Claude George Exercises in Integration

Problem Books in Mathematics Series Editor: P.R. Halmos

Unsolved Problems in Intuitive Mathematics, Volume I: Unsolved Problems in Number Theory by Richard K. Guy 1981. xviii, 161 pages. 17 illus.

Theorems and Problems in Functional Analysis by A.A. Kirillov and A.D. Gvishiani 1982. ix, 347 pages. 6 illus. Problems in Analysis by Bernard Gelbaum 1982. vii, 228 pages. 9 illus.

A Problem Seminar by Donald J. Newman 1982. viii, 113 pages.

Problem-Solving Through Problems by Loren C. Larson 1983. xi, 344 pages. 104 illus.

Demography Through Problems by N. Keyfitz and J.A. Beekman 1984. viii, 141 pages. 22 illus.

Problem Book for First Year Calculus by George W. B/uman 1984. xvi. 384 pages. 384 illus. Exercises in Integration by Claude George 1984. x. 550 pages. 6 illus. Exercises in Number Theory by D.P. Parent 1984. x. 541 pages.

Problems in Geometry by Marcel Berger et al. 1984. viii. 266 pages. 244 illus.

Claude George

Exercises in Integration

With 6 Illustrations

Springer-Verlag New York Berlin Heidelberg Tokyo

Translator J.M. Cole

Claude George University de Nancy I UER Sciences Mathematiques Boite Postale 239 54506 Vandoeuvre les Nancy Cedex France

17 St. Mary's Mount Leybum, North Yorkshire DL8 5JB U.K.

Editor Paul R. Halmos Department of Mathematics Indiana University Bloomington, IN 47405 U.S.A.

AMS Classifications: OOA07, 26-01, 28-01

Library of Congress Cataloging in Publication Data George, Claude. Exercises in integration. (Problem books in mathematics) Translation of: Exercices et problemes d`int6gration. Bibliography: p. Includes indexes.

1. Integrals, Generalized-Problems, exercises, etc. I. Title. II. Series. QA312.G39513

1984

515.4

84-14036

Title of the original French edition: Exercices et problemes d'integration, © BORDAS, Paris, 1980. © 1984 by Springer-Verlag New York Inc. All rights reserved. No part of this book may be translated or reproduced in any form without written permission from Springer-Verlag, 175 Fifth Avenue, New York, New York, 10010, U.S.A.

Printed and bound by R.R. Donnelley & Sons, Harrisonburg, Virginia. Printed in the United States of America.

987654321 ISBN 0-387-96060-0 Springer-Verlag New York Berlin Heidelberg Tokyo ISBN 3-540-96060-0 Springer-Verlag Berlin Heidelberg New York Tokyo

Introduction

Having taught the theory of integration for several years at the University of Nancy I, then at the Ecole des Mines of the same city, I had followed the custom of the times of writing up detailed solutions of exercises and problems, which I used to distribute to the students every week.

Some colleagues who had had

occasion to use these solutions have persuaded me that this work would be interesting to many students, teachers and researchers. The majority of these exercises are at the master's level; to them I have added a number directed to those who would wish to tackle greater difficulties or complete their knowledge on various points of the theory (third year students, diploma of education students, researchers, etc.).

This book, I hope, will render to students the services that this kind of book brings them in general, with the reservation that can always be made in this case: that certain of them will be tempted to look at the solution to the exercises which are put to.them without any personal effort.

There is hardly any need to

emphasize that such a use of this book would be no benefit. the other hand, the student who

On

after having worked seriously

upon a problem, seeks some pointers from the solution, or compares it with his own, will be using this work in the optimal way.

V

INTRODUCTION

vi

Teachers will find this book to be an important, if not exhaustive, list of exercises, certain of which are more or less standard, and others which may seem new. I have also noted (and this is what led me to edit these sheets)

that from one year to another one sometimes forgets the solution of an exercise and that one has to lose precious time in redisThis is particularly true for those solutions of

covering it.

which one remembers the heuristic form but of which the writing up is delicate if one wishes to be clear and precise at the same time.

Now, if one requires, quite rightly, that students

write their homework up correctly, then it is befitting to submit impeccable corrections to them, where the notations are judiciously chosen, phrases of the kind "it is clear that ... " used wittingly, and where the telegraphic style gives way to conciseness.

It is often the incorporation of these corrections which

demands the most work; I have therefore striven to take pains with the preparation of the proposed solutions, always remaining persuaded that perfection in this domain is never attained.

If

this book encourages those who have to present (either orally or in writing) correct versions of problems to improve the version they submit, the object I have set myself will be partly realised. In this book researchers will find some results that are not always treated in courses on integration; they are either properties whose use is not as universal as those which usually appear and which are therefore found scattered about in appendices in various works, or are results that correspond to some technical lemmas which I have picked up in recent articles on a variety of subjects: group theory, differential games, control theory, probability, etc.,

...

.

In presenting such a work it is just as well to make explicit those points of the theory that are assumed to be known.

This is

the object of the brief outline which precedes the eleven chapters of exercises.

INTRODUCTION

vii

In view of the origin of this book, it is evident that I took as a reference point the course that I gave at the time.

After

having taught abstract measure theory one year, I opted the next for a course expounding only the Lebesgue integral.

This is not

the place to discuss the advantages and inconveniences of each of the two points of view for the first year of a master's programme. I will say only that I have always considered the course that I gave to be more a course in analysis in which it is decided to use the Lebesgue integral than as a dogmatic exposition of a particular theory of integration.

The choice of exercise reflects

this attitude, especially in the emphasis given to trigonometric series, thereby paying the hommage due to the theory which is the starting point of the works of Cantor, Jordan, Peano, Borel, and Lebesgue.

From this it results that, except for the seven exer-

cises of Chapter 2 concerning a-algebras, all the others deal with Lebesgue measure on]Rr.

The advantage that has to be conceded to

this point of view is that it avoids the vocabulary of abstract measure theory, which constitutes an artificial obstacle for those readers who might not yet be well versed in this theory.

As for

students who might have followed a more sophisticated course, I can assure them that by substituting du for dx and u(E) fore meas(E) they will essentially rediscover the problems as they are commonly put to them, except for pathological examples about measures that are not a-finite and the applications of the RadonNikodym Theorem.

Furthermore, on this latter point the more per-

spicacious amongst them will not fail to see that the chapter treating the relationships between differentiation and integration is not foreign to this theorem.

Truthfully, there is another

point that is not tackled in this book, namely the matter of Fourier transforms of finite positive measures and Stone's Theorem, which to my mind is better suited to a course on probability. As was mentioned above, numerous exercises are devoted to trigonometric series, which provides an important set of applications

Viii

INTRODUCTION

of Lebesgue's theory.

This has led me to include some exercises

on series, summation processes, and trigonometric polynomials. Other exercises use the theory of holomorphic functions.

In

particular, some results of the PhrUgmen-Lindeltff type arise on

two occasions; in each instance I have given its proof under the hypotheses that appear in the exercise.

Quite generally, I have

included in the solutions, or in an appendix to them, the proofs of certain points of analyis, topology, or algebra which students may not know.

I have chosen to make each solution follow immediately after the corresponding problem.

The other method, which consists of

regrouping the former in a second part of the work, seemed to me (from memories I have retained from my student days) much less

manageable, especially when the problem is long, for it then becomes necessary to return often to the back of the book in order to follow the solution. I find it difficult to cite the origin of these exercises.

Many are part of a common pool of knowledge, handed down, one might say, in the public domain.

Others are drawn from different

classic works where they are proposed without proof or followed by more or less summary indications (in this respect it is interesting to note that in forcing oneself to write down the solutions one discovers a certain number of errors -just as many in the questions as in the suggestions offered).

Certain of the exer-

cises in this book were communicated to me orally by colleagues; I would thank them for their help here.

Lastly, others are, as

I have already said, lemmas found here and there, and which I have sometimes adapted.

Table of Contents

INTRODUCTION

...

...

...

...

...

V

CHAPTER 0: OUTLINE OF THE COURSE ...

...

...

...

...

1

CHAPTER 1: MEASURABLE SETS ... ... 1.21) (Exercises 1.1

...

...

...

...

37

CHAPTER 2: a-ALGEBRAS AND POSITIVE MEASURES Exercises 2.22 - 2.28)

...

...

...

79

CHAPTER 3: THE FUNDAMENTAL THEOREMS. (Exercises 3.29 - 3.72)

...

...

...

89

CHAPTER 4: ASYMPTOTIC EVALUATION OF INTEGRALS... (Exercises 4.73 - 4.78)

...

... 177

CHAPTER 5: FUBINI'S THEOREM... ... (Exercises 5.79 - 5.99)

...

...

...

... 199

... CHAPTER 6: THE LP SPACES ... (Exercises 6.100 - 6.125)

...

...

...

... 225

CHAPTER 7: THE SPACE L2. ... ... (Exercises 7.126 - 7.137)

...

...

...

... 285

...

...

...

...

CHAPTER 8: CONVOLUTION PRODUCTS AND FOURIER TRANSFORMS (Exercises 8.138 - 8.162) CHAPTER 9: FUNCTIONS WITH BOUNDED VARIATION: ABSOLUTELY CONTINUOUS FUNCTIONS: DIFFERENTIATION AND INTEGRATION (Exercises 9.163 - 9.173)...

ix

... 325

... 405

x

TABLE OF CONTENTS

CHAPTER 10: SUMMATION PROCESSES: TRIGONOMETRIC POLYNOMIALS.. 429 (Exercises 10.174 - 10.184) CHAPTER 11: TRIGONOMETRIC SERIES ... ... (Exercises 11.185 - 11.230)

...

ERRATUM TO EXERCISE 3.45

...

...

...

...

BIBLIOGRAPHY

...

...

...

...

...

...

...

... 547

NAME INDEX..

...

...

...

...

...

...

...

... 549

...

... 451

... 545

CHAPTER 0

Outline of the Course

a-ALGEBRAS AND MEASURES

0.1

DEFINITION: A family A of subsets of a set X is called a a-ALGEBRA ("sigma algebra") if 0 e A, and if A is closed under complementation and countable union.

From this it follows that the set X itself belongs to the a-algebra A, and that the a-algebra A is closed under countable intersecFor two sets A,B e A let us denote A - B = {x:x a A,x $ B};

tion.

then we have (A - B)e A. The smallest a-algebra containing the open sets of ]R a-algebra of BOREL SETS of ]R

;

is the

this a-algebra is also the small-

est a-algebra which contains the closed (resp. open) rectangles

of]R

.

DEFINITION: A (positive) MEASURE on a a -algebra A is a mapping u of A into [0,oo] such that if E is the disjoint union of a sequence of sets En e A, then u(E) = I u(En).

It follows that u(o) = 0, and then, if E is the union (not necessarily disjoint) of the sets En, U(E) 5 L u(En).

An equi-

valent definition is the following: If E is the union of a fin-

1

CHAPTER 0: OUTLINE

2

ite number of Ei's, each of which is in A and which are pairwise disjoint, then p(E) = p(E1) +

+ p(Ep); and furthermore p(A) _

limu(An) when A is the union of an increasing sequence of sets An of A.

If p is a measure and A is the intersection of a decreas-

ing sequence of sets An e A and if u(A1) < -, then p(A) = limp(An) There exists one and only one positive measure v on the a-algebra of Borel sets of ]R

such that, for every rectangle P, its

measure v(P) is equal to the volume of P.

DEFINITION: A set E of]R is called a NEGLIGEABLE SET if there exists a Borel set A such that E C A and v(A) = 0.

This definition is equivalent to the existence, for every e> 0, of a sequence of rectangles covering E, the sum of the volumes of which is less than c.

A countable union of negligeable sets is

negligeable, and every affine sub-manifold of iRp that is of dimension less that p is negligeable. DEFINITION: A set of ]R

is called a LEBESGUE MEASURABLE SET (or

simply a MEASURABLE SET) if it belong to the smallest a-algebra containing the Borel sets and negligeable sets of]R1. In order that E C ]R

be measurable it is necessary and suffic-

ient that there exist the Borel sets A and B such that A C E C B and v(B - A) = 0; upon then setting meas(E).= v(A) one unambiguously defines a positive measure on the a-algebra of Lebesgue measurable sets of ]Rp.

SURE ON I(.

This measure is called the LEBESGUE MEA-

A set is negligeable if it is measurable and of

(Lebesgue) measure zero.

This is why one also uses the expres-

sion SET OF MEASURE ZERO to denote a negligeable set.

The Lebesgue measure is invariant under translation as well as under unimodular linear transformations (i.e., those with determinant equal to ±1).

A homothety of ratio A multiplies the

Lebesgue measure by JAJp (where p is the dimension of the space).

OF THE COURSE

3

DEFINITION: If A is a o-algebra of subsets of X and B is a Q algebra of subsets of Y, a mapping f:X + Y is said to be an A B-MEAS-URABLE MAPPING if f 1(B) e A for a Z When Z B e Y= B.

Ilzp and B is

the a-algebra of Borel sets, one says, simply, that f is an A-MEASURABLE MAPPING.

In this case the definition is equivalent to re-

quiring f_1(V) eA for every open set V of Y.

Furthermore, when

X = n2q, the mapping f is said to be a BOREL MAPPING or a LEBESGUE-

MEASURABLE MAPPING according as A is the a-algebra of Borel sets or the Lebesgue-measurable sets of X. If f:X -;Ill, in order that f be A-measurable it is sufficient

that (f < a) = {x:f(x) < a} e A for all a em (and even for a e

This condition is taken as the definition of the A-measurability of an ARITHMETIC FUNCTION, that is to say of a mapping of X into [-co,+m] =3-R.

If f is an A-measurable mapping of X into Iltp and g

a Borel mapping of Ilzp into zzq, then gof is A-measurable. note that every continuous mapping of Ilzp into Ilzq is Borel.

Let us If

(fn) is a sequence of A-measurable arithmetic functions, the functions supfn,inffn,limsupfn,liminffn are also A-measurable. DEFINITION: A function is called a SIMPLE FUNCTION (with respect to the a-algebra A) if it is a linear combination of characteristic functions of sets of the a-algebra A.

For every A-measurable positive arithmetic function f there exists an increasing sequence of positive simple functions which converges to f at every point of X.

DEFINITION: A property holding on the points of a set A of7Rp is said to be true ALMOST EVERYWHERE ON A SET A if the set of points

of A for which this property is not satisfied has measure zero.

If f and g are two mappings from z

into zzq (or m) such that

f is measurable and f = g almost everywhere, then g is measurable.

CHAPTER 0: OUTLINE

4

This allows the notion of measurability to be extended to functions that are defined only almost everywhere.

DEFINITION: A function defined on]RP is called a STEP FUNCTION if it is a linear combination of characteristic functions of rect-

angles of]RP. Every measurable arithmetic function on]R

is the limit almost

everywhere of a sequence of step functions.

THEOREM: (Regularity of the Lebesgue Measure): For every measurable set E ofiRP one has: sup{meas(K):K compact K C E}; meas(E) = inf{meas(V):V open V D E}.

THEOREM: (Egoroff): Let X be a measurable set of]R

such that

meas(X) < co and (fn) a sequence of measurable functions such that

fn - f almost everywhere on X.

For every e > 0 there exists a

measurable set A C X such that: (i): meas(X - A) < ci (ii): fn -> f uniformly on A.

0.2

INTEGRATION OF MEASURABLE POSITIVE FUNCTIONS

NOTATION: If cp is a simple function on Min that takes the positive

values a1,...,ap on the (disjoint) measurable sets Al....)Ap, we set

I

n

q,(x)dx =

aimeas(Ai),

9 _

i=1

OF THE COURSE

5

with the convention that a.(+-)

or 0 according as a > 0 or

a = 0. DEFINITION. With the above notation, if f is a positive measurable arithmetic function on ]Rn there exists an increasing sequence ((pi) of positive simple functions which tends towards f One then sets:

at every point.

f(x)dx = if = limf

J

. .

This element of [0,+-]=]K +, which does not depend upon the sequence

(Ti) selected, is called the (LEBESGUE) INTEGRAL of f on7Rn.

This (Lebesgue) integral possesses the following properties (where f and g denote measurable positive arithmetic functions): PROPERTY (1): If f = g almost everywhere, then if = Jg;

PROPERTY (2): Jf = 0 if and only if f = 0 almost everywhere;

PROPERTY (3): if < - implies f < m almost everywhere;

PROPERTY (4): f 4 g almost everywhere implies if

PROPERTY (5):

PROPERTY (6)

1<

Jg;

J(f + g) = if + fg;

If A e]-R+., then JAf = AJf.

One can prove the following two fundamental results:

THEOREM: (Lebesgue's Monotonic Convergence): If (fn) is an increasing sequence of measurable positive arithmetic functions, then

Jlimf

n

n

=

limif-' n

CHAPTER 0: OUTLINE

6

LEMMA: (Fatou): If (fn) is a sequence of positive arithmetic functions, then

Jliminffn " liminf Ifn. n n

These two essential properties of the Lebesgue integral are equivalent to the following statements: Let (fn) be a sequence of positive arithmetic functions:

(a):

f(I fn) = E (Jfn);

(b):

If fn -* f almost everywhere, and if there exists A eIt+

such that

Jfn 5 A

for all n,

then

Jf , A. Property (1) allows the definition of the integral to be extended to measurable arithmetic functions that are defined only almost everywhere. And lastly:

NOTATION: If f is a measurable positive arithmetic function, and if E is a measurable set of ]R , we set

fEf(x)dx=JE =fiv, where 1lE denotes the CHARACTERISTIC FUNCTION of E.

The mapping E + J f defines a positive measure on the 'a-algebra

E

OF THE COURSE

7

of (Lebesgue) measurable sets.

0.3

INTEGRATION OF COMPLEX MEASURABLE FUNCTIONS

NOTATION: For every real function u, we set: u+ = sup(u,O) _ I(luI + u), u- = sup(-u,0) _ i(lul - u), so that

u = U+ - u_,

= u+ + u-,

lu l

u+u- = 0.

Let f be a complex measurable function on7Rn, and let u and v be its real and imaginary parts.

The function f is measurable

if and only if u+U_,v+,v- are measurable. DEFINITION: f is (LEBESGUE) INTEGRABLE ON]R if it is measurable

and if

11fl < The INTEGRAL is then defined by setting

Jf

=

Ju+

- Ju

+

iJv -

iJv,

(which has a meaning, because u+,u_,v+,v_ are majorised by lfl).

The Lebesgue integral possesses Properties (1) and (5) of the preceding Section; it also possesses Property (4) when f and g are real, as well as Property (6) with A e T.

The sum, and the

pointwise maximum and minimum of a finite number of integrable

CHAPTER 0: OUTLINE

8

functions are integrable.

If f is integrable, then

ff1 Dk--* K ->LWc

c

c

UCW

t

It will be noted that D

-> E for any space E from the table

above, and also E -> Lloc for every E; note also that the Lp spaces

are not mutually comparable. NOTATION: For every function f and every a e Rn, we define the

TRANSLATION OF f BY a by f.(x) = f(x - a).

Sometimes the notation Taf is used to designate fa.

If f and g

are convolvable, then

(f*g)a = fa,:g = ff:ga. DEFINITION: If E is one of the function spaces defined above, one says that E is INVARIANT UNDER TRANSLATION if f e E implies that

fa e E for every a eRn. Furthermore, if ai -> a implies that f,'. -f I in E, one says that the TRANSLATIONS OPERATE CONTINUOUSLY ON E.

OF THE COURSE

21

Cm,UC, CO,Ek and

THEOREM: (1): The spaces LP,LlOC,LC (1 5 p are invariant under translation;

Dk (0 5 k

(ii): The translations operate continuously on all these spaces,

except upon LW,L

loc

,L and Cam. c

NOTATION-DEFINITION: If F,GH are three of the function spaces

defined above, the notation F::G C H expresses that if f e F, g e G, then f and g are convolvable and f*g e H,

Furthermore, if fixgi -)-

fs:g in H whenever fi + f in F and gi -*-g in G, one writes F::G C H (continuously).

In the case where F = G = H = A it is said that

A is a CONVOLUTION ALGEBRA.

Lastly, if A is a convolution algebra

and A*E C E, one says that A OPERATES IN E; A is said to OPERATE CONTINUOUSLY IN E if A^E C E continuously.

THEOREM: (i): L1*L C UCW (continuously); furthermore,

IIffgII < (ii): Lp*Lq C C0 (continiously) if 1 < p,q < -, p + q = 1; furthermore,

IIf*gL s IIfIIpIIgIIq; (iii): L1s:Lp C LP (continuously) if 1 , p , m; furthermore,

11f* g11

, IIfII1IIgIIp.

In particular, L1

(iv): L

1

is a convolution algebra

operates continuously in Lp (1 4 p ,

(v). Dk*Lloc C Ek (continuously), 0 < k

(vi): L1 it a convolution algebra;

oo),Cm,UC

and CO;

CHAPTER 0: OUTLINE

22

(vii): L1 operates continuously in Lioc,L0 (1 5 p

W) Ek and

k D;

(viii): Dk is a convolution algebra.

Lastly, we have the formula

DS(f*g) = ff:Dsg,

REMARK: Since D -> L1 3 L1 (continuously), the convolution alge-

bras D and L1 operate continuously in all the function spaces which have previously been defined.

NOTATION-PROPOSITION: For every function f defined on1Rn we set: f(x) = f(-x),

.'(x) = f(-x).

When f and g are convoZvable, so are fand I (resp. f and g), and

(f*g)" = f*.,

(f g)- = f"g.

Furthermore : = 1, we set

NOTATION-PROPOSITION: If g e LP, h e Lq, p + 4

(g,h) = Jgh = (g*)(O) (g1h) = Jgh = (g*h)(O).

When f eL1, geLP, heLq, p + q = 1,

(f*g,h) - (g,f*h),

(f,eg1h) = (gl?*h).

OF THE COURSE

23

NOTATION: We denote by LPQ1r) the set of measurable functions on iR

that have period 2nand are such that

If 11p =

[

p

J2nIfIP)1/P 0

IIfII,, = ess sup I f(x) I

< Co,

05x62n

The set of k-fold continuously differentiable functions on ]R with period 2n is denoted Ek(W) (here 0 < k 4 Co, and E0(i') is

also denoted C(ur)). DEFINITION: The CONVOLUTION PRODUCT OF TWO MEASURABLE FUNCTIONS f and g WITH PERIOD 21E is defined by the formula: 2n

f(x - y)g(y)dy.

(f*g)(x) = 2nJ 0

Defining fi -> f in Ek(IF) to mean that f(S) f(S) uniformly for every integer s with 0 4 s < k, the following hold: LP(a)&rLq(a) C C(a)

(continuously), if

L1(a)*LP(a) C LP(Ir)

(continuously),

L'(T.')*E (a) C Ek(a)

(continuously),

= 1,

+

p

4

as do the inequalities:

IIf* IIC 5 IIfIIpIIgIIq, if feLP(a), geLcl(r), p +

= 1, q

IIf*IIp -1 IIfIIlIIgIIp,

if feL'(l), geLP(a).

By defining J`,J as above, and

24

CHAPTER 0: OUTLINE 21

(f,9) = 2nJ

2n f9,

f9

(.fig) = f1J 0

0

one obtains the same formulae as above.

Let us note that in the

formulae defining f*g,(f,g) and (fig), one can replace the range of integration (0,2n) by any interval of length 2n.

Clearly one can consider functions having an arbitrary period T > 0, it is then just a matter of replacing it by T/2 everywhere.

0.8

REGULARISATION OF FUNCTIONS

DEFINITION: One calls an approximate identity in L1 every sequence (rpi) of integrable functions that satisfies the following conditions:

(i): There exists a constant M such that 11,Pi11l < M for all i; (ii): lim J'Pi = 1; i

(iii): For every a > 0,

limJ i

m = 0, JxJ>,a i

An approximate identity (ml) is said to be compact if all the functions cp. vanish outside the same compact set of]Rn. i

In L1 there exist compact approximate identities consisting of

functions belonging to V ; these are called REGULARISING SEQUENCES. THEOREM: Let (rp

be an approximate identity in L1.

If E is one

of the spaces LP (1 < k < .o) or UC , then f o r every function f e E

one has 9 i *f -> f in E. If the approximate identity (W.) is compact this property extends to the spaces LPoc,Lp (1 < p < o), Ek,Vk (0 < k From this one deduces the following corollaries:

THEOREM: (Density); D

!1 < p < c ), Ek, Vk.

is dense in each of the spaces Lp;LPoc, LP c

OF THE COURSE

25

LEMMA: (Calculus of Variations): If f e L10c is such that Jf9 = 0 for any rpe D , then f = 0 almost everywhere.

DEFINITION: An approximate identity in L1(1r) is a sequence (p.) of integrable functions with period 27t, such that

(i):II(piII1,M; (

(ii) : lim 2l -71 i = 1; (iii): For all a, 0 < a < n, liml

i

p.(x)dx = 0.

a 0, there exists 6 > 0 such that for every finite sequence of mutually disjoint sub-intervals ]ai,si[ of [a,b] one has

i

If(si) - f(ai)I < e

whenever

i

(R. - ai) < 6.

THEOREM: (Lebesgue): If f is absolutely continuous on [a,b] f is differentiable almost everywhere, f' is integrable and

b

f(b) - f(a) = J.f'(x)dx, a

b V(f;a,b) = J If'(x)Idx. a

Conversely, if F is integrable on [a,b], and if

f(x) = JF(t)dt, x a

a < x < b,

CHAPTER 0: OUTLINE

30

then f is absolutely continuous and f' = F almost everywhere. The theory of differentiation makes use of the two following Lemmas, which are interesting in their own right. LEMMA (1): Let K be a compact set of atn covered by a family of open cubes.

From this family there can be chosen a finite sequence

C1.$---.$Cp of mutually disjoint cubes such that

meas(K) < 3n

meas(Cp). k=1

LEMMA (2): (The Setting Sun Lemma): Let f be a real continuous function on [a,b]., E the set of points x of this interval for

which there exists a y such that x < y < b and f(x) < f(y).

Then

the set E is the disjoint union of a sequence of intervals with end points an < bn such that f(an)

0.11

f(bn)

TRIGONOMETRIC SERIES If (un)nea is a sequence of complex numbers indexed by some

positive or negative integers, one sets

W

+W

-

un = u0 +

I

N (un + u_n) = lim

E

un,

N- n=-N

n=1

when this last limit exists.

luni < - one also has

When I00

u

n

=lim N-

M E

n=-N

u.n

M-

A trigonometric series is a formal series of the type

OF THE COURSE

tm c e n

31

m

a

inx

=

2

(ancosnx + bnsinnx),

+

n=1

where, upon agreeing to set b0 = 0,

an = cn + c-n, b

n

C

= i(c n - c -n ),

- ib ), n = Y(a n n

n > 0.

c-n = '(an + ibn),

In what follows it is assumed that f e L1(7r) (L1(w) has been

defined in Section 0.7).

For every n e 2z we set

(2n e-inxf(x)dx.

?(n) = I

0

The ?(n) are called the FOURIER COEFFICIENTS of f, and the formal series

inx

is the FOURIER SERIES of f.

We use the notation

L oneinx

f(x) ti

-W

to indicate that the second member is the Fourier series of f, that is to say that cn = ?(n) for every n ez.

It will be noted

that:

12n r

an

f(x)cosnxdx,

n 0

bn =

1 n

2n

f(x)sinnxdx.

On setting en(x) = einx we have the following FORMULAS

32

CHAPTER 0: OUTLINE

.'(n) _ (f l en), fsaen = f(n)en,

f(n) = ?(n)

and f(n) = ?(-n),

f*g(n) = f(n)g(n),

P (n) = in(n)

if f is absolutely continuous.

For the explicit calculation of the Fourier series of a function f the following result, which generalises (5) in the Formulas above, is useful:

PROPOSITION: If f is a pieeewise continuously differentiable function, that is to say, if there exist points.

-u 4 al < a2 < ... < a p

it

<

such that f coincides on each open interval ]as,as+l[' 1 5 s S p and ap+l = al + 21E, with the restriction of a function fs contin-

uously differentiable on the closed interval [as,as+1]' upon denoting by [f'] a function with period 27[ that coincides with f'

s

on each ]as'as+1[, one has:

ina

{f(as + o) - f(as - o)}e

in?(n) = [f'](n) + 2n s=1

EXAMPLE 1.

f(x)

If f(x) = x when -n < x <

,, 2

n=1

(-1)n+1 sinnx n

it,

S.

OF THE COURSE EXAMPLE 2.

33

If f(x) = -1 when -n < x < 0 and f(x) = 1 when 0 < x

< it,

f(x) 4 4

sin(2n + 1)x

2n+1

nn=

EXAMPLE 3.

If f(x) = it - Ixi when -n

f(x)ti'-`+4 2 it

x < it,

cos(2n + 1)x n=0

On + 1)2

THEOREM: (Riemann-Lebesgue):

lim ?(n) = 0.

Ini-'°° THEOREM: (Fourier): If f(n) = 0 for all n ea, then f = 0 almost everywhere.

If f is p-fold continuously differentiable, f(n) = o(Inl-P); if f has bounded variation on [0,2n], f(n) = 0(1/Inj).

Converse-

ly, if for an integer p one has I InLPIf(n)I < =, then f coincides almost everywhere with a p-fold continuously differentiable function.

NOTATION: We set:

N SN(f;x) =

I

f(n)einx

(FOURIER SUMS),

n=-N

aN (f x)= 3

s0(f;x) + ... + SN(f;x)

N+1

(FEJER SUMS),

then:

SN(f;x) = (f*DN)(x), with

aN(f;x) = (f'N)(x),

34

CHAPTER 0: OUTLINE

DN (x) = sin(N + Z)x

(DIRICHLET'S KERNEL),

sin Zx

FN(x)

1(sisN

(FEJER'S KERNEL).

+ 1)jx inrx

N +

THEOREM: (Localisation): If f(x + 0) and f(x - 0) exist at a point

x, we set

cpx(t) = f(x + t) + f(x - t) - f(x + 0) - f(x - 0). For every 0

d < it one has:

sN(f;x) - '-z(f(x + 0) + f(x - 0))

1

d

(X(t)

sin(Nt+ zt

dt + r, (X,6).,

IO

with lime N(x,5) = 0. N-

(0.11.6)

Furthermore, if f is continuous on I = ]a,b[, the convergence in (0.11.6) is uniform for x belonging to a compact set contained in I.

THEOREM: (Jordan-Dirichlet): If f is of bounded variation on every compact interval contained in an open interval I, its Fourier series converges at every point x of I to 2(f(x + 0) + f(x - 0)). Furthermore, if f is continuous on I, its Fourier series converges uniformly towards f on every compact set of I. If f is of bounded variation on [0,2n] there exists a constant A such that IsN(f;x)I < A for every integer N >, 0 and for all

x e]R.

OF THE COURSE

35

THEOREM: If one of the functions f or g is of bounded variation

on [0,27E], then:

1r2n 2n

f(x)g(x)dx =

1

f(n)g(-n)

n=-W

0

.(n)(2neinxg(x)dx.

=

E

n=-0

0

In other words: In order to integrate f with respect to g(x)dx one can integrate its Fourier series term by term. THEOREM: (Fejer): If f(x + 0) and f(x - 0) exist at a point x, then lima N(f;x) = j(f(x + 0) + f(x - 0)).

N Whenever f is continuous on a compact interval, one has aN(f;x)-> f(x) uniformly on this interval.

The Fejer kernels form an approximate identity, so that if

f e LP(T), 1 < p <

then aN(f) -> f in LP(a).

THEOREM: (Fejer-Lebesgue): 1imoN(f;x) = f(x)

(0.11.7)

at every point x such that

lim LJhlf(x + t) - f(x)ldt = 0. 0 h+0

In particular, Equation (0.11.7) is true for almost all x. THEOREM: (Plancherel): In order that f e L2('T) it is necessary and

sufficient that

CHAPTER 0: OUTLINE OF THE COURSE

36

E

If(n)I2

<

and in this case 27E

if(x)I2dx

2n0

=

L

I?(n)I2.

NOTE

In the statements of the exercises in the following chapter,

the functions considered are always complex valued when no further indication is given.

Similarly, unless otherwise indicated, the

sets considered are measurable sets of ]R

.

CHAPTER 1

Measurable Sets

Let E1,...,E n be a finite sequence of sets of fin-

EXERCISE 1.1: ite measure.

For every integer p (1 . p _< n) set

meas(E. P

(a):

Z1<

< p

1P

11

Show that n

meas(E1 U

U En) =

(-1)p-1aP

(POINCARE'S FORMULA).

1

p=1

(b):

For every integer s (1 s s < n) we denote by GS the

set of points which belong to exactly s of the sets E1,...,En Show that

n

meas(G) =

I

(

(-1)P-5

p=s

(c):

l

l s la l

11

.

p

Let HS be the set of points which belong to at least

s of the sets E1,-..,En.

Show that

37

CHAPTER 1:

38

n

(_1)p-s(s-11

I

meas(HS) =

P=S

Qp.

l

A0t = VA V = AVA = VAV = AV For every non-empty subset A of {l,...,n} let us set

SOLUTION:

EA = EA - U Ei.

EA = n Ei,

i+A

ieA

The EA are mutually disjoint and it is clear that

E

A

= U EB', BDA

(-1)po

(_1)CardAmeas(EA).

I

=

CardA=p

P

One has:

SOLUTION: (a):

n (-1)pa

p=1

(-1)CardA meas(EA) _

=

p

(-1)CardA

(_1)CardA.

I meas(EB) B

ACB

Note that if p = CardB, then

(_1)CardA

(_1) r(P 1

=

x

r=1

ACB

l

=

- 1.

J

From this it follows that

n (-1)Pa

p=1

meas(EB)

= p

B

meas(E1 U

L

BD A

A

A

UEn.

meas(EB)

MEASURABLE SETS One has:

SOLUTION: (b): n

39

(

l

(-1)p 8Ja p=s

p

l

(-1)CardA(CarsdA lmeas(EA)

=

I

CardA3s

J

(-1)CardA(CardA 1

Card4 s

s

l meas(EB)

E

CardB;s

I

meas(E')

BDA

J

B

(-1)CardA(CardA)

ACB

J

ll

CardA=s On setting CardB = p , s again, one has:

C

ACB

( -1)CardA [ CardA) l s J

( -1)r (r 1 (p )

=

ls)lr

r=S

CardA3s

=

l

s ) I ( -1)r (r - s ) P-8

(-1)s

s)

(-1)rlprs1

0

ifp> s,

(-1)S

if p = a.

From this it results that: n

SOLUTION: (c):

s

meas(E') = (-1)smeas(G ).

= (-1)S

(-l)PI p)Q

p=s

,

CardB=s

s

s

The formula is true for s = 1 (for this is none

other than that obtained in (a)). HSt1 = HS - GS,

Also,

CHAPTER 1:

40

whence, proceeding by induction and taking account of (b),

n

(-i)P-s-1 j(

meas(Hs+l) = I

l

lsJ

p=s n

-1 [P-111ap

(-1)P-s-1(p $ 1 l

p=s+1

l

op.

J

For all c > 0 construct an open set U everywhere

EXERCISE 1.2:

dense in R, and such that meas(U) < e.

ovo = vov - ovo = vov = AVA

SOLUTION:

Let (rn) be the sequence of rational numbers, and for

all n let us denote by In the open interval with centre rn and length e2-n.

The union U of the In is an everywhere dense open

set of 3R (for it contains all the rational numbers) and furthermore,

meas(U) S

e2-n

1

= e.

n=1

EXERCISE 1.3:

Let (En) be a sequence of measurable sets such

that

E meas(E n

<

Show that the set of points which belong to an infinity of En's has measure zero (The Borel-Cantelli Lemma). AVA = VAV = AVA = VAV = ova

FIRST SOLUTION:

The set considered is

MEASURABLE SETS

41

A n [UEJ p mead U E (pan

]

I

-<

p

pan

meas(E ), p

and consequently meas(A) = lim mead U Epn

J

(pan lim

I

n pan

meas(E ) = 0. P

Let cpn be the characteristic function of En.

SECOND SOLUTION:

By virtue of the theorem on the term by term integration of series of positive functions, one has

i cn = I

Jcpn = I meas(En) < -.

J

From this it follows that the set of points where

Pn = W has

measure zero.

But this set if precisely A.

EXERCISE 1.4:

Let (En) be a sequence of measurable sets such

that

G

meas(En) <

n For every integer s, H

is denoted as the set of points which s

belong to at least s of the sets En.

meas(Hs) 4

G meas(En). n s

Show that

CHAPTER 1:

42

First, suppose we had a finite sequence E1,...,

FIRST SOLUTION:

En and for every non-empty subset A of (1,...,n) let us denote by EA the set of points belonging to the Ei for which i e A and not

belonging to those for which i4 A (in Exercise 1).

smeas(Hs) =

Then

smeas(EA)

E

CardA>,s

n

CardAmeas(E') _ A

CardA>,1

meas(E.). i=1

1

In the general case one considers, for n > s, the sets Hs n formed by the points which belong to at least s of the sets E1,.. -,nIt is clear that Hs

n C Hs n+l

and that

Hs = U Hs,n.

n=s Consequently

meas(H ) = Jim meas(H ) n s,n S

n Jim

JH S

EXERCISE 1.5:

sets one sets

For every finite sequence E1,...,En of measurable

MEASURABLE SETS

43

n

n

l

D (E1,...,En) =

E1l

Eil -

I F]

,

J

and if the E. are not all negligeable,

meas(D(E1,...,En) a(E1,...,En) =

meas

Show that if none of the E. is negligeable one has

a(E1,...,E1) E

n

1

1

1

i. 1 the number

ap(x) = u(2p-1 x)

is the p-th term in the binary development of x e [0,1[.

Whatever may be the numbers el,...,ep (ei = 0 or 1), the set of x's such that 0 4 x < 1 and al(x) = ell

has measure 2-P.

ap(x) = cp,

From this it follows that the set Bp of the x's

such that 0 4 X < 1 and

MEASURABLE SETS

49

a1(x) = 0,

a3(x) = 0,

. ,

a 2p-1 (x) = 0,

has as its measure 2-2p+1 x

2p-1

= 2-p.

Since B is the intersection of the Bp one certainly has meas(B) =0.

SOLUTION: (b):

The set A' = 2B is also negligeable.

Now A' is

the set of numbers of the type

Y

a 2-(2p+1)

p

p=0

a p = 0 or 1.

As every real number may be written

n+

ap=0or1,

ap2p,

p

it follows that ]R = A + A'.

EXERCISE 1.10:

Let f be a complex measurable function on ]R such

that f(x + 1) = f(x) for almost all x.

Show that there exists a

function g such that f = g almost everywhere and g(x + 1) = g(x) for all x.

AVA = DA4 = ADA = DAD = ADA

Let E = {x,f(x) + f(x + 1)} and let us set

SOLUTION:

+co

(E t n),

F= n=-co

as well as

CHAPTER 1:

50

- f(x) if xeF. It is clear that g has all the properties desired.

EXERCISE 1.11:

Let A be a bounded set of ]R

Sip = meas{x: I Ix II

and let

r 1},

where lixll denotes the Euclidian norm of x. Let (xi)i>.1 be a sequence of points of A; we set

do = inf{Ilxi - xi II1 1 4 i < ,j 4 n}. Prove that

liminfndP n n <

1 meas(A)

ap

Qp

where (

p

1

cp = 2 -P{ 1 + p10 ( 1 + t l

dt}

For this one will consider a number y such that ndP > y for n n sufficiently large, as well as the balls B. with center xi and radius ri, where

if 1i y1/Pn - 1/P whenever n is large enough, and

yl/pn-1/P.

ri t rj =

n, one has

When 1 4 i < j E 2pn and j

- 1/p, IIxi - xi II > dj > yl/Pj and we also have

ri + r.

yl/Pn- 1/P

zyl/P(2j-1/p

+

- n-1/P)

=

yl/Pj -1/p.

From this it follows that the balls Bi are mutually disjoint. If one sets A(e) = {x:d(x,A) .< e), one will then have

urn - 1 2

LL

n

Pn (2i-1/P + 2C

- n 1/P)Pl

< meas(A(e )), n

J

i==n

where

e

n

=

2yl/Pn - 1/p

We have 2Pn 1 (2i-'/P -

lnm

i=n

n-1/p)P

Pn [2rnl -1/p - 1]p = 2C = lim n i==n

Il

J

(Contd)

CHAPTER 1:

52

r2P (2,-l /P - 1)Pdx

(Contd) JI

1

=

(1 - t)P

1tt

p1

dt,

(the last integral is obtained by setting x = (1 t t)P).

On the

other hand, the sets A(En) are decreasing and have A as intersection; furthermore, as A is bounded they are of finite measure. From this it follows that

meas(A) 1

c

St

p

P

and consequently that

liminfndp 4

n

n

EXERCISE 1.12: = 1.

1 meas(A)

c

P

SZ

P

Let X be a measurable set of ]RS such that meas(X)

Let u be a bijection of X onto itself such,that for every

subset E of X, E is measurable if and only if u(E) is measurable, and then meas(E) = meas(u(E)).

Furthermore, assume that if N is

a measurable subset of X and if u(x)e N for almost all the points of N, then N or X - N is negligeable. Let E be a measurable subset of X such that meas(E) > 0, and

if x e X if uP(x)$ E for all p > 1, inf(p:p >, 1,uP(x)e E}

(one sets u1 = u, uPt1 = u0up).

Show that

otherwise

MEASURABLE SETS

53

1 f n(x)dx=1. E

A00 = 0A0 = A0A = VAV = A00

SOLUTION:

E

n

For 1 .< n 6 W let us set

= {x:x e E and n(x) = n}.

Let us also set GO = E,

n

u_P(E),

Gn = E - U

n . 1,

p=1

where u -p = (U-1 )P).

It is clear that

14n 1 one has n-1 un(Gn)

=

un(E)

- U

up(E).

P=O From this it results that the sets un(G ), n 3 0, are mutually n

disjoint, and that

y = l_J un(G ) n=0

= lJ un(E). n=0

Since u(y) C -y and meas(y) > meas(E) > 0, the second hypothesis

54

CHAPTER 1:

made on u implies that meas(y) = 1; in other words, since u preserves the measure, m

Go

1 =

meas(G.). E E measun(G ) = n=O n=0 n n

Let us note that the Gn are decreasing; consequently meas(E ) = lim meas(G ) = 0.

n

n

Thus one has

N n(x)dx =

E

n=1

nmeas(E ) = lim n

N

X

n=1

N

(N-1

= liml I

N- In=O

(n + 1)meas(G )

n

rN-1

-

I

n=O

l

nmeas(G )J

l

= liml I meas(Gn) - Nmeas(GN)J N n=O 1

It remains to be observed that

meas(Gn) = 1

L

and

meas(Gn) 3 meas(Gn+1)

implies Nmeas(GN) -} 0.

Thus

I n(x)dx = 1. J

meas(G )} n

n{meas(G n-1

E

REMARK: We have used the following classical result:

n

MEASURABLE SETS If un

55

> un+l >

0 and G Un < m, then nun -> 0.

We shall briefly recall the proof: let c > 0 and let p be such that

(n - p)un S up+1 + ... + Un S E

for all n

p; then

liminfnun s E. QED n

EXERCISE 1.13:

One says that a set A of]Rp is ALMOST OPEN if

almost all the points of A are interior points of A. Let f be a real function defined on an open set U of Iltp.

Prove that the following conditions are equivalent: (a): f is continuous at almost all the points of U;

(b) :

For all a e]R the sets (f > a) and (f < a) are almost open sets. AVA = VAV = AVA = V0V = AVA

SOLUTION: (a):

Let E be a negligeable set in U such that f is

continuous at every point of -U - E.

If x e (f > a) - E one' will

have f(y) > a for all the points y of a neighbourhood of x, therefore x is interior to (f > a).

This set is therefore almost open.

One argues similarly for (f < a). (b):

f is continuous at x if for every rational number

r < f(x), x is interior to (f > r), and if, for every rational number s > f(x) x is interior to (f > s).

If r is rational, let

us denote by Ar (reap. Br) the set of points of (f > r) (reap. (f < r)) not interior to this set.

If these sets are neglige-

able then so is their union, which, by the preceding, contains the set of points of discontinuity of f.

56

CHAPTER 1:

EXERCISE 1.14:

One says that a bounded real function f defined

on ]R is ALMOST EVERYWHERE CONTINUOUS if the set of its points of discontinuity is negligeable. (a):

Give an example of a function that is almost everywhere

continuous and such that there exists no continuous function coinciding with it almost everywhere. (b):

Show that in order for a bounded real function f to be

almost everywhere equal to an almost everywhere continuous function, it is necessary and sufficient that there exists a set A of R such that ]R - A is negligeable, and that the restriction of f to A is continuous. (c):

Deduce from (b) that f is measurable and that there ex-

ists a sequence of continuous functions fn which is convergent at every point of R and whose limit is almost everywhere equal to f. (d):

Show that a right-continuous function is continuous

except at the points of a set that is at most denumerable, and therefore is almost everywhere continuous. A0A = V AV = A0A = 0M4 = AVI

SOLUTION: (a):

If f(x) = 0 for x < 0 and f(x) = 1 for x > 0, f

cannot coincide almost everywhere with a continuous function g, for with the complement of a negligeable set being everywhere dense in R, one would be able to find two sequences xi < 0 < yi tending to zero, and such that g(xi) = f(xi) = 0,

g(yi) = f(yi) = 1.

On passing to the limit one would have g(0) = 0 and g(O) = 1 at the same time, which is absurd. SOLUTION: (b):

The condition is evidently necessary.

show that it is sufficient.

To do that let us set:

Let us

MEASURABLE SETS

57

g(x) = Iim{sup(f(y):y e A,Iy - xI < a)}. a-*0 a>0

This definition has.a meaning, for A is everywhere dense. one has f = g.

On A

Furthermore, if x e A and if c > 0 there exists

a > 0 such that f(x) - c .< f(y) . f(x) + c if yeA and I y -xI < a. Then if ix - x '

l y - x' I

l

< a one has

l y - x l

< a' = a - Ix - x' I

< a for all y e ll such that

.

From this it follows that g(x) - e .< g(x') .< g(x) + c,

which proves that g is continuous at each point of A. SOLUTION: (c):

For every x and all c > 0 there exists a > 0 such

that f(y) .< g(x) + c if y e A, ly - xI < a.

As above, from this one deduces that

g(x') . g(x) + e if Ix' - xl < a.

The function g is therefore bounded and upper

semi-continuous.

There then exists (cf., a course on Topology)

a sequence fn of continuous functions that converges everywhere towards g.

SOLUTION: (d):

Let us assume that f is right-continuous.

If A

is a non-empty set of 3t we shall denote by e(A) the diameter of A, that is to say the upper bound of the numbers la - bl for a e A,

b e A.

For all x e]R let us then set w(x) = inf{8(f(V)):V a neighbourhood of x},

58

CHAPTER 1:

(one says that w(x) is the OSCILLATION OF f AT x).

The set of

points of discontinuity of f is then:

{w> 0} = U JW >n} n=l

JJJJ

I t therefore suffices to prove that for all a > 0 the set A = {w > a} is denumerable.

By reason of the right-continuity of f,

for all x e A there exists a

x

> 0 such that

a(f(]x,x + ax[)) < a. But then:

]x,x+ax[(nA=0. Let us choose a rational number rx in ]x,x + ax [ .

If y e A and

x < y, one therefore has

rx < y < r y , which proves that x e A - rx is an injection of A into Q, and con sequently that A is denumerable.

EXERCISE 1.15: < -.

Let A be a measurable set of ]R such that meas(A)

Show that the function x y meas(A(1]-co,X]) is continuous.

AVA = V AV = AVA = V AV - AVl

SOLUTION:

If xn is a decreasing sequence, and tends to x, one

has

A(1]-co,x] = n {An]-.,x n]}, n

and consequently

MEASURABLE SETS

59

meas(Afl]--,x]) = 1im meas(Afl]-m,xn]) n If xn is a strictly increasing sequence that converges towards x then

Ar)]-.,x[ = U {Afl]-W,xn]}, n and consequently, because meas({x}) = 0, meas(A r)]-W,x]) = meas(Af)]-o,x[) = lim meas(A r)]-m,x ]),

n

n

which proves that the function x 1+ meas(Afl]-oo,x]) is continuous.

EXERCISE 1.16:

Let 0 < A < 1.

For any measurable sets A,B C[0,1]

of positive measure, do there exist 0 < x < y < 1 such that

meas(Ar)[x,y]) = Ameas(A), meas(Br)[x,y]) = Ameas(B)?

AVA = V AV = AVA = VAV = A VA

The answer is affirmative if and only if A = 1/n, where

SOLUTION:

n = 2,3,...

First of all let us assume that A = 1/n.

.

Since the

function f(x) = meas(A r) [O,x])

is continuous and increasing on [0,1], there exist points 0 =t t1 <

< to = 1 such that

f(ts) =

meas(A).

n

< 0

60

CHAPTER 1:

Under these conditions one has:

meas(A), meas(A 0 [ts,ts+1 ]) = n

0 .<

s F n - 1.

Furthermore, the n numbers meas(B n [ts,ts+1]) have meas(B) as

Therefore they cannot all be strictly less than or all

sum.

strictly greater than (1/n)meas(B).

Therefore there exists an s

(1 < s 6 n - 1) such that one of the numbers meas(BI)[ts-l' ts]) and meas(Bf) [ts'ts+l]) is less than (1/n)meas(B) and the other is greater.

The idea of the proof is to 'vary x continuously' from ts_1 to is and y from is to ts+l in such a way that f(y) - f(x) = meas(Af)[x,y]) always remains equal to (1/n)meas(A).

meas(B n [x,y] ) will then vary continuously between two values on

opposite sides of (1/n)meas(B), and must therefore take this value at least once.

The rigorous proof reduces to proving that in the rectangle is-1 <

x < ts,

is 6 y < ts+l the set E of points such that

meas(A)

f(y) = f(x) +

n is connected.

Let us note that if ts_1 < x < is one has

f(ts) = f(ts_1) +

meas(A) s f(x) +

meas(A)

n n

E f(x) + f(ts+1) - f(ts)

E f(ts+1). From this it follows that meas(A)}

EX = {y:ts 4 y 4 ts+l'f(y) = f(x) + n

MEASURABLE SETS

61

is a non-empty compact interval, and that E is the union of the Ex's (if EX is identified with {x} x Ex).

Note that E is compact;

let us assume that it is the disjoint union of two non-empty comHaving seen that EX is compact and connected,

pact sets E1,E2.

one has either EX C E1 or EX C E2.

In other words, the projec-

tions of E1 and E2 onto the x-axis are two non-empty compacts sets that are disjoint and have the union [ts_l,ts] which is absurd.

Let us now assume that A is not of the above form; then there exists an integer n such that

Let a,$ > 0 be two numbers such that

(n+1)a+ns= 1, and let us decompose [0,1] into 2n + 1 contiguous intervals, al-

ternatively of length a and B, the two extreme intervals having length a.

Let A be the union of intervals of length a, and B

that of intervals of length S.

If

meas(Afl [x,y]) = Ameas(A), then [x,y] contains at least one of the intervals of length S, as otherwise one would have

meas(Afl[x,y]) < a =

n + 1

meas(A) < Ameas(A).

But then,

meas(Bf)[x,y]) >, B =

meas(B) > Ameas(B). n

EXERCISE 1.17:

Let I be a compact interval of 3R such that meas(I)

> 0 and 0 < S < 1.

We shall say that the operation T(s) is carried

62

CHAPTER 1:

out on I if one subtracts from I the open interval having the same centre as I and of length Smeas(I).

More generally, if I is

a disjoint union of a finite number of compact intervals of nonzero lengths, to apply T(S) to I consists in carrying out this operation on each of the intervals forming I.

Now let (0n) be a sequence of real numbers 0 < Sn < 1; we shall denote by In the compact set obtained by successively carrying out the operations T(S1),T(S2),...,T(Sn)

starting from the interval

[0,1].

Show that

(a):

In+1 C In,

meas(In)

2)...(1 -

1)(1 -

n

and that every interval contained in In has a length less than 2-n

From this deduce that

(b):

K

R In n

is compact, non-empty, nowhere dense, has no isolated point, and that

meas(K) = lim(l n (c):

1

S) n

)(1 - a

Assume that Sn = 1/3 for all n.

Show that meas(K) = 0

and that X e[0,1] belongs to K if and only if it can be written in base three uniquely using only the digits 0 and 2.

From this

deduce that K is not countable.

(d):

If Bn = 1 -

al/n(n+l),

0 < a < 1, show that meas(X) = a

(which proves the existence in [0,1] of nowhere dense compact sets whose measure is arbitrarily close to 1).

MEASURABLE SETS (e):

63

Deduce from part (d) the existence in [0,1] of a se-

quence An of sets of the first category that are mutually disjoint and such that: (i): meas(An) = 2-n n

(ii): Kn = U A

is a nowhere dense compact set;

i=1

(iii): Every interval contiguous with n (that is to say, every connected component of the complement of Kn in [0,1]) contains a set in An+1 of measure greater than zero. From this dedua

A

that

U An n=1

is of first category, has measure one, and that its complement in [0,1] is a set of second category of measure zero.

(f) :

Let

E nU-0 A2nt1'

Show that for every interval I contained in [0,1] and of nonzero length there holds 0 < meas(E()I) < meas(I).

(g):

From part (f) above deduce the existence of Borel sets

E C IR such that for every interval I of non-zero length one has 0 < meas(E(lI) < meas(I). (h):

Can one have meas(E) < m for such sets?

Deduce from the preceding that there exist positive

functions that are Lebesgue integrable, but that are not limits almost everywhere of increasing sequences of positive step functions.

CHAPTER 1:

64

SOLUTION: (a):

It is clear that

In+l C In

and

meas(In) _ (1 - S1)...(1 - On).

Moreover, n is formed by 2n mutually disjoint intervals of equal lengths; when the latter property. SOLUTION: (b):

The set K is compact and non-empty by virtue of a

well known theorem in Topology.

If I is an interval contained in

K one has I C I for all n; by Question (a) one thus has meas(I) < 2-n and consequently meas(I) = 0.

In other words the interior

of K is empty, which is the definition of a nowhere dense compact set.

If x e K and e > 0, for large enough n one of the intervals

forming In will be contained in ]x - e,x + e[; for this it is sufficient that 2-n < e.

Now, the two endpoints of this interval

belong to K, which shows that x is not an isolated point of K. Lastly,

meas(K)= lim meas(In) = lim (1 -

Sn).

n

SOLUTION: (c):

In this case, one has: ( ln

meas(K) = liml3J

n

= 0.

Furthermore, I1 is equal to the set of x's which are written in base three as

x = O.ala2...an...'

with a1 = 0 or 2 (it will be noted that 1/3 = 0.0222 and that

1 = 0.222 ).

Similarly it is seen that x e In if and only if

ai = 0 or 2 for 1 i i < n.

From this one deduces that x e K if for

all i ai = 0 or 2; the expansion of x in this form is then unique. If with every set A C iN one associates xA =

0

MEASURABLE SETS

65

if i eA and ai = 2 if i4 A, a bijection between P(v) and K is realised; K is therefore not denumerable. a1/n(n+l),

SOLUTION: (d):

If Bn = 1 -

0 < a < 1, one has

a

meas(K) = lima n, n

with

1 _

1

n+1

(n

whence meas(K) = a. SOLUTION: (e):

By the preceeding it is seen that in every non-

empty open interval I there exists a nowhere dense compact set whose measure is imeas(I).

The let Al be a nowhere dense compact

set of [0,1] such that meas(A1) = z.

In each interval contiguous

to Al let us choose a nowhere dense compact set the measure of which is half that of this interval, and let us denote by A2 the union of these compact sets; A2 is of first category (for the set of intervals contiguous to a compact set is denumerable) and its measure is 1.

Furthermore, K2 = Al U A2 is closed; in fact,

if x is a limit point of K2 and does not belong to Al it belongs to an interval I contiguous to A

1

and is therefore a limit point

of IflA2, which is compact, whence x e A2.

On the other hand, it

is clear that [0,1] - K2 is dense in [0,1] - K1, which itself is dense in [0,1], which proves that K2 is nowhere dense.

Quite

generally, An+1 will be constructed by choosing in every interval contiguous to the compact set K

n

a nowhere dense compact set of

CHAPTER 1:

66

measure equal to half the length of this interval, and by taking the union of these compact sets. is of first category and that Kn+l

As above, it is seen that An+1 = Kn U An+1 is

Further-

closed.

more,

n

2-i

meas(Kn) _

= 1 - 2-n, i=1

whence:

z[1 - (1 - 2-n)] = 2-(n+1)

meas(An+l)

Finally, Condition (iii) is satisfied by construction.

A

If

U An n=1

this set is of first category, for it is the countable union of sets of first category, and

-

W

meas(A) =

2-n = 1.

1

n=1

Its complement is not of first category, by a theor m of Baire, and its measure is zero. SOLUTION: (f):

E

n

Let

A2n+1'

F

i A2n' l

If I is an interval of length greater than zero-contained in [0,1], its intersection with E or F is of measure greater than zero, since meas(E U F) = 1.

meas(If1E) > 0.

Let us assume, for example, that

Then ICE contains at least two points x < y.

Let n be such that these two points belong to Al U A2 U

U A2n+1'

MEASURABLE SETS

67

They therefore belong to K2n+1, and as this set is compact and nowhere dense there exists an interval J contained in [x,y] which is contiguous to it; one then has meas(If)F) > meas(JflA2n+2) > 0.

Since

meas(I) = meas(IfE) + meas(If)F), it follows from this.that meas(Ef)I)

< meas(I).

If E is the set studied in Question (f) above,

SOLUTION: (g): then

E= U (E + n) nea

answers the question.

By going back to the proof of Question (f)

again one sees that for E one can take the set

EN

N

n

A2n+1'

If

E = U (EINI + N), ne2Z

then for every interval I of positive length one has: 0 < meas(If1E) < meas(I), and furthermore, W

m

m

2-(2n+1) = 10

2-(2n+1) + 2

meas(E) _ n=0

N=1 n=N

9

CHAPTER 1:

68

SOLUTION: (h):

Let f be the characteristic function of the set

E defined above, and let cp be a positive step function such that rp 4 f almost everywhere.

If I is an interval of length greater

than zero on which 9 is equal to a constant, since meas(I - E) > 0 this constant is zero.

Thus cp = 0 almost everywhere.

From this

it follows in particular that f cannot be the limit almost everywhere of an increasing sequence of step functions. REMARK:

To prove the existence of a nowhere dense compact set of

[0,1], the measure of which is arbitrarily close to unity, one may also consider the set E = [0,1] - Q.

There holds

1 = meas(E) = sup{meas(K):K compact, K C E}, and every compact set contained in E is evidently nowhere dense.

EXERCISE 1.18:

Consider a double sequence (fm n) of measurable

complex functions on X = [0,1] such that for all m the sequence converges almost everywhere towards a function gm and

(fm

that the sequence (gm) converges almost everywhere towards a function h.

Show that there exist two sequences of strictly increasing inconverges n s' S Generalise this result to the case where

tegers (ms) and (ns) such that the sequence (fm almost everywhere to h.

X =P. (orEP).

wA=vw=w0=0AV =MA SOLUTION: set E C X.

First assume that every convergence is uniform on a Then there exists ml < m2 <

such that on E there holds

h - g ms Thus

0,

and, on the other hand, one has limgm = h

uniformly on E.,

Iimfm,n

uniformly on En if m > 1.

= gm

On setting

E=nm m=O

the desired result is obtained.

Let us now use an argument known as the "diagonal process". By applying the preceding result to e = 1 one first determines a measurable set E1 and two strictly increasing mappings T1,-5 1 of

N* into itself, such that meas(X - E1) < 1 and

f91(n),O1(n) ' h

on E1.

Now applying the same result to e = Z and the double sequence

f

(m) 0 (n) one obtains a measurable set E2 and two strictly in1

creasing mappings 92102 of 1V* into itself such that meas(X - E2) <

I and

CHAPTER 1:

70

f1Plo(P 2(n),91082(n) -

h

on E2.

Proceeding thus repeatedly, one obtains finally a sequence (Er) of measurable sets and two sequences (pr) and (0r) of strictly increasing mappings of]N* into itself, such that meas(X - E ) < r 1/r and

fY10...oq) r(n),O1o...o8r(n) -> h

on Er.

Let us then set

E = U r, r mS = 910...op (a),

ns = 010...o0s(a).

S

The set E is measurable and meas(X - E) = 0. every r 3 1 the sequence (fm

Furthermore, for

)s-r is a subsequence of the se--

s,n

s

From this it follows

quence that

fm in s s

-> h

on E.

It will be noted that in order to use Egoroff's Theorem it suffices to assume that X C 1R

and meas(X) < -.

By writing ]R

as

the union of a sequence of such sets, a new application of the diagonal process allows this result to be generalised to the case

where X = ]R

.

EXERCISE 1.19:

sets of]R set K of ]R

A mapping t -- F(t) ofIR into the set FP of closed

is called a MEASURABLE MAPPING if for every compact the set {t:F(t)f1K 4 01 is measurable.

MEASURABLE SETS

71

(a):

Show that {t:F(t) = O} is measurable.

(b):

Show that t -; F(t) is measurable if and only if

{t:F(t)f1 B * 0} is measurable for every open ball B of ]RP, or

again if t - F(t)1K is measurable for every compact set K of ]R Show that if t -

(c):

.

F1(t) and t - F2(t) are measurable

mappings of]R into the closed sets of)R

and]Rq respectively,

then t - F1(t)x F2(t) is measurable. Show that if t - K(t) is a measurable mapping of ]R into

(d):

the compact sets of ]R , and if f is a continuous mapping of IlRP into 3R q, then t - f(K(t)) is measurable.

Show that if t -> Fn(t) are measurable, then

(e):

t - n Fn(t) n is also measurable. (f):

Show that t - F(t) and t -

of Et into the sets of ]R

K(t) are measurable mappings

which are respectively closed and compact,

then t -; F(t) + K(t) is measurable.

1V = VV =Ova=V V =AVA SOLUTION: (a): n.

Let Bn be the closed ball with center 0 and radius

We have: Co

{t:F(t) = O} =]R - U {t:F(t)f1Bn * 0}. n=1

SOLUTION: (b):

If t -; F(t) is measurable and if V is an open set

of]RP, there exists a sequence of compact sets Kn of which V is the union, and:

CHAPTER 1:

72

{t:F(t) n V 4 0) = U {t:F(t) n n # 0}. n=1 If now {t:F(t)n B 4 0} is measurable for every open ball B, as every open set V is the countable union of such balls, the set {t:F(t)n V 4 0) is measurable.

If K is compact, let us consider

the open sets Vn = {x:d(x,X) < 1/n} where d(x,K) denotes the distance from x to K for a norm on]R .

One has:

{t:F(t)f1K 4 0) = U {t:F(t)f1Vn 4 o}. n=1

In fact it is clear that the first set is contained in the second; moreover, if for all n there exists xn a F(t) n Vn one can find a

subsequence x

n

such that x .

It is clear that x e K (as

-> X. .

K) = 0), and that x e F(t) also, for F(t) is

d(x,K) = linrl(x Z

n

1

closed.

If t -> F(t) is measurable and K is compact, t - F(t)n K is

evidently measurable (this is true, moreover, if K is closed). If t + F(t)n K is now measurable for every compact set K, by considering afresh the balls Bn (cf., Question (a) above), we have

{t:F(t) n K # 0} = U {t:(F(t)f1K)n Bn # 0}, n=1 which proves that t SOLUTION: (c):

F(t) is measurable.

Note that we have not specified the norm on ]R P.

If we choose the norm II(xl,...,x P of ]Rp ]R

)II

= Maxlxil, every open ball

is of the form B1 xB2, where B1 and B2 are open balls of

and itq respectively.

Then:

{t:(F1(t)xF2(t))n(B1xB2) 4 0} = n {t:Fi(t)nB. 4 O}. i=1,2

MEASURABLE SETS SOLUTION: (d):

73

For every open set V of Rp we have:

{t:f(K(t))nv # f } = {t:K(t)nf 1(V) # 0}. SOLUTION: (e):

Consider first the case of two measurable mappings

t - F1(t) and t - F2(t).

Let us denote the 'diagonal' of RpxRq

For every compact set K of Rp,

by A.

{t:F1(t)nF2(t)nK = 0} = {t:(F1(t) x (F2(t)nK))nt + f}. By what has gone before, t + F1(t)x (F2(t)n K) is measurable, and on the other hand 0 is a countable union of compact sets.

that t -} F1(t)n F2(t) is measurable.

the result for a finite intersection.

This shows

By recurrence one obtains In the general case one

writes n

{t: ( n F (t))nK + y} = n {t:(n Fr(t))nK + 0}, n=1

n=1

r=1

an equality that results because the decreasing compact sets

n

n (Fr (t)nK) r=1

have a non-empty intersection-if and only if each of them is nonempty.

SOLUTION: (f):

It is known (see a course on Topology) that

F(t) + K(t) is closed if F(t) is closed and K(t) is compact. Moreover, if u denotes the mapping (x,y) -; x + y of Rp x Rq into

Rp, we have u(F(t)x K(t)) = F(t) + K(t) By the preceding, t -r F(t) x K(t) is measurable, and it is seen,

as in Question (d) above, that u(F(t)x K(t)) is also.

CHAPTER 1:

74

Let Fp be the set of closed sets of R

EXERCISE 1.20:

set of non-empty compact sets of i2P, x ->

l

ix1

1

,

KP the

a norm on ]R

and

,

for every non-empty set F of Fp d(F) = Min{ IjxII:x a F}. For F E Fp write:

(a):

cp0(F) _

{x:x eF, jjxjj = d(F)}

if F # 0,

{0}

if F = 0.

Show that t u cp0(F(t)) is measurable if t - F(t) is measurable

(for the definition of the measurability of a one-parameter family of closed sets see the preceding Exercise).

Let ei(x) =X. if x = (x1,. .. ,xp) e:IR

(b) :

.

For K e K0 set

e.(K) = Min{e.(x):x a K} and cpi(K) = {x:x a K,ei(x) = ei(K)}.

Show that t -

(K(t)) is measurable if t - K(t)E K0 is measurP

able.

From this deduce that there exists a mapping 9:F

(c):

->iRp

P

such that:

(i) : ip(F) e F if F e Fp and F $ 0; (ii): t - q(F(t)) is measurable if t y F(t) is.

Let f:jRp +R be continuous.

(d):

a 2Rq x F

->

P

Show that there exists

PP such that :

(i) : If F e F and x e f(F), then a(x,F) e F and f(a(x,F)) = x; P

(ii): If t H x(t)e]i

,

t i+ F(t)e Fp are measurable, then

t * a(x(t),F(t)) is.

(e):

Assume that the mapping t f* Fi(t)e FP (i = 1,2) and

MEASURABLE SETS

75

t -> x(t) eF1(t) + F2(t) are measurable. Show that there exist t -> xi(t)e Fi(t) that are measurable and

such that x(t) = x1(t) + x2(t) for all t. A0A = V AV = A0A = 0A0 = A0A

SOLUTION: (a):

Let K be a non-empty compact set of ItP.

The set

{t:g0(F(t))nK # 0} is equal to

{t:F(t)f1K # 0,d(F(t)) = d(F(t)f1K)}

(*)

when 0 4K, otherwise it would be necessary to add {t:F(t) = O}. The latter set is measurable (cf., Exercise 1.19(a)).

Since the

set {t:F(t)n K + 0} is measurable by definition, and as

K

is measurable it suffices to prove that t -} d(F(t)) is measurable on its defining set.

Now, for all a 3 0

{t:d(P(t)) < a) = {t:F(t)n Ba # 0},

where B

a

SOLUTION:

denotes the closed ball with centre 0 and radius a.

(b):

Similarly, on the measurable set {t:K(t)n K # 0}

one has:

{t:ei(K(t))n K # 0} = {t:ei(K(t)) = ei(K(t)f1K)},

and one sees, as above, that t - e .(K(t)) is measurable (by note

ing that {t:K(t)n F # y} is measurable for F closed). SOLUTION: (c):

1op0)(F) is reFor every F e F the set (cp P P It is clear that F -> q(F) has all the re-

duced to a point p(F). quired properties.

76

CHAPTER 1:

SOLUTION: (d):

Let us, for x ERq and F e Fp, write:

a(x,F) = p(f 1(x)t F). It is clear that Condition (i) is satisfied.

To prove that (ii)

is also satisfied it suffices to show that t -> f 1(x(t))r)F(t) is

Now, for every compact set K of Rp

measurable.

{t:f 1(x(t))r)K 4 O} = {t:x(t)e f(K)}, and f(K) is compact.

SOLUTION: (e) :

Let f 2R x]R -> Rp be defined by (x,y) -> x + y,

and let a be defined as above. Let us set a(x(t),F1(t)x F2(t)) = (x1(t),x2(t)).

Then x1(t) and x2(t) satisfy the properties required.

EXERCISE 1.21:

In this Exercise it is proposed, by consideration

of the Axiom of Choice, to prove the existence of non-measurable The Axiom of Choice appears in the following form:

sets of R.

Given a non-empty family (B ) of mutually disjoint non-empty sets i

of ]R there exists a set E of 3R which contains one and only one

point of each Bi. (a):

Show that there exists a set E C [0,1] such that for

every x eR there exists an unique y e E such that x - y is rational. (b):

Let S be the union of the sets E + r, where r runs over

the set of rational numbers lying between -1 and 1. Show that [0,1] C S C [-1,2],

and that if r,s are two distinct rational numbers, then E + r and

MEASURABLE SETS

77

E + s are disjoint. (c):

Deduce from this that E is not measurable.

ovo = vov = ovo = vov = ovo

SOLUTION: (a):

Let Q be the set of rational numbers, and if x,y

are real numbers let us express that x - y eQ by writing x ti y. This defines an equivalence relation on]R which all the equivalence classes intersect [0,1].

The existence of E follows from

this and the axiom of choice. SOLUTION: (b):

It is clear that

E + r C [0,1] + [-1,1] = [-1,2], and therefore S C [-1,2].

On the other hand, if 0 _< x s 1 there

exists y e E and re Q such that x = y + r.

1 which proves that [0,1] C S.

One has Iri = Ix - yJ

Lastly, if r # s one has (E + r)

fl (E + s) _ 0, otherwise there would exist z eIIZ and y1 a E, Y2 e E such that

z =y1+r=y2+s, and consequently y1 4 y2, which contradicts z being equivalent to a single element of E. SOLUTION: (c):

If E were measurable S would also be measurable,

and

meas(S) =

I

meas(E + r).

reQfl [-1,1] Now, all the numbers meas(E + r) are equal to meas(E), so that meas(S) = 0 if meas(E) = 0, and meas(S) = - if meas(E) > 0.

This

is absurd, because by virtue of the inclusions proved in (b) one would have to have 1 4 meas(S) 4 3.

CHAPTER 2

v-Algebras and Positive Measures

EXERCISE 2.22:

Let (X,C) be a measurable space, (xi) a sequence

of points of X, and (mi) a sequence of real numbers mi > 0.

For

every set E e C set: V(E) =

I

m..

x.eE i

(a):

Show that u is a measure on C.

(b):

Show that if {xi}e C for all i then one has C,1 = P(X),

and conversely.

AVA = VAV = AVA = VAV = AVA

SOLUTION: (a):

It is clear that u(O) = 0 (as usual, the conven-

m = 0). Let (E ) be asequence of mutuE i n ieo For every inally disjoint sets of C, and let E be their union. tion is adopted that

teger N one evidently has: N

11 (E1U...UEN) =

1 n=1

(E),

ir(En)

79

80

CHAPTER 2: a-ALGEBRAS

whence

u(En)

E

u(E).

n=1 Moreover, for every finite set A of 1N one has

E

m

X eE

= 1

i

m. 4

I

G

n=1 x.eE

n

i

1

1 n=1

u(En

ieA

ieA

whence M

u(E) = sup

X

m. <

G

1

n=1

A x ieE

U(E ) .< u(E). n

ieA

SOLUTION: (b):

A set E C X is measurable if and only if there

exist A,B e C such that A C E C B and xi $ B -A for all i.

If for

an index i one has {xi} $ C then {x.} is not u-measurable, because A can only be the empty set and so B D {xi} implies that xi e B -A On the other hand, if {xi} e C for all i and if E is an ar-

= B.

bitrary set of X, let

A = {xi:xi a E},

C = {xi:xi * E}.

Then A and B = X - C belong to C, A C E C B and {xi} $ B - A for all i.

From this it follows that E is u-measurable.

EXERCISE 2.23:

Let a be the set of positive, nul or negative

integers. (a):

Show that the set of subsets.A of a such that for

every integer n 3 1 one has 2n e A if and only if 2n + 1e A is a a-algebra.

AND POSITIVE MEASURES

81

Show that the mapping f of a into itself defined by

(b):

f(n) = n + 2 is a measurable bijection, but that f_1 is not meas-

urable. AVA = DAD = AVA = VAV = ADA

SOLUTION: (a):

It is clear that 0 possesses the property, and

that the property is preserved under complementation.

If the Ai

have the property for i = 1,2,..., it is impossible that for an n > 1 one of the two integers 2n,2n + 1 belongs to one of the A. and that the other belongs to none of them.

This shows that the

union of the Ails also possesses the property. SOLUTION: (b):

It is clear that f is a bijection.

Furthermore,

if A has the property and if n 3 1 one has 2n e fl(A) if and only if 2(n + 1)e A, hence if and only if 2(n + 1) + 1 e A, that is to say, if and only if 2n + 1 e f 1(A). urable.

This proves that f is meas-

Finally note that A = {0} has the property, but that

f(A) = {2} does not, since 2e f(A) and 3 * f(A), so f_1 is not measurable.

EXERCISE 2.24:

Let C be a family of subsets of a set X.

If

M C X, set: CM = (Mf)E;E a C}. (a):

Show that if C is a a-algebra on X, CM is a a-algebra

on M (CM is called the a-algebra INDUCED on M by C). (b):

If Me C give a simple characterisation of CM.

(c):

If C is generated by a family A of subsets of X, show

that CM is generated by AM. (d):

Deduce from part (c) that if M is a subset of a topo-

CHAPTER 2: a-ALGEBRAS

82

logical space X, the Borel a-algebra associated with the topology induced by X on M is equal to the a-algebra induced on M by the Borel a-algebra of X.

Consider in particular the case where M is

a Borel set.

AVA = VAV = AVA = VtV = tVt

SOLUTION: (a):

0 = MflO, and:

M - (MnE) = Mfl (X - E),

(*)

U (MnEn) = Mfl(U En), n n which shows that CM is a a-algebra on M.

SOLUTION: (b) :

If M e C, then

CM = {E:E e C and E C M}.

SOLUTION: (c):

Since AM C CM, C(AM) C CM.

subsets E C X such that MflEe C(AM).

Let C

be the set of 0

Evidently one has 0 e Coy

and equalities (*) and (**) show that C0 is a a-algebra on X. Since A C CO, one has C C CO, which proves that CM C C(AM).

SOLUTION: (d):

This follows from TM being the topology induced

on M by the topology T of X.

If M is a Borel set, the Borel sets of M can be interpreted either as the Borel sets of X contained in M or as the Borel sets of the topological

EXERCISE 2.25:

ability on C.

subspace M of X.

Let C be a a-algebra on X and let p be a probLet M C X be such that E e C and E D M implies

p(E) = 1. Show that a probability 11

M

is defined on CM by setting

83

AND POSITIVE MEASURES

PM(MnE) = u(E) for all E e C (the induced a-algebra CM was defined in the preceding Exercise).

Let us first prove that if E e C, F e C, and Mr) E = Mf1 F, then u(E) = u(F). Now, in this case one has SOLUTION:

(X - E) U (EnF) > M,

(X - F)U(EfF) > M, whence

1-u(E)+u(EnF)=1, 1 - u(F) + u(E n F) = 1,

and consequently u(E) = u(F). thus defined unambiguously.

The mapping

uM It is clear that:

of CM into [0,1] is

PM(0) = uM(Mflf6) = i(f) = 0, um(M) = PM(Mf1X) = u(X) = 1.

Moreover, if E e C, F e C, and (MUE) fl (MUF) = Mn(EUF) _ 0,

one has:

u(EnF) = 0, whence

uM{(MnE)U(MnF)} = uM{Mn(EUF)} = U(EUF) _

(Contd)

84

CHAPTER 2: a-ALGEBRAS = p(E) + p(F)

(Contd)

= 1M(MfE) + uM(Mf)F).

Hence

is additive. It remains to prove that it is continuous. uM To do this let us consider a sequence (En) of elements of C such that

Mfl En C Mf1En+1

Let us set Fn = El U M(lEn = MnFn,

UEn, so that

Fn Fn+1'

If F is the union of the F 's, then F e C and n

Iimll (MfFn) = 1imu(Fn) = u(F) = uM(MnF)

= M(U (MnFn)). n

EXERCISE 2.26:

Let X be a non-empty set.

Show that the a-alge-

bra generated by the sets {x}, x e X consists of the sets E C X such that E or X - E is countable.

Prove that a positive measure

is defined on this a-algebra by setting p(E) = 0 or 1 according as E is countable or not.

AVA = VtV = 1Vt = VAT = AVA

SOLUTION:

It is clear that the a-algebra generated by the {x},

x e X contains all the sets indicated.

these sets form a a-algebra.

It remains to prove that

The family of them is closed under

complementation and contains the empty set.

If (En) is a sequence

of such sets their union is countable if all of them are; if the

AND POSITIVE MEASURES

85

complement of one of them is countable the complement of their union will be, a fortiori.

This proves the first part of the

Exercise.

Let us now note that if the En are mutually disjoint there can be at most one of them that is not countable.

The countable

additivity of it follows from this.

EXERCISE 2.27:

Let N be the set of natural number, P(N) the a-

algebra of all subsets of N.

For every natural number n denote

by nN the set of multiples of n. Show that there cannot exist a probability p on P(ri) such that for every integer n , 1 there holds:

u(nO = n1 . d0A

SOLUTION:

0L0

AVA

V AV

LVL

Since {0} C n N for all n , 1 one would have 1j({0})4 1/n,

and consequently p({0}) = 0.

Furthermore, since 0 is the unique

integer that is divisible by an infinite number of prime numbers, one would have:

{o}

r=1 i=r

{piNJ

where (pi) denotes the sequence of prime numbers.

From this it

would result that

0 = limp{ U pi3N r i=r 1

(*)

By virtue of an elementary property of arithmetic one has:

pi w n ... npi 3N = pi ...pi N 1

a

1

a

< ia)

CHAPTER 2: a-ALGEBRAS

86

Therefore one would have

u(p. Nn ... np. i) = 11

1

pi , ..pi

la

a

1

r 4 s Poincare's Formula (cf., Exercise 1.1) would give

If 1

l

s

u(U

pi1VJ

i=r

p1a r4i1 1 one would have

ll

ti-r

ll

((

i1VJ =

o[

s->-

i=r

piJ = 1,

which would contradict (*). Now for the proof of (**).

Let As be the set of non-zero in-

tegers that do not have prime factors greater than ps. s

iT

1

i=1 1 -

1

Pi

k,

=IT{E i=1

n=0 pi

keAs

Then:

AND POSITIVE MEASURES

87

whence:

s 1im

s

1 1

i=1 1 -

EXERCISE 2.28:

= lim

I

s- keA S

pi

_

I

k= .

k=1

Show that there does not exist a a-algebra having

a countably infinite number of elements.

AVA = VOV = AV1 = VAV = AVA

Let X be a set and let C be a countable family of sets

SOLUTION:

of X which is a a-algebra on X.

For all x e X the set of the E e C

such that x e E is countable; therefore the intersection Ex of

these sets belongs to C, and this is the smallest set of C that Since for all E e C the point x c (Ex - E) or x e ExflE

contains x.

one has either Ex - E = E{ or ExflE = Ex, that is to say ExflE = O In particular, for two arbitrary points x,y e X one

or Ex C E.

has ExflEy = 0 or Ex = Ey.

Let I be a countable set such that

{Ex}xeX - {EiIieI'

with E. # E

.

if i + j.

For every subset A C I,

7

EA = U B. e C, ieA

and EA $ EB if A # B. P(I) onto C.

It is clear that A ' EA is a bijection of

Then if I is finite C is finite, and if I is infin-

ite C is not countable.

CHAPTER 3

The Fundamental Theorems

Calculate

EXERCISE 3.29:

(1

xnlogxdx 0

for every integer n 3 0, and deduce from it the value of

1 10

x dx,

1

given that 1

1 n=1 n2

12 6

Av4 s vov ° AVA - vov - VA

SOLUTION:

By setting x = e-t one has:

Jixnlogxdx = - J-te-(nt1)tdt = 0 0

89

(Contd)

CHAPTER 2: THE

90 W

_-

(Contd)

1 2J (n + 1) 0

te-tdt = -

1 2 (n + 1)

Moreover, if 0 < x < 1, OD

logx

1-x

_

- xnlogx.

I n=0

As the functions x + -xnlogx are positive, one can integrate term

by term, which gives

J

l logx dx = O 1 x

C

(n +

n==O

EXERCISE 3.30: (a):

= - n2

1

-

1)2

Let a > 0.

6

For what values of s em is the

function x + xse-ax integrable on gt+?

Also calculate the value

of its integral with the aid of the r function. Show that for Re(s) > 1,

(b):

n n=1

-s

=rs

x-1 s x

J

0e -1

dx.

ADA = VAO = ADA = DAD = AVA

SOLUTION: (a):

Since

Ix se-axl = xRe(s)e-ax

it is clear that this function is integrable only when Re(s) > - 1, and that then:

rxse-axdx

a-(s+1)( xse-xdx = a-(s+l)r(s =

0

0

+ 1).

FUNDAMENTAL THEOREMS SOLUTION: (b):

xs-1

91

For x > 0

xs-1

= e -x

ex - 1

e-x =

1 -

xs-1e -nx

L

(*)

n=1

Since C

xRe(s)-le-nxdx

I

n=1

n=1 J0

0

F(Re(s))

00

E

n-Re(s) < w

n=1 if Re(s) > 1, equation (*) can be integrated term by term, giving

r- x s-1

ds =

EXERCISE 3.31:

C

n=1

0 es - 1

r xs-le-nxdx = r(s)

I n-s. n=1

0

For every integer n 3 0 calculate:

0 2

and from this deduce that for all z e c the function t

e-t coszt

is integrable on [0,Co], and calculate its integral. AVA

SOLUTION:

V AV

AVA

X04

By making the change of variable t - ti one has:

t2ne-t2dt =

0

J'o

000

otn-e-tdt

Jo

=

r(n +

) _

(Contd)

CHAPTER 3: THE

92

(Contd)

= z(n -

... ''r(Z)

'-z)

1.3.

(2n - 1) 2n+1

-

(2n)!

22n+1n

since r(z) = u. Therefore 2n

2

2 t2ne-t

e t coszt =

n=0

(-1)n (2n

,

!

and 2n G

(-1)n

J

n-0

0

T

2

Cw

2

t2ne-t

(2n)!

I

dt = n G n=O 2

Thus, 2n

°°

!t2ne-t

n (2n

e-t cosztdt n=O

2

dt

0

2 n

(z l

(-1)n

= 2 E

n4

'

n0 e

-z2/4

2

EXERCISE 3.32:

snax

J Oex-1

Establish the relation:

dx

12n

2n+1

a

n=1n+a 2 2

AVA = OAV = AVA = V1V = t1V4

n!

/Tr

FUNDAMENTAL THEOREMS SOLUTION:

93

If x > 0,

s in=

Co

=

ex - 1

e I's inax .

E

(*)

n=1

Furthermore, by using the inequality Isinul < lul, one has for

n

1

le nxsinaxldx < aJ xe-

J

0

0

rix

dx = a n2

so that:

J0ie-nx sinaxldx <

n=1

Equation (*) can therefore be integrated term by term, which gives the stated result, since

a

Joe-nxsinaxdx = 0

0

EXERCISE 3.33: c n=1

n2 + a 2

Find a method of calculating the series

(-1) n+1

n

by the integration of a series of integrable functions. A VA - VAV = tVO = VAV = tVo

SOLUTION:

Evidently there is a large number of possible methods.

But it is first necessary to reduce it to the calculation of an absolutely convergent series.

One can proceed as follows:

CHAPTER 3: THE

94

nl

( -1) n

n+1

1

W

n 0 [T -+1

Furthermore, for 0 4 x <

1

1

2n+

°C°

2,

=n

0

1

2n+1 2n+2

1

x2n+1

1+x

21og1-x and consequently

1 +x dx

21J1l0 0 gl-x

=

1

c

n=0 2n+1)(2n 7-2

Now, the integral is equal to

[(1 + x)log(1 + x) + (1 - x)log(1 - x)]X_ = 21og2,

so we have proved that: n+1 (-1)

n=1

REMARK:

= log2.

n

Recall that the most natural method is to apply Abel's

Theorem to the series development of log(1 t x).

EXERCISE 3.34:

Show that if a 3 0 then

s i n a x

f 0x(x + 1)

dx =

(1 - e-a). 2

(Consider the function of a defined by the left hand side). AVA = VAV = 40A = VAV = OVA

FUNDAMENTAL THEOREMS SOLUTION:

95

Let cp(a) be this left hand side.

cosax I

Since

1

x2 +1

x +1

the function (p is continuously differentiable on Ft

and

x2 W cosax dx.

(p, (a) _

fo

+1

Furthermore, if a > a0 > 0, then for A > 1:

xsinax

2

Aa0

JAx2+1

(by the second mean-value formula, since x(x2 +

1)-1

is decreas-

This proves the uniform convergence of the im-

ing for x z 1).

proper integral

xsinax 1

0x2+1

on [a0,=[.

dx

Consequently 9 is twice continuously differentiable

for a > 0, and f0mxsinax fW xsinax

Ox2+1

dx

rW sinax dx + 0

x

On [0,oo[ rp is therefore a solution of the system

9(0) = 0,

9'(0) = 2

(x2+1)

dx

96

CHAPTER 3: THE

(Indeed, as 9 coincides on ]0,-[ with a solution of gyp" - cp = -n/2

and as, furthermore, it is continuous on [0,co[, it satisfies this From this it follows that

differential equation on [0,oo[). (1-e-a).

cp(a)=2

EXERCISE 3.35:

For every integer n 3 1 and all real x, let 2e-2nx

fn(x) = e- T'x -

Show that the series with the general term fn(x) is convergent for all x > 0, and calculate its sum f(x).

Next, show that each

fn, as well as f, is integrable on (0,co), and compare

ff(x)dx

and

n1

J:ffl(X)d.X.

A0A = V AV = AVA - VAV = A00

SOLUTION:

L

n=1

For x > 0 one has:

f (x)= n

e

-x

-2

1 - e -x 1

ex-1 1

ex + 1 On the other hand,

_

e

1-

-2x

e-2x

2

e2x-1

FUNDAMENTAL THEOREMS

10fn(x)dx =

97

- 2 2n = 0,

n so that:

0.

n=1 JJ 0 However, x

Jm

dx

Oex +1

ex

[log

=

e

x

+1x=0

= log2.

J

Hence one must have:

00,

n=1

0

which is immediately verified by observing that Jm

1

olfn(x)Idx=nJol1-2uldu=n

EXERCISE 3.36: (a):

Let (fn)nal be the sequence of functions de-

fined on ]R by

if 04x4n

2

n x

fn(x)

=

- n21x -

n 0

if n s< x s n,

ifx>' 2n '

Calculate the four numbers: A = lim infJJJfn ,

n

A' = Ilim inff , n n JJJ

CHAPTER 1: THE

98

B' = Juim supfn.

B = lnm supJfn,

(b):

The same question for the sequence (gn)n>0 defined by:

gn = 810,4] if n is even,

otherwise.

gn = IL11 1]

(c):

If (hn) is a sequence of positive measurable functions,

what can be said of the four numbers A,A',B,B', and more particIs the Fatou Lemma true for sequences of

ularly about B and B'?

real measurable functions with arbitrary sign?

AV = 000 = 00A = VAV = 00A

SOLUTION: (a):

For all n one has

Jfn = 1,

so that A = B = 1.

On the other hand,

lim inffn = lnm supfn = 0,

and A' = B' = 0. SOLUTION: (b):

In particular, A' < A and B' < B. In this case Jgn is alternately equal to I and 4, Furthermore,

so that A = a, B = 4. lim infgn = 0

and

lim supfn

= Il[0,1]'

whence A' = 0, B' = 1. Thus A' < A < B < B'. SOLUTION: (c):

A' < A < B

There always holds: and

A' < B'.

(The inequality A' 4 A is Fatou's Lemma, the two others are clear)

99

FUNDAMENTAL THEOREMS

The two preceding examples prove that in general nothing can be said about the relative values of B and B'. However, one can extend Fatou's Lemma in the following way: let us assume that the fn's are real and measurable, and that there exists a measurable real function g such that: for all n,

9- 4f n

jg_ < . In fact 0 .< fn - g, and

One then has A' s A.

Jfn'

- °° < Jg <

therefore we can suppose that ig < W, and in this case

J (fn - g) = Jfn - jg'

The classical Fatou Lemma shows that

Ilim inf(fn - g) = Ilnm inffn - Jg n

.< lim inf Jfn -

1g,

n whence the result.

Jg

=

(Recall that if Jg- < - one sets

Jg_

-

which is an element of

g < h,

fg-

< -

and that

imply

Similarly one shows that if:

Jh_ < -

and

Jg < Jh).

100

CHAPTER 3: THE

fn < g

for all n,

jg+ < W, then

lnm supJfn

fuim supfn.

(Consider the functions g - fn and recall that

lnm inffn.

lnm sup( - fn) Here one again sets

if

Jg = Jg+ - Jg_

Jg+ < W,

and

g < h,

< M

jh

+

EXERCISE 3.37: (a):

imply

Jg+ < -

and

Jg

'C

Jh).

Show that if f is integrable on

,

and if

K is a compact set of this space, then

lim zI I-

(b):

lf(x)ldx = 0.

J K+z

Show that if f is uniformly continuous on]Rm, and that

if there exists p > 0 such that IfIp is integrable, then lim f(x) = 0. IIxII-)W

ovo = VAV = ovo - vov = ovo

FUNDAMENTAL THEOREMS

101

SOLUTION: (a):

If(x)Idx = 0,

liml

r-* since

0

fr

and

If(x)Idx < J

x+z

IfI

If S = sup{IIyII:y a K}

(Lebesgue's Theorem).

J

Ifrl

,

If(x)Idx, IIxIIIIxII-a

whence the result.

+ 0 as IIx II e > 0 and a sequence xn such that (b) :

SOLUTION:

Assume that

If(xn)I

There there exist

>. e

As the function f is uniformly continuous there exists a closed ball B, with centre 0 and radius greater than zero, such that s

If(y) - f(x)I <

2

if

y e B + x.

In particular, IfI > e/2 on the balls B + xn, and consequently P IfIP > (2) meas(B) > 0,

J

B+x

l

n

which contradicts the first part of the Exercise.

EXERCISE 3.38:

Let G be a continuous function on 3R such that

G(0) = 0 and G(x) > 0 if x # 0. Show that if f is an uniformly continuous bounded real func-

tion on 3t and

CHAPTER 3: THE

102

mG(f(x))dx < -,

j

then

1im f(x) = 0.

IIxlIH A0, = V AV = tot = VMv = LVA

If f(x)-J 0 as x

SOLUTION:

there would exist c > 0 and a

sequence xn such that

If(xn)I z E.

IlxnII - -,

As the function f is uniformly continuous there exists a closed ball B, with centre 0 and radius greater than zero, such that

I f(y) - f(x) I< 2 if y e B + x. In particular, if M is the upper bound of IfI on e, one would

have

24IfISM on the balls B + xn.

By virtue of the assumptions made about G,

one has inffG(u):

juI < M) = u > 0,

2 so that

JG(f(x))dx 3 umeas(B), B+x

n

which would contradict that G(f) is integrable (cf.,the preceding Exercise).

FUNDAMENTAL THEOREMS

103

Let f be an integrable function on ]R

EXERCISE 3.39: (a):

.

Show that

meas(IfI > a) = o11)

(b):

as a --* W.

Show that if f is measurable (and almost everywhere

finite) on F? then it is integrable if and only if

2nmeas(2n-1

< IfI i 2n)

X

100 = VAV = MMA = VAV = MMA

This results from

SOLUTION: (a):

I f(x)Idx = 0, a-J lim lfl >a and the inequality

ameas(IfI > a) E J

If(x)Idx. Ifl

Let us set

SOLUTION: (b):

An =

>a

(2n-1

< Ifl 6 2n),

B = (f = 0),

These sets are mutually disjoint, with union ]R ure zero.

J

R

whence

Then

IfI =

T

J" If

n=-W A

n

C=(IfI ,

and C has meas-

104

CHAPTER 3: THE +m

(

2

2nmeas(An)

n=-m

1

1A s

I

2nmeas(An),

n=-m.

IR

which proves that the condition is necessary and sufficient.

EXERCISE 3.40:

Let f be an integrable positive real function on

Pp. Show that there exist measurable sets An C Rp (n ea) with finite measure such that

f(x) _

n=-

for all x e]R

2')LA (x) n

.

AVA = DAD = AV4 = 4AV = DAD

Observe that if for all t elR+ and n e2z one sets:

SOLUTION:

an(t) = IL[z,11(2-(n+1)t - [2-

where [x] denotes the integral part of x, then

2nan(t).

t

n=(Note that an(t) = 0 whenever 2n > t; the an(t) are simply the digits of the binary expansion of t).

In particular,

2nan(f(x)).

f(x) _

n=-m If one sets A urable, and

= {x:x ep

and a (f(x)) = 1), then A

n is meas-

FUNDAMENTAL THEOREMS

f(x) =

105

2)LA (x).

.1

n

n= Also,

2nmeas(An) <

so A

n

fRpf(x)dx,

has finite measure.

EXERCISE 3.41:

Let X be a measurable set of SRp such that meas(X)

= 1.

Show that if f is integrable on X then

f(x)dx = 0 JX

if and only if

JIi + zf(x)Idx

1

for every complex number z. AVA = DAD = AVA = V AV = AVA

SOLUTION:

The condition is necessary, because

+ zf(x))dx= Ii + zl fdxI = 1.

+zf(x)ldx Jxl

X

Let usnow show that it is sufficient. written in the form:

11 + peiof(x)I - 1 dx 3 0, J

X

p

X

The inequality can be

106

CHAPTER 3: THE An easy calculation shows that:

where p > 0, 8 eat.

11 + pzI p

- 1 _ 2Re(z) + pIzI2

1+pz + 1

if p > 0, z e0, so that

lim

- 1 = Re(eiof(x)).

11 + peI f(x)I p

p-)-0

Moreover,

11+ p egs f(x) I- 1 I< I f(x) I. p

Therefore, by Lebesgue's Theorem,

Re(eloJ f(x)dx)

=

JRe(f(x))dx X

x

limJ

I1 + pe10f(x)I - 1

p->0 X

dx

p

>. 0,

and this holds for all e em.

el$Jxf(x)dx = -

By choosing 0 so that

Jf(x)dxl X

one obtains

f f(x)dx = 0.

JX

EXERCISE 3.42:

Let f be an integrable function on at and let a'>0.

Show that for almost all x eat the series

FUNDAMENTAL THEOREMS

107

+W

n=- f[.+nI

(*)

is absolutely convergent, and that its sum F(x) is periodic with period a and is integrable on (O,a).

tVA = VOV = AVA = VAT - OVA

SOLUTION:

Setting u =

+ n on obtains

a

I J:If l

+

n]

=a

I

dx

n=-w

C'fxI

=a

JI

.f (x) I dx

so that the series in (*) converges absolutely at almost all points of (O,a).

As this series does not change when x is replaced by

x + a, it follows from this that it is absolutely convergent at almost all points of ]R, and that its sum coincides with a function

F of period a (setting, for example, F(x) = 0 at the points x where the series is not absolutely convergent) that is integrable on

(0,a). Since one can integrate term by term, one obtains by proceeding as above: fa

i

F(x)dx =

J(x)dx.

0

EXERCISE 3.43:

Let f be an integrable function on Ir, and let a > 0.

Show that for almost all x e 1R:

limn af(nx) = 0.

n-*-

CHAPTER 3: THE

108

DOA - VAV = X00 = VAV = 400

SOLUTION:

t

r(

(x)Idx,

so

in-CL

n=1

f(nx) I dx <

m

I

From this it follows that for almost all x eat

I

n-ajf(nx)I

<

n=1

and in particular that

n-af(nx) - 0. EXERCISE 3.44:

Let f be a measurable complex function oniR with

period T > 0 and such that:

T

I f(x)l dx < m

A= J

(a):

0

Show that for almost all x,

limn-2f(nx) = 0.

n

(b):

From this deduce that for almost all x,

limlcosnxll/n = 1.

n

FUNDAMENTAL THEOREMS

109

MVA = VAV = AVA = VAV = MVA

SOLUTION: (a):

Set

ppn(x) = 2 f(nx). n Then:

1T

T

1 nT

Jk(x)dx = 2J If(nx)Idx = 3f 0 n 0 n 0

If(x)Idx

A

= n2 Hence

T Ipn(x)Idx <

n

0

The series I Pn (x) therefore converges for almost all x, and in

particular

n(x) = n-2f(nx) -> 0

for almost all x. SOLUTION: (b):

Consider the function:

(log1cosxl)2.

f(x) _ It has period it and is integrable on [0,n], for in the neighbour-

hood of n/2 it is equivalent to (loglx - 271)2.

lim(n 1loglcosnxl)2 = 0, n i.e.,

By the preceeding,

CHAPTER 3: THE

110

limlcosnxll/n = 1 n for almost all x.

EXERCISE 3.45:

Let (xn)nal be a sequence of points of [0,1].

If

0 < a < b 4 1 and N a 1 is an integer, one denotes by v(N;a,b) the number of integers n such that 1 4 n 4 N and a S xn < b.

The se-

quence (xn)n31 is called an EQUIDISTRIBUTED SEQUENCE if, for any

0s as b5 1, lim v(N;a,b) = b - a. N

N-

Show that the following conditions are equivalent:

(a):

(1): The sequence (xn)na1 is equidistributed; (ii): For every continuous function f on [0,1] 1 J

f(x)dx = 1imN 0 N-?-

N

-1

Y

f(xn);

(*)

n=1

(iii): For every integer p a 1, -1

limN N-

N I

2nipxn e

= 0.

n=1

APPLICATIONS: Investigate whether the sequences xn = na - [na]

(a irrational)

and

x

n

= logn - [logn]

are equidistributed (here [a] denotes the integral part of a).

FUNDAMENTAL THEOREMS (b):

113

Show that if (xn)n31 is equidistributed and if f is

Riemann integrable, then (*) still holds. ovo = vov = ovo = vov = ovo

SOLUTION:

We shall prove part (b) first of all, which will also

prove the step (i) => (ii) of part (a).

We may assume that f is

Let us note that if (xn) is equidistributed then (*) holds

real.

for every characteristic function of an interval [a,b], 0 4a 4b4 1.

This formula therefore also holds, by linearity, for every step function.

Now for every e > 0 there exists a step function 9 such

that: 1

J TSe+

f5 cp,

J0

J0

f

From this one deduces that:

-1

N

l im supN

E

n=1

N

f (x n )

.<

l imN

-1

1

N

I

9_(X

n=1

N

)1 S E +

J

n

f,

0

and consequently

1 im supN

N-

t

1

N

n=1

f (Xn ) <

J

f.

0

Replacing f by -f, this yields:

N lim infN 1

N--

1

f(x ) > J f,

n=1

n

0

which shows that (*) holds for f. That (ii) => (iii) is trivial upon setting f(x) =

e2nipx

in

N. Finally, let us show that (iii) => (i).

We first show that

CHAPTER 3: THE

112 (iii) => (ii)t.

Now, if (iii) holds, then (*) is true for every

trigonometric polynomial.

If f is continuous on [0,1], and if

e > 0, there exists such a trigonometric polynomial 9 for which If - 91

To simplify the

E (the Stone-Weierstrass Theorem).

ensuing calculation, set

uN(f) = N-

((1 CN

f(xn) - J f

=1

0

Then IUN(f)I < IuN(9)I + IuN(f - q))I < IuN((P)I + 2E,

and consequently, since uN(P) -> 0,

lim supluN(f)I .< 2E, N-

which proves that (ii) is satisfied. let

Now let 0 < a < b 4 1 and

be the characteristic function of [a,b].

For every E > 0

there exist two continuous functions gE and fE such that

0< f

E

. 1 is equidistributed and if the boundary of E has measure zero, then

limN 1v(N;E) = meas(E),

N-

where v(N;E) denotes the number of integers n such that 1 < n < N

and xn a E.

EXERCISE 3.46:

Let f be a continuous function on ]R2 such that:,

f(x,y) = f(x + 1,y) = f(x,y + 1)

FUNDAMENTAL THEOREMS

115

for any x,y. Show that for every irrational number

f(x,y)dxdy = lim

J1

T-

0' 0

there exists a step function h such that

E I f - hI < E. Setting

TI g(x)dx, 0

one has (taking. account of lyl < M)

((+00

II

+m

f(x)[g(nx + an) - y]dxl < Jh(x)[g(nx+a)Y]dxI+2Mc ,

and consequently,

117

FUNDAMENTAL THEOREMS tW

lnmssupl l

f(x)[g(nx + an) - Y]dxI 4 We, _W

which proves (*).

When g(x) = eix and an = 0 for all n, one obtains the

REMARK:

Riemann-Lebesgue Lemma: +m f(x)e1nx

lima

n-. _,

dx = 0

if f is integrable.

EXERCISE 3.48:

Let (pn) and (an) be two sequences of real numbers

such that

I Ipncos(nx + an)I < n for all the x's of a set A of measure greater than zero.

Show that L

Ipnl <

n AV1 = VAV = AVO = VAV = LVA

SOLUTION:

For every integer n let AN be the set of x's such that

I Ipncos(nx t an)I 4 N. n

Since A is the union of the increasing sequence of AN's, meas(AN) - meas(A) > 0.

Hence there exists an N such that meas(AN) > 0.

Let E C AN, 0 < meas(E) <

Then (Con td) r

IpnIJElcos(nx + an)Idx

Ipncos(nx + an)I I dx c

CHAPTER 3: THE

118

Nmeas(E) <

(Contd)

Furthermore, the function x -> Icosxl has period it and rn n

cosx dx =

2

0

so that Fejer's Formula (cf. the preceding exercise) applied to the characteristic function of E gives

meas(E) > 0.

Icos(nx + an)Idx =

limJ

n

it

E

From this it follows that

EIpnI<

n

EXERCISE 3.49:

Let n1 < n2 <

< nS <

be a strictly in-

creasing sequence of positive integers and (0 s)sa1 be a sequence of real numbers.

Prove that for every measurable set E of [0,2n] and

(a):

for every integer p

1

S + S)dx S S ° Cos2P(nx = 2-2p1 p)meas(E).

l

E

(b):

Deduce from this that for almost all x e]R

liSmssuplcos(nsx t 8s)I = 1. A4A = V AV = A00 = VMv = 10A

SOLUTION: (a):

An elementary calculation shows that:

119

FUNDAMENTAL THEOREMS

cos2pz =

2-2p(eiz + e-'z )2p

2-2p(p )

+ 21-2p krl

i

2pk Icos2kz.

Furthermore,

IJEcos2k(nsx + 8s)dxl <

IJE cos2kn sxdxI + IJEsin2kn sxdxl,

and the two latter integrals tend towards zero by virtue of the Riemann-Lebesgue Theorem. Let 0 < a < 1 and let Ea be the set of x's (0 4 x < 2n)

(b):

such that limsuplcos(nsx + 9S)I ' a. S

By virtue of Fatou's Lemma and of part (a),

a2pmeas(Ea) a

limsup cos2p(nsx + 6 )dx

I

S

E

a

>

limsupfE

cos2p(nsx + 0S)dx

S-'°

a

Note that 22p

-2p(2p)1

2p lI (

P J

(2p)2p+1/2e-2p ti

= 2

1 r2 -,E

(p!)2

ti

2-2p

p2p+1e-2p

120

CHAPTER 3: THE

Since a2pPVT-> 0 when p meas(E(1

it follows from the preceding that

The set of x's from [0,2i] such that

= 0.

limsupjcos(n x + s 34w

)I 4 1

S

being the union of the E1

(n 3 1), the proposition is proved.

- 1/n

Let f be a positive integrable function on an open

EXERCISE 3.50:

such that meas(X) < w.

set X of ]R

Show that there exists a function g, lower semi-continuous Qn X, such that g >. 1/f and J fg < W.

X 401 = 000 = 404 = V1V = 001

Let us set:

SOLUTION:

An

n) + 1 < f 4

fn

(n = 0,1,2,... ),

and

o). There exist open sets VD and n contained in X such that

Vn D A_,

JV n

f(x)dx <

Wi J An,

2-n

f(x)dx < (n t JW

n

-A

.

n

The function

0

g=

E

n=0

(n + 1)ILW

+

ILV

n

n=0

1)-12-n

n

FUNDAMENTAL THEOREMS

121

is lower semi-continuous and g a 1/f.

°

JX

n=0

J

f+ V

n

4+

f + JA

(n+

f+

1)fW

n=0

C

Furthermore,

n

n + l

n==1

-A

(n1)JA n=0

n

f

n

meas(A) n <

0

for I meas(An) < -, since the An's are mutually disjoint and meas(X) < -.

If meas(X) = m, the result may not hold.

REMARK:

Indeed, if X

is the disjoint union of sets Xn such that meas(Xn) = n z and if

1

f=

IL

n=1 n

Xn

one will have, if g >. 1/f:

J1fg

==n=1

X

n

EXERCISE 3.51:

n=1

Let (fn) be a 'sequence of measurable functions

on X = [0,1].

Show that the following conditions are equivalent: (i): There exists a subsequence-(fn ) of the sequence S

which converges to zero almost everywhere; (ii): There exists a sequence (tn) of real numbers such the limsupltnI > 0 and the series E tnfn(x) converges for almost all x;

(iii): There exists a sequence ItnI of real numbers such that

LltnI

and the series I tnfn(x) is absolutely

convergent for almost all x.

122

CHAPTER 3: THE ova = vav = eve = vev = ave

SOLUTION: (i) => (ii),(iii):

Without loss of generality fn -> 0

By Egoroff's Theorem there exists a sequence

almost everywhere.

of measurable sets A

C A2 C

1

such that meas(X - As) < 1/s and

f - 0 uniformly on each of the A n S of integers n1 < n2 <

Hence there exists a sequence

.

such that for all n 3 n8 one has Ifnl <

2-s

on A

.

S

Let A be the union of the A

s

so that meas(X - A) = 0.

Set to = 0 if n 4 ns for all s, and let to

= 1.

Then

S

I tnfn(x) = E fn cx), n

s

s 2-s

and for all x e A one will have Ifn (x) I <

whenever s 3 s0,

s

where s0 is chosen so that x e AS

This proves (ii) and (iii)

.

hold. (ii) => (i):

quence n1 < n2 <

If limsuptn > 0 there exists a > 0 and a sesuch that Itn

I

> a for all s.

If for al-

s

most all x the series E tnfn(x) converges, then in particular

to fn (x) -> 0, and consequently fn (x) -> 0. s

s

s

(iii) => (i):

Letting g(x) = I Itfn(x)I, by hypothesis one

has g(x) < - almost everywhere.

Then if A is the union of the

As = {g < s} one has meas(X - A) = 0.

I Itnl JA IfnI = JA 9 <

n

S

S

Since X Itnl = -, one thus has:

liminfJA

Ifn1 = 0.

n-o

S

Furthermore,

FUNDAMENTAL THEOREMS

123

It is then possible to determine a sequence n1 < n2 <

such

that:

IfJARs

2-s

<

I

S

For every integer a one has:

S1

fA a

I fn

s

because As D Aa whenever s 3 a.

From this it follows that at al-

converges, and fn - 0 in

most all. points of Aa the series E fn

s particular.

But then fn

s

--> 0 at almost all points of A, hence

s

also at almost all points of X, because meas(X - A) = 0.

EXERCISE 3.52: (a):

Let f be an integrable function on X =]R

.

Show that for all c > 0 there exists a measurable set

with finite measure such that f is bounded on A and

(b):

From this deduce that:

Ifl = 0

lim E

(which means that for all E > 0 there exists S > 0 such that if E is measurable and meas(E) < d then

jE

Ifl < E. AVO = VAV = AVA = VAV = AV!

CHAPTER 3: THE

124 SOLUTION:

0 (since only 1A appears in

We may assume that f

the statement of the exercise). SOLUTION: (a):

Let:

A0= (f=0),

An=

II<

f f uniformly on A.

Hence there exists no such that

for n 3 n0 Ifn - fI

and

on A,

< meas(A)

- f)I < E. IJX(fn

By writing

JX-A

if n

fn=J

f +JX (fn - fJAn

B-A f+JX-B

-f),

n0 one obtains:

JX_Afnl

< Mmeas(B - A) + E + E +

mess A

meas(A) < 4E.

Furthermore, if

g = IfI + meas A)

on A

and

g = 0

on X - A,

it is clear that g is integrable and that Ifnl < g on A for n 3 n0. SOLUTION: (b):

By Fatou's Lemma,

JX_AIfI < E,

and by the Lebesgue Dominated Convergence Theorem,

JA If - fnI } 0.

FUNDAMENTAL THEOREMS

127

Then by the inequality

fx If - fnl < JX-A Ifl + JXA Ifnl + Jlf - fl one concludes

li pJX If

- fnj c 2e,

which proves that

J/n J[ SOLUTION: (c):

Here are three counter-examples where all the

functions considered are zero outside [0,1], which comes down to taking this interval as X. (i): Let

fn(x)=-(n 22) if1 x< for n ? 2.

1

and fn(x)=1 if

Then fn -> 1 almost everywhere.

If A =

nO ax a>0

exists, for every e > 0 there is an a > 0 such that if A = [a,l] one has

IJX-Afnl < E.

On A the Ifnl are uniformly bounded.

Now, fn(x) - (l/x)sin(l/x)

almost everywhere and this function is not Lebesgue integrable. (iii): Finally, let

fn(x) = x sin x

if 2 < x s

and fn(x) = 0 otherwise. n

Then fn - 0 and

- 0.

JX f

n

If the conditions of part (b) were satisfied by the f 's they n would also be satisfied by the Ifnl, and one would have: 2n

lin

IsixnxI

lim= 0. fXlfnl

dx =

n

Now, by Exercise 3.48,

1im12n Isinnxl dx = limJ2 Isinnxl ax

n.=i n

x

n-

=

1

x

1 sinxdxl Xk 0

1

= 21o g2

FUNDAMENTAL THEOREMS EXERCISE 3.54:

129

Let (fn) be a sequence of measurable functions

on a measurable set X of ]R

.

One says that this SEQUENCE CON-

VERGES IN MEASURE towards a measurable function f if for all a >0 lim meas(If - f

n

n-).-

I

>

a) =

0.

Show that if fn - f in measure there exists a subse-

(a):

quence (fn ) which converges to f almost everywhere.

Give an ex-

S

ample showing that the sequence (fn), itself, need not converge to f almost everywhere. (b):

Show that if the fn's are positive, converge in meas-

ure to f, and if

(Fatou's Lemma for convergence in measure). (c):

Show that if the fri converge in measure to f, and if

there exists a positive integrable function g such that IffI < g for all n, then f is integrable and

n-'° X if l imJ

- fn I= 0

(Dominated Convergence Theorem for convergence in measure). (d):

Assume that meas(X) < -.

Show that if the fn's con-

verge towards f almost everywhere, they also converge towards f in measure.

Give an example showing that the condition meas(X)

< - is indispensible.

CHAPTER 3: THE

130

tVA = VAV = AVA = VAV = AVA

SOLUTION: (a):

Choose a sequence of integers n1 < n2 <

such

that

ll

< 2-s,

>

measlIf - fn

I

s)

S

Then the set of points that belong to an infinite number of sets (If - fn

I

> 11s) has measure zero (cf., Exercise 1.3).

Now,

s

this set contains the one formed of the points where (fn ) does s

For every integer n

not converge to f.

1 let fn be the charac-

teristic function of the interval [r2-s,(r + 1)2-s], where n and s are the integers such that n = 2s + r, 0 < r < 2s.

The sequence

(fn) converges in measure on [0,1] but does not converge at any point of this interval. SOLUTION: (b):

By part (a) there exists a subsequence which con-

verges to f almost everywhere, and the classic Fatou's Lemma immediately furnishes the result. SOLUTION: (c):

Let e > 0.

There exists an integrable set A C X

such that

g E/meas(A)) and note that by part (a) one has n IfI < g almost everywhere; then, by decomposing the integral on X I

into a sum of integrals on X - A, A - A n, and Af1An, one obtains:

If - f I < 2e + e +

limsupJ

n-

X

n

limsupJA

2g.

n7

Now, meas(An) - 0 by hypothesis, so (cf., Exercise 3.5M ,

FUNDAMENTAL THEOREMS

131

2g = 0,

1imJ

n-'°° A

n

SOLUTION: (d):

By Egoroff's Theorem, for all c > 0 there exists

a measurable set E such that meas(X - E) < c and fn -> f uniformly

Then, for all a > 0 one has (If - fnl > a) C X - E when-

on E.

ever n is large enough, which proves that fn -> f in measure.

The

functions fn = n[n n+1] converge to zero almost everywhere, but not in measure.

EXERCISE 3.55:

Prove that if f and the fi's are pos-

Let a > 0.

itive measurable functions and if fi -; f in measure, then fi - fa in measure.

AVD = VAV = AVA = VIxV = AV6

SOLUTION:

If 0 < a s 1 this results immediately from the inequal-

ity: Ifa

- fil , If - files.

When a > 1 one has (by the mean value theorem):

lfa fil

aI f - fil (f V

fi)a-

Let e > 0 and let

where f V fi denotes the maximum of f and fi. M > 0.

Then

meas(Ifa - fll > c)

mead If - fil > l

meas(fV fi > M). > M, then

Note that if f v f i

E

1)

CHAPTER 3: THE

132

If - fiI >

or

f > -zM

M,

for otherwise one would have

fi = f + (fi - f) < zM + ,M = M. Thus:

meas(lf°` -l > E) < mead if - fil >

E-1

"Ma

+ meas(If - fil > ZM) + meas(f > 2M), so that

limsup meas(lf°` - fal > e) < meas(f > 15M). Z-

Since lim meas(f > M/2) = 0,

M+=

lim meas(Ifa

> e) = 0.

Z-

EXERCISE 3.56:

0 < meas(X) < m.

Let X be a measurable set of Iltp such that

Denote by M the vector space of measurable comIf f e M, set

plex functions on X.

p(f) =

j

+

Ifl)-1

(a):

Show that fn - 0 in measure on X if and only if p(fn)

(b):

Show that (f,g) y p(f - g) is a metric on M if one

+ 0.

agrees to identify two functions that are equal almost everywhere

133

FUNDAMENTAL THEOREMS

Show that p is not a norm on M, but that the sum and

(c):

the product of two functions of M are continuous operations in the metric defined by p.

Show that M is complete in the metric defined by p.

(d):

AVA = 0A0 = A00 = VAT = MMA

This follows from the fact that if Ac = (IfI > c)

SOLUTION: (a):

1 + e

meas(Ac) < p(f) < meas(Ac) + 1 +

meas(X).

If f,g a M, from the inequality

SOLUTION: (b):

1 + If

c

1+ f+ 1+ g

g

one deduces that p(f + g) < p(f) + p(g).

From this it results

that p(f - g) satisfies the Triangle Inequality.

Furthermore,

if p(f - g) = 0 then f - g = 0 almost everywhere, that is to say that f = g in M. SOLUTION: (c):

p is not a norm, for in general p(Af) + IXlp(f)

Nevertheless, if fn - f and gn -> g in measure, then

P(f + g - fn - gn) < p(f - fn) + p(g - gn) + 0,

so f + g -> fn + gn in measure. Assume now that fn - 0 in measure and that for all c > 0 there exists a > 0 such that

m meas(Ig limsup

I

n

> a) < c

(which is the case if gn + 0 in measure, or if gn = g for. all n).

Since, for all a > 0,

CHAPTER 3: THE

134

(IfngnI > B) C (IgnI > 0)U(IffI

>

one will have limsup meas (I fngn I

> S) < e,

which shows that fngn - 0 in measure.

From this it follows imme-

diately that if fn - f and gn - g in measure, then fngn - fg in measure.

SOLUTION: (d):

Let (fn) be a Cauchy sequence in M.

There exists

a subsequence (fn ) such that s

p(fn

- fn ) < 4-s. s+1

S

By the inequality from the answer to part (a), for all s one has:

- fn

meas(Ifn s+1

I

> 2-s) s (1 +

2s)4-s.

S

By exercise 1.3, for almost all x one has

Ifn

s+1

(x) - fn W1 <

2-s

s

for s sufficiently large.

converges almost

In other words, fn s

everywhere to a function f.

Applying Lebesgue's Bounded Converg-

ence Theorem to the integral which gives p(f - fn ) shows that s

p(f - fn ) + 0.

Thus the Cauchy sequence (fn) possesses a con-

s

vergent subsequence; it is therefore itself convergent.

EXERCISE 3.57:

Let (fn) be a sequence of positive integrable

functions that converge in measure to an integrable function f.

FUNDAMENTAL THEOREMS

135

Show that if

n

Jfn = Jf

then

n-Jlf-fnl=0. (See Exercise 6.114 for a generalisation of this result to LP spaces).

ovo = vav = ovo - vov = ovo

SOLUTION:

It is clear that

(f - min(f,f) > a) C (If - ffI > a), so

min(f,f ) -> f in measure. n On the other hand, 0 4 min(f,fn)

f. '<

Hence, by the Dominated Convergence Theorem for convergence in measure (cf., Exercise 3.54)

Jmin(f ,f) - Jf.

Then

Jmax(f,f) so that :

=

Jf

+

Jfn - Jmin(fif

)-

Jf1

CHAPTER 3: THE

136

I f - fnl = Jmax(fif ) - Jmin(fif

)

-* 0.

J

EXERCISE 3.58:

Let H be a family of positive integrable functions

on a measurable set X of 30. (a):

(i):

Show that the following conditions are equivalent:

limJ

C-}°° (f>c ) (ii):

lim

f(x)dx = 0

uniformly for f e H;

j f(x)dx = 0

uniformly for f e H.

meas(E)-+0 E (A family satisfying these conditions is called UNIFORMLY INTEGRABLE). (b):

Show that if H is uniformly integrable and addition-

ally satisfies the condition: (iii): For any c > 0 there exists a measurable set

A C X

such that

meas(A) <

and

jX-A f(x)dx < E for all f e H; then

supJ f(x)dx < °°. fell X (It is said that H CONSERVES MASS if the Condition (iii) is satisfied). (c):

Show that if there exists a positive measurable func-

137

FUNDAMENTAL THEOREMS

G defined on yt such that limt-1G(t) = m t-- =

and

sup

G(f(x))dx <

f eAJX

then the family H is uniformly integrable.

Convers9ly, if H is

uniformly integrable there exists a function G that satisfies the above conditions and that can further be chosen to be convex and (We refer the reader to Exercise 6.113 for an appli-

increasing.

cation of the notion of uniform integrability to the problem of convergence in Lp spaces). A0A = 000 = A00 = V AV = AVA

SOLUTION: (a):

(i) => (ii):

f+ cmeas(E).

1 f 4 J

E

For every set E of finite measure,

(f>c)

For every E > 0 one can choose c such that the integral of f on (f > c) is less than e/2 for all f e H.

Then, if meas(E) < £/2c,

f < e for any feH. E (ii) => (i): First let us prove that

lim meas(f > c) = 0 uniformly for f e H.

(*)

c-

Otherwise, there would exist a > 0, a sequence ek

and a se-

quence fk e H, such that

meas(fk > ck) > a

for all k.

Let 0 > 0 be such that meas(E) 6 0 implies for all f e H:

f < 1. JE

CHAPTER 3: THE

138

For every k one can find Ek C (fk > ek) such that:

meas(Ek) = min(a,s).

One would then have:

1 >

1

fk > ekmin(a,s), Ek

which contradicts ek ->

Now let e > 0 and 6 > 0 be such that

meas(E) < 6 implies

f e) < 6 for all f e H, and consequently:

f < E. (PC) SOLUTION: (b):

In fact the conclusion still follows if Condition

(iii) is weakened to requiring only the existence of a set A of finite measure such that

sup{

f<

feH X-A which in essence reduces us to the case where meas(X) = a < Assume the conclusion is false, i.e.,

sup f = . feHfX Let (X1,12) be a partition of X into two measurable sets such that meas(X1) = meas(X2) = a/2.

Then

FUNDAMENTAL THEOREMS

139

sup f = co fEH J Xi

for at least one of the integers i = 1,2.

Repeating this parti-

tion argument, one constructs a sequence of measurable sets (En )n>0 such that

meas(En) = a2-n,

f

feHJ En

But this manifestly contradicts Condition (ii). SOLUTION: (c):

Assume first that there exists such a function G.

Let e > 0 and set

M = sup J G(f).

feH X There exists c0 such that t-1G(t) > M/E if t > c0.

Then for all

c > c0 and all fEH J

f s L G(f) s E. Mx

(f>c)

Now assume that H is uniformly integrable and look for G in the form

G(t) =

rt

g(u)du,

J 0

where g(0) = gn = constant on [n,n + 1[, with 0 = go 4 gl 6 and g w. Then G will-be increasing (and hence measurable), n convex (since g is increasing), and will satisfy

,

140

CHAPTER 3: THE

limt-1G(t) t

for, g(u) -> - as u

and this implies that the mean of g on

[O,t] tends to infinity when t i

It remains to choose the g

n

suitably in order that

supJ {G(f) < feH X

(1)

Define

an(f) = meas(f > n); by noting that G(f) = 0 if 0 .< f < 1 (since g0 = 0) one obtains:

G(f) _

X

(as G is increasing).

G(n + 1)(an(f)

G(f) <

J

n=1 (n 0 and c > 1/e one has meas(fn > c)

= 0 if n . c

meas(fn>c)=n1

p

We are going to show that the A(t) realise some sort of "homotopy"

between A10 and Al

1,1 ,

,

in the sense that they satisfy the follow-

ing properties: (1):

A(0) = A10 and ,

(2):

,

meas(A(t)) = 2meas(A);

(3):

f1 .

= 0,

JAW (4):

A(1) = A11;

1 4 i F n;

For every function f integrable on A the function e+

t

A(t)f

is continuous.

CHAPTER 3: THE

144

This will imply the property P+1, for by virtue of the Intermediate Value Theorem there exists a t (0 4 t S 1) such that

A(t)fntl =

J

2( JA(O)f'+l

=

(/

+

fn+1 +

JA(l)fn+l)

f

Al,l

A1,0

fn+1)

JAfn+l = 0.

It remains to prove properties (l)-(4).

A1(0) = Al

First of all,

0, ,

and if p >. 1,

A

p+l

(0) = U A p+3-,q,q

=

U

05q 0 for all a e A, then for every integrable mapping f of X into A

Z(J f(x)dx) = 1 t(f(x))dx : 0, X

X

which proves that:

1AdxCTA. X

Now assume that, with f being as above,

y=1 f(x)dx6FA)-P(A).

CHAPTER 3: THE

150

Since the interior of 1T A is equal to that of r(A) (a classical result about convex sets), y must be a boundary point of r A

.

There would then exist a non-constant affine function such that

.2(y) = JxI(f(x))d3 = 0,

and 1(a) > . 0 for all a e t A , and in particular t(f(x)) >. 0 for all x e X.

By modifying f on a negligeable set, one would then

have 1(f(x)) =.O for all x e X, or f(x)e.2-1(0)n A, so then y 4 r(.2-1(0)nA).

Since it can always be assumed that A affinely

generates R , and consequently that the affine dimension of .1-1(o)n A is strictly less than that of A, one is led to a contradiction.

REMARK: More generally, one can show analogously that if g is measurable and positive on IItn, and

Jg(x)ds = 1,

the set of points

Jf(s )g(s )dx,

where f:3tn N. A is integrable, is equal to F(A). EXERCISE 3.63: (a):

Let f be a mapping of X =

into Y = IItn

Show that there exists a smallest closed set Af such

that f(x) a Af for almost all x e X, and that Af is non-empty.

Next show that if f is measurable, then y e Af if and only if meas(f 1(V)) > 0 for every open neighbourhood V of y. (b):

Now assume that f is integrable (or measurable and

bounded) and consider the set If formed of points of Y of the type:

FUNDAMENTAL THEOREMS

151

1

mess E

f(x)dx,

E

where E is an arbitrary measurable part of X such that 0 < meas(E) < -.

Show that If is convex.

What are the relations between Af and

If?

AVA = VIV = AVA = VAV = AVA

SOLUTION: (a):

In order to prove the existence of Af it suffices

to prove that of a largest open set Vf of Y such that f 1(Vf) has measure zero; Af will then be the complement of Vf.

To do that

it suffices to prove that if (Va)aei is a family of open sets of

such that f 1(V

has measure zero for all a e I, then the union

Y of the Va's possesses the same property.

By virtue of LindelSf's

Theorem (cf., the end of the Exercise) there exists a sequence (an )n

1

of elements of I such that W

V= U V n=1

n

and consequently,

f 1(V) = U f 1(Va n=1

n

Certainly has measure zero.

Furthermore, as X = f1(Y), one has

1'f + Y, and consequently Af $ 0.

Now assume that f is measurable.

If y e Af and if V is an open

neighbourhood of y, Vf u V is an open set which strictly contains Vf, so that:

f 1(Vfuv) = f 1(Vf)uf 1(V)

CHAPTER 3: THE

152

has positive measure; since f 1(Vf) has measure zero, meas(f 1(V)) > 0.

Conversely, if y4 Af and V = Vf one has meas(f 1(V)) = 0.

SOLUTION: (b):

Let E. (i = 0,1) be two measurable sets of X such

that 0 < meas(Ei) <

E = E0 U E1, and

1

yi

mess TE-i7f

E.f I

which may also be written as:

(f - yi) = 0.

JEEI Assume first that meas(E0flE1) > 0.

By the remark at the end of

Exercise 3.59 there exists a family (Ft)0Ct41 of measurable sets such that

Ft C E,

F1=E,

F0 = O,

t * meas(Eif1Ft)

is continuous,

and JF IlE1(f - yi) = 0,

which we may write as:

f = meas(EifFt)y,.

J Ei f1 Ft Set:

Gt = (E0 - Ft) U(E1f1Ft);

then

FUNDAMENTAL THEOREMS

153

f = meas(E0 - Ft)y0 + meas(E1 n Ft)y1. J/1-

t

Now note that Gt D EOflE1, so that meas(Gt) > 0 for all t; the function

meas(E0 - Ft) t .-

meas( t)

is therefore continuous and varies from 1 to 0.

For all a, 0 < a

,< 1, there therefore exists a t for which it takes the value a, and then

mess Gt

f = ay0 + (1 - a)y1. G

When meas(E0f)E1) = 0 one can assume that E0f1E1 = 0.

It is poss-

ible to determine two families (Fi,t)0, 0 for y e Af, and

UP=o, JE

one would be able to assume that Af C £-1(0), and one would end by arguing by induction on the affine dimension of Af.

Let us

note that if f is bounded, then Af is compact, and consequently

If = r(Af).

LINDEZAF'S THEOREM: In a topological space possessing a countable

basis of open sets (U)-.which is the case for

Rm

-the union of

an arbitrary family of open sets (V ) is the union of a countable a

subfamily of them.

This is proved by considering initially the subset J of integers n for which Un is contained in at least one of the Va. with every n e J one associates an index an such that Un C Va n Then,

U Va =U V a

neJ an

Then,

155

FUNDAMENTAL THEOREMS Indeed, if x e Va there exists n such that x e Un C V

of a basis of open sets).

(definition

But then n e J and x e Va , which proves n

the Theorem.

Let f be an integrable function on at.

EXERCISE 3.64:

Show that if b

f(x)dx = 0 for all real numbers a and b, then f = 0 almost everywhere. AVA = VAV = AVA = VAV = OVA

Every open set V of ]R being the countable union of in-

{SOLUTION:

ttervals, for all such V one has:

Lf E is measurable there exists a decreasing sequence of open Sets (Vn) such that

E C n V n

meas(E) = lim

and

n

ffo that :

jE

=limfV n

f=0.

n

Lbus the set If of all the numbers

1 meas E

Jf E

Mtluces to {0}.

0 < meas(E) < 0, '

Now if contains the smallest closed set Af such

CHAPTER 3: THE

156

that f(x)e Af for almost all x (cf. the preceding exercise). Hence one has Af = {0}, or in other terms f = 0 almost everywhere.

EXERCISE 3.65: (a):

Let p be a norm on 3kn.

Show that if E is a subset of measure zero of IIz+, then

{x:p(x) a E} is a subset of measure zero of ]Rn. (b):

Let V = meas(p c 1) and f a measurable function on at+.

Show that:

nVJtn-lf(t)dt

= 12 f1f(p(x))dx

0

if f 3 0, or if J0t1t)dt < AVA = VAV - OVA = ViV = AVA

SOLUTION:

Note first that:

meas(p = r) = lim meas(r 4 p < r + 6+0 = meas(p < 1)lim((r +

0n

- rn) = 0.

ey0

This allows us to write

rb

meas(a 6 p . b) = V(bn - an) = nVJ to-ldt, a and consequently the formula nVJtn-lf(t)dt

= 3t f'f(p(x))dx

0

(1)

FUNDAMENTAL THEOREMS

157

is valid for every step function. Let E be a subset of measure zero of 3R

contained in [0,R].

There exists a seqeunce (fi) of step functions, zero on [2R,m[, such that

04fi4fi+l,

and fi -* -DonE.

sUpJ0fi m on {x:p(x)e E}, which proves that this set has measure zero.

If E has measure zero without being bounded one has

(p e E) = U (p e E fl [0,Ri] )

i

and therefore has measure zero.

if Ri Hence if fi - f almost everywhere

am ]R+, then fiop + fop almost everywhere on ]Rn.

Now assume that

Jt'If(t)Idt < m iad let (fi) be a sequence of step functions such that

fi --> f (a.e.),

J0tflhIf(t) - fi(t)Idt

0.

(2)

lien

lim J nIfiop - f.opl = nV lim

i, j-t°° IR

i,j mJ 0

tn-1If.(t) 1

- f (t)Idt + 0,

158

CHAPTER 3: THE

and consequently fiop converges in L1ORn). most everywhere, fiop -> fop in L1(]R ).

Since fiop - fop al-

Replacing f by fi in

Equation (1), and passing to the limit, one obtains the desired result.

Finally, if f is measurable and greater than or equal

to zero, then fK = inf(f,k)IIl[O,k] - f on 3R., 0 6 fK < fKtl'

Pass

ing to the limit in Equation (1) where f is replaced by fK, it is seen that this Formula is still true even if to-1f(t) is not integrable.

REMARK: To obtain Equation (2) we have used the following proposition: If g >, 0 is locally integrable, if f is measurable, and

if

J ifIg < -, there exists a sequence of step functions (f.) such that

(f - fi)g -

0

(a.e.),

jIf - filg -

0.

This is a particular case of a general proposition in measure theory.

It can be proved directly by observing that if h e L

and

Jgh = 0

for every rectangle P, then gh = 0 almost everywhere.

In other

words the annihilator in L°' of the set of functions IlPg (which

belongs to L1 because g is locally integrable) is the vector subspace {h:h a

this subspace.

0 a.e.}.

Now fg a L1 and is annihilated by

Hence fg is the limit in L1 of functions fig,

where the fi's are step functions (by the Hahn-Banach Theorem), and by taking a subsequence one can assume that fig -

fg almost

everywhere.

If one wants to avoid using the Hahn-Banach Theorem (as well,

as the property (L1)' = L-) one can argue in the Hilbert space

FUNDAMENTAL THEOREMS

159

L2, and note that if f >. 0 then g e L2.

If g were not to be-

long to the closed vector subspace generated by the functions

IlP/, the projection theorem would assure us of the existence of

a function he L2 such that

JhV>

Jh1

0,

=0

P a rectangle of ]R' .

P This is absurd, for the second condition implies that hV = 0 alHence there exists a sequence (fi) of step func-

most everywhere.

tions such that fiV- g in L2; but then the functions f2 are also step functions, and

If - filg

<

II ( f = u + iv, u = u+ - u-, V = v+ - V_.

f

This result generalises to the case of a function f such that:

Jlflpg

P % 1,

for, on reducing to the case where f >. 0 there exists a sequence

(fi) of positive step functions (cf., the second proof of the case p = 1) such that fig -> fpg in L1

,

and consequently fi pg

liP

1'

fg1/p in Lp (cf., Exercise 6.105). Instead of considering sequences of step functions one can, for example, consider infinitely differentiable functions with compact support.

EXERCISE 3.66:

Let V be a convex bounded set of tn.

Show that:

meas(V - V) .< I nnJmeas(V). (If p(x) = inf(X > O:x e X(V - V)) is the gauge of the symmetric

}

160

CHAPTER 3: THE

convex set V - V, p is a norm on in. there exists y e]R

Show that for all x e V - V

such that:

(1 - p(x))V + y C Vn (V + x) and use the preceding exercise).

OVA = VAV = AVA - VAV = AVA

SOLUTION:

If x e V - V then p(x) < 1 and x = p(x)z, where z beLet (zi) and (z2) be two sequences

longs to the boundary of V - V.

of points of V such that zi - zi -* z.

one can assume that z1

X=

p(x)(z1 - z2 ),

-*

As the open set V is bounded

z1, z2 ; z2, so that:

z1 a V, z2 a V.

Let y = p(x)z1 and observe that as V is open and convex (1 - t)a +

tb e V if 04 t < 1, a e V, bet. Consequently, for a e V (1 - p(x))a + y = (1 - p(x))a + p(x)z1 a V,

(1 - p(x))a + y = (1 - p(x))a + p(x)z2 + x e V + x,

which proves that

(1 - p(x))v + y c vn (v + x). Denoting the characteristic function of V by cp one obtains:

(1 - p(x))nmeas(V) = meas{(1 - p(x))V + y)

s meas(Vr(V + x)) = whence meas(V)I 11 V-V

(1 - p(x))ndx c

t f

n(q)*)(x) = meas(V) 2,

161

FUNDAMENTAL THEOREMS and consequently, because meas(V) > 0,

(1 - p(x))ndx < meas(V).

V-V By the preceding exercise (1

(1 - p(x))ndx = nmeas(V - V)J to-1(1 - t)ndt

f

V-V

0

_

nr(n)r(n + 1) meas(V - V) r(2n + 1

(n!)2

(2n)!

meas(V - V),

rhence, at last, meas(V - V) <

EXERCISE 3.67:

[]meas(v).

Let p be an even, positive function on ]R, decreas-

ing on [0,00[, and such that:

+0 f- p(t)dt = 1,

+W f- t2p(t)dt = 02 < 00

(a):

Set:

G(x) = 2f'XP (t)dt,

x - 0.

Show that G is convex, and calculate:

G(x)dx.

I

00 0

162

CHAPTER 3: THE Let OAB be a right tri-

(b):

angle, M a point on its hypotenuse,

and P and Q the orthogonal projec-

Q

tions of this point onto the other two sides of the triangle.

Find

the maximum area of the rectangle

0

P

OPMQ as M varies along AB.

From the preceding considerations deduce that for all

(c):

A > 0

p(t)dt < 12 2a

tI,acr

Prove the following result (F. Gauss (1821): an im-

(d):

provement, under the given condition, of the Bienyame-Chebycheff Inequality):

1 2a2

p(t)dt 1

23

3

if 0< a<

AVA = DADA= AVA = VAV= AVA

SOLUTION: (a):

Because p is decreasing and positive on (0,-),

the function

X Er J p(t)dt x is convex and decreasing on this interval.

Since x is concave,

it follows that the functions x 's G(x) and x H G(x2) are convex and decreasing.

Furthermore,

163

FUNDAMENTAL THEOREMS

j0

r G(x)dx = 2J dxl p(t)dt = 2 o

J2

p(t)J 0

V"X

dx 0

2J0t2p(t)dt = a2. 0

If OA = a, OB = b, OP = x, OQ = y, then

SOLUTION: (b):

and consequently:

b

xy =

2]

[1

-

(Q

-

K)

The maximum area of the rectangle OPMQ is achieved when a =

,

that is to say when x = a/2, y = b/2, and this maximum is equal to half the area of the triangle OAB. SOLUTION: (c):

Consider the area A of the triangle bounded by

the coordinate axes and a line of support for the curve y = G(x) at the point (A 2a2,G(A2a2)).

A2a2G(a2a2)

< A <

By the preceding,

2a2,

whence:

,,,.2 2,

1

2A

2

'

which is the desired inequality.

Since the function A 'r G(A2a2) is convex and G(O) = 1, for

(d) :

O 0

168

CHAPTER 3: THE

for all x e X (otherwise replace X by the set of x e X such that

g(x) > 0), one therefore has f(x) e.C 1(0)(1I for all x eX (after modifying f if necessary on a set of measure zero).

One then

argues by induction on the affine dimension of I. SOLUTION: (b): Formula.

Let us assume that the equality holds in Jensen's

Using again the proof above it is seen first of all

that one can assume that y0 is interior to I, and next that

(P(f(x)) _ (P(y0) + u(f(x) - y0) for almost all x.

Now, if cp is strictly convex this equality can

only hold if f(x) = y0. SOLUTION: (C):

It suffices to examine the case where I is not

bounded from above, T(-) > -- and

Jf(x)(x)dx = +m. X

If p(o') = - there exist a > 0 and 0 egt such that T(y) 3 ay + for all y e I, whence

J(f(x))g(x)dx

>>

J(cf(x) t s)g(x)dx = t-.

X If

cp(m) = a < m, then cp(y) . a for all y e I, and consequently

J (p(f(x))g(x)dx X

Jag(x)dx = a. X

Here, briefly expounded, are the proofs of the properties of convex functions used above. Let us, in Y xYt, consider the set,

FUNDAMENTAL THEOREMS

169

J = {(y,t):yeI, t 3 (,(y)}. It is easy to see that J is convex, and that if y0 a I, the point (y0'(P(y0)) belongs to the boundary of J.

Hence there exists an

affine function on Y X1R which vanishes at this point, is positive on J, and is not constant.

This is expressed by the existence of

a linear form u on Y and of two numbers a,B such that u and a are not simultaneously zero, and

u(y0) + a9(y0) + B = 0,

u(y) + at + B

0,

y e I and t > q)(y).

The second condition is equivalent to a . 0 and

yel,

u(y) + acp(y) + B > 0, So

a(cp(y) - (p(y0)) + u(y - y0) 3 0.

(1)

Now note that if y0 is an interior point of I then necessarily a > 0; otherwise one would have u + 0, u(y0) + B = 0 and u(y) + B a 0 for all y e I, which would contradict the fact that y0 is an interior point.

Dividing both-members of Formula (1) by a and

`replacing -u/a by u yields Formula (*). If for y e I, y # y0, one has

sp(y) = 9(y0) + u(y - y0);

Orall0 O,q > 0,

SECOND APPLICATION:

(1

p(l -

B(p + l,q) = J 17X---X)

o

(1

=

=

p

x)p+q-1dx

+ q 0 (1 -

x)p+q

p-1

x 1

x

dx

(1 - x)2

B(p,q).

p

From this it follows that for every integer n 3 1, B(p,q) _ (p + q)(p + q + 1)...(p + q + n) B(p + n + 1,q). n) p(p +

Now it is known that as n

a(a +

n)

nl

na

- r(a)

,

So

(p +

q + n) p...(p + n)

,,

r(p) nq r(p + q)

On the other hand,

B(p + n + l,q) =

x)penlog(1-x)

11xq-1(1 -

dx.

J0

Applying the first part with a = A = 1 and S = q - 1,

B(p + n + l,q) % r(q)n q, So

EVALUATIONS OF INTEGRALS

185

B(p,q) = r(p)r(q) r(p + q)

Prove that for n

EXERCISE 4.76:

, n -x`` sinx4dx

x e

0,

= 0,

0

and then that as t -> -,

1

f eitx-x4sinx"dx

r()ein/8t-5/4

ti 4

J0

004 - 040 = 400 - V AV = A4A

SOLUTION:

The function

x4n+3e-zxdx

F(z) = J

0

is holomorphic for Re(z) > 0.

F(z) =

When z is real one has:

(4n + 3)! 4n+4 z

By analytic continuation this formula is valid for every z with Re(z) > 0.

In particular, if z = 1 + i,

x4n+3e-(i+i)xdx

=

(-1)n+i On + 3)!

T

22n+2

Taking the imaginary parts of both sides yields:

J x4n+3e-xsinxdx 0

= 0,

CHAPTER 4: ASYMPTOTIC

186 1

and carrying the change of variable x i x" gives

(Wn-xIT J x e

, sinx'dx = 0.

0

Now set: 1

1

f(z,t) = exp(itz - z')sinz",

which for fixed t is holomorphic for Re(z) > 0 and continuous 1

(One chooses the principal branch of z4 in this

for Re(z) > 0. half-plane).

Taking into account the majorisation

Isin(x + iy)I < ey,

which is valid for y > 0, then for R > 0 and t > 0 one has:

I

If(Re l$, t)

I

I

< exp[- tRsina - R4(cos 4 - sin-!)), 4

< exp

2t7 8

-

,2R"sin i

-

(1)

Moreover, for z = x + iy, y 3 0:

Iz "et"f(z,t)I < min(IzI 4,Iz "sinz"I), and there therefore exists a constant M such that

(2)

Y A 0.

If(z,t)I < MIzI"e-ty,

Therefore, using (1), the integral of f(z,t) along the circular quadrant 8 - z = Re'

(0 < 8 < 7t/2) is majorised in modulus by

I-

Rexp( 0

l

2t exp( - v

sin 8) ,

LUATIONS OF INTEGRALS

187

consequently tends to zero as R -

By Cauchy's Theorem

therefore has

J f(x,t)dx = if f(iy,t)dy 0 0

=

it-5/4ein/8 OtIe-it[/B flit-1y,t)dy. J

0

is easily verified that

limt e-in/8f(it-1y,t) = y e y

by (2) It4e-in/8f(it-1y,t)I < My4e y.

refore by Lebesgue's Theorem

limt5/4JWf(x,t)dx t-3

ieln/BrWy4e-ydy

=

0

0

iein 8r [T5) = i_ r(4)ein B. 4

RCISE 4.77:

For every integer n

1 denote by do the number

partitions of a set with n elements. (a):

Set d0 = 1.

Show that for all complex numbers z:

n n=0 n n, =

(b):

exp(ez - 1).

Deduce from this that for every real number u > 0:

CHAPTER 4: ASYMPTOTIC

188 u+2co

i

do

= 2nieJu_i z-(ntl)exp(eZ)dz.

Let u

(c):

n

be the unique real root of the equation

zeZ=n+1. Show that

d , n

u U exp(e n - u e nlogu - lu n n n

n'

eJ

and deduce from this that logdn '' nlogn.

Ava = vev = ove = vov - ovo

SOLUTION: (a):

To determine a partition of En+1= {1,2,...,n + 1}

one may first fix the part of En+1 which contains n + 1; if this

part contains p + 1 elements (0 < p < n) there are (p) possible choices.

Next it remains to choose a partition of the n - p re-

maining elements, which gives, taking account of the convention do = 1, n

do+1

p0

n lP ,dn-p

(1)

This relation can be written as do+l (n + 1)

-

(n + 1)!

n

1

d -p

p)! p-_0 p! (n -

.

If the series n

f(z) _

do n, n=0 L

(2)

EVALUATIONS OF INTEGRALS

189

has a radius of convergence R > 0, then taking the derivative and using the above relation shows that for jzj < R

f'(z) = ezf(z), and consequently, since f(0) = 1,

f(z) = exp(ez - 1).

This function is entire, and if its Taylor expansion is written in the form of Equation (2) the coefficients do have to satisfy Equation (1), so, since d0 = 1, the proposition follows. SOLUTION: (b):

By Cauchy's Theorem:

z-(n+1)exp(ez)dz

IN

C

n = 2nieI

- M

where r denotes the rectangle indicated by the figure.

Note that, on

B

O

D -IN

A

this rectangle, if z = x + iy, lexp(eZ)I = exp(excosy) .< exp(eu),

and consequently

.< exp(e

JBCDA

u

2u + 4M )

0+1

On the other hand: iy)-(n+l)exp(eu+iy)l

j(u +

(U2 =

+

y2)-(n+l)/2exp(eu)

From this one deduces that for n 3 1: =

do

n1

2x_f+ (u + iy)-(n+l)exp(eu+iy)dy,

(3)

190

CHAPTER 4: ASYMPTOTIC

the integral being absolutely convergent. SOLUTION: (c):

Introducing the principal branch of log z in the

half-plane Re(z) > 0, and if u is chosen so that ueu = n + 1,

+W do

n!

exp(eu - ueulogu)J- g(y,u)dy,

(4)

where

g(y,u) = exp{eu[ely - 1 - ulogI 1 + mil]}. Note that

l g(y,u) I = exp{ -

2

((

eu[2sin2 + 2 log I 1 +

2 ] }. U,

111111

Therefore

if ly l >7E

+ u 2y2)-jue

9(y,u)dy1 < 2J (1

u

dy

n

2

dy Jn 1 + -u Ie u y 2

The last integral has the value e-u/242-utan-i( a-u/2 fu)

whence

fly,>ng(y,u)dy

= 0(ue-u).

If lyl < it then Isin(y/2)I

lg(y,u)I < explI

-o

lyl/n, so that:

2 euy2) It

(5)

191

EVALUATIONS OF INTEGRALS Setting y = to-u/2 yields:

+x

xeu/2 e-u/2J-xeu/2

g(y,u)dy f-x

g(te,u)dt, -u/2

=

and for t > 0, 2

limg(te-u/2, u) = e-t /2

Since for 0 < ItI 4 xeu/2 2

a- 2t /x

2

Ig(te-u/2,u)I .<

Lebesgue's Theorem implies that:

tie-4/2Je_t2/2dt =

Jx g(y,u)dy

e-4/2

x.

x

Taking account of Equations (4) and (5), u

d

n

ti

n!

e/27

exp(e n- u eunlogu

n

-

n

u ).

n

Bence u u logdn = logn! - (une nlogun - e n + jun) + 0(1).

Now logn! ti nlogn and

u u e nlogu

n

u

n

- e n + '2u

n

'

nlogu

n

= o(n(u

n

+ logu ))

= o(nlogn),

Which proves that logd N nlogn.

n

CHAPTER 4: ASYMPTOTIC

192

Show that for every real number t the integral

EXERCISE 4.78:

33Jdx 1

(

fi(t) _J cos)tx + 0

l

is convergent, and that the function c, so defined (called the

AIRY FUNCTION) is a solution of the differential equation,

4,"-to=0. Next prove that as t -> = t3/2 (1

1

4,(t) =

7.7 2

expi- 3

+ 0(t-3/4)),

0(-t) =

01*J t1T

Note

SOLUTION:

J°°expi{tx

3 Jdx.

If z =x+iy, (

Iexpiltz +

3

3

II

= exp( - ty -

x2p + 1 y3),

so that if 0 < y < a and Iti < A, 2

3l

iexpi(tz + 3 JI < Be -X y,

where B = explaA +

a31. 3

Therefore

EVALUATIONS OF INTEGRALS x+ia

193

3l

expi+ zJdzl Itz

fx+i0

3

2

(a

4 BJ a-x ydy 4 B 0 x2

This shows that the integral which defines 0 is convergent, and that if La is the straight line (-« + ia,m + ia)(a > 0) then

ll

2/(t) =

JL{

tz + 3 Jdz.

(1)

a

From this it follows that 3

2/0"(t)

z 2explltz +

Idz )

JL

a

the differentiation under the integral sign being legitimate be-

cause on La one has for Itl 6 A

2 ( 3 )e-ax Iz2expi ltz + 3JI 6 B(x2 + a2

Hence (

3

(t + z)expiltz 2 + 3Jdz l JL

2,r7,-(,"(t) - ti(t))

(

3l1II

[iexpiltz + 3J]

-+2'.a

z=-°+ia =

0.

When t > 0, carrying out the change of variable z + t2z in Equation (1) and setting R = t la,a = t3/2 yields (

2

4(t) =

expixlz +

J

LS

l

3 -) dz.

CHAPTER 4: ASYMPTOTIC

194

Since the derivative of z + z3 /3 vanishes for z = i, let us take S = 1 in the formula above.

This yields

+m ( 3l 'r O(t) = e-2X/3+- expal- u2 + 23 Idu 1t

2

l

1 e -2a/3

(+mexp

111

I-

u2 + 3 -Idu.

The latter integral can be written

V 1,

+

- 1)du.

J+We-CO

The absolute value of integral above is majorised by 2

JIuI3e_u

du.

-m

From this, one deduces that as t -*

¢(t) = 't "exp(- 3 +3/21(1 t O(t-3/ )).

Setting A = t3/2 as before, 3

(

rit"

(-t) = J

)dx. 0

As the derivative of x3/3 - x vanishes for x = 1, one is led to

the change of variable x = 1 + u//, which gives

(

(-t) = 1-

Re{e-2ia/3T

3

(

xpi 1u2

+ 3

/T) d.)

EVALUATIONS OF INTEGRALS

195

Set ((

f(z) = expila2 + 3x) = Rei*, where, as z = pale, 3

R = exp( - p2sin28 - - _ sin30), 3

= p2cos20 +

p COS33.

Assume that 0 < 0 4 n/4 and p

. - v.

Then sin20 3 0 and sin38 3 0.

From this it follows that for every number F such that 0 4 E F 1:

sin20 + .

n8 - 8 = a8,

sin30

Where a = 4/n - 1 > 0.

Therefore

2 e-'p

IRI <

$

(2)

By the Mean Value Theorem

R - e-p2sin28

where 0 < E < 1.

IR -

3

3A

sin38exp (l

e-p2sin281 <

IMI

a-ap28 (3)

3/

-rei8, 0 < 0 6 n/4.

"f

(u)du =

,

Consequently:

[.et L be the half-line pain/4,

J

p sin28 - CPI sin30 J 2 3/A

f

p >

Then (2) shows

f(z)dz, t+L

and r the circular arc

CHAPTER 4: ASYMPTOTIC

196

and then that

J

rf(z )dz J

4e-ax8dO

Q

Note next that for 0 = n/4,

_

=O

lfJ

-p3/3 I. Then (2) and (3) show

that in/4) If(pe

- e -p2

<

IRe"

- RI + IR - e-pI 2

I

+le-anp2/4 3 3X Consequently:

J f(z)dz = e L

ix/4 f

f

J

e_p 2

dp + H,

with

7

H S constant

p

I3e-anp /4dp

= 01J

Since, furthermore

r-F_e p2dp

e "I

-A

2F

'

one has

- =

e1rz/4J,rE + O{

1

J

Finally, x

n

Re e

-2iA/3

in/4

1

EVALUATIONS OF INTEGRALS

-

(-t) = t 4Cos

197

l

lI

t3/2 + 41 + 0

-

3

I

CHAPTER 5

Fubini's Theorem

EXERCISE 5.79:

For each of the functions f below, calculate

Jdxf1(x,y)dy, 0

JpyJf(x,y)dx, 0

0

2

f(x,y) = x - y

Ii. O 0.

lf(x,y)!dxdy.

CHAPTER 5:

200

000 - VAV = Ova = VAV = A0A

SOLUTION: (a):

Y0 dy J0

J0 (x2 + y2)2

1

dx [-x2+

=

fl fl x2 0 (x2 + y2)2

_

1

2

JJ Otx,y0,b>0

1 + (x +

y)3

x+y0 u2 + v2+ W2 +

u2v2'W2

v>o w>0

Changing to spherical coordinates yields

/2d3f/2dTf

0

I = J0

1 + r4sin4Ocos28cos2

0

2

Finally, setting

r = t4(sin8)cos

3cos cpsin cp,

we have _

(n/2

I

_1

_1

(n/2

cos 1sdolo

(W

cos 'gsin 2gdgj0

= f0

i r(i)r(i) r(1)r(1) 4

r(

)4,

r() sin it

7E /2-

=

4

,4-1

1 + u

du

r(3 r(4)

Noticing that

r(;)r(4) =

sin 4n =

nom,

the answer can be written

I = 4x(4)44

EXERCISE 5.92: ence:

Use a double integral to represent the differ-

214

CHAPTER 5: J+_

f(x) 2 dxfg(x)2dx - lIJ f(x)g(x)dxl2

-00

ava = vav = ova = vov = ave

SOLUTION:

It will be found that

2JJ2(f(x)g(y) - f(y)g(x))2dxdy

satisfies the requirement.

EXERCISE 5.93:

Show that the centre of gravity G of a homo-

geneous cone satisfies:

= loo, where 0 denotes the vertex of the cone, and G

0

the centre of

gravity of its base.

AVA = VAV = 1v6 = Vtv - pvt

FIRST SOLUTION:

On placing the coordinate origin at 0 and

making the plane xOy parallel to the base of the cone C, the coordinates E,n,r of G are given by:

JJJ'ydz

dxdydz

'SIC and two analogous formulae.

If

are the coordinates of

G0,S0 the area of the base, and CZ the cross-section of the cone cut by the plane parallel to xOy with height z, then:

215

FUBINI'S THEOREM 1

r

dxdy =

JJC

1

z

JJ xdxdy =

ydxdy

V0,

3 n0S0,

1

2 = zdxdy = zICoJS0

11

JJ

s0,

3

z

Jcz

0J

2

z

0Jc6S0, 3

1

ll

so that

J 00

NZ-013 C0S0dz

3

=E , 4

0

etc.....

2 S0dz J00N-E-O)

SECOND SOLUTION:

G is the centre of mass of the segment OG0

weighted with density ku2, where u = OM and k is a constant. It follows immediately that

EXERCISE 5.94:

Show that the volume bounded by a ruled surface

and two parallel planes is equal to

V = 6 Sl + S2 + 4S3 where h denotes the distance between the two planes, S1 and S2 are the areas of the cross-sections cut out by these two planes, and S3 is the area of the cross-section cut out by the plane parallel to the other two and located at a distance h/2 from

216

CHAPTER 5: (The Pile of Sand Formula).

each of them.

t0E = V AV = M1L - 010 = AVA

SOLUTION:

As the z-axis is perpendicular to the planes under

consideration, the cross-section cut out by the horizontal plane with height z is bounded by a curve given parametrically by equations of the type x = a(t) + zb(t),

rz

{l

y = c(t) + zd(t),

where a,b,c,d are periodic functions that we shall assume to be piecewise continuously differentiable.

The area S(z) of the

corresponding section, given by

xdb - bdx

S(z) = 2J r

Z

is therefore a second-degree polynomial in z.

From this it fol-

lows that (Simpson's Formula) h V =

I

11

S(z)dz = 6 (S1 + S2 + 4 S3). 0

Show that if

EXERCISE 5.95:

then the order of the integration can be inverted in

0

0

217

FUBINI'S THEOREM From this deduce that J0(y)

afm

I

sinax

dx =

0 a2+y2

+x`

1

d y, y,

where JO is the Bessel function:

2

rn/2 cos(ycose)d8.

J0(y) = nJ 0

AVL = VAV = LVA = V1V = OVA

If 0 < E < X < w then

SOLUTION:

Jsinaxdxjf(y)e-x3'dy = JW f(y)dyje-x3'sinaxdx, 0

E

0

E

since

x r Isinaxf(y)e-xyldy < (X - E) Je- Eylf(y)

dxJ fE

0

dys -.

0

The Second Mean Value Theorem gives

r

x e-xysinaxdx

J

e

2

a

and the inequality Isinaxl

.

ax implies

X

J_XYid

E

Y

Consequently,

X f(y)f e-xysinaxdx E

4 a Il(0,1)(y)If(y)I + an(11-)(y)lf(y)Iy-2

CHAPTER 5:

218

The Dominated Convergence Theorem leads to the formula

J sinaxdxJ f(y)e-xydb = 0 0

J

f(y)dyJ e-xysinaxdx. 0 0

Now,

e-xysinaxdx =

a

,

y 2 + a2

0

so

f (y)

of

sinaxdxJf(y)e-xydy.

dy =

o a2 + y2

0

0

Since J0 is continuous and IJ01 < 1, it is clear that the conMoreover:

ditions of the problem are satisfied if f = J0.

JJ0(y)e XYdY =

Icos(ycos9)d8

fee-xydy

J0

0

0

n/2 =

W

nd8J e-xycos(ycose)dy fo

=

2

0

(n/2 x2

0

2

-

xd8 +

cos28

tan

dt 0 1 + x2 + x212

2x rm n

t

1

W

11

f

`n 1

7,77j=: xt + x t=0

(Switching order of integration is legitimate, since

W

n/2

e-xyjcos(ycos8)Id8 < x < ").

2J dyJ n

n

1

+ x-

FUBINI'S THEOREM EXERCISE 5.96:

219

Let H be a continuously differentiable function

For all r > 0 set

on [0,°[.

m(r) = sup(xlogr - H(x)).

show that if

Joe-H'(x)

<

T 0

then m(r)

T 1 + r2

dr <

AVA - OAV m AVA - 0A0 - AVA

SOLUTION:

We shall first prove the following property:

If cp is

a measurable function on [0,m[ such, that q(t) > 0 for t 3 0 and:

(t)dt < -,

I

0

and if for r > 0 one sets

(x

u(r) = supJ log(rcp(t))dt, x>.0 0

then

I' (r) dr < f0

r2

Notice that u(r) > 0 for all r > 0.

On the other hand,

log(rp(t))dt < J1og+(rq(t))dt, x

I

0

0

CHAPTER 5:

220 so

Jlog+(r(t))dt.

0 < u(r) <

0

Consequently,

log(r(p(t))

f

r2

0

0

r

(J'OW(t

To return to the problem, put p = e-H'.

Then:

(x

xlogr - H(x)

H(o) + J

log(rpp(t))dt, 0

so

m(r) = - H(o) + P(r), which implies the stated result.

EXERCISE 5.97:

Let X be a measurable set of ]Rn with 0 < meas(X)

< -, and let f be an integrable function on X. every complex number z

JlogIl + zfj = 0,

then f(x) = 0 for almost all x.

Show that if for

FUBINI'S THEOREM

221

Show that

Jlot'l = 0 for all p > 0 by using the formula 2n

Jo logll + zeitldt = log+lzl.

AVA = VM0 = t01 = VLV - AVA

SOLUTION:

It may be assumed that meas(X) = 1.

Let p > 0.

Then:

2n

dtJ log 11 + Peitf(x)Idx = 0. f0,

X

If Fubini's Theorem can be applied then 2n 0

dx

X

log 11 + peitf(x)ldt = 21

log+lpfl.

X

JO

From this it follows that IfI < p-1 almost everywhere, hence that f = 0 almost everywhere.

It remains to be seen that Fu-

bini's Theorem actually can be applied.

JX log+11 + zfl =

By hypothesis

log 11 + zfI JX

for every complex number E.

Therefore

J0dtJ1H1 + Peitf(x)Ildx = 2J I

dtJ lo11 + pitf(x)ldx 0

X

4 4n(1 t p JX If 1) < m

CHAPTER 5:

222

(We have used the inequality log+Il + z1 s 1 + IZ1).

EXERCISE 5.98:

With every function f that is positive on E = 3Rn

associate the set Df C E x3R formed by the points (x,t) such that

0 c t '< f(x). (a):

Show that f is measurable if and only if Df is measur-

(b):

Show that if f is measurable and p > 0:

able.

Jf(x)Pdx = pJWtP-lmeas(f > t)dt.

E

0

Show that if f is measurable its graph is a set of

(c):

measure zero in E x]R.

000 - VAV = 000 - VtV = 000

SOLUTION: (a): is also.

If f is measurable, the function p(x,t)= f(x)- t

Since Df = (p

0) it follows that Df is

0)fl(t

measurable.

Now assume that Df is measurable.

Then x I-r meas((Df)x)= f(x)

is measurable (Fubini's Theorem; here, for A C E x]R and x e E,

AX denotes the set of is such that (x,t) e A).

SOLUTION: (b):

J fP = E

Fubini's Theorem also shows that

pJEdx(0(x)tp-ldt = pJOtp-1dtJ(flt)dx

M =

pJtP-1meas(f > t)dt. 0

SOLUTION: (c);

One can show, as in part (a), that the set D'f

FUBINI'S THEOREM

223

of (x,t)'s such that 0 < t < f is measurable, and the calculation carried out in part (b) proves that

f f = meas(Df) = meas(D'f).

J

E

Consequently Df - D'f, which is the graph of f, has measure zero

EXERCISE 5.99:

Let (Dn) be a sequence of closed discs, contain-

ing in the unit disc D, of radii rn > 0, and mutually disjoint.

show that if meas(D - U D ) = 0 then G r = . n n

n

AVO = VAV = AVO = VLV = AVo

SOLUTION:

Let In be the orthogonal projection of Dn on the meas(In) = 2 1 rn, and consequently if E rn <

Then

x-axis.

then almost all x's belong to only a finite number of the In's; that is to say that the vertical line L

x

with abscissae x meets

only a finite number of discs Dn, say Dn ,...,Dn

.

If IxI < 1

k

none of the intervals L

x

f1D

n

.

1

can be equal to L 0 D, for in that x -

= D, which is absurd, because r > 0 for n n.

case one would have D

all n and because the Dn's are mutually disjoint and contained It is then clear that:

in D. k

meas(Lxf1Dn ) < meas(Lxf)D) i=1

(1)

1

for almost every x e]-1,1[.

If i

is the characteristic func-

tion of D - U D , then (1) means that for almost all x (IxI < 1) n

JP(x)dy > 0,

CHAPTER 5: FUBINI'S THEOREM

224

and consequently:

meas(D - U Dn) = JdxJP(xiy)dy n

> 0.

CHAPTER 6

The LP-Spaces

EXERCISE 6.100:

Prove Holder's Inequality by using Jensen's In-

equality with the function P(x) = xP, x >. 0, p > 1.

MMA = 0M = MMA = VAT = AVA

SOLUTION:

Since cp is convex, for a positive measurable f and a

positive g with integral equal to one,

(Jfg)p < JfPg

(*)

Now assume that f 3 0, g , 0,.and: jgq JfP

-

1, =

.where q is such that

1

p

+ 1 = q

.

Replacing f in (*) by fgl q and

g by gq yields:

Jfg' 1. EXERCISE 6.101:

Prove Minkowski's Inequality using Jensen's In-

equality and the function 'p(x) = (1 - xl'P)P, 0 6 x s 1, p , 1.

225

CHAPTER 6: THE

226

AVO = VAV = AVO = VAV = OVA

The function p is convex, for it is continuous on

SOLUTION:

[0,1], and on ]0,1[ its derivative is equal to

1/PlP-1

(1 l

x l/p

)

which is increasing in x.

Consequently, if 0 < f E 1, g 3 0, and

Jg = 1, then

1 < (Jfg)1/P + (J(1 - ?/p)pg)1/p Now assume that f

0, g

(*)

0, and

J (f + op = 1. Replacing g in (*) by (f + g)p and f by 0 where f + g = 0 and by fP(f + g)-P otherwise, yields

1 . 0, g > 0, and

Jg = 1, then

1 + (Jfg)1/p < (J(1 + fl/P)Pg)l/P.

(*)

If now f >. 0, g > 0, and

JgP

1, =

then replacing g by gp in (*) and f by 0 when g = 0 and by

fPg-P

otherwise, yields

1+

<

(Jfp)1/p

EXERCISE 6.103:

(J(f + op) 1/p.

Let 0 < p < 1.

If f,g e LP set:

d(f,g) = Jif - gIP.

Show that this defines a metric on LP (with the condition that two functions equal almost everywhere are identified), and that, when provided with this metric, Lp is complete.

Is the mapping

f > d(f,0) a norm? AVA- = VOV = AVO = VLV = VAV

SOLUTION:

We have (x + y)P .5 xP + yp if x >. 0, y > 0, and 0< p-41,

CHAPTER 6: THE

228

as can be seen by studying the function

x->xp+1- (x+l)p. From this it follows that if f,g,h e L

d(f,g) = Jif - gIP < J(If - hl + <

I h - gI )P

Jf -hIP + Jig - hIP

= d(f,h) + d(h,g). Furthermore, d(f,g) = 0 implies that If - gIP = 0, that is to say f = g almost everywhere. To show that LP is complete it suffices to prove that if fn e LP and

Y d(f ,fn+i) = A <

n

NW

then fn converges to an element of L.

gN =

E

n=1

Ifn - fn+1l,

g =

E

n=1

Set

Ifn - fn+ll'

Then gN - g and

JgN

A

By Fatou's Lemma

JgP < A

Consequently g < m almost everywhere; but then the series

i (fn - fn+1) n

LP-SPACES

229

is almost everywhere absolutely convergent, which implies the existence of a measurable function f such that fn - f almost everywhere.

If r < s then

[Ifr - fslp ¢

I

d(fn,fnt1).

n=r

Making s - - and using Fatou's Lemma again yields

Ilfr -

f I p

<

n=r

d(fn'fn+1),

which proves that f = fr + (f - fr) e LP and that fn -> f in LP.

Since d(Af,0) = Ialpd(f,0), the mapping f H d(f,O) is not a norm

onLpif0 R.

These two sets are mapped into each

CHAPTER 6: THE

230

other by the symmetry x - a - x, and the first is contained in the 'annulus' R <

I x I

< R +

a

meas(BRtr)

Ilfa - flip E 2 (

meas(BR)

.

Thus when a e X

- 1)

1/p

Since meas(BR) = Rnmeas(B1) the right side tends to zero as R-;

and consequently can be made less then c by choosing R large enough.

EXERCISE 6.105:

If p > 0 denote by LP the set of positive func-

tions whose p-th powers are integrable.

Show that for all a > 0 the mapping f * fa is a topological isomorphism of LP onto LP/a

AVA = V AV = AVA = VAV = AVA

SOLUTION:

This is a matter of proving that

J If - file -->

0

Jjfa - file/a

implies

--> 0.

When 0 < a 4 1 we use the inequality

Ixa - ya'I ' Ix - yla,

x > 0, y a 0,

which gives

Jlfa - falPla < JIf - filP.

Now let us consider the case a > 1.

IJ

- Til

If - fiI (f V

fi)a-1,

By the Mean Value Theorem,

f V fi = max(f,fi),

and by HUlder's Inequality for the pair a,a(a -

1)-1

LP-SPACES

f I fa

231

iIp/a

-

<

If - f1Ip)1/a(((fV f1)p)1-1/a

ap/a((I

It remains to observe that

(fvf.) , (f + If-fiI)p 0 such that II9n11q < 1

and Jfngn

>-

IIfnIIP - 1.

Since Jfgn > Jfgn and IIfnIIP -; IIfIIP

the result is again true in this case.

Let E

SOLUTION: (c):

AE .

n = {x:lxl .< n and If(x)I < n}, and let

fn

=

If g is such that 11g11q < 1 there exists a sequence of sim-

n

ple functions (gi) such that IIg - gillq -

Jfg

0.

Then, because fn a LP,

= limJfngi = limJf(g 1E ).

i

i

n

The functions g nE are simple, II9iIlE Iiq < 1, and f(g IlE n tegrable.

IJfn9I

n

)

is in-

n

From this it follows that

< M.

By part (a) this implies that IIfnIIP < M, and therefore that

IIfIIP = 1nmllfnllP < M. Let fn be as above, and set

SOLUTION: (d):

u(g) = Jfg,

un(g) = Jfg,

9 e Lq.

The un's are continuous linear forms on Lq.

fng -; fg and Ifn9l E If9I, JIfI < un(g) -; u(g).

Moreover, for g e Lq,

By Lebesgue's Theorem

By the Banach-Steinhaus Theorem u is a continuous

239

LP-SPACES

linear form on Lq, that is to say there exists a constant M such Using part (c), it follows from

that IJfgI s M if IIgIIq < 1.

this that IIfIIp < M. Denote by En the set of step functions on]Rn.

EXERCISE 6 108:

Let f be a locally integrable function on]R

.

Show that if g e E, g > 0, then

Jfg = sup{ Jfh:heEn and 0 < h < g} if f is real.

(*)

From this deduce that in this case

J If g = sup{Jfh:h e En and -g < h < g}.

Finally, show that when f is complex

J IfIg = sup{IJfhl:h e En and IhI < g}.

AVA = VAV = tWA = VAV = tWA

By replacing f with zero outside the support of g one

SOLUTION:

can assume f is integrable.

< g, then since f

Jfh

<

Notice that if f is real and 0 < h

>. f,

Jfg.

When f is complex and IhI < g,

I Jfhl , JIfIi and, furthermore, if f and h are real:

Jfh< IJfhI.

CHAPTER 6: THE

240

All of this shows that the left side of each of the equalities to be proved is greater than or equal to the right side. Let c > 0, and let fl a En be such that

J If - f1I9 < E. We may further assume that fl is real if f is.

Set

fl hl = If g 1

1

(where we have adopted the convention that

f1If1I-1

= 0 when fl = 0),

and if f is real

f1+ h

= 2

f

9 1

(with an analogous convention).

It is clear that hl e En, Ih1I

< g, and that if f is real, -g < hl < g, h2 a En, 0 .< h2 .< g.

Furthermore:

Jflhl = JIf1I 9, and if f is real,

ffig

Jf1h2

=

-

When f is real,

J IfI9 -

ffhl = J(IfI - Ifll)g + J(f1 - f)hl

c 2J If - f1I9 < 2E.

Replacing IfI,1f11,h1 by f+,fl,h2, and noticing that If+ - fI

LP-SPACES

241

If - f1l, one obtains

Jftg- Jfh2 0,

and an integer N such that

lflp < e2 p,

JX_B

JIfIp < e2 -p

if meas(E) < ,

CHAPTER 6: THE

246

IIf-fntIp< 111p if n > N. By (1), if n > N and meas(E) < s:

JX-B

IfnIp < 6,

JE 'fn'

p 0 numbers such that

X-B n

IfnIp N, and consequently

11f - fnIIP < (JX-A IfIP)11P + ( JX-A Ifnlp)1/P

+ (J

IfI

)1/P +

JE

n

+

(J A-E

Ifl)n

If - fn1P)1/P

n

< 5c 1/p which proves that f e Lp and fn , f in LP.

REMARK: If fn e Lp(X), fn - f almost everywhere, and Condition

(ii) is satisfied, one again has f + f in Lp. n

It suffices to

replace the sets En in the proof above by a set E (Z A such that meas(E) < 6 and fn -> f uniformly on A - En, which is possible by Egoroff's Theorem.

Let (fn) be a sequence of functions in Lp(X),

EXERCISE 6.114:

1 f in Lp(X). (b):

Show that the conclusion above is still valid if the

convergence almost everywhere of f

n

to f is replaced by conver-

gence in measure.

avn - vov - ovo a vov - ovo

CHAPTER 6: THE

248 SOLUTION:

Let e > 0.

(a):

There exists a set B of finite meas-

ure and a number 6 > 0 such that

JXB Iflp < 2

< 2

j If IP

if meas(E) < 6.

E

By Egoroff's Theorem there exists a set A C B such that meas(B - A) < 6 and fn -+ f uniformly on A.

Therefore, by Fatou's

Lemma

f

X

lflp < e t fA Iflp < e t liminffA If Ip.

J

n

X

n

IfnIp = j Iflp, X

this yields

j Iflp

E+ J

X

Iflp - limsupj

X

i.e.

limsupfnIP t C. n-

J

Furthermore:

X-A

n- X-A Ifnlp,

249

LP-SPACES

If - ffIIP 4

IfIP)11P + (

( 1

X-A

LA ffIP)1/P I

+ suplf - fnl.meas(A)1/P, A

whence

l ymsupllf - ffIIP

2e1/p

which proves that fn -* f in LP(X).

SOLUTION: (b):

By the preceding exercise it suffices to show

that the sequence (IfnIP) conserves mass and is uniformly inteAs Fatou's Lemma is valid for convergence in measure,

grable.

the argument used in part (a) above shows that

JX-A IfPI4E implies that

limsup1 IfnIP n'°° X-A

6 e,

which proves that the sequence (If

P) conserves mass.

If this

sequence were not uniformly integrable there would exist a > 0,

a sequence (Ek) of sets of finite measure and some integers n1 < n2 <

< nk <

such that

meas(Ek) -* 0, (1)

JEk

If

nk

IP > a.

250

CHAPTER 6: THE

By considering a subsequence one would be able to assume addition-

ally that fn - f almost everywhere (cf., Exercise 3.54).

By part

k

(a) above fn

would tend to f in LP, and consequently the fn k k would be uniformly integrable (cf., the preceding exercise), which contradicts (1).

EXERCISE 6.115:

Let fn be a sequence of integrable functions on

a measurable set X of Rp.

Assume that meas(X) = 1.

Show that if

(a):

nil IRe(fn)I = 1, 11lX

limmJ I1-I.fnlI=0, X

then

IIm(fn)I = 0.

n JX

(b):

Now assume that

IfnIi

lim Re(fn) = n X n

1XIf

Show that lim n

JX

Ii - fnl = 0.

I

n

JX

2 = 1.

251

LP-SPACES 1Vt = VAV = 1VA = VLtV = EVts

SOLUTION: (a):

Set fn = un + ivn (un,vn real).

And

en = JXI1 - IffII

Now,

IfnI 4 1 + 11 - IfnII,

IvnI '<

and consequently

JXIVn1 4 1 + en.

Therefore the second hypothesis implies

= l YmssupJx IV n S 1. I

Assume that 0 < L 6 1 and choose a A such that 0 < A < £C.

By

considering a subsequence, one may assume that

JXIvnI > A

for all n.

Now let a,3 > 0 be numbers which we will specify

how to choose later.

Set

An = (11 - Iffii > a),

Bn = (IvnI 1 $).

CHAPTER 6: THE

252

Then lira meas(An) = 0.

nFurthermore

Ivn'

X

+

B

n

(1

J

X-B

n

+

n

4 Smeas(Bn) + 1 - meas(Bn) + En. If 0 < a < 1 this would imply

1 - A + E

meas(B) <

Set D

n

= A

n

U B

n

1 - B

On X - D

.

n

Ifnl < 1 + a,

IvnI

n

we have simultaneously

> a.

Hence

lung < d =

(1 + a)2 - S2.

Since lunI < 1 + 11 - Ifnll it would follow that

J Iun1 < 6(1 - meas(Dn)) + meas(Dn) + En

x

6 + (1 - 6)meas(Dn ) + c n . If 0 < 6 < 1, taking into account that

meas(Dn) < meas(An) + meas(Bn),

one would have

LP-SPACES

253

1=n

1

X

=1- (1-d)1-S

lunI 6 d+(1-d) 1_S

One is therefore led to a contradiction if O < 0 < A.

It remains

to show that a,$ can actually be chosen so that 0 < S < A (which S2

This is always

will imply y < 1)

and d =

possible choosing

a first and then a so small such that

(1 + a)2 -

< 1.

d < 1. With the same notations as in part (a),

SOLUTION: (b):

lira u= limJ Ifn

I

= 1.

(1)

The inequalities

IJX nI < JXlunI

<

JXIfn1

imply

lima lu

n

Xn

I

= 1.

(2)

Furthermore, the Cauchy-Schwar-z Inequality gives

(JXI1

- Ifn11)2 < JX(1 - I.fn12)21 (1 + Ifn12)2

With E = ±1,

ix(1

+ ElfnIj)2 = 1 +

2EJXIfn1

+

JXIfnI,

which, as n -> -, tends to 4 or 0 according as e = +1 or -1.

this it follows that

From

CHAPTER 6: THE

254

(3)

X Relations (2) and (3) allow us to use part (a), so

n-oJXIvnI = 0.

(4)

But then, from the inequalities:

Il - IffII + (IfnI - fnI

Il - fnI

J- - Ifn I

I

+ IvnI + Ifn I - un

and from (1),(3) and (4), it follows that rl-imm1X I 1 - =fn 0. I

EXERCISE 6.116:

Let X be a measurable set of 32m and f a measur-

able function on X.

IIfI

I

P=

(

As usual we write

J Iflp)11p if 0 < p < ., X

IIfiIm = ess sup I f I X

Assume that f is not equal to zero almost everywhere and set

if = {P:IIfIIp < W}. (a):

Show that If is an interval.

Can one choose f so that I.

is an arbitrary interval in ]0,co]?

(b) :

If If is not empty, show that a -} loglIf II l1a is convex

LP-SPACES

255

on its interval of definition (the convention 1/0 = W is used). From this deduce that p ** IIfIIp is continuous on If.

More pre-

cisely, prove that even if r is an endpoint of If and 0 < r < W,

then

I

IfI

I

p

+ IIfIIr when p e If' p + r, whether IIfIIr is finite or

not.

(c): Show that: p ** plogIIfIIP is convex on if - (W}. (d):

Deduce from the preceding that if 0 < r < p < s <

then

IIfIIp < max(IIfIIr,IIfhIs) Conclude that Lr(X)r)Ls(X) C LP(X).

AVA = DAD = ADA = VA0 = 000

SOLUTION: SOLUTION:

One may assume that f >. 0.

Let A = (f

1).

If r < p < s and r e If, s e If, when s < m one has

(a):

fP < fs on A and fP < fr on x - A, which proves that fP is integrable, and therefore that p e If.

If s = -, fP 0

when x + W, then If =] - a,W],

However, if 1/a f is continuous on ]0,oo[, f(x) = 0 for x >. 1, and f(x) v xbut If = [a,-] if f(x) v x-1/a(logx)-b with ab > 1.

when x + 0, then If = ]0,a[, but If =]0,a] if f(x) v x-1/a(logx)-b'

ab > 1. Lastly, if f(x) =x, If = 0. Considering the sums of two functions of the above type, any integral in ] 0,oo] can be obtained for if. SOLUTION: (b):

This is a question of proving that if r e If, s e If

CHAPTER 6: THE

256

and

p=n+ (ls A)

0< At 1,

then:

llfllp < llfllrllflls-A

(1) f(1-A)p

If r,s are finite, it suffices to make h = fAp g =

in

Holder's Inequality c ( J JX

JX

X

When 0 < r < s = -, one has p = r/A > r, whence

llfllp <

IIf(r/A)-rllWI

JX fr

=

IlfurAllfIIP(1-A)

,

which shows that (1) is still valid in this case. By a classical property of convex functions, from this it follows that a '+ logllfll1/a is continuous on the interior of its defining interval, and consequently that p -> llfllp is continuous on

the interior of If.

It remains for us to prove that if p e if tends

to an endpoint r of If, finite or not but different from zero, then

llfllp -. llfllr whether llfllr is finite or not. r is finite.

In this case fp i fr, and does so monotonically on

each of the sets A and X - A.

JAf p

-I

Assume first that

From this it follows that

JX-Af p ' JX-Afr

(the Monotonic Convergence Theorem; it will be noticed that as

the fp's are integrable, this Theorem can still be used in the case of decreasing sequences). case.

This proves the result in this

257

LP-SPACES

If r = m and 0 < Y < IIfllm then and

meas(f > Y) > 0

IIfIIp > Y(meas(f > Y))1/P

whence

lipnfllfllp

Y,

and consequently, in view of the arbitrariness of y,

limiflIfllp

IIfiL.

On the other hand, if p > r, with r e if, when

II f 1l

m < W one has

-r/Pllfll/p, r

IIfIIp < IIf IIl whence

limspllfllp 4 IIfil SOLUTION: (c):

This is a matter or proving that if 0 < r < s < =

and 0 < A < 1, then ,Xr+(1-a)s

(

f)A(( fs)1-A

(J

.<

X

X

X f1r,

It suffices to make h = rX Jfg < ((X ?/A X )

g = f(1-a)s

in Holder's Inequality:

fl/(1-a))1-A ( fX

J

SOLUTION:

(d) :

The inequality is trivial if Hf Ilr = m or IIfAI S = °°.

Assume, therefore, that these two numbers are finite.

If s is fin-

ite the inequality results from the convexity of logIlfIIp as a function of 1/p.

If s = -, for p < a < o one has

258

CHAPTER 6: THE

IIfIIp < max(IIfIIr,IIfIIc), and by making a

m the inequality sought is obtained because of It is then clear that

part (b) above.

LrnLS C Lp.

The notations are the same as in the preceding

EXERCISE 6.117: exercise.

It is further assumed that meas(X) = 1.

Show that IIfIIr have IIfIIr = IIfIIs < °°? (a) :

(b):

IIfil

if 0 < r < s < -.

Can one

Assume that IIfIIr < 00 for one r > 0.

Show that:

limjjfjjP

=

exp(JXlogIfl ),

where by convention exp(-m) = 0.

ovo = VAV - ovo = vov = ovo

SOLUTION: (a):

Let a be such that

+

=

r

s

I. Then a

IIfglIr < IIfIISIIgIIa

(1)

On making g = 1 the desired inequality

(cf., Exercise 6.106).

Since (1) is obtained by applying Holder's Inequal-

is obtained.

ity to fr and gr, equality can be had with g = 1 only if f is equal to a constant almost everywhere. By Jensen's Formula,

SOLUTION: (b):

loglIfIl

P

=

1 logJ fp " Jlogf. P

X

X

LP-SPACES

259

On the other hand, since logu 6 u - 1,

1 logJ fP <

[1 fp - 1] = J

P 1

X

X

up - 1). X

On A = (f > 1) p-1(fp - 1) decreases to logf, and on X - A it increases to the same function.

limJ P-1 (fp - 1) =

' X

Consequently

logf,

I

X

which proves that

lim

logIIfIIp =

Jlogf.

NOTE: As in the preceding exercise, we have assumed that f >. 0.

EXERCISE 6.118:

The notations are the same as in the preceding

exercise (in particular, meas(X) = 1).

Find all the functions 0 on ]0,-[ such that

a(l ollflip) = J (f ) for every bounded measurable function f > 0. 00A = V AV _ AVA = 000 - MMA

SOLUTION:

Let 0 < c < 1, and let A C X be such that meas(A) = c.

The function f, equal to x > 0 on A and to 1 on X - A, is such

that : l

llfllp = expJXlogf = exp(clogx) = xc

(cf., the preceding exercise).

As (f) = fi(x) on A and ¢(f) =

CHAPTER 6: THE

260

4(1) on X - A, one must therefore have: O(xc) = co(x) + (1 - c)O(i) for x > 0 and 0 4 c < 1 (the formula is in fact trivial if c =0 or c = 1).

If we set ct(x) _ iy(logG), where p is defined on at, the

preceding relation becomes

0 4 c 6 1,

ip(cx) = c,y(x) + (1 - e)V(0),

x eat.

From this it follows that there exist three constants al,a2,b such that:

ip(x) =alx+b

ifx>.0,

4,(x) =a2x+b

ifxt 0.

Let us then consider u,v, 0 < u < 1 < v such that uv > 1 and B C X such that meas(B) = Z.

For the function f equal to u on B

and tov on X - B

lIIfIIP = exp(ilogu + Zlogv) = u > 1. Consequently:

4(,IIfIIp) = jal(logu + logy) + b. On the other hand,

0(f) = 2(a2logu + b) + "(allogy + b).

J

X From this it follows that al = a2.

Hence 4 must be of the form

O(x) = aloge + b.

Conversely, if 0 is so defined, then

LP-SPACES

261

w( mII.fIIp) = alog(limll flip) + b p;G (

= a1 logf + b = X

EXERCISE 6.119:

(alogf + b) 1

X

For any function * increasing on [l,m[, such

that 4(l) > 0 and limi(p)

show that there exists on [0,1] a

p-,

measurable function f such that limilfilp =

and. Il,fllp < *(p)

P-1for all p. 400 = V AV = AVA = VAV - AVA

SOLUTION:

Let L

be the set of functions f, measurable on [0,1],

such that:

IIfII,y = sup

Ilfll 1f 11 <

It is clear that L norm on this space.

°°

is a vector space and that f - Ilfil

4 IIfiL. (cf., exercise 6.117)

Since IIfiL Y

Lm

it is clear that

is a

C L

Everything reduces to proving that

L- # LV for then there exists f such that IIf IL by exercise 6.116 one has limllfIIp

< 1;

= °°, IIfII

and on the other hand the

p'°° definition of IIfII

implies that IIfIIp s ,y(p) for all p

First, let us show that L complete.

provided with the norm f > IIf11

For this, consider a sequence (fn) of elements of L

such that:

E Ilfnli <

n

1.

is

CHAPTER 6: THE

262

For p > 1,

E IlfnllP < *(P) E Ilfnll IP

<

From this it follows that the series:

g(x) = E fn(x) n converges for almost all x.

Ilg - SOP <

Furthermore, if sN = fl +

+ fN,

IlfnllP < *(p) nIN Ilfn%,

which proves that g - sN e L., hence that g e L.J. and furthermore

Ilg - SNIIC'

E

n>N

IIffII*,

and consequently that

"M 119 - SN'I* = 0. N-

Assume that Lm = L of generality.

Ilfll

'

W) = 1 can be assumed without any loss

Since

= sup p>.1

Ilf1IP

*'p ¢ Ilfi

the norms 11.11, and 11.11- would be equivalent by virtue of a

theorem of Banach; that is to say, there would exist a constant

M such that for every f e L

Ilfli', < MlIfII*' Now this is absurd, as for every A > max(M,l) there exists p0 such that iy(p) >, A for p > p0; then if

LP-SPACES

263

f=AIL

-P [O,A

0l

one has

IIfil = A,

if p > p0,

IIfIIP < IIflIW = A F (p)

and:

when l 1 one has

IF(x + h) - F(x) I

=

x+fI

If x

h1 -1/P( x+hl fX

1/p. f I p)1/P < eh1 -

CHAPTER 6: THE

264

By part (a) above, the function f is uniformly

SOLUTION: (b):

0 as jxl -; W (cf., Exercise

continuous, and consequently f(x) 3.37).

EXERCISE 6.121: meas(X) <

Let X be a measurable set of IlRn such that 0 <

Moreover, let f and F be two functions defined on

X that are measurable and positive. Show that if for all A > 0

meas(F > A) .<

1JX(F>X)f,

then for all p (1 < p < 00)

(1 FP)l/P `

p

X

(1

1

fP)1/P X

A00 - V AV - eve - VAV = OVA

One can clearly assume that 1I fP < °D, as otherwise

SOLUTION:

there would be nothing to prove.

that (F

Let Fn = min(F,n); it is clear

> A) C (F > A) and that (F

> A) _ 0 if A

>, n.

From

this it follows (cf., Exercise 5.98) that

( I

Fn - p1(n0 AP-lmeas(F n > a)da

p(nAp-2da1

f(x)dx

4<

J

0

(F>A)

(min(n,F(x)) ( = pJ f(x)dxl

X

aP-2da

=1

0

By Ht5lder's Inequality, this yields, if

t

p

q

f(x)Fn(x)p-ldx.

X

= 1

LP-SPACES

265

fP)1/P((F np-1 )q)1/q.

XFp <

p - 1

J

J

Now (p - 1)q = p,

JX n

(J FP)1/P 6

X n

J

1

< - since 0 t Fn 6 n and meas(X) < m, so

(J

fP)1/P, X

p - 1

The desired result is now obtained by making n ->

and using the

Monotonic Convergence Theorem.

EXERCISE 6.122:

Let 1 < p < -.

Show that if f is locally inte-

grable the following conditions are equivalent:

(i) :

f e Lp; There exists a constant M such that for every sequence

(ii):

P1,...,Pn of disjoint rectangles that are of non-zero measure

n 1

meas(P.)p_111PfIP E M.

(*)

JP.

i=1

If these conditions hold, the smallest constant M that can be

taken in (*) is equal to

II f lip.

ove - vov - ovo = VAV = ovo

SOLUTION:

= 1, then

+

If f e LP and

q p if

fIP 4 meas(P .)p/q( P. 1

1

IfIp,

JP. 1

and consequently, because p/q = p - 1

CHAPTER 6: THE

266

meas(P.)p-11JP.fIp

i 1

<

JP.IfIp < IIfIIP

(1)

.=i

Let g =

Now assume that condition (*) holds.

c IlP. be a i

i

step function, with ci # 0, and the Pi being disjoint non-negligeable rectangles.

JfI = Ii

Then

ciJ fl P.

i

< (1

1

meas(P.)p-1

i

IJ

fIp)1/P(G IciI%ieas(Pi)q(p-l)/P)l/q P.

i

M1/pq Icil'meas(Pi))1/g

=

M1/pIIQIIq

i

By Exercise 6.109 we therefore have

IIfIIP < M.

(2)

Comparing (1) and (2) yields the last part of the problem.

Denote by En the vector space of step functions

EXERCISE 6.123: on ]R

.

Let T be a linear mapping from En into Lloc(l ).

that for every f e En IITfllgl < A111fllp1 ,

liTfilgo < AollfllPOJ

where 1 < pi,gi < -, AO > 0, Al > 0. Prove that if 1

1 - t

P

P0

t

+ P1

'

1

1- t

q

q0

t

+ q1

9

0< t c 1,

Assume

LP-SPACES

267

for all feE IITfIIq, AO-tA4IIfiIP.

(RIESZ-THORIN THEOREM)

(Set ai = 1/pi, a = 1/p, B. = l/qi, B = l/q and, for all z e C,

a(z) = (1 - z)a0 + zal, o(z) _ (1 - z)00 + zBl.

+ q = 1.

geEn be such that IIfIIP = IIgIIq = 1, where function f may then be written: U

f=

rke

E

k=1

i8

kIl

Let f e En and

The

4

, Ak

where rk > 0, 0k eR, and the Ak are disjoint rectangles of Rn;

similarly

ip

v

g= I

PRe

nB ,

R=1

with analogous conditions.

If a > 0 and a < 1 then set

a(z)/a l0kb rk e

fz =

Ak,

k

gz

P(1-B(z))/(1-B) 1Tp e

k

F(z) = 0

AzAz 1

IlBR

RmT(fz)gz.

Show that F is an entire function, bounded on the strip 0 4 Re(z) E 1, and that for all y eR IF(iy)I 4 1,

IF(1 + iy)I < 1. 2

By considering for e > 0 the functions G (z) = eEz F(z) to which e

CHAPTER 6: THE

268

one can apply the maximum principle, deduce that IF(z)I S 1 if Next examine the case where all - 0) = 0).

0 5 Re(z) S 1.

Use the Riesz-Thorin Theorem to prove that if f e Lp, g e Lq,

r

=

1 + 1 - 1

0, then f and g are convolvable, and IIf*gIlr

IIfIIPpIIgI

q (cf., Exercise 6.106). AVA = DAD = AVA = VAV = 0VA

SOLUTION:

Note first of all that

a(t) = a,

0(t) = 0,

(1)

and that if y eat, Rea(iy) = a0,

Re6(iy) = 00, (2)

Res(1 + iy) = Sl.

Rea(1 + iy) = al,

Also notice that if z belongs to the strip 0 4 Re(z) < 1 then so do a(z),R(z),l - z, and that if a > 0 then min(l,a) 4 IaZI s max(l,a).

F(z) =

Since

Al zAz kLk 1 ' 0

rka (z)/ap(9

1-S(z))/(1-S)ekk

T(IlAk), 1

it follows that F is an entire function bounded by a constant M for 0 4 Re(z) 6 1.

Hence by (2), upon setting qi = 1/(1

q' = 1/0 - 0), one has

Ilflyllpo = Ilfl+iyllpl

rk/ameas(Ak) = IIfAIP = 1,

=

k

Ilgiyllgo

= II gl+iyllg1 = R ameas(BQ) = Ilgllq = 1,

that is to say:

LP-SPACES

269

Ilfiyllpo = IIf1+iyllpi = Ilgiyllgo = IIg1+iyllgi = 1 (if one of the numbers p0,p1,q

or qi is equal to =, these form-

ulae are still true; for example, if p0 =

Ilfiyllm =

then Rea(iy) = 0 and

= 1).

maxlrk(iy)/al

HSlder's Inequality then gives

IF(iy)I

A0IIT(fiy)IIg0llgiyllgo

1

F(1 + iy) I . AiI I T(f1+iy) I I q1 l I g1+iy ll ql . 1.

If e > 0 and G(z) =

IG(x + iy)I =

eez

2 F(z), if 0 0.

Fin-

ally, let T be a mapping of E into the set of measurable functions on Y such that

JT(f + g)l < lT(f)l + IT(g)I for any f e E, g e E.

(1)

273

LP-SPACES (a):

Assume that p1 < m and that there exist numbers AO > 0,

Al > 0 such that for all f e E and all A > 0

AiIIfIIP Pi meas(ITfI > A) <

1

i = 0,1.

,

(2)

Show that for all p,p0 < p < p1, there exists a constant Ap,

that depends only upon pO9p19p,A09A1, such that

IITfIIp , APIIfAIP,

fe E.

(3)

(Use Exercise 5.98 and the decomposition f =

f + f

and evalu-

ate the quantities meas(IfXI > t),meas(IfAI > t)).

(b):

Show that (3) still holds when p1 = W, if the inequal-

ity in (2) for i = 1 is replaced by

f e E.

IITfll0 < (c):

(21)

Assume in addition that meas(Y) < W and that if f e E,

A > 0, and

fi(x) _

f(x) if I f(x)I 0

then fA e E.

A,

otherwise,

Set f = f

Show that if 1 = p0 < p1 t m and (2) is satisfied (and (2') if p1 = co), then if C < p < 1 there exists a constant Ap such that

IITfIIp . APIIfII1,

f e E.

(4)

(Evaluate the integrals:

J0pxPlmeas(TfI > A)da,

JP_lmeas(ITfI > A)da,

a

274

CHAPTER 6: THE

then make a = (d):

11fII1)

The hypotheses are as for Part (c), and assume further

than meas(X) < oo.

Show that there exist constants B,C such that

I I TfI I14 B + CJX IfIlog+lfI

(Use the decomposition f =

VT

+

(5)

, then evaluate the inte-

grals: 1

rm

meas(ITfI > A)da, f0

meas(ITfI > A)da. f

1

Also note that there exists a constant M such that u 6 M(1 + ulog+u) for all real u).

NOTE: Formula (3) is a weakened form of a more general theorem of Marcinkiewicz (cf., for example, R.E. Edwards: Fourier Series, Vol. II, (Holt, Rinehart and Winston, Inc.), pp. 157 et seq.).

/VA = VAV = AVA = vtv = A4t

SOLUTION: (a):

Let

p(a) = meas(IfI > A),

Since JTfI 5 ITf

W) = meas(ITfI > A).

+ ITffI by (1), it is clear that

(ITfl > A)C (IT?i > 2A)U(ITfXI > jX), and consequently:

275

LP-SPACES

W) 2A) + meas(ITfAI > IX)

(6)

Furthermore, by (2)

2A0IIfAIIP

PO O

meas(IT? I >

,

a

P1

(7)

111faIIP meas(ITfXI >

1

A)

From (6),(7) and Exercise 5.98 it follows that

IITfIIP = 0

p(2A0)PO0aP-PO

dal If(x)IPOdx X

(W

1

+ p(2A1)Pll 1P P1

daj IfA(x)IPIdx.

X

0

(IfI > t) if 0 < t < 1,

if t

0

a,

and that If"(x)I > t > 0 if and only if If(x)I > A, and If(x)I(1-AIf(x)I-1)>t,

or in other words

(III>t)_(IfI>t+A).

(8)

276

CHAPTER 6: THE

Thus

meas(I? I > t) = q(t + A),

meas(IfA I

> t)

-

(p(t)

if0 -i and t - A 5 t, JoAp-pO-1d)LJot1) 0-1c(t

+ A)dt 5

t

1 =

J

WtPO

dA

Jo

0

=

1

Ap PO

q,(t)dt

1

JtP_1(t)dt

p-p00 1

=p(p-P 0 )

IIfIIP.

Similarly, because p - p1 - 1 < -1, pl-1,(t)dtJtAp-pl-idA

AP-pl-1dA(otPl-1q(t)dt

J

= 0

J- t

=

LP-SPACES

277

1

tP-1cp(t)dt

pJO

pl

p pl-Z-

l )-llfllP.

(12)

(10),(11) and (12) yield

Pi

p0

IITfIIP .

P0(2A 0) I

pp,lp ) IIflIP.

+

P - P0

1

which proves (3). SOLUTION:

(b):

When p1 =

' and (2') holds, write

A/2A

meas(ITfI > A)

mI > -)L) + meas(ITfa/2AWI

-,-c meas(ITf

> #L)'

Now the set (ITf1AAml > A/2) has measure zero, since almost every-

Tfa/2A

-1 < A.

2..

11fa/2A_I1- .

From this it follows that

'A)dX

IITfIIP : 0

W ppO1 pOp(2AO) P0 J a 0

m p0-1 A/2A. meas(If dal t

I

> t)dt

0

p0p(2A0)poJXp-pO-1 dXJ0tP0

=

O 0

l

o

+

( P

2Am)dt

Ilt

(Contd)

CHAPTER 6: THE

278

p-p0-1

p0 , pop(2A0) 1

(Contd)

o

p0(°° p0-1

= pop(24 )

t

J

p0- 1

Q(t)dt

t dxJW A/2AW

A

(2A t p-p0-1

c(t)dtJ

A

dA

0

0

_

p

-

po

p0(2A0)P0(2Am)P-P0

HA P.

p0

P

SOLUTION: (c):

0

P

It can be assumed that a = IIfIIl > 0.

also be assumed that 1 = p0 < pl < -; indeed, if p1 if p0 < pi <

It can and

by part (b) P

meas(ITfI > A) S lr xPl

ITfI ITfI>A

IITfIIP1

- S P1

Pl

APl

P;

a

This being so,

a J0PAP1AdA

1

meas(Y)J padA = ameas(Y). 0

On the other hand

Vi(a) 4 meas(IT? I

> 2A) + meas(IT."

I

>

'fA) 5

(13)

279

LP-SPACES

2A0I

II

1

2A

1 Il a II

Pi

Pi

whence:

ja

2pA0JaXp-2dXjIf(x)I>aIf(x)Idx

pAP-1iP(A)da

a

Pi °° P-P1 1 + p(2A1)

P1 dA J

1 A

If(x) I

dx

a

1

In the last integral If(x)Ipi 4

JpAPhlp(A)dX

< 2pA0 J

a

+

If(x)I, and consequently

API-

If(x)Idxj X a,T

1

If(x)Ipldx.

+ (2A )P1 l

da 1 xp1fIf(x)I5y1-x(P1 -1)/2

If(x)IP1

In the last integral

JP(A)dA : 2A 1 If(x)IdxJ 1

If(x)I, whence

5 A

X

a-1da 1 0, y >. 0, 2

xy 0.

Let d > 0

By Egoroff's theorem there exists a

set A of X such that

meas(X - A) < 6

and

En = suplanfn(x)l -> 0. A

Then lan12 = J

lanfnl2

X

= fA

lanfnl2 +

fnl2 s

JX_A'

(Contd)

289

.< en + M21an126,

2

lanl

En

2

2

1-Md

which proves that an - 0. SOLUTION (b): I

b

Choose a sequence (bn) such that bn -+ 0 and

(for example, bn = 1//n).

almost everywhere.

Assume that E Ibnfn(x)I2<

Choosing 6 as in part (a), there exists a set

A such that meas(X - A) < 6, and

L

n>n0

Ib f (x)I2 < 1 n n

for all x e A whenever n0 is large enough (Egoroff's Theorem). Now, for every n.

1=JAIfnl2+JX-AIfn 12 < JAIfn12 + M26,

so

1 3 meas(A) >,

which is absurd.

X n>,n0

lb

12J

n

lbn1i2, E A Ifn 12 % (1 - M26) n,n0

290

CHAPTER 7:

EXERCISE 7.129:

Let U be an open set of ]Rp and H = L2(U).

Re-

call that the Gram determinant associated with elements f1,..., fn of the Hilbert space L2(U) is defined by

G(f1,...,fn) = detll(.fil.fj)II1, -m, and therefore that snpIsn(x)l < - almost everywhere.

SOLUTION (I):(b):

Here again it may be assumed that the an's are

Let n1 < n2 <

real.

be a sequence of integers such that

2

a- 4 k-4 n>nk

Set n0 = 0, wn = l if 0 1.

Then 2

2

w a = G a2 + k2 a2 nk4n (ii). Now assume that (ii) is satisfied.

f1(x) =

g(t)dt,

Let a < x0 < b, and

a < x < b.

Jx0

By what has just been proved,

rb

rb

a

a

I fl9'+J gro=0, f

9 e D-(-T).

Comparing this with (ii) yields

rb I

(f - f1)p' = 0,

'p

e DP(I).

JJJa

When 9 runs over such that Ji = 0.

9' runs over the set of functions *e D-(I)

By exercise 8.141 there exists a constant C

such that f - f1 = C almost everywhere, which proves (ii) _> (i). Lastly, if the pairs (f,g),(f,gl) satisfy (ii), by substraction one obtains

CHAPTER 8: CONVOLUTION PRODUCTS

344

b

9 e D(I),

(g - g1)T = 0, a

whence g = g1 almost everywhere.

The Formula defines a positive alternating form

SOLUTION (b):

on H2(I), and (fIf) = 0 implies that

2(I , whence f = 0.

1IfII

L

It

)

2

remains to prove that H (I) is complete in the norm associated 2

with this scalar product.

If (fi) is a Cauchy sequence in H (I),

(f ) and (Df ) are Cauchy sequences in L2(I).

Hence there exist

f,g e L2(I) such that fi - f, Dfi - g in L2(I). If p e DP(I), then b

b

Dfi.W = 0,

Jafi.cp' +

fa

and on passing to the limit: b

rb

fp' + J

a

gcp = 0. 1

a

Consequently g = Df and fi - f in H2(I).

SOLUTION (c):

If cpe D00(JR), then by Plancherel's Formula and the

relation ,'(y) = iyV(y),

+0

_ JP?(Y)TTdY

r+0 = 1 _ Df(x)Zxsdx

+W

Jf(x)c'(x)dx

=-

J](Y)iy(y)d J

=

J00Y(Y)EYYdY. _

As the y e DW(G) are dense in L2(It), and as the Fourier transform-

ation is an isometry of L2(R) onto itself, the $ are dense in

AND FOURIER TRANSFORMS

345

Hence bf-(y) = iyf(y) almost everywhere.

L2OR).

First Proof: Note that if A =

SOLUTION (d) :

IIDf II 2

L (I)

and

a < x < y, then

4 A.

If (y ) - f(x) I

If f(x) were not to tend to zero as x ->

there would exist E > 0

and a sequence of points xn such that 2

xn+l > xII

+

E

2 ,

If(xn)I 3 E.

4A

Consequently, for xn < x < xn + E2/4A2

I f(x) I a

I f(xn) I- If(X) - f(xn) I

E- A 2A =

2

But then

x +c 2

W

J If(x)I2dx > I j n

a

n xn

4A

2 E

If(x)I2dx

2

2 E

= W.

n 4 4A2

Second Proof: We can reduce to the case where I =]R by the folartifice: if a > -m, f is extended by continuity at a (cf.

(i)

of part (a)); let a < a, and let S be a continuous function with support contained in [a,a] and such that

a 0(t)dt = f(a). fa

By extending Df to [--,a] by setting Df(t) = 8(t) for t E a, E±nC setting

x f(x) = f(a) + J Df(t)dt, a

x e ]R,

CHAPTER 8: CONVOLUTION PRODUCTS

346

an extension of f is obtained that belongs to H2(]R).

Now, for

f eH2(JR) the functions ?(y) and iy7(y) belong to L2(B) by part (c).

From this it follows that (1 + y2)27(y) e L2(]R);

(1 + y2) 2 e L2(kt) we have

since

But then f is the inverse

'e L1(R).

Fourier transform of ?, and therefore tends to zero at infinity.

SOLUTION (e):

Let us show that f + f(c) is a continuous linear

To do that, write

functional on H2(I).

(c

f(c) = f(x) + 1 Df(t)dt,

a 4 x < b,

x so ,

If(c)I 4 (b - a)2IIDfII

2

L (I)

+ If(x)I,

or

If(c)I2 4 2(b - a) IIDfII22 L (I)

21f(x)12.

+

(*)

Integrating with respect to x from a to b yields

I f(c)

12

, 2(b - a)

112

112

+

b-a

L2(I)'

which proves our assertion.

The existence and uniqueness of i

c

results from this.

In par-

ticular, if W e V (I),

c

b

cp'(x)dx = a

b

J(p(x)*c(x)dx + a

a

which can again be written, by introducing the characteristic function Xc of the interval [a,c],

AND FOURIER TRANSFORMS

347

b

b

Jaq,'(x){Dhc - Xc(x)}dx + JaV(x)*c(x)dx = 0.

Comparison with (ii) of part (a) shows that this implies

DOC - Xc e H2(I)

Since *

c

and

D(D*c - Xc) _ *C.

is continuous Dpc - Xc is continuously differentiable

and its derivative is * c.

(1)

c

In other words,

is continuous;

(2)

p" = i on [a,b] - {c};

(3)

y' (c - 0) - *C, (c + 0) = 1.

Furthermore, if a < y < c and if f(x)

x)+, then Df = -X

,

and consequently,

0 = aJ(Y - x)*c(x)dx - j

Y

- x)*(x)dx.

c(Y) - c(a) = Ja

If M is the maximum of I'Pci on I, it follows from this that Y

IVC(Y) - c(a)I < MJ (Y - x)dx < 'M(Y - a)2, a

and consequently *'(a) = 0.

Proceeding analogously for the other

end point, to conditions (1),(2),(3) can be added

(4)

*c(a) = *c(b) = 0

CHAPTER 8: CONVOLUTION PRODUCTS

348

(this is in the case where a < c < b).

From (2) and (4) one ob-

tains

Acosh(x - a),

a < x < c,

Bcosh(b - x),

c < x C b.

Conditions (1) and (2) imply the system Acosh(c - a) - Bcosh(b - c) = 0, Asinh(c - a) + Bsinh(b - c) = 1. From this it is not difficult to deduce that

c (x)

cosh(b - c)cosh(x - a) sinh(b - a)

a 4 x 4 c,

cosh(c - a)cosh(b - x) sinh(b - a

c

IIDf JJ

2 LI)

is, on

H2(I), a norm equivalent to that induced by the norm of H2(I).

The existence (and uniqueness) of 8c follows from this. The orthogonal projection of Xc onto the orthogonal of 1 in L2(I) is:

b-c b Xc(x) - b l a JXc(t)dt = a

ba

a4x4c,

c - a

and consequently

(b - c)(x - a) -

a)

(c - a)(b - x)

-a REMARK:

a

4x4c ,

c.., n, fn(x) = 0 otherwise.

Show that f*G is defined everywhere and that there exists a continuous function hn such that

f*G = hn + fn*G.

Conclude from this that f*G is continuous.

AVA = vov = AVO = VAV = AVo

SOLUTION (a):

JIfI

= J

Noting that log3 , 1 gives

IfI2IfI

< 4n +

Jlfllog(l + Ifl ),

Furthermore, if a,b , 0 then

ab < alog(1 + a) + eb - 1.

(YOUNG'S INEQUALITY)

Consequently, if A > 1

J If(x)g(y - x)ldx < xJIfIlog(1 + AI.fi) +

JceA'I

Now log(1 + alfl) < loga + log(1 + IfI), so

If*gI < AlogAJIfl + AJIfIlog(1 + Ifl) + J(eA-1Igl - 1).

SOLUTION (b):

G(x) ti log

If x0 is a root of cosx, then as x -. x0

Ix -ix01

AND FOURIER TRANSFORMS

353

and consequently eA-1G(x) ti

Ix

- x

0I-1/A

This shows that G e L1 and exp(A-1G)e L1 if A > 1.

The function

fxg is therefore defined everywhere, as is fn*G, moreover, since fn also satisfies (*).

As for hn = (f - fn)*G, this is also a

function defined everywhere, since f - fn is bounded. cisely, hn is continuous, as LW *L1 C C.

More pre-

For A > 1 and every

integer n

IIf*G - hnil. 1

< AlogAJlfnl

Since Ifl l

>.

+ Xflfnllog(l

+ Ifnl) + J(ea_

G

- 1).

>. --- and I fnI - 0, this yields

l f21

-1

limsupllf.G - hnll 4 J(eXG n-Making A } W shows that hn -> f*G uniformly, and therefore that f*G is continuous.

EXERCISE 8.149:

Prove that the algebra L1(Rn) does not have a

unit element.

AVA = V1V = AVM = VtV = OVo

FIRST SOLUTION:

If f e L1 were such that f*g = g for all g e L1

then taking the Fourier transform gives

=

Now, for all

x e32n there exists g e L1 such that g(x) + 0, and consequently

?,(x') = 1, which is impossible, because f must tend to zero at infinity.

CHAPTER 8: CONVOLUTION PRODUCTS

354

There would exist a > 0 such that

SECOND SOLUTION:

JIf(x)Idx -1

1.

Let p be the characteristic function of the ball with centre zero

and radius a/2.

f* (P (x) =

Then if IxI 4 a/2 one would have

f(x - y)dy .

J

If(y)Idb < 1,

J

Iy I < a

IHI -< a

a contradiction.

EXERCISE 8.150:

Prove that the algebra L1(,Rn) possesses divisors

of zero.

AVO = vov = ove = VAV = ovo

SOLUTION:

First Proof: Let V1 and V2 be two non-empty and dis-

joint bounded open sets of Mn.

Let p1,(P2 be non-zero indefinite-

ly differentiable functions compactly supported in V1,V2 respect-

If fl = F(wl), f2 = F(cp2), then fl eLl(R' ), f2 eL'(Itn),

ively.

fl 4 0, f2 4 0, and f1*f2 = 0, since F(f1''f2) _ 9192 = 0.

Second Proof: We shall assume that n = 1.

If a e]R, denote by Ta

the operator of translation by a, that is to say, that Taf(x) = f(x - a).

It is known that Ta is an operator on L1 with norm 1,

and that

TaOTb = Ta+b'

Ta(f*g) = Taf*g.

For every summable sequence a = (an)nex' +M T

a

=

I n=-m

an T

nn

(1)

AND FOURIER TRANSFORMS

355

is a continuous operator of L1 (the series (1) converges normally in L(L1,L1)).

If

n)ne2Z

Tao Ts = Ty,

is another summable sequence, then

apq.

Yn =

(2)

p+Q-n Now assume that we have determined a,s so that

TOT0 = 0,

a 4 0,

S 4 0.

(3)

If m e L1, then (TQ*Ta9) = (ToT$)((p*c) = 0.

It will be possible to ensure that Tag # 0, T9 # 0 by taking for 9 a function zero outside ]0,1[ and such that 11N111 > 0; then the functions Tncp will have disjoint supports, so that, for example,

11Tam 111 =

11m 111

G 1%1 > 0. n

In order to determine a,s so that (3) is satisfied, note that if

u(x) _

anell"{,

v(x)

Snel n

n then:

uv = 0,

(4)

and conversely (4) implies (3).

Consider the function Isinxl;

its Fourier coefficients are

en =

n

(n 1,R1

_ e -ice n

It can be shown that

l sinx ldx

=

nJ cosnxsinxdx. 0

CHAPTER 8: CONVOLUTION PRODUCTS

356

c2n+1

2

C2n-n

0, =

1

The function Isinxl being continuous and piecewise continuously differentiable, the elementary theory of Fourier series ensures that +00

cneinx

I

Isinxl =

n=-m But then

(sinx)+

4

=

+co

(eix - e-ix) +

cneinx

n=-(5)

- (sinx)_ = 42

+m

(e ix - e-ix) -

C

cneinx

z

n=G-w

so that the sequences a and S defined by

a1

= s1

1

=1,

a-1 = 0-1

4 , 1

a2n

- s2n

1

n

1 - 4n2

a2n+1 = 82n+1 = 0 satisfy (3).

if n + 0 and -1,

One can, if one wants, obtain real sequences on re-

placing x by x + 1/2 in (5), which gives a1=S1=a-1=S-1,

a2 n

- - stn

1 n

(

- 1)n

1 - 4n2

a2n+1 = 02n+1 =

0

if n4 0 and -1

AND FOURIER TRANSFORMS EXERCISE 8.151:

357

Let L+ be the set of functions locally integrable

on ]R and zero on ]--,0[.

Two functions of L+ that are equal al-

most everywhere are identified. (a):

Show that if f,g a L+, then they are convolvable, and

fiegeL+. (b):

Show that if f e L+ and if f*f = 0 almost everywhere on

[0,2a], a > 0, then for every integer n >,

a enxf(a

-

x)dxI2 4

1

If(a - u)f(a - v)Idudv. JJ

a

u>-a v>-a u+v>O

Using Exercise 3.72, deduce from this that f = 0 almost everywhere on [O,a]. (c):

Show that if f,g e L+ and f*g = 0, then, setting f1(x)

= xf(x), g1(x) = xg(x),

f*gl + f1*g = 0, and consequently f*g1 = 0.

(d):

Conclude from the preceding that the algebra L+ does

not possess divisors of zero (TITCAMARSH'S THEOREM). tVt = VAV ° AVA s VtV = AV1

SOLUTION (a):

Let M > 0 and let fM(x) = f(x) if x < M, fM(x) = 0

if x > M, gM being define analogously. Then if x < M,

x

tom

J

-WIf(y)g(x

If(y)g(x - y)Idy =

- y)Idy = j 0

(Contd)

358

CHAPTER 8: CONVOLUTION PRODUCTS

(Contd)

= Jx

fM(y)gM(x - y) dy

0

= f

fM(y)gM(x - y) dy.

Since fM,gM are integrable, it follows first that (f?,g)(x) is de-

fined for almost all x < M, and consequently for almost all the

x eat, and then that x f(y)g(x - y)dy.

(f*g)(x) = 0

This shows that (fig)(x) = 0 if x < 0.

Furthermore, for almost

all x e [0,M] (frcg)(x) = (fM*gM)(x),

which proves that f*g is locally integrable, and hence in L.

SOLUTION (b):

a

fJ en(u+v)f(a

enxf(a - x)dx)2 =

(

-a

- u)f(a - v)dudv

(ul'a IV1,-a

v>,-a u+v50

+

ff u., 0.

Let

be a sequence of compactly supported continuous positive functions

i

such that f + f in Lp.

Then F. -' F in LP, and so by exercises 1

6.105 FP-1 + FP-tin Lp/(p-1) = Lq.

Replacing F by Fi in (1) and

passing to the limit, shows that this formula is still valid for f.

As (2) results from Holder's Inequality applied to the right

AND FOURIER TRANSFORMS

365

side of (1), equality can hold in (2) only if it holds in this Holder inequality, that is to say, if there exists a constant A >, 0 such that

FP = F(p-1)q = A/p, Let us note that if

that is to say F = Bf, B >, 0 a constant.

f $ 0, we necessarily have F 4 0, whence B > 0.

Since F is to

be continuous f would be also, and by differentiation one would obtain

Bxf'(x) = (1 - B)f(x), that is to say:

C > 0,

f(x) = Cxa,

a =

1 - B B '

This is absurd, because whatever a may be such a function does not belong to Lp(O,m). If f e LP(O,W) without being positive, but if f 4 0 almost

everywhere, then setting

G(x) = xf

lf(t)Idt 0

yields

-]IF IIp < IIGIIp <

REMARK:

p p 1 IllfI Ilp = p p

l

Ilfllp-

The second proof given for part (a) above is more ele-

mentary than the first proof, which uses the generalised Minkowski inequality.

Further, the proof of part (b) is based upon

formula (1), which is the essential point of the second solution.

One can avoid using exercise 6.105 to establish the validity of (1) whenever f e LP(O,m) is only assumed to be positive, by pro-

CHAPTER 8: CONVOLUTION PRODUCTS

366

ceeding in the following manner, which furnishes a third proof of Hardy's inequality. Let

(P(x) = eX/Pf(eX),

e-x/q

if x

I

0,

ifx P,

a generlization of Hardy's inequality which corresponds to the case -g = p.

To conclude this remark, let us indicate to the readers who know about the notion of convolution product on the multiplicative group ]R

that the preceding is expressed more simply in the

following way: If 9(x) = x1/Pf(x), and

368

CHAPTER 8: CONVOLUTION PRODUCTS

x1/q

for x >, 1,

0

if 0 0. Denoting by f*g the function thus defined almost everywhere on ]0,oo[, show next that

370

CHAPTER 8: CONVOLUTION PRODUCTS

IIf'=gII

r

`

-1

L (t

IIf1I

dt)

-

p

11g 11

L (t

q L (t

dt)

if r = p + q - 1,

-1 dt)

= 1, then f*g is continuous, and

Lastly, show that if p + Q

limf*g(x) = lim(ffg)(x) = 0. x-

X-+0

AVA = VAV = AVA = 0E0 = A0t

SOLUTION:

This is a matter of developing the last part of the If we set F(x) = f(ex), G(x) _

remark in the preceding exercise.

g(ex), it is not hard to verify that f*g is defined almost everywhere if and only if` F,G are convolvable in the usual sense, and that then (F*G)(x) = (feg)(ex).

Then it suffices to use the ele-

mentary properties of the convolution product on ]R, exercise

6.106, and the equality IIFIILp R)

= IIf1ILp(t-1dt)

EXERCISE 8.154:

< a < 2 - p ; for f e Lp(0,co)

Let p >. 1 and 1 -

set

p

F(x)=Tf(t) sinxtdt. to

Showthat for all r >, p there exists a constant A(r,p,a) such that

(

f xr(l - 1/p - a)-1IF(x) Irdx)1/r

J0

Show, further, that if p > 1

< A(r,p,a) IIf IIp.

AND FOURIER TRANSFORMS

371

=o(xa-1+1/p)

F(x)

asx -> 0 or x

Deduce from this that o(Ihla-1 + 1/p)

suplF(x + h) - F(x)I =

X ovo = vov = ovo = vov = ovo

SOLUTION:

Note that

F(x) = xl/p + a-1( tl/p f(t) J0

sinxt

dt

(xt )1/p + a-1 t

'

Let

W(t) = tl/pf(t),

X(t) =

tl -1/P -a

sint,

so that

F(x) = x1/p + a-1

d' J(t)x(rt)

.

0

- xl/p +a-1

tt

.

in

Setting ap(t) = rp(t-1) and using the notations of the previous

exercise, this becomes x1/p+a-1('*X)(x).

F(x) = Now observe that

IN 11

Lp(t

-1

=11w1I

dt)

Lp(t-1dt)

=Ilf1I p

CHAPTER 8: CONVOLUTION PRODUCTS

372

and that

r

UXII

L

0 0, which ensures the result for r

for

r 1 O*x is continuous and tends to zero as x -> 0

or x - m, which proves that

F(x) = 0

(2-1+1/P).

Finally,

F(x + h) - F(x) = 2f(t)

E

cos(x + Zh)tsinj'ht dt,

to

AND FOURIER TRANSFORMS

373

o(jhla-1 +11P).

suplF(x + h) - F(x)I < 2F(2'Ihl) =

x

EXERCISE 8.155:

Set

Let p > 1 and f e LP(O,m).

F(x) = J0e_Xtf(t)dt,

x > 0.

Show that for r >, p

(f J

- 1I F(x) Irdx)1/r < 0

1/qr (411/S

l-I

11f 11p, Il

J

p+q= 1, Show also that

F(x) = O(x-1/q)

as x -* 0 or x -* -.

ove = vev = evo = vev = ovo

SOLUTION:

Proceed as in the preceding exercise, observing that

if tl/qe-t,

W(t) = t1/Pf(t),

X(t) =

then F(x) =

x-1/q(,*f)(x).

Furthermore, if a > 0

Ilx

IILS(t-tat) =

(jo S/q -1 e- St dt) 1/s

=

I

s

l 1/qr

1411/S.

CHAPTER 8: CONVOLUTION PRODUCTS

374

Show that a subset H of Lp(]R ), 1.4 p < -, is

EXERCISE 8.156:

relatively compact if and only if it satisfies the following conditions: (i):

There exists a number M such that for all f e H:

IIf1Ip-C M; (ii):

For all e > 0 there exists a compact set X of stn

such that for all f e H: <

JnIf -K (iii):

;

For any e > 0 there exists a neighbourhood V of zero

in ]tn such that for all a e V and all f e H: Ilfa -

fllp¢e.

V. denotes the translation of f by a; to show that the conditions are sufficient prove that for every compact set K of Iltn and every

continuous and compactly supported function 9 the set of functions (AK )*q), f e H, is equicontinuous).

A0A = 0A0 = A00 = 0A0 = A0A

SOLUTION : The Conditions are Necessary: If H is relatively compact in Lp, H is bounded, which proves (i). For every e > 0 there exist functions f1,...,fr e Lp such that

sup( min IIf - fS II P )

t 6.

feH l*ssr

There is a compact set K of Ltn such that

J]Rfl-KSIp 4 ep,

1 4 s 4 r.

AND FOURIER TRANSFORMS

375

Then for all f e l l Iflp < (2e)p,

1

and ,(Pn > 0;

1; 1 uniformly on every compact set of ntp.

(Let p > 0 be such that

Jf(x)dx > 0.

Ixk2p For every integer n > 1 let On a

0 4 On < 1, 0 (x) = 1 if

jxj 4 2np and 0n (x) = 0 if jxi > (2n + 1)p.

consider some x e]R n (f*8n)(xn)

I.1

such that

-

1a.

Show that if hn(x) =

a-1

t xn

Set an =

hhf*$n11. and

378

CHAPTER 8: CONVOLUTION PRODUCTS

one has Iif*hn II = 1 and when x < p (

11 an

(f*h ) (x) a I1 n nJJan+1

(c):

Deduce from this that if g e Ll(a) is such that f*g

then

Jg(x)dx 3 0. AVO = VAV = AVA = VAV = avo

SOLUTION (a):

If

a limsup

n

n

an+1

one would have, starting from some n0, an+l > )an, and consen-n 0 when n 3 n0, which contradicts the hypoquently an > an0 x thesis an = 0(na).

SOLUTION (b):

With P.0n,an being as indicated in the question,

note that if IzI < p, then for all x e]i

(")

0n(x) < 0n+1(x + z).

In fact this inequality is clear if x

(2n + 1)p, and in the

contrary case

Ix + zI < (2n + 1)p + p = 2(n + 1)p, and therefore

0,

AND FOURIER TRANSFORMS

379

On+1(x + z) = 1 >, 8n(x).

In particular On < $n+1' whence an < anti.

Now note that

f(x)dx > 0,

a1 >. (ffcq) 1)(0) >1

I r . I1 - 11 n ` nJJa n n n+l an+l

By (**) and part (a), for all e > 0 there exists an integer n0 such that

(1- it

a n

0 >1-E.

n0 an0+1 Setting 9 = hn

one has therefore determined for every p large

enough and all c > 0 a function p e D", c 3 0, such that Ilf*c L = 1 and

380

CHAPTER 8: CONVOLUTION PRODUCTS 1 - e .<

(f*q,)(z) 4 1

the function corresponding to p = n n and e = 1/n, one obtains the desired sequence. Denoting by w

if Izi < p.

SOLUTION (c):

This follows from

) Jg = fg = lim(g,f*p n) = lim(g*f,p n} n n

and the fact that g*f

EXERCISE 8.158:

IJf(x)dxl

0 implies that (g*f,wn)

Let f e Li(R ).

=

0 if Tn > 0.

Prove that

infJIA1f(x - a1) + ... + Anf(x - an)Idx,

where the infimum is taken over all the systems of elements a1, ...,an e]R

and positive numbers ai,...,A

such that Al + tan = 1.

(Use the result proved in the preceding exercise).

AvA = vAv - AvA = vov = AvA

SOLUTION:

First let us fix some notations.

As usual set f(x)=

f(x - a); in addition, if f e L1 we shall set

I(f) = Jf(x)dx.

Let P be the set of positive functions on]Rp that are zero except at a finite number of points and are such that:

E

A(a)=1.

aeatp If A e P, set:

(A*f)(x) =

I

adRp

A(a)f(x - a).

AND FOURIER TRANSFORMS

381

There is no difficulty in verifying that if al,a2 e P then

Al*(A2*f)

=

(al*A2)*.f,

where

(A1*A2)(a) =

I

1 Aba2(a - b)

bdRP (the reader well versed in measure theory will recognise the notion of the convolution product of f with discrete probability measures).

If A e P, f e L1, it is clear that

Ila*fIll <

Ilf1I1.

Lastly let us set, for f e L1,

p(f) = inflla*fIll,

A eP.

It is then a matter of proving that II(f)I = p(f).

(1)

If a e Q and f,g a L1, then the following properties hold:

I(f) = p(f)

if f 3 0,

p(af) = Ialp(f),

II(f)I < P(f), p(f + g) < p(f) + p(g).

The first two are evident; the third follows from

CHAPTER 8: CONVOLUTION PRODUCTS

382

II(f) I

I I(x*f) I , Ila*fll1,

=

and the last from the fact that for all c > 0 there exist X1,a2 e P such that

Ila1*fll1 < p(f) + I1a2*9111

'c p(9) +

so that p(f + g) < II(xl*x2)*(f + g)111 11X2*(a1*f) II1 + llal*(a2*9) 111 11X1*fI11 + Ila2*9111 p(f) + p(g) + 2e.

Now assume that one had proved that f e L1, f real, and I(f)= 0 implies p(f) = 0.

Then if f e L1 is real and if cp >, 0, I((p) = 1,

one would have II(f)I s p(f) 4 p(f - I(f)T) + II(f)l = II(f)I, which would prove (1) in this case. observe that there exists a e Q,

To pass to the general case

lad = 1, such that II(f)l = I(af),

if of = u + iv, where u,v are real, then p(u) = lI(u)I = 1(u),

p(v) = II(V) I = 0,

AND FOURIER TRANSFORMS

383

and consequently:

p(f) > II(f)I = I(u) = p(u)

>. p(u + iv) = p(af) = p(f).

Therefore let us assume that f e L1, f real, and I(f) = 0.

Note that p(f) is the distance in L1 from zero to the closed convex envelope of the fa's, a e ]R

.

If one had p(f) > 0 the Hahn-

Banach Theorem would assure us of the existence of a number a > 0 and of a non-zero continuous linear form on L

bounded from be-

low by a on this convex envelope, and in particular on the set of In other words there would exist g e L , g non-zero almost

fa's.

everywhere, and such that

Jf(x - a)g(x)dx > a > for all a e]R

(2)

Since I(f) = 0 the left side of (2) is not changed

.

by adding a constant to g, which allows us to assume that g > 0. Let h(x) = g(-x); then h > 0, h e Lm, and (2) can be written (3)

Let M =

IIhII..

Note that by (3) M > 0.

(f - M q,)*h > a - M M = 0. By the preceding exercise one would have

I(f-M9)=I(f)-M> 0, which is absurd.

Let p be as above.

Then

384

CHAPTER 8: CONVOLUTION PRODUCTS

EXERCISE 8.159:

is x

b

f

If f is integrable on [a,b], and if

n dx = 0

f(x)e a

for a sequence (an) of complex numbers having at least one finite limit point, then f = 0 almost everywhere on [a,b].

AVn = VAV = AVA = VAV = evn

SOLUTION:

For all z e a: let

b

F(z) = 1 f(x)eizxd a

This defines an entire function which vanishes at each a the latter have a finite limit point, then F(z) = 0.

.

If

n In partic-

ular the Fourier transform of the function equal to f on [a,b] and zero elsewhere is zero.

Consequently f = 0 almost everywhere

on [a,b]. EXERCISE 8.160: (a):

Let (an) be a sequence of real numbers such

that

lime

n-9

is x n

exists for all x's belonging to a measurable set A of 3R with meas(A) > 0.

Show that the sequence (an) is convergent. (b):

Let (cn) be a sequence of complex numbers and (an) a

sequence of real nubmers such that

lime e

n- n

is x n

AND FOURIER TRANSFORMS

385

exists for all the x's belonging to a measurable set A of ]R with meas(A) > 0.

Show that either

n-,m n

lime

= 0,

lime

= c $ 0

or

n. ,

n

lima

and

n._,a, n

= a.

ADA = DAD = ADA = DAD = ADA

SOLUTION

(a) :

The set of the x eat for which lime

is nx

exists is

an additive subgroup that is measurable and of strictly positive measure.

From this it follows that

g(x) = lime n-),-

is x n

(*)

exists for all x eat (cf. exercise 8.138).

Then for every inte-

grable function f on at, +W

f_

( -+Q

is x

f(x)g(x)dx = limJ f(x)e n dx n-t°°

m

(Dominated Convergence Theorem).

E

Assume that for a subsequence

By the Riemann-Lebesgue lemma,

(On) of (an) IsnI ->

0

for every integrable function f, and consequently g(x) = 0 for

almost all x, which is absurd because Ig(x)I = 1 for all x. sequence (an) is therefore bounded. points of the sequence.

The

Let a,8 be two accumulation

By considering subsequences of (an) con-

386

CHAPTER 8: CONVOLUTION PRODUCTS

verging towards a and B, one deduces from (*) that eiax

eiBx

-

This proves that the

for all x, whence by differentiation a = B. sequence (an) is convergent.

Since

SOLUTION (b): is x I

icne

n

= Icn1,

we have that IimIc

exists.

I

= p

Ti

n-),w

Assume that p > 0.

Then from some point on, cn 4 0.

is (x-y) is therefore convergent. x,y e A, the sequence e n

If

Because

meas(A - A) > 0 it follows from part (a) above that a = lima exn-).w n

is x

is x

ists.

Finally, if x e A the sequences ene

vergent, which proves that lime

n-).w n

EXERCISE 8.161:(a):

f+(x) =

Let f e L1(T), and

suph-1fo If(x

+ u)Jdu,

h>o

f(x)

exists.

= max(f+(x),f (x)).

Show that for all A > 0,

n

and e

n

are con-

AND FOURIER TRANSFORMS

387

meas{ (f+ > A) n [-n, n] } <

Ln

(*)

11f111,

and some analogous relations for f and ?.

(Use the "Setting

What does the application of Marycinkiewicz's Theo-

Sun Lemma").

(Cf. exercise 6.124).

rem give in this case?

Let (Kn) be a sequence of functions of L1(T) and (H ) n a sequence of continuously differentiable functions on [-n,n] (b):

such that (i):

IKn(x)I < Hn(x),

-n < x < n;

n

(ii):

A = sup 2nj

Hn(x)dx < -n

(+W (iii):

B = sup

1 2n

IxH'(x)Idx < n

If f e L1(T) set: X*f (x) = sup I (Kn f) (x) I

.

Show that K*f(x) < (A + 2B)f'(x).

What conclusion can be drawn from this? 000 = vov = ovo = vov = ovo

SOLUTION (a):

Let P(A) be the left side of inequality (*).

0 < A < 211fp1, then sp(a) < 2% <

11fD D.

Furthermore, if IxI 4 1 and h > 0 are such that

If

CHAPTER 8: CONVOLUTION PRODUCTS

388

211f111 < A <

h-1Jhi f(x + u)I du, 0

and if N is the integer such that 2n(N - 1) < h < 20, then because f has period 2n,

JhIf(x + u)Idu < NJ0 If(u)Idu = 2nNI1f111 < (h + 2n)IIftl1, 0

0

and consequently h < 2n.

From this it follows that for A > 211f1I1,

z

(A) = meas(x:lxl < n,

z

1 xJ

> A for a z such that

IfI

x < z A for a z such that

x

x < z < 3n}.

continuous function

x F(x) = J lf(u)ldu - Ax 0

the last inequality may be written p(a) < meas(V), where

V = {x:-n < x < 3n,F(z) > F(x) for a z such that x < z < 30 By the "Setting Sun Lemma" V is the disjoint union of a sequence of intervals (a.,b.) such that F(a.) < F(b.), and consequently

b lfI i

J

n

f

AND.FOURIER TRANSFORMS

389

which accomplishes the proof.

Proceeding similarly for f and observing that

(f° > A) _ (f+ > A)U(f > A), one obtains

meas{(f> a)fl< !

11f111}

Let us note that: 10h

?(x)

=

suplh-1

lf(x + u)ldul,

so

Furthermore, if f e L-(T) then

Ilf 11. S IIfll.. By Marcinkiewicz's Theorem (cf. exercise 6.124) for all p, 1 < p 4 m, there exists a constant AP such that

Ilf Ilp < APIIf11P,

f eLW(T).

The relation above is still valid if f e Lp(T).

In fact, if 9i

is an increasing sequence of simple functions that tends to Ifl at every point, then for all x and all h > 0

h

h

suph 11 gi(x + u)du = h-11 If(x + u)ldu, i

so

o

0

390

CHAPTER 8: CONVOLUTION PRODUCTS

supgi(x) = f(x). 2

From this it is deduced that

supgi() = ?(X), i

and therefore

Ilf lip = Sup Ilg" IIp < Apsup 1191 lip = AP Ilf IIp

i

2

(Actually the argument above is valid for every measurable function with period 2%).

In particular, fA e Lp if f e Lp, 1 < p 4 -.

Using the last part of Exercise 6.124, one can show similarly that there exist constants A,B,Ap (0 < p < 1) such that

II? ll t A + Bj, Ifi log+Ifl n

0 < p < 1.

Ilf°Ilp , Apllf111,

SOLUTION (b):

As each function Hn is bounded, so is each Kn, and

the convolution product K *f is therefore defined everywhere, n moreover

((n

I(Kn*f)(x)l < ?nJ- Hn(y)lf(x - y)ldy. E

Writing

n(y) = n(n) - Jll'(u)du y

(1)

AND FOURIER TRANSFORMS

391

and changing the orders of integration, this yields x

7c

JE(Y)If(x

- y)Idy = Hn(n)JOIf(x - y)Idy

(n

u

- J H'(u)duJ If(x - y)Idy. 0n

0

Foru>.0 u If(x - y)Idy 6 of (x),

J

0

so

JH(Y)If(x - y)Idy 6 00 (n) + JQIuH'(u)Idu)f (x).

Similarly, 0

_x n

f(x - y)Idy

(iH(-n) + n

(JH(y)I

so that by (iii) and (1)

I(Kn*f)(x)I < '1(Hn(n) + Hn(-n) + 2B)?(x).

Furthermore, the relation:

Y n )du + Jb'uIc(u)du yH (y) = fo H(u

x

n

s J:Hfl(U)dU + f0juU1(u)jdu,

(2)

CHAPTER 8: CONVOLUTION PRODUCTS

392 0

r0

n n( - n) 4 I -n

H m

du + I

JuH'(u)ldu,

_n

m

and consequently, by (ii) and (iii): J( n(n) + Hn(-n)) < A + B.

Returning to (2) one obtains

I(K *f)(x)l 4 (A + 2B)fA(x).

n

One can then deduce for the operator f + K*f the same properties

as for f

f".

EXERCISE 8.162:

In what follows set z = x + iy, where x eat and

y > 0, and

P(x,y) = n

2

x

1

y f

Q(x,y) =

2

f

K(x,y) = P(x,y) + iQ(x,y) = nz

(a):

Show that the functions

form, as y -+ 0, an ap-

proximate identity in L1(gt).

(b) :

If f e LP(It), 1 < p < m, set:

Pf(z) =

JP(x

- t,y)f(t)dt.

Determine limPf(x + iy) when f is the characteristic funcy;0 tion of an interval [a,b]. (c):

Drawing inspiration from the preceding exercise, show

AND FOURIER TRANSFORMS

393

that if f e L1(R) and A > 0, then

meas{x:supIPf(x + iy)l > Al c 2a-1iIfII

From this deduce that limQf(x + iy) = f(x) for almost all x. Y-'-0

(Use part (b) above).

Generalise this result to the case where

f e LP(R), 1 s p< (d):

Let g be a function bounded and holomorphic in the

half-plane II

= {z:Im(z) > 0}.

Show that there exists g e L %R) such that g = P.

(Use the

Lm

fact that from every bounded sequence of

one can extract a sub

sequence which converges weakly in this space).

(e):

When f e Lp(Ik), 1 6 p < -, set:

Kf(z) =

Qf(z) =

JK(x

- t,y)f(t)dt,

+ 1mQ(x - t,y)f(t)dt.

Show that Kf is holomorphic in the half-plane H.

Next show

(Reduce to the case that liiKf(x + iy) exists for almost all x. y-*o where f < 0, set g(z) = expKf(z), and use parts (c) and (d) above.

Deduce from this that Hf(x) = limQf(x + iy). y->0

exists for almost all x. f)-

(Hf is called the HILBERT TRANSFORM of

CHAPTER 8: CONVOLUTION PRODUCTS

394 (f):

Calculate the Fourier transform Q0,y) of Q(-,y).

From this deduce that 1IHf 112 (g) :

If 112 when f e L2 (M).

=

Let f e L1Ot), f 3 0. Show that for all e > 0 and all

y > 0

logjl t pKf(t + ie)I

f-m

dt = logll + tiKf(z + ie)I.

(x - t)2 + y

Deduce from this that

JlogIQfct

+ ie)Idt < uIIfIll.

Next show that for every f e L1cR)

meas{t:IQf(t + ie)I > a} < ae IIfII'. (h):

Using parts (f) and (g) and Marcinkiewicz's Theorem

(cf. exercise 6.124) show that if 1 < p < 2 there exists a con-

stant Ap such that (IHfIIp < ApIIfIIp

for all feLp(gt).

Extend this result to the case where 2 < p <

AVA = VA0 - AVA = 010 = Lot

SOLUTION (a):

(b

This follows from P(x,y) > 0 and

P(x,y)dx = 1 a

SOLUTION (b) :

(tan-1b

- tan-1 a ).

Note that P( ,y) a Lq(,t) for 1 4 q <

(1)

which as-

AND FOURIER TRANSFORMS

395

sures the existence of Pf(z).

Moreover, (1) shows that if f is

the characteristic function of [a,b] then

ifa. 0, and that

SOLUTION (c):

I -uP'(u,y)du = 2( 0 X 0

u2du

(u2 + 1)2

-_

1

2

.

Using the notations of the preceding exercise, this yields

JOP(t,y)lf(x ± t)ldt = JIf(x ± t)IdtJt-PX(u,y)dy

=

J0_uPX(u,y)IuJOIf(x ± t)Idt]du

< if-(x).

Since

Pf(z) = J P(t,y){f(x + t) + f(x - t)}dt, 0

it follows that

JPf(z)l < ?(X). Proceeding as in the preceding exercise, one shows that for all A > 0

meas(f± > A) < A-1IIf ilI,

CHAPTER 8: CONVOLUTION PRODUCTS

396

from which the desired inequality follows.

This being so, let

(Tk) be a sequence of step functions such that

f = I

Tk

OR-4)

and

II(Pk111 =

in LThere exists a set N1 of measure zero such

that

f(x) =

x4N1.

(pk(x),

(2)

k Furthermore, for all z e R,

Pf(z) = I Prpk(z).

(3)

k Finally, by part (b) there exists a set N2 of measure zero, such that 1imPpk(x + iy) _ (P k(x) y->o

for all k and all x 4N

(4)

2'

Since

OR-2)

I meas{x:suplPcpk(x + iy)I > k-2} = I

y>o

k

<

k

there exists a set N3 of measure zero such that if x *N

:

3

supIPpk(x + iy)I 4 k-2 y>o starting from some k

when x 4N

3

0

(which may depend on x).

In other words,

the series (3) is uniformly convergent with respect

to y. It then follows from (2),(3) that if x+N1UN2UN3, limPf(x + iy) = X limPq) (x + iy) = X k k y-Yo k y->o

k

(x) = f(x).

When f e LPOR), 1 < p 4 -, for every integer N 3 1 let us set

AND FOURIER TRANSFORMS

397

fN(x) = f(x) or 0 according as IxI < N or not, and fN = f - fN.

Since fN a L1(IR),

limPfN(x + iy) = f(x) Y+0 for almost all x such that IxI < N.

IPfN(x

+ iy)l

Furthermore,

It 1;.N

(t)I dt. (x - t) 2

From this one concludes that limPf(x + ig) = f(x) Y-0 for almost all x e]-N,N[, and consequently for almost all x e]R.

For c > 0 and R > 0 let rc R be the loop formed by

SOLUTION (d):

the segment [-R + ic,R + ic] and the semi-circle E * is + Rein For all z e n, whenever 1/c and R are large enough,

0 < 8 t n.

g(z) =

g() -

2n1iJ

d

l Jr 0 = 2ni e,R

r c,R

so by subtraction it is deduced that

g(z) =

r c,R

- z- z

d.

As the function g is bounded, the integral along the semi-circle is 0(1/R), and consequently

g(z)

E

_ V-

n

.

(t + ic) t + is - z t + 2c -

dt.

Now, let (ck) be a sequence that decreases to zero.

(5)

Extracting

CHAPTER 8: CONVOLUTION PRODUCTS

398

a subsequence, one can assume that the functions t ' g(t + ick) converge weakly in L OR) to a function g.

One can also assume

Since

that co < y.

It + iEk - zI a It + iE0 - zI

It + iEk - 21

,

> It - 51,

Lebesgue's Theorem shows that the functions

t '* (t + iEk - Z)-1(t + ick -

8)-1

tend in norm in L1OR) to the function

t y ((x - t)2 +

y2)-1,

and consequently one deduces from (5) that

g(z) = Pg(z).

Note that by the preceding part of the exercise, for the sequence under consideration, limg(x + iy) = g(x) y+0 for almost all x.

The functions

SOLUTION (e):

belong to Lq(nt) for 1 < q <

which ensures the existence of Kf(z) when f e Lp(1R), 1 < p < -.

Furthermore, if C is a compact set contained in R there exists 0 < R < a such that the disc with centre is and radius R contains C.

But then

z(t)

1 I

1

((a 2 + t )Z - R)

If(t)I

when zeC.

The function of t which appears in the right side of this inequal-

AND FOURIER TRANSFORMS

399

ity is integrable, for the first factor belongs to Lq(R2) for 1 < q 4 -.

From this it follows that Kf is holomorphic in H.

Further, if f < 0 then ReKf = Pf < 0,

and the function g = expKf is holomorphic and bounded in H.

By

what we have just seen, ling(x + iy) exists for almost all x, as y-*O

well as limPf(x + iy), moreover. y40

Since

expiQf = gexp(-Pf),

it follows from this that limQf(x + iy) = Hf y40 exists for almost all x.

When f is not negative it is written

as a linear combination of four negative functions.

The function

SOLUTION (f):

belongs to L2(kt).

Its Fourier

transform is, by definition, the limit in L2(1R) of the functions:

r

E J

e M -M

-2nil;

- -in-M xsin2nlx x2+y2 M

Q(x,y)dx

dx

as M 3 m.

As the integral on the right has a limit as M 3 -, one deduces from this that f+C0

Q(E,y) = - i

xsin2niax ax = - if+" xsin2nyEx d,. + y2 n1 x2 + 1

_ ,Xx2

It is a classic result that for u >, 0,

1

+W

nJ -m

cosux dx =

x2 + 1

e-u.

CHAPTER 8: CONVOLUTION PRODUCTS

400

One easily justifies the differentiation with respect to u under the integral sign, which, taking into account that

(-E,y) = - (E,y), yields

(E,y)

isign(E)e-2t[yIEI.

Let

Y(E) = - isign(E).

When f e L2(JR) we have

and therefore

Q(-'Y?' functions which tend to Y? in L2(JR) as y - 0.

As the Fourier

transformation is an isometry of L2(It), it follows that tends in L2 OR) to the inverse Fourier transform of Y}'.

Hf

almost everywhere,

it follows that Hf a L2OR), Hf = Y7, and therefore that

IIHf1I2 = II?II2 = SOLUTION (g):

Since

IK(x,y)I s n y

it is clear that

IIYYII2 = 117112 = IIfI12.

Since

AND FOURIER TRANSFORMS

401

nylKf(z)I < IIfII1 If, furthermore, f >. 0, then

ReKf = Pf > 0, and the function 1 + pKf(z) takes its values in the half-plane By considering the principal value of the logarithm

Re(z) - 1.

in this half-plane, the function log(l + pKf(z)), which is holomorphic in IT, is defined unambiguously; moreover, its modulus is

bounded by

2+logll+pxf(z)I 0 and meas(f > y) > x} = ]O,fh(x)[). (d):

Show that for every measurable set E C ]0,1[

J f(x)dx 4 J

E

To

(Show that

BOUNDED VARIATION etc.

421

and that for every positive and decreasing function w on ]0,1[ 1

1

f*(x)q(x)dx.

f(x)cp(x)dx 4 1

J 0

0

(e):

If f is positive and integrable on ]0,1[, then for

0 0; therefore it contains all the Borel sets of [0,o[.

If s is positive and Borel on [0,o[,

then for all y > 0

meas(f 1(s-1(]y,='[))) =

meas(g-1(s-1(]y,°[))),

which proves that sof and sog are equimeasurable. SOLUTION (c):

Let us begin with some considerations about de-

creasing functions.

Let F be the set of positive functions on

]0,u[ that are decreasing, right-continuous, and tend to zero at infinity.

If (p e F and x > 0, the set ((p > x) is a bounded inter-

val which, if it is not empty, has 0 as its left end point, and is open on the right. that

In other words there exists T*(x) >. 0 such

BOUNDED VARIATION etc.

423

(cp > x) = ]O,9*(x)[.

(1)

We have T* a F; in fact q* is decreasing and right continuous,

for xn > xn+l' xn - x implies

(m>x)= U (c>xn), n

i.e.

]0,cp*(x)[ = U ]0,p*(xn)[, n which means that T*(x n) -> cp*(x).

Finally, if one had T* > c > 0

then, for 0 < y < e, one would have p(y) 3 x for all x > 0, which is absurd.

Note that (1) can be expressed in the following man-

ner: for x > 0 and y > 0, the conditions p(y) > x and c*(x) > y are equivalent.

From this it follows that

(q,*)* = (P.

(2)

Now return to our problem.

If we set T(y) = meas(f > y) for

y > 0, it is easily verified that T e F.

Let g be a positive, de-

creasing, right-continuous functions on ]0,1[, and let gl be the function obtained by extending g by 0 on [1,oo[.

Then gI a F, and

in order that g be equimeasurable with f it is necessary and sufficient that for all y > 0 cp(y) = meas(f > y) = meas(g > y) = meas(gl

that is to say, by (2), that gl = T*. SOLUTION (d):

31 JE = J0E.f

JQE*.

=

> y) = g*(y),

424

CHAPTER 9:

For all y > 0 meas(ILEf > y) s meas(E),

and consequently (IlEf)*(x) = 0

if x >. meas(E).

Also,

meas(ILEf > y) < meas(f > y), so that

(if)* < P. Therefore eas(E)

f-

JJ E

fineas(E)

Ef)ot

0

J 0

f

In particular, if 0 < c < 1, then

f

O E JO rt

From this it is immediately deduced that 1

1

J fW< J0 f*,

(3)

0

for every positive decreasing step function 9.

If cp

is positive

and decreasing on ]0,1[ it is the limit on this interval of an increasing sequence of positive decreasing step functions, and (3) is therefore still valid in this case.

BOUNDED VARIATION etc. SOLUTION (e):

425

Let

X F(x) = Jf(t)dt,

0 < x < 1.

0

The set (9f > y) is formed of the x's for which there exists u such that 0 < u < x and F(u) - yu < F(x) - yx.

By the "Setting

Sun lemma", this set is the disjoint union of a sequence of intervals ]an,bn[ such that F(an) - yan 4 F(bn) - ybn.

Thus

meas(of > y) = E (b - a ) n n n

1b y

an =

1

meas(8f>y)

yJ(8f>y)f

I

f*.

y 0

Therefore when meas(Of > y) > 0 1

rmeas(8f>y) (4)

meas 8f > y J0 a

decreases as a increas-

Note that since f* is decreasing, a-1J 0

es.

Consider, then, an x such that (8f)*(x) > 0 (otherwise there

would be nothing to prove).

Since

{y:meas(sf > y) > x) = ]0,(sf)*(x)[,

426

CHAPTER 9:

it follows from (4) and from the remark above that (0 x

y < (8f)*(x) implies y .< _- f*, x1 which shows that xrx

(8f)*(x) s

f* 0

SOLUTION (f):

As the functions soBf and so(8f)* are equimeasur-

able,

JosoBf

=

SOLUTION (g):

Joso(8f)*

S

oo

Js{.Jf*}dx

.

In particular, if s(x) = xP, p > 0, then

PAP -< JOjo

Pdx.

(5)

J

After extending f* by zero on [1,W[ and setting

Jx

G(x) = x-1 f*, 0

(5) and Hardy's inequality gives

IIof1Ip 4 IIGUP s p

since f

p

p

IlfllP

and (f*)P are equimeasurable.

SOLUTION (h): 1

Ilf llp =

If 0 < p < 1 then Holder's inequality for the pair

T-1 p1 applied to (5) yields

BOUNDED VARIATION etc.

110f 11P F

427

O xP 0

x

P-- p)

-11-PLJ0

rr10

xPJp J < IIf IIiJ

O xP

so

1 1 pl 1/P IIf Ill.

II$f IIP <

SOLUTION (i):

Finally, if p = 1, then starting from (5), one

obtains

OOf 6 J

10

J

o

=

I of*(t)dt1 t

= 10f*(t)log dt.

t

Let e > 0 and choose a such that 0 < a < 1 and a

log1 s <

.

f0

Then a

1

of f0

1

1

1

f*(t)log t dt +

1

f(t)dt.

log a 1

0

0 i

Let E be the set of t 's such that 0 < t < a and f*(t) 4 t Then

Jf*(t)log..dt <

Jiog t

VT

+ 21 f*(t)log+f*(t)dt < (Contd) o

428

CHAPTER 9 1

e + 2Jf(t)log+f(t)dt,

(Contd)

0

whence, at last,

1

1

Of

J 0

2

+

1

1

f+ flog f + log -JO a

Jo

CHAPTER 10

Summation Processes: Trigonometric Polynomials

EXERCISE 10.174: (a):

If the series

W

Iun

(1)

n=0

converges, and if

0

n0 I

(2)

Ixn - An+ll <

show that the series 00

L

n=o

(3)

Anun

converges. (b):

Conversely, prove that if the series (3) converges

whenever the series (1) converges, then (2) is satisfied.

ovp = vov - nvo - vev = ovn

429

CHAPTER 10: SUMMATION PROCESSES

430

(2) implies that An = A t en, with

SOLUTION (a):

L len - entll < °°

and

suplu0 +

cn

3 0.

As

+ unI <

the Abel-Dirichlet Theorem shows that the series E enun is convergent.

The same then holds for the series I 'nun.

SOLUTION (b):

First Proof: Let E be the Banach space formed by

convergent sequences s = (sn), with the norm IIslI

= suplsnl.

For every integer N >. 1 define a linear function fN on E by

N-1 fN(s) =

I

()n - An+1)sn t ANsN.

n=0 This linear functional is continuous, and it is a classical result that its norm is

N-1 IlfN ll =

E

n=o

l An - Xntl I

+ I AN1'

Since

N fN(s) = A0s0 t

=an(sn - sn-1),

n1 and as the series I (sn - 8 n-1) is convergent, limfN(s) exists N_

for every s e E.

By the Banach-Steinhaus Theorem one therefore

has

Su "fNII< which implies that:

TRIGONOMETRIC POLYNOMIALS

G

n=0

431

Ian - xn+1I <

Second Proof: Let An - an+1 = rJn,

rn > 0,

IEnI = 1,

sn+rO+r1+ +rn En

un = sn - sn-1

U 0 = 3 0 ,

if n >. 1.

If it is assumed that

G

n=0 then sn -

G

Ixn - xn+1I

n=0

rn

0, and the series E un is therefore convergent.

Fur-

thermore,

NC

N-1

(an - an+1)sn

Anun = aNsN +

nI.

n I0

0

The first term of the right -side is bounded by IA0I + r0 + ... + rN-1

1 + r0 +

'

+ rN

a quantity that remains bounded as N - ; the second term is equal to

N-1 C

r

n-01+r0+

n

CHAPTER 10: SUMMATION PROCESSES

432

and tends to infinity with N by a classical property of positive It follows that the series

divergent series.

Anun is divergent:

Let (un)na0 be a decreasing sequence of positive

EXERCISE 10.175: real numbers.

Show that if

then

un = of

n). ovo = V1V = tV1 = VAV = ovo

SOLUTION:

Let

EN

n=N+1

u

If p > N then (p - N)uP < uN+1 + ... + up 4 EN,

and consequently

li ppuP

EN,

which implies the desired result upon making N

EXERCISE 10.176:

fan

Let (an)n30 be a sequence of real numbers.

_

= an - an+1,

2

1 a

_ =

n =

a

4a n+1'

Set:

TRIGONOMETRIC POLYNOMIALS

433

The sequence is called a CONVEx SEQUENCE if

02an > 0

for all n > 0.

Let c be a convex function on [0,-[.

(a):

Show that the

sequence an = 9(n) is convex.

Show that if the sequence (an )n>0 is convex and bound-

(b):

ed, then It is decreasing;

(i):

(ii):

L1an = o(1/n);

(iii):

(n + 1)t2a. = a0 - lima..

L

n

n=0

(c):

Show that for every sequence (cn)n>0 which tends to

zero there exists a convex sequence (an)n>0 which tends to zero, and which is such that Ic.I < an for all n a 0. ADA - VAV = A0A = V AV = DOA

SOLUTION (a): 2

This follows from

an = an - 2a n+1 + a

n+2

= 2['9(n) + 3(p(n + 2) - p('n + L(n + 2))]. SOLUTION (b):

The condition A2a

sequence (da ) is decreasing.

n

0 for all n means that the

If, for some integer r Aa

then for every integer n > r

an = ar -

n-1 E

s=r

Aas

ar + (n - r)(

Aar),

r

< 0,

434

CHAPTER 10: SUMMATION PROCESSES

and so the sequence (an) would tend to +W, which is absurd. Therefore Dan 3 0 for all n, which proves (i). But then the sequence (an) is decreasing and bounded, and thus has a limit.

Since:

N-1 E

Aan = a0 - aN,

n=0 the series I Aan is therefore convergent, and since the Aan's are positive and decreasing, (ii) follows by the preceding exercise. Finally,

N-1

N-1

2

Aan - NAan = a0 - aN - NAaN,

(n + 1)A an = n I=O

n I=O

which implies (iii), since by (ii) the last term tends to zero. SOLUTION (c):

On replacing cn by sup(IcSl:s i n) it can be as-

sumed that the sequence (cn) is positive and decreases to zero. By part (a) it suffices to construct a convex function q on [0,W[ that tends to zero at infinity, and that is such that 9(n) > cn for all n.

To do that, consider a sequence (q)k) such that

9k ' qk+l

if k

0,

q)

1 = c0,

-+ 0.

(1)

lpk

Let no = 0, and let n1 be such that

ni > n0,

cn1 <

q) 2,

(2)

then construct by induction some integers nk such that for k 3 2

nk > nk-1'

(3)

TRIGONOMETRIC POLYNOMIALS

435

< (P

(4)

k+l'

enk (Pk-1

(Pk

(Pk-2

(Pk-1

nk-i

nk-2

<

nk - nk-1

(5)

This is possible because the sequence (cn) tends to zero. that (4) is valid for all k > 0. that cp(nk) =

Note

This being so, let m be such

and that is linear on each of the intervals

k

This function is decre sing and tends to zero at

[nk,nk+l].

infinity by (1), and it is convex by (5).

If nK

< n < nK+1,

then by virtue of Relation (4)

en < enk 5 9(nk+1) < (P(n). EXERCISE 10.177:

Let (u

)

n,p n30,p

be a double sequence of com-

plex numbers, such that W (i):

lu

X

n=0

4 M for all p >. 0;

1

n,P

W (ii):

l}i.)m

00 P+00

I

n=0

un

,P

(iii): un,P = 0

= 1; for all n -> 0.

Show that if sn + s, then

lim

I

un p8n = s.

P" n=0

'

1Vt - V AV - AVA a VAV = ovo

SOLUTION:

If an -> a, then on setting an = s + cn it is seen that

436

CHAPTER 10: SUMMATION PROCESSES

by (ii) one is reduced to the case where s = 0.

Let c0 be the

Banach space formed by the sequences c = (en) that tend to zero, with the norm 11C11 = sup f cn I

n

and let us consider on e0 the linear functionals

cc

fp(e) _

un Pen.

I

n=0

'

It is clear that these latter are continuous and that cc

IIfP II S

I un, p

E

I

.

M.

(1)

n=o

It is a question of proving that fp(c) -> 0 for all c ec0.

en be the sequence such that em = 6nm.

Let

From the relation

N

I1c -

C

cnen II = sup I en I

n>N

n=0

one deduces that the eats form a total set in c0. therefore suffices to prove that for all n limfp(en) = 0.

But this is nothing other than (iii), because

f (en) = u P

n,p

By (1) it

437

TRIGONOMETRIC POLYNGd'4IALS

In the following six exercises the following notations are used:

is a sequence of complex numbers, set:

If (cn)ne2Z

80+81+. . .+8N CC

sN

Ini s.

Show that sN -

s if and only if

tN = o(N). (c):

Show that if for some real

Again assume that aN -)- s.

number p , 1

Inlp-llcnIP G

<

then sN -), s. (d):

Prove the following result (HARDY'S LEMMA):

0(1/n) then aN -> a implies sN -> s.

N1 < N2 <

tN

If cn =

(Assume that for a sequence

, BNS, B > 0, for example, and show that s

tN +v >' 2 S

BNS

if 0 4 v 4

Ns

2A

where A is a constant such that Icn + c_nI

.

A/n for n >, 1).

ta0 = VAV = MMA = VAT = 00A

SOLUTIONS (a);(b): N

n cn =

tN = InI,N

I

N

n(c

+ c_n) =

n=1

E

n(sn

sn-1

n=1

from which it is easily deduced that tN = N(sN - aN-1),

(1)

TRIGONOMETRIC POLYNOMIALS which proves (b).

439

Now

sN = (N + 1)GN - NoN-1'

and substituting this expression in (1) yields

tN = AN + 1)(aN

- 'N-1

(2)

).

Therefore, aN converges if and only if the series

NL

N=i AN + 1)

(a N - a N-1 )

converges.

Since the sequences

aN = N(N + 1) -1

and

aN =

N-1

(N + 1)

satisfy

N=1 IAN -

<

N+1

the convergence of aN is equivalent to that of the series

L

N 2t,

(cf. exercise 10.174). SOLUTION (c):

E

If p = 1 the condition becomes

ICnI < 00,

in which case sN converges to a value that can only be s.

There-

440

CHAPTER 10: SUMMATION PROCESSES

fore assume that p > 1, and let q be such that

+ q = 1; it is

now a matter of proving that N 1tN -+ 0 (cf., the p part (b)). if r < N then

Now,

It

Inll/P InI1 - 1/pcnl

N

< NI

nlcnl + NI

I

rr InIP-1IC lp)1/p n

By making r + - the desired result is obtained.

SOLUTION (d):

Assume that aN -+ s and that

Icn + c-nI E n

Let us show that N

1

,

n

- 0 under these conditions, which by part N (b) will prove that sN - s. Otherwise there would exist a B > 0 t

and an increasing sequence of integers N

s

such that

TRIGONOMETRIC POLYNOMIALS tN

441

> BNs. S

If VS is the integral part of BNs/2A, then for 0 < v < vs

tN +v

= tN

s

+

>. BN

>

s

S

- vA

S

BN BN

n(cn + c-n

I

N 0. N necessarily liminfN 1tN < 0, and it would suffice to change the we supposed that limsupN

signs of all the cn's.

t

442

CHAPTER 10: SUMMATION PROCESSES

EXERCISE 10.180:

Set

8P + 8 =

ap,N

(a):

+ p+1 N + 1

+

8p+N

Express ap,N as a function of the aN's.

Deduce from this that if a

N

-> s then a

p ,NP

-* s for every

sequence of integers (Np) such that NP > ap, a > 0.

(b):

Prove that if ian! < Alni

Iap,N - spI < A

(c):

N

for n >, 1, then

.

Deduce from the preceding a new proof of Hardy's Lemma

(cf. the preceding exercise).

AVA - VAV = AVA = VAV = AVA

SOLUTION (a):

(p + N + 1)ap+N

pop-1

N + 1

ap,N

ap+N + NN p+N - ap-1 If the sequence (Np) is such that Np > ap, a > 0, then

N< a< so that if a

N

1Q

-> s then a

p,Np

-> B.

(1)

TRIGONOMETRIC POLYNOMIALS

443

SOLUTION (b):

p+N

(N + I)ap,N = (N + 1)8

+

G

(N + p + 1 - n)(cn + c-n),

n=p+l so that

p+N Iop,N - spI

N +

11

(N + p + 1 - n)(cn + c-n

G

n=p+1 N

p N2+ 1 r=1

SOLUTION (c):

E p

P

If Icnl

.

r=

p

AIn(-1 for n a 1 and aN - s, let

sup IaN - 0 1I, P N>p-1

so that ep -> 0.

Let N be the integer such that

N 4pp 0 and that en = 0 except when

EXERCISE 10.181:

n = 0,±n1,±n2,... with

p+l>, q > 1. n p

Show that if aN 4 s then sN -> B. and show that for np .< N <

N 1tN < 2

(Let ap = max(Icn l,ic-n 1) p p

np+l

¶ argr p; r=1

then use exercises 10.177 and 10.179).

ADA = VAV - AVA - DAD = ADA

SOLUTION:

n

-E n

if p >, r,

>, qp-r

r

so that if np < N < np+1' then

IN 1tNl = IN 1 ¶

nr,(cn

r=1

<

nr

a

2

r=1

+ c_n )I

r

r 2

r np

Now note that ar + 0, and that

l im I Per=l

qr-p =

q

q-1'

¶ a qr-p. r=1

r

TRIGONOMETRIC POLYNOMIALS

limgr-P

445

= 0.

P4°°

By exercise 10.177 it follows that

lim

L argr P = 0,

P-'°° r=1 and therefore that N

1

t

N

-> 0.

Then by exercise 10.179 sN -> s if

aN ->s.

EXERCISE 10.182:

Let (an)n>0 be a sequence of strictly positive

real numbers that is convex and tends to zero. Show that if

'ON' < A,

SN = 0 [a N)

then the series +00

alnlgn

is convergent.

SOLUTION:

Using the notations Aan,42an of exercise 10.176, we

have N I

N

alnlcn = a0c0 +

n=-N

I

an(cn + c-n

n=1

an(sn - sn-1)

= a0s0 +

n=1

(Contd)

446

CHAPTER 10: SUMMATION PROCESSES N-i = a 8

(Contd)

N N

+

Aa.8

I

n=0 N-i

= a 8 + N N

Aan((n + 1)an - no n-1)

E

n=0 N-2

= a a + NAaN-1aN-1 + N N

By hypothesis a a N N and bounded

E

(n + 1)A2anan

n=0

Moreover, as the sequence (an) is convex

-r 0.

and further (cf. exercise 10.176)

00

E

(n + 1)IA2an.an1 s A

(n + 1)A2an = a0A.

n=0

n=0 From this it follows that

N lien

a

N-

EXERCISE 10.183:

c

n

=

(n + i)A2a

a

n

.

Denote by En the set of trigonometric polynom-

ials of the form

f(x) = e0 + e1cosx +

+ cncosnx,

where

c0 a c1 > ... > cn > 0. Set m(f) = sup{ If(x)I:Zn s x -C n},

M(f) = sup{If(x)I:0 < x 4 2n}.

TRIGONOMETRIC POLYNOMIALS

447

Prove that

I

fmf

1 i - 21 n n+ 1

inf M

:fe

(1 E'J < l2 +

11 J

1

n+ 1

tVt - VtV - AV = VAV = t4A

SOLUTION:

We can restrict ourselves to the case where

cp + c1 +

+ cn = 1.

We then have M(f) = 1

and

cp

(n + 1)-i.

f(x)dx = 2 cp - c1 + 3 - 5 +

n

J71/2

B y noticing that

it

CO - c1

;.,

(2 - iJcp 3 12 - 1)n + 1

and that :

we obtain:

m(f)

Ii l

21

1

nJn+1

On the other hand, if:

P x) = 1 + cosx n+ + 1 + cosnx

<

2

m(f).

448

CHAPTER 10: SUMMATION PROCESSES

an elementary calculation shows that:

f(x)

1

1

=n+12+

j

sin(n + Z)x

2s in x

-

It follows from this that if n/2 < x <

2n+1 (ltsi-n n)

If(x)I

EXERCISE 10.184:

ir:

_

Let f be a trigonometric polynomial of degree

at most N.

(a):

Show that if f 4 0 then f has at most 2N zeros in the

interval [0,27r[, counting multiplicity. (b):

Assume that f is real and that

= f(x0).

A = IlfII

Show that if Ixl < n/N then

f(x0 + x) 3 AcosNx.

(c):

f is no longer to be assumed to be real.

Ilf'II. < NRfIIW.

(BERNSTEIN'S INEQUALITY)

AVA = VAO = AVA = DAD = AVA

SOLUTION (a):

If

f(x) _

cne1 lnlkN

consider the polynomial

Show that

TRIGONOMETRIC POLYNOMIALS

449

C zn

P(z) = zN

InI,N Then f(x) = 0 if and only if P(e1x) = 0.

The result follows from

the fact that P + 0 and deg(P) < 2N. SOLUTION (b):

By translation we reduce to the case where x0 = 0.

If the property were not true, then possibly after replacing f(x) by f(-x), there would exist a y such that

0 < y < N ,

f(y) - AcosNy < 0.

For a >. 0 let ge(x) = f(x) - (A + E)cosNx.

If a is small enough then gE(y) < 0.

Furthermore, if xr = rn/N From

then gE(xr) >, e if r is odd and gE(xr) < -e if r is even.

this it follows that for e > 0 gE has at least one zero in each of the intervals ]y,xl[,]xl,x2[,...,]x2N_2,x2N_1[.

By making

e -> 0 one concludes from this that g0 has at least 2N - 1 zeros in the interval [y,2n[.

Moreover, g0(0) = g;(0) = 0; g0 would

therefore be a trigonometric polynomial of degree less than or equal to 2N which would have at least 2N + 1 zeros in the interval [0,2n[, which is absurd (because g0 # 0, since g0(y) < 0).

SOLUTION (c):

First assume that f is real.

Replacing f by -f,

if necessary, it may be assumed that for some x

m = 11f,11. = f'(x0). By part (b)

n/2N 211f 11

> f (x0 +

2N)

- f (x0

2N, = J-

(Contd)

f' (x0 + x )dx >. n/2N

450

CHAPTER 10 rn/2N

(Contd)

mJ_

cosNxdx =

2m

n/ 2N

N

,

which proves the result in this case. In the general case let x be a real number. a such that lal = 1 and If'(x)I = af'(x). where u and v are real.

Then

If'(x)I = u'(x) 6 Nlu I ' NIIafIIm = NIIfil..

There exists an

Let us set of = u + iv,

CHAPTER 11

Trigonometric Series

EXERCISE 11.185:

ezx

=

Let z e a, z $

e2nz - 1(2z 1 + n

e

zx _ eTIz - 1 nz

+

X n=1

zcosnx - nsinnxl z2 + n2

)

0 < x < 21; I

[(-1)nenz - 1] z2osn2

2 n

Prove the following formulas

.

z +n

n=1

,

0 < x < n;

W

ezX = n E

n=1

[1 -

(-1)nenz] nsinnx

0 < x < R.

z +n

What are the values of these series when x = 0? AVO = VMV = AVA = VOV = AVo

SOLUTION:

Let f be a function of period 27E such that f(x) = ezx

if 0 < x < 21.

Then [f'] = zf, and consequently

in f at 0 is 1 - e21z).

451

Therefore

452

CHAPTER 11: e2nz - 1

(n)

1

n $ 0.

z n

2n

=

Since

2tcz

(0)

=

2nz

the first formula is proved (Jordan-Dirichlet Theorem).

For

x = 0 the sum of the series is

(e2nz + 1)

=

z(f(x + 0) + f(x - 0)

from which it is easily deduced that W V

ncothnz = 1 +

2z

n=1 z2 + n2

z

Now consider the functions g and h, with period 2n, respectively even and odd, and which coincide with f on ]0,n[. [g'] = zh,

[h'] = zg,

so

ing(n) = A(n), inTi(n) = zg(n) + 2n (2 - 2e1Ze-

Therefore if n 4 0 then

g(n) _

di(n) =

(-1)nenz - 1 n

z

z2

+

z2

+ n2

1-(1-(T Zit

n2

n

,

nn).

Then

TRIGONOMETRIC SERIES

453

Moreover

e

nz

9(0) =

7Z(0) = 0,

1

nz

,

which proves the two other formulas. for x = 0.

In fact the second is valid

From this it is easily deduced that m

(_1) n

C

sinhnz

z +

n=1

n2

z

+ EXERCISE 11.186:

I

Find the sums of the following series:

acosnx 2

n=1 n+ a

SOLUTION:

nsinnx c

2

n=1 n2 + a

2,

(a real, a + 0).

Let

W

g(x) _

acosnx

(1)

I

n=1 n2 + a inz

G(z)

n=1 n

2e

2

z = x + iy,

y >, 0.

(2)

+ a

The series in (2) converges absolutely for y > 0 and defines on the half-plane y > 0 a holomorphic function that satisfies the differential equation

ae

G" - a2G = e

iz

Re(z) > 0.

(3)

- 1

But the G can be analytically continued to a minus the two real

454

CHAPTER 11:

From this it follows that g is

half-lines ]--,0] and ]2n,+m[.

infinitely differentiable (and even analytic) on ]0,2n[ and that on this interval satisfies the differential equation: 2

g,, - a g = aRe

eix

le ix

a

2

Hence

g(x) = acosha(x - n) + Ssinha(x - n) -

1

2a

for 0 < x < 2n, and hence also, by continuity, for 0 < x < 2n.

Since g(0) = g(20, 0 = 0. Now,

r212asinhan it

0

so that

acosnx = ncosha(n - x) 2sinhna n=1 a2 + n2

-

0 0;

(i):

f is the restriction to]R of a function holomorphic

(ii):

on the strip IIm(z)I < 6, 6 > 0; The series

(iii):

+00 ?(n)einz

(*)

n=-oo converges on a strip IIm(z)I < 6, 6 > 0; The series (*) converges at two points, z1 = x1 t i61

(iv):

and z2 = x2 - i62, with 61,62 > 0.

4V4 = 040 = 4MA = VAV = t0A

SOLUTION:

Let

an(z) = ?(n)einz +

?(-n)e-inz

Then 2inz (e

1

2inz

- e

inz

2)j(n) = an(z1)e

inz

1

- an (z2 )e

2

TRIGONOMETRIC SERIES

461

If (iv) is satisfied there exists a constant M such that lan(z1)I .< M,

Ian (z2)I < M,

n >. 0,

so that if n j 0 -n61 + e 1?(n)l < M le2n62 - e-2n611 n62

e

if c = min(61,62).

-

0(e-elnl),

This proves that (iv) _> (i).

If (i) is satisfied, the series (*) converges in norm on every strip IIm(z)I < 6, 6 < e

Therefore (i) => (ii).

Since it is trivial that (iii) => (iv), it remains to prove that (ii) => (iii).

Let F be a function holomorphic on the strip Im(z) < 6, 6 > 0, and such that f(x) = F(x) if x is real.

As the relation

F(x + 2n) = F(z) is true on at, it holds on the entire strip IIm(z)I < 6 (principle of analytic continuation).

Hence there exists a function G de-

fined on the annulus _6 e

<

J k I

< e6 ,

such that

F(z) =

if

= eiz

As the function e1z is locally invertible, it is clear that G is holomorphic on its domain of definition.

anEn

Let

462

CHAPTER 11:

Since the circle IzI = 1 is contained

be its Laurent expansion.

in the annulus of definition of G, +m E

lanl < W, +cm

f(x) =

G

anel

x R.

,

and consequently an = f(n).

It follows that the series (*) con-

verges (and even converges absolutely) on the strip IIm(z)I < 6.

EXERCISE 11.191: (a):

IIf(S)II0

?(n) (b):

O(Rsr(as + 1)),

Show that if

R>0, a>0,

0(InI1/20'e-(InI/R)1/a).

=

Now assume that (*) is satisfied.

11f(s)II

(c):

=

Let f e C00(T).

=

O(ssRsr(as + 1)),

0 = (a -

Show that

)+.

Deduce from this that f is analytic if and only if O(e-elnl)

for some e > 0.

If(n)I =

AVA = VAV - OVA = VAV = AVA

SOLUTION: (a):

If A is such that

IIf(S)II < ARSr(as + 1), then for all n 4r 0

s a o,

TRIGONOMETRIC SERIES

463

I(in)-sf^ (n)I < AI IRIJsr(as + 1).

1 (n)I =

n

Let us take for s the integral part of (1/a)(Inl/R)1/a. IS

Il lnlr(as + 1) < (as)-asr(as + 1) ti

e-as

Since

1/a I

- a < as < [J..L]

I

/a

J

(*) certainly holds.

SOLUTION: (b): +00

f(s)(x)

=

s > 0,

(in)sf(n)e11"`,

E

-00

and consequently

IIf(s) II00

= 0( 1 ns +1/2a e- (n/R)

1/a

).

n=1

Let a = a + 1/2a and

xae-(x/R)1/a

a

9

(x) =

x

0.

attains its maximum at

xo = R(aa)a.

Let Na be the integral part of xa.

Then

2.

Then

464

CHAPTER 11:

Ja()dx + q) a(NQ) + 9 N

spa(n) F n=1

a

(N

a

+ 1) +

+1Ta(x)dx

1N a

S

(Pa(x)dx + 2(p a(xa).

Now,

(x)dx = aRo+lr(aa + a), J0

Ra(aa)aae-aa

cpa(xa) =

Since r(x + h) ti xhr(x) as x -} 00,

aRa+1r(aa + a) = 0(aaRar(aa)). Furthermore,

Ra(aa)aae-aa

= O(a'Rar(aa))

Therefore 00

ma(n) = 0(aYRar(aa)),

y = max(a,2),

n=1 so

11f(s)JI = 0(syRsr(as + 2)) = O(sy ZRsr(as + 1),

which is precisely (**).

SOLUTION (c):

It is known that f is analytic if and only if

TRIGONOMETRIC SERIES

IIf(s)II

Co

465

= O(RSal),

R > 0.

(1)

By part (a) above, (1) implies that 0(e-£InI)

I?(n)I = 0(InI2e-InI/R) =

if 0 < e < R

On the other hand, if

IJ(n)I = O(e-EInI),

then by part (b),

0(s2E-SS!) = O(Rss!)

= IIf(s)II0o

EXERCISE 11.192:

when R > 1g .

Show that if f e L1(T) and:

Co f(n)einx

g(x) _

almost everywhere,

n=-then f = g almost everywhere.

A0A = VAV = AVA = VAV = A0A

SOLUTION:

By the Lebesgue-Fejer theorem

aN(x) + f(x)

almost everywhere,

But SN(x) - g(x) implies that aN(x) - g(x), so f = g almost everywhere.

EXERCISE 11.193:

I Ifcx 0

Let f e L1(T).

Show that if

+ t) + f(x - t) - 2sI tt <

466

CHAPTER 11:

then

= s.

?(n)elnx

(DINT'S TEST)

n=nvn = vnv - nvn = vnv = nvn

SOLUTION:

In fact

r

a (f;x) - s

n

=1 J

(f(x + t) + f(x - t) - 2s)

sin(Nt+ J)t

dt

0

+ eN(x),

with eN(x) - 0.

t

Since the function

t-1(f(x + t) + f(x - t) - 2s)

is integrable, the integal tends to zero by the Riemann-Lebesgue Theorem.

EXERCISE 11.194:

Let f be a complex function with period 2n.

It is said that f e Lip(a) if there exists a constant M such that

If(x) - f(y)l 4 mix - Y10,

for all x and all y.

(Evidently, 0 < a 6 1.

Show that then

?(n) = 0(n-a). nvn = vov - nvn - vnv = nvn

SOLUTION:

In fact

Why?).

TRIGONOMETRIC SERIES

467

2n

(

0

l

4nj OX) - flx + n,

=

)e-inxdx,

J

so

?(n)I

2(rc)a.

EXERCISE 11.195: (a):

Let f e L. (T).

If at a point x where f(x + 0) and f(x - 0) exist, one

has for0 D

27E

If(x t h) - f(x)Idx =

0(Ihla).

0

(a):

Show that if

f1 = f

and

fk = f"fk-1'

then

fk e L2(T)

when k >

2a

k 3 2,

TRIGONOMETRIC SERIES (b):

471

Show that fk coincides almost everywhere with a con-

tinuous function when k > 1/a.

AV4 = VtV = 4VA = VAV - 4VA

SOLUTION:

Note that 2n

f(n)

= 1

{f(x) - f+ 0

IX

n)}e-inxdx, J

so

0(Inl-a).

If(n)I =

SOLUTION: (a):

(1)

It follows from (1) that

0(Inl-ak),

Ifk(n)I =

and consequently that

E Ifk(n)I2 < W n

if k > 21 a

But then fk e L2(T).

SOLUTION: (b):

If k > 1/a, then

G I?k(n)I <

n

in which case fk coincides almost everywhere with the sum of its Fourier series, which is absolutely convergent, and hence continuous.

CHAPTER 11:

472

Let f eL1(T) be such that f(n) =

EXERCISE 11.199:

O(Inl-a),

a > 0. Show that if p is an integer such that a - p > 2, then f(p) is defined almost everywhere and belongs to L2(T).

AVA = VAV = AVA = via = Ovp

SOLUTION:

It suffices to carry out the proof for p = 1.

Indeed,

the result is evident if p = 0, and if p > 2 then lnlp-ilf(n)I E

< -,

which ensures that f is (p - 1) times continuously differentiable, and that

If(P-1)(n)I

0(InI-(0'-p+1))

=

with (a - p + 1) - 1 >

Assume, therefore, that a > 3/2, and let us prove that f' exists almost everywhere and belongs to L2(T).

We have

Y Inf(n)12 < M.

Hence there exists g e L2(T) such that

g(n) = in?(n),

n ea.

Since it is possible to integrate a Fourier series term by term,

rx

g(t)dt = 0

Since

Y If(n)I < -,

(f(n)el"x - f(n)).

TRIGONOMETRIC SERIES

473

the right side is equal to f(x) - f(0), which proves that f'(x)= g(x) almost everywhere.

EXERCISE 11.200:

If f is absolutely continuous and f' e

L2,

then

I I?(n) 14 IL I11 + " IV' II2 .

n

ovo = vov = ovo = vov = ovo

In fact, 11(0)1 : If II1, and

SOLUTION:

E

I1(n)I 4 (I n-2)2(1

n40

n40

Inf(n)I2

n40

t Ilf' 112

ExERCISE 11.201:

For every finite set AC 2Z and all e > 0 show

that there exists f e L1 such that

(i) :

0 4 f(n) 4 1 for all n ea;

(ii): f(n) = 1 if n e A;

(iii):

11f111r+ 2N + 1.

Using the Cauchy-Schwarz inequality and Bessel's equation, it follows that

211

If(x)I dx < ON + 1)-1(2N + 1)2 UN + 2r + 1)2

27E 1

0

2N+2r+1 (

2N+ 1

If N is chosen sufficiently large, then

0,

E cn

=

c2 <

W

n

n

on

and let g eL2(T) be such that g(n) = cn. If I

I1pi111 <

°° were to

hold, then we could write

E cn = I I n i n

_

i n

_

(g*pilgi)

= G (g I Ai,ec"pi) Z

l1q'i111,

I19112

which is absurd.

Let 1 < n1 < n2 <

EXERCISE 11.203:

< n

<

p

increasing sequence of integers.

inx p

N

fN(x) =

Set

e N

P=1

Show that for almost all x

limfN(x) = 0. N

AVA - V AV - t0A = VAV - t0A

be a strictly

476

CHAPTER 11:

By Plancherel's Formula

SOLUTION: Cm

2n

L

21[

m=1

J

If

0

2

2(x)I2dx =

< °°,

m=1 m

m

and consequently for almost all x: limf 2(x) = 0.

(1)

M m

If m2 c N < (m + 1)2, then

N

2

fN(x) - N f 2(x) = m

I

NI

e

in x p

2 p=m+1

(m+1)2-1- m2` 2 N

rN-

Since m2/N -' 1 as N -* -, it follows that

limfN(x) = 0

N for every x for which (1) holds.

EXERCISE 11.204:

Let E be a subset of ]0,21r[ with measure zero.

Show that there exists a function f e Lp(T) for all p < and such that for all x e E

limaNf(x) _ .

N AVD = VAV = AVA = VAV = AVA

SOLUTION: that

For every integer k > 1 let Vk be an open set such

TRIGONOMETRIC SERIES

477

meas(Vk) E 2-k

E C Vk C ]0,2n[, Let

be the characteristic function of Vk and f the function,

9k

with period 2n, equal to G k9k on [0,2n].

If 0 < p < - then

r2n1 1/p

k2-k/p

IIkWk IIp 4

k

<

k

which shows that f e Lp(T) for 0 < p < -.

As Fejer's kernel is

positive, the inequality f > kcp implies that k aNf > koNfk.

By Fejer's Theorem, if x e E C Vk, then

liminfaNf(x) > k,

N so

lima f(x)

N

EXERCISE 11.205:

Let (pi) be -a sequence of positive elements of

L1(T) such that

limcpi(n) =

1

for all n e 7L.

2

(a):

Show that if f e C(T) then qi*f - f uniformly.

(b):

Deduce from this that (Ti) is an approximate identity

for L1(T).

AVA = VAV = AVA = vov = ovo

478

CHAPTER 11:

SOLUTION: (a):

If f e C(T) let Ti(f) = w.*f.

The Ti's are con-

tinuous linear operators on C(T), and

M = sup 11T111 = suPII(piII1 = su0i(o) <

i

i

i

If en(x) = exp(inx), then

T.(en) = r. Wen }

en

in the sense of convergence in C(T).

As the en's form a total

set in C(T), it follows that Ti(f) -> f in C(T) for all f e C(T).

SOLUTION: (b):

Let 0 < a < n, and let f be the function in C(T)

that is zero on [-Ja,'-za], equal to unity on [-n,-a] and [a,n], and linear on [-a,-'-2a] and ['za,a].

Then

r

pi(t)dt 1

atlxI

1t

pi(t)f(t)dt = 2n(9i*f)(0) J_n

2nf(0) = 0.

This shows that (pi) is an approximate identity, because, furthermore, Ti > 0 and

IIcpiII1 = ?i(o) ->- I.

ExERCISE 11.206: Let (an)n0 be such that an > 0 for all n, and A } W. i n Show that there exists a continuous function f such that

liml upxInl I f(n)I

I

_W

al", I f(n)I2 = .

TRIGONOMETRIC SERIES

479

in x (Consider a series I An'Ie s , where the integers ns increase s

quickly enough).

nvn = vov = ovo = vov = ovo

SOLUTION:

Let n1 < n2 <

be a sequence of integers such that

The series in x

1

f(x) _

ante

s

s

is then absolutely convergent, and thus f is continuous.

If n 4

ns for all s then 7(n) = 0, while

`(ns) = A. Therefore

li is

PXlnl

f(n)

mans = m,

=

I?(ns)I2

E a1n1If(n)12 = I X" n

EXERCISE 11.207:

s

S

=E1

Let (an)ne2Z be a sequence of complex numbers

such that

E 1Xnf(n)I < n

=.

s

for all f e L1(T).

480

CHAPTER 11: Show that

AV1 = VtV = AVO = V1V = AVo

SOLUTION:

Consider the linear mapping of L1(T) into Q1(2Z) which

associates to f e L1(T) the sequence

f = (anf(n))nea' If fi + 0 in L1(T) and f i -; y in t1(2), then Yn = limfi(n) = limxnfi(n) = 0. Z

1.

By the Closed Graph Theorem the mapping f -+ f is continuous.

Hence there exists a constant A > 0 such that:

E IAnf(n)1 < Allf111, n

f e L1(T).

If in the preceding inequality we take for f the Fejer kernel FN, one obtains

I1 lnl W this yields

L Ian1 < A. n

EXERCISE 11.208: that

Let (), n)ne2Z be a sequence of real numbers such

TRIGONOMETRIC SERIES

X

481

EX2

E X=-

a0,

n

n

Show that there exists a function f e L2(T) such that (i):

(ii):

If(n)I = o(an);

For any a < b, ess suplf(x)I = . asxxb

ovo = VtV = ova = VAV = AVA

SOLUTION:

Let c0 be the vector space of sequences u = (un)nea

of complex numbers such that Inlm IunI = 0,

provided with the norm

Hull = S1PIunl .

If u e c0, there exists u e L2(T) such that

co is a Banach space.

u (x) '\

anune",

X

n and

g112

2 = E anlunl2 < IIuI12 E an

n

n

The mapping u -; u is therefore continuous from c0 into L2(T).

I is a compact interval of length greater than zero, let

pI(u) = ess sup I u(x) I .

xeI

If

482

CHAPTER 11:

It is clear that pI is a semi-norm on co; this semi-norm, furthermore, is lower semi-continuous.

In fact, for every function

g, continuous on I, the mapping

is continuous on c0 and

sup{IJIugl:g

pI(u) =

continuous, J1ii $ 1}.

For every integer s a 1 let us consider the closed set AS = {u:pl(u) < s}.

We are going to show that AS = 0, which will prove that pl(u) for all the u belonging to an everywhere dense Gs set (Baire's O

Theorem).

If one had AS +O one would deduce, taking the convex-

ity of AS into account, that for some p > 0, huh < p => pI(u) '< s.

(1)

Let (ar), 1 < r 4 k, be a sequence of real numbers such that the intervals I + ar cover [0,2n].

hull < 1,

and

If

u(r) _ (une is n r nea'

then also 11U(r) 11

F 1,

and by (1)

ess sup Iu(x)I = ess sup I c(x + ar) I I+a r

I

=

pl(u(r))

4 P

TRIGONOMETRIC SERIES

483

(since u(x+ar)=u(r)(x)).

s/p would hold if Ihu1141.

Hence

For every function f e L1(T) the mapping

2n

u

2nI

z7(x)f(-x)dx 0

would be a continuous linear functional on c0.

Thus there would

exist a sequence (an)ne7G such that

((2n

E lanI <

and

unan = 2n10 u(x)f(-x)dx,

u e c0.

By taking for u the sequence whose n-th term equals l and all of

whose other terms are zero, one obtains an = anf(n),

and consequently

X andf(n) < n

and thus would hold for all f e L1(T).

this would imply

IX n

n

By the preceding exercise

-

<

contrary to the hypothesis.

Let (1k) be a sequence of compact intervals with lengths greater than zero such that every interval I of length greater than zero contains an Tk.

For every k there exists a Gd everywhere

dense in c0, say Ek,,such that pI (u) = m if u e Ek. k

The set

484

CHAPTER 11:

E = nk Ek is then a Gd everywhere dense in c0, and if us E then

u e L2(T), u(n) = anun = o(A ),

pI(u) = ess sup Iu(x)I >. ess sup Iu(x)I _ .

I

EXERCISE 11.209:

Ik

Assume that

en < n ,

cn ' cn+1 3 0,

n > 1.

Show that

(a):

N I I

c sinnxl < 2Av.

n=1

(b):

n

Deduce from this that the functions

W f(x) =

E

cnsinnx

n=1

belongs to L1(T), and that the right side is its Fourier series. 000 = vov = ovo = vov = ovo

SOLUTION: (a):

One can assume that 0 < x <

ger between 1 and N - 1.

sinnxI < nx

and

it.

Let M be an ipte-

Noting that

Isinpx +

+ singxl <

1 si- x < x 9

TRIGONOMETRIC SERIES

485

one obtains N

M

1 n=1

csinnx1.1 111+1 n=1

N E

15AMx+

n=M+1

An

M+1x.

If x > / let M = 0, so that the first sum disappears and a bound

A' is obtained.

If x < //N let M = N, so the second sum disap-

pears, which again gives the bound AI.

Finally, if 1 4 I/x < N

make M = [I/x], which leads to the bound 2AVi. SOLUTION: (b):

It is known that the series giving f(x) converges

for all x (Abel-Dirichlet Theorem).

By Lebesgue's Dominated Con-

vergence theorem, f e L1(T) and the partial sums of the series con-

verge to f in L1(T), so ?(n) = en.

f is continuous on ]0,n], since

REMARK 1:

ensinnxi 4

I

n=N+l

x aN+l'

Plancherel's Theorem shows directly that there exists

REMARK 2:

a function g e L2(T) C L1(T) such that:

M

g(x) =

E

Cnsinnx,

n=1

in the sense of convergence in L2(T).

By Remark 1 f = g almost

everywhere, whence the result.

EXERCISE 11.210:

Consider the trigonometric series

W

ansinnx, n=1

(*)

486

CHAPTER 11:

where (an) is a sequence of positive numbers which decreases to zero. (a):

Show that the series (*) converges for all x.

(b):

Show that the following conditions are equivalent:

(i):

(ii):

The series (*) is uniformly convergent; The series (*) is the Fourier series of a continuous function;

(iii):

an = o(1/n).

Show that the following conditions are equivalent:

(c): (i):

There exists a constant M such that:

N E

ansinnx I c M,

x eat,

N>, 1;

n=1 (ii):

The series (*) is the Fourier series of a function of L-(T);

(iii):

(d): (i):

an = 0(1/n).

Show that the following conditions are equivalent: The series (*) is the Fourier series of a function of L1(T);

(ii):

(iii) :

The sum of the series (*) is a function in L1(T);

E n-1 a

< -.

Show that in this case the partial sums of the series (*) converge in L1(T).

000 = Vov - AVO = vav = ovo

TRIGONOMETRIC SERIES

487

First recall that the formula:

SOLUTION:

2sinnx.sin-x = cos(n - J)x - cos(n t J)x shows that for N < M and 0 <

N D sinnx = (x) = N,M n=N

IxI 4 it

cos(N - ')x - cos(M + })x 2sin

1 )

(

-

and consequently that

IDN,M(x)I

i

<

since

T

IxI 5 n.

2)

Aan.DN n(x),

(3)

0 <

,

On the other hand,

M I

n=N

M-1 a sinnx = aMDN M(x) + n

E

n=N

'

where 4an

= an - anti

3 0.

From this one deduces that naN

Mp

ansinnxl < 'T

,

0 <

IxI < it.

(4)

InLN

SOLUTION: (a):

(4) shows that the series (*) is uniformly con-

vergent for 0 < d s IxI < it.

It clearly converges for x = 0.

Thus its sum is an odd function, continuous for 0 < IxI 4 it. SOLUTION: (b):

If the series (*) is uniformly convergent, then

488

CHAPTER 11:

lim{supl

N-' x

ansinnxl} = 0.

I

gN+1,,N

If 0 <

IxI < it, there exists an integer p such that

it

FT-1 Using (4) with N replaced by N + p and letting M

one then

has

I

I n>.N

ansinnxl 4

ansinnxl +

1

1

IxI

I

Non'N+p

na

+

naN+

n

ix

.pIXICN+(p+1)xN+

P

so that, because

:

I

N4n;(i).

If the series (*) is uniformly convergent, its sum is a continuous function f whose Fourier coefficients are obtained by integrating term by term, which shows that the Fourier series of f is certainly (*).

Thus (i) => (ii).

If (*) is the Fourier series of a continuous function, then the Fejer sums aN(x) of (*) converge uniformly.

Since aN(0) = 0, it

follows that

N02N(2'N) = 0. If x = 7t/2N then nx < it, hence sinnx 3 0, when n < 2N and nx < 7E12, 2nx Thus when n < N. hence sinnx a = N,

n1

a2N(77E

N)

;' n1N (1

2N n (iii). SOLUTION: (c):

Inequality (5) proved above again shows that (i)

=> (iii), and in the same way (iii) => (i) follows from (6). Furthermore, (i) implies that the sum of the series (*) belongs to LW(T), and Lebesgue's Theorem justifies obtaining the Fourier

490

CHAPTER 11:

coefficients of this sum by integrating term by term, so that (*) is certainly the Fourier series of its sum.

It has thus been

On the other hand, if (ii) is satisfied,

proved that (i) => (ii).

the aN are uniformly bounded, and the inequality

> 4 (N + 1)aN

a2N (2'N,

proved in part (b) shows that (ii) => (iii). Letting N = 1 and M -

SOLUTION: (d):

f(x) _

ansinnx = n=1

in (3) yields

Aanb1 n(x). n=1 L

'

Note that by (1)

cosix - cos(n + D1,nx_-

g(x) =

2sinx

n=1

h(x) = 2

)x

sin2 x , + "2sinnx, tangy

Dansin2'nx tanx 1

ansinnx.

n=1

Since Aan > 0

and

E Aan = a0,

the function h is continuous (note, furthermore, that as the series which gives h is absolutely convergent, this series is cer-

491

TRIGONOMETRIC SERIES Furthermore, because

tainly the Fourier series of h).

sin 22nx ' 0

for 0 < x 4 n,

tanix g >. 0 and

(n

'

cWc

Og(x)dx =

n=1

n

°anj0

(7)

ax

tan x

It follows that f e L1(T) if and only if the series which appears in the right side of (7) converges.

Furthermore, in this case

the Fourier series of f certainly is (*), for the integrals of g(x)sinnx and h(x)sinnx are obtained by integrating term by term, and therefore lrn

J71

n

n

f(x)sinnxdx =

Da

k=1

k

nJI

n

1,k

(x)sinnxdx =

k=n

Aa

k

= a

.

n'

Thus (ii) => (i).

As the fact that (i) => (ii) follows from exercise 11.192, in order to prove that (ii) (iii) it remains to prove that the convergence of the right side of (7) is equivalent to I n-1a Now, since

1

tan x -

2 X

is integrable on (0,n), we have

ji7c

JO sinntx dx = 2J1 s x z n logn,

dx + 0(1) =

+ 0(1)

n

<

CHAPTER 11:

492

and because

lognti1+2+ to, (ii) is therefore equivalent to

I1 t

t .1 Aan <

+

(8)

n

2

n=1

11

Since N

I n=

I1 +

+ ... + N)aNtl +

2

n=1

+ ... +

I1 +

1

n=1 l

2

nlpan 11

The converse is also true,

it is clear that (iii) implies (8).

because

+

I1 t

[i++

+ NJaN+l =

... +

tan

n:Ntl

2 1

n>.N+1

{1 +

+

+ nldan.

2

Finally, if sN is the n-th partial sum of the series (*), then

f(x) - aN(x) _

Aan.D1 n(x)

n>N (cf.

'

(3)), and consequently

Ilf - SN II1 <

I

Aan IID1

n>N

But by the preceding

II51 nIll ti 1 logn,

'n

Ill.

TRIGONOMETRIC SERIES

493

hence by (8)

IIf-SNII130. EXERCISE 11.211: (a):

If (an)n30 is a sequence of complex num-

bers such that 00

an -' 0,

E

IA2anI < 00,

n=0

prove that the series Co

f(x) = la0 +

ancosnx

(1)

n=1 converges for 0 < x < 2n, and that Co

f(x) = 2

1

(n + 1)A2anFn(x),

(2)

n=0 where Fn denotes the n-th Fejer kernel. (b):

Show that if

an -> 0,

1

(n + 1)Io2anI <

n=0

then the function f e

an + 0,

L1,

and that if:

n&an -> 0,

1

(n + 1)IA2anI < m,

n=0 then the right side of (1) is the Fourier series of f. evo - vev - 4aL - vov = pv4

CHAPTER 11:

494

We have (where Dn is the n-th Dirichlet kernel):

SOLUTION: (a):

N

N

a

8N(x) = 2 +

ancosnx =

E

n=1

2 n=0 =

an(

n

(x) - Dn_1(x))

N-Ci

2 aNDN(x) +

AanDn(x).

G

2

n=o

Since

+ Dn = (n + i)Fn,

D0 +

this yields N-1

sN(x) = 'aNDN(x) + jN aN_1FN-1(x) +

E

(n + i)A2an.Fn(x).

n=0 For 0 < x < 2n:

1 sin x

IDN(x)

,

IF (x)l <

n

1

(n +

-12X

which gives the result. SOLUTION: (b):

If

an - 0,

G

(n + 1)IA2anI < °°,

then (2) shows that f e L1(T), since IIFn1I, = 1.

Moreover, the

convergence in (2) holds in L1(T), so for every integer k:

(n + 1)62anFn(k)

2f(k) _

n=0

m =

I

n=k

(n + i - k)t2an.

TRIGONOMETRIC SERIES

495

Now,

N (n + 1 - k)a2an = ak

E

n=k

- aN+l

+ (N + 1 - k)DaN+1'

Consequently, if, in addition, NAa

n

-> 0

we obtain that ?(k) = jak,

which proves the last assertion.

EXERCISE 11.212: (a):

Let f e L1(T), f >, 0, and

0.

Show that for 0 < a < it,

r

a

n

f(t)dt < 1 1

aJ-f(t)dt. a

- it

(b):

Let (cn)ne2Z be a sequence of positive real numbers.

Suppose that there exists d > 0 and he L2(-S,S) such that a

+m E

n ccp(-n) =

n=-m

71-h(x)(p(x)dx

(*)

S

for every function cp that is infinitely differentiable and has

compact support contained in ]-S,S[. Show that

C2

n=-w

aJ a

Ih(x)I2dx.

< S

nvn - VAV - tV - vov = nvn

496

CHAPTER 11:

SOLUTION: (a):

Let 8 be the characteristic function of [-a/2,

a/2] and W = 8*8.

a 0 1 a

f(t)c(t)dt = 2n

f(n)s(- n)

=

n

n

I8(0)I2n

2nf(0)$(0) =

fn

f(t)dt

2 n

(2 n) f _nf(t)dt, and the result follows SOLUTION: (b):

(f19) =

Set, as usual

J_1r.

For s = 1,2 ... let 85 a C'(T), 85

0,

11-5sll

l = 1, and 85(x) = 0

if 1/s 0, and choose d > 0 so that Ie(u)I < a

if

0 E u < d.

The Riemann-Lebesgue Theorem and the fact that

AN(u)>.0

if 0su4 R.

show that n

r

n

n

iJ eA*1 < a A* + J e(u)cotiudu + o(1). N J d 0 oN

lui 4 it,

TRIGONOMETRIC SERIES

501

Since

JOAN = JODN + 0(1)

ti 2logN,

from this one concludes that n

limsup(logN)-11E

J

fo

N

E 2a,

an d, since a is arbitrary, that

EAN = 0(logN). J

0

By (3) we also have: n

J0CD

= o (logN).

SOLUTION: (d):

As the function DN is odd, it is easily seen that n

N(x) _ - 2nJ (f(x + u) - f(x - u))DN(u)du, 0

If

f(x + u) - f(x - u) = C + e(u)

with lirE(u), u-*0

from parts (b) and (c) above one deduces that

lim(logN)-1sN(x) _ - n N

(4)

CHAPTER 11:

502

If f has left and right hand limits at all points

SOLUTION: (e):

then £ exists at every point x and is equal to f(x + 0) - f(x - 0).

Further, if

f(n) = a

11n

J

'

then

IsN()I 6

I

(11(n) + f(-n)) = of I I = o(logN).

(n=1

n=1

From (4) it is then deduced that £ = 0, which proves that f is continuous.

EXERCISE 11.214: (a):

Let (un)n)1 be a sequence of positive num-

bers such that

U

n

c

A

n

Set:

u1 + 2u2 +

an =

+ nun

n

,

2 2

u2 + 4u2 +

bn =

+ n u

n

,

W

yn = n

L

s=1

ussin2

2n

Show that the conditions an -> 0, bn - 0, yn -* 0 are equiv-

TRIGOROMETRIC SERIES

503

(Prove that

alent.

an2 < bn

Aan,

bn

yn

and that for every integer v

n

2

vb

2

+A

Yn

4

(b):

Let f be a real function with left and right hand

nv

v

limits at all points, and with period 2n.

For every integer

n > 1 set 2n-1

9n(u) _

If(u + xr+l)

f(u + xr)I2

r=0 where xr = nr/n.

Show that if

If(C + 0) - fU - 0)j > d > 0

at some point E.

then whenever n is large enough

rp n(u) > d

for almost all u.

Show also that if f is continuous and has bounded variation, then (pn -> 0 uniformly on R.

(c):

Calculate:

r2n

pn(u)du,

J

0

using Plancherel's Formula. (d):

With the aid of what has gone before, prove the fol-

CHAPTER 11:

504

A necessary and sufficient con-

lowing theorem (owed to Wiener):

dition for a function f with bounded variation to be continuous

is that: 2f(2) + ... + nf(n) = 0. lim f(1) +

n

n

AVA = V/V = AVA = VAV = AVt

SOLUTION:

2 2

Since s us

(a):

.<

Asus, it is clear that b n

.<

Aan.

The Cauchy-Schwarz Inequality shows that

Iu2

+

+ n 2u 2

4u2 +

a

n

l

2 = J

b12.

n

Furthermore, noting that sinu . 2u/n if 0 .< u < n/2,

yn > n

us(;8 )2

I

= bn.

s=1

yn < n

X

s=1

"

us l(2n) 2+ A2n r

2

4

bnv + A2nl

L

s>nv s 2 2

a2 _ n4v bnv + A

1nv x

2

v

All this shows that an - 0 is equivalent to bn - 0, then that

yn

-)- 0 implies

bn + 0, and finally that if b n

integer v > 0 2 l i nsupyn - A

and consequently yn + 0.

-* 0 then for every

TRIGONOMETRIC SERIES SOLUTION: (b):

505

If

there exists an integer n0 such that:

n) - f(a)I > d if n a n0 and a < E < a + n

(1)

If u t C + nQ there exists an integer k such that

u< C+2kn n0

and

u $ E + nip.

If it is now assumed that f is continuous and has bounded variation, then for all e > 0 there exists n > 0 such that If(u) - f(v)I < e

if

Iu - VI < n,

and consequently if n/n < n, 2n-1 qpn(u) < e

I

r=0

If(u + xr+1) - f(u + xr)J e EV(f;0,2n).

CHAPTER 11:

506

This shows that n i 0 uniformly on1R.

SOLUTION: (c): (eisxr+1 - eisxr)f(s)eisu

f(u + xr+l) - f(u + xr) ti s

2ie

ti

is(x + n/2n) r

s

((

ll sinl2nJf(s)eisu.

ll

Plancherel's Formula gives m

(2 2nl 0

gn(u)du = 8n

Y

s=

IJ'(s)I2sin212nJ l

and as f is real, ?(-s) = f(a), so that M

(27[

2j

pn()du = 16n

SOLUTION:

=

(d):

By the results proved in parts (b) and (c),

If(s)I2sin2[In-

n

Ij(s)I2sin2[22n-

s=1

0

s=1

0

JJJJ

if and only if f is continuous.

Furthermore, as f has bounded

variation,

Using part (a), Wiener's Theorem is obtained.

EXERCISE 11.215: (a):

Show that there exists a sequence

such that 0 4 as 4 1 and that for every integer p 3 1:

TRIGQNOMETRIC SERIES

507

(1 t cost)(1 + cosst). (1 t cos4p-1t)

(4P-1)/3 _

ascosst.

I

(1)

S=O

(b):

Denote by yp the number of indices s such that:

0 v a '

- 8 v a an

av

> (n - v)aloga

Since

n

n

avioga - 1 > a ologa - z > as

0

> 0,

it follows that

IF(an-8 V

)I

1 <

n(avloga - 1)

if n > v.

> avioga - 2.

TRIGONOMETJ?IC SERIES

513

If it is now assumed that n0 E n < v then

sv - an > av - an -

> (v - n)anloga - ' >. anloga - 'z > aan,

'7

because

a

n

n0

>, a

>

2 logo - A

Thus

IF(an - sv)I <

1

naa

SOLUTION: (d):

if n0 4< n < V.

n

Taking into account part (c) and the inequalities

IF(x)l < 1 for all x and IF(x)l

In0 1 and-a)n

F(an

>. 2/n if jxj < 'zf we obtain

- sv)I .

n=0 a+ s v a-an

n=n0

n=v+1

= 0(a-v),

a-an = 0(a-v),

n(avloga - Z) n=0

and

(1-a)v a

av+sv

(1-a)v

F(av

a(l-a)n

v n=0

1

s

r 1

v n=0

v-1 n s

_n

- s v )>,2a 71 2av+ 1 ti

-av

7Ca

It follows that if 0 < 6 < 1/rz then

-

0(a-v),

CHAPTER 11:

514

av1f(s)I

> a

when v is large enough.

SC, l?(8V)

Since sv ti aV we also have

>

for v large enough, which proves that the .W's are not o(s-a).

EXERCISE 11.217: (a);

If -1 < r < 1 set

n inX I rlle

Pr(x) =

1 - r2 1 - 2rcosx + r2 (b):

For every function f e L1(T) set +00

Ar(f;x) _

rlnl?(n)e"x.

L

Prove that if f(x + 0) and f(x - 0) exist at a point x, then lim A(f;x) r = '2(f(x + 0) + f(x - 0)).

r-*1(c) :

F(x) =

Set

(x J

f(t)dt. n

Prove that if F(n) = 0, then

TRIGONOMETRIC SERIES

( x) = r f'

A

n

1 2nj_n

515

F(x + t) - F(x - t) Q (t)dt r 2sint

'

where Qr is a function to be determined. Prove that as r -> 1 (r < 1) the Qr's form an approx-

(d):

imate identity in L1(T).

Prove that if at some point x the "symmetric deriva-

(e):

tive of F", Dsf(x) = lim[F(x + t) - F(x - t)]/2t, exists and is t->0 finite, then limA (f;x) = D F(x).

r+1_ (f):

r

s

Compare this result with the Lebesgue-F6jer theorem. AVl = V AV = p0A = OAV = A0A

SOLUTION: (a):

re ix

P(x)=1+

1 - relx

r

SOLUTION: (b):

+

re -ix 1 - relx

-

1-r

2

1 - 2rcosx + r2

Replacing the f(n) by their defining expressions

in the formula for Ar(f;x) shows that Ar(f;x) = (f*P )(x).

Note that the r form an approximate identity in L1(T) as r -} 1, since

Pr > 0,

and for all 6 > 0

1

2n

n

J- P (x)dx = 1, n

r

516

CHAPTER 11:

2n

P (x)dx < P (d),

dslxl 0; if d is chosen so that if 0 < t < d,

IQ(t)I < e then

IAr(f;a) - sI . e + Pr(d)IIQIII, which proves that

limAr (f;x) = s.

r+1-

Integrating by parts and taking into account

SOLUTIONS: (c);(d):

that Pr'(-t) = -Pr(t) yields

it

1

1

n

Ar(f;x) = 2J- f(t)Pr(x - t)dt - 2nJ- F(t)Pr(x - t)dt n

n

n = 2nJ-

F(x n

so

t)P'r (t)dt

n = - 2nJ- F(x + t)P'(t)dt, n

r

TRIGONOMETRIC SERIES

517 n

j(F(x + t) - F(x - t))P'(t)dt

Ar(f;x) = - 2n n

setting Qr(t)

sint.P (t),

we have

F(x +

Ar(f;x) = 2If

- t) Qr(t)dt. (1)

2sintF(x

n

It is clear that

Qr(-t) = Qr(t)

and

Qr >. 0.

Also, if f(x) = cosx then F(x) = sinx; therefore taking f(x) _ cosx and y = 0 in (1) yields n

2nI

Qr(t)dt = r. n

Finally an elementary calculation shows that

(

l1

- 2rcost + r

attains its maximum when

cost =

2r

1 + r2 Therefore, if 0 < d < it then sup Qr(t) 4 Qr(d),

66t67[

2

sint

Qr(t) = 2r(1 - r2 )I

2) )

CHAPTER 11:

518

as long as

r2)-1

cosd < 2r(1 +

Since Qr(d) -> 0 as r -> 1, the Qr form an approximate identity in L1(T).

SOLUTION: (e):

By adding a constant to f, F(n) may be assumed

Setting

to 0.

fi(t) = F(x + t) - F(x - t) - D F(x), 2sint S we have n

Ar(f;x) - D(x) = 2nJ (t)Qr(t)dt + (r - 1)DF(x), -n

since i(t) -> 0 as t - 0, one can show as in (b) that the integral

on the right side tends to 0 as r - 1, from which the result follows.

(Note that there is no problem at ±n since at these points

Qr(t) " (constant)sin2t).

SOLUTION: (f):

In particular

Ar(f;x) + f(x)

at all points where f(x) = F'(x), i.e. almost everywhere.

Now,

the condition f(x) = F'(x) can be written rh

limh_1 h-*O

0

+ u) - f(x))du = 0.

Therefore this result is better than that of Lebesgue-Fejer, which says aN(f;x) -

f(x) if

519

TRIGONOMETRIC SERIES

h limb-11 If(x + U) - f(x)ldu = 0. h->0 J0

The series

EXERCISE 11.218:

en converges to s in the sense of

Borel (written

(B)

c _m

n

s)

if

s r

r

r-

lime

n

ni

= s

n=0

where e

sn = IpI

. 0;

n,i

w

(ii) :

u

E

n=0 (iii):

limu

n,i = 1;

ni

ti

= 0.

'

By exercise 10.177 this implies

s rn

limer

X1->-

ni

SOLUTION: (b):

ifsn->s.

In this case

1-zn+1 sn

s

n=0

1-z

and consequently

522

CHAPTER 11:

e -r

s rn n n=0 n! L

= 1 1

z

(1 - ze

from which the result follows. Because

SOLUTION: (c):

f*Dn it is clear that

Br(f; ) = f*Br, where

Br

= e

-r

C

n Dr n

n=O

nt

.

Recall that i(n+ )x

si-fix

Dn(x) = ImIe

0 < lxI

z

and consequently

x

i

Br(x) = se. =

Im(ejx+reix

)

e-r(1-cosx) sin(x + rsinx) sinix

SOLUTION: (d):

First of all, if n > 0 is an integer, then

n-1 J(nn) =

n

J

8=0

0

sinx X

sn

dx.

The integral that appears in this expression is equivalent to 2/ns are s -> -, which proves that

TRIGONOMETRIC SERIES

J(ma) ti

523

loge.

Therefore, when x -> m:

0 < J(x) - J(1[x/71]) <

[x/1]

J(n[x/n]) ti n log[x/n] tin logx,

and consequently

J(x) ti n logx.

SOLUTION: (e):

2 1/2 e-2r sin2u IIBrIIl = nj

sinu

0

1.1

As the function

U

-

1

s inu

I

Isin(u + rsin2u)Idu.

is bounded on [0,1/2], it is clear

that as

2 n/2 e-2rsin2u Isin(u + rsin2u)Idu + o(1). nj u _

IIBrIIl =

0

Using the mean value theorem and the inequalities

sinu 32n ,

0,0 n!

e-r

Ln I n. n>.O

n

21J-1Dn

= 1.

If, for 0 < Iti 4 it, we set

526

CHAPTER 11:

w(t) = (f(x + t) - f(x))log1--1 t

,

Br(t)

Qr(t) =

log l t l then U

Br(f;x) - f(x) = 2nf-X cp(t)Qr(t)dt.

(1)

Note that for 0 < 6 < ItI E n:

-2rsin2'z6

IQr(t)1

log2 sin 6

On the other hand, some calculations similar in every way to those carried out in (e) show that

n/2 a-2ru

2

Isin(2r + 1)uldu + o(1).

IIQr1I1 = jcJ

0

ulogu

Decomposing the integral above into an integral taken between 0

and 1/1, and another taken between l/V and n/2, it is seen to be bounded by:

J((2r + 1-)r-') +

logn

IIQx,111 4< A,

a2u2

1

log2

T

u

du.

A = constant.

But then if E > 0 and if 6 > 0 is such that ItP(t)I < e for 1t1.< 6.

TRIGONOMETRIC SERIES

527

by (1) we shall have

- 2rsin2d

IB(f;x) r - f(x)I < eA +

l0 2 sin d.l

d4Itkit

Icp(t)Idt,

which proves that

limB(f;x) r = f(x). r-+W

Let (an) be a sequence of numbers an >. 0, and

EXERCISE 11.219:

An = a0 +

+a

.

Suppose that

a

n-> 0.

An

A

n The series +W L

en

is said to converge to s in the sense of NBrlund, written +w (N)

E Cn - 8,

_m

if

vn =

where

a s + ... + a 8 n 0 0 n

A n

528

CHAPTER 11: 8

=

C

X

IP14n

(I):

.

P

Show that if a series converges in the ordinary

(a):

sense it converges in the Narlund sense. (I):

(Use exercise 10.177).

Prove that if 0 < a0 .< a1 <

(b):

Cesaro converg-

ence implies Narlund convergence. (II):

In the whole of this Part (II) assume that:

a0>a1>,.., >,an>,**,

an-*0.

9

(*)

For f e L1 set n Nn(f) = A 1 apse-p(f), n p=0 where s(f) denotes the n-th Fourier sum of f. (II):

(a):

Show that

Nn(f) = f,t(Un - n), where

Un(x) = A

sinsn x X

V

1 cos(n + 1)x = A

n

Show that 1/n

n(x)Idx J

p acos(p +

)x,

(1)

a sin(p + 7)x.

(2)

p=0

n

sin ix

n p=0

P

TRIGONOMETRIC SERIES

529

is bounded for n >. 1. (II):

n

Isin(p t 1)x dx 4 B(p + 1)log p + C(p + 1),

J1/n n

1/n

Show that there exist constants B,C,D such that:

(b):

14 p < n,

2sin2ix

slnxI

dx < Dlogn,

n > 2.

2sin2Ix

(II): (c):

By carrying out some suitable Abel transforma-

tions on (1) and (2), prove that

IIVn Ill < 1, and:

IUn(x)Idx < C + f 1/n

Aplogll +

(Elogn + B

p=1

n

`

where E is a constant. (II): (d):

Now assume that

SnPIIUnII, By carrying out another Abel transformation on (1) for Un and observing that

sin2(n + 1)x < Isin(n + 1)x I,

show first of all that the quantity

71

JO(sin2(n

n-l

x )x

+ 1)W){cossnnt l

+ An

Asin(p + 1)xfdx p= nL

2

JJJ

CHAPTER 11:

530

stays bounded, then by studying the behaviour of the integrals:

s infix

+ 1)x.cos(n + z) x

7E sin2(n f0

dx'

lTE

cos(2n + 2)x.sin(p + 1)xdx, 0

deduce that

nA

1

sup n

< I -n p=1 P

A

(II):

.

Conclude from the preceding that when (*) is

(e):

satisfied the following conditions are equivalent:

For all f e L1(T)

(i) :

f(n)e

inx (N)

= f(x)

at every point x where f is continuous;

n

sup A n

A P

p=1 P

< . V,

For every real number4 set

(III):

na = (a +

and assume that

a

_

n

n)

n! -

n

Aa-1

n

TRIGONOMETRIC SERIES (III): (a):

A

n

531

Show that

= Aa

n

and that if a > 0 one obtains a summation method in the style of NSrlund.

Next show that if 0 < a < 1 one is in the case studied in

part (II) above, if a > 1, in that of part (I)(b); and that finally, if a = 1 Cesaro's method is recovered. (III): (b):

g0 An1 + ... + S n

a a

n

It is said that:

=

A

Aa n

is the CESARO SUM OF ORDER a of the series I en, and that the latter is (C,a)-CONVERGENT if 6n is (a > 0).

Prove that for every a > 0 the Fourier series of a function f e L1(T) is (C,a)-convergent to f at every point where f is continuous.

ovo = VAV - ovo = vov = ovo

SOLUTION: (I): (a):

vn

P10 un,pap, =

where a

A-p n U

n,p

if 0 < p 6 n,

=

0

if p > n.

CHAPTER 11:

532

It is clear that

= 1.

u p>0

n,P

Moreover,

a n_p un,P { An-p

and consequently = 0. limu n n,P

By exercise 10.177 sn -> s implies vn -> s. SOLUTION: (I): (b):

Writing an for the n-th Cesaro sum of the

series X Cn,

n

an-p((p + 1)ap -

vn

PCP-1

n p0

n-1 (n + 1)a A 0 an + A (p + 1)(anP - an- P 1)a p n n p=0

p=O

an,paP,

where

(P (ann

(n + 1)a0 n,p

A

0

P

- an-P

1)

if 0 4 p < n - 1, if p = n,

n

ifp>n.

TRIGONOMETRIC SERIES As the sequence (a

n

)

533

is increasing, all the An,p >. 0.

It is

clear that = 0.

lima

n

n,P

Furthermore, when all the sn's are equal to one, the same holds for all the a 's and v 's. so n n

= 1.

A

n,P

p=0

By Exercise 10.177 an -* s implies vn - s.

SOLUTION: (II):

(a):

Nn(f) is equal to the convolution product

of f with 1

n I

n p=0

n

1

p aDn-p(x) = A

=

+ sin(n sin x

ap

E

n p=0

sin(n ± 1)x

1

n

apcos(p +

A

cossn 1

-7-

i IX

1)x

n 0

p which implies the first result desired. Note that

sin(n + 1)x s in x

and consequently IUnI < 2(n + 1),

so

)x

p=0

n

An

)x

2(n + 1) ,

apsin(p + J)x,

534

CHAPTER 11:

1/n

f0

l

(

4 211 + l

JU

SOLUTIONS: (II): (b);(c):

4.

Denoting by Fn the n-th Fejer kernel

and setting A

= ap - ap+i'

we obtain

V (x) =

1

An

n-1

cos(n + 1)x{(n + 1)anFn(x) +

(p + 1)h F (x)}. P P

E

p=0

(1)

Since

and

AP;0, 0

IIFPIII=1,

it follows that

11711

A {(n + 1)an +

n-1

p + i)AP} = 1.

E

p=0

n

Similarly, noting that

n I

cos(p + ')x =

sin(n + 1)x

2sinx

p=o we obtain

Un(x) = A sin(n + 1)x

(2)

n

x

Ja

n

sin(n + 1)x 2sin21x

n,l A + p=0

p

sin(p t 1)x) 2sin2.,x

J

TRIGONOMETRIC SERIES

535

Now note that fit

2

sinx

xdx

(n

dx 4

1/n 2sin tx

=

it

log(nn) ` Dlogn.

1/n (2x )

for all n >. 2 if D is chosen large enough.

When 14p