem) Consider C(G, IR) and suppose that {In} is a sequence in C(G, IR) which is monotonically increasing (i.e., fn(z) s 1.+1(Z) for all z in G) and lim/,,(z) = fez) for all z in G wherefE C(G, IR). Show that/" -+ f 7. Let {In} C C(G, Q) and suppose that {/,,} is equicontinuous at each point of G. If fE C(G, Q) and fez) = limfn(z) for each z then show that/" -+ f

151

Spaces of analytic functions 00

8. (a) Let f be analytic on B(O; R) and let fez)

J,.(z) =

n

L akzk , show that J,. ~ fin

q.

C(G;

k-O

= L anzn for Izl < n-O

R. If

(b) Let G = ann (0; 0, R) and letfbe analytic on G with Laurent series development fez) C(G; q.

=

L anzn. 00

=

Put J,.(z)

L ak zk "

and show that fn ~ fin

k=-oo

"=-00

§2. Spaces of analytic functions Let G be an open subset of the complex plane. If H (G) is the collection of analytic functions on G, we can consider H (G) as a subset of C(G,C). We use H(G) to denote the analytic functions on G rather than A (G) because it is a universal practice to let A (G) denote the collection of continuous functionsf:G-~C that are analytic in G. Thus A(G)'I: H (G). The letter H is used in reference to "analytic" because the word holomorphic is commonly used for analytic. Another term used in place of analytic is regular. The first question to ask about H(G) is: Is H(G) closed in C(G,C)? The next result answers this question positively and also says that the functionf~f' is continuous from H(G) into H(G). 2.1 Theorem./f{J,.} is a sequence in H(G) andfbelongs to C(G, In ~ f then f is analytic and f~k) ~ j 0 such that for all I in :F,

]/(z)] :::; M, for ]z - a] < r. Alternately, :F is locally bounded if there is an r > 0 such that sup {]/(z)]: ]z-a] < r,JE:F} <

00.

That is, :F is locally bounded if about each point a in G there is a disk on which :F is uniformly bounded. This immediately extends to the requirement that :F be uniformly bounded on compact sets in G. 2.8 Lemma. A set :F in H(G) is locally bounded KeG there is a constant M such that

iff lor

each compact set

]/(z)] :::; M lor all I in :F and z in K. The proof is left to the reader.

2.9 Montel's Theorem. Alamity:F in H(G) is normal iff :F is locally bounded. Proof Suppose :F is normal but fails to be locally bounded; then there is a compact set KeG such that sup {]/(z)!: z E K,JE:F} = 00. That is, there is a sequence {In} in :F such that sup {]/n(z)]: z E K} 2: n. Since :F is normal there is a function I in H(G) and a subsequence {Ink} such that Ink -+ f But this gives that sup {]/nk(z) - I(z)]: z E K} -~ 0 as k -+ 00. If ]/(z)] :::; M for z in K, nk :::; sup {llnN)-/(z)]: z E K}+M;

since the right hand side converges to M, this is a contradiction. Now suppose :F is locally bounded; the Arzela-Ascoli Theorem (1.23) will be used to show that :F is normal. Since condition (a) of Theorem 1.23 is clearly satisfied, we must show that :Fis equicontinuous at each point of G. Fix a point a in G and ( > 0; from the hypothesis there is an r > 0 and M> 0 such that B(a; r) c G and ]/(z)] :::; M for all z in B(a; r) and for all 1 in .'F. Let ]z - a] < !r and 1 E:F; then using Cauchy's Formula with yet) = a + re il , 0 :::; t:::; 277, I/(a) - I(z)]

=---1

21T

\J' -----lew) (a-z)

(IV-a) (w-z)

dw I

y

2M :::; -]a-z] r Letting is < min

<

E"

{lr'4~ £}

for all/in:F . •

it follows that la-z] < is gives ]/(a)-/(z)]

154

Compactness and Convergence

2.10 Corollary. A set ff' c H(G) is compact iff it is closed and locafly bounded. Exercises I. Let f, 11,12' ... be elements of H(G) and show that f" ---* I iff for each closed rectifiable curve y in G,ln(z) ---* I(z) uniformly for z in {y}. 2. Let G be a region, let a E IR, and suppose that I: [a, 00] x G ---* C is a continuous function. Define the integral F(z) = S: I(t, z)dt to be uniformly convergent on compact subsets of G if lim S~ l(t, z)dt exists uniformly for z b-HO

in any compact subset of G. Suppose that this integral does converge uniformly on compact subsets of G and that for each t in (a, co), I(t, .) is analytic on G. Prove that F is analytic and 00

F(k)(Z) =

f ~kl(t,31 dt a

cz k

3. The proof of Montel's Theorem can be broken up into the following sequence of definitions and propositions: (a) Definition. A set ff' c C(G, C) is locally Lipschitz if for each a in G there are constants M and r > 0 such that I/(z)-/(a)1 :0; Mlz-al for alliin ff' and Iz-al < r. (b) If ff' c C(G, C) is locally Lipschitz then ff' is equicontinuous at each point of G. (c) If ofF c H(G) is locally bounded then ,CY; is locally Lipschitz. 4. Prove Vitali's Theorem: If G is a region and Un}C:.H(G) is locally bounded and f E H(G) that has the property that A = {z E G: limfn(z) = fez)} has a limit point in G then fn-+(. 5. Show that for a set §- c H(G) the following are equivalent conditions: (a) ff' is normal; (b) For every E > 0 there is a number c > 0 such that {c/: IE ff'} c B(O; E) (here B(O; E) is the ball in H(G) with center at 0 and radius E). 6. Show that if ff' c H(G) is normal then ff" = {f':/E ff'} is also normal. Is the converse true? Can you add something to the hypothesis that ff" is normal to insure that ff' is normal? 7. Supposeff' is normal in H(G) and Q is open in C such that/(G) c Q for every I in ff'. Show that if g is analytic on Q and is bounded on bounded sets then {g I: IEff'} is normal. 8. Let D = {z: Izl < I} and show that ff' c H(D) is normal iff there is a sequence {Mn} of positive constants such that lim sup I:./ Mn :0; 1 and if 0

I(z)

=

a:>

Lo anzn is in ff' then rani

:0;

Mn for all n.

9. Let D=B(O; I) and for O 0 such that Boo(oo; p) c C",---K. The proof is left to the reader. The first observation is that M(G) is not complete. In fact if j,,(z) == Il then U~} is a Cauchy sequence in M(G). But {fn} converges to the function which is identically (fJ in C(G, Coo) and this is not meromorphic. However this is the worst that can happen. 3.4 Theorem. Let {f,,} be a sequence in M(G) and supposej" ~ fin C(G, C",J. Then either f is meromorphic or f == 00. each j" is analytic then either f is analytic or f == 00.

rr

Proof Suppose there is a point a in G withf(a) # XJ and put M = If(a)1 + 1. Using part (a) of Proposition 3.3 we can find a number p > 0 such that Bx(f(a); p) c B(f(a); M). But since j,,+ f there is an integer flo such that d(j,,(a)j(a» < 1p for all n ? 11 0 , Also UJIJb ... } is compact in C(G, Cx» so that it is equicontinuous. That is, there is an r > 0 such that Iz - al < r implies d(fn(z), .!n(a» < 1p. That gives that d(j,,(z), f(a» ::; p for Iz-al ::; r and for fl ? no. But by the choice of p, Ij,,(z) 1 ::; Ij,,(z) - f(a) I + If(a) 1 ::; 2M for all z in B(a; r) and n ? flo. But then (from the formula for the metric d) L

(1 + 4M2) Ij,,(z) -

fez)! ::; d(j,,(z),J(z»

for z in BCa; r) and n ? 110' Since d(j,,(z) , fez») ~ 0 uniformly for z in B(a; r), this gives that if~(z)-f(z)1 -+ 0 uniformly for z in BCa; r). Since the tail end of the sequence {j,,} is bounded on BCa; r), j" has no poles and must be analytic near z = a for Il ~ flo. It follows thatfis analytic in a disk about a. Now suppose there is a point a in G withfCa) = 00. For a function g in C(G, Coo) define! by g 00;

fn

and

----7

(g/1)-- (z)

= 00

(1) g

(z)

=L if g(z) # g(z)

0 or 00;

1

if g(z) = O. It follows that g

E

(1) g

(z) = 0 if g(z)

C(G, C w )' Also, since

I fin C(G, C"J it follows from formulas (3.1) and (3.2) that ;~

C(G, C",). Now each

1

function.f~

=

I

~f

in

is meromorphic on G; so the preceding

1 1 paragraph gives a number r > 0 and an integer no such that f and j" are

analytic on BCa; r) for

11 ~ flo

and

/"

~

1

f

uniformly on B(a; r). From

Spaces of meromorphic functions

157

I

1

Hurwitz's Theorem (2.5) either j == 0 or j has isolated zeros in B(a; r). So if I ¢

00

then

j

¢ 0 and I must be merom orphic in BCa; r). Combining this

with the first part of the proof we have that I is meromorphic in G if I is not identically infinite. 1 If each is analytic then - has no zeros in B(a; r). It follows from

f:

I.

Corollary 2.6 to Hurwitz's Theorem that either; == 0 or

But since I(a) =

00

~

never vanishes.

1 we have that I has at least one zero; thus

f ==

B(a; r). Combining this with the first part of the proof we see that I

is analytic.

III

==

00

00

in

or I

3.5 Corollary. M(G) u {oo} is a complete metric space. 3.6 Corollary. H(G) U {oo} is closed in (,(G, C oc ). To discuss normality in M(G) one must introduce the quantity

2il'(z)1

1+11\;)1

2 '

for each meromorphic function f However if z is a pole off then the above expression is meaningless since I'(z) has no meaning. To rectify this take the limit of the above expression as z approaches the pole. To show that the limit exists let a be a pole off of order m 2': I; then

I(z) = g(z) +

Am_ + ... + . Al

(z-a)m

(z-a)

for z in some disk about a and g analytic in that disk. For z =ft a

Thus

211'(z)1

mAm _ 2 I (z _ a)m +l

1+ I/(zW

- ~I-'- ~m

AI,

+ ... + ~-;;-)2 _

(z-a)'"

Iz - al 2m

I

+ ... +1.1_ + g(Z)1 2 (z-a)

+ Al(Z - a)ml - g'(z)(z - a)m+ll + lAm +- ... +- Al(Z - a)ml +- g(z)(z - a)ml2

21z - aim-limA," +- ...

-----------------

g (z)

158

Compactness and Convergence

So if m :?: 2 lim z-'a

If m

=

~!.f'(=-2L

1+[f{z)[2

0

=

1 then

. 21f'(z)[ lIm --------z~a 1 + [f(zW

=

2

-----

[All

3.7 Definition. Iff is a meromorphic function on the region G then define f1-(f): G -----3>- [R by

whenever z is not a pole of J, and

. 2if'(z)1 f1-(f) (a) = h m - - z~a 1 + if(zW if a is a pole off It follows that f1-(f) E C(G, q. The reason for introducing f1-(f) is as follows: If f: G -----3>- Coo is meromorphic then for z close to z' we have that d(f(z), fez')) is approximated by f1-(f) (z) Iz-z'l. So if a bound can be obtained for f1-(f) thenfis a Lipschitz function. Iff belongs to a family of functions and /1-(f) is uniformly bounded for fin this family, then the family is a uniformly Lipschitz set of functions. This is made precise in the following proof. 3.8 Theorem. Afamity :F f E :F} is locally bounded.

c

M(G) is normal in C(G, C",) iffA:!F) == {f1-(f):

0;:. T hus:f/' = {'f,}' n. IS 1+n normal in C(G, CG(J and f1-(,SF) is locally bounded. However, .'F is not normal in M(G) since the sequence {fn} converges to the constantly infinite function which does not belong to M(G).

Note. If f,,(z)

=

nz for n :?: 1 then f1-(f,,) (z)

=

- - -21121-;'

z,-

Proof of Theorem 3.8. We will assume that f1-(:F) is locally bounded and prove that .SF is normal by applying the Arzela-Ascoli Theorem. Since C", is compact it suffices to show that ,SF is equicontinuous at each point of G. So let K be an arbitrary closed disk contained in G and let M be a constant with f1-(f) (z) ::; M for all z in K and all f in :F. Let z and z' be arbitrary points in K. Suppose neither z nor z' are poles of a fixed function f in :F and let c< > 0 be an arbitrary number. Choose points Wo = Z, WI, . . . , Wn = z' in K which satisfy the following conditions: 3.9 win

[W k - 1 ,

wd implies w is not a pole of /;

159

Spaces of meromorphic functions

3.10

L

k=1

3.11

IWk-Wk-d

:s:

2Iz-z'l;

+ 1/(Wk_1)1 2 ) [(1 + 1/(wk)1 2 )(1 + 1/(Wk-1)lZ))' (1

I

3.12I/(Wk)-/(Wk-l) - /'(Wk-I)I < Wk-W k - I

IX,

-

1 < a,

1 ::s: k::s: n;

1:S: k :S: n.

To see that such points can be found select a polygonal path Pin K satisfying (3.9) and (3.10). Cover P by small disks in which conditions similar to (3.11) and (3.12) hold, choose a finite subcover, and then pick points W o, ••• , W II on P such that each segment [Wk - I , wk] lies in one of these disks. Then {wo, ... , wII } will satisfy all of these conditions. If 13k = [(I + I/(Wk-1W) (l + I/(wkW)]t then

2 d(f(Wk-l),f(Wk» II

d(/(z),f(z'») :S:

k-I

=

2 2 13k-1/(wk)-/(wk-I)1 II

*

k _I

:S: L.. -21 / (W k)-/(Wk- 1) - j"( Wk- 1)\1 Wk-Wk- 11 k -1 13k Wk - Wk- 1

+

2 2 13k-//'(Wk-I)//wk-w.-d II

Ic= 1

Using the fact that 211'(wk)1 :S: M(I + /1(w.W) and the conditions on wo, ••. , W II this becomes

2 M/w.-w._ I/ n

:S: (4IX+2IXM) Iz-z'/ +

k-I

:S: (4cx+2cxM+2M)/z-z'l Since 3.13

IX

> 0 was arbitrary this gives that if z and z' are not poles of.f then d(f(z),f(z'» :S: 2M/z-z'/.

Now suppose z' is a pole of 1 but z is not. If then it follows from (3.13) that

W

is in K and is not a pole

d(/(z), (0) :S: d(/(z),f(w»+d(f(w), (0)

:S: 2M/z-w/ + d(f(w) , (0).

Compactness and Convergence

160

Since it is possible to let w approach z' without wever being a pole of f (poles are isolated!), this gives thatf(lI') -lo- fez') = CI) and IZ-Wl-lo-Iz-zl Thus (3.13) holds if at most one of z and z' is a pole. But a similar procedure gives that (3.13) holds for all z and z' in K. So if K = B(a; r) and E > 0 are given then for [; < min {r, E/2M} we have that Iz - al < [; implies d(f(z) , J(a» < E, and [; is independent offin ff. This gives that ff is equicontinuous at each point a in G. The proof of the converse is left to the reader. • Exercises

1. Prove Proposition 3.3. 2. Show that if ff c M(G) is a normal family in C(G, Coo) then p,(ff) is locally bounded. §4. The Riemann Mapping Theorem

We wish to define an equivalence relation between regions in C. After doing this it will be shown that all proper simply connected regions in C are equivalent to the open disk D = {z: Izl < I}, and hence are equivalent to one another. 4.1 Definition. A region G 1 is coriformally equivalent to G z ifthereis an analytic function f: G 1 -lo- C such that l is one-one and l(G) = Gz• Clearly, this is an equivalence relation. It is immediate that C is not equivalent to any bounded region by Liouville's Theorem. Also it is easy to show from the definitions that if G 1 is simply connected and G 1 is equivalent to G z then G z must be simply connected. If l is the principal branch of the square root then f is one-one and shows that C - {z: z :s; o} is equivalent to the right half plane. 4.2 Riemann Mapping Theorem. Let G be a simply connected reginn which is not the whole plane and let a E G. Then there is a unique analytic function f: G -lo- C having the properties:

(a) f(a) = 0 andf'(a) > 0; (b) l is one-one; (c) f(G) = {z: Izl < I}. The proof that the function f is unique is rather easy. In fact, if g also has the properties off and D = {z: Iz 1< I} then fog - ) : D -lo- D is analytic, one-one, and onto. Also f 0 g-l(O) = lea) = 0 so Theorem VI. 2.5 implies there is a consta'1t c with lei = 1 and f g-l(Z) = cz for all z. But then fez) = cg(z) gives that 0 < f'(a) = cg'(a); since g'(a) > 0 it follows that c = 1, orf= g. To motivate the proof of the existence of J, consider the family ff of all analytic functions f having properties (a) and (b) and satisfying 1f(z) 1 < 1 for z in G. The idea is to choose a member of ff having property (c). Suppose 0

The Riemann Mapping Theorem

161

UKn = G and a 00

{Kn} is a sequence of compact subsets of G such that

E

Kn

n~l

for each n. Then {f(Kn)} is a sequence of compact subsets of D = {z: [z[ < I}. Also, as n becomes larger f(Kn) becomes larger and larger and tries to fill out the disk D. By choosing a function f in ff with the largest possible derivative at a, we choose the function which "starts out the fastest" at z = a. It thus has the best possible chance of finishing first; that is, of having 00

Uf(Kn) = D. n~l

Before carrying out this proof, it is necessary for future developments to point out that the only property of a simply connected region which will be used is the fact that every non-vanishing analytic function has an analytic square root. (Actually it will be proved in Theorem VIII. 2.2 that this property is equivalent to simple connectedness.) So the Riemann Mapping Theorem will be completely proved by proving the following. 4.3 Lemma. Let G be a region which is not the whole plane and such that every non-vanishing analytic function on G has an analytic square root. If a E G then there is an analytic function f on G sllch that: (a) f(a) = 0 andf'(a) > 0; (b) f is one-one; (c) f{G) = D = {z: [z[ < I}.

Proof Define ff by letting .SF

=

{I,

H(G):fis one-oneJ(a)

= O,f'(a) > O,f(G)

c D}

Since f(G) c D, sup {[f(z)!: z E G} :s:: 1 for f in ff; by MonteI's Theorem ff is normal if it is non-empty. So the first fact to be proved is

4.4 It will be shown that

4.5

.SF- = ff u {O}.

Once these facts are known the proof can be completed. Indeed, suppose (4.4) and (4.5) hold and consider the function f ---?- f'(a) of H(G) ---?- C. This is a continuous function (Theorem 2.1) and, since ff- is compact, there is an fin ff- with f'(a) 2: g'(a) for all g in ff. Because ff "# 0, (4.5) implies that fE:7. It remains to show that f(G) = D. Suppose WED such that w rtf(G). Then the function fCz)-w l-cvf(z) is analytic in G and never vanishes. By hypothesis there is an analytic function h: G -;.. :c such that

4.6

[h(z)J

2

f(z)-w 1-wf(z)

= ---

.

162

Compactness and Convergence

n

Since the Mobius transformation

~-w

l~=~i; maps D onto D, h(G) c D.

=

Define g: G -)0 IC by

g(z) = Ih'(a)[ ~~i~)-IJ(a) h'(a) I-h(a)h(z) Then g(G) c D, g(a) =c 0, and g is one-one (why?). Also g'(a)

= !!l'(~J . h'(a) [l_-:-jh(a)I~~] [1-lh(a)12]2

h'(a)

~~L

l-lh(aW

But [h(aW = I-wi = Iwl and differentiating (4.6) gives (sincef(a) = 0) that

2h(a)h'(a) = Therefore

rea) (l-lwI2).

rCa) (~-::J(UL2 . _J

g'(a) =

2y'l wl

l-Iwl

=

rea) (~JI~i)

>

rCa)

This gives that g is in :F and contradicts the choice of f Thus it must be thatf(G) = D. Now to establish (4.4) and (4.5). Since G i= IC, let bE IC-G and let g be a function analytic on G such that [g(Z)]2 = z-b. If Z1 and Z2 are points in G and g(Zj) = ±g(zz) then it follows that ZI = Z2' In particular, g is one-one. By the Open Mapping Theorem there is a number r > 0 such that

4.7

g(G)

::>

B(g(a); r)

So if there is a point z in G such that g(z) E B( - g(a); r) then r > Ig(z) + g(a) I = l-g(z)-g(a)l. According to (4.7) there is a IV in G with g(w) = -g(z); but the remarks preceding (4.7) show that }\' = z which gives g(z) = O. But then z-b = [g(Z)]2 = 0 implies b is in G, a contradiction. Hence 4.8

g(G) n

g:

1~+g(a)1

< r} = D.

Let U be the disk g: i~+g(a)i < r} = B( --g(a); r). There is a Mobius transformation T such that T(ICw-U-) = D. Let gj = Tog; then gj is analytic and gj(G) c D. If C( = gj(a) then let gz(z) = CF. gj(z); so we still have that g2(G) c D and gz is analytic, but we also have that g2(a) = O. Now it is a simple matter to find a complex number c, lei = 1, such that g3(Z) = cgzCz) has positive derivative at z = a and is, therefore, in ~. This establishes (4.4). 0

The Riemann Mapping Theorem

Suppose {in} is a sequence in ff and In and since f:(a) -i> f'(a) it follows that

4.9

...,.. fin

163

HeG). Clearly f(a)

=

0

f'(a) ::?:: O.

Let z 1 be an arbitrary element of G and put' = fez 1); let ,. = .f~(z 1.). Let z 2 E G, z 2 -.f. Z I and let K be a closed disk centered at z 2 such that z t ¢ K. Thenfn(z) -~" never vanishes on K sincef. is one-one. Butfl::) -~. -i> fez) - ~ uniformly on K, so Hurwitz's Theorem gives that fez) - ~ never vanishes on K or fez) == ~. If fez) == ~ on K then f is the constant function' throughout G; since.f(a) = 0 we have thatf(z) == O. Otherwise we get thatf(z2) -.f. f(zl) for Z2 -.f. z 1; that is,fis one-one. But if fis one-one thenf' can never vanish; so (4.9) implies thatf'(a) > 0 andfis in ff. This proves (4.S) and the proof of the lemma is complete. III 4.10 Corollary. Among the simply connected regions there are only two equivalence classes; one consisting of C alone and the other containing all the proper simply connected regions.

Exercises 1. Let G and 0 be open sets in the plane and let f: G -i> 0 be a continuous function which is one-one, onto, and such thatf- I : 0 -i> G is also continuous (a homeomorphism). Suppose {zn} is a sequence in G which converges to a point z in 8G; also suppose that w = lim f(z.) exists. Prove that WEan. 2. (a) Let G be a region, let a E G and suppose thatf: (G- {a}) -i> C is an analytic function such that f{G - {a}) = n is bounded. Show that f has a removable singularity at z = a. If f is one-one, show that f(a) E 80. (b) Show that there is no one-one analytic function which maps G = {z: < Izl < I} onto an annulus 0 = {z: r < Izl < R} where r > O. 3. Let G be a simply connected region which is not the whole plane and suppose that Z E G whenever z E G. Let a E G () IR and suppose that f: G -i> D = {z: [zl < I} is a one-one analytic function with/(a) = O,f'(a) >0 and /(G) = D. Let G+ = {z E G: Im z > O}. Show that /(G+) must lie entirely above or entirely below the real axis. 4. Find an analytic function f which maps {z: Izl < 1, Re z > O} onto B(O; I) in a one-one fashion. S. Let f be analytic on G = {z: Re z > O}, one-one, with Re fez) > 0 for all z in G, andf(a) = a for some real number a. Show that If'(a) 1 ::; 1. 6. Let G 1 and G 2 be simply connected regions neither of which is the whole plane. Let f be a one-one analytic mapping of G 1 onto G 2 • Let a E G 1 and put ex = f(a). Prove that for anyone-one analytic map h of G 1 into G 2 with Ii(a) = ex it follows that Ih'(a)1 ::; 1f'(a)l. Suppose Ii is not assumed to be one-one; what can be said? 7. Let G be a simply connected region and suppose that G is not the whole plane. Let .6. = {~: !~I < I} and suppose that f is an analytic, one-one map of G onto .6. with f(a) = 0 and f'(a) > 0 for some point a in G. Let g be any other analytic, one-one map of G onto .6. and express g in terms off

°

164

Compactness and Convergence

8. Let '" '2' R" R 2 , be positive numbers such that R,f" = R 2 f'2; show that ann (0; '10 R I ) and ann (0; '2' R 2 ) are conformally equivalent. 9. Show that there is an analytic function f defined on G = ann(O; 0, 1) such that f' never vanishes and f(G) = B(O; 1).

§ 5. The Weierstrass Factorization Theorem The notion of convergence in H(G) can be used to solve the following problem. Given a sequence {ak } in G which has no limit point in G and a sequence of integers {mk}. is there a function J which is analytic on G and such that the only zeros of J are at the points ak • with the multiplicity of the zero at ak equal to m k? The answer to the question is yes and the result is due to Weierstrass. If there were only a finite number of points, ai' ... , an then J(z) = (z-an)m, ... (z-an)m n would be the desired function. What happens if there are infinitely many points in this sequence? To answer this we must discuss the convergence of infinite products of numbers and functions. Clearly one should define an infinite product of numbers Zn (denoted

n zn) as the limit of the finite products. Observe, however, that if one of 00

by

n-I

the numbers Zn is zero, then the limit is zero, regardless of the behavior of the remaining terms of the sequence. This does not present a difficulty, but it shows that when zeros appear, the existence of an infinite product is trivial.

5.1 Definition. If {zn} is a sequence of complex numbers and if z exists, then z is the infinite p,oduct of the numbers

Zn

= lim

n n

Zk

k -I

and it is denoted by

TI 00

Suppose that no one of the numbers and is also not zero. Let Pn

=

Zn

is zero, and that

=

Zn

exists

n=1

n

TI

Z

Zk

for

11

k=1

C 1; then no Pn is zero and

~ = Zn' Since Z # 0 and Pn -+ Z we have that lim Zn = 1. So that except Pn-I for the cases where zero appears, a necessary condition for the convergence of an infinite product is that the n-th term must go to 1. On the other hand, note that for Zn = a for all nand lal < 1, = 0 although lim Zn = a # O. Because of the fact that the exponential of a sum is the product of the exponentials of the individual terms, it is possible to discuss the convergence of an infinite product (when zero is not involved) by discussing the convergence of the series L log Zn' where log is the principal branch of the logarithm. However, before this can be made meaningful the Zn must be

TIz.

165

Weierstrass Factorization Theorem

restricted so that log Zn is meaningful. If the product is to be non-zero, then Zn ~ I. So it is no restriction to suppose that Re zn > 0 for all n. Now n

suppose that the series

L log Zn converges. If Sn = L log Zk and Sn ~ S then

n Zk so that n Zn is convergent to Z = e' i= O. > 0 for all n ~ 1. Then n Zn converges to a non n

00

k~1

n~1

k~1

exp Sn ~ exp s. But exp Sn =

5.2 Proposition. Let Re Zn

00

n~1

00

zero number iff the series L log Zn converges. n~1

Proof Let Pn = (Zl·· ·zn), Z = re i9 , -7T < 8 :s; 7T, and t(Pn) = log IPnl+i8n where (}-7T < 8n :s; 8+7T. If Sn = log Zl + ... +log Zn then exp (sn) = Pn so that Sn = t(Pn) + 27Tikn for some integer k n. Now suppose that Pn ~ z. Then sn-sn-l = log zn~O; also t(Pn)-t(Pn-l)~O, Hence, (kn-kn-l)~O as n ~ 00. Since each k n is an integer this gives that there is an no and a k such that k m = k n = k for m, n ~ no. So Sn ~ t(z) + 27Tik; that is, the series L log Zn converges. Since the converse was proved above, this completes the proof. • Consider the power series expansion of log (I-t-z) about Z = 0:

10g(l+z) =

zn 2: (_l)n-l_ = n=l n 00

Z -

Z2 -

2

+ ... ,

which has radius of convergence 1. If Izi < 1 then 11 _IOg(;+Z)1 = l-!-z-tz 2 +···1 :s; -!-(Izi + IzI2 + ...)

Izi

= -!- l-lzl· If we further require Izi < -t then 11 - log (; +Z)I :s;

t.

This gives that for Izi < -t

5.3

-tizi :s; Ilog (1 + z)1 :s; 1-14

This will be used to prove the following result. 5.4 Proposition. Let Re zn > -1; then the series

absolutely iff the series L zn converges absolutely.

L

L

log (1 +zn) converges

Proof If IZnl converges then Zn ~ 0; so eventually IZnl < t. By (5.3) Ilog (1 +zn)1 is dominated by a convergent series, and it must converge Ilog (1 +zn)1 converges, then it follows that IZnl < -!also. If, conversely, for sufficiently large n (why?). Again (5.3) allows us to conclude that L IZnl converges . •

L

L

166

Compactness and Convergence

We wish to define the absolute convergence of an infinite product. The first temptation should be avoided. That is, we do not want to say that IT IZnl converges. Why? If IT IZnl converges it does not follow that IT Zn converges. In fact, let Zn = -1 for all n; then IZnl = 1 for all n so that n

IT IZnl converges to 1. However IT Zk is ± 1 depending on whether 11 is even k-I or odd, so that IT Zn does not converge. Thus, if absolute convergence is to

imply convergence, we must seek a different definition. On the basis of Proposition 5.2 the following definition is justified.

5.5 Definition. If Re Zn > 0 for all n then the infinite product IT Zn is said to converge absolutely if the series L log Zn converges absolutely. According to Proposition 5.2 and the fact that absolute convergence of a series implies convergence, we have that absolute convergence of a product implies the convergence of the product. Similarly, if a product converges absolutely then any rearrangement of the terms of the product results in a product which is still absolutely convergent. If we combine Propositions 5.2 and 5.4 with the definition, the following fundamental criterion for convergence of a infinite product is obtained. 5.6 Corollary. If Re Zn > 0 then the product the series

L (zn -

IT

Zn converges absolutely

iff

1) converges absolutely.

Although the preceding corollary gives a necessary and sufficient condition for the absolute convergence of an infinite product phrased in terms with which we are familiar, it does not give a method for evaluating infinite products in terms of the corresponding infinite series. To evaluate a particular product one must often resort to trickery. We now apply these results to the convergence of products of functions. A fundamental question to be answered is the following. Suppose {J,,} is a sequence of functions on a set X and J,,(x) -+ f(x) uniformly for x in X; when will exp (fn(x) -+ exp (f(x» uniformly for x in X? Below is a partial answer which is sufficient to meet our needs. 5.7 Lemma. Let X be a set and let f, f1' f2' ... be functions from X into C such that j~(x) -+ f(x) uniformly for x in X. If there is a constant a such that Re f(x) :0; a for all x in X then exp J,,(x) -+ exp f(x) uniformly for x in X. Proof If € > 0 is given then choose 0 > 0 such that Ie' - 11 < E e - a whenever Izl < o. Now choose no such that Ifn(x) - f(x) I < 0 for all x in X whenever

n ;;::

110'

Thus

a-a> lexp [fn(x)-f(x)]-II

=

lexp fn(x) - 1/ expf(x)

It follows that for any x in X and for n ;;:: no,

lexpJ,,(x)-expf(x)1 <

E

e- a lexpj(x)1

:0;

E• •

5.S Lemma. Let (X, d) be a compact metric space and let {gn} be a sequence

Weierstrass Factorization Theorem

167

of continuous functions from X into IC such that and uniformly for x in X. Then the product

L gn(x) converges absolutely

n (1 + g.(x)) .=1 co

f(x) =

converges absolutely and uniformly for x in X. Also there is an integer no such that f(x) = 0 iff gn(x) = -1 for some n, 1 ~ n ~ no'

L

Proof Since g.(x) converges uniformly for x in X there is an integer no such that [g.(x) [ < t for all x in X and n > .no. This implies that Re [1 + g.(x)] > 0 and also, according to inequality (5.3), [log (l + gn{x» [ ~ t [gn(x) I for all n > no and x in X. Thus 00

hex) =

I

n =no

+1

log (1 + g.(x»)

converges uniformly for x in X. Since h is continuous and X is compact it follows that h must be bounded; in particular, there is a constant a such that Re hex) < a for all x in X. Thus, Lemma 5.7 applies and gives that

n 00

exp hex) =

n:;;no+ 1

(1 + gn(x»

converges uniformly for x in X. Finally, f(x) = [1 + g 1 (x)] .. . [1 + gnJx)] exp hex) and exp hex) #- 0 for any x in X. So iff(x) = 0 it must be that gn(x) = -1 for some n with 1 ~ n ~ no. III We now leave this general situation to discuss analytic functions. 5.9 Theorem. Let G be a region in IC and let Un} be a sequence in H(G) such that no fn is identically zero. If I. [J,,(z) -1] converges absolutely and uniformly 00

on compact subsets ofG then Ilfn(z) converges in H(G) to an analyticfunction n~1

f(z). If a is a zero off then a is a zero of only a finite number of the functions J", and the multiplicity of the zero off at a is the sum of the multiplicities of the zeros of the functions fn at a. Proof Since [j~(z) -1] converges uniformly and absolutely on compact subsets of G, it follows from the preceding lemma that fez) = nj~(z) converges uniformly and absolutely on compact subsets of G. That is, the infinite product converges in H(G). Suppose fCa) = 0 and let r > 0 be chosen such that B(a; r) c G. By hypothesis, I [J,,(z) -1] converges uniformly on BCa; r). According to Lemma 5.8 there is an integer n such that fCz) = fl Cz) .. .. f,.(z)g(z) where g does not vanish in B(a; r). The proof of the remainder of the theorem now follows. III Let us now return to a discussion of the original problem. If {an} is a sequence in a region G with no limit point in G (but possibly some point may be repeated in the sequence a finite number of times), consider the

I

168

Compactness and Convergence

functions (z - an). According to Theorem 5.9 if we can find functions gnCz) which are analytic on G, have no zeros in G, and are such that LI(z ---an)gn(z)---- 11 converges uniformly on compact subsets of G; thenf(z) = II(z - an)gn(z) is analytic and has its zeros only at the points Z = an' The safest way to guarantee that gn(z) never vanishes is to express it as gn(z) =exp hn(z) for some analytic function hn(z). In fact, if G is simply connected it follows that gn(z) must be of this form. The functions we are looking for were introduced by Weierstrass. 5.10 Definition. An elementary factor is one of the following functions Eiz) for p = 0, 1, ... :

Eo(z)

=

l-z,

Ep(z) = (1-z) exp (z

~ + ... + ~) ,p :2:

+

The function Eiz/a) has a simple zero at z

if

=

1.

a and no other zero, Also

bis a point in IC - G then Ep(az-b - b) has a simple zero at z

=

a and is

analytic in G. These functions will be used to manufacture analytic functions with prescribed zeros of prescribed multiplicity, but first an inequality must be proved which will enable us to apply Theorem 5.9 and obtain a convergent infinite product.

5.11 Lemma.

If Izl ::; 1

and p:2: 0 then II-Eiz)1 ::; IzlP+l,

Proof We may restrict our attention to the case where p :2: 1. For a fixed p let Eiz) = 1

GO

I akz k=l

+

k

be its power series expansion about z = O. By differentiating the power series as well as the original expression for Eiz) we obtain

I

OCJ

E-~(z) =

kakzk-l

k~1

= - zP exp

(z + .. , + ~)

Comparing the two expressions gives two pieces of information about the coefficients ak' First, a 1 = a z = , . , = a p = 0; second, since the coefficients of the expansion of exp

(z + ... + ~)

are all positive, ak

Thus, lakl = -ak for k :2: p+ 1; this gives

o=

Ell)

or 00

L

k~p+

I

la k = !

=

1

+

if;

I ak' k=p+l

::;

0 for k :2: P + I,

Weierstrass Factorization Theorem

Hence, for

izl ::;

169

I,

IEp{z) -

11

=

I

00

L

akzkl

k =p+ 1

=lzIP+li L akzk-P-!I 00

k=p+l

L lakl 00

::; IzIP+ I = IzIP+!

k=p+l

which is the desired inequality. III Before solving the general problem of finding a function with prescribed zeros, the problem for the case where G = C will be solved. This is done for several reasons. In a later chapter on entire functions the specific information obtained when G is the whole plane is needed. Moreover, the proof of the general case, although similar to the proof for C, tends to obscure the rather simple idea behind the proof.

5.12 Theorem. Let {an} be a sequence in C such that lim la,,1 = 00 and an # 0 for all n 2:: 1. (This is not a sequence of distinct points; but, by hypothesis, no point is repeated an infinite number of times.) If {Pn} is any sequence of integers such that 5.13

i

n= 1

(~)pn+ 1 1"1

<

00

for all r > 0 then f(z)

=

n EPn (z/a.) 00

n=1

converges in H(C). The function I is an entire function with zeros only at the points an' II z 0 occurs in the sequence {an} exactly m times then f has a zero at z = z 0 of multiplicity m. Furthermore, if Pn = n -I then (5.13) will be satisfied. Proof Suppose there are integers Pn such that (5.13) is satisfied. Then, according to Lemma 5.1 I,

whenever Izl ::; rand r ::; lanl. For a fixed r > 0 there is an integer N such that lanl ;;:: r for all n 2:: N (because lim la"1 = (0). Thus for each r > 0 the series L II - EpJ z / an)1 is dominated by the convergent series (5.13) on the disk 11(0; r). This gives that [l-Ep.(z/a n )] converges absolutely in H(C).

L

n Ep.(z/a 00

By Theorem 5.9, the infinite product

n

~

I

n)

converges in H(C).

To show that {P.} can be found so that (5.13) holds for all r is a trivial

170

Compactness and Convergence

matter. For any r there is an integer N such that lalll > 2r for all n ~ N. This gives that

C~:I)

<

!

for all n

~

N; so if Pn

= n-l for all n, the tail

L

end of the series (5.13) is dominated by CD". Thus, (5.13) converges. 18 There is, of course, a great latitude in picking the integers PII' If Pn were bigger than /1- I we would have the same conclusion. However, there is an advantage in choosing the Pn as small as possible. After all, the smaller the integer Pn the more elementary the elementary factor Epn{z/an ). As is evident in considering the series (5.13), the size of the integers Pn depends on the rate at which {:anl} converges to infinity. This will be explored later in Chapter XI. 5.14 The Weierstrass Factorization Theorem. Let f be an entire function and let {an} be the non-zero zeros off repeated according to multiplicity; suppose f has a zero at z = 0 of order m ;;. 0 (a zero of order m =0 at z =0 means f(O) *0). Then there is an entire function g and a sequence of integers {Pn} such that

Proof According to the preceding theorem integers {Pn} can be chosen such that

has the same zeros as f with the same multiplicities. It follows that f(z)/h(z) has removable singularities at z = 0, a j , a z , .... Thus flh is an entire function and, furthermore, has no zeros. Since IC is simply connected there is an entire function g such that

fez) h(z) The result now follows.

=

ef/(Z)

18

5.15 Theorem. Let G be a region and let {a j } be a sequence oj distinct points in G with no limit point in G; and let {m be a sequence oj integers. Then there is an analytic function f defined on G whose only zeros are at the points a j ; furthermore, aj is a zero off of multiplicity mj'

J

Proof We begin by showing that it suffices to prove this theorem for the special case where there is a number R > 0 such that 5.16

{z:

Izl

> R}

C

G and

lajJ

~ R

for allj ~ 1.

f in R(G) = the multiplicity of the zero at z = a j ;

It must be shown that with this hypothesis there is a function

with the a/s as its only zeros and mj

171

Weierstrass Factorization Theorem

and with the further property that

lim /(z)

5.17

z-+oo

= 1.

In fact, if such an f can always be found for a set satisfying (5.16), let G 1 be an arbitrary open set in C with {a j } a sequence of distinct point~ in G 1 with no limit point, and let {mj} be a sequence of integers. Now if B(a; r) is a disk in G 1 such that aj fE B(a; r) for all j:;:,: 1, consider the Mobius transformation Tz = (z - a)-I. Put G = T(G 1) \ {oo}; it is easy to see that G satisfies condition (5.16) where aj = Taj = (a j - a)-I. If there is a function f in H(G) with a zero at each aj of multiplicity mi' with no other zeros, and such that f satisfies (5.17); then g( z) = f(Tz) is analytic in G I - {a} with a removable singularity at z = a. Furthermore, g has the prescribed zero at each aj of multiplicity m j. So assume that G satisfies (5.16). Define a second sequence {z.} consisting of the points in {a j}, but such that each aj is repeated according to its multiplicity m j ' Now, for each n :;:,: 1 there is a point w" in C - G such that

Iw.-z.1

= d(z",

C-G).

Notice that the hypothesis (5.16) excludes the possibility that G = C unless the sequence {aj} were finite. In fact, if {aj} were finite the theorem could be easily proved so it suffices to assume that {a)} is infinite. Since laJi :s R for allj and {aj} has no limit point in G it follows that C-G is non-empty as well as compact. Also,

lim Iz. - w.1 Consider the functions

=

o.

E,,(Z.-W,,) ; z-w"

each has a simple zero at z = Zn' It must be shown that the infinite product of these functions converges in H(G). To do this let K be a compact subset of G so that d(C-G, K) > O. For any point z in K

:s Iz,,-w.1 [d(w", K)r IZn-Wnl z-w" :s

1

Iz.-w.1 [d(C-G, K)r 1

It follows that for any 8, 0 < 8 < 1, there is an integer N such that



n

o.

n [1 + ~J and nil + ~I.

n=2 n 2 8. For which values of z do the products (I-zn) and (1 +z2n) converge? Is there an open set G such that the product converges uniformly on each compact subset of G? If so, give the largest such open set.

n

n

174

Compactness and Convergence

9. Use Theorem 5.15 to show there is an analytic function fan D = {z: G which properly contains D. 10. Suppose G is an open set and {J,,} is a sequence in H(G) such that /(z) = converges in H(G). (a) Show that

Iz I < I} which is not analytic on any region

llt..(z)

k~l ~k(Z) JJkt..(Z)] converges in H(G) and equals J'(z). (b) Assume that f is not the identically zero function and let K be a compact subset of G such thatf(z) # 0 for all z in K. Show that J'(z) = ~ f~(z) fez) J,,(z)

.:S

and the convergence is uniform over K. 11. A subset J of H(G), G a region, is an ideal iff: (i) f and g in J implies af+bgis in Jfor all complex numbers a and b; (ii)fin J andg any function in H(G) impliesfg is in J. J is called a proper ideal if J # (0) and J # H(G); J is a maximal ideal if J is a proper ideal and whenever J is an ideal with J c J then either J = J or J = H(G); J is a prime ideal if whenever f and g e H(G) and /g e J then either fe J or g e J. If fe H(G) let ~(.f) be the set of zeros of/counted according to their multiplicity. So ~«z-a)3) = {a, a, a}. If!/ c H(G) then ~(!/) = n {~(f):fe!/}, where the zeros are again counted according to their multiplicity. So if!/ = {(z-a)3 (z-b), (z-a)2} then ~(!/) = {a, a}. (a) Iff and g e H(G) then / divides g (in symbols, fIg) if there is an h in H(G) such that g = fh. Show that fIg iff ~(f) c ~(g). (b) If Yc H(G) and Y contains a non-zero function then f is a greatest common divisor of Y if: (i) fig for each g in Y and (ii) whenever hlg for each g in Y, hi/, In symbols, f = g.c.d.Y. Prove that f = g.c.d.Y. iff !Ir(f) = !Ir(Y) and show that each non-empty subset of H(G) has a g.c.d. (c) If A c G let J(A) = {fe H(G): ~(f) ::l A}. Show that J(A) is a closed ideal in H(G) and J(A) = (0) iff A has a limit point in G. (d) Let a E G and J = J( {a}). Show that J is a maximal ideal. (e) Show that every maximal ideal in H(G) is a prime ideal. (f) Give an example of an ideal which is not a prime ideal. 12. Find an entire function / such that fen + in) = 0 for every integer n (positive, negative or zero). Give the most elementary example possible (i.e., choose the Pn to be as small as possible). 13. Find an entire function/such that f(m + in) = 0 for all possible integers m, n. Find the most elementary solution possible.

§6. FIlctoriZlltion of the sine function In this section an application of the Weierstrass Factorization Theorem to sin 1TZ is given. If an infinite sum or product is followed by a prime

175

Factorization of the sine function

(apostrophe) (i.e., 2:' or TI'), then the sum or product is to be taken over all the indicated indices 11 except n = 0. For example, co

00

00

2:' an = n=1 L a_ n + "==1 Lan" n=-oo The zeros of sin

7TZ =

1 .. h . 21 (e mZ - e - InZ) are precisely t e Integers; moreover,

each zero is simple. Since

n~~ GY < for all r > 0, one can (5.13) choose fl. Factorization Theorem. Thus sin

7TZ =

[exp g(z)] z

=

00

1 for all n

111

the Weierstrass

J~I~ (1 - ~) ex!n;

or, because the terms of the infinite product can be rearranged,

6.1

sin

7TZ =

[exp g(z)] z

for some entire function g(z). If fez) = sin 'IT

D(1 -::) 7TZ

then, according to Theorem 2.1,

Fez)

cot 'lTZ = -.-fez) 1

=

2z ;2_n 2

00

+ -z + L, '"

g'(z)

n~l

and the convergence is uniform over compact subsets of the plane that contain no integers (actually, a small additional argument is necessary to justify this-see Exercise 5.10). But according to Exercise V. 2.8, 'IT

cot 'lTZ

1 =

Z

+

2z 2: .=1z-n 00

-2--2

for z not an integer. So it must be that g is a constant, say g(z) = a for all z. It follows from (6.1) that for 0 < Izi < 1 sin

TTZ =

7TZ

::

fr (1 _;z~).

Letting z approach zero gives that ea =

6.2

sin 'lTZ =

n

7T n= 1

'lTZ

'IT.

This gives the following:

1:':'r (1 _z:) n= 1

fl

and the convergence is uniform over compact subsets of C.

176

Compactness and Convergence

Exercises 1. Show that cos

D[

(in4~2I)2 J.

1-

TTZ =

2. Find a factorization for sinh z and cosh z. 3. Find a factorization of the function cos I \ • ,

4. Prove WallIs s formula:

7T

:2

xo n·

:.J -sin(

7TZ).

. 4

!

(2n)2

= n~ 1 (2n -1)(2~+ 1) .

§7. The gamma/unction Let G be an open set in the plane and let {/,.} be a sequence of analytic functions on G. If {/,.} converges in H(G) tofandfis not identically zero, then it easily follows that {In} converges to f in M(G). Since d(ZI, Z2) =

d(~, ~), where dis the spherical metric on Coo (see (3.1)), it follows that '" I

Z2

converges to! {~} /,. f verges uniformly to

in ll'I(G). It is an easy exercise to show that

.!

f

{~} /,.

con-

on any compact set K on which no /,. vanishes. (What

does Hurwitz's Theorem have to say about this situation?) Since, according to Theorem 5.12, the infinite product

converges in H(C) to an entire function which only has simple zeros at Z = - I, - 2, ... , the above discussion yields that

1"1 ro

7.1

11=

1

(

1

+~

)-1

e z/ n

converges on compact subsets of C - { - I, -- 2, ... } to a function with simple poles at z = -1, -2, .... 7.2 Definition. The gamma function, r(z), is the meromorphic function with simple poles at z = 0, -1, ... defined by

7.3

r(z) =

e~YZ(1(1

+

where y is a constant chosen so that r(l)

~rlez/n,

= I.

Oil

C

The gamma function

177

The first thing that must be done is to show that the constant y exists; this is an easy matter. Substituting z = 1 in (7.1) yields a finite number C

D + ~1)-1 00

=

(

1

e l/n

which is clearly positive. Let y = log c; it follows that with this choice of y, equation (7.3) for z = 1 gives r(l) = 1. This constant y is called Euler's constant and it satisfies

7.4

=

eY

D + ~1)-1 00

(

1

e lln

Since both sides of (7.4) involve only real positive numbers and the real logarithm is continuous, we may apply the logarithm function to both sides of (7.4) and obtain y

~ 10g[(1 + O-le

=

=

i [~ -

k~1

= lim

k

log (k+ 1)+log k]

.i: [~ k

" .... OOk=1

= lim [(I + " .... 00

log (k+ 1)+log k]

~2 +

Adding and subtracting log

11

fact that lim log ( 11 :

yields

7.5

y

1lk ]

1) = 0

... +

!)

n

-IOg(n+1)].

to each term of this sequence and using the

= lim [(1 + n .... oo

!+

2

... +

~)

11

- log

n] .

This last formula can be used to approximate y. Equation (7.5) is also used to derive another expression for r(z). From the definition of r(z) it follows that

0

- yz r(z) = _e_ lim " ( 1 + ~ z n .... ook=1 k - yz

= _e_ lim z

lim

n .... oo

)-1

ezlk

o_en

n .... oo k=1

k

zlk

z+k

I))

e-yzn' . exp ( z( 1 +1+ ... + z(z+ I) ... (z+n) n

178

Compactness and Convergence

However e-- YZ exP [(1+1+ ...

+~)zJ = nzexp[z( -y+I+± + ... + ~ -IOgn)].

So that the following is obtained 7.6 Gauss's Formula. For z #- 0, -1, ...

n' n- - fez) = lim --------------. "--->00 z(z+ 1) ... (z+n) Z

The formula of Gauss yields a simple derivation of the functional equation satisfied by the gamma function. 7.7 Functional Equation. For z #- 0, -1, ... f(z+ 1)

=

zf(z)

To obtain this important equation substitute z + 1 for z in (7.6); this gives n! n 0 there is a number K > 1 such that 12c(e--!-a- e--!-P)1 < E whenever a, f3 > K, giving part (b) . • The results of the preceding lemma embody exactly the concept of a uniformly convergent integral. In fact, if we consider the integrals

182

Compactness and Convergence

for 0 < IX < 1, then part (a) of Lemma 7.16 says that these integrals satisfy a Cauchy criterion as IX -+ O. That is, the difference between any two will be arbitrarily small if IX and f3 are taken sufficiently close to zero. A similar interpretation is available for the integrals

for

IX

> 1. The next proposition formalizes this discussion.

7.17 Proposition. If G

= {z: Re z > O} and

f e-tt n

/,,(z) =

Z- 1

dt

lin

for n ~ 1 and z in G, thell each J.. is analytic vergent ill B(G).

011

G and the sequence is

Proof. Think of/,,(z) as the integral of cp(t, z) = e - tt z segment

[~,

l

COIl-

along the straight line

11] and apply Exercise IV. 2.2 to conclude that J.. is analytic.

Now if K is a compact subset of G there are positive real numbers a and A such that K c {z: a :::; Re z :::; A}. ,Since

f e-tt

lin

fm(z)-J..(Z) =

z-

11m

l dt

+

f e-tt m

z- l

dt

n

for m > n, Lemma 7.16 and Lemma 1.7 imply that {J..} is a Cauchy sequence in B(G). But B(G) is complete (Corollary 2.3) so that {J..} must converge. • If f is the limit of the functions {J..} from the above proposition then define the integral to be this function. That is,

f e-tt O.

o

To show that this functionf(z) is indeed the gamma function for Re z > 0 we only have to show that f(x) = rex) for x ~ 1. Since [1, 00) has limit points in the right half plane and both f and r are analytic then it follows thatfmust be r (Corollary IV. 3.8). Now observe that successive performing of integration by parts on (1- tjlltt,,-l yields

f( n

o

1 - ~). t X n

1

dt = _ _ _ Il_!n_"_ __ x(x+ 1) ... (x+n)

which converges to rex) as 11 -+ 00 by Gauss's formula. If we can show that the integral in this equation converges to SO' e-tt x - 1 dt = f(x) as 11 -+ 00 then

183

The gamma function

Theorem 7.15 is proved. This is indeed the case and it follows from the following lemma.

+

7.19 Lemma. (a) {( 1

(b)

Iff

~

0 then

~)"}

converges to e in H(C). Z

(1 -~)" :S:

~

e-'forall n

t.

Proof (a) Let K be a compact subset of the plane. Then lzj < K and 11 sufficiently large. It suffices to show that

lim /llog

n-+oo

(1 +=)

for all z in

z

=

11

11

uniformly for z in K by Lemma 5.7. Recall that

L (-It-

wk

00

logO+w) =

1

k

k=!

for

IlVl

< J. Let n > II

So 7.20

Izi

log

for all

( + -z) 1

n

( z) -

n log 1 + ~

z

z in

K; if

I Z2 2 n

= z -

=

z is

any point in K then

1 Z3 3n

+ - - 2 - . .. .

[I2(z) +:31(Z)2 -...] ; ~

z -

~

taking absolute values gives that

In log (1 +~) - z\:S: Izi ~~I~r-l :S: Iz/

~ I~r

k=l

Izl2

1

--_.-----

n I

-Iz/nl

R2 - n-R

1 and let £ > O. According to Lemma 7.16 (b) we can choose K > 0 such that r

f e-tt

7.21

x- 1

dt <

~

whenever r > K. Let II be any integer larger than defined in Proposition 7.17. Then

K

and letfn be the function

J[e- (1 + ~YJ t

-

tx -

1

dt

lin

Now by Lemma 7.19 (b) and Lemma 7.16 (a) 7.22

for sufficiently large II. Also, if II is sufficiently large, part (a) of the preceding lemma gives

for tin [0, K] where M = So

tx-

1 dt. Thus

7.23

Using Lemma 7.19 (b) and (7.21)

for

II

>

K.

If we combine this inequality with (7.22) and (7.23), we get

185

The gamma function

for n sufficiently large. That is

-X(-X-+-l-;-~~-~-(x-+-nJ

= lim [/"(X) = I(x) - rex).

This completes the proof of Theorem 7.15 . • As an application of Theorem 7.15 and the fact that 2) notice that

reD =

.;:;

(Exercise

0 by setting it equal to [r(z)] - 1 times the right hand side of (8.7). In this manner ~ is meromorphic in the right half plane with a simple pole at z = 1 n- I diverges) whose residue is 1. Now suppose 0 < Re z < I; then

(I

co

(Z-1)-1

Jt

= -

z-

2 dt.

I

Applying this to equation (8.7) gives

f(_,_l- - 1) co

8.8

~(z)f(z)

=

o

e -1

t

tz -

I

dt,

0< Re z < 1.

Again considering the Laurent expansion of (e Z -1) - 1 (8.6) we see that [ee' - 1) -1 - r 1+ -!l :::; ct for some constant c and all t in the unit interval [0, IJ. Thus the integral

is uniformly convergent on compact subsets of {z: Re z > -I}. Also, since lim t (}:- I~oc

_f

_1_)' 1 e -1 =

t

191

The Riemann zeta function

there is a constant c' such that

(~

1- ) -1 with a simple pole at z = 1. Now if -1 < Re z < 0 then

nr

f 00

t,-1 dt =

z

1

inserting this in (8.9) gives

1 1+ -I)" f( e -1 2 00

t(z)f(z) =

8.10

-

But

-.- t

o

1

et _ 1

1

+i

t

t .-1 dt

+ 1) i i1 (e;t=1 = i

'

-1 < Re z < O.

t

=

cot (Yt).

A straightforward computation with Exercise V. 2.8 gives

2

cot (-!it) = --:- - 4it It

for t

=1=

O. Thus

1 2'"= -2--22 + 4n

n

1

t

1T

Compactness and Convergence

192

Applying this to (8.10) gives 1, where den) IS the number of

2:"

divisors of n. 4. Prove that

nZ

n~l

~(z)~(z-l)

=

. 2:'" a(n) - , for Re z > 1, where a(n) IS the sum of n Z

n~l

the divisors of n. Uz-I)

5. Prove that - - - = ~(z)

2:"" cp(n) - for Re z > I, where cp(n) is the number of

n~l nZ

integers less than n and which are relatively prime to n. 1 ~ p.(n) . 6. Prove that = L. - for Re z > 1, where p.(n) IS defined as follows. ~(z) n~l nZ pftp~2 •.. p~m be the

Let n = factorization of n into a product of primes PI' •.• ,Pm and suppose that these primes are distinct. Let p.(l) = 1; if k I = ... = k m = I then let p.(n) = ( _l)m; otherwise let p.(n) = o. nz) ~ A(n) . 7. Prove that 1(z) = - L. 7 for Re z > 1, where A(n) = log P If n = n~l

pm for some prime P and m ;::: I; and A(n) = 0 otherwise. 8. (a) Let 7](z) = ~'(z)/~(z) for Re z > 1 and show that lim (z-zo)7](z) is z-+zo

always an integer for Re z 0 ;::: I. Characterize the point z 0 (in its relation to 0 in terms of the sign of this integer. (b) Show that for E > 0 Re 7] (1 +E+it) =

L 00

n:;;;; 1

A(n)n-(1+ 0, 3Re 7] (1 +E)+4Re 7] (1 + E+ it) + Re 7] (1 +€+2it) ::; (d) Show that

~(z)

¥ 0 if Re z

=

1 (or 0).

o.

Chapter VIII Runge's Theorem In this chapter we will prove Runge's Theorem and investigate simple connectedness. Also proved is a theorem of Mittag-Leffler on the existence of meromorphic functions with prescribed poles and singular parts. §1. Runge's Theorem In Chapter IV we saw that an analytic function in an open disk is given by a power series. Furthermore, on proper subdisks the power series converges uniformly to the function. As a corollary to this result, an analytic function on a disk D is the limit in H(D) of a sequence of polynomials. We ask the question: Can this be generalized to arbitrary regions G? The answer is no. As one might expect the counter-example is furnished by G = {z: o < 1=1 < 2}. If {Pn(z)} is a sequence of polynomials which converges to an analytic functionj on G, and y is the circle Izl = 1 then JJ = lim JyPn = O. But Z-1 is in H(G) and SyZ-l 1= O. The fact that functions analytic on a disk are limits of polynomials is due to the fact that disks are simply connected. If G is a punctured disk then the Laurent series development shows that each analytic function on G is the uniform limit of rational functions whose poles lie outside G (in fact at the center of G). That is, each j in H(G) is the limit of a sequence of rational functions which also belong to H(G). This is what can be generalized to arbitrary regions, and it is part of the content of Runge's Theorem. We begin by proving a version of the Cauchy Integral Formula. Unlike the former version, however, the next proposition says that there exist curves such that the formula holds; not that the formula holds for every curve. 1.1 Proposition. Let K be a compact subset oj the region G; then there are straight line segments YI"'" Yn in G---- K such that Jor every Junction J in H(G),

j(z) =

~ _1. f£(~~ dw k=

12m

w-z

Yk

Jor all z in K. The line segments Jorm a Jinite number oj closed polygons. ProoJ. Observe that by enlarging K a little we may assume that K = (intKf· Let OO then there is a rational junctiun R(z) whose only poles lie in E and such that Ij(z)-R(z)lO; that is, such that j = u -lim Rn on K. Let B(E)=all functions j in C(K,C) such that there is a se4uence {Rn} of rational functions with poles in E such that j = u····1im Rn on K. Runge's Theorem states that if j is analytic in a neighborhood of K then flK, the restriction of jto K, is in B(E). 1.8 Lemma. B (E) is a closed subalgebra oj C (K, C) that contains every rational junction with a pole in E. To say that B(E) is an algebra is to say that iff and g are in B(E) and ex E C then rxf, j + g, and jg are in B (E). The proof of Lemma 1.8 is left to the reader.

1.9 Lemma. Let V and U be open subsets of C with V c U and dV /j H is a component of U and H n V7'=O then He V.

n U = O.

Proof. Let a E H n V and let G be the component of V such that a E G. Then HuG is connected (II.2.6) and contained in U. Since H is a component of U, C c fl. But dC c av and so the hypothesis of the lemma says that dC n H =0. This implies that H - C= H

n [(c -

) u DC]

C

=Hn(C-G

)

so that H-C is open in H. But G is open implies that H-G=Hn(cG) is closed in H. Since H is connected and G 7'= 0, H- G=O. That is, H=Gc V.JIII

199

Runge's Theorem

1.10 Lemma. If a e C - K then (z - a)-I e B (E).

Proof. Case 1. 00 , E. Let U=C-K and let V={a e C:(z-a)-I e B(E)}; so Ec V cU.

1.11

If a

E

Vand 1 b - al < dCa, K) then

bE

V.

The condition on b gives the existence of a number r, O 0 then there is a g in R(G) with /I(z) - g(z) I < EO for all z in K; (b) If D is a bounded component of G - K then D - () 8G =1= 0; (c) If z is any point in G-K then there is a function/in R(G) with componentDofG-KhasD- () 8G

I/(z)/ > sup {/I(w)l: win K}. 5. Can you interpret part (c) of Exercise 4 in terms of K? 6. Let K be a compact subset of the region G and define KG = {z E G: /I(z)/ :::; II/IIK for all I in R(G)}. (a) Show that if Coo - G is connected then KG = K. (b) Show that d(K, C-G) = d(KG' C-G). (c) Show that RG c the convex hull of K == the intersection of all convex subsets of C which contain K.

202

Runge's Theorem

(d) If KG c G 1 c G and G1 is open then for every g in B(G 1 ) and € > 0 there is a functionfin B(G) such that /f(z)-g(z)/ < € for all z inKG' (Hint: see Exercise 4.) (e) KG = the union of K and all bounded components of G-K whose closure does not intersect aG. §2. Simple connectedness Recall that an open connected set G is simply connected if and only if every closed rectifiable curve in G is homotopic to zero. The purpose of this section is to prove some equivalent formulations of simple connectedness. 2.1 Definition. Let X and n be metric spaces; a homeomorphism between X and U is a continuous map f: X -* n which is one-one, onto, and such thatf-l: n -* X is also continuous. Notice that if f: X -* n is one-one, onto, and continuous then f is a homeomorphism if and only if f is open (or, equivalently, f is closed). If there is a homeomorphism between X and n then the metric spaces X and n are homeomorphic. We claim that C and D = {z: /z/ < I} are homeomorphic. In fact fez) = z(l + /zl) - I maps C onto D in a one-one fashion and its inverse, f-l(w) = w(l-Iwi)-I, is clearly continuous. Also, iffis a one-one analytic function on an open set G and n = f(G) then G and n are homeomorphic. Finally, all annuli are homeomorphic to the punctured plane. 2.2 Theorem. Let G be an open connected subset of C. Then the following are equivalent: (a) G is simply connected; (b) n(y; a) = 0 for every closed rectifiable curve y in G and every point a in C-G; (c) C",-G is connected; (d) For any f in B (G) there is a sequence of polynomials that converges

tofinB(G); (e) For anyf in H(G) and any closed rectifiable curve y in G, Srf = 0; (f) Every function f in B (G) has a primitive; (g) For any fin B(G) such that fez) #- 0 for all z in G there is afunction g in B(G) such that fez) = exp g(z); (h) For any fin B(G) such that fez) #- 0 for all z in G there is a function

ginB(G)suchthatf(z) = [g(Z)]2;

(i) G is homeomorphic to the unit disk; (j) If u: G -* R is harmonic then there is a harmonic function v: G such that f = u+iv is analytic on G.

Proof. The plan is to show that (a) = (b) = ... = (i) = (a) and (h) = (g). Many of these implications have already been done.

-*

=

R

(j)

Simple connectedness 203 (a) => (b) If y is a closed rectifiable curve in G and a is a point in the complement of G then (z - a) - 1 is analytic in G, and part (b) follows by

Cauchy's Theorem. (b) => (c) Suppose Ceo -G is not connected; then Ceo -G = Au B where A and B are disjoint, non-empty, closed subsets of Coo. Since 00 must be either in A or in B, suppose that 00 is in B; thus, A must be a compact subset of C (A is compact in Coo and doesn't contain (0). But then G 1 = G u A = Coo - B is an open set in C and contains A. According to Proposition 1.1 there are a finite number of polygons Yl' ."•. ,Ym in G 1 -A = G such that for every analytic function f on G 1

fez) =

i

~.J·f(W)

k=12m

w-z

dw

7k

for all z in A. In particular, if fez) == 1 then

1=

m

L n(Yk; z) k=l

for all z in A. Thus for any z in A there is at least one polygon Yk in G such that n(Yk; z) "# O. This contradicts (b). (c) => (d) See Corollary I.lS. (d) => (e) Let Y be a closed rectifiable curve in G, let f be an analytic function on G, and let {Pn} be a sequence of polynomials such thatf = lim Pn in B(G). Since each polynomial is analytic in C and Y '" 0 in C, LPn = 0 for every n. But {Pn} converges to f uniformly on {y} so that L f = lim LPn = o. (e) => (f) Fix a in G. From condition (e) it follows that there is a function F: G -+ C defined by letting F(z) = Lfwhere Y is any rectifiable curve in G from a to z. It follows that F' = f (see the proof of Corollary IV. 6.16). (f) => (g) Iff(z) "# 0 for all z in G then/'/fis analytic on G. Part (f) implies there is a function F such that F' = /,If It follows (see the proof of Corollary IV. 6.17) that there is an appropriate constant c such that g = F + c satisfies fez) = exp g(z) for all z in G. (g) => (b) This is trivial. (b) => (i) If G = C then the function z(1 + Izi) -1 was shown to be a homeomorphism immediately prior to this theorem. If G "# C then Lemma VII. 4.3 implies that there is an analytic mapping f of G onto D which is one-one. Such a map is a homeomorphism. (i) => (a) Let h: G -+ D = {z: Izl < I} be a homeomorphism and let Y be a closed curve in G (note that Y is not assumed to be rectifiable). Then a(s) = h(y(s» is a closed curve in D. Thus, there is a continuous function A: [2 -+ D such that A(s, 0) = a(s) for 0 :s; s :s; I, A(s, I) = 0 for 0 :s; s :s; 1 and A(O, t) = A(I, t) for 0 :s; t :s; 1. It follows that r = h- 1 0 A is a continuous map of [2 into G and demonstrates that Y is homotopic to the curve which is constantly equal to h- 1(0). The details are left to the reader. (b) => (j) Suppose that G "# C; then the Riemann Mapping Theorem implies there is an analytic function h on G such that h is one-one and h(G)

204

Runge's Theorem

= D. If u: G ---J. R is harmonic then U 1 = U h - I is a harmonic function on D. By Theorem III. 2.30 there is a harmonic function VI: D ---J. R such that 11 = U 1 + iv, is analytic on D. Let I = 11 h. Then I is analytic on G and u is the real part off Thus v = 1m I = VI 0 h is the sought after harmonic conjugate. Since Theorem m. 2.30 also applies to C, (j) follows from (h). (j) (g) Suppose I: G ---J. C is analytic and never vanishes, and let u = Re J, v = 1m f If V: G ---J. R is defined by Vex, y) = log I/(x+iy)1 = log [u(x, y)2 + vex, y)2]t then a computation shows that V is harmonic. Let V be a harmonic function on G such that g = V + iV is analytic on G and 0

0

=

let h(z) = exp g(z). Then h is analytic, never vanishes, and

I/(Z)I = h(z)

1 for all

z in G. That is,f/h is an analytic function whose range is not open. It follows that there is a constant c such that/(z) = c h(z) = c exp g(z) = exp [g(z) + CI]' Thus, g(z)+c l is a branch of log/(z). This completes the proof of the theorem. III This theorem constitutes an aesthetic peak in Mathematics. Notice that it says that a topological condition (simple connectedness) is equivalent to analytical conditions (e.g., the existence of harmonic conjugates and Cauchy's Theorem) as well as an algebraic condition (the existence of a square root) and other topological conditions. This certainly was not expected when simple connectedness was first defined. Nevertheless, the value of the theorem is somewhat limited to the fact that simple connectedness implies these nine properties. Although it is satisfying to have the converse of these implications, it is only the fact that the connectedness of Coo - G implies that G is simply connected which finds wide application. No one ever verifies one of the other properties in order to prove that G is simply connected. For an example consider the set G = C - {z = re ir : 0 :::; r < ro}; that is, G is the complement of the infinite spiral r = (), 0 :::; () < roo Then CO() - G is the spiral together with the point at infinity. Since this is connected, G is simply connected.

Exercise I. The set G={re il : ----ro R for j =I- k. Thusf(z) = Sk(Z)+g(z) for 0 < iz-akl < R, where g is analytic in B(ak ; R). Hence, z = ak is a pole off and Sk(Z) is its singular part. This completes the proof of the theorem. II Just as there is merit in choosing the integers Pn in Weierstrass's Theorem (VII. 5.2) as small as possible, there is merit in choosing the rational functions RnCz) in (3.3) to be as simple as possible. As an example let us calculate the simplest meromorphic function in the plane with a pole at every integer n.

at

The simplest singular part is (z - n) - 1 but

co

I

(z -n) -1 does not converge in

-- co

M(C). However (z_n)-1+(z+II)-1 = 2Z(Z2_ n 2)"1 and 1 Z

+ ~'_. ~z_ L.,

"=1

Z2

_n 2

does converge in M (C). The singular part of this function at Z = n is (z - n) -1. In fact, from Exercise V. 2.8 we have that this function is 7r cot 7TZ. Exercises 1. Let G be a region and let {an} and {b m} be two sequences of distinct points in G without limit points in G such that an =I- bn for all n, m. Let Sn(z) be a singular part at an and let Pm be a positive integer. Show that there is a meromorphic functionfon G whose only poles and zeros are {an} and {b m} respectively, the singular part at z = an in Sn(z), and z = b m is a zero of multiplicity Pm' 2. Let {an} be a sequence of points in the plane such that lani--+- co, and let {b n } be an arbitrary sequence of complex numbers. (a) Show that jf integers {k n } can be chosen such that 3.4 converges absolutely for all r > 0 then

3.5 converges in M(C) to a function/with poles at each point z

=

an'

Mittag-Leffler's Theorem

kn

207

(b) Show that if lim sup Ibnl < 00 then (3.4) converges absolutely if n for all n. (c) Show that if there is an integer k such that the series

=

3.6 converges absolutely, then (3.4) converges absolutely if k n = k for all n. (d) Suppose there is an r > 0 such that Ian - ami ;::: r for all n # m. Show that L lanl- 3 < 00. In particular, if the sequence {b n } is bounded then the series (3.6) with k = 2 converges absolutely. (This is somewhat involved and the reader may prefer to prove part (f) directly since this is the only application.) (e) Show that if the series (3.5) converges in M(I[) to a meromorphic function f then

fez)

=

[b

b{ ( '7)

(

~ ----::- + -" ] + -"'- + ... + ---- )kn L

n= 1

Z

an

an

'7

an

an

-I}]

(f) Let wand w' be two complex numbers such that 1m (w'lw) # O. Using the previous parts of this exercise show that the series

where the sum is over all w = 2nw+2n'w for 1'1, n' = 0, ± I, ±2, ... but not w = 0, is convergent in Mel[) to a meromorphic function, with simple poles at the points 2nw+2n'w'. This function is called the Weierstrass zeta function. (g) Let &,0(Z) = - nz); fp is called the Weierstrass pe function. Show that gJ(z)

=

;12

+ L' ((~~W)2 -

:2)

where the sum is over the same w as in part (f). Also show that fp(z) = gJ(z+2nw+2n'w') for all integers nand 1'1'. That is, gJ is doubly periodic with periods 2w and 2w'. 3. This exercise shows how to deduce Weierstrass's Theorem for the plane (Theorem VII. 5.12) from Mittag-Leffler's Theorem. (a) Deduce from Exercises 2(a) and 2(b) that for any sequence {an} in C with lim an = 00 and an # 0 there is a sequence of integers {k n } such that h(z) =

(z) + ... + ---I (z-- )kn - J

~,[ --1 + -1 + -1 L

n=1

z-a n

an

an an

1

an an

is a meromorphic function on C with simple poles at aI' a 2, .. , . The remainder of the proof consists of showing that there is a functionf such that h = f'lf This function f will then have the appropriate zeros.

Runge's Theorem

208

(b) Let z be an arbitrary but fixed point in I[; - {aI' a z, ••• }. Show that if Y1 and yz are any rectifiable curves in 1[;- {aI, a2,"'} from 0 to z and his the function obtained in part (a), then there is an integer m such that

Jh - Jh 1'1

=

27Tim.

Y2

(c) Again let h be the meromorphic function from part (a). Prove that for z i= aI' a z, .•. and y any rectifiable curve in 1[;- {ai' a z , .• •},

fez) = exp

(f h)

defines an analytic function on I[; - {aI, a2, ... } with j'lf = h. (That is, the value of fez) is independent of the curve y and the resulting function f is analytic. (d) Suppose that z E {ai' a 2 , •• • }; show that z is a iemovable singularity of the function f defined in part (c). Furthermore, show that fez) = 0 and that the multiplicity of this zero equals the number of times that z appears in the sequence {a l , a z , ... }. (e) Show that

3.7

fez) =

n 00

n= I

(

1 - -z ) exp [ -z an an

+ 1- (Z)2 . . . . + ... + -1 (z- )knJ 2 an

k n an

Remark. We could have skipped parts (b), (c), and (d) and gone directly from (a) to (e). However this would have meant that we must show that (3.7) converges in H(C) and it could hardly be classified as a new proof. The steps outlined in parts (a) through (d) give a proof of Weierstrass's Theorem without introducing infinite products. 4. This exercise assumes a knowledge of the terminology and results of Exercise VII. 5.11. (a) Define two functionsfandg in H(G) to be relatively prime (in symbols, (I, g) = 1) if the only common divisors off and g are non-vanishing functions in H(G). Show that (I, g) = 1 iff !!t'(f) n !!t'(g) = O. (b) If (I, g) = 1, show that there are functionsfl' gl in H(G) such that ill + gg 1 = 1. (Hint: Show that there is a meromorphic function 'P on G such that II = 'Pg E H(G) and gi(i-fJ1)') (c) Let fl' ... ,f" E H(G) and g = g.c.d {II,' .. ,fn}. Show that there are functions 'Pl"'" 'Pn in H(G) such that g = 'Plf1 + .. '+'Pn/". (Hint: Use (b) and induction.) (d) If {.J'~J is a collection of ideals in H(G), show that .J' = ,i'a is

n a

also an ideal. If Y' c H(G) then let .J' = n {)"": )"" is an ideal of H(G) and Y' c ),,"}. Prove that .J' is the smallest ideal in H(G) that contains Y' and .J' = {'Pdl + ... +'P,,/": 'Pk E H(G),jk E Y' for 1 ~ k ~ n} . .J' is called the ideal generated by Y' and is denoted by .J' = (Y'). If Y' is finite then (Y') is called a finitely generated ideal. If .7 = {f} for a single function f then (f) is called a principal ideal.

Mittag-Leffler's Theorem

209

(e) Show that every finitely generated ideal in H(G) is a principal ideal. (f) An ideal ..1' is called a fixed ideal if :2"(..1') D; otherwise it is caned a free ideal. Prove that if ..1'= (Y') then :2"(..1') = :2"(,sP) and that a proper principal ideal is fixed. (g) LetJiz) = sin (2-·z) for all n ~ 0 and let ..1' = ({ft,J2' ... D. Show that J is a fixed ideal in H{C) which is not a principal ideal. (h) Let J be a fixed ideal and prove that there is an J in H(G) with :!l'(f} = :!l'(..1') and J c (f). Also show that ..1' = (f) if ..1' is finitely generated. (i) Let Jt be a maximal ideal that is fixed. Show that there is a point a in G such that Jt = «z - a». (j) Let {an} be a sequence of distinct points in G with no limit point in G. Let J = {f E H(G): J(an ) = 0 for all but a finite number of the an}. Show that J is a proper free ideal in H(G). (k) ILf is a proper free ideal show that for any finite subset Y' off, :2"(Y') :F O. Use this to show that J can contain no polynomials. (1) Let ..1' be a proper free ideal; then f is a maximal ideal iff whenever g E H(G) and :2"(g) n :2"(/) D for all f in f then g E f. 5. Let G be a region and let {a.} be a sequence of distinct points in G with no limit point in G. For each integer n ~ 1 choose integers k n ~ 0 and constants A~k), 0 ::; k ::; k n • Show that there is an analytic function J on G such thatJ(k)(an) = k!A~k). (Hint: Let g be an analytic function on G with a zero at an of multiplicity k n + 1. Let h be a meromorphic function on G with poles at each an of order k n + I and with singular part Sn(z). Choose the Sn so thatJ = gh has the desired property.) 6. Find a meromorphic function f with poles of order 2 at 1, 12, 13, ... such that the residue at each pole is 0 and lim (z-.Jn)2J(z) = 1 for all n.

'*

'*

z-v'n

Chapter IX Analytic Continuation and Riemann Surfaces Consider the following problem. Letfbe an analytic function on a region G; when can f be extended to an analytic function fl on an open set G I which properly contains G? If G 1 is obtained by adjoining to G a disjoint open set so that G becomes a component of G 1 , f can be extended to G I by defining it in any way we wish on G 1 - G so long as the result is analytic. So to eliminate such trivial cases it is req uired that G 1 also be a region. Actually, this process has already been encountered. Recall that in the discussion of the Riemann zeta function (Section VII. 8) ~(z) was initially defined for Re Z > 1. Using various identities, principal among which was Riemann's functional equation, ~ was extended so that it was defined and analytic in IC - {I} with a simple pole at z = 1. That is, ~ was analytically continued from a smaller region to a larger one. Another example was in the discussion that followed the proof of the Argument Principle (V.3.4). There a meromorphic function f and a closed rectifiable curve y not passing through any zero or pole of f was given. If z = a is the initial poi!'t of y (and the final point), we put a disk Dl about a on which it was possible to define a branch tl of log f Continuing, we covered y by a finite number of disks D 1 , D z, . .. , Dn, where consecutive disks intersect and such that there is a branch tj of logf on D j ' Furthermore, the functions tj were chosen so that t/z) = tj-1(z) for z in D j - 1 n D j , 2 ::; j ::; n. The process analytically continues t1 to D 1 U D 2, then DIU D z U D 3 , and so on. However, an unfortunate thing (for this continuation) happened when the last disk Dn was reached. According to the Argument Principle it is distinctly possible that tn(z) i= t 1(z) for z in Dn n D j • In fact, t"(z)-t1(z) = 27TiK for some (possibly zero) integer K. This last example is a particularly fruitful one. This process of continuing a function along a path will be examined and a criterion will be derived which ensures that continuation around a closed curve results in the same function that begins the continuation. Also, the fact that continuation around a closed path can lead to a different function than the one started with, will introduce us to the concelJt of a Riemann Surface. This chapter begins with the Schwarz Reflection Principle which is more like the process used to continue the zeta function than the process of continuing along an arc. §1. Schwarz Reflection Principle If G is a region and G* = {z: Z E G} and if fis an analytic function on G, then f*: G* --+ IC defined by f*(z) = fez) is also analytic. Now suppose that 210

Schwarz Reflection Principle

211

G = G*; that is, G is symmetric with respect to the real axis. Then g(z) = fez) - fez) is analytic on G. Since G is connected it must be that G contains an open interval of the real line. Suppose f(x) is real for all x in G () IR; then g(x) == 0 for x in G () IR. But G () IR has a limit point in G so thatf(z) = fez) for all z in G. The fact that f must satisfy this equation is used to extend a function defined on G () {z: 1m z ~ O} to all of G. If G is a symmetric region (i.e., G = G*) then let G + = {z E G: 1m z > O}, G_ = {z E G: 1m z < O}, and Go = {z E G: 1m z = O}. 1.1 Schwarz Reflection Principle. Let G be a region such that G = G*. If f: G + u Go -+ C is a continuous function which is analytic on G + and if f(x) is real for x in Go, then there is an analytic function g: G -+ C such that g(z) = fez) for z in G + U Go. Proof For z in G _ define g(z) = fez) and for z in G + u Go let g(z) = fez). It is easy to see that g: G -+ C is continuous; it must be shown that g is analytic. It is trivial that g is analytic on G + U G _ so fix a point x 0 in Go and let R > 0 with B(xo; R) c G. It suffices to show that g is analytic on B(xo; R); to do this apply Morera's Theorem. Let T = [a, b, c, a] be a triangle in B(xo; R). To show that Jrg = 0 it is sufficient to show that Jpg = 0 a

-------?'------------~----G()

b~--------~G~--------whenever P is a triangle or a quadrilateral lying entirely in G + U Go or G _ u Go. In fact, this is easily seen by considering various pictures such as the one above. Therefore assume that T c G + u Go and [a, b) c Go. The proof of the other cases is similar and will be left to the reader. (See Exercise 1 for a general proposition which proves all these cases at once.) Let Ll designate T together with its inside; then g(z) = fez) for all z in Ll. By hypothesisfis continuous on G+ u Go and sofis uniformly continuous on Ll. So if E > 0 there is a 8 > 0 such that when z and z' Ell and \z-z'\ < 8 then If(z)-f(z')\ < E. Now choose 0: and (:3 on the line segments [c, a] and [b, c) respectively, so that \o:-a\ < 8 and 1(:3-bl < 8. Let T J = [0:, (:3, c, 0:] and Q = [a, b, (:3, 0:, a]. Then hf = JTJ+ JQf, but T J and its inside are contained in G + and f is analytic there; hence 1.2

Jf= Jf

T

Q

Analytic Continuation and Riemann Surfaces

212

0.

r------

------>.. B

- - - - a - L . . . . - - - - - - - - - - - - - ' - h- - 0 "

But if 0 :,;

t :,;

1 then i[t;3+(I -t)01]-[tb+(1-t)a]1 < S

so that !/(t/l+(l-t)01)-I(tb+(l-t)a)! <

If M

=

max {i/(z)l: z Ell} and

t=

E.

the perimeter of T then

J1/ = ICb - a) II(tb + (1 ... t)a)dt _. (;3····01) I j{t;3 + (1- t)01)dtl I

1.3

I

I 1+

[a,b]

1

0

[~,C(J

0

Ij, [f(tb+(l-t)a)-/(I(3+(l-t)01)dl I 1

:,; Ib-a!

l

o

1

+ I(b -

a) -

«(3 - a)IIJ(I(3 + (1- t)a)dtl o

:,; €lb-a!+M!Cb-.B)+(a-a)! :,; d+2M8 Also

I JII :,; Mia-a! :,; MS [Ct,

aJ

and

I IIlJ II:,; MS. [b,

Combining these last two inequalities with (1.2) and (l.3) gives that

IIII :,; d+4MS. T

Since it is possible to choose 3 < € and since E is arbitrary, it follows that h/= 0; thuslmust be analytic . • Can the reflection principle be generalized? For example, instead of requiring that G be a region which is symmetric with respect to the real axis, suppose that G is symmetric with respect to a circle. (Definition III. 3.17).

Analytic Continuation Along a Path

213

Can the reflection principle be formulated and proved in this setting? The answer is provided in the exercises below. Exercises 1. Let y be a simple closed rectifiable curve with the property that there is a point a such that for all z on y the line segment [a, z] intersects {y} only at z; i.e. [a, z] n {y} = {z}. Define a point w to be inside y if [a, w] n {y} = 0 and let G be the collection of all points that are inside y. (a) Show that G is a region and G- = G u {y}. (b) Letf: G- --J>- IC be a continuous function such thatfis analytic on G. Show that J = O. (c) Show that n(y; z) = ± 1 if z is inside y and n(y; z) = 0 if z - IC is analytic, define 1*: G* --J>- IC by f*(z) = fO/2). Show that f* is analytic. (c) Suppose that G = G* and f is an analytic function defined on G such that fez) is real for z in G with Izl = 1. Show that f = f*. (d) Formulate and prove a version of the Schwarz Reflection Principle where the circle 1fl = 1 replaces ~. Do the same thing for an arbitrary circle. 3. Let G, G +, G _, Go be as in the statement of the Schwarz Reflection Principle and letf: G+ u Go --J>-lC ao be a continuous function such thatfis rneromorphic on G +. Also suppose that for x ill Go f(x) E ~. Show that there is a merom orphic function g : G -+ IC 00 such that g(z) = f(z) for z in G + u Go. Is it possible to allow fto assume the value 00 on Go? §2. Analytic Continuation Along a Path Let us begin this section by recalling the definition of a function. We use the somewhat imprecise statement that a function is a triple (I, G, D) where G and n are sets andfis a "rule" which assigns to each element of G a unique element of D. Thus, for two functions to be the same not only must the rule be the same but the domains and the ranges must coincide. If we enlarge the range Q to a set D1 then (I, G, .nJ is a different function. However, this point should not be emphasized here; we do wish to emphasize that a change in the domain results in a new function. Indeed, the purpose of analytic continuation is to enlarge the domain. Thus, let G = {z: Re z > -1} and

Analytic Continuation and Riemann Surfaces

214

fez) = log (l +z) for z in G, where log is the principal branch of the logarithm. Let D = B(O; 1) and let g(z)

=

.i (-__

1).-1

~

n for z in D. Then (J, G, C) =F (g, D, q even thoughf(z) = g(z) for all z in D. Nevertheless, it is desirable to recognize the relationship between f and g. This leads, therefore, to the concept of a germ of analytic functions. "=1

2.1 Definition. A function element is a pair U; G) where G is a region and fis an analytic function on G. For a given function element (J, G) define the germ of f at a to be the collection of all function elements (g, D) such that a E D and fez) = g(z) for all z in a neighborhood of a. Denote the germ by [fla. Notice that [fla is a collection of function elements and it is not a function element itself. Also (g, D) E [fla if and only if (J, G) E [gJa. (Verify!). It should also be emphasized that it makes no sense to talk of the equality of two germs [flo and [gh unless the points a and b are the same. For example, if (J, G) is a function element then it makes no sense to say that [fla = [f]b for two distinct points a and b in G. 2.2 Definition. Let y: [0, lJ --+ C be a path and suppose that for each t in [0, 1] there is a function element U;, D t ) such that: (a) yet) EDt; (b) for each t in [0, 1] there is a 0 > 0 such that yes) E Dr and

Is - tl < 0

implies

2.3 Then (fl' Dd is the analytic continuation of (fo. Do) along the path y; or, (fl' D j ) is obtained from (fa, Do) by analytic continuation along y. Observe that the sets Dr in the definition can be enlarged or diminished without affecting the fact that there is a continuation. This will be useful in the proofs below. Before proceeding, examine part (b) of this definition. Since y is a con tinuous function and y( t) is in the open set DI' it follows that there is a 8 > 0 such that y(s) c D f for Is - [I < 8. The important content of part (b) is that (2.3) is satisfied whenever Is - tl < (i. That is,h(z) = j,(z), whenever Is - II < (i and z belongs to the component of Ds n Dr that contains yes). Whether for a given curve and a given function element there is an analytic continuation along the curve can be a difficult question. Since no degree of generality can be achieved which justifies the effort, no existence theorems for analytic continuations will be proved. Each individual case will be considered by itself. Instead uniqueness theorems for continuations are proved. One such theorem is the Monodromy Theorem of the next section. This theorem gives a criterion by which one can tell when a continuation along two different curves connecting the same points results in the same function element.

Analytic Continuation Along a Path

215

The next proposition fixes a curve and shows that two different continuations along this curve of the same function element result in the same function element. Alternately, this result can be considered as an affirmative answer to the following question: Is it possible to define the concept of "the continuation of a germ along a curve?" 2.4 Proposition. Let y: [0, 1]-+ IC be a path from a to b and let {eft, D ,): o ~ t ~ I} and {(gr' B,): ~ t ~ I} be analytic continuations along y such that (fO]a = (g O]a' Then [fdb = [g th·

°

Proof This proposition will be proved by showing that the set T = {t E [0,1]: (f,L(,) = [g']Y(I)} is both open and closed in [0, 1]; since T is non-empty (0 E T) it will foHow that T = [0, 1] so that, in particular, 1 E T.

The easiest part of the proof is to show that T is open. So fix t in T and assume t i= 0 or 1. (If t = 1 the proof is complete; if t = then the argument about to be given will also show that [a,a + 0) c T for some I) > 0.) By the definition of analytic continuation there is a I) > such that for Is-tl < 0, yes) E Dr n B, and

°

°

2.5

{ffs]y(s) :

[1.]y(S)· [gs]y(S) - [g,]y(S)' Let II be a connected subset of Dr n B, which contains yes) and yet). But since t E T,J,(z) = g,(z) for all z in II. Hence [J,]r(s) = [g,]y(s) for aU yes) in D, n B,. So it follows from (2.5) that [fsL(s) = [gsL(s) whenever Is - tl < b. That is, (t - b, t + b) c Tand so Tis open. To show that T is closed let t be a limit point of T, and again choose I) > 0 so that yes) E D/ n B, and (2.5) is satisfied whenever Is- tl < o. Since t is a limit point of T there is a point s in T with Is - tl < b. Let G be a region such that y«t - b, t - b)) £; G £; D/ n B,; in particular, yes) E G. Thus, !s(z) = gs(z) for all z in G by the definition of T. But, according to (2.5), j~(z) = /,(z) and gs(z) = g,(z) for all z in G. So /,(z) = g,(z) for all z in G and, because G has a limit point in D, n B" this gives that [I.]y(l) = [gIL(t). That is, t E T and so T is closed. II 2.6 Definition. If y: [0, 1) --+ IC is a path from a to band {(J" D,): 0 ~ t ~ I} is an analytic continuation along y then the germ [fdb is the analytic continuation of [fola along y.

The preceding proposition implies that Definition 2.6 is unambiguous. As stated this definition seems to depend on the choice of the continuation {(/" Dr)}' However, Proposition 2.4 says that if {(g" B,)} is another continuation along y with [foJa = [golD then (flh = [glh. So in fact the definition does not depend on the choice of continuation. 2.7 Definition. If (f, G) is a function element then the complete analytic function obtained from (f, G) is the collection ff of all germs [gJo for which there is a point a in G and a path y from a to b such that (gh is the analytic continuation of [f]a along y.

216

Analytic Continuation and Riemann Surfaces

A collection of germs :7 is called a complete analytic function if there is a function element (I, G) such that ff is the complete analytic function obtained from (I, G). Notice that the point a in the definition is immaterial; any point in G can be chosen since G is an open connected subset of IC (see II. 2.3). Also, if :7 is the complete analytic function associated with (I, G) then [flz E ff for all z in G. Although there is no ambiguity in the definition of a complete analytic function there is an incompleteness about it. Is it a function? We should refrain from calling an object a function unless it is indeed a function. To make ff a function one must manufacture a domain (the range will be q and show that:7 gives a "rule". This is easy. In a sense we let .fF be its own domain; more precisely, let

Sf

=

{(z, [flz): [f]z

E

:7}.

Define .fF: Sf -J> IC by ff(z, [fJz) = fez). In this way ff becomes an "honest" function. Nevertheless there is still a lingering dissatisfaction. To have a satisfying solution a structure will be imposed on j f which will make it possible [0 discuss the concept of analyticity for functions defined on 212. In this setting, the function ,'IF defined above becomes analytic; moreover, it reflects the behavior of each function element belonging to a germ that is in ff. The introduction of this structure is postponed until Section 5. Exercises 1. The collection {Do, D 1 , ••• ,Dn} of open disks is called a chain of disks if D j - 1 (\ D j ¥ for I ::; j ::; n. If {(ij' D j ): 0 ::; j ::; n} is a collection of function elements such that {Do, D l , ... , Dn} is a chain of disks andfj ___ I(z) = fiz) for z in D j - I n D j ' I ::; j ::; n; then {(fj, D): 0 ::; j ::; n} is called an analytic continuation along a chain of disks. We say that (I., Dn) is obtained by an analytic continuation of (fa, Do) along a chain of disks. (a) Let {(fj, D): 0 ::; j ::; n} be an analytic continuation along a chain of disks and let a and b be the centers of the disks Do and Dn respectively. Show that there is a path y from a to b and an analytic continuation {(go B t )}

=:

n

along y such that {y}

cUD j'

[fola = [g o]a> and

Lfnh =

[g Ih·

j~O

(b) Conversely, let {(/" D t ): 0::; t::; I} be an analytic continuation along a path y: [0, 11 -J> IC and let a = yeO), b = y(l). Show that there is an analytic continuation along a chain of disks {(g j ' B j): 0 ::; j ::; n} such that n

{y}

C

U Bj ,

[JoL

=

[gaL, and

U~h

=

[gnh·

j~O

2. Let Do

B(l; 1) and let fo be the restriction of the principal branch of Let yet) = exp (2-rrit) and aCt) = exp (41Tit) for 0 ::; t ::; 1. (a) Find an analytic continuation {(/" Dr): 0 ::; t ::; I} of (j~, Do) along y and show that [Jdl = [-fol1.

.Jz to Do.

=

Monodromy Theorem

217

(b) Find an analytic continuation {(g" B,): 0 :::; t :::; I} of (fo, Do) along u and show that [gl]l

= [gO]l'

3. Letfbe an entire function, Do = B(O; 1), and let y be a path from 0 to h. Show that if {(/" D,): 0 :::; t :::; I} is a continuation of (f, Do) along y then fl{Z) = fez) for all z in D l . (This exercise is rather easy; it is actually an exercise in the use of the terminology.) 4. Let y: [0, 1] ~ C be a path and let {(/" D,): 0 :::; t :::; I} be an analytic continuation along y. Show that {(f:, D,): 0 :::; t :::; I} is also a continuation along y. 5. Suppose y: [0, 1] ~ C is a closed path with yeO) = y(I) = a and let {(/" D,): 0 :::; t :::; I} be an analytic continuation along y such that [fl]a = [f~]a andfo ~ O. What can be said about (fo, Do)? 6. Let Do = B(I; 1) and let fo be the restriction to Do of the principal branch of the logarithm. For an integer n let yet) = exp (27Tint), 0 :::; t :::; 1. Find a continuation {(f" D,): 0 :::; t :::; I} along y of (fo, Do) and show that [fdl = [fo+27Tin]1' 7. Let y: [0, 1] ~ C be a path and let {(/" D,): 0 :::; t :::; I} be an analytic continuation along y. Suppose G is a region such that /,(D,) c G for all t, and suppose there is an analytic function h: G ~ C such that h{fo(z» = z for all z in Do. Show that h(f,(z» = z for all z in D, and for all t. Hint: Show that T = {t: h(/,(z» = z for all z in D,} is both open and closed in [0, 1]. 8. Let y: [0, 1] ~ C be a path with yeO) = 1 and yet) ~ 0 for any t. Suppose that {{/" D,): 0 :::; t :::; I} is an analytic continuation of fo(z) = log z. Show that each/, is a branch of the logarithm. §3. Monodromy Theorem Let a and b be two complex numbers and suppose y and a are two paths from a to b. Suppose {{It,D,)} and {{gI'B,)} are analytic continuations along y and a respectively, and also suppose that [fO]a = [gola' Does it follow that [fdb = [g,h? If y and a are the same path then Proposition 2.4 gives an affirmative answer. However, if y and a are distinct then the answer can be no. In fact, Exercises 2.2 and 2.6 furnish examples that illustrate the possibility that [fdb [g .lb' Since both of these examples involve curves that wind around the origin, the reader might believe that a sufficient condition for [fdb and [gdb to be equal can be couched in the language of homotopy. However, since all curves in the plane are homotopic the result would have to be phrased in terms of homotopy in a proper subregion of C. For the examples in Exercises 2.2 and 2.6, this sought after criterion must involve the homotopy of the curves in the punctured plane. This is indeed the case. The origin is discarded in the above examples because there is no germ [h]o centered at zero that belongs to the complete analytic function obtained from (fo, Do). If (f, D) is a function element and a E D then f has a power series expan-

*

218

Analytic Continuation and Riemann Surfaces

sion at z = a. The first step in proving the Monodromy Theorem is to investigate the behavior of the radius of convergence for an analytic continuation along a curve.

3.1 Lemma. Let y: [0, 1] --.. IC be a path and let {(I" D/): 0 :s; t :s; I} be an analytic continuation along y. For 0 :s; t :s; 1 let R(t) be the radius of convergence of the power series expansion of1, about z = yet). Then either R(t) == 00 or R: [0, 1] --.. (0, (0) is continuous. Proof If R(t) = 00 for some value of t then it is possible to extendJ, to an entire function. It follows that fs(z) = J,(z) for all z in D. so that R(s) = 00 for each s in [0, 11; that is R(s) == 00. So suppose that R(t) < 00 for all t. Fix t in [0, 11 and let T = y(t); let

J,(z) =

co

L TnCZ-T)n n= 0

be the power series expansion of 1, about T. Now let Sl > 0 be such that < 8 1 implies that yes) E D t n B(T; R(t» and [fsly(s) = [J,ly(s)' Fix s with Is-tl < 8 1 and let a = yes). Now 1, can be extended to an analytic function on B(T; R(t». Sincefs agrees withJ, on a neighborhood of a,!. can be extended so that it is also analytic on B(T; R(t» U Ds. If I. has power series expansion

Is-tl

fs(z) =

00

L

n= 0

aiz-a)"

about z = a, then the radius of convergence R(s) must be at least as big as the distance from a to the circle Iz - TI = R(t): that is, R(s) ~ dCa, {z: Iz - 71 = Ret)}) = R(t) - IT - al· But this gives that R(t) - R(s) :s; Iy(t) - y(s)l. A similar argument gives that R(s) - R(t) :s; Iy(t) - y(s)l; hence IR(s)-R(t)1

$

Jy(t)-y(s)J

for Js- tj < 8 1 , Since y: [0, 1] --.. IC is continuous it follows that R must be continuous at t . • 3.2 Lemma. Let y: [0, 1]->- C be a path from a to b and let {(I" :s; I} be an analytic continuation along y. There is a number € > if a: [0, 1] --.. IC is any path from a to b with Iy(t) - a(t)1 < € for {(gt, B t): 0 $ t $ I} is any continuation along a with [g ola = [glh = [fdb'

Dr): 0 :s; t 0 such that all t, and if [fola; then

Proof For 0 $ t :s; 1 let R(t) be the radius of convergence of the power series expansion of ft about z = yet). It is left to the reader to show that if R(t) == 00 then any value of to will suffice. So suppose R(t) < co for all t. Since, by the preceding lemma, R is a continuous function and since R(t) > 0 for all t, R has a positive minimum value. Let

3.3

o<

€

<

t

min {R(t): 0 :s; t :s; I}

and suppose that a and {(gr' B,)} are as in the statement of this lemma.

Monodromy Theorem

219

Furthermore, suppose that Dr is a disk of radius R(t) about y(t). The truth of the conclusion will not be affected by this assumption (Why?), and the exposition will be greatly simplified by it. For the same reason it will also be assumed that each Br is a disk. Since la(t)-y(t)1 < E < R(t), a(t) E B t n D t for all t. Thus, it makes sense to ask whether gtCz) = J,(z) for all z in B t n D t. Indeed, to complete the proof we must show that this is precisely the case for t = 1. Define the set T = {t E [0, 1]: J,(z) = gt(z) for z in B t n D t }; and show that 1 E T. This is done by showing that T is a non-empty open and closed subset of [0, 1].

From the hypothesis of the lemma, 0 E T so that T oF O. To show Tis open fix I in T and choose S > 0 such that

3.4

Ins) - y(t)1 < { la(s) - a(t)1 <

E,

[J.]y(S) = [J,]y(s),

E,

[g.]u(S) = [gt]u(S)' and

a(s) E B t

whenever Is-tl < S. We will now show that Bs n B t n Ds n D t oF D for Is-tl < S; in fact, we will show that a(s) is in this intersection. If Is-tl < S then la(s)-y(s)1 < E < R(s) so that a(s) E D •. Also la(s) - y(t)1 :::;; la(s) - y(s) I+ Iy(s) - y(t) I < 2E < R(t)

by (3.3); so a(s) EDt. Since we already have that a(s) E Bs n B t by (3.4), a(s) E B. n B t n Ds n D t = G. Since t E T it follows thatJ,(z) = gt(z) for all z in G. Also, from (3.4)J.(z) = J,(z) and gs(z) = gt(z) for all z in G. Thus J.(z) = g.(z) for z in G; but since G has a limit point in Bs n Ds it must be that sET. That is, (t-S, t+S) c T and T is open. The proof that T is closed is similar and will be left to the reader. • 3.S Definition. Let (J, D) be a function element and let G be a region which contains D; then (f, D) admits unrestricted analytic continuation in G if for any path y in G with initial point in D there is an analytic continuation of (f, D) along y., If D = {z: Iz -11 < I} and f is the principal branch of .jz or log z then (f, D) admits unrestricted continuation in the punctured plane but not in the whole plane (see Exercise 2.7). It has been stated before that an existence theorem for analytic continuations will not be proved. In particular, if (f, D) is a function element and G is a region containing D, no criterion will be given which implies that (f, D) admits unrestricted continuation in G. The Monodromy Theorem assumes that G has this property and states a uniqueness criterion. 3.6 Monodromy Theorem. Let (f, D) be a function element and let G be a region containing D such that (f, D) admits unrestricted continuation in G.

220

Analytic Continuation and Riemann Surfaces

Let a E D, bEG and let Yo and Y1 be paths in G from a to b; let {(fl' D, ): t :$ I} and {( g" B, ): 0 :$ t :$ I} be analytic continuations of (f, D) along Yo and Y1 respectively. If Yo and Y1 are FEP homotopic in G then

o :$

Proof Since Yo and YI are fixed-end-point homotopic in G there is a continuous function r: [0, 1J x [0, I] -+ G such that

r(t, 1) = YI(l)

ret,O) = YoU)

1'(1, u) = b

reO, u) = a

for all t and u in [0, 1]. Fix u, 0 ::; u ::; I and consider the path Yu, defined by yit) = r(t, u), from a to b. By hypothesis there is an analytic continuation

{(hr,u, D, ..): 0 ::; t ::; I} of (f, D) along YU' It follows from Proposition 2.4 that [g Jb = [h 1, Ih and lfdb = [hI, 01&· So it suffices to show that

[hl,olb

=

[hj,lh·

To do this introduce the set

u=

°

{u

E

[0,1]: [h1,uh

=

[hi, olb}'

and show that U is a non-empty open and closed subset of [0, IJ. Since E U, U O. To show that U is both open and closed we will establish the following.

*

°

3.7 Claim. For u in [0, 1J there is as> such that if [u -1'[ < S then [11 1. uh = [hl,vh. For a fixed u in [0, IJ apply Lemma 3.2 to find an E > 0 such that if a is any path from a to b with [Yu(t) -aCt)[ < € for all t, and if {(k " E,)} is any continuation of (f, D) along a, then

3.8 Now r is a uniformly continuous function, so there is as> 0 such that if [u - v[ < S then [YuC!} - yJt)[ = [r(t, u) - {'(t, v)[ < E for all t. Claim 3.7 now follows by applying (3.8). Suppose u E U and let 8 > 0 be the number given by Claim 3.7. By the definition of U, (u - 8, u + 8) c U; so U is open. If Ii E U- and 5 is again chosen as in (3.7) then there is a u in U such that lu - vi < 5. But by (3.7) [hl.uJb = [hl.vh; and since u E U, [hl.vh = [hl,ok Therefore [h1,ulb = [hl.OJb so that u E U, that is, U is closed. II The following corollary is the most important consequence of the Monodromy Theorem.

3.9 Corollary. Let (f, D) be a function element which admits unrestricted continuation in the simply connected region G. Then there is an analytic function F: G -+ IC such that F(z) = fez) for all z in D. Proof Fix a in D and let z be any point in G. If y is a path in G from a to z

Topological Spaces and Neighborhood Systems

221

and HI" Dr): 0 ~ t ~ I} is an analytic continuation of (f, D) along y then let F(z, y) = II (z). Since G is simply connected F(z, y) = F(:, 0) for any two paths y and a in G from a to z. Thus, Fez) = F(z, y) gives a well defined function F: G --+ C. To show that F is analytic let z F G and let y and {(f;, Dr)} be as above. A simple argument gives that F(w) = Ii (II') for w in a neighborhood of z (Verify!); so F must be analytic. II Exercises

1. Prove that the set T defined in the proof of Lemma 3.2 is closed. 2. Let (f, D) be a function element and let a E D. If y: [0, 1] --',> IC is a path with yeO) = a and y( 1) = b and {(I" Dr): 0 ~ t ~ I} is an analytic continuation of (f; D) along y, let R(t) be the radius of convergence of the power series expansion of fr at z = yet). (a) Show that R(t) is independent of the choice of continuation. That is, if a second continuation {(g" B t )} along y is given with [g oL = [lL and ret) is the radius of convergence of the power series expansion of g, about z = yet) then ret) = R(t) for all t. (b) Suppose that D = B(l; 1), I is the restriction of the principal branch of the logarithm to D, and y(t) = 1 +a t for 0 ~ t ~ 1 and a > O. Find R(t).

(c) Let (f, D) be as in part (b), let 0 < a < 1 and let yet) = (1·-- a I) exp (27Tit) for 0 ~ t ~ 1. Find R(t). (d) For each of the functions R(t) obtained in parts (b) and (c), find min {R(t): ~ t ~ I} as a function of a and examine the behavior of this function as a --+ CIJ or a --+ 0. 3. Let 1': [0, 1] x [0, 1] ....>- G be a continuous function such that 1'(0, 1/) = a, 1'(1, u) = b for all u. Let yu(t) = ret, u) and suppose that {Ut,u, D t ..,): ~ t ~ I} is an analytic continuation along y" such that [!O,,,]a = [Io,,,1o for all u and v in [0, 1]. Let R(t, u) be the radius of convergence of the power series expansion of.fr." about z = u). Show that either R(t, u) == CIJ or R: [0, 1] x [0, 1]+ (0, co) is a continuous function. 4. Use Exercise 3 to give a second proof of the Monodromy Theorem.

°

°

ret,

§4. Topological Spaces and Neighborhood Systems The notion of a topological space arises by abstracting one of the most important concepts in the theory of metric spaces-that of an open set. Recall that in Chapter II we were given a metric or distance function on a set X and this metric was used to define what is meant by an open set. In a topological space we are given a collection of subsets of a set X which are called open sets, but there is no metric available. After axiomatizing the properties of open sets, it will be our purpose to recreate as much of the theory of metric spaces as is possible.

4.1 Definition. A topological space is a pair (X, 3') where X is a set and .'1 is a collection of subsets of X having the following properties:

222

Analytic Continuation and Riemann Surfaces

(a) U E:Yand XE:Y; (b) if U j ,

• •• ,

Un are in:Y then j

(c) if {U i : i

E

n U E:Y;

n

j

~J

J} is any collection of sets in /Y then

U U i is in :Y.

iEl

The collection of sets :Y is called a topology on X, and each member of :Y is called an open set. Notice that properties (a), (b), and (c) of this defimtion are the properties of open subsets of a metric space that were proved in Proposition II. 1.9. So if (X. d) is a metric space and c'T is the collection of all open subsets of X then (X"Y) is a topological space. When it is said that a topological space is an abstraction of a metric space, the reader should not get the impression that he is merely playing a game by discarding the metric. That is, no one should believe that there is a distance function in the background, but the reader is now required to prove theorems without resorting to it. This is quite false. There are topological spaces (X, :T) such that for no metric d on X is ,Y the collection of open sets obtained via d. We will see such an example shortly, but it is first necessary to further explore this concept of a topology. The statement "Let X be a topological space" is, of course, meaningless; a topological space consists of a topology :Y as well as a set X. However, this phase will be used when there is no possibility of confusion. 4.2 Definition. A subset F of a topological space X is closed if X - F is open. A point a in X is a limit point of a set A if for every open set U that contains a there is a point x in An U such that x cf- a. Many of the proofs of propositions in this section follow along the same lines as corresponding propositions in Chapter n. When this is the case the proof will be left to the reader. Such is the case with the following two propositions. 4.3 Proposition. Let eX, .r) be a topological space. Then: (a) [J and X are closed sets; (b) if F J , ••• , Fn are closed sets then FI U ... U Fn is closed; Fi is a closed set. (c) ({ {Fi: i E J} is a collection of closed sets then

n

f

E

I

4.4 Proposition. A subset of a topological space is closed iff it contains all its limit points. Now for an example of a topological space that is not a metric space. Let X = [0, I] = { [ : O:s; t:S; I} and let:Y consist of all sets U such that: (i) if 0 E U then X - U is either empty or a sequence of points in X; (ii) if 0 ¢ U then U can be any set. It is left to the reader to prove that eX, :Y) is a topological space. Some of the examples of open sets in this topology are: the set of all irrational numbers in X; the set of all irrational numbers together with zero. To see

Topological Spaces and Neighborhood Systems

223

that no metric can give the collection of open sets :Y, suppose that there is such a metric and obtain a contradiction. Suppose that d is a metric on X such that U E :Y iff for each .x in U there is an € > 0 such that B(x; €) = {y:d(x,y) < €} C U. Now let A = (0, 1);ifUE:YandOE Uthenthereisa point a in UnA, a i= 0 (in fact there is an infinity of such points). Hence, o is a limit point of A. It follows that there is a sequence {t n } in A such that d(tn' 0)-)-0. But if U= {XEX: x i= tn for any n} = X-{tl' fz, .. . }then o E U and U is open. So it must follow that tn E U for n sufficiently large; this is an obvious contradiction. Hence, no metric can be found. This example illustrates a technique that, although available for metric spaces, is of little use for general topological spaces: the convergence of sequences. It is possible to define "convergent sequence" in a topological space (Do it !), but this concept is not as intimately connected with the structure of a topological space as it is with a metric space. For example, it was shown above that a point can be a limit point of a set A but there is no sequence in A that converges to it. If a topological space (X, :Y) is such that a metric d on X can be found with the property that a set is in :Y iff it is open in (X, d), then (X, :Y) is said to be metrisable. There are many non-metrisable spaces. In addition to inventing non-metrisable topologies as was done above, it is possible to define processes for obtaining new topological spaces from old ones which will put metrisable spaces together to obtain non-metrisable ones. For example, the arbitrary cartesian product of topological spaces can be defined; in this case the product space is not metrisable unless there are only a countable number of coordinates and each coordinate space is itself metrisable. (See VII. 1.18.) Another example may be obtained as follows. Consider the unit interval 1= [0, 1]. Stick one copy of I onto another and we have a topological space which still "looks like" 1. For example, [1, 2] is a copy of I and if we stick it onto I we obtain [0, 2]. In fact, if we "stick" a finite number of closed intervals together another closed interval is obtained. What happens if a countable number of closed intervals are stuck together? The answer is that we obtain the infinite interval [0, (0). (If the intervals are stuck together on both sides then u-;g is obtained.) What happens if we put together an uncountable number of copies of J? The resulting space is called the long line. Locally (i.e., near each point) it looks like the real line. However, the long line is not metrisable. As a general rule of thumb, it may be said that if a process is used to obtain new spaces from old ones, a non-metrisable space will result if the process is used an uncountable number of times. For another example of a space that is non-metrisable let X be a set consisting of three points-say X = {a, b, c}. Let /:1 = {D, X, {a}, {b}, {a, b} }; it is easy to check that ff is a topology for X. To see that (X, /T) is not metrisable notice that the only open set containing c is the set X itself. There do not exist disjoint open sets U and V such that a E U and c E V. On the other hand if there was a metric d on X such that ..'T is the collection of open sets relative to this metric then it would be possible to find such

224

Analytic Continuation and Riemann Surfaces

open sets (e.g., let U = B(a; t-) and V = B(c; €) where € < dCa, c». In other words, (X, 51 fails to be metrisable because .r does not have enough open sets to separate points. (Does the first example of a non-metrisable space also fail because of this deficiency?) The next definition hypothesizes enough open sets to eliminate this difficulty. 4.5 Definition. A topological space (X, .r) is said to be a Hausdorff space if for any two distinct points a and b in X there are disjoint open sets U and V such that a E U and b E V. Every metric space is a Hausdorff space. As we have already seen there are examples of topological spaces which are not Hausdorff spaces. Many authors include in the definition of a topological space the property of a Hausdorff space. This policy is easily defended since most of the examples of topological spaces which one encounters are, indeed, Hausdorff spaces. However there are also some fairly good arguments against this combining of concepts. The first argument is that mathematical pedagogy dictates that ideas should be separated so that they may be more fully understood. The second, and perhaps more substantial reason for not assuming all spaces to be Hausdorff, is that more examples of nDn-Hausdorff spaces are arising in a natural context. Even though there will be no non-Hausdorff space which will appear in this book, this separation of the two concepts will be maintained for a while longer. The next step in this development is the definition, in the setting of topological spaces, of certain concepts encountered in the theory of metric spaces and the stating of analogous propositions.

.r) is connected if the only nonempty subset of X which is both open and closed is the set X itself.

4.6 Definition. A topological space (X,

4.7 Proposition. Let (X, .r) be a topological space; then X = U {Ci : i E I} where each C i is a component of X (a maximal connected subset of X). Furthermore, distinct components of X are disjoint and each component is closed. 4.8 Definition. Let eX, .r) and (n, 9') be topological spaces. A function f: X -+ is continuous iff-l(~) E . r whenever ~ E 9'.

°

4.9 Proposition. Let (X, .r) and (0,9') be topological spaces and letf: X -+ be a function. Then the following are equivalent: (a) f is continuous; (b) if A is a closed subset ofn thenf-I(A) is a closed subset of X; (c) if a E X and if f(a) E ~ E 9' then there is a set U in . r such that a andf(U) c ~.

E

° U

4.10 Proposition. Let (X, 51 and (0, 9') be topological spaces and suppose that X is connected. Iff: X -+ n is a continuous function such that f(X) = 0, then n is connected. 4.11 Definition. A set K c X is compact if for every sub-collection (!) of

.r

Topological Spaces and Neighborhood Systems

U {U: U E (9} there are a n (9 such that K c U Uk'

such that K in

C

225

finite number of sets U 1,

••. ,

Un

k~1

4.12 Proposition. Let (X, :T) and (n, ,9") be topological spaces and suppose K is a compact subset of X. Iff: X --i>- n is a continuous function then f(K) is compact in n. If (X, d) is a metric space and Y c X then (Y, d) is also a metric space. Is there a way of making a subset of a topological space into a topological space? We could, of course, declare every subset of Y to be open an~ this would make Y into a topological space. But what is desired is a topology on Ywhich has some relationship to the topology on X; a natural topology on a subset of a topological space. If (X, :T) is a topological space and Y c X then define :T y

It is easy to check that

::r y

=

{Un Y: UE:T}.

is a topology on Y.

,r

4.13 Definition. If Y is a subset of a topological space (X, :T) then y is called the relative topology on Y. A subset W of Y is relatively open in Y if WE:T y; W is relatively closed in Y if Y - WE :T y. Whenever we speak of a subset of a topological space as a topological space it will be assumed that it has the relative topology unless the contrary is explicitly stated. 4.14 Proposition. Let (X, :T) be a topological space and let Y be a subset of X. (a) If X is compact and Y is a closed subset of X then (Y, :T y) is compact. (b) Y is a compact subset of X iff (Y, :T y) is a compact topological space. (c) If(X, :T) is a Hausdorff "'pace then (Y, :T y) is a Hausdorff space. (d) rr(X, :T) is a Hausdorff space and (Y, :T y) is compact then Y is a closed subset ofX. Proof The proofs of (a), (b), and (c) are left as exercises. To prove part (d) it suffices to show that for each point a in X - Y there is an open set U such that a E U and Un Y = O. So fix a point a in X - Y; for each point y in Y there are open sets Uyand Vy in X such that a E UY' Y E V y, and U y n Vy = o (because (X, ,r) is a Hausdorff space). Then {Vy n Y: y E Y} is a collection of sets in :T y which covers Y. So there are points yj, ... , Yn in Y such

that Y c

U(V)"

i~l

n Y) c

UV y,

i~l

=

V. Since a E U y, for each i, a E U = n

Up,; also U E:T. It is easily verified that Un V = U (U n V y ,) that Un Y = O. • ;; 1 The proof of this proposition yields a stronger result.

n

i~1

=

0 so

4.15 Corollary. Let (X, ,.71 be a Hausdorff space and let Y be a compact subset of X; then for each point a in X - Y there are open sets U and V in X such that a E U, Y c V and Un V = O. If we return to the consideration of metric spaces we can discover a new

Analytic Continuation and Riemann Surfaces 226 way to define a topology. The sequence of steps by which open sets are obtained in a metric space are as follows: the metric is given, then open balls are defined, then open sets are defined as those sets which contain a ball about each of their points. What we wish to mimic now is the introduction of balls; this is done by defining a neighborhood system on a set.

4.16 Definition. Let X be a set and suppose that for each point x in X there is a collection JVx of subsets of X having the following properties: (a) for each V in JVx , x E V; (b) if V and V E.AI", there is a Win JVx such that We V n V; (c) if V E JVx and V E JVy then for each z in V n V there is a Win JV;, such that W c V n V. Then the collection {JVx : x E X} is called a neighborhood system on X. If (X, d) is a metric space and if x E X then JVx = {B(x; €): € > O} gives a collection {u-V:: x E X} which is a neighborhood system. In fact, this was the prototype of the above definitions. Notice that condition (c) relates the neighborhood systems of different points. If only conditions (a) and (b) were satisfied, it would not follow that these neighborhoods would be open sets in the topology to be defined. For example, if X is a metric space and JVx is the collection of closed balls about x then {JVx : x E X} satisfies conditions (a) and (b) but not (c). Moreover, it is easy to verify that by letting x = y = z condition (b) can be deduced from (c) (Verify!). The next proposition relates neighborhood systems and topological spaces. 4.17 Proposition. (a) If (X, Y) is a topological space and JVx = {V E Y: x E U} then {.AI,,: x E X} is a neighborhood system on X. (b) If {JVx : x E X} is a neighborhood system on a set X then let Y = {V: x in V implies there is a V in .AI" such that V c V}. Then Y is a topology on X and JVx c Y for each x. (c) If(X, Y) is a topological space and {JVx : x E X} is defined as in part (a), then the topology obtained as in part (b) is again Y. (d) If {JVx : x E X} is a given neighborhood system and Y is the topology defined in part (b), then the neighborhood system obtained from Y contains {JVx : x E X}. That is, if V is one of the neighborhoods of x obtained from Y then there is a V in .AI" such that V c v.

Proof. The proof of parts (a), (c), and (d) are left to the reader. To prove part (b), first observe that both X and 0 are in Y(O E Y since the conditions are vacuously satisfied). Let VI' ... , Vn

E

Y and put V

=

nV n

j •

If x

E

V

j~l

then for each) there is a Vj in.Al" such that Vj

C

V j . It follows by induction

on part (b) of Definition 4.16 that there is a V in.Al" such that V c

n Vj . n

.i~1

Thus V c V and V must belong to Y. Since the union of a collection of

The Sheaf of Germs of Analytic Functions on an Open Set

227

sets in Y is clearly in Y, Y is a topology. Finally, fix x in and let U E JVx • If y E U it follows from part (c) of Definition 4.16 that there is a V in JVy such that V c U. Thus U E Y . • 4.18 Definition. If {,Ai'x: x E X} is a neighborhood system on X and Y is the topology defined in part (b) of Proposition 4.17, then Y is called the topology induced by the neighborhood system. 4.19 Corollary. If {JVx : x E X} is a neighborhood system on X and Y is the induced topology then (X, Y) is a Hausdorff space iff for any two distinct points x and y in X there is a set U in JY" and a set V in JVy such that Un V

= O.

Exercises 1. Prove the propositions which were stated in this section without proof. 2. Let (X, Y) and (0, sP) be topological spaces and let Y c X. Show that if f: X --+ a is a continuous function then the restriction off to Y is a continuous function of (Y, Y y) into (0, sP). 3. Let X and a be sets and let {JVx : x E X} and {.Aw: W EO} be neighborhood systems and let Y and sP be the induced topologies on X and a respecti vel y. (a) Show that a function f: X --+ a is continuous iff when x E X and w = f(x), for each Ll in .A(i) there is a U in JVx such thatf(U) c Ll. (b) Let X = a = C and let ~ = .Az = {B(z; E): E > O} for each z in Co Interpret part (a) of this exercise for this particular situation. 4. Adopt the notation of Exercise 3. Show that a function f: X --+ a is open iff for each x in X and U in JVx there is a set Ll in.Aw (where w = f(x» such that Ll c f( U). 5. Adopt the notation of Exercise 3. Let Y c X and define 'Ply = {Y n U: U EJVy } for each yin Y. Show that {'Ply: y E Y} is a neighborhood system for Yand the topology it induces on Y is Y y. 6. Adopt the notation of Exercise 3. For each point (x, w) in Xx a let 'PI(x,W)

= {UxLl: UEJY",LlE.Aw }

(a) Show that {ott(X,W): (x, w) EXXO} is a neighborhood system on Xx a and let flJ be the induced topology on Xx O. (b) If U E Y and Ll E sP, call the set Ux Ll an open rectangle. Prove that a set is in flJ iff it is the union of open rectangles. (c) Define PI: Xx a --+ X and P2: Xx a --+ a by PI(X, w) = x and P2(X, w) = w. Show that PI and P2 are open continuous maps. Furthermore if (Z, fJ€) is a topological space show that a functionf: (Z, fJ€) --+ (Xx 0, flJ) is continuous iff PI 0 f: Z --+ X and P2 0 f: Z --+ a are continuous.

§5. The Sheaf of Germs of Analytic Functions on an Open Set This section introduces a topological space which plays a vital role in complex analysis. In addition to the topological structure, an analytical

228

Analytic Continuation and Riemann Surfaces

structure is also imposed on this space (in the next section). This will furnish the setting in which to study the complete analytic function as an analytic function. If (f, D) is a function element, recall that the germ of (f, D) at a point z = a in D is the collection of all function elements (g, B) such that a E B and g(z) = fez) for all z in the component of B n D that contains a. This germ is denoted by [fl •. 5.1 Definition. For an open set G in C let .'l'(G) = {(z,[f]z):zEG,fisanalyticatz}. Define a map p: .9"(G) -'>- C by p(z, [f]z) = z. Then the pair (.9"(G), p) is called the sheaf of germs of analytic functions on G. The map p is called the projection map; and for each z in G, p -I(Z) = P -I( {z}) is called the stalk or fiber over z. The set G is called the base space of the sheaf. Notice that for a point (z, [fL) to be in .9"(G), it is not necessary thatfbe defined on all of G; it is only required that f be analytic in a neighborhood of z. How do we picture this sheaf? (There are, of course, too many dimensions to form an accurate geometrical picture.) One way is to follow the agricultural terminology used in the definition. On top of each stalk there is a collection of germs; the stalks are tied together into a sheaf. A better feeling for .9"(G) can be obtained by examining the notation for points in .9"(G). When we consider a point (z, [f]z) in.9"( G), think of a function element (f, D) in the germ [f]. instead of the germ itself. For every point win D there is a point (w, [f]w) in .9"(G). Thus about (z, [f]z) there is a sheet or surface few; [f]w): wED}. In fact, .9"(G) is entirely made up of such sheets and they overlap in various ways. Alternately, we can think of .9"(G) as the union of graphs; each point (z, [fJ.) in .9"(G) corresponding to the point (z, fez»~ on the graph of (f, D). (The graph of (f, D) is a subset of C 2 or ~4.) Two function elements are equivalent at a point z if their graphs coincide near z. A topology will be defined on .9"(G) by defining a neighborhood system. For an open set D contained in G and a function f: D -'>- C analytic on D define

5.2

N(f, D) = {(z, [f]z): zED}.

That is, N(f, D) is defined for each function element (f, D). If we think of .9"(G) as a collection of sheets lying above G which are indexed by the germs, then N(f, D) is the part of that sheet indexed by f and which lies above D. 5.3 Theorem. For each point (a, [f].) in .9"(G) let .A(.,[f)a)

= {N(g, B): a E B and [g]. = [f].}.

then {.A(. ,[JJa): (a, [f].) E .9"( G)} is a neighborhood system on .9"( G) and the induced topology is Hausdorff. Furthermore. the induced topology makes the map p: .9"(G) -'>- G continuous. Proof. Fix (a, [f].) in .9"(G); since it is clear that condition (a) of Definition

The Sheaf of Germs of Analytic Functions on an Open Set

229

4.16 holds, it remains to verify that condition (c) holds. (Condition (b) is a consequence of (c).) Let N(g!, B I ) and N(g2' B 2 ) EJY(a,[fla) and let

5.4 It is necessary to find a function element (k, W) such that N(k, W) EA''(b,[hl b ) and N(k, W) c N(gl, B 1 ) n N(gz, B2)' It follows from (5.4) that bE BI n B2 and [hh = [gdb = [gzh. If bE WeB! n B z and h is defined on W then N(h, W) c N(g!, B I ) n N(g2' Bl)' To show that the induced topology is Hausdorff, use Corollary 4.19. So let (a, [fl.) and (b, [g]b) be distinct points of 9'(G). We must find a neighborhood N(f, A) of (a, [f],,) and a neighborhood N(g, B) of (b, [g]b) such that N(f, A) n N(g, B) = O. How can it happen that (a, [fl.) =f. (b, [gh)? There are two possibilities. Either a =f. b or a = band [fla =f. [g]". If a =f. b then let A and B be disjoint disks about a and b respectively; it follows immediately that N(f, A) n N(g, B) = O. If a = b but [fla =f. [g]a, we must work a little harder (but not much). Since [fl. =f. [g]a there is a disk D = BCa; r) such that bothfand g are defined on D andf(z) =f. g(z} for 0 < Iz-al < r. (It may happen that f(a) = g(a) but this is inconsequential.)

Claim. N(f, D) n N(g, D) = O. In fact, if (z, [hl z) E N(f, D) n N(g, D} then zED, [h]z = [f]., and [hl: = [g]z. It follows thatfand g agree on a neighborhood of z, and this is a contradiction. Hence the induced topology is Hausdorff. Let U be an open subset of G; begin the proof that p: .9"'(G) -7 G is continuous by calculating p -I( U). Since p(z, [flz) = Z, p-l(U) = {(z, [flz): z

E

U}.

So if (z, [f]z) E P-I( U) and D is a disk about z on whichf is defined and such that D c U, N(f, D) c p U). It follows that p must be continuous (Exercise 4.3). I11III Consider what was done when we showed that the induced topology was Hausdorff. If a =f. b then (a, [fl.) and (b, [glb) were on different stalks p -lea) and p -l(b); so these distinct stalks were separated. In fact, if a E A, bE B and An B = 0 then p-l(A) n p-I(B) = D. If a = b then (a, lIJa) and (a, [g]a) lie on the same stalk p -lea). Since [f1o # [gL we were able to divide the stalk. That is, one germ was "higher up" on the stalk than the other. In the remainder of this section some of the properties of 9'(G) as a topological space are investigated. In particular, it will be of interest to characterize the components of 9'(G). However, we must first digress to study some additional topological concepts.

-Ie

5.5 Definition. Let (X, :Y) be a topological space. If Xo and Xl E X then an arc (or path) in X from X () to Xl is a continuous function y: [0, 1]-+ X such that yeO) = x 0 and y( I) = X l' The point x 0 is called the initial point of y and x 1 is called the final point or terminal point. The trace of y is the set {y} = {y(t): S t S; I}.

°

230

Analytic Continuation and Riemann Surfaces

A subset A of X is said to be arcwise or path wise connected if for any two points Xo and Xl in A there is a path from Xo to Xl whose trace lies in A. The topological space (X, ff) is called locally arcwise or path wise connected if for each point X in X and each open set U which contains X there is an open arcwise connected set V such that X E V and V c U. For each X in X let ~ be the collection of all open arcwise connected subsets of X which contain x. Then X is locally arcwise connected iff {~: x E X} is a neighborhood system which induces the original topology on X. (Verify!) The proof of the following proposition is left to the reader. 5.6 Proposition. Let (X, ff) be a topological space. (a) If A is an arcwise connected subset of X then A is connected. (b) If X is locally arcwise connected then each component of X is an open

set. The converse to part (a) of the preceding proposition is not true. For example let

X =

{t+isin~

: t>

o}

U

{si: -1

~s~

1}.

Since X is the closure of a connected set it is itself connected. Howcrver, there is no arc from the point 1/7T to i which lies in X. X is also an example of a topological space which is connected but not locally arcwise connected. Suppose X is connected and locally arcwise connected; does it follow that X is arcwise connected? The answer is yes. In fact, this is an abstract version of a theorem which was proved about open connected subsets of the plane. Since disks in the plane are connected, it follows that open subsets of IC are locally arcwise connected. Recall that in Theorem II. 2.3 it was proved that for an open connected subset G of the plane, any two points in G can be joined by a polygon which lies in G. Hence, a partial generalization of this (the concept of a polygon in an abstract metric space is meaningless) is the following proposition whose proof is virtually identical to the proof of II. 2.3. 5.7 Proposition. If X is locally arcwise connected then an open connected subset of X is arcwise connected. We now return to the sheaf of germs of analytic functions on an open set G. 5.S Proposition. Let G be an open subset of the plane and let U be an open connected subset of G such that there is an analytic function f defined on U. Then N(j, U) is arcwise connected in Y(G).

Proof Let (a, [f]a) and (b, [fh) be two generic points in N(f, U); then a, b E U. Since U is a region there is a path y: [0, 1] --+ U from a to b. Define a: [0, 1] --+ N(j, U) by aCt)

= (y(t), [f]Y(I).

231

The Sheaf of Germs of Analytic Functions on an Open Set

Clearly a(O) = (a, [f]a) and a(1) = (b, [fh); if it can be shown that a is continuous then a is the desired arc. Fix t in [0, 1] and let N(g, V) be a neighborhood of a(t). Then yet) E= V and [fL(t) = [gL(t). So there is a number r > 0 such that B(y(t); r) c Un V and fez) = g(z) for Iz-y(t)1 < r. Also, since y is continuous there is a 8 > 0 such that Iyes) -y(t)1 < r whenever Is- Ii < 8. Combining these last two facts gives that (t-8, t+8) c a- 1(N(g, V». It follows from Exercise 4.3 that a is continuous.

III

5.9 Corollary. 9"(G) is locally arcwise connected and the components of 9"(C) are open arc wise connected sets. Proof The first part of this corollary is a direct consequence of the preceding proposition. The second half follows from Proposition S.6(b) and Proposition 5.7. III In light of this last corollary, it is possible to gain insight into the nature of the components of 9"(G) by studying the curves in 9"(G).

5.10 Theorem. There is a path in 9"(G) from (a, [flo) to (b, [g]b) iff there is a path y in G from a to b such that [gh is the analytic continuation of[f]a along y. Proof Suppose that a: [0, 1] - ? ,51'(G) is a path with a(O) = (a, [flo), a(l) = (b, [gh). Then y = po a is a path in G from a to b. Since a(t) (o9"(G) for each t, there is a germ [ftly(!} such that

aCt) = (y(t), [It],(t».

°: :;

We claim that {[J,]y(t): t :::;; I} is the required continuation of [fla along y. Since [/la = [fo]a and [g]b = [/lh, it is only necessary to show that {[J,]y(t)} is a continuation. For each t let D t be a disk about z = yet) such that D t C G and J, is defined on D t • Fix tin [0, 1]; since NUt, D,) is a neighborhood of a(t) and a is continuous, there is a 8 > 0 such that (t-8, 1+8) c a- 1(N(J" D,».

That is, if Is - tl < 8 then (y(s), [j;')y(s» = a(s) E N(J" D,). But, by definition, this gives that yes) E= D t and [Is]'(s) = [j~]y(S); and this is precisely the condition needed to insure that {(ft, D ,): 0 :::; t :::; I} is a continuation along y (Definition 2.2). Now suppose that y is a curve in G from a to band (JJ,],(t): t :::;; I} is a continuation along y such that [fO]a = [fla and [fl]b = [g]b' Define a: [0, 1]->- Y(G) by aCt) = (y(t), [J,ly(t»; it is claimed that a is a path from (a, [fla) to (b, [gh). Since the initial and final points of a are the correct ones it is only necessary to show that a is continuous. Because the details of this argument consist in retracing the steps of the first half of this proof, their execution is left to the reader. III

°: :;

5.U Theorem. Let 9"(G) (ff

C(f

c 9"(G) and let (a, [fla)

E

C(f;

then

C(f

is a component of

232

ce =

Analytic Continuation and Riemann Surfaces

reb, [glb): [glb is the continuation of[fla along some curve in G}.

ce

ce

Proof Suppose is a component of 9"(G); by Corollary 5.9 is an open arcwise connected subset of 9"(G). So by the preceding theorem, for each point (b, [gh) in [gh is the continuation of [fla along some curve in G. Conversely, if [glb is the continuation of [fla along some curve in G then (b, [gh) belongs to the component of 9"(G) which contains (a, [fJ.); that is, (b, [gh) E Now suppose that consists of all points (b, [gh) such that [glb is a continuation of [fla. Then is arcwise connected and hence is connected. If is the component of 9"(G) containing then by the first half of this proof it follows that = I' • Notice that the point (a, [f]a) in the statement of the preceding theorem has a transitory role; any point in the component will do. Fix a function element (f, D) and recall that the complete analytic function :F associated with (J, D) is the collection of all germs [glz which are analytic continuations of [fJ. for any a in D (see Definition 2.7). Let

ce,

ce.

cel

5.12

ce

ce

ce

ce

ce

G = {z: there is a germ [g]z in:F];

it follows that G is open. In fact, if Z E G then there is a germ [glz in :F and a disk B about z on which g is defined. Clearly BeG so that G is open. It follows from Theorem 5.11 that

5.13 is a component of 9"«(:) and that p(~) 5.14 Definition. Let :F defined in (5.13) and p pair (Jr, p) is called the (5.12) is called the base

= G. (It is also true that ~

c 9"(G).)

be a complete analytic function. If /J£ is the set is the projection map of the sheaf 9"«(:) then the Riemann Surface of :F. The open set G defined in space of ::;;.

5.15 Theorem. Let::;; be a complete analytic function with base space G and let (Jr, p) be its Riemann Surface. Theil p: P-£ ~ G is an open continuous map. Also, if(a, [f]a) is a point in [j£ then there is a neighborhood NC.f, D) of (a, [fla) such that p maps N(J, D) homeomorphically onto an open disk in the plane. Proof Consider ,'?/t as a component of 9"«(:). Since p: 9"«(:) ~ C is continuous it follows that p: ~ ~ G is continuous. To show that p is open it is sufficient to show that p(NU; U» is open for each (J, U) (Exercise 4.4). But p(N(J, U» = U which is open. If (a, [fla) E.~ let D be an open disk such that (J, D) E [fla. Then p: N(J, D) ~ D is an open, continuous, onto map. To show that p is a homeomorphism it only remains to show that p is one-one on N(J, D). But if (b, [fh) and (c, [fle) are distinct points of N(J, D) then b =1= c which gives that p is one-one. • Remarks. In its standard usage the Riemann surface of a complete analytic function consists not only of what is here called a Riemann Surface but also

Analytic Manifolds

233

some extra points-the branch points. These branch points correspond to singularities and permit a deeper analysis of the complete analytic function. Exercises 1. Define F: 9'«(;) -+ C by F(z, [f]z) = fez) and show that F is continuous. 2. Let :F be the complete analytic function obtained from the principal branch of the logarithm and let G = C- {O}. If D is an open subset of G andf: D -+ C is a branch of the logarithm show that [f]a E :F for all a in D. Conversely, if (J, D) is a function element such that [f]a E :F for some a in D, show thatf: D -+ C is a branch of the logarithm. (Hint: Use Exercise 2.8.) 3: Let G = C - {O}, let :F be the complete analytic function obtained from the principal branch of the logarithm, and let (3l, p) be the Riemann surface of:F (so that G is the base of 3l). Show that 3l is homeomorphic to the graph r = {(z, e z E G} considered as a subset of C x C. (Use the map h: 3l -+ r defined by h(z, [f).) = (f(z), z) and use Exercise 2.) State and prove an analogous result for branches of zl/n. 4. Consider the sheaf 9'«(;), let B = {z: Jz-IJ < I}, let t be the principal branch of the logarithm defined on B, and let tl (z) = t(z) + 21ri for all z in B. (a) Let D = {z: JzJ < I} and show that (t, [t]t) and (t, [ttlt) belong to the same component of p -I(D). (b) Find two disjoint open subsets of 9'«(;) each of which contains one of the points (t, [t1t) and (t, [ttlt). Z ):

§6. Analytic Manifolds In this section a structure will be defined on a topological space which, when it exists, enables us to define an analytic function on the space. Before making the necessary definitions it is instructive to consider a previously encountered example of such a structure. The extended plane Coo can be endowed with an analytic structure in a neighborhood of each of its points. If a E Coo and a =I- 00 then a finite neighborhood V of a is an open subset of the plane. If fP: V -+ C is the identity map, fP{z) = z, then fP gives a "coordinatization" of the neighborhood V. (Do not become confused over the preceding triviality. The introduction of the identity function fP seems an unnecessary nuisance. After all, V is an open subset of C so we know what it means to have a function analytic on V. Why bring up fP? The answer is that for the general definition it is necessary to consider pairs such as (V, fP); trivialities appear here because this is a trivial example.) If a = 00 then let Voo = {z: IzJ> l}u {oo} and define fPoo: Voo-+C by fPoo(z) = Z-I for z =I- 00, fPoo( 00) = o. So fPoo is a homeomorphism of V 00 onto the open disk B(O; 1). Hence to each point a in Coo a pair (Va' fPa) is attached such that Va is a neighborhood of a and fPa is a homeomorphism of Va onto an open subset of the plane. What happens if two of the sets Va and Vb intersect? Suppose for example that a =I- 00 and Va n V =I- O. Let Goo = B(O; 1) = fPoo( V 00) and let Ga = fPa(Va) (=Va)' Then fP;;;t(z) = Z-I for all z in Goo; thus fPa fP;;;;I(Z) = Z-I for all z in fPoclVoo n Va). Since 00 tVa, 0 tfPoo(Voo n Va) 00

0

234

Analytic Continuation and Riemann Surfaces

so that 'fa () 'P-;' 1 is analytic on its domain. Similarly, if both a and bare finite then 'Po 'Pb- 1 is analytic on its domain. This is the crucial property. 0

6.1 Definition. Let X be a topological space; a coordinate patch on X is a pair ( U, 'f) where U is an open subset of X and 'f is a homeomorphism of U onto an open subset of the plane. If a E U then the coordinate patch (U, 'P) is said to contain a.

6.2 Definition. An analytic manifold is a pair (X, - n be a continuous function; let a E X and a = f(a). The function f is analytic al a if for any patch (.\, .j;) in tj? which containsCi there is a patch (U, 'P) in - IR be a function with continuous second partial derivatives and define UCr, 0) = u(r cos 0, r sin 0). (a) Show that

256

So if 0 ¢ G then u is harmonic iff 8

(8U) + 88£]2U= o. 2

r iir r a;.

(b) Let u have the property that it depends only on Izl and not arg z. That is, u(z) = 9'(lzl). Show that u is harmonic iff u(z) = a log Izl +b for some constants a and b. 9. Let u:G --* IR be harmonic and let A = {z E G: u,,(z) = u,(z) = O}; that is, A is the set of zeros of the gradient of u. Can A have a limit point in G? 10. State and prove a Schwarz Reflection Principle for harmonic functions. II. Deduce the Maximum Principle for analytic functions from Theorem 1.6.

§2. Harmonic functions on a disk Before studying harmonic functions in the large it is necessary to study them locally. That is, we must study these functions on disks. The plan is to study harmonic functions on the open unit disk {z:lzl < I} and then interpret the results for arbitrary disks. Of basic importance is the Poisson kernel. 2.1 Definition. The function Pr(O)

=

L 00

n=

rlnleino,

-00

for 0 ::; r < 1 and - 00 < 0 < 00, is called the Poisson kernel. Let z == re iO , 0 ::; r < 1; then 1 +re iO -1--;0 = (1 +z)(1 +Z+Z2 + ...) -re

00

=

1 +2 2: r"e inO n=1

Hence, 00 1 +reiO) Re ( --iO = 1 +2 2: r" cos nO l-re n=1

=

1+

2: r"(einO+e-inO) 00

n= 1

Harmonic functions on a disk

1+re i6 l-re,e

Also - - . =

257

1+rei8_re-iO_r2 16 2

Il-re I

2.2

P(O) r

-

so that

(1

iB ) 1-r2 +re - Re 1-2r cos 8+r2 l-re i8

2.3 Proposition. The Poisson kernel satisfies the following:

(a)

~ 217

f "

P,(O) de = 1;

(b) Pr(8) > 0 for ai,

e, Pre -8) = P,(8), and P

r

is periodic in 8 with period

217; (c) Pr(O) < PreS) if 0 < S < 101 :::;; 17; (d) for each S > 0, lim p,(e) = 0 uniformly in e for 17 ;::: lei;:::

s.

Proof. (a) For a fixed value of r, 0 :::;; r < 1, the series (2.1) converges uniformly in O. So

-"

-"

(b) From equation (2.2), P.(O) = (l_r 2 )11_re i8 1- 2 > 0 since r < 1. The rest of (b) is an equally trivial consequence of (2.2). (c) Let 0 < S < e : :; 17 and define f: [S, e] -+ IR by f(t) = PrCt). Using (2.2), a routine calculation shows that F(t) < 0 so that f(S) > fee). (d) We must show that lim [sup{P,(8):S :::;; 101:::;; 17}] = 0

,. ...... 1-

But according to part (c), p,(e) :::;; P,(S) if 3 :::;; 181 :::;; 17; so it suffices to show that lim PreS) = O. But, again, this is a trivial consequence of equation (2.2).• ,.-.1-

Before going to the applications of the Poisson kernel, the reader should take time to consider the significance of Proposition 2.3. Think of P,(fJ) not as a function of rand e but as a family of functions of e, indexed by r. As r approaches 1 these functions converge to zero uniformly on any closed subinterval of [-17, 17] which does not contain e = 0 (part d). Nevertheless, part (a) is still valid. So as r approaches 1, the graph of P, becomes closer to the 8 axis for 8 away from zero but rises sharply near zero so that 2.3(a) is maintained. The next theorem states that the Dirichlet Problem can be solved for the unit disk.

2.4 Theorem. Let D = {z: Izi < I} and suppose that f: aD -+ IR is a continuous function. Then there is a continuous function u:D - ~ IR such that (a) u(z) = f(z) for z in aD; (b) u is harmonic in D.

258

Harmonic Functions

Moreover u is unique and is defined by the formula

2.5 for 0

u(re''0 ) $

r < I, 0

$

f)

$

"

ft)f(e't) ' dt = - IP,(f)27T

27T.

Proof Define u:D ~ IR by letting u(re iO ) be as in (2.5) if 0 $ r < 1 and letting u(e iO ) = f(e io ). Clearly u satisfies part (a); it remains to show that u is continuous on D- and harmonic in D. (i) u is harmonic in D. If 0 $ r < 1 then

, 1f" Re [1 +rei(O-t)] " "(O-t) f(e't) dt

u(re'o) = 27T

I-re'

1t

{ If

,[I

+rei(O-t)] } = Re 27T f(e") 1- rei(O-t) dt . -1t

{If " 1t

it

iO

= Re 27T fCe't) [eeit +re _ reiO ] dt } . So define g:D

~

IC by 1t

I ff(e't) ' [eit+z] eit _ z dt.

g(z) = 27T

-" Since u = Re g we need only show that g is analytic. But this is an easy consequence of Exercise IV.2.2. (ii) u is continuous on D-. Since u is harmonic on D it only remains to show that u is continuous at each point of the boundary of D. To accomplish this we make the following

2.6 Claim. Given IX in [ -7T, 7T] and E" > 0 there is a p, 0 < P < I, and an arc A of aD about e iot such that for p < r < 1 and e iO in A, [u(re iO ) - f(e iot)[ <

E"

Once claim 2.6 is proved the continuity of u at e iot is immediate since f is a continuous function. To avoid certain notational difficulties, the claim will only be proved for IX = O. (The general case can be obtained from this one by an argument which involves a rotation of the variables.) Since f is continuous at z = I there is a 0 > 0 such that 2.7

<

1

-E"

3

Harmonic functions on a disk

259

if 101 < o. Let M = max {If(eiB)I: 101 ::; 7T}; from Proposition 2.3 (d) there is a number p, 0 < P < 1, such that €

2.8

Pr(O) < 3M

for p < r < 1 and 101 ~ and p < r < 1,

u(reiB)-f(l) =

~ 27T

= -1

27T

"

f

oand 101

Let A be the arc {e iB : 101 < 10}. Then if e iB

"

f

E

A

P.(O-t)f(e it) dt-f(1)

-"

III 0, and suppose h is a continuous real valued function on {z: Iz - al = p}; then there is a unique continuous function w:B(a; p) -+IR such that w is harmonic on B(a; p) and w(z) = h(z) for Iz-al = p.

Proof Considerf(e iB ) = h(a+ pe i8 ); thenfis continuous on oD. Ifu:D- -+ IR is a continuous function such that u is harmonic in D and u(e iB) = f(e i8 ) then

Harmonic Functions

260

it is an easy matter to show that w(z)

=

u(

z~a) is the desired function on

B(a; p) . • It is now possible to give the promised converse to the Mean Value

Theorem. 2.11 Theorem. If u: G -r III is a continuous function which has the MVP then u is harmonic. Proof Let a E G and choose p such that B(a; p) c G; it is sufficient to show that u is harmonic on B(a; p). But according to Corollary 2.10 there is a continuous function w: B(a; p) -rill which is harmonic in B(a; p) and w(a+pe i9 ) = u(a+pe i9 ) for all 8. Since u-w satisfies the MVP and (u-w)(z) = 0 for Iz-al = p, it follows from Corollary 1.9 that u == w in B(a; p); in particular, u must be harmonic. • To prove the above theorem we used Corollary 2.10, which concerns functions harmonic in an arbitrary disk. It is desirable to derive a formula for the Poisson kernel of an arbitrary disk; to do this one need only make a change of variables in the formula (2.2). If R > 0 then substituting rlR for r in the middle of (2.2) gives R2_r2 .. _ - R2 -2rR cos 8+r2

2.12

~ r < R and all B. So if u is continuous on B(a; R) and harmonic in B(a; R) then

for 0

2.13

u(a+re i9 )

I =-21T

f[ "

2

R -

2

R2_r2 ] (8) 2 u(a+Re it ) dt - t +r

r R cos

Now (2.12) can also be written R2 _

r2

IR - re i9 12

and R-r ~ IRe it -re i8 ~ R+r. Therefore 1

R-r

R2 _r2

R+r

--O} ~ IR by 00 . 1 xf(it) U(X+lY) = 2 ( )2 dt. 7T x + y-t -00 Show that u is a bounded harmonic function on the right half plane such that for c in 1R,J(ic) = lim u(z).

f

z-+ic

7. Let D= {z : /z/ < I} and suppose f: aD~1I! is continuous except for a jump discontinuity at z = I. Define u: D~IR by (2.5). Show that u is harmonic. Let v be a harmonic conjugate of u. What can you say about the behavior of vCr) as r~I -? What about v(re iO ) as r~I - and f}~O? §3. Subharmonic and superharmonic functions In order to solve the Dirichlet Problem generalizations of harmonic functions are introduced. According to Theorem 2.11, a function is harmonic exactly when it has the MVP. With this in mind, the choice of terminology in the next definition becomes appropriate.

264

Harmonic Functions

3.1 Definition. Let G be a region and let cp: G -+ IR be a continuous function. cp is a subharmonic function if whenever B(a; r) c G,

Lf

2"

cp(a) :s;

cp(a+re iB ) dB.

o

cp is a superharmonic function if whenever B(a; r)

c

G,

2"

cp(a)

~ 21T ~f cp(a+re

iB )

dB.

o

The first comment that should be made is that cp is super harmonic iff - cp is subharmonic. Because of this, only the results on subharmonic functions will be given and it will be left to the reader to state the analogous result for superharmonic functions. Nevertheless, we will often quote results on superharmonic functions as though they had been stated in detail. In the definition of a subharmonic function cp it is possible to assume only that cp is upper semi-continuous. However this would make it necessary to use the Lebesgue Integral in the definition instead of the Riemann Integral. So it is assumed that cp is continuous when cp is subharmonic even though there are certain technical advantages that accrue if only upper semicontinuity is assumed. Clearly every harmonic function is subharmonic as well as superharmonic. In fact, according to Theorem 2.11, u is harmonic iff u is both subharmonic and superharmonic. If CP! and CP2 are subharmonic then so is a!CP! + a 2CP2 ifa!, a2 ~ o. It is interesting to see which of the results on harmonic functions also hold for subharmonic functions. One of the most important of these is the Maximum Principle. 3.2 Maximum Principle (Third Version). Let G be a region and let cp: G -+ IR be a subharmonic function. If there is a point a in G with cp(a) ~ cp(z) for all z in G then cp is a constant function. The proof is the same as the proof of the first version of the Maximum Principle. (Notice that only the Minimum Principle holds for superharmonic functions.) The second version of the Maximum Principle can also be extended, but here both subharmonic and superharmonic functions must be used. 3.3 Maximum Principle (Fourth Version). Let G be a region and let cp and if; be bounded real valued functions defined on G such that cp is subharmonic and '" is superharmonic. If for each point a in a 0 and

v(x+iy) = arctan (Y:f3k) -arctan (Y:f3} so if x+ iy ~ ic, c < 13k < 13, then v(x+ iy) ~ o. Similarly u(x+iy) ~ 0 as x+iy ~ ic, with oc < OCk < c. Hence, Claim 4.13 yields 4.14

lim [h(z)-hiz)] = O.

But

hk(X+iy) = arctan e~OCk) -arctan e:f3k). so lim hk(z) z-+ic

= ?T;

this combined with (4.14) implies Equation (4.12).

It remains to substantiate Claim 4.13; we argue geometrically. Recall (Exercise III.3.19) that hiz) is the angle in the figure. Consider all the intervals (ioc j' if3 j) lying above (iock' if3k) and translate them downward along the imaginal')' axis, keeping them above (iock' if3k) until their endpoints coincide and such that one of the endpoints coincides with if3k. Since r,(f3 J - oc j) ~ 2?T there is a number 13 < (13k + 2?T) such that each of the translated intervals lies in (if3k' if3). Now if z = x+iy, x > 0 and OCk < Y < 13k,

z

Harmonic Functions

274

then the angle h j(z) increases as the interval (icx j' ifl) is translated downward. Hence CXk < 1m z < flk implies

"[.h j(z)

4.15

:::; v(z),

j

where v is as in the statement of the claim and the sum is over all j such that cx j ;::.: (3k' By performing a similar upward translation of the intervals (icx j. i(3j) with (3j :::; CX k, there is a number cx < (cx k -21T) such that the translates lie in the interval (icx, icxk)' So if u is as in the claim and CXk < 1m z < (3k'

"[.h /z)

4.16

:::; u(z)

j

where the sum is over allj with flj :::; CX k. By combining (4.15) and (4.16) the claim is established . • 4.17 Corollary. Let G be a region such that no component of Coo - G reduces to a point; then G is a Dirichlet Region. 4.18 Corollary. A simply connected region is a Dirichlet Region. Proof If G #- C the result is clear since Coo - G has only one component. If G = C then the result is trivial. • Theorem 4.9 has no converse as the following example illustrates. Let 1 > rl > r2 > ... with rn -? 0; for each n let Yn be a proper closed arc of

the circle Izl = ro with length V(Yn)' Put G = B(O; 1) suppose that lim V(Yn)/r n = 2n. So Coo - G

00

[U {Yn} n=1 00

= {O} u U

n=l

{Yn}

U

U

{O}] and

{z:

Izl;;::': I}.

According to Theorem 4.9 there is a barrier at each point of oooG = oG with the possible exception of zero. We will show that there is also a barrier at zero. If rn-

m

1

> r > rn and if m > n, let Bm = B(O; r)-U {Yj}. Let h m be the j=n

continuous function on B;;' which is harmonic on Bm with hm(z)

Izl

=

rand hm(z)

=

0 for

z

in

m

U {y j}. Then

=

1 for

{h m} is a decreasing sequence of

j~n

positive harmonic functions on G(O; r); by Harnack's Theorem {h m } converges to a harmonic function h on G(O; r) which is also positive (Why?) Since lim h(z) = 1 for Iwl = r, we need only show that lim h(z) = O. Let k m z-w

z-o

be the harmonic function on B(O; rm) which is 0 on {Ym} and 1 on {z: Izl = rm} - {Ym} (this does not have continuous boundary values, only piecewise continuous boundary values which are sufficient-see Exercise 2.2). Then o :::; h :::; k m on B(O; rm) and

f

2n

km(O) =

~

21T

km(r mew) de

o

= ~ (21T _ V(Ym») 21T

rm

Green's Function

Since V(Ym] ~ 211, km(O) rm a barrier at zero.

275

~ 0; it follows that h(z) ~ 0 as z ~ O. Thus, G has

Exercises l. Let G = B(O; 1) and find a barrier for G at each point of the boundary. 2. Let G = C - (00, 0] and construct a barrier for each point of 0exp. 3. Let G be a region and a a point in 000G such that there is a harmonic function u: G ~ IR with lim u(z) = 0 and lim inf u(z) > 0 for all w in oooG, w -# a. Show that there is a barrier for G at a. 4. This exercise asks for an easier proof of a special case of Theorem 4.9. Let G be a bounded region and let a E oG such that there is a point b with [a, b] n G- = {a}. Show that G has a barrier at a. (Hint: Consider the transformation (z-a)(z-b)-I.)

§5. Green's Function In this section Green's Function is introduced and its existence is discussed. Green's Function plays a vital role in differential equations and other fields of analysis.

5.1 Definition. Let G be a region in the plane and let a E G. A Green's Function of G with singularity at a is a function ga : G - {a} ~ IR with the properties: (a) ga is harmonic in G- {a}; (b) G(z) = ga(z)+log Iz-al is harmonic in a disk about a; (c) lim ga(z) = 0 for each win 0ooG. For a given region G and a point a in G, ga need not exist. However, if it exists, it is ,unique. In fact if ha has the same properties, then, from (b), ha - ga is harmonic in G. But (c) implies that lim [ha(z) - ga(z)] = 0 for z-w

every win 8rJ:JG; so ha = ga by virtue of the Maximum Principle. A second observation is that a Green's Function is positive. In fact, ga is harmonic in G - {a} and lim ga(z) = + 00 since ga(z) + log Iz - al is z-+a

harmonic at z = a. By the Maximum Principle, ga( z) > 0 for all z in G - {a}. Given this observation it is easy to see that C has no Green's Function with a singularity at zero (or a singularity at any point, for that matter). In fact, suppose go is the Green's Function with singularity at zero and put g = -go; so g(z) < 0 for all z. We will show g must be a constant function, which is a contradiction. To do this, it is sufficient to show that if 0 -# Zl -# Z2 -# 0 then g(Z2) :; g(zl)' If £ > 0 then there is as> 0 such that

276

Harmonic Functions

Ig(Z)-g(Zl)1 < e if IZl -zi < 8; so g(z) < g(zl)+e if IZ-Zll < 8. Let r > IZl -z21 > 8; then

hr(z)

=

(8)

g(Zl) + e log -

IZ-Z11 log - r -

r

is harmonic in C - {Z l}' It is left to the reader to check that g(z) :::; hr(z) for Z on the boundary of the annulus A = {z:8 < jz-zll < r}. By the Maximum Principle, g(z) :::; hr(z) for Z in A; in particular, h.(z2) ;;:: g(zz). Letting r _ 00 we get r-+ '"

since e was arbitrary, g(Z2) :5 g(Zl) and g must be a constant function. When do Green's Functions exist?

5.2 Theorem. If G is a bounded Dirichlet Region then for each a in G there is a Green's Function on G with singularity at a. Proof Define f: 8G -IR by fez) = log .jz-aj, and let u:G- _IR be the unique continuous function which is harmonic on G and such that u(z)=f(z) for Z in oG. Then ga(z) = u(z)-log jz-al is easily seen to be the Green's Function . • This section will close with one last result which says that Green's Functions are conformal invariants.

5.3 Theorem. Let G and Q be regions such that there is a one-one analytic function f of G onto Q; let a E G and a. = f(a). If ga and y~ are the Green's Functions for G and Q with singularities a and a. respectively, then ga(z) = yif(z». Proof Let cp: G - IR be defined by cp = y~ f To show that cp = ga it is sufficient to show that cp has the properties of the Green's Function with singularity at z = a. Clearly cp is harmonic in G - {a}. If w EO", G then lim cp(z) = 0 will follow if it can be shown that lim cp(zn) = 0 for any sequence 0

z-+w

{zn} in G with Zn -+ w. But {f(zn)} is a sequence in Q and so there is a subsequence {znJ such thatf(znk) -+ w in Q- (closure in C"')' So 'Y~(f(znJ) -+ O. Since this happens for any convergent subsequence of {f(zn)} it follows that lim cp(zn} = lim yif(zn» = O. Hence lim cp(z) = 0 for every w in oooG. z-+w

Finally, taking the power series expansion off about z fez) = a.+A 1(z-a)+A 2(z-a)2+ ... ;

or f(z)-a.

=

(z-a)[Al +A 2 (z-a)+ .. .].

Hence

5.4

loglf(z) - a.J = 10gJz-aJ +h(z),

=

a,

Green's Function 277 where h(z) = log IAI +Aiz-a)+ ... is harmonic near z = a since AI =1= 0. Suppose that y~( w) = ~(w) -logl IV - 0:1 where ~ is a harmonic function on Q. Using (5.4) 1

'F{z) =

~{f{z»-loglf(z)-o:l

= [~(r{z)-h(z)]-loglz-al·

Since ~ f - h is harmonic near z z = a. • 0

=

a, 'F(z) + log Iz - al is harmonic near

Exercises 1. (a) Let G be a simply connected region, let a E G, and let f: G -+ D = {z: Izi < I} be a one-one analytic function such that f(G) = D and f{a) = 0. Show that the Green's Function on G with singularity at a is ga{z) = -log If(z)l· (b) Find the Green's Functions for each of the foHowing regions: (i) G = C-(oo, 0]; (ii) G = {z:Re z > o}; (iii) G = {z:o < 1m z < 21T}. 2. Let ga be the Green's Function on a region G with singularity at z = a. Prove that if!f is a positive superharmonic function on G - {a} with lim inf

[!f(z)+log Iz-all > - 00, then ga(z) .:s:; !fez) for z =1= a. 3. This exercise gives a proof of the Riemann Mapping Theorem where it is assumed that if G is a simply connected region, G =1= C, then: (i) Coo - G is connected, (ii) every harmonic function on G has a harmonic conjugate, (iii) if a 0, for a suitable choice of 0:. (b) Let G be a simply connected region with G =1= C, but assume that G is unbounded and 0, 00 E 0ocP. Let t be a branch of log z on G, a E G, and ex = tea). Show that t is one-one on G and ((z) =1= ex+21Ti for any z in G. Prove that 'F(z) = [t(Z)-0:-21Tirl is a conformal map of G onto a bounded simply connected region in the plane. (Show that t omits all values in a neighborhood of 0: + 21Ti.) (c) Combine parts (a) and (b) to prove the Riemann Mapping Theorem. 4. (a) Let G be a region such that 8G = y is a simple continuously differen-

°

278

Harmonic Functions

tiable closed curve. If !:iJG ->- IR is continuous and g(z, a) Green's Function on G with singularity at a, show that

5.5

Il(a)

=

f!CZ)

=

gaCz) is the

~-Gn (z, a) ds

is a formula for the solution of the Dirichlet Problem with boundary values!; where

_~!I is the derivative of g in the direction of the outw:lfd normal to y and en

ds indicates that the integral is with respect to arc length. (Note: these concepts are not discussed in this book but the formula is sufficiently interesting so as to merit presentation.) (Hint: Apply Green's formula

fJ'~[WlV-VdUJdxdy = n

J'[

;,n -

(Iv

II,

Gn

GU] ds -u -;_. en

with n = G- {z: I,:-al s 8}, (3 < dCa, {y}), lie, h, Ii = gaCz) = g(z, a).) (b) Show that if G = {z: 121 < I} then (5.5) reduces to equation (2.5).

Chapter XI Entire Functions To begin this chapter, let us recall the Weierstrass Factorization Theorem for entire functions (VII. 5.14). Let f be an entire function with a zero of multiplicity m ~ 0 at z = 0; let {an} the zeros off, an of- 0, arranged so that a zero of multiplicity k is repeated in this sequence k times. Also assume that JaIl :::; la 2 1:::; .... If {Pn} is a sequence of integers such that 0.1 for every r > 0 then P(z) =

0.2

00

TI

Epn(z/an)

n=1

converges uniformly on compact subsets of the plane, where Ep(z) = (l-z)exp ( z

0.3 for P

~

+ 2 + ... + Z2

zP) p

1 and Eo(z) = l-z

Consequently 0.4

fez)

=

zm e 9 (Z) P(z)

where g is an entire function. An interesting line of investigation begins if we ask the questions: What properties off can be deduced if g and P are assumed to have certain "nice" properties? Can n:strictions be imposed on f which will imply that g and P have particular properties? The plan that will be adopted in answering these questions is to assume that g and P have certain characteristics, deduce the implied properties off, and then try to prove the converse of this implication. How to begin? Clearly the first restriction on g in thIs program is to suppose that it is a polynomial. It is equally clear that such an assumption must impose a growth condition on e 9 (Zj. A convenient assumption on Pis that all the integers Pn are equal. From equation (0.1), we see that this is to assume that there is an integer P ~ 1 such that 00

I lanl-

n;l

P

< 00;

that is, it is an assumption on the growth rate of the zeros off In the first 279

280

Entire Functions

section of this chapter Jensen's Formula is deduced. Jensen's Formula says that there is a relation between the growth rate of the zeros of I and the growth of M (r) = sup {1/(reie)l: 0 :5 8 :5 217} as r increases. In succeeding sections we study the growth of the zeros ofland the growth of M(r). Finally, the chapter culminates in the beautiful factorization theorem of Hadamard which shows an intimate relation between the growth rate of M(r) and these assumptions on g and P. §1. Jensen's Formula

If I is analytic in an open set containing B(O; r) and I doesn't vanish in B(O; 1') then log III is harmonic there. Hence it has the Mean Value Property eX. 1.4); that is

f ' 2n

1.1

log

1/(0)1 = --1

217

log i/(re''0 )1

de.

o

Suppose I has exactly one zero a = rei> on the circle Izi (z - a) -1 then (1.1) can be applied to g to obtain

f;. f [log I/(re )1-log ire

= r. If g(z) = I(z)

21t

log ig(O) I =

i8

iB -

rei'l]

de

o

Since log ig(O)1 = log I/(O)I-Iog r, it will follow that (Ll) remains valid where I has a single zero on iz' = r, if it can be shown that

j'

21t

2;1

' d8 log Ire'' e -re"1

= log r;

o

alternately, if it can be shown that 2"

flog II-e i8 1 d8 = O.

o

But this follows from the fact that

Jlog (sin2 28) dB = 2"

417 log 2

o

(Exercise V. 2.2(h)). So (1.1) remains valid if I has a single zero on Izl = r; by induction, (1.1) is valid as long as/has no zeros in B(G; r). The next step is to examine what happens if I has zeros inside B(O; r). In this case, log !/(z)1 is no longer harmonic so that the MVP is not present.

1.2 Jensen's Formula. Let I be an analytic lunction on a region containing

281

Jensen's Formula

S(O; r) and suppose that a1, ... , all are the zeros of fin B(O; r) repeated according to multiplicity. If f(O) 0 then

*"

log If(O) I =

-

L

~ log C;kl) +

2"

flOg If(reiB)1 dB.

o

Proof If Ibl < 1 then the map (z-b) (l_hZ)-1 takes the disk B(O; 1) onto itself and maps the boundary onto itself. Hence r 2(z- ak) r2-iikz maps B(O; r) onto itself and takes the boundary to the boundary. Therefore

-akz f(z) TI r(z-ak) n

F(z) =

r

2

-

k=1

is analytic in an open set containing B(O; r), has no zeros in B(O; r), and IF(z)1 = If(z) I for Izi = r. So (1.1) applies to F to give 2"

log IF(O) I =

2~ flOg If(rei6) I d8. o

However

F(O) = f(O)

D(-;J

so that Jensen's Formula results . • If the same methods are used but the MVP is replaced by Corollary X. 2.9, log If(z) I can be found for z oft ak' 1 :$ k :$ n. 1.3 Poisson-Jensen Formula. Let f be analytic in a region which contains B(O; r) and let ai' ... , an be the zeros off in B(O; r) repeated according to multiplicity. if Izl < r andf(z) oft 0 then

log If(z) I = -

-ii zj 1 f2" (reIB+z) 2:" log Ir2r(Z-ak k) + 2- Re -I-B- log If(re IB )I d8. 71" re -z

k=1

o

Exercises 1. In the hypothesis of Jensen's Formula, do not suppose that f(O) oft O. Show that if f has a zero at z = 0 of multiplicity m then

log

f ro

If(z) I < exp (1XIzl"+ I) Proo! Since f is an entire function of genus fL fez) = zm eg(z)

n E,,(z/a 00

n ),

n=1

where g is a polynomial of degree :::;; fL. Notice that if Izi < 2.7

log IE,,(z) I

=

Re {log(l-z)+z+ .. . +Z"/fL}

=

1 +2 } I +I Re { - fL + 1 zIt - fL + 2 Zll - ...

:::;; Izl/+1

{_1_ +K fL+l fL+2

+ ... }

:::;; Izl,,+1(1+-t+(!)2+ ... )

= 2Izl"+1 Also

IE,.(z) I :::;; (1 + IzJ) exp (izi + ... + Izl"/fL), so that

1- then

log jE,.(z)1 :::;; log (1 + Izl) + Izi + ... + Izl"/fL·

Entire Functions

284

Hence,

r

z~n;,

1£/z)1 Izll'+ 1 -

log

0

So if A > 0 then there is a number R > 0 such that 2.8

But on {z: t ~ Izl ~ R} the function Izl-(I'+I) log I£I'(z) I is continuous except at z = + I, where it tends to - 00. Hence there is a constant B > 0 such that 2.9

Combining (2.7), (2.8), and (2.9) gives that log I£/z) I ~ Mlzll'+ 1

2.10

for all z in C, where M Since Ilan l- r 1•

r 1 } then

I ~ ~ Izll'+ 1 for Izi

> r 2.

n;1

Combining this with (2.11) gives that 2.12

log

IP(z)1 =

2: log co

1£I'(zjan) I .:s;

~lzll'+1

n;1

for

Izi

> r2. Since g is a polynomial of degree ~IL, .

11m %-00

Izl + Ig(z) I IZ 11'+1 = o.

m log

So there is an r3 > 0 such that m log Izl + Ig(z)1 < Together with (2.12) this yields log If(z) I < ex Izll'+1

t IX Izll'+1

for

Izl > r3.

The genus and order of an entire function

285

for Izl > ro = max {fl, r3}' By taking the exponential of both sides, the desired inequality, is obtained. III The preceding theorem says that by restricting the rate of growth of the zeros of the entire function fez) = zm exp g(z)P(z) and by requiring that g be a pofynomiai, then the growth of M(r) = max {If(re i8 )1: 0 :0; 8 :0; 27T} is dominated by exp (~lzIJi+l) for some 11- and any ~ > O. We wish to prove the converse to this result. 2.13 Definition. An entire function f is of finite order if there is a positive constant a and an ro > 0 such that If(z) 1< exp (lzla) for Izi > roo Iffis not of finite order then f is of infinite order. If f is of finite order then the number A = inf {a: If(z)1 < exp (lzl") for Izl sufficiently large} is called the order off Notice that if If(z) 1< exp (Izla) for Izl > ra > I and b > a then If(z)1 < exp (IzJh). The next proposition is an immediate consequence of this observation. 2.14 Proposition. Let f be an entire function of finite order A. If E > 0 then If(z) 1< exp (lzIH£) for all z with Izl sufficiently large; and a z can be found, with Izl as large as desired, such that If(z)1 ;?: exp (lzI;'- 0, and if I has order ,\ then A=ex- I . 6. Find the order of each of the following functions: (a) sin z; (b) cos z; (X

(X

I

CD

(c) cosh

z; (d)

n-D"zn where a > O. (Hint: For part (d) use Exercise 5.)

n=l

7. Let II and 12 be entire functions of finite order AI' A2 ; show that I = 11/2 has finite order A :S max (AI' '\2)' 8. Let {an} be a sequence of non-zero complex numbers. Let p = inf {a: I lanl- a < (J)}; the number p is called the exponent 01 eom-ergence of {an}.

Hadamard Factorization Theorem

287 (a) Iff is an entire function of rank p then the exponent of convergence p of the non-zero zeros off satisfies: p ::; p ::; p + 1. (b) If p' = the exponent of convergence of {an} then for every E" > 0, lanl-(P+E) < 00 and lanl-(p-E) = 00. (c) Letfbe an entire function of order,\ and let {ai, a2, ... } be the nonzero zeros of f counted according to multiplicity. If p is the exponent of convergence of {an} prove that p ::; ,\. (Hint: See the proof of (3.5) in the next section.)

L

L

n co

(d) Let P(z)

=

£/z/an) be a canonical product of rank p, and let p

n~1

be the exponent of convergence of {an}. Prove that the order of P is p. (Hint: If,\ is the order of P, p ::; '\; assume that lall ::; la 2 1::; ... and fix z, Izl > 1. Choose N such that lanl ::; 2 Izl if n ::; Nand lanl > 2 Izi if n ~ N + l. Treating the cases p < p + 1 and p = p + 1 separately, use (2.7) to show that for some E > 0

Prove that for Izi ~ 1, log 1£/z)1 < B of z. Use this to prove that

IzlP where B is a constant independent

for some constant C independent of z.) 9. Find the order of the following entire functions:

(a) fez) =

n (l-anz), co

D( :J

0 <

lal

< 1

n~1

(b) fez) =

1-

(c) fez) = [r(z)r l . §3. Hadamard Factorization Theorem In this section the converse of Corollary 2.16 is proved; that is each function of finite order has finite genus. Since a function of finite genus can be factored in a particularly pleasing way this gives a factorization theorem.

3.1 Lemma. Let f be a non-constant entire function of order A with f(O) #- 0 and let {a 1, a 2 , ••• } be the zeros off counted according to multiplicity and arrang~d so that la 11 ::; la 2 1 ::; . . . . If an integer p > ,,- 1 then

288

Entire Functions

Proof. Let n = nCr) = the number of zeros off in BCO; 1'); according to the Poisson-Jensen formula

. = log If(z)1

z Jog Ir---iikz k=! r(z-ak)

Ln. ,

I+

1

271"

f2n

o

i8 Re (re - , -+ z) log If(re''0 )1 dB re,e_ z

.

for Izl < r. Using Exercise 1 and Leibniz's rule for differentiating under an integral sign this gives

2"

+ for

Izl

~f

271"

2re iO (re i8 -z)-Zlog If(re i8 )1 dB

o

< rand z of- ai' ... , an" Differentiating p times yields:

21<

+(p+ I)!

~ J' 2re i8(re i8 _Z)-p-2 log If(re

217

i8 )

1

dB.

o

Now as I' --+ 00, n(r)>- 00 so that the result will follow if it can be shown that the last two summands in (3.2) tend to zero as r ~"" 00. To see that the second sum converges to zero let r > 2Jzl; then !akl ~ I' gives Ir2-(lkzl :2:!r 2 so that (Iakl Ir 2 -ak zl!)p+! ~ (2/r)P+!. Hence the second summand is dominated by nCr) (2/r)P+!. But it is an easy consequence of Jensen's Formula (see Exercise 1.2) that log 2n(r)::s; log M(2r). Sincefis of order A, for any E > 0 and r sufficiently large (log 2)n(r)r(p+1)::s; log[M(2r)]r-(P+!)

But p+ 1 > ,\ so that E may be chosen with ('\+E)-(P+ 1) < O. Hence nCr) (2/rY+ 1 --+ 0 as r --+ 00; that is, the second summand in (3.2) converges to zero. To show that the integral in (3.2) converges to zero notice that

r reiB(reiO_z)-P-Z dO

2"

o

=

0

since this integral is a multiple of the integral of (w-z)-p-2 around the circle Iwi = I' and this function has a primitive. So the value of the Integral in (3.2) remains unchanged if we substitute log Ifl-Iog M(r) for log If I·

289

Hadamard Factorization Theorem

So for 21z1 < r, the absolute value of the integral in (3.2) is dominated by 2"

3.3

(p+ I)! 2 P +3 r-(p+l)

~f

21T

[log M(r)-log If(re i9 )1] dB.

o

But according to Jensen's Formula, 2"

~ flog If(re i 9) I dB ;:::

21T

0

o

since f(O) = I. Also log M(r) :::; rH' for sufficiently large r so that (3.3) is dominated by (p+l)! 2 P +3 r H .-(p+l) As before, £ can be chosen so that rHo -(p+ I) ~ 0 as r ~ 00 • • Note that the preceding lemma implicitly assumes that f has infinitely many zeros. However, iff has only a finite number of zeros then the sum in Lemma 3. I becomes a finite sum and the lemma remains valid. 3.4 Hadamard's Factorization Theorem. order A then I has finite genus fL :::; A.

If f is an entire function of finite

Proof Let p be the largest integer less than or equal to A; so p :::; A < p + 1. The first step in the proof is to show that I has finite rank and that the rank is not larger than p. So let {ai' a 2 , ••• } be the zeros off counted according to multiplicity and arranged such that la II :::; la21 :::; .... It must be shown that

3.5

L la.r(p+ .;1 00

I)

<

00.

There is no loss in generality in assuming that 1(0) = 1. Indeed, if I has a zero at the origin of multiplicity m and M(r) = max {If(z)l: Izl = r} then for any £ > 0 and Izl = r log If(z)z-ml :::; log [M(r)r- m] :::; rH