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 Theoryby Richard K. Guy1981. xviii, 161 pages. 17 illus.

Theorems and Problems in Functional Analysisby A.A. Kirillov and A.D. Gvishiani1982. ix, 347 pages. 6 illus.

Problems in Analysisby Bernard Gelbaum1982. vii, 228 pages. 9 illus.

A Problem Seminarby Donald J. Newman1982. viii, 113 pages.

Problem-Solving Through Problemsby Loren C. Larson1983. xi, 344 pages. 104 illus.

Demography Through Problemsby N. Keyfitz and J.A. Beekman1984. viii, 141 pages. 22 illus.

Problem Book for First Year Calculusby George W. B/uman1984. xvi. 384 pages. 384 illus.

Exercises in Integrationby Claude George1984. x. 550 pages. 6 illus.

Exercises in Number Theoryby D.P. Parent1984. x. 541 pages.

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

Claude George

Exercises inIntegration

With 6 Illustrations

Springer-VerlagNew York Berlin Heidelberg Tokyo

Claude George Translator

University de Nancy I J.M. ColeUER Sciences Mathematiques 17 St. Mary's MountBoite Postale 239 Leybum, North Yorkshire DL8 5JB54506 Vandoeuvre les Nancy Cedex U.K.France

EditorPaul R. HalmosDepartment of MathematicsIndiana UniversityBloomington, IN 47405U.S.A.

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

Library of Congress Cataloging in Publication DataGeorge, Claude.

Exercises in integration.(Problem books in mathematics)Translation of: Exercices et problemes dìnt6gration.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 formwithout 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 TokyoISBN 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 de-

tailed solutions of exercises and problems, which I used to dis-

tribute 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. On

the other hand, the student who 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

vi INTRODUCTION

Teachers will find this book to be an important, if not ex-

haustive, 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 redis-

covering it. This is particularly true for those solutions of

which one remembers the heuristic form but of which the writ-

ing 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 sub-

mit impeccable corrections to them, where the notations are jud-

iciously chosen, phrases of the kind "it is clear that ... " used

wittingly, and where the telegraphic style gives way to concise-

ness. 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 proper-

ties 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, prob-

ability, 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 par-

ticular 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 mea-

sures that are not a-finite and the applications of the Radon-

Nikodym 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 integra-

tion is not foreign to this theorem. Truthfully, there is another

point that is not tackled in this book, namely the matter of Four-

ier 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 trig-

onometric 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 be-

comes 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 inter-

esting 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 ... ... ... ... ... ... 37

(Exercises 1.1 1.21)

CHAPTER 2: a-ALGEBRAS AND POSITIVE MEASURES ... ... ... 79

Exercises 2.22 - 2.28)

CHAPTER 3: THE FUNDAMENTAL THEOREMS. ... ... ... ... 89

(Exercises 3.29 - 3.72)

CHAPTER 4: ASYMPTOTIC EVALUATION OF INTEGRALS... ... ... 177(Exercises 4.73 - 4.78)

CHAPTER 5: FUBINI'S THEOREM... ... ... ... ... ... 199

(Exercises 5.79 - 5.99)

CHAPTER 6: THE LP SPACES ... ... ... ... ... ... 225

(Exercises 6.100 - 6.125)

CHAPTER 7: THE SPACE L2. ... ... ... ... ... ... 285

(Exercises 7.126 - 7.137)

CHAPTER 8: CONVOLUTION PRODUCTS AND FOURIER TRANSFORMS ... 325

(Exercises 8.138 - 8.162)

CHAPTER 9: FUNCTIONS WITH BOUNDED VARIATION: ABSOLUTELYCONTINUOUS FUNCTIONS: DIFFERENTIATION ANDINTEGRATION (Exercises 9.163 - 9.173)... ... 405

ix

x TABLE OF CONTENTS

CHAPTER 10: SUMMATION PROCESSES: TRIGONOMETRIC POLYNOMIALS.. 429(Exercises 10.174 - 10.184)

CHAPTER 11: TRIGONOMETRIC SERIES ... ... ... ... ... 451(Exercises 11.185 - 11.230)

ERRATUM TO EXERCISE 3.45 ... ... ... ... ... 545

BIBLIOGRAPHY ... ... ... ... ... ... ... ... 547

NAME INDEX.. ... ... ... ... ... ... ... ... 549

CHAPTER 0

Outline of the Course

0.1 a-ALGEBRAS AND MEASURES

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-alge-

bra A, and that the a-algebra A is closed under countable intersec-

tion. For two sets A,B e A let us denote A - B = {x:x a A,x $ B};

then we have (A - B)e A.

The smallest a-algebra containing the open sets of ]R is the

a-algebra of BOREL SETS of ]R ; 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

2 CHAPTER 0: OUTLINE

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-alge-

bra 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 dimen-

sion 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 unambig-

uously defines a positive measure on the a-algebra of Lebesgue

measurable sets of ]Rp. This measure is called the LEBESGUE MEA-

SURE ON I(. 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 de-

terminant 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 alge-

bra 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 Z B e B.When Y= Ilzp and B isthe a-algebra of Borel sets, one says, simply, that f is an A-MEAS-

URABLE 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. Let us

note that every continuous mapping of Ilzp into Ilzq is Borel. If

(fn) is a sequence of A-measurable arithmetic functions, the func-

tions 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 ex-

ists 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.


This allows the notion of measurability to be extended to func-

tions 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 measur-

able 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 q,(x)dx = 9 _ aimeas(Ai),n 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 meas-

urable arithmetic function on ]Rn there exists an increasing

sequence ((pi) of positive simple functions which tends towards f

at every point. One then sets:

J f(x)dx = if = limf . .

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 1< Jg;

PROPERTY (5): J(f + g) = if + fg;

PROPERTY (6) If A e]-R+., then JAf = AJf.

One can prove the following two fundamental results:

THEOREM: (Lebesgue's Monotonic Convergence): If (fn) is an in-

creasing sequence of measurable positive arithmetic functions,

then

Jlimf = limif-'n n n


LEMMA: (Fatou): If (fn) is a sequence of positive arithmetic func-

tions, then

Jliminffn " liminfn n

Ifn.

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 ex-

tended 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-algebraE

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_,

lu l = u+ + u-,

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


functions are integrable. If f is integrable, then

ff1 <J1f11

with equality holding only if there exists a e 0 such that f =

alfl almost everywhere.

The two essential properties of the Lebesgue integral are the

following:

THEOREM: (Lebesgue's Dominated Convergence): Let (fn) be a sequence

of integrable functions that converges to f almost everywhere. If

there exists a measurable positve arithmetic function g such that

lfnl 5 g for all n, and Jg < W,

then f is integrable, and

Jf= limn Jfn.

THEOREM: (Term by Term Integration of Series of Functions): If

(un) is a sequence of integrable functions such that

I JUj

<

n

then the series

u(x) = I un(x)n

is absolutely convergent for almost all x, the function u (which

is defined almost everywhere) is integrable, and

Ju = fUnn

OF THE COURSE 9

If f is integrable and if E is a measurable set, we again set

JE = JLEf

(the integral of f over E). In fact this integral depends only

upon the values of f on E and can be defined whenever f itself is

only defined on the set E; it suffices, for example, that the

function obtained by extending f by zero outside E be integrable

over]R". In this case one says that f is INTEGRABLE OVER E. All

the properties of the Lebesgue integral over3Ru extend to this

case. Furthermore, if f is integrable over E then it is integrable

over every measurable set contained in E, and if E is the disjoint

union of a sequence (En) of measurable sets,

fEf

= L JE f.nn

Similarly, if E is the union of an increasing sequence (En) of

measurable sets,

f.J f = l imfEE n

n

This formula is still valid if E is the intersection of a de-

creasing sequence (En) of measurable sets and if f is integrable

over E1.

In the case of integration on ]R, it is convention to write

when I is an interval with endpoints a and b(-o 4 a 4 b 4+-),

and f is either a measurable positive arithmetic function on I,

or a complex integrable function on this interval. When a > b


we write

when f is integrable over [a,b], and a "Chasles' Formula" can then

be written. If -- < a < b < +-, then f is Lebesgue-integrable

over [a,b] whenever it is Riemann-integrable over this interval,

and the two integrals are equal. However, when the interval is

infinite the Lebesgue integral only constitutes an extension of

the notion of absolutely convergent (generalized) Riemann integral.

The GENERALIZED (or SEMI-CONVERGENT) LEBESGUE INTEGRALS can be

defined in the following way. For example, let us assume that f

is (Lebesgue) integrable over every interval [0,M], 0 : M <

one then sets

Mf

J_f

M}oj0

when this limit exists. In this respect let us note the follow-

ing Proposition:

PROPOSITION: (Second Mean Value Formula): If f is a decreasing

positive function on [a,b], and g an integrable function on this

interval, then

IJbfgl : f(a) sup IJxgl.

a a,<xsb a

To close this Section let us indicate that if f is a mapping

from3RP into3Rq of which the q coordinates are integrable, the

integral of f is the element 0f] of which each component is

equal to the integral of the corresponding component of f. Every-

thing that has been said above remains valid when the absolute

OF THE COURSE 11

value is replaced by a norm on U .

0.4 FUBINI'S THEOREM

THEOREM: (Fubini) : Let X = Iltp, y = ]R then the formula

jjXxf (x,y)dxdy = fXdxjf(x,y)dy =fy

dyfXf(x,y)dx

is valid in each of the following two cases:

(1): f is a measurable positive arithmetic function on X x Y;

(2): f is an integrable function over X x Y.

Amongst other things the validity of the formula means (accord-

ing to the case) that for almost all x e X the function y Fa f(x,y)

is measurable (resp., integrable) on Y, and that the function

x ' J f(x,y)dy, defined almost everywhere on X, is measurable

X(resp., integrable) on X.

In particular, in order to have the rule of "interchangeability

of the order of integration" it suffices to be assured, when f is

a measurable complex function on Xx Y, that

(x,y)ldy <JJ1If

Whenever f is positive and measurable this interchange is al-

ways legitimate.

Certain proofs of Fubini's Theorem use the following Lemma,

which is interesting in its own right:

LEMMA: In order that a set have measure zero it is necessary and

sufficient that there exist an increasing sequence (qn) of posi-

tive step functions such that

sup 9 < lima (x) if x e E.n n n n


In this statement one can replace "step functions" by "compact-

ly supported continuous functions".

0.5 CHANGE OF VARIABLES

Let V be an open set of ]R and u a diffeomorphism of V onto

u(V), that is to say a bijection of V onto u(V) such that u and

u-1 are continuously differentiable. The JACOBIAN of u is denoted

J(u).

PROPOSITION: The formula

x))IJ(x)ldxf(u(J

f(x)dx = Ju(V) V

is valid in each of the following two cases:

(1): f is a measurable positive arithmetic function on u(V);

(2): f is an integrable function over V.

Amongst other things the validity of the formula means that

fou is measurable (resp. (fou)IJI is integrable) on V.

Spherical Coordinates in ]R

e with the half-hyperplane defined by x1 < 0, x2 = 0 removed,

possesses a proper parametric representation:

x1 = rsinen_2 sin82sino1coscp,

x2 = rsinOn_2 sinO2sin81sin9,

= rsin sin8 cosx n_2 2 1,3

x4 = rsin4n_2 coso2,

xn = rcos8n_2,

OF THE COURSE 13

where r > 0, 0 < of < it, ITI < R. The formula for the change of

variables in this case comes down to replacing the xi by their

expression as a function of r, the oils, and cp, and dx = dx1dx2...

dxn by

rn-lsinn-ton-2... sin2o2sinoldrdSn-2...de2doldq.

Let us also recall that the volume of the ball x2 + +X2

<, 1n

is

n/2v_ n

n r 2 + 11

0.6 THE LP SPACES

In what follows we make the convention of setting 1/co = 0,

a.- = 0 or - according as a = 0 or a > 0, and coo = m if a > 0.

HOLDER'S INEQUALITY: Let 1 < p,q < - be such that

p1 +q =1,

and let f and g be measurable positive arithmetic functions on

31n; then:

Jfg `[JfPJ

l/p[Jgq/

1/q

MINKOWSKI'S INEQUALITY: Let 1 S p < c, and let f and g be measur-

able positive arithmetic functions on ]R", then

If (f + g)PJ 1/p <

[JfPJ1/p +

[JgPJ'/P.

When p = q = 2, Holder's Inequality is known as the (CAUCHY-


SCHWARZ INEQUALITY. For 0 < p < 1 one still has an inequality

which is obtained by replacing 4 with : in Minkowski's Inequality

(it is sometimes called MINKOWSKI'S SECOND INEQUALITY).

In order to have equality in Hblder's Inequality it is neces-

sary and sufficient that there exist a 3 0 such that fP = agq al-

most everywhere. For Minkowski's Inequality when 1 < p < -, equal-

ity is only obtained if f = ag almost everywhere.

NOTATION: For every measurable function f on IItn one sets

where 0<p<-.llfll = (Jlfll1/p

Also, there exists M e- such that lfl 4 M almost everywhere,

and such that this does not hold for any M' < M. The element M

is called the ESSENTIAL SUPREMUM Of lfl and one defines

lifilm = ess sup If I .

DEFINITION: With the preceding notations, for 0 < p << - one de-

fines the following functional spaces:

Lp(]Rn) { f: f is measurable on IItn; ll f llp < -).

These spaces are vector subspaces of the space of functions

on] .When 1 : p the mapping f i+ llflip is a semi-norm on LP (where

we write LP for LP(]R )); the kernel of this semi-norm coincides

with the vector subspace N of functions zero almost everywhere, so

that LP/N is canonically provided with a norm. In this work we

shall make not distinction between Lp and LP/N, which amounts to

identifying two functions that are equal almost everywhere. Let

us point out that certain authors denote by XP that which we de-

note by LP, and they keep the latter notation for .P/N. Having

taken into account the convention we have just indicated, in the

OF THE COURSE 15

remainder we shall consider f ' If 11 as a norm on LP (1,<p<,-).

THEOREM: Let 1 5 p 5 00 and let (un) be a sequence of elements of

LP such that

G iiunilp<n

Then for almost all x the series

u(x) = I u(x)n

is absolutely convergent; the function u thus defined almost

everywhere belongs to LP and one has

in the sense of convergence in the norm of L.

COROLLARY: For 1 f in LP there exists a sub-sequence fn

.

such that fn

-r f

almost everywhere.

It will be noted that if fn -> f in L , then fn -> f almost every-

where. This property may fail if fn - f in LP, 1 p < 00.

When f e LP, g e Lq, 1 : p,q <

p

+

Q

= 1, the function fg is

integrable and

Jfgl 6 IIfIIpIIgIIq (Holder's Inequality).

The spaces LP (1 S p 5 00) are BANACH SPACES; L2 is even a Hil-

bert space if the scalar product is defined by

2(fig)= Jf0 f,g e L.


For every measurable set E of Rn one similarly defines the

spaces Lp(E) by everywhere replacing the integrals taken over Rn

by integrals over E. The space Lp(E) is identified with the

closed vector subspace of Lp(IZn) formed of the functions that

vanish outside E. When meas(E) < W one has the inclusions

Lq(E) C Lp(E) if 0 < p < q < -.

DENSITY THEOREM NO. 1: Let E be a measurable set of Rn; then the

characteristic functions of the measurable sets A C E are total

forLp(E) for 1SpSW.

DENSITY THEOREM NO. 2: Let V be an open set of Rn, then the char-

acteristic functions of the rectangles P such that P C V are total

in LP(V) for 1 4 p < The set of continuous functions compactly

supported in V are dense in Lp(V), 1 5 p <

Let us recall that a set A of a metric space E is said to be

DENSE in E if A = E; if E is a normed space, A is said to be TOTAL

in E if the linear combinations of elements of A are dense in E.

0.7 CONVOLUTION

DEFINITION: Two measurable functions f,g on Rn are said to be

CONVOLVAELE if the function y H f(x - y)g(y) is integrable for

almost all x; in this latter case one can define almost every-

Where the CONVOLUTION PRODUCT OF TWO FUNCTIONS f and g by

(f*g)(x) = Jf(x - y)g(y)dy.

We have:

f*g = g*f.

If f is convolvable with g a n d h, then it is with Ag + uh (A,1,

e ¢), and

OF THE COURSE

f*(Ag + uh) = A(f*g) '+ u(f*h).

17

If IfI 1, fl, IgI 1, gl, f and g are convolvable whenever fl and

5l are; from this it follows that if f = u + iv, g = r7+ is (u,v,

r,s real), f and g are convolvable if and only if each of the

functions u+,u_,v+,v_ is convolvable with each of the functions

r+,r_,s+,s_.

Whenever f and g are convolvable, and are zero respectively

outside measurable sets A and B, f*g is zero outside A + B. In

particular, if f and g are compactly supported, f*g is also com-

pactly supported.

In addition to the spaces Lp (i.e., Lp(]R )) introduced in the

preceding Section it is useful to define the following functional

spaces:

DEFINITION: If 1 < p < m we define LPoc to be the vector space of

measurable functions f such that for every compact set K of Rn one

has

HAIp,k =[JKIf1i'/p

<_-

The functions belonging to LPoc are said to be LOCALLY p-INTEGRABLE.

If p Lloc is the space of LOCALLY BOUNDED MEASURABLE FUNCTIONS,

that is to say that for every compact set K one has

Ilfi, K = ess sup If (X) I <

xeK

One can be satisfied with looking at compact sets of the type

KN = {x:Ixl <, N}, where N 3 1 is an integer and x -> IxI a norm on

RR, We then set

IIfIIp,K =n

IIfIIp,N,

and the formula


W IIf - gIIP,Nd(f,g) =

12-

N=11 +

If - gIIp,N

defines a metric on LPoc (with the usual condition of identifying

two functions; which are equal almost everywhere). This metric is

compatible with the vector space structure on LPoc, fi ; f if and

only if If - fiIIP,K 0 for every compact set K, and LPoc is com-

plete in this metric.

DEFINITION: The space of p-INTEGRABLE COMPACTLY SUPPORTED MEASUR-

ABLE FUNCTIONS is written LP for 1 0 such that for all i one has fi(x)

= 0 whenever IxI > A. This latter condition is expressed by say-

ing that the fi have their SUPPORT contained in a fixed compact

set.

DEFINITION: The SPACE OF k-FOLD CONTINUOUSLY DIFFERENTIABLE FUNC-

TIONS on ]Rn

is written Ek (0 s k s co). (For k = 0, E0 is defined

to be the space of continuous functions on 3zn; in this case we

may also write C instead of E0).

For every n-tuple s = (s1,...,sn) of integers greater than or

equal to zero one sets

Isl = s1 + .. + sn,

and we define

Ds =D I S I,ff

axs1 i...axnn '

feEk, IsI << k.

If 0 S k < -, for every integer N 3 1 and f e Ek we set:

pN(f) = sup{IDsf(x)I:Isj : k,lxl 4 N),

OF THE COURSE

and ifk= -,

pN(f) = sup{IDsf(x)I:Isj < N,jxj 4< N}.

With these notations the formula

19

W -N pN(f - g)

d(f,g) =N11

2- 1 +pN

f -g

defines a metric on Ek compatible with the vector space structure,

and for which Ek is complete. Furthermore, fi - f in Ek if for

every s such that Isl < k one has limDsf. = Dsf uniformly on every

compact set of ]R' .

DEFINITION: The vector space C is the vector SPACE OF BOUNDED

CONTINUOUS FUNCTIONS on ]R provided with the NORM

IIfL = suplf(x)I

It is a Banach space.

DEFINITION: The closed vector subspace of CW formed of UNIFORMLY

CONTINUOUS BOUNDED FUNCTIONS on ]R' is denoted UC`°.

DEFINITION: The space of CONTINUOUS FUNCTIONS of ]R" WHICH TEND TO

ZERO AT INFINITY is denoted CO. It is a closed vector subspace

ofUC .

DEFINITION: The vector subspace of Ek consisting of COMPACTLY

SUPPORTED k-FOLD CONTINUOUSLY DIFFERENTIABLE FUNCTIONS is written

Dk (0 5 k co). If k = 0 the,space D0, that is to say the space

of compactly supported continuous functions, is also written K.

We shall say that fi -> f in Dk if for all s (Isl s k) one has

limDsfi = Dsf uniformly on ]n and if the fi have their supporti

contained in a fixed compact set.


If E and F are two of the spaces that have just been defined

we shall write E - F if E C F and if fi - f in E implies that

fi -> f in F. It is clear that E - F and F -> G implies E -* G.We have the following diagram, where 0 < k < - and 1 Dk--* K ->LWc c c

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,'. -fI

in E, one says that the TRANSLATIONS OPERATE CONTINUOUSLY ON E.

OF THE COURSE 21

THEOREM: (1): The spaces LP,LlOC,LC (1 5 p Cm,UC,

D

CO,Ek and

k (0 5 k are invariant under translation;

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

except upon LW,L ,L and Cam.loc 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 AÊ 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; fur-

thermore,

IIf*gL s IIfIIpIIgIIq;

(iii): L1s:Lp C LP (continuously) if 1 , p , m; furthermore,

11f* g11 , IIfII1IIgIIp.

In particular, L1 is a convolution algebra

(iv): L1operates continuously in Lp (1 4 p ,

oo),Cm,UC

and CO;

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

(vi): L1 it a convolution algebra;


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

kD;

(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 :

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

4

= 1, we set

(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= [ J2nIfIP)1/P 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*g)(x) = 2nJ f(x - y)g(y)dy.0

Defining fi -> f in Ek(IF) to mean that f(S) f(S) uniformlyfor every integer s with 0 4 s < k, the following hold:

LP(a)&rLq(a) C C(a) (continuously), if

p

+

4

= 1,

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

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

as do the inequalities:

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

= 1,

IIf*IIp -1 IIfIIlIIgIIp, if feL'(l), geLP(a).

By defining J`,J as above, and


21 2n

(f,9) = 2nJ f9, (.fig) = f1J f90 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 condi-

tions:

(i): There exists a constant M such that 11,Pi11l < M for all i;

(ii): limi

J'Pi = 1;

(iii): For every a > 0, limJ m = 0,

i JxJ>,a i

An approximate identity (ml) is said to be compact if all the

functions cp.ivanish outside the same compact set of]Rn.

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 ex-

tends to the spaces LPoc,Lp (1 < p < o), Ek,Vk (0 < k

From this one deduces the following corollaries:

THEOREM: (Density); D is dense in each of the spaces Lp;LP LPoc,c

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

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 p.(x)dx = 0.i a<IxI<<1[

If E denotes one of the spaces LP('T) (1 f in E for every function f e E. In

L'(T.') there exist approximate identities consisting of functions

belonging to Em(u); from this it follows that Em(r) is dense in

each of the spaces LP(,r), 1 < p < m and Ek(a).

0.9 FOURIER TRANSFORMATIONS

DEFINITION: For every f e L1 (= L1(R)) we give the name FOURIER

TRANSFORM of f to the function

tme-2%ixy f(x)dx,y e3R.

E f(y)=- J-W

The symbol f is also employed to designate Ff. The function

F f(y) _ (tme21Eixyf(x)dx,m

y e 3R,

is called the ADJOINT FOURIER TRANSFORM of f.

Upon setting


e (x) =e2itiax

a a eat, x eat.

one obtains the following Formulas, where f and g belong to L1:

f'':ea = ?(a)ea$ (0 9.1)

F(f) = Tf-, F(f) = F(f), (0 9.2)

F(.f) = (Ff)-, F(f) = (Ff)-, (0.9.3)

F(f*g) _ Ff.Fg, F(f*g) _ f. Fg., (0.9.4)

F(fa)e-aF(f),

F(fa) = eaF(f), (0.9.5)

F(eaf) = (Ff)a, F(eaf) _ (ff)_a, (0.9.6)

(Fflg) = (flpg), (0.9.7)

f::g = F(Ff.g). (0.9.8)

Furthermore, if f is piecewise continuously differentiable,

and if f and f' belong to L1, then

f'(y) = 2uiyf(y).

Whenever f and x } g(x) = xf(x) belong to L1, f is continuously

differentiable and

g(y) = 2nif' (y).

THEOREM: (Riemann-Lebesgue): F is a linear mapping of L1 into C0

such that I j Ff l l 11f111-

THEOREM: (Fourier's Inversion): If f and Ff belong to L1, then f =

F(Ff) almost everywhere. In particular, if Ff = 0, then f = 0

almost everywhere.

OF THE COURSE 27

THEOREM: (Fourier-Plancherel) : If f e L1 f1 L2, then 11 Ff 11 2 = 11f112-

It follows from the Fourier and Fourier-Plancherel Theorems

that the restriction of F and F to L1f1L2 extend in an unique

way into two isometries of L2 onto itself, which are inverses of

each other; these will still be denoted F and F. If f e L2 and

f = Ff, then

M

f (Y) e-2nixyf(x)dx,JMM 2nixy-f(x) = lim e f(y)db,

M}-m-M

in the sense of convergence in L2. Formulas (2),(3),(5),(6),(7)

are still valid for f,g e L2; Formula (4) generalises to the case

where f e L1 and g e L2; and Formula (8) can be replaced by

F(fg) = F f*Fg

whenever f,g eL2.

Lastly, let us note that if f e L1 and Ff e L2,

then f e L2

0.10 INTEGRATION AND DIFFERENTIATION

NOTATION: We provide]R with the norm lxl -- Max(lx1l,...,lxnl).

By the letter C we denote a CUBE containing the origin, that is

to say, a set such that 0 e C and C = {x:!x - xol 4 a}, where

x0 ein and a > 0. The number 2a is called the diameter of the

cube C and it will be denoted by 6(C).

THEOREM: (Lebesgue): Let f e L1 There exists a negZigeable

nloc

set N of 3R such that if x $ N then for every complex number a one

has

Zi?n6( )#O

mess C Clf(x + t) - aldt = If(x) - a1 J l


In particular, for almost all x,

f(x) = limmess C J

f(x + t)dt.d(C)+0 C

DEFINITION: A function f is a FUNCTION Of BOUNDED VARIATION ON AN

INTERVAL [a,b] if

n-1

V(f;a,b) = sup E If(xi+1) - f(xi)I <A i=0

the supremum being taken over all decompositions

A _ (a = x0 < x1< ... < xn = b)

of the interval [a,b].

If f is of bounded variation on [a,b] and on [b,c], then it is

on [a,c], and

V(f;a,c) = V(f;a,b) + V(f;b,c).

THEOREM: (Jordan): A real function is of bounded variation on

[a,b] if and only if i-t is the difference of two increasing func-

tions on [a,b].

A complex function is of bounded variation if its real and

imaginary parts are. Every function f with bounded variation is

regular, and upon setting V(x) = V(f;a,x), a << x b, one has

V(x + 0) - V(x) = If(x + 0) - f(x)I,

V(x) - V(x - 0) = If(x) - f(x - 0)I)

at every point where these symbols have a meaning. Therefore, in

Jordan's Theorem, if f is real, continuous, and of bounded vari-

OF THE COURSE 29

ation on [a,b], then on this interval it is the difference of two

continuous increasing functions. Jordan's Theorem extends to

real functions defined on an open interval I and with bounded

variation on every compact interval contained in I.

DEFINITION: A function f is said to be a FUNCTION WITH BOUNDED

VARIATION on]R if V(f;O,x) and V(f;-x,O) have finite limits when

x the sum of these limits is called the VARIATION OF THE

FUNCTION f on u .

THEOREM: (Lebesgue): Every function with bounded variation on

[a,b] is differentiable almost everywhere, its derivative is in-

tegrable, and

b

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

DEFINITION: A function f is called ABSOLUTELY CONTINUOUS on [a,b]

if, for all c > 0, there exists 6 > 0 such that for every finite

sequence of mutually disjoint sub-intervals ]ai,si[ of [a,b] one has

If(si) - f(ai)I < e whenever (R. - ai) < 6.i i

THEOREM: (Lebesgue): If f is absolutely continuous on [a,b] f is

differentiable almost everywhere, f' is integrable and

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

a

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

a

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

xf(x) = JF(t)dt, a < x < b,a


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) f(bn)

0.11 TRIGONOMETRIC SERIES

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

positive or negative integers, one sets

+W W Nun = u0 + I (un + u_n) = lim E un,- n=1 N- n=-N

when this last limit exists. WhenI

luni < - one also has00

M

u =lim E u.n N- n=-N n

M-

A trigonometric series is a formal series of the type

OF THE COURSE

tm a minx

= 2cne + (ancosnx + bnsinnx),

n=1

where, upon agreeing to set b0 = 0,

an = cn

+ c-n,

b = i(c - c ),n n -n

C = Y(a - ib ),n n n

c-n = '(an + ibn),

n > 0.

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

?(n) = I(2n e-inxf(x)dx.

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

f(x) ti L oneinx-W

31

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:

12nan

n

r

f(x)cosnxdx,0

2n1

bn =n

f(x)sinnxdx.

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

32

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

fsaen = f(n)en,

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

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

CHAPTER 0: OUTLINE

P (n) = in(n) if f is absolutely continuous.

For the explicit calculation of the Fourier series of a func-

tion f the following result, which generalises (5) in the Formu-

las above, is useful:

PROPOSITION: If f is a pieeewise continuously differentiable

function, that is to say, if there exist points.

-u 4 al < a2 < ... < ap < 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 de-

noting by [f'] a function with period 27[ that coincides with f's

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

ina

in?(n) = [f'](n) + 2n {f(as + o) - f(as - o)}e S.

s=1

EXAMPLE 1. If f(x) = x when -n < x < it,

f(x) ,, 2(-1)n+1 sinnx

n=1n

OF THE COURSE 33

EXAMPLE 2. If f(x) = -1 when -n < x < 0 and f(x) = 1 when 0 < x

< it,

f(x) 4 4 sin(2n + 1)xnn= 2n+1

EXAMPLE 3. If f(x) = it - Ixi when -n x < it,

f(x)ti'-`+4 cos(2n + 1)x2 it

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 coin-

cides almost everywhere with a p-fold continuously differentiable

function.

NOTATION: We set:

NSN(f;x) = I f(n)einx (FOURIER SUMS),

n=-N

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

N 3 N + 1

then:

(FEJER SUMS),

SN(f;x) = (f*DN)(x), aN(f;x) = (f'N)(x),

with


D (x) = sin(N + Z)xN sin Zx (DIRICHLET'S KERNEL),

FN(x)+ + 1)jx (FEJER'S KERNEL).N

1(sisNinrx

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))

IO

d 1

(X(t)sin(Nt+ zt

dt + r, (X,6).,

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 ser-

ies 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 con-

verges 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:

1r2n1 f(x)g(x)dx = f(n)g(-n)

2n0 n=-W

= E

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

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 f in LP(a).

THEOREM: (Fejer-Lebesgue):

1imoN(f;x) = f(x)

at every point x such that

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

(0.11.7)

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

36 CHAPTER 0: OUTLINE OF THE COURSE

E If(n)I2 <

and in this case

27E if(x)I2dx

= L I?(n)I2.2n0

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

EXERCISE 1.1: Let E1,...,En be a finite sequence of sets of fin-

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

meas(E.P Z1< < p 11 1P

(a): Show that

nmeas(E1 U U En) = 1

(-1)p-1aP

(POINCARE'S FORMULA).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 ( l

meas(G) = I (-1)P-5 l s la .

p=s l 11 p

(c): Let HS be the set of points which belong to at least

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

37

38

nmeas(HS) = I (_1)p-s(s-11

Qp.P=S l

A0t = VA V = AVA = VAV = AV

CHAPTER 1:

SOLUTION: For every non-empty subset A of {l,...,n} let us set

EA = n Ei, EA = EA - U Ei.ieA i+A

The EA are mutually disjoint and it is clear that

E = U EB',A BDA

(-1)po = I(_1)CardAmeas(EA).

P CardA=p

SOLUTION: (a): One has:

n(-1)pa = (-1)CardA meas(EA) _ (-1)CardA L meas(EB)

p=1 p A A BD A

I meas(EB) (_1)CardA.

B ACB

Note that if p = CardB, then

x(_1)CardA = (_1)r(P 1 = - 1.

ACB r=1 l J

From this it follows that

n(-1)Pa = -

p=1 pmeas(EB) meas(E1 U UEn.

B

MEASURABLE SETS 39

SOLUTION: (b): One has:

n( l

(-1)p 8Ja = (-1)CardA(CarsdA lmeas(EA)

p=s lp

CardA3s IJ

(-1)CardA(CardA 1 I meas(E')Card4 s l s J BDA B

meas(EB) E(-1)CardA(CardA)

CardB;s A C B ll JCardA=s

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

C ( -1)CardA [ CardA) = ( -1)r (r 1 (p )ls)lrACB l s J r=S

CardA3s

=l s ) I ( -1)r (r - s )

P-8 -(-1)ss)

(-1)rlprs1

0 ifp> s,(-1)S if p = a.

From this it results that:

n(-l)PI p)Q = (-1)S meas(E') = (-1)smeas(G ).

p=ss

, CardB=ss s

SOLUTION: (c): The formula is true for s = 1 (for this is none

other than that obtained in (a)). Also,

HSt1=HS - GS,

40 CHAPTER 1:

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

nmeas(Hs+l) = I

(-i)P-s-1 j(

l s Jl

[P-111ap- 1p=s

n(-1)P-s-1(p $ 1 l op.

p=s+1 l J

EXERCISE 1.2: For all c > 0 construct an open set U everywhere

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 further-

more,

meas(U) S 1e2-n

= 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

A n [UEJp

mead U E ] -< I meas(E ),(pan p pan p

and consequently

meas(A) = lim mead U Epn J

(pan

lim I meas(E ) = 0.

n pan P

SECOND SOLUTION: Let cpn be the characteristic function of En.

By virtue of the theorem on the term by term integration of

series of positive functions, one has

J

i cn = I Jcpn = I meas(En) < -.

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

41

For every integer s, Hs

is denoted as the set of points which

belong to at least s of the sets En. Show that

meas(Hs) 4

s

G meas(En).n

42 CHAPTER 1:

FIRST SOLUTION: First, suppose we had a finite sequence E1,...,

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). Then

smeas(Hs) = E smeas(EA)CardA>,s

nCardAmeas(E') _ meas(E.).

CardA>,1A

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,.. -,n-

It is clear that Hsn C Hs n+l

and that

Hs = U Hs,n.n=s

Consequently

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

n<<

s

Jim meas(E.) = s meas(E.).n i=1 i=1

SECOND SOLUTION: Let q) be the characteristic function of En and

m = I pn. One has

smeas(H).meas(En) = Jp >JH

S

EXERCISE 1.5: For every finite sequence E1,...,En of measurable

sets one sets

MEASURABLE SETS

En

l

D(E1,...,En) =n

il - I F]E1l

J

,

and if the E. are not all negligeable,

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

meas

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

a(E1,...,E1) En

11 1

i<j

MMA = VM0 = A VA = VAV = MMA

43

SOLUTION: Let us show, by induction on n, that if x e D(E1,...,En)

then x e D(E1,E)'for at least n - 1 pairs i < j. This is clear

if n = 2. Let us assume that it be true up to n - 1 and that

x e E2 U UEn, x e El. One then has x e D(Ei,Ei) for at leastn - 2 pairs such that

(l,i) with i Z 2.

2 < i < j 4 n and for at least one pair

From this it follows, by Exercise 4, that

meas(D(E1,...,En) n 1 L meas(D(Ei,Ej)).i<<j

Since

meas(E1U UE) 3 meas(EiUEj ),

from this one deduces that

1C meas(D(E.,E

a(E1,"'' n n - 1 i<J meas E1U UEn

4 n 11 E

i<j

44 CHAPTER 1:

EXERCISE 1.6: Consider a sequence (En) of measurable sets such

that

mead( U E n) <n

and

inf meas(E ) = a > 0.n n

Show that the set A of points that belong to an infinity of

sets E is measurable and that meas(A) > a.n

ovo = vov = ovo = vov = AV

SOLUTION: We have

A=ni(UEsln=1 s=n

which shows that A is measurable. Now we also have

m

mead U ES] . meas(En) > a.s=n

As the sets UU ES form a decreasing sequence and have finitesin

measure, we have

meas(A) = lim measi U ES1 > a.n s=n

REMARK: The first condition is essential; for example, consider

the sequence En = [n,n + 1[.

EXERCISE 1.7: Let A be the set of real numbers x for which there

MEASURABLE SETS 45

exists an infinite number of pairs (p,q) of integers such that

q;i, 1and

x Pq

Show that A is negligeable.

4V4 = VAV = AVA = VAV =-000

SOLUTION: Since

x + n -ng =x -q ,

by setting

B = An [0,1]

from the former relation one deduces that

A = U (B + n).n=-w

Hence it suffices to prove that B is negligeable.

For p and q integers, and q la let us set

I = LP- - 1 P- + 1Jp,q q q3 ' q q3 '

and let us note that x e Ip,q

is equivalent to

(1) qx-2<p.qx+ 2q q

Since in an interval of length 2/q2 there can be only at most one

or three integers according as q >. 2 or q = 1, from this it

46 CHAPTER 1:

results that x e B if and only if x belongs to an infinite number

of sets

Bq = [0,1[n (U Ip,q).

p

By Exercise 1.3 it therefore suffices to show that

(2) G meas(B ) <q q

Now by virtue of Equation (1) above, the integers for which

Ip,gfl[0,l[ + O are such that

1 1-2. p. q + 2q q

When q > 2 this is equivalent to

0<p q.

Consequently

meas(B ) < 2(q + 1)q 3

q

which proves Equation (2) above.

EXERCISE 1.8: Let (Dn) be a sequence of discs in the plane, of

unit diameter and mutually disjoint. The number of discs that

are contained within the disc with center 0 and radius n will be

denoted v(n).

Show that if

liminf v(2) = a > 0n n

MEASURABLE SETS 47

there exists a half-line issuing from 0 that meets an infinite

number of these discs Dn

AVA = V AV = AVL = V AV - AVA

SOLUTION: Let us consider the cones Cn with vertex 0 generated

by the Dn, and let An be the intersection of Cn with the circle

with center 0 and unit radius. For some integer p and q, 1p P,q

Everything reduces to proving that the BP, which form a de-

creasing sequence, have a non-empty intersection. The union of

the radii of the disc with center 0 and radius q that meet AP,q

covers (v(q) - v(p)) discs of unit diameter and mutually disjoint.

From this it follows that if meas(AP,q

) denotes the Lebesgue mea-

sure (evaluate in radians) of Ap ,q

the unit circle, one has,q

iq2meas(AP,q)

it

(v(q) - v(p)).

Since Ap,q

C A p,gt1 , one therefore has

meas(B ) = lim meas(A )

P q p,q

2 liminfv(q)

2 v(p) 2q q


measl n BP) = lim meas(BP) 2l p P

48 CHAPTER 1:

which proves that the intersection of the Bp is non-empty.

EXERCISE 1.9: Let A be the set of real numbers of the form

n + E at-p,p=1

p

where n ea and ap = 0 or 1.

(a): Show that A is negligeable.

(b): From this deduce that there exist two negligeable sets,

the sum of which is fit.

evo = VtV = AVA = VAV = eve

SOLUTION: (a): Let B = Afl[0,1[. Then

A = fg, (B + n).n=-w

Hence it suffices to show that B is negligeable. Let u be the

function, with unit period, equal to zero on [0,2'[ and to unity

on [12,1[. For every integer p >. 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 ap(x) = cp,

has measure 2-P. 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,

has as its measure

2-2p+1 x 2p-1 = 2-p.

. , a2p-1 (x) = 0,

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) a = 0 or 1.p=0 p

p

As every real number may be written

n+ ap2p, ap=0or1,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

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

+co

F = (E t n),n=-co

as well as

50 CHAPTER 1:

- 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 and let

Sip = meas{x: I Ix II 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 <1 meas(A)

n n ap Qp

where

-P(

1 pcp = 2 { 1 + p10 ( 1 + t dt}l

For this one will consider a number y such that ndP > y for nn

sufficiently large, as well as the balls B. with center xi and

radius ri, where

r. _

Y1/Pn - 1/P

i11/P(2i-1/P - n-1/p) if n y1/Pn - 1/P

whenever n is large enough, and

ri t rj =yl/pn-1/P.

When 1 4 i < j E 2pn and j n, one has

IIxi - xi II > dj > yl/Pj - 1/p,

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

Pnurn - 1 + 2C (2i-1/P - n 1/P)Pl < meas(A(e )),2 LL n

i==nJ n

where

e = 2yl/Pn - 1/pn

We have

2Pn Pn -1/plnm

1

(2i-'/P -n-1/p)P

= lim n2C [2rnl - 1]p =

i=n i==n Il J

(Contd)

52

(Contd)

CHAPTER 1:

r2P

JI

(2,-l /P - 1)Pdx11

= p1(1 - t)P

dt,1 t t

(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 intersec-

tion; furthermore, as A is bounded they are of finite measure.


meas(A) 1St cp P

and consequently that

liminfndp 41 meas(A)

n n cP

SZ

P

EXERCISE 1.12: Let X be a measurable set of ]RS such that meas(X)

= 1. 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} otherwise

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

Show that

MEASURABLE SETS

1 f n(x)dx=1.E

A00 = 0A0 = A0A = VAV = A00

SOLUTION: For 1 .< n 6 W let us set

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

Let us also set

GO = E,

nGn = E - U

u_P(E),

n . 1,p=1

where u -p = (U-1 )P). It is clear that

En=Gn-1Gn, 14n<W,

G.E = o G

For n > 1 one has

n-1un(Gn) = un(E) - U up(E).

P=O

From this it results that the sets un(Gn), n 3 0, are mutually

disjoint, and that

y = l_J un(G ) = lJ un(E).n=0 n=0

53

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 pre-

serves the measure,

Go m

1 = E measun(G ) = E meas(G.).n=O n n=0 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 = nmeas(E ) = lim X n{meas(G

n-1meas(G

nE n=1

n N n=1

(N-1 N l

= liml I (n + 1)meas(G ) - I nmeas(G )JN- In=O n n=O n

rN-1 l

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

It remains to be observed that

L meas(Gn) = 1 and meas(Gn) 3 meas(Gn+1)

implies

Nmeas(GN) -} 0.

Thus

JI n(x)dx = 1.E

REMARK: We have used the following classical result:

MEASURABLE SETS 55

If un> 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. QEDn

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 almostopen 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, there-

fore 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 coin-

ciding 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 func-

tion, 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. Let us

show that it is sufficient. To do that let us set:

MEASURABLE SETS 57

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

This definition has.a meaning, for A is everywhere dense. On A

one has f = g. 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 < a one has l y - x l < a for all y e ll such that

l y - x' I < a' = a - Ix - x' I .


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 > 0 such thatx

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 < ry ,

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

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

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 =t0

<

t1 < < to = 1 such that

f(ts) =

n

meas(A).

60

Under these conditions one has:

CHAPTER 1:

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

meas(A), 0 .< s F n - 1.

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

sum. Therefore they cannot all be strictly less than or all

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

f(y) = f(x) +

n

meas(A)

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

f(ts) = f(ts_1) +

n

meas(A) s f(x) +n

meas(A)

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

E f(ts+1).


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

n

meas(A)}

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 com-

pact sets E1,E2. Having seen that EX is compact and connected,

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 ab-

surd.

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 =

n

meas(B) > Ameas(B).

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 non-

zero 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].

(a): Show that

In+1 C In,meas(In) 1)(1 - 2)...(1 - n

and that every interval contained in In has a length less than2-n

(b): From this deduce that

K R Inn

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

that

meas(K) = lim(l - )(1 - a S)1 nn

(c): 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 63

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

quence An of sets of the first category that are mutually dis-

joint 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 that

A U Ann=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

EnU-0 A2nt1'

Show that for every interval I contained in [0,1] and of non-

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

(h): 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 func-

tions.

64 CHAPTER 1:

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 suf-

ficient 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 = 0.n

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 re-alised; K is therefore not denumerable.

SOLUTION: (d): If Bn = 1 -a1/n(n+l),

0 < a < 1, one has

a

meas(K) = lima n,n

with

(1 _ 1

n n + 1

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 A1

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 Kn

a nowhere dense compact set of

66 CHAPTER 1:

measure equal to half the length of this interval, and by taking

the union of these compact sets. As above, it is seen that An+1

is of first category and that Kn+l= Kn U An+1 is

closed. Further-

more,

nmeas(Kn) _

2-i= 1 - 2-n,

i=1

whence:

meas(An+l) z[1 - (1 - 2-n)] = 2-(n+1)

Finally, Condition (iii) is satisfied by construction. If

A U Ann=1

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

sets of first category, and -

W

meas(A) = 1 2-n = 1.n=1

Its complement is not of first category, by a theor m of Baire,

and its measure is zero.

SOLUTION: (f): Let

E A2n+1' F

l

i A2n'n

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. Let us assume, for example, that

meas(If1E) > 0. 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).

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

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 nN 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

meas(E) _ 2-(2n+1) + 22-(2n+1) = 10

n=0 N=1 n=N 9

68 CHAPTER 1:

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 every-

where 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

(fmconverges almost everywhere towards a function gm and

that the sequence (gm) converges almost everywhere towards a func-

tion h.

Show that there exist two sequences of strictly increasing in-

tegers (ms) and (ns) such that the sequence (fmn

convergess' S

almost everywhere to h. Generalise this result to the case where

X =P. (orEP).

wA=vw=w0=0AV =MA

SOLUTION: First assume that every convergence is uniform on a

set E C X. Then there exists ml < m2 < and n1 < n2 < and

such that on E there holds

h - gm

<1s s

gmS- fmsons

Thus

MEASURABLE SETS 69

Ih- fm ,n I <

son E,

s s

which proves that fm n - h on E.' s

We are going to show that for all e > 0 there exists such a

set E, which, moreover, is measurable and satisfies meas(X - E)

< S. In fact, by Egoroff's Theorem there exist measurable sets

EO,E12... such that

meas(X - E ) < e2-(m+1) > 0,m

and, on the other hand, one has

limgm = h uniformly on E.,

Iimfm,n= gm

uniformly on En if m > 1.

On setting

E = n mm=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 in-1

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

< I and

70 CHAPTER 1:

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 - Er) <

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. Furthermore, for

every r 3 1 the sequence (fm )s-r is a subsequence of the se--s,n

s

quence From this it follows

that

fm in-> h on E.

s s

It will be noted that in order to use Egoroff's Theorem it suf-

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

sets of]R is called a MEASURABLE MAPPING if for every compact

set K of ]R 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 .

(c): Show that if t - 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.

(d): Show that if t - K(t) is a measurable mapping of ]R into

the compact sets of ]R , and if f is a continuous mapping of IlRP in-

to 3R q, then t - f(K(t)) is measurable.

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

t - n Fn(t)n

is also measurable.

(f): Show that t - F(t) and t - K(t) are measurable mappings

of Et into the sets of ]R which are respectively closed and compact,

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

1V = VV =Ova=V V =AVA

SOLUTION: (a): Let Bn be the closed ball with center 0 and radius

n. 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:

72 CHAPTER 1:

{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 dis-

tance 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 xn

such that xn

-> X. It is clear that x e K (as. .

d(x,K) = linrl(x K) = 0), and that x e F(t) also, for F(t) isZ 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 con-

sidering 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 - F(t) is measurable.

SOLUTION: (c): Note that we have not specified the norm on ]R P.

If we choose the norm II(xl,...,xP

)II = Maxlxil, every open ball

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

]R 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 73

SOLUTION: (d): 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

by A. For every compact set K of Rp,

{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. This shows

that t -} F1(t)n F2(t) is measurable. By recurrence one obtains

the result for a finite intersection. 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

r=1

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

empty.

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)

nn (Fr (t)nK)

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.

74 CHAPTER 1:

EXERCISE 1.20: Let Fp be the set of closed sets of R , KP the

set of non-empty compact sets of i2P, x ->l i x 1 1

a norm on ]R , and

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

(a): For F E Fp write:

{x:x eF, jjxjj = d(F)} if F # 0,cp0(F) _

{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 fam-

ily of closed sets see the preceding Exercise).

(b) : Let ei(x) =X. if x = (x1,. .. ,xp) e:IR . 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 measur-P

able.

(c): From this deduce that there exists a mapping 9:FP

->iRp

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.

(d): Let f:jRp +R be continuous. Show that there exists

a 2Rq x F -> PP such that :P

(i) : If F e FP

and x e f(F), then a(x,F) e F and f(a(x,F)) = x;

(ii): If t H x(t)e]i , t i+ F(t)e Fp are measurable, thent * 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 Ba

denotes the closed ball with centre 0 and radius a.

SOLUTION: (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 - ee.(K(t)) is measurable (by not-

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

SOLUTION: (c): For every F e FP

the set (cpP

1op0)(F) is re-

duced to a point p(F). It is clear that F -> q(F) has all the re-

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

measurable. Now, for every compact set K of Rp

{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

sets of R. The Axiom of Choice appears in the following form:

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

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 equival-

ence 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. One has Iri = Ix - yJ

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

(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-

tion is adopted that E mi

= 0). Let (En) be asequence of mutu-

ieoally disjoint sets of C, and let E be their union. For every in-

teger N one evidently has:

N11 (E1U...UEN) = 1 ir(En) (E),

n=1

79

80 CHAPTER 2: a-ALGEBRAS

whence

E u(En) u(E).n=1

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

E m = G I m. 4 1 u(EXieE 1 n=1 x.eE 1 n=1

ni n

ieA ieA

whence

M

u(E) = sup X m. < G U(E ) .< u(E).A x

ieE 1 n=1

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

= B. On the other hand, if {xi} e C for all i and if E is an ar-

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

(b): Show that the mapping f of a into itself defined by

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). This proves that f is meas-

urable. 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-


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. Let C0be the set of

subsets E C X such that MflEe C(AM). 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 subspace M of X.

EXERCISE 2.25: Let C be a a-algebra on X and let p be a prob-

ability on C. Let M C X be such that E e C and E D M implies

p(E) = 1.

Show that a probability 11M

is defined on CM by setting

AND POSITIVE MEASURES

PM(MnE) = u(E)

83

for all E e C (the induced a-algebra CM was defined in the preced-

ing Exercise).

SOLUTION: 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

(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). The mappinguM

of CM into [0,1] is

thus defined unambiguously. It is clear that:

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)


(Contd) = p(E) + p(F)

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

HenceuM

is additive. It remains to prove that it is continuous.

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 UEn, so that

M(lEn = MnFn, Fn Fn+1'

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

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. It remains to prove that

these sets form a a-algebra. 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 = 1 .n

d0A 0L0 AVA V AV LVL

SOLUTION: 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} {piNJr=1 i=r

where (pi) denotes the sequence of prime numbers. From this it

would result that

0 = limp{ U pi3Nr i=r 1

By virtue of an elementary property of arithmetic one has:

(*)

pi w n ... npi 3N = pi ...pi N < ia)1 a 1 a


Therefore one would have

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

pi ..pi

1 a

If 1 r 4 s Poincare's Formula (cf., Exercise 1.1) would give

s lu(U pi1VJ

i=r )) r4i1<...<1 8 pilp1a

= 1 -sir (i - 1-

Now, it is known (see the end of this solution) that

slim 1 1 = 0,s i=1 pi

and therefore that for r >, 1

s

lim IT 1 - l = 0.

s- i=r pi

Therefore for all r > 1 one would have

ti-rll (( ll

i1VJ = o [ piJ = 1,s->- i=r

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. Then:

siT 1 =IT{E k,

i=1 n=0 pi keAsi=1 1 -1

Pi

1

AND POSITIVE MEASURES

whence:

s1im 1 1 = lim I _ I k= .s i=1 1 - s- keA k=1

pi S

87

EXERCISE 2.28: Show that there does not exist a a-algebra having

a countably infinite number of elements.

AVA = VOV = AV1 = VAV = AVA

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

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

contains x. Since for all E e C the point x c (Ex - E) or x e ExflE

one has either Ex - E = E{ or ExflE = Ex, that is to say ExflE = O

or Ex C E. In particular, for two arbitrary points x,y e X one

has ExflEy = 0 or Ex = Ey.

Let I be a countable set such that

{Ex}xeX - {EiIieI'

with E. # E7. if i + j. For every subset A C I,

EA = U B. e C,ieA

and EA $ EB if A # B. It is clear that A ' EA is a bijection of

P(I) onto C. Then if I is finite C is finite, and if I is infin-

ite C is not countable.

CHAPTER 3

The Fundamental Theorems

EXERCISE 3.29: Calculate

(1xnlogxdx0

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

1

1

1x dx,

0

given that

1

112

n=1 n26

Av4 s vov ° AVA - vov - VA

SOLUTION: By setting x = e-t one has:

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

0 0(Contd)

89

90 CHAPTER 2: THE

W

(Contd) _ - 1 2 J te-tdt = - 1 2(n + 1) 0 (n + 1)

Moreover, if 0 < x < 1,

OD

logx_

I - xnlogx.1 - x n=0

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

by term, which gives

J

l logx dx = C - 1 = - n2

O 1 xn==O (n +

1)2 6

EXERCISE 3.30: (a): Let a > 0. 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.

(b): Show that for Re(s) > 1,

s-s x-1n = r s

Jx dx.

n=1 0 e - 1

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+ 1).

0 0

FUNDAMENTAL THEOREMS 91

SOLUTION: (b): For x > 0

xs-1-x

xs-1s-1 -nx= e = L x e

ex - 1 1 -e-x

n=1

Since

CI xRe(s)-le-nxdx

n=1 J0 n=1 0

F(Re(s)) En-Re(s) < w00

n=1

(*)

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

s-1r- x ds = C r xs-le-nxdx = r(s) I n-s.0 es - 1 n=1 0 n=1

EXERCISE 3.31: 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 V AV AVA 000 X04

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

otn-e-tdt = r(n + ) _ (Contd)J'o

t2ne-t2dt=

Jo0

92

(Contd) = z(n - '-z)... ''r(Z)

CHAPTER 3: THE

1.3. (2n - 1) - (2n)! /Tr2n+1 22n+1n

since r(z) = u. Therefore

2 2n 2

e t coszt = (-1)n (2n !

t2ne-t ,

n=0

and

n-0 0G J

2n 2

(-1)n(2n)!

t2ne-tCw

I 12n

dt = n G 2n+1n=O 2 n!

Thus,

T 2 2n °° 2

e-t cosztdt n (2n !t2ne-tdt

n=O 0

2 n(z l

'

= 2 E (-1)nn4

n0

-z2/42

e

EXERCISE 3.32: Establish the relation:

snaxdx

a

J 2 2Oex-1 n=1n+a

AVA = OAV = AVA = V1V = t1V4

FUNDAMENTAL THEOREMS

SOLUTION: If x > 0,

Co

s in=ex - 1 n=1

93

(*)

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

n 1

J le nxsinaxldx < aJ xe-rixdx = a

20 0 n

so that:

J0ie-nx sinaxldx <

n=1

Equation (*) can therefore be integrated term by term, which gives

the stated result, since

Joe-nxsinaxdx =a

20 0 n2 + a

EXERCISE 3.33: Find a method of calculating the series

c (-1) n+1

n=1n

by the integration of a series of integrable functions.

A VA - VAV = tVO = VAV = tVo

= E e I's inax .

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:

94 CHAPTER 3: THE

( -1)n+1 W 1 1 °C° 1

n l n n 0 [T-+1 2n+ 2, =n 0 2n+1 2n+2

Furthermore, for 0 4 x < 1

1 1 + x x2n+121og1-x

and consequently

1J1l0 1 +x dx = c 12 0 gl-x 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:

(-1)n+1

= log2.

n=1n

REMARK: 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

fs i n a x

dx =

2

(1 - e-a).0x(x + 1)

(Consider the function of a defined by the left hand side).

AVA = VAV = 40A = VAV = OVA


SOLUTION: Let cp(a) be this left hand side. Since

cosax I 1

2x +1 x +1

the function (p is continuously differentiable on Ft and

(p, (a) _ W cosax dx.fo

x2

+ 1

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

xsinax 2

JAx2+1 Aa0

95

(by the second mean-value formula, since x(x2 +1)-1

is decreas-

ing for x z 1). This proves the uniform convergence of the im-

proper integral

1

xsinaxdx

0x2+1

on [a0,=[. Consequently 9 is twice continuously differentiable

for a > 0, and

fW xsinaxdx

Ox2+1rW sinax dx +

f0mxsinaxdx

0x (x2+1)

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

9(0) = 0, 9'(0) = 2

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

differential equation on [0,oo[). From this it follows that

cp(a)=2 (1-e-a).

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 J:ffl(X)d.X.n1

A0A = V AV = AVA - VAV = A00

SOLUTION: For x > 0 one has:

Ln=1 n 1 - e -x 1- e-2x

1 _ 2

ex-1 e2x-1

1

ex + 1

e e-x -2x

f (x)= -2

On the other hand,


10fn(x)dx =

n

- 2 2n = 0,

so that:

0.

n=1 JJ 0

However,

Jmdx ex

x

x= [log

xJ

= log2.Oe +1 e +1x=0

Hence one must have:

00,

n=1 0

which is immediately verified by observing that

1Jm

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

97

EXERCISE 3.36: (a): Let (fn)nal be the sequence of functions de-

fined on ]R by

2n x

fn(x)=

if 04x4n

- n21x -n

if n s< x s n,

0 ifx>' 2n '

Calculate the four numbers:

A = lim infJJJfnJJJ

, A' = Ilim inff ,n n n

98

B = lnm supJfn, B' = Juim supfn.

CHAPTER 1: THE

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

gn = 810,4] if n is even, gn = IL111]

otherwise.

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

what can be said of the four numbers A,A',B,B', and more partic-

ularly about B and B'? Is the Fatou Lemma true for sequences of

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. In particular, A' < A and B' < B.

SOLUTION: (b): In this case Jgn is alternately equal to I and 4,

so that A = a, B = 4. Furthermore,

lim infgn = 0 and lim supfn= Il[0,1]'

whence A' = 0, B' = 1. Thus A' < A < B < B'.

SOLUTION: (c): There always holds:

A' < A < B and A' < B'.

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


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:

9- 4f n for all n,

jg_ < .

One then has A' s A. In fact 0 .< fn - g, and

- °° < Jg < Jfn'

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 - Jgn

.< lim infn

Jfn - 1g,

whence the result. (Recall that if Jg- < - one sets

Jg = - Jg_

which is an element of and that

g < h,fg-

< - imply Jh_ < - and Jg < Jh).

Similarly one shows that if:

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 sup( - fn) lnm inffn.

Here one again sets

Jg = Jg+ - Jg_ if Jg+ < W,

and

g < h, jh+

< M imply Jg+ < - and Jg 'C Jh).

EXERCISE 3.37: (a): Show that if f is integrable on , and if

K is a compact set of this space, then

lim lf(x)ldx = 0.z I I - J K+z

(b): 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


SOLUTION: (a):

liml If(x)Idx = 0,r-*

since

fr 0 and Ifrl IfI

(Lebesgue's Theorem). If S = sup{IIyII:y a K} ,

J If(x)Idx < J If(x)Idx,x+z IIxIIIIxII-a

whence the result.

SOLUTION: (b) : Assume that + 0 as IIx II There there existe > 0 and a sequence xn such that

If(xn)I >. e

As the function f is uniformly continuous there exists a closed

ball B, with centre 0 and radius greater than zero, such that

If(y) - f(x)I <s

if y e B + x.2

In particular, IfI > e/2 on the balls B + xn, and consequently

J IfIP > (2)Pmeas(B) > 0,

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

102 CHAPTER 3: THE

j mG(f(x))dx < -,

then

1im f(x) = 0.

IIxlIH

A0, = V AV = tot = VMv = LVA

SOLUTION: If f(x)-J 0 as x there would exist c > 0 and a

sequence xn such that

IlxnII - -, If(xn)I z E.

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):2

juI < M) = u > 0,

so that

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

n

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

Exercise).


EXERCISE 3.39: (a): Let f be an integrable function on ]R .

Show that

meas(IfI > a) = o11) as a --* W.

103

(b): Show that if f is measurable (and almost everywhere

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

X2nmeas(2n-1

< IfI i 2n)

100 = VAV = MMA = VAV = MMA

SOLUTION: (a): This results from

limI f(x)Idx = 0,a-J lfl >a

and the inequality

ameas(IfI > a) E J

Ifl

If(x)Idx.

>a

SOLUTION: (b): Let us set

An =(2n-1

< Ifl 6 2n), B = (f = 0), C=(IfI

These sets are mutually disjoint, with union ]R , and C has meas-

ure zero. Then

J IfI = T J" IfR n=-W An

whence

104 CHAPTER 3: THE

(+m

2 2nmeas(An) 1 1A s I 2nmeas(An),n=-m IR n=-m.

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 fin-

ite measure such that

f(x) _ 2')LA (x)n=- n

for all x e]R .

AVA = DAD = AV4 = 4AV = DAD

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

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

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

t 2nan(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,

f(x) _n=-m

If one sets A

2nan(f(x)).

= {x:x ep and a (f(x)) = 1), then An is meas-

andurable,


f(x) = .1 2)LA (x).n= n

Also,

2nmeas(An) < fRpf(x)dx,

so A has finite measure.n

105

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

Jf(x)dx = 0

X

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

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

X X

Let usnow show that it is sufficient. The inequality can be

written in the form:

J

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

106 CHAPTER 3: THE

where p > 0, 8 eat. An easy calculation shows that:

11 + pzI - 1 _ 2Re(z) + pIzI2p 1+pz + 1

if p > 0, z e0, so that

lim 11 + peI f(x)I - 1 = Re(eiof(x)).

p-)-0 p

Moreover,

11+ p egs f(x) I- 1 I0 X p

>. 0,

and this holds for all e em. By choosing 0 so that

el$Jxf(x)dx = - Jf(x)dxlX

one obtains

Jf f(x)dx = 0.X

EXERCISE 3.42: Let f be an integrable function on at and let a'>0.

Show that for almost all x eat the series


+W

f[.+nIn=- (*)

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 =

a

+ n on obtains

I J:If+ n] I

dx= a

C'fxI = aJI .f (x) I dxl n=-w

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 proceed-

ing as above:

ifa

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-*-

108 CHAPTER 3: THE

DOA - VAV = X00 = VAV = 400

SOLUTION:

r( t(x)Idx,

so

I

in-CLf(nx) I dx <n=1 m

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:

TA =

JI f(x)l dx < m

0

(a): Show that for almost all x,

limn-2f(nx) = 0.n

(b): From this deduce that for almost all x,

limlcosnxll/n = 1.

n


MVA = VAV = AVA = VAV = MVA

SOLUTION: (a): Set

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

Then:

Jk(x)dxT 1 T 1 nT

= 2J If(nx)Idx = 3f If(x)Idx0 n 0 n 0

A= 2n

Hence

TIpn(x)Idx <

n 0

109

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:

f(x) _(log1cosxl)2.

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. By the preceeding,

lim(n 1loglcosnxl)2 = 0,n

i.e.,

110

limlcosnxll/n = 1n

for almost all x.

CHAPTER 3: THE

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

(a): Show that the following conditions are equivalent:

(1): The sequence (xn)na1 is equidistributed;

(ii): For every continuous function f on [0,1]

J

1 -1 Nf(x)dx = 1imN Y f(xn);

0 N-?- n=1

(iii): For every integer p a 1,

-1 N 2nipxnlimN I e = 0.N- n=1

APPLICATIONS: Investigate whether the sequences

xn = na - [na] (a irrational)

and

xn

= logn - [logn]

(*)

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


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

Riemann integrable, then (*) still holds.


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

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

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:

f5 cp,J

1

J TSe+J0

f0

From this one deduces that:

-1 N -1 N 1

l im supN E f (x ) .< l imN I 9_(X )1 S E + f,N n=1

nN n=1

J

n 0

and consequently

N 11 im supN t f (X ) < f.n

0N- n=1J

Replacing f by -f, this yields:

N 1

lim infN 1 f(x ) > J f,N-- 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

112 CHAPTER 3: THE

(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 E (the Stone-Weierstrass Theorem). To simplify the

ensuing calculation, set

CN((1

uN(f) = N- 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. Now let 0 < a 0

there exist two continuous functions gE and fE such that

0< fE. <g

E< 1,

fe(x) = 1 if a + E .< x .b+ E.

Then

( 1 N 1 Nb - a - 2e < Jf =

limN-Xf (x ) < him

infN-X (xnE N- n=1 e nN-)oo n=1

-1 N -1 Nb - a + 2E Jg = limN X g (x ) > lim supN I ',(x ).

N-°n=1 E n

N-n=1 n

t see Erratum, p. 545.


Letting a -> 0 yields:

N

limN 1 I V+(xn) = limN 1v(N;a,b) = b - a.N-- ' n=1 N-

APPLICATIONS: If xn = an - [an], and if p 1 is an integer,

N 1 N e2nipxn = N1 N e2nipan = Nle2nipa 1 e21Eipa

n=1 n=1 1 -e2nipa

since because a is irrational, e2nipa $ 1. Hence one clearly has

N 2nipx1imN

1 e n = 0N n=1

in this case. When xn = logn - [loge] it is necessary to study

the sums:

N 1 N e21iplogn = N1 N n2nip.

n==1n=1

Now

2nip N N nt2nipd(t

- n + 1),1 N = N1f t2nipdt

= N-1CJn_i

0 n=1

N 2nip N jnNN1 C n2nip =

1+

2n p+ 2nipN 1 I (t - n + 1)t2nip-ldt.

n=1 n=1 n-1

When n a 2

In

(t - n + 1)t2nip-ldtlJn

t-idt= log[;, n 1, .

fn-1 n-1

114 CHAPTER 3: THE

From this one deduces that:

N 1 c n2nip =

n=1

N2nip

+

rlogN)

1+2npol{N

which shows that in this case the sequence xn is not equidistrib-

uted.

REMARKS: Formula (*) is not satisfied by all the Lebesgue-inte-

grable functions. In fact, if f is the characteristic function

of the set {xn} one has

1

J f(x)dx = 0,0

but

NN1 E f(xn) = 1 for all n.

n=1

Let us again note that the characteristic function of a set E

is Riemann integrable if and only if the set of points of discon-

tinuity of this function has measure zero, or, in other words, if

the boundary of E has measure zero. From this one deduces that

if (xn)n>1 is equidistributed and if the boundary of E has meas-

ure 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)


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

J1 f(x,y)dxdy = lim jf(t,at)dt.

0'<x,y<1 T- 0

AVA = 000 = AVA = VAV = 000

SOLUTION: For p,q a 0 integers, the Formula is verified without

difficulty for the function:

(x) = e2ni(px+gy)P,q

by observing that p + aq # 0 if p2 + q2 4 0 (for a is irrational).

By linearity the Formula holds for the linear combinations of

functionsP,4'

The Stone-Weierstrass Theorem assures, for every

continuous function f with period unity at x and at y, and for

all e > 0, the existence of such a linear combination 9 satisfy-

ing If - 91 a e. Denoting by uT(f) the difference between the

two sides of the Formula to be proved, one has:

IuT(f)I < IuT((P)I + IuT(f - p)I < IUT(a)I + 2e.

As we have just seen that uT(9)-; 0, the Formula is proved.

EXERCISE 3.47: Let g be a measurable function on 3R, bounded, and

of period T > 0.

Show that for every integrable function f, and for every sequence

(an) of real numbers,

(+W{Jgx)dx]

T r (+011 f(x)g(nx + an)dx = Jf(x)dxJ

0(* )

(AJER'S FORMULA)AVA = VAV - AVA - VAV - AVA

116 CHAPTER 3: THE

SOLUTION: Assume first that f is the characteristic function of

the interval [a,B]. Then

nB+aJf(x)g(na

B 1Jflg(x)dx.+ an)dx = 1 g(nx + an)dx =

a na+aa

In view of the periodicity of g,

JnB+ang(x)dx .

rn(B

Ta)1(

gT

(x)dx + en

na+anlL

J J 0

with lenI < MT, where M = supIgI. Consequently,

((+m T

limf(x)g(nx + an)dx = B

Ta1 g(x)dx,

0

n

which proves (*) in this case. By linearity the formula holds

for every step function. In the general case, for all e > 0

there exists a step function h such that

f - hI < E.E I

Setting

TI g(x)dx,0

one has (taking. account of lyl < M)

+m((+00 Jh(x)[g(nx+a)Y]dxI+2McII f(x)[g(nx + an) - y]dxl < ,

and consequently,


tW

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

which proves (*).

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

Riemann-Lebesgue Lemma:

+m

limaf(x)e1nxdx = 0n-. _,

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

td)r

(Con

IpnIJElcos(nx + an)Idx Ipncos(nx + an)I I dx c

118

(Contd) Nmeas(E) <

CHAPTER 3: THE

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

rn

ncosx dx =

2

0

so that Fejer's Formula (cf. the preceding exercise) applied to

the characteristic function of E gives

meas(E) > 0.limJ Icos(nx + an)Idx =itn E


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.

(a): Prove that for every measurable set E of [0,2n] and

for every integer p 1

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

S Sl

(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:


cos2pz =2-2p(eiz + e-'z )2p

2-2p(p ) + 21-2p krl i 2pk Icos2kz.

Furthermore,

IJEcos2k(nsx + 8s)dxl < IJEcos2knsxdxI + IJEsin2knsxdxl,

and the two latter integrals tend towards zero by virtue of the

Riemann-Lebesgue Theorem.

(b): Let 0 < a < 1 and let Ea be the set of x's (0 4 x < 2n)

such that

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

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

a2pmeas(Ea) a I limsup cos2p(nsx + 6 )dxE Sa

> limsupfE cos2p(nsx + 0S)dxS-'°

a

Note that

22p 2p lI

= 2-2p(2p)1 ti 1 2-2p

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

( P J (p!)2r2 -,E p2p+1e-2p

ti

120 CHAPTER 3: THE

Since a2pPVT-> 0 when p it follows from the preceding that

meas(E(1 = 0. The set of x's from [0,2i] such that

limsupjcos(n x + )I 4 134w s S

being the union of the E1- 1/n

(n 3 1), the proposition is proved.

EXERCISE 3.50: Let f be a positive integrable function on an open

set X of ]R such that meas(X) < w.

Show that there exists a function g, lower semi-continuous Qn X,

such that g >. 1/f and JXfg < W.

401 = 000 = 404 = V1V = 001

SOLUTION: Let us set:

Anfn + 1 < f 4 n) (n = 0,1,2,... ),

and

o).

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

Vn D A_,

Wi J An,

The function

Jf(x)dx < 2-n

Vn

n nJ

f(x)dx < (n t1)-12-n

W -A .

0g = E ILV + (n + 1)ILW

n=0 n n=0 n


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

fJX

°J

f + (n+ 1)fW f+ (n1)JAn=0 V n=0 -A n=0

n n n n

4+ f +C n +

nl meas(A) <

JA0

n==1

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

meas(X) < -.

REMARK: If meas(X) = m, the result may not hold. Indeed, if X

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

f = 1IL

n=1 n Xn

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

J1fg ==n=1X n=1n

EXERCISE 3.51: 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 sequenceS

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

almost everywhere. By Egoroff's Theorem there exists a sequence

of measurable sets A1C A2 C such that meas(X - As) < 1/s and

fn - 0 uniformly on each of the AS

. Hence there exists a sequence

of integers n1 < n2 < such that for all n 3 n8 one has Ifnl <2-s

on A . Let A be the union of the A so that meas(X - A) = 0.S s

Set to = 0 if n 4 ns for all s, and let to = 1. ThenS

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

and for all x e A one will have Ifn (x) I <2-s

whenever s 3 s0,s

where s0 is chosen so that x e AS . This proves (ii) and (iii)

hold.

(ii) => (i): If limsuptn > 0 there exists a > 0 and a se-

quence n1 < n2 < such 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. Furthermore,

I Itnl JA IfnI = JA 9 <n

S S

Since X Itnl = -, one thus has:

liminfJA Ifn1 = 0.n-o

S


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

that:

IfJARs I <2-s

S

For every integer a one has:

S1 fAI fn

sa

because As D Aa whenever s 3 a. From this it follows that at al-

most all. points of Aa the series E fn converges, and fn - 0 ins s

particular. But then fn --> 0 at almost all points of A, hence

s

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

EXERCISE 3.52: Let f be an integrable function on X =]R .

(a): 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:

lim Ifl = 0E

(which means that for all E > 0 there exists S > 0 such that if

E is measurable and meas(E) < d then

jEIfl < E.

AVO = VAV = AVA = VAV = AV!

124 CHAPTER 3: THE

SOLUTION: We may assume that f 0 (since only 1A appears in

the statement of the exercise).

SOLUTION: (a): Let:

A0= (f=0), An= II< f<nJ (n31), AW= (f

Since Al C A2 C . and as

M

F1 (X - An) = A0UA.,n=1

one has, because meas(A.) = 0 since f is integrable,

ni lX-Anf = 1

A 0f+ JAmf=0.

Hence there exists an integer n0 such that, letting A = A n ,

0

Furthermore, f is bounded on A by n0. Finally A has finite meas-

ure, for

1meas(A) ' f <

n0 JA

SOLUTION: (b): Let A be as in part (a) with e/2 instead of c.

Set M = sAupf. Then if E is measurable and meas(E) < c/2M, one

has

JE fl_/ +JEfl/

+Mmeas(E)


EXERCISE 3.53: Let (fn) be a sequence of integrable functions on

X = R such that fn - f almost everywhere.

(a): Show that if f is integrable and if

JXn - Ix f,

then for all e > 0 there exists a measurable set A with finite

measure, an integrable function g > 0, and an integer n0, such

that for all n >, n0

X-AfnI< e and I g on A.

(b): Now assume that for all e > 0 there exists a measur-

able set A, an integrable function g > 0, and an integer not such

that for all n >, n0 one has:

JX_A11l < e and IfnI < g on A.

Show that under these conditions f is integrable and that

(c): Show by examples that the conditions of part (a) are

not sufficient, and those of part (b) are not necessary in order

that f be integrable and

fX fn -r fX f .

AVA = VAV = X04 - 040 = 000

SOLUTION: (a): Let e > 0. By part (a) of the preceding exercise

126 CHAPTER 3: THE

there exists a measurable set B with finite measure such that IfI

is bounded on B by a constant M. and

JXBIfI < E.

By Egoroff's Theorem there exists A C B such that meas(B - A) <

e/M and fn -> f uniformly on A. Hence there exists no such that

for n 3 n0

Ifn - fI < meas(A)on A, and

IJX(fn

- f)I < E.

By writing

JX-Afn=J

B-Af+JX-Bf +JX(fn - fJAn -f),if n n0 one obtains:

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

mess Ameas(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.


Then by the inequality

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

one concludes

If - fnj c 2e,li pJX

which proves that

J/n J[

127

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< 1 and fn(x)=1 if n<x41

for n ? 2. Then fn -> 1 almost everywhere. If A = one has

IfnI < 1 on A and

fn = 0.X-A

Nevertheless,

J/n = 2

(ii): Let

fn(x) = x sin x if n< x c 1 and fn(x) = 0 if 0 4 x 4n

128 CHAPTER 3: THE

Since

lint 1 sin 1 dxa->O ax X

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

n

and fn(x) = 0 otherwise.

Then fn - 0 and

JX fn - 0.

If the conditions of part (b) were satisfied by the fn's they

would also be satisfied by the Ifnl, and one would have:

lin2n

IsixnxIdx =

n fXlfnllim= 0.

Now, by Exercise 3.48,

1im12n Isinnxl dx = limJ2 Isinnxl ax.=in n x n- 1 x

= 1 sinxdxl k = 21o g2

0 1X


EXERCISE 3.54: 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

a) =I

> 0.lim meas(If - fnn-).-

(a): Show that if fn - f in measure there exists a subse-

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

l imJ if - fn I= 0n-'° X

(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.

130 CHAPTER 3: THE

tVA = VAV = AVA = VAV = AVA

SOLUTION: (a): Choose a sequence of integers n1 < n2 < such

that

measlIf - fn I

>s)

< 2-s,ll S

Then the set of points that belong to an infinite number of sets

(If - fnI> 11s) has measure zero (cf., Exercise 1.3). Now,

s

this set contains the one formed of the points where (fn ) doess

not converge to f. For every integer n 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 im-

mediately furnishes the result.

SOLUTION: (c): Let e > 0. There exists an integrable set A C X

such that

g<E.X-A

Set An = (If - fn

I > E/meas(A)) and note that by part (a) one has

IfI < g almost everywhere; then, by decomposing the integral on X

into a sum of integrals on X - A, A - A n, and Af1An, one obtains:

2g.limsupJ If - f I < 2e + e + limsupJAn- X n n-

7Now, meas(An) - 0 by hypothesis, so (cf., Exercise 3.5M ,


1imJ 2g = 0,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

on E. Then, for all a > 0 one has (If - fnl > a) C X - E when-

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: Let a > 0. Prove that if f and the fi's are pos-

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-

where f V fi denotes the maximum of f and fi. Let e > 0 and let

M > 0. Then

meas(Ifa - fll > c) mead If - fil > E

1)l

meas(fV fi > M).

Note that if f v fi> M, then

132

f > -zM or If - fiI > M,

for otherwise one would have

fi = f + (fi - f) < zM + ,M = M.

Thus:

meas(lf°` -l > E) < mead if - fil > E-1"Ma

CHAPTER 3: THE

+ 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: Let X be a measurable set of Iltp such that

0 < meas(X) < m. Denote by M the vector space of measurable com-

plex functions on X. If f e M, set

p(f) = j +Ifl)-1

(a): Show that fn - 0 in measure on X if and only if p(fn)

+ 0.

(b): Show that (f,g) y p(f - g) is a metric on M if one

agrees to identify two functions that are equal almost everywhere


(c): Show that p is not a norm on M, but that the sum and

the product of two functions of M are continuous operations in

the metric defined by p.

(d): Show that M is complete in the metric defined by p.

AVA = 0A0 = A00 = VAT = MMA

SOLUTION: (a): This follows from the fact that if Ac = (IfI > c)

1 + emeas(Ac) < p(f) < meas(Ac) + 1 +

cmeas(X).

SOLUTION: (b): If f,g a M, from the inequality

1 + If g 1+ f+ 1+ gone 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

mlimsup meas(Ig I > a) < cn

(which is the case if gn + 0 in measure, or if gn = g for. all n).

Since, for all a > 0,

134 CHAPTER 3: THE

(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 thats

p(fn - fn ) < 4-s.

s+1 S

By the inequality from the answer to part (a), for all s one has:

meas(Ifn - fn I > 2-s) s (1 +2s)4-s.

s+1 S

By exercise 1.3, for almost all x one has

Ifn (x) - fn W1 < 2-s

s+1 s

for s sufficiently large. In other words, fn converges almosts

everywhere to a function f. Applying Lebesgue's Bounded Converg-

ence Theorem to the integral which gives p(f - fn ) shows thats

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.


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,fn

) -> f in measure.

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) = Jf + Jfn - Jmin(fif ) - Jf1

so that :

136 CHAPTER 3: THE

JI f - fnl = Jmax(fif ) - Jmin(fif ) -* 0.

EXERCISE 3.58: Let H be a family of positive integrable functions

on a measurable set X of 30.

(a): Show that the following conditions are equivalent:

(i): limJ f(x)dx = 0 uniformly for f e H;C-}°° (f>c )

(ii): lim j f(x)dx = 0 uniformly for f e H.meas(E)-+0 E

(A family satisfying these conditions is called UNIFORMLY INTE-

GRABLE).

(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 sat-

isfied).

(c): Show that if there exists a positive measurable func-


G defined on yt such that

limt-1G(t) = m and supJ

G(f(x))dx <t-- = f eA X

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

increasing. (We refer the reader to Exercise 6.113 for an appli-

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): For every set E of finite measure,

1 f 4 J f+ cmeas(E).E (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:

JEf < 1.

138 CHAPTER 3: THE

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

jEf < E

for all f e H. There exists e0 such that e a c0

implies that

meas(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


supJ

f = co

fEH 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,

feHJ E fnBut 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.(f>c) Mx

Now assume that H is uniformly integrable and look for G in the

form

rtG(t) =

Jg(u)du,

0

where g(0) = gn = constant on [n,n + 1[, with 0 = go 4 gl 6,

and gn

w. Then G will-be increasing (and hence measurable),

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 gn

suitably in order that

supJ {G(f) <feH X

Define

an(f) = meas(f > n);

(1)

by noting that G(f) = 0 if 0 .< f < 1 (since g0 = 0) one obtains:

G(f) _ J G(f) < G(n + 1)(an(f)- an+l(f))

X n=1 (n<f<n+1) n=1

(as G is increasing). Since G(1) = 0 and G(n + 1) - G(n) = g n,

one has:

N N

L

gnan(f) - G(N + 1)aN+1

Li

G(n + 1)(an(f)- an+1(f)) =

n in=1(f),

and consequently:

J

G(f)gnan(f)X n=1

(2)

Let (es) be an increasing sequence of integers tending to infin-

ity, and such that for all feH:

1f s 2-S

Then


fI fs=1 (f>cs) s-1 n=cs (n<-4n+1)

00 W

X X n(an(f) - an+l(f)).s=1 n=c

s

Noting that nan(f) -*0 (cf. Exercise 3.39), an Abel transformation on

the summation over n leads to the inequality, valid for all f e H:

W 00

I E a (f) 1,s=1 n=c n

s

which can also be written, letting gn denote the number of inte-

gers s such that cs s n,

W

G gan(f) r 1.n=1

(3)

This choice of the gn's in the definition of the function g

clearly makes Equation (1) hold, on account of Equations (2) and

(3).

REMARK: Condition (*) does not. imply the uniform integrability

of If. In fact, if:

fn = nIl [0,/n]'

the sequence (fn) is not uniformly integrable, since

Jf = 1[0,1/n] n

contradicts condition (ii). Nevertherless,

lim meas(f > c) = 0Cya, n

142 CHAPTER 3: THE

uniformly in n, because if e > 0 and c > 1/e one has meas(fn > c)

= 0 if n .<< c, and when n > c

meas(fn>c)=n<I<e.

EXERCISE 3.59: Let A be a measurable set of 3Rk such that

0 < meas(A) < m, and let fl,...,fn be real functions integrable

on A. Assume that

I fi=0, 1< i4 n.A

Show that for every number a, 0 4 a 4 1, there exists a meas-

urable set B such that B C A, meas(B) = ameas(A) and

JB1 =0, 1.<i4n.

(First of all consider the case a = 1).

A00 = VAV = AVA = 0A0 = AVA

SOLUTION: First Step: Assume a = 2': Let us denote the property

to be proved by Pn (n > 1). Note that for every measurable set

A there exists a measurable set B C A such that meas(B) =

Jmeas(A); call this property P0. Now show that Pn => Pn+l for

n >. 0. Noticing that if B has the property Pn (relative to A)

then the same holds for A - B (since a = z one can construct

step by step some measurable sets Apiq (p >. 0, 0 < q < 2P) such

that:

(i): A0,0 = A;

(ii) : Ap, 2q

f1Ap,2q+1

= 0, Ap, 2q

VAP,2q+1

= AP-l,q,

if p >. 1, 0 4 q < 2P-1;


(iii): meas(Ap,q

) = 2 pmeas(A);

(iv):

J

= 0, 1 0, 0 S q < 2p;

then meas(Ap,q

) _ £(Ip,q

)meas(A), where £(Ip,q

) denotes the length

of the interval Ip,q. Finally, for 0 < t 4 1 and p a 1 set:

It- [t ttii

2 2

Ap(t) = U Ap,q'

p, q CIt

A(t) = U A (t).P>1 p

We are going to show that the A(t) realise some sort of "homotopy"

between A10 and Al1,1

in the sense that they satisfy the follow-, ,

ing properties:

(1): A(0) = A10 and A(1) = A11;, ,

(2): meas(A(t)) = 2meas(A);

= 0, 1 4 i F n;(3):

JAWf1 .

(4): For every function f integrable on A the function

t

e+

A(t)f

is continuous.

144 CHAPTER 3: THE

This will imply the property P+1, for by virtue of the Intermed-

iate Value Theorem there exists a t (0 4 t S 1) such that

JA(t)fntl = 2( JA(O)f'+l + JA(l)fn+l)

= (/ fn+1 + f fn+1)A1,0 Al,l

JAfn+l = 0.

It remains to prove properties (l)-(4). First of all,

A1(0) = Al 0,,

and if p >. 1,

A (0) = U Ap+3-,q = U (A UA

p+l06q<2p

,q 05q<2p_1 p+1,2q p+1,2q+1

U Ap,q =Ap(0),

OGq<2p-1

which shows that A(0) = Al 0. Similarly one sees that A(1) =

Al 1. From the equation

meas(A (t)) _ meas(A ) = meas(A) .2(i ),

P I CI p,q I Cl p,qp,q it p,q it

one deduces that

lim meas(Ap(t)) = Z(It)meas(A) = 'meas(A).

P+_

Since AP(t) C Ap+1(t) one clearly has property (2). If 1 4 i n,

0,J

fi= I 1 fiAp(t) Ip,gClt Ap,q


whence

Jfi=limJ fi=0.

A(t) P Ap(t)

145

It remains to prove (4). Let e > 0; there exists 6 > 0 such that

E C A and meas(E) < 6 imply

(cf., Exercise 3.52). If 0 < t < T < 1,

JAp(T)f -1Ap

fl .jEE(t,T)

Ifl,(t)

where Ep(t,T) is the union of the A

p,qis for the q's such that

Ip , q

C It or Ip , q

C IT

, but Ip,q

(t It ('IT

. The sum of the lengths

of these intervals is less than T - t + 2 -P, so that:

meas(E(t,T)) 4 (T - t + 21-p)meas(A).

If p0 is the smallest integer such that21-p

meas(A) < 6/2, then

for T -- t< 6/2meas(A) and p > p0

IfA (T)f - fA (t)fl ¢

e,

p p

and on taking p -+ w,

IJA(T)f-1A(t)fl<e.

Second Step: a Is Arbitrary: Let us again use the notations

above, and also set:

146 CHAPTER 3: THE

B (a) = lJ A , B(a) = U B (a).

P Ip,q C [O,a[p,q pal P

As before, one sees easily that:

Jf1 . = 0 (1 .< i < n), lim meas(BPW()) = ameas(A),BP(a) p+

whence:

Jf. = 0 (1 4 i s n), meas(B(a)) = ameas(A).B(a) 1

REMARK: It will be noted that B(O) = 0, B(1) = A, B(a1) C B(a2)

if a1 3 a2, and that if f is integrable on A, the function

ayB((%)f

is continuous.

EXERCISE 3.60: Let X be a measurable set of ]R such that meas(X)

1 and let f be an integrable mapping of X into]R

Show that:

if f(x)dx:E C X and E measurable)E

is a convex set of]Etq (Lyapunov's Theorem).


SOLUTION: Let E1 and E2 be two measurable sets of X, let

Yi=1 fi (i=1,2),E.


and 0 < a < 1. Since

J(LEf - yi) = 0 (i = 1,2),

147

by Exercise 3.59, there exists a measurable set E of X such that

meas(E) = a,

yi) = o,

which implies

JE(lEf=ayi..i

Set F = (Ef1E1)U ((X - E)f1E2). Then

JFf = JEf1E1f + JE - JEf1E2f

=ay1+ (1-a)y2.


1, and for all x e X let A(x) be a set of 3Rq. Denote by

JX

A(x)dx

the subset of itq formed by the integrals

JX

f(x)dx,

where f is an integrable mapping of X into lR such that f(x) e

4(x) for all x e X.

148 CHAPTER 3: THE

Show that this subset is convex (Richter's Theorem).

DOA = VAV = OVA = 0AV = AVA

SOLUTION: Let fi (i = 1,2.) be integrable mappings at X into]Rq

with fi(x) a A(x) for all x e X. Let

yi= JXfi

and let 0 < a < 1. Since

JX(fi - yi) = 0,

there is a measurable subset E of X, with meas(E) = a, such that

JE' -Yi) =

0

(cf., exercise 3.59). Hence

JE1 = ayi

Set

-fl(x) x e E,

f(x)f2(x) x e X - E.

Then f(x) a A(x) for all x e X and

J

X = J f l + J Xf2 - J E 2 = ayl + (1 - a )y 2.

EXERCISE 3.62: Let X be a measurable subset of ]R such that

meas(X) = 1 and let A be a subset 0 f ]


Show that the set

AdxfX

is equal to the convex hull of A. (See the previous exercise for

the notation).

ADA = QAa = AQA = DAD = ADA

SOLUTION: We shall argue by induction on the dimension of the

affine subspace generated by A. If this dimension is zero, that

is to say, if A is a point, the proposition is trivial. Let r(A)

denote the convex hull of A. If al,...,an belong to A and al,...,

an are positive numbers adding up to one, there exists a partition

of X into measurable subsets Xi,...,Xn such that meas(Xi) = ai.

The function which takes the value ai on I. has alai + + an an

as its integral, which shows that

r(A) C 1 Adx.X

If t is an affine function on Rq such that t(a) > 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).

150 CHAPTER 3: THE

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 con-

tradiction.

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: Let f be a mapping of X = into Y = IItn

(a): 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:


1

mess Ef(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 unionY 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 )n1

of elements of I such that

W

V= U Vn=1 n

and consequently,

f 1(V) = U f 1(Van=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)

152 CHAPTER 3: THE

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 messTE-i7fE.fI

which may also be written as:

JEEI(f - yi) = 0.

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, F0 = O, F1=E,

t * meas(Eif1Ft) is continuous,

and

JF IlE1(f - yi) = 0,

which we may write as:

Jf = meas(EifFt)y,.

Ei f1 Ft

Set:

Gt = (E0 - Ft) U(E1f1Ft);

then


J/1-

f = meas(E0 - Ft)y0 + meas(E1 n Ft)y1.

t

153

Now note that Gt D EOflE1, so that meas(Gt) > 0 for all t; the

function

t .-

meas(E0 - Ft)

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 GtG

f = ay0 + (1 - a)y1.

When meas(E0f)E1) = 0 one can assume that E0f1E1 = 0. It is poss-

ible to determine two families (Fi,t)0<t4l such that:

Fit C Ei, meas(Fi t) = tmeas(Ei)

JF(f-yi)=0,i,t

and then if Gt =FO t U (E1 - Fl t)'

, ,

(i = 0,1),

1 tmeas(E0)y0 + (1 - t)meas(E1)yl

meas Gt JG f tmeas EO + (1 - t meas E1t

rso that the first member runs over the segment [y1,y0] when t

varies from 0 to 1. We have thus proved that If is convex.

Let us now show that Af C If. If y e Af and c > 0, the set

{x:IIf(x) - y11 < c} does not have measure zero; it therefore con-

tains a measurable set E such that 0 < meas(E) < m. But then

154 CHAPTER 3: THE

C.Ily meas E Jfll Ilmeas E JE(f Y) 11 4

Finally, let us show that if c r(Af), where r(Af) denotes the

convex hull of Af. Proceed as in exercise 3.62, showing first of

all that if £ is an affine function on Y such that 2(y) 3 0 for

all y eAf, then:

meas(E1 1

JE ) meas(E JE (f) > 0'

which proves that If C r Af . Next, if there existed a non-con-

stand affine function t such that l(y) >, 0 for y e Af, and

JEUP=o,

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 forRm

-the union of

an arbitrary family of open sets (Va) is the union of a countable

subfamily of them.

This is proved by considering initially the subset J of inte-

gers n for which Un is contained in at least one of the Va. Then,

with every n e J one associates an index an such that Un C Van

Then,

U V =U Va

a neJ an


Indeed, if x e Va there exists n such that x e Un C V (definition

of a basis of open sets). But then n e J and x e Va , which proves

the Theorem. n

EXERCISE 3.64: Let f be an integrable function on at.

Show that if

bf(x)dx = 0

for all real numbers a and b, then f = 0 almost everywhere.

AVA = VAV = AVA = VAV = OVA

{SOLUTION: Every open set V of ]R being the countable union of in-

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 and meas(E) = limn n

ffo that :

f=0.jE

=limfVn

n

Lbus the set If of all the numbers

1 Jfmeas E

E'

0 < meas(E) < 0,

Mtluces to {0}. Now if contains the smallest closed set Af such

156 CHAPTER 3: THE

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: Let p be a norm on 3kn.

(a): 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:

f1f(p(x))dx=

nVJtn-lf(t)dt

12 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

rbmeas(a 6 p . b) = V(bn - an) = nVJ to-ldt,

a

and consequently the formula

f'f(p(x))dx=

nVJtn-lf(t)dt(1)

3t 0


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, sUpJ0fi<M, and fi -* -DonE.i

Then fiop . f i+1op,

tn-1f(t)dt)supJf.op = sup(nVTOi 1 i 1

nV(2R)n-1 supJ f1(t)dt < m on {x:p(x)e E}, which proves that this set has meas-

ure zero. If E has measure zero without being bounded one has

(p e E) = U (p e E fl [0,Ri] ) if Rii

and therefore has measure zero. 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.), - fi(t)Idt 0. (2)J0tflhIf(t)

lien

lim J nIfiop - f.opl = nV lim tn-1If.(t)- f (t)Idt + 0,

i, j-t°° IR i,j mJ 0 1

158 CHAPTER 3: THE

and consequently fiop converges in L1ORn). Since fiop - fop al-

most everywhere, fiop -> fop in L1(]R ). 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' Passing 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 in-

tegrable.

REMARK: To obtain Equation (2) we have used the following propos-

ition: 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 sub-

space {h:h a 0 a.e.}. Now fg a L1 and is annihilated by

this subspace. 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


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> 0, Jh1 = 0 P a rectangle of ]R' .P

This is absurd, for the second condition implies that hV = 0 al-

most everywhere. Hence there exists a sequence (fi) of step func-

tions such that fiV- g in L2; but then the functions f2 are

also step functions, and

If- filg < II ( -

f f = u + iv, u = u+ - u-, V = v+ - V_.

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 theli

1'1P }pgcase p = 1) such that fig -> fpg in L , and consequently fi

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) . 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. Show that for all x e V - V

there exists y e]R 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 be-

longs to the boundary of V - V. Let (zi) and (z2) be two sequences

of points of V such that zi - zi -* z. As the open set V is bounded

one can assume that z1 -* z1, z2 ; z2, so that:

X = p(x)(z1 - z2 ), 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 (1 - p(x))ndx c f n(q)*)(x) = meas(V) 2,11 V-V t


and consequently, because meas(V) > 0,

(1 - p(x))ndx < meas(V).V-V

By the preceding exercise

(1

f (1 - p(x))ndx = nmeas(V - V)J to-1(1 - t)ndtV-V 0

_ nr(n)r(n + 1)meas(V - V)

r(2n + 1

(n!)2meas(V - V),(2n)!

rhence, at last,

meas(V - V) < []meas(v).

EXERCISE 3.67: Let p be an even, positive function on ]R, decreas-

ing on [0,00[, and such that:

+0 +W

f- p(t)dt = 1, f- t2p(t)dt = 02 <00

(a): Set:

(t)dt, x - 0.G(x) = 2f'XP

Show that G is convex, and calculate:

00

I G(x)dx.0

162

(b): Let OAB be a right tri-

angle, M a point on its hypotenuse,

and P and Q the orthogonal projec-

tions of this point onto the other

two sides of the triangle. Find

the maximum area of the rectangle

OPMQ as M varies along AB.

Q

0 P

CHAPTER 3: THE

(c): From the preceding considerations deduce that for all

A > 0

p(t)dt < 12tI,acr 2a

(d): Prove the following result (F. Gauss (1821): an im-

provement, under the given condition, of the Bienyame-Chebycheff

Inequality):

1

2a2

p(t)dt

1 2 3 3if 0< a<

AVA = DADA= AVA = VAV= AVA

SOLUTION: (a): Because p is decreasing and positive on (0,-),

the function

X Er J p(t)dtx

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,


jr J2

G(x)dx = 2J dxl p(t)dt = 2 p(t)J dxV"X0 o 0 0

2J0t2p(t)dt = a2.

0

SOLUTION: (b): If OA = a, OB = b, OP = x, OQ = y, then

and consequently:

163

xy = b[1 - (Q - K)

2]

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)). By the preceding,

A2a2G(a2a2) < A < 2a2,

whence:

,,,.2 2, 1

2A2 '

which is the desired inequality.

(d) : Since the function A 'r G(A2a2) is convex and G(O) = 1, forO<a 0

l lG(A2a2) < 1 -

J

+ 1 - 1 -121

X0 0 2A0))) 0

164 CHAPTER 3: THE

The best approximation will be obtained by choosing A0 so that

the line

is tangent at the point (A0,1/2a0) to the curve

u= 1

2A2

For that to happen it is necessary that:

- 2(1- 12J =- 1A

0 2a X00

A0 - 2

EXERCISE 3.68: Let X be a measurable set of Mm such that meas(X)

= 1, and f a real measurable function on X such that:

v= J f2<M, f=0.X JX

1iminfe-2(1 - I !1 + Efia) = 21a(1 - a)V.

C-0 X

AVA = VAV = AoA = VAV = AVA

SOLUTION: Now,

e-2(1 -Jx

1 + fa=J

1 + af -2

1 +

X


Note that

ml1 + au - 11 + u'l alll J = Ya(1 - a),

U- *O u2

and also that

Jim (1+au- 11+u1al -0lul- U2

J

so that there exists a constant B such that

I1 + au - 11 + ulaI < Bu2 for all u.

Then

(1 + aef - l l + of lalliml 2 JI = la(1 - a)f2,E-}0 E

11 + aef - l1 + of f a l< Bf2.

2

165

One can therefore apply Lebesque's Theorem, which gives the result.

EXERCISE 3.69: Let X be a measurable set of stn, g a positive

measurable function on X such that

f g(x)dx = 1,X

and let rp be a measurable convex real function on a convex set I

of Y = e.

(a): Let f be a measurable mapping of X into I and such

that fg is integrable.

Show that:

166 CHAPTER 3: THE

@(J f(x)g(x)dx) < J(f)dx. (JENSEN'S FORMULA)X

(b): Further assume that g(x) > 0 for all x e X and that p

is strictly convex.

Show that the inequality in Jensen's Formula is strict unless

f is equal to a constant almost everywhere.

(c): Show that Jensen's Formula is still true if, with I

an interval of at,

1

f(x)g(x)dx > -W.X

(If I is not bounded from above, one sets

W(-) = N(y)).


SOLUTION: Recall that for a real measurable function h it is

said that

if

and one then sets

Jh=Ih+_1h_e

It is easy to see that if


h1 < h2, Jh1 > - 00,

then

Jh1, Jh2.

ova = vov = ova = vov = ovo

SOLUTION: (a): By exercise 3.62

y0 = 1 f(x)g(x)dx

is a point of I. Assume first that y0 is an interior point of I.

There then exists a linear form u on Y such that:

c(y) > c(y0) + u(y - y0), y e l.

In particular,

4P(f(x)) .> Vy0) + u(f(x) - y0), x e X.

(*)

Multiplying both sides by g(x) and then integrating yield Jensen's

Formula, since

JX (f(x) - y0)g(x)dx = u(`X(f(x) - y0)g(x)dx) = 0.

If y0 belongs to the boundary of I there exists a (non-constant)

affine function £ on Y such that P(y0) = 0 and £(y) a 0 for all

y e I. Then

f(f(X))9(x)dx.2= £(y0) = 0,X

and £(f(x)) 0 for all x e X. Since it can be assumed that g(x)> 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 (aftermodifying f if necessary on a set of measure zero). One then

argues by induction on the affine dimension of I.

SOLUTION: (b): Let us assume that the equality holds in Jensen's

Formula. 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

I f cp(m) = a < m, then cp(y) . a for all y e I, and consequently

J (p(f(x))g(x)dx Jag(x)dx = a.X X

Here, briefly expounded, are the proofs of the properties

of convex functions used above.

Let us, in Y xYt, consider the set,


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

u(y) + acp(y) + B > 0, yel,

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<a<1,

acp(y0) + (1 - M)9(y) cp(ay0 + (1 - a)y)

3 cp(y0) + (1 - a)u(y - yo) = acp(y0) + (1 - a)(p(y),

170 CHAPTER 3: THE

so

9(ay0 t (1 - a)y) = av(y0) + (1 - a)(P(y),

and w is not strictly convex.

If I is an interval of 3t not bounded from above, there exists

an increasing function g on I such that:

m(z) - w(y) = Jg(t)dtzy

when y e°I and z y. If 9(co) is finite, the integral:

J g(t)dt = m(-) - w(y)

y

is convergent, and consequently g < 0; but then m(o') - m(y) .< 0.

This inequality is still valid if y = infl a I, for then

pa(y) > 1imc(z).z--yz>y

EXERCISE 3.70: Let X be a measurable set of atn such that meas(X)

1 and f be a positive measurable function on X.

Show that if one sets

A = Jf(x)dx.X

then:

1 + A`J

1 + f (x)dx < 1 + A.X

If X = [0,1] and f = F', where F is continuously differentiable,


the preceding inequalities have a simple geometric interpretation.

Use this interpretation to discover under what conditions equal-

ity holds, and then prove your answer.

AVA = VAV - AVA = VAV = AVA

2SOLUTION: If W(x) _ (1 + x )2, (p is continuous on 7R and cp"(x) _(1 t x2)-3/2

W is therefore convex. If f is integrable, then by

Jensen's Inequality

1 + A` 5 J 1 +f2. (*)

If A = equality holds, for 1 + f s f. Finally, the inequality

1 1 +f2E1+AY

follows from 1

-+f2.< 1 + f. If

X = [0,1] and f = F' 3 0, one has

A = F(1) - F(0), and the integral

ofI,-+-f

is the length of the

arc AC (cf. the Figure).

Inequalities (*) and (**) ex-

press that:

AC S AC F AB t BC.

Fca

One will therefore have equality in (*) if f is constant, and in

(**) if f is zero. In fact, if:

J/1+f2=1

1 + A2,0

172 CHAPTER 3: THE

then on the one hand

1 t f`>. 1 +A`+AA1 + A`

,

and on the other hand the integrals of both sides are equal; since

f is continuous, it folllows that f = A. Similarly, if

J1

=1+A=1 (1+f),0 0

then since 1 + f` < 1 + f, one has

1 +f2 = 1 + f,

and so

f = 0.


1, and let f,g be two positive measurable functions on X.

Show that if fg , 1, then:

Jx f.1Xg >. 1.


SOLUTION: The inequality fg >. 1 in fact implies that f(x) > 0

and g(x) > 0 for all x e X, so that

Jxf > 0,

Jxg > 0.

We therefore need only consider the case when


Then Jensen's Inequality applied to f and to 9(x) = l/x, x > 0,

gives

((Xf)-1 < Jf-' $ J

EXERCISE 3.72: Let f be a bounded measurable function on E =3RP

such that f(x) > 1 for all x e E.

(a): Show that if g is integrable on E, and

IfT gI s M, n = 0,1,2,...E

then:

= 0, n = 0,1,2,... .

JE

(b): From this deduce that if g is integrable on [0,a] and

ifaentg(t)dtl.< M, n = 0,1,2,...

0

then g = 0 almost everywhere.

AVO = vov = AVO = vov = ovo

SOLUTION: (a): Dividing gby M reduced us to the case where M= 1.

There exists a constant A such that 1 < f < A; for all x . 0, one

therefore has, uniformly on E:

eXf =xn

n-0 n

174

and consequently

n

fexf91 = IX niJf I ¢ x.n

CHAPTER 3: THE

Setting F = f - 1, one therefore has 0 < F . B = A - 1, and

I JexFgl - 1, x >. 0.

For all z = x + iye Q set

4)(z) =JeZFg.

(1)

(2)

ThenIeZFgl

< eBx IgI, which proves that ¢ is an entire function

and that

IIgII1eBx+

By Inequality (1) it is also true that

I0(x)I .< 1, x 0.

(3) and (4) imply, on setting C = max(IIgII1,l), that

C if Re(z) < 0 or z eR+.

We will show that in fact:

Io(z)I 4 C, zea.

(3)

(4)

(5)

(6)

It will then follow from Liouville's Theorem that 0 is constant.

Since it is clear that 4)(-x) - 0 as x + +-, this constant is zero.

Differentiating relation (2) n times and setting z = 0 gives

J F

1 ) nn >, 0.


But

fnCn(f - 1)Ss

s

which implies the desired result.

We now next prove (6). This follows from the PhrXgmen-Lindeldf

Theorem. We shall give the proof of it in this particular case.

Without loss of generality g is real, so that 0(z) = OZz T. It is

therefore sufficient to prove that (6) holds when Re(z) > 0,

Im(z) > 0.When Re(z) >. 0, Im(z) >, 0, define

G(z) = exp( - ee-ian/4a )4(a)

where c > 0 and 1 < a < 2. If a = pe18, p >, 0, 0 8 . n/2, then

IG(z)I = exp(- epacosa(8 - 4n))I0(z)

so that by (5):

IG(x)I < C, IG(iy)I s C, x >, 0, y >, 0. (7)

By (3), since x+ 1 the right side of (9) tends to zero as p there-

'pre there exists p0 such that

IG(z)I c C; Re(z) >, 0, Im(z) >, 0, Izl >, p0. (10)

Ipw by (7) and (10) IGI 6 C on the boundary of the domain

176 CHAPTER 3: THE FUNDAMENTAL THEOREMS

Re(z) > 0, Im(z) > 0, IzI < p0, by the Maximum Principle one also

has IGI .< C in this domain. Finally, if z = pelf, p - 0, 0 < 0

< n/2,

IG(z)I = exp(-E1[acosa(e - 4n))1 0(z) I < C.

On taking e + 0 this yields:

I0(z)I < C, Re(z) 0, Im(z) y 0.

SOLUTION: (b): By the preceding,

J0tgt)dt= 0, n - 0.

It follows that for every polynomial P,

ae

J P(u)g(logu ) uu = 0.1

If p is a continuous function on [l,ea] and if (P.) is a se-

quence of polynomials which tends uniformly to (p on this inter-3

val, then since u-1g(logu) is integrable, one obtains

ae

c(u)g(logu) du = 0.1

But then u-1g(logu) = 0 almost everywhere on [l,ea], i.e., g(t)

= 0 almost everywhere on (0,a].

CHAPTER 4

Asymptotic Evaluation of Integrals

3

EXERCISE 4.7,#'' Let cp be a continuous real function on [O,a].

'Assume that W(x) > 0 if 0 < x -< a and that p(x) ti Axr as x ; 0

(A > O,r , 0). For all t > 0 set

F(t) =f

adx

ot+W(x)

(a): Show that if 0 4 r < 1,

ra dx

t->0(t)

0 q(x)

(b): Show that if r > 1, then as t } 0

F(t) tiIT 1

rAl rsin(n/r) tl - 1 r

(c): Show that if r = 1, then as t -> 0

F(t) ti

A

log 1

178 CHAPTER 4: ASYMPTOTIC

(For part (c), show that this can be reduced to the study of the

integral

a

(t + (p(x))-1dx,at

and make the change of variable x = aty).

000 = 0A0 = A0A = voo = ova

SOLUTION: (a): As t decreases to zero, (t + T(X))-1 increases to

1/y(x) for all x e]O,a], whence the result.

SOLUTION: (b): Make the change of variables x = (ty)1/r. Then

rt y1/r - 1F(t) = rtl - 1/rJ0 1 + t-1cP(t1/ry1/r)

dy.

Set:

1/r-1y if 0 < y 4 ar/t,

ft(y) 1 + tqt1/ryl/r)

0 if y > ar/t.

Since, for y > 0 fixed and t - 0, t 1p(t1/ry1/r) -Ay,

1/r - 1lTO mft(y) =

1y+ Ay

Furthermore, x rp(x) extends to a strictly positive continuous

function on [O,a]; hence there exists a constant B > 0 such that

p(x) >. Bxr. Therefore

1/r - 1ft(y 1+By .

EVALUATION OF INTEGRALS 179

The Dominated Convergence Theorem can therefore be applied, yield-

ing

(ft(y)db =

lY

dy =1 it

+oJ

1 + l r sin(77r)t-r0

0 0 A

SOLUTION: (c): Note that the relation proved in part (a) is true

for all r 0, so that lim F(t) = W if r >, 1. Now,t+0

j0 t +d-p(x<

10t

t d+xBx =B log(1 + Ba),

So that

F(t) v a dxfat t + wp x

The change of variable x = aty gives

x Y

fat t+cx =alogt0t+9(aty)

Men 0 < y < 1 and t -> 0 one has 9(aty) v aAty and t = o(ty).

Consequently

lim ty -t->0 t + p(aty)

Furthermore, for 0 4 y s 1 and t > 0,

t + P(aty) t + aBty< aB

from this it follows that


1lim f t}' dy = - .t->0 0t+ip(ats')

EXERCISE 4.74: Let p,q,r be three positive real numbers. Find

the necessary and sufficient condition that

fxP+0 1 + xqjsinxxjr

dx < 0.

What type of counter-example does this furnish?

AVA - V AV = A0A = ono = AVA

SOLUTION: The integral is equal to

n=0

where

I = fn+1xp

_J1 (x + n)p

dx.

n n 1 + xq, sin= Ir o 1 + (x + n)q(sinnx)r'

For t > 0, set

(p(t) _

then

dx

0 t + (sinnx) r

7 (2np m((n + 1)-q) < I <2(n + 1)P (p(n q) .

n + 1)q n nq

By the results of the preceding exercise, as t + 0 one has

;EVALUATION OF INTEGRALS

cp(t) +J

(sinnx)-rdx = Ar if 0 4 r < 1,0

p(t)tinlog1 if r = 1,

1 - 1/r1

(P(t)ti rsin(7 r)(t)

if r > 1.

Consequently, if q > 0,

I ti2Anpq if04r<1,n r

nti 21 np glogn if r = 1,

2 p - q/r

n rsin(irt/r)n if r > 1.

181

It follows that if q > 0 the integral converges if and only if

> max(r,l).p

For q = 0 the integral is never convergent, so that the preceding

Condition is valid in all cases.

If p > O,r > 0 and q > (p + l)max(l,r) then for x = n the in-

tegrand has the value np; it is therefore not bounded as x -}

Although its integral is finite.

EXERCISE 4.75: Let f,g be two real functions on ]O,a[. Assume

that

(i): f(x) ti Axa and g(x) ti xs when x -+ 0 (a > O,A > 0,

0 > -1) ;

(ii): f is strictly positive and increasing on ]O,a[;

(a -f(x)(iii): J l9(x)le dx <

0

E82 CHAPTER 4: ASYMPTOTIC

Show that as t -> -

+ 1l -(S+1)/aI

g(x)e-tf(x)dx ' i rP-1-)(At)J0 a

(2): Prove that

B(p,q) =r(p)r(g)r(p+q)

(Begin by showing that

B(p,q) =p q+ 4 B(p + 1,q).)

SOLUTION: Let 0 1; there exists b such that 0 f(b) > 0 for b 4 x < a, when t > 1 one has

t(S+1)/alag(x)e-tf(x)dx

1b

4 t(B+1)/ae-(t-1)f(b) a lg(x) le-f(x)d.,J0

which proves that the left hand side tends to zero as t 3 .

Moreover, the change of variable x .+ xt-1/a gives

t(0+1)/a j bg(x)e-tf(x)dx= JF(x)dx

0

herew

'EVALUATION OF INTEGRALS

Ft(x) =

By (i) ,

is/°g(xt-1/a)e-tf(xt-1/a)

if 0 < x bt1/a

0

limFt() =xse-Ax

a

t-and by (*) ,

a0 < Ft(x) < yxse-B"

if x > btl/a.

183

Applying Lebesgue's Theorem, and recalling that the integral from

b to a tends to zero

limt(S+1)/a(a

g(x)e-tf(x)dx = J-X'e -Axadx

t

10

0

1 A-(Otl)/ar(B + 11aa

FIRST APPLICATION: Setting x = a(u + 1) one obtains

r(a + 1) = J xae-xdx =aa+1e-a(W e-a(u-log(u+l))du.

0 1

0

The above result can be applied to each of the integrals and

with a = 2, A = , B = 0, giving0

Fe-a(u-log(u+1))du

ti2xr()(2)- =

an1 l

184

r(a + 1) ,aa+gi e-aV

SECOND APPLICATION: For p > O,q > 0,

CHAPTER 4: ASYMPTOTIC

(1B(p + l,q) = J

17X---X)

p(l -x)p+q-1dx

o

=(1

x)p+q xp-1

dx(1 -p + q 0 1 x (1 - x)2

=p

B(p,q).

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).

p(p + n)

Now it is known that as n

a(a + n) nanl - r(a) ,

So

(p + q + n),,

r(p) nqp...(p + n) r(p + q)

On the other hand,

B(p + n + l,q) = 11xq-1(1 -x)penlog(1-x)

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

B(p,q) = r(p)r(q)r(p + q)

EXERCISE 4.76: Prove that for n 0,

n -x`` ,x e sinx4dx = 0,

0

and then that as t -> -,

1

f eitx-x4sinx"dxti 4

r()ein/8t-5/4

J0

004 - 040 = 400 - V AV = A4A

SOLUTION: The function

F(z) =J

x4n+3e-zxdx

0

is holomorphic for Re(z) > 0. When z is real one has:

F(z) = (4n + 3)!4n+4

z

185

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


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 continuous1

for Re(z) > 0. (One chooses the principal branch of z4 in this

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:

II < exp[- tRsina - R4(cos 4 - sin-!)),

4

l$, -2t7 8

- , i

If(Re t) I < exp 2R"sin (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

If(z,t)I < MIzI"e-ty, Y A 0. (2)

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

Rexp( - I - 2t exp( - v sin 8) ,0 l

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)dy0 0

= it-5/4ein/8 OtIe-it[/B -1J flit y,t)dy.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 =ieln/BrWy4e-ydy

t-3 0 0

= i_ r(4)ein B.iein 8r [T5)

4

RCISE 4.77: For every integer n 1 denote by do the number

partitions of a set with n elements. Set d0 = 1.

(a): Show that for all complex numbers z:

n

n n, = exp(ez - 1).n=0

(b): Deduce from this that for every real number u > 0:

188

i u+2co

do= 2nieJu_i z-(ntl)exp(eZ)dz.


(c): Let unbe the unique real root of the equation

zeZ=n+1.Show that

u U

d , n' exp(e n - u e nlogu - lun eJ n n n

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,

nn

do+1p0 lP ,dn-p

This relation can be written as

(n + 1)do+l

-n

1 d-p

(n + 1)! _0 p! (n - p)!.

p-

If the series

(1)

n

f(z) _L

d n, (2)

n=0o

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:

n = 2nieIz-(n+1)exp(ez)dz

where r denotes the rectangle indi-

cated by the figure. Note that, on

this rectangle, if z = x + iy,

C

- M

D

lexp(eZ)I = exp(excosy) .< exp(eu),

and consequently

JBCDA.< exp(e

u)

2u + 4M0+1

On the other hand:

IN B

O

-INA

j(u +iy)-(n+l)exp(eu+iy)l

=(U2

+y2)-(n+l)/2exp(eu)

From this one deduces that for n 3 1:

= n12x_f+ (u + iy)-(n+l)exp(eu+iy)dy, (3)do

190

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,

+Wdo

n! exp(eu - ueulogu)J- g(y,u)dy,

where

g(y,u) = exp{eu[ely - 1 - ulogI1 + mil]}.

Note that

((2

l g(y,u) I = exp{ - eu[2sin2 + 2 log I111111

1 + 2 ] }.U,

Therefore

if9(y,u)dy1 < 2J (1 + u 2y2)-jue

udy

ly l >7E n

2dy

Jn 1 + -uIeuy2

The last integral has the value

e-u/242-utan-i( a-u/2 fu)

whence

g(y,u)dy = 0(ue-u).fly,>n

If lyl < it then Isin(y/2)I -o lyl/n, so that:

(4)

(5)

lg(y,u)I < expl-It

2 euy2)I


Setting y = to-u/2 yields:

f

+x xeu/2-u/2

-g(y,u)dy

=

e-u/2J-xeu/2g(te,u)dt,

x

and for t > 0,

2

limg(te-u/2, u) = e-t /2

Since for 0 < ItI 4 xeu/2

2 2

Ig(te-u/2,u)I .<a- 2t /x

Lebesgue's Theorem implies that:

Jx g(y,u)dytie-4/2Je_t2/2dt

= e-4/2 x.x

Taking account of Equations (4) and (5),

u

d ti n! exp(e n- u eunlogu - u ).n e/27 n n n

Bence

u ulogdn = logn! - (une nlogun - e n + jun) + 0(1).

Now logn! ti nlogn and

u uu e nlogu - e n + '2u ' nlogu = o(n(u + logu ))n n n n n n

= o(nlogn),

191

Which proves that logd N nlogn.


EXERCISE 4.78: Show that for every real number t the integral

( 1

fi(t) _J cos)tx +33Jdx

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 -> =

1

7.7 expi- 3t3/2 (1

+ 0(t-3/4)),4,(t) =

2

0(-t) = 01*Jt1T

SOLUTION: Note

3 Jdx.J°°expi{tx

If z =x+iy,

( 3

Iexpiltz + II = exp( - ty -x2p + 1 y3),3

so that if 0 < y < a and Iti < A,

3l 2

iexpi(tz + 3 JI < Be -X y,

where B = explaA +

3

a31. Therefore


f

x+iaItz

3l (a 2

expi+ zJdzl 4 BJ a-x ydy 4 Bx+i0 3 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

2/(t) = tz + 3 Jdz. (1)JL{ll

a


3

2explltz +)Idz2/0"(t)

JLz

a

the differentiation under the integral sign being legitimate be-

cause on La one has for Itl 6 A

(Iz2expi

3

ltz + 3JI 6 B(x2 + a2)e-ax

2

Hence

( 3

22,r7,-(,"(t) - ti(t)) (t + z)expiltz + 3JdzJL l

(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

3(

2 4(t) = J expixlz + -) dz.LS l


Since the derivative of z + z3 /3 vanishes for z = i, let us take

S = 1 in the formula above. This yields

+m ( 3l

2'r O(t) =

e-2X/3+-expal- u2 + 23 Idu

1tl 111

1 -2a/3 (+mexp I - u2 + 3 du.e -I

The latter integral can be written

V 1, + J+We-CO- 1)du.

The absolute value of integral above is majorised by

2JIuI3e_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

(-t) = J0

rit")dx.

As the derivative of x3/3 - x vanishes for x = 1, one is led to

the change of variable x = 1 + u//, which gives

( (3

(-t) = 1-Re{e-2ia/3T

xpi 1u2 + 3/T) d.)


Set

((

f(z) = expila2 + 3x) = Rei*,

where, as z = pale,

3R = exp( - p2sin28 - - _ sin30),

3

= p2cos20 + p COS33.

195

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 + . sin30 n8 - 8 = a8,

Where a = 4/n - 1 > 0. Therefore

2

IRI <e-'p

$

By the Mean Value Theorem

R - e-p2sin28 3 sin38expl(- p sin28 - CPI sin30 ,

3A 2 3/A J

where 0 < E < 1. Consequently:

IR -e-p2sin281 < IMI

a-ap28

3/

(2)

(3)

[.et L be the half-line pain/4, p > and r the circular arc

-rei8, 0 < 0 6 n/4. Then (2) shows

J

"f(u)du = f f(z)dz,

t+L

196

and then that

Jrf(z )dz J Q

4e-ax8dO = O lfJ


Note next that for 0 = n/4, _ -p3/3 I. Then (2) and (3) show

that

If(pein/4)

- e-p2 2I

< IRe" - RI + IR - e-pI

+le-anp2/43 3XConsequently:

2

J f(z)dz = eix/4

J

fe_p

dp + H,L f

with

7H S constant pI3e-anp /4dp = 01J

Since, furthermore

F_ 2 -Ar- e p dp"I

e2-F

one has

- = e1rz/4 E + O{J

J,r-1

Finally,

'

x n -2iA/3 in/4 1Re e


- l

I

lI

(-t) = t 4Cos -

3t3/2 + 41 + 0

CHAPTER 5

Fubini's Theorem

EXERCISE 5.79: For each of the functions f below, calculate

yJf(x,y)dx,Jdxf1(x,y)dy,Jp 00 0

2 2

f(x,y) = x - y(x2 + y2)

2

(b):

f(x,y) =

Ii.O<x,y61

(x - )-3 if 0 < y < Ix-1,0 otherwise.t

f(x,y) = x - y(x2 + y2)3/2 ,

(d) :

f(x,y) = (1 - xy)-p, p > 0.

199

lf(x,y)!dxdy.

200

SOLUTION: (a):

CHAPTER 5:

000 - VAV = Ova = VAV = A0A

J

Y0

0 J0 (x2 + y2)2

dy=

I dx [-x2+ 1

_

fl fl x2 2 1 x x=1

JO

1 (_d )n

0 (x2 + y2)2 0 ix2 + 2]= b _ - 4

JJOtx,y<1

2 2x -Y2)2(x +Y

1 :2 _ 2dxdy = 2 2dxJ dy

0 0 (x2 + y2 ) 2

1 y =x 1 dx2 fo [x2 +

y2]y=0 =0

= W.x

SOLUTION: (b):

1

J f(x,y)dy =x

0(x - )3

and this function of x is not integrable on (0,1). Moreover,

if O<y4i,

f

1 (-y 1

f(x,y)dx = J (x - )-3dx + J (x - )-3dx = 0,0 0 i+y

and if 1 < y < 1 this integral is again trivially zero. Thus:

1 1

0dyj0 f(x,y)dx = 0.

J

Finally,

1f0lf(x,y)ldy = Ix - I-2,

FUBINI'S THEOREM 201

so

J if0) 0If(x,y)Idxdy = -.

SOLUTION: (c):

f

1

fo

1 x_ by

1 1 + x y=lf

0 Lx (x2 + y2)Jy=00 x(x2 + y2)3/2 d=

Clearly

J117X2- 1dx = flog x + x/ +11 X-o = log2.

+1 J `

1 (1 x - ydy1 dx = - log2,

0 0 0 (x2 +y2)3/2

and

if x - b dy =lax x - b

djoJ 2 + y 2)3/2 yj J I 2 + y 2

)3/2I o (x2

0 0 (x

f0ldxlx(x2 Jy=O= (f2-1)J xu

SOLUTION: (d): As the function is positive the three integrals

are equal. For example, if p 4 1:

f

1 _ P _dxJ1(1

- xy) Pdy = 1 1(1

(1 - x 1 dx,

o o p o

an integral which converges only if p < 2. When p = 1:

r1dxr1(1 - xy)-1dy (1log(lx - x) dx < + =.0 0 Jo

202 CHAPTER 5:

EXERCISE 5.80: Let 0 < a < b. By applying Fubini's Theorem to

the double integral:

JJxydxdy,0<x<1a,cysb

prove that:

J 1 xb - xa ax = log 1 + b0 logx 1 + a

000 - VAV = AVA = vov = AVA

SOLUTION: On the one hand

f

b ('1 bdyJ xydx = J yyl = log 1 + b 'a O a

and on the other hand,

f

10dxfab 1 xy y-b 1 b

ogx

a

xydy = JOdx [l]ya = J0

xdx.

EXERCISE 5.81: Let R be the region in the quadrant x >. O,y 3 0

bounded by the curves y - x = 0, y2 - x2 = 1, xy = a, xy = b

(0 < a < b). Calculate the integral

J J (y2 - x2)Xy(x2 + y2)dxdy.R

by using a change of variables that transforms R into a rectangle.

AVA - DAD = AVA - VAV - AVA


SOLUTION: Set u = y2 - x2 and v = xy; R is then transformed in-

to a rectangle 0 6 u S 1, a .< v< b. Now

D(u,y)- - 2(x2 + y2

D(x,y)

so

(x2 - y 2)Xy(x2 + y2

b 1 b

)dxdy = fadvf0uvdu = fa v + 1

ffR J

'lo l+b l+b2gIta -21og1+a

EXERCISE 5.82: Let T be the interior of the tetrahedron def-

ined by x 3 O,y O,z >. 0, and x + y + z < 1. Calculate the

integral:

JJJT xyz(1 - x - y - z)dxdydz

by carrying the change of variable x + y t z = X, y t z = XY,

and z = XYZ.

AVA - TAV - AVA=VAV=AVA

SOLUTION: We have

x = X(1 - Y), Y = XY(1 - Z ), XYZ.

The open tetrahedron T is thus transformed into the open rec-

tangle 0 < X,Y,Z < 1. Also,

dxAdyAdz = (dx + dy t dz)A(dy t dz)Adz

= dXA(YdX t XdY)A(YZdX + ZXdY t XYdz)

= X2YdXAdYAdZ,

204

so

fffT

xyz(1 - x - y - z )dxdydz =

= JJJ

CHAPTER 5:

X5Y3Z(1 - X)(1 - Y)(1 - Z)dXdYdZ =0<X,Y,Z<1

r(6)r(2)r(4)r(2)r(2)r(2)=

1

r(8)r(6)r(L) 7!.

EXERCISE 5.83: Calculate:

dxdy

x>O,y>O (1 + y)(1 + x2y)

Deduce from this the value of

flow' dx0

x2 - 1

00A = 000 - A0A = V V = AOA

SOLUTION: As the integrand is positive:

+ y1 +x2y 2J0 1y+ y

dyJL>0- f0 1

f

it it n2

2 sin(n/2) 2

From this one deduces that

n2

yr

dx =2 = J d

o 0 (1 + y)(1 + x2y)(Contd)

FUBINI'S THEOREM

Jo

dx

1

jm

2 )dyo`1 + x2y

-f-1-

=J° dx Ilog11ylyco

ox2-1 1+y y=0

2logx dx,

0 x2 - 1

so

l ,x dx=n

J00

2

0 x 12- 4


3ydxdy

Jx>0,b>0 1 + (x +y)3

.

x+y<a


205

SOLUTION: Make the change of variables u = x + y, V = x - y,

for which D(u,v)/D(x,y) = -2 and y = J(u - v). The domain of in-

tegration becomes 0 O

and deduce from this the value of

To

(tan izJ2dz.Il

OVA = VOV = tVt - VOV = AV&

SOLUTION:

dxdydz

JO<x,y<1 (1 + x222)(1 + y2z2)z>0

a

'1O<x,y<dxdYJ dz

22 221 0 (1 + xz)(1 + yz)

CHAPTER 5:

IJO<x,y<1 1 +xx2z2 1

+yy2z2Jdz

2 2

dxdy=nff =

2 0<x,y<1x+ y 2 0 0 x+ y

n

2

1(log(x+ 1) - logx)dx = nlog2.

0

It follows that

nlog2 = Fdzffdxdy

O<x,y<l (1 + x2z2)(1 + y2z2)

FUBINI'S THEOREM

1 2 2 2dz-Jo{J

1+dxz

= r0(tan-1z12dz.

JO zJ

207

EXERCISE 5.86: Let q be a positive definite quadratic form on

in, Prove the formula

q(x)dx

_

it

n/2

where A is the determinant of q.

AVA - Vnv - AVA - vov - AVA

SOLUTION: There exists an orthonormal basis with respect to

which

q(x) = alx2 + ... + AIx2, A. > 0.

Then

p = a1...anp

so

e q(x)=

7n7- +e-asu2du=

nR X n/2

JIltn 8 1 J s i" S

EXERCISE 5.87: Determine the values of a for which:

cosx

IJ0<x,y<n/21 -sinxsiny)a

dxdy <

AVA - VOV - AVA - VAV - OVA

208 CHAPTER 5:

SOLUTION: By carrying the change of variables x - jn - x,

-> 2n - y, one is led to study

sinxa

I = ffO<x,y<7c/2 (1 - cosxcosy) day'

The integrand is continuous on the square 0 S x,y < it/2, ex-

cept at the point (0,0). If r = (x2 + y2), then in a neigh-

bourhood of (0,0)

sinx 2x + 0(x3 ) 2x 2cosxcosy = 2 4 - 2 (1 + 0(r ) ),

r + 0(r ) r

whence:

sinxa = 2axa + xa

(1 - cosxcosy) 2a 0( 2a-21r r

Note that changing to polar coordinates gives

11O<rl xardxdy =(1rdr.J n/2

cosxdx,JO 0

x>O,y>O

a quantity which is finite if and only if a > -1 and $ < a + 2.

Hence one has I < - only in the case where -1 < a < 2.


IP(a) = xlx2...xpdx1dx2...dxp

P<a

ovo - vov = vav = ovo = vov

SOLUTION: The change of variables xi - axi shows:

FUBINI'S THEOREM

I(a) = a2pI(1).

Ifp32:

IP(1) = J x1dx1J x2...xpdx2...dxP(0 1 rxi

0x2+...p<1-xl

(1

= Ip-1(1)J x1(1 - xl)2P-2dx10

Since

r(2)r(2p - 1)= Ip-1(1) r(2p + 1)

r1I1(1) =

J

xdx0

it follows that

1I(1) r(3)

= 2 r(2p +p

1) - (2p)!

so that

T(a)=a2p

p (2p) 1 .

EXERCISE 5.89: Prove the formula(DIRICHLET'S INTEGRAL):

al-1 ari 1

xl ... xn dx1...dxn

209

1l P1 rnPn

xi>0,J

+...+J sl a a all (Yanl

n a11...ann r 11 ... r.p a a1 n 1+...+ n+ll

1 pn

210 CHAPTER 5:

ove = vov = AVA - vov = eva

SOLUTION: The change of variable Xi = (xi/ai)pi

transforms the

integral into

..a a-1 - nal an1 n_

a . - 1Xlp1 dXl...dXnpl...pn Jf

JX.>0iX1+...+n

l

Thus it suffices to prove the formula when a. = p. = 1. For

1<s4nand A 30 set

JI

a 1

IS (A) = J ... fxIxl1-

...xss dx1...dxs.> 0

x.+---+x 4XI S

a

It is clear that IS(A) = A SIs(1). Consequently,

1 a -i a

In(1) = In-1(1) xnn (1 - xn) 1n-1dx

00

r(an)r(a1 + + an-1 + 1)

- In-1(1) r a1 + + an + 1)

Since

(l a -1Ii(1) = J x11 dxl = a ,

0

it follows that

1 r(an)...r(a2)r(al + 1)

In(1) = a1 r(a1 + ... + an + 1)

FUBINI'S THEOREM

r(al)r(a2)...r(an)

r(a1 + + an + 1)

211

EXERCISE 5.90: Determine the values of the real numbers a,s,y

for which:

dxdydz

Jx>0 1 + xa + ys + zyy>Oz>O

and then calculate this integral.

AVA = VAV = AVA = VAV = eve

SOLUTION: First of all it is clear that one must have a > 1,

0 > 1, and y > 1, for if, for example, a s 1:

dx

JO1+xa+ys+zy

for all y > 0 and all z > 0. Set

2/a 2/$ 2/yx = u , Y = v ,

The integral may be written

z=w

_ Su2/a - lv2/S - lw2/y - 1

I aOy 111 2 2 2dudvdw.

u>0 l+u +v +wv>ow>o

Now change to spherical coordinates

u = rsinecoscp, V = rsinesincp, w = rcose,

which yields

212

1=

Thus if

CHAPTER 5:

8 rm r2(1/a + 1/0 + 1/Y)-1drJ ic/2cos2/a - 19sin2/5 - 1(pdW

aSO 0 r2+1 0 X

I-a/2sin2(1/a+ 1/B)-l6cos2/Y

- 13d8

0

1 r a r s r a t s P Y rm ul/adu

aRY r(a + r(q + 2 +Y)

0 1 + U

1a + a +

y< 1

then

1 = n rlalr[oil r[1]aay

r {cx + B +yj

sinn l a+ 0+ Y)

which can also be written

aBy r (a) r (S) r Y) I aY)


fffdxdydz

0<x,y,z «1 - cosxcosycosz

OVA = VAV = MVA = VAV = OVA

SOLUTION: Set

FUBINI'S THEOREM

u = tanix, V = z = taniz,

so that the integral becomes:

I 4dudydw

JJJ>0 u2 + v2+ W2 + u2v2'W2

v>ow>0

Changing to spherical coordinates yields

I =J

/2d3f/2dTf

0 0

0

1 + r4sin4Ocos28cos22

Finally, setting

r = t4(sin8)cos 3cos cpsin cp,

we have

(n/2 _ (n/2 _1 _1 (W ,4-1

I= f0

cos 1sdolo cos 'gsin 2gdgj01 + u du

i r(i)r(i) r(1)r(1) it 7E /2- r(3= 4 r(4)4 r( )4, r() sin

Noticing that

r(;)r(4) =sin 4n =

nom,

213

the answer can be written

I = 4x(4)44

EXERCISE 5.92: Use a double integral to represent the differ-

ence:

214 CHAPTER 5:

J+_

f(x)2 dxfg(x)2dx - IJ f(x)g(x)dxl2l-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 G0the 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

'SICdxdydz

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:

FUBINI'S THEOREM

2r 1

JJC dxdy =1 0J

s0,z

3JJ xdxdy = V0,

z

Jc1ydxdy

n0S0,3z

11 2 3zdxdy = zICoJS0 = 1 0Jc6S0,JJ z ll

so that

3J 00

NZ-01 C0S0dz 3

=E , etc.....

J00N-E-O)

2S0dz4 0

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

215

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:

each of them. (The Pile of Sand Formula).

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 equa-

tions of the type

x = a(t) + zb(t),r {lz

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

S(z) = 2J xdb - bdxrZ

is therefore a second-degree polynomial in z. From this it fol-

lows that (Simpson's Formula)

hV = I S(z)dz = 6 (S1 + S2 + 4

0

S3).11

EXERCISE 5.95: Show that if

then the order of the integration can be inverted in

0 0

FUBINI'S THEOREM

From this deduce that

sinax dx =afm J0(y)

d y,y,I1 +x` 0 a2+y2

where JO is the Bessel function:

2 rn/2J0(y) = nJ cos(ycose)d8.

0

AVL = VAV = LVA = V1V = OVA

SOLUTION: If 0 < E < X < w then

Jsinaxdxjf(y)e-x3'dy = JW f(y)dyje-x3'sinaxdx,E 0 0 E

since

f

x rdxJ Isinaxf(y)e-xyldy < (X - E) Je- Eylf(y)

E 0 0

The Second Mean Value Theorem gives

xr

J

e-xysinaxdxe

2

a

and the inequality Isinaxl . ax implies

XJ_XYid

E Y

Consequently,

dys -.

217

Xf(y)f e-xysinaxdx

4 a Il(0,1)(y)If(y)I + an(11-)(y)lf(y)Iy-2E

218 CHAPTER 5:

The Dominated Convergence Theorem leads to the formula

J sinaxdxJ f(y)e-xydb =J

f(y)dyJ e-xysinaxdx.0 0 0 0

Now,

so

0 y + ae-xysinaxdx =

2

a2

,

off (y) dy = sinaxdxJf(y)e-xydy.

o a2 + y2 0 0

Since J0 is continuous and IJ01 < 1, it is clear that the con-

ditions of the problem are satisfied if f = J0. Moreover:

JJ0(y)e XYdY = fee-xydy Icos(ycos9)d8

0 0 J0

fo

n/2 W

= nd8J e-xycos(ycose)dy0

=2 (n/2 xd8 - 2x rm dt

0 x2 + cos28 n 0 1 + x2 + x212

tW 11

f

2tan

1

xt`n 1 7,77j=:+ x t=0 1 + x-

(Switching order of integration is legitimate, since

W n/22J dyJ e-xyjcos(ycos8)Id8 < x < ").

n n


EXERCISE 5.96: Let H be a continuously differentiable function

on [0,°[. For all r > 0 set

m(r) = sup(xlogr - H(x)).

show that if

<T

Joe-H'(x)

0

then

Tm(r)

2

dr <

1 + r

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:

I (t)dt < -,0

and if for r > 0 one sets

(xu(r) = supJ log(rcp(t))dt,

x>.0 0

then

f

I' (r) dr <

0 r2

Notice that u(r) > 0 for all r > 0. On the other hand,

xlog(rp(t))dt < J1og+(rq(t))dt,I

0 0

220 CHAPTER 5:

so

0 < u(r) < Jlog+(r(t))dt.0

Consequently,

flog(r(p(t))

r2 0 0 r

OW(t(J'

To return to the problem, put p = e-H'. Then:

(xxlogr - 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. Show that if for

every complex number z

JlogIl + zfj = 0,

then f(x) = 0 for almost all x.

FUBINI'S THEOREM

Show that

Jlot'l = 0

for all p > 0 by using the formula

2n

Jo logll + zeitldt = log+lzl.

AVA = VM0 = t01 = VLV - AVA

221

SOLUTION: It may be assumed that meas(X) = 1. Let p > 0. Then:

f

2n

dtJ log 11 + Peitf(x)Idx = 0.X0,

If Fubini's Theorem can be applied then

2n

0 dx log 11 + peitf(x)ldt = 21 log+lpfl.X JO X

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. By hypothesis

JXlog+11 + zfl =JXlog 11 + zfI

for every complex number E. Therefore

J0dtJ1H1 + Peitf(x)Ildx = 2J dtJ lo11 + pitf(x)ldxI 0 X

4 4n(1 t pJX If 1) < m

222 CHAPTER 5:

(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-

able.

(b): Show that if f is measurable and p > 0:

Jf(x)Pdx = pJWtP-lmeas(f > t)dt.E 0

(c): Show that if f is measurable its graph is a set of

measure zero in E x]R.

000 - VAV = 000 - VtV = 000

SOLUTION: (a): If f is measurable, the function p(x,t)= f(x)- t

is also. Since Df = (p 0)fl(t 0) it follows that Df is

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): Fubini's Theorem also shows that

EP = pJEdx(0(x)tp-ldt = pJOtp-1dtJ(flt)dxJ f

M

= pJtP-1meas(f > t)dt.

0

SOLUTION: (c); One can show, as in part (a), that the set D'f


of (x,t)'s such that 0 < t < f is measurable, and the calcula-

tion carried out in part (b) proves that

Jf f = meas(Df) = meas(D'f).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 nn

AVO = VAV = AVO = VLV = AVo

SOLUTION: Let In be the orthogonal projection of Dn on the

x-axis. Then meas(In) = 2 1 rn, and consequently if E rn <

then almost all x's belong to only a finite number of the In's;

that is to say that the vertical line Lx

with abscissae x meets

only a finite number of discs Dn, say Dn ,...,Dn . If IxI < 1k

none of the intervals Lxf1D

ncan be equal to L

x0 D, for in that

. -

1

case one would have D = D, which is absurd, because rn

> 0 forn.

all n and because the Dn's are mutually disjoint and contained

in D. It is then clear that:

kmeas(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,

224 CHAPTER 5: FUBINI'S THEOREM

and consequently:

meas(D - U Dn) = JdxJP(xiy)dy > 0.

n

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:

JfP -

jgq

=1,

.where q is such that1

+ 1 =

g by gq yields:

Jfg' 1.

p q. Replacing f in (*) by fgl q and

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

226 CHAPTER 6: THE

AVO = VAV = AVO = VAV = OVA

SOLUTION: The function p is convex, for it is continuous on

[0,1], and on ]0,1[ its derivative is equal to

(1 - 1/PlP-1

l l/p )x

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 << (JfP)1/P + (JgP)1/P.

EXERCISE 6.102: Prove Minkowski's Second Inequality using Jen-

sen's Inequality and the function p(x) = (1 + x1/P)P, x A 0,

0 . 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< (J(f + op) 1/p.

EXERCISE 6.103: 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,

228 CHAPTER 6: THE

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

then fn converges to an element of L. SetNWgN = E Ifn - fn+1l, g = E Ifn - fn+ll'

n=1 n=1

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 ex-

istence of a measurable function f such that fn - f almost every-

where. 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

< d(fn'fn+1),n=r

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<p<1.

EXERCISE 6.104: Let K be a compact set of IItn, e > 0, and 1 , <p <Show that there exists f e Lp(ntn) such that f 3 0, 11f11p = 1,

and:

Ilfa - flip<e, a e K.

AVA = 0A0 = AVA = DAD - a0A

SOLUTION: Let us denote by Br the ball with centre 0 and radius

r, and by Xr is characteristic function. Assume that K C Br, and

for R > r let:

1f

meas(BR)1/pXR.

We have f 3 0, IIfI{p = 1 and ifa

- fI is, to within a factor

meas(BR)-1/p, the characteristic function of the union of the set

of points x for which Ix - al .4 R, IxI > R, and of that defined

by IxI < R, Ix - al > R. These two sets are mapped into each

230 CHAPTER 6: THE

other by the symmetry x - a - x, and the first is contained in

the 'annulus' R < I x I < R + a . Thus when a e X

E 2meas(BRtr)

- 11/p

Ilfa - flip ( meas(BR) )

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 implies Jjfa - file/a --> 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. By the Mean Value Theorem,

IJ - Til If - fiI (f V fi)a-1, f V fi = max(f,fi),

and by HUlder's Inequality for the pair a,a(a -1)-1

LP-SPACES

f I fa - iIp/a < ap/a((IIf - f1Ip)1/a(((fV f1)p)1-1/a

It remains to observe that

(fvf.) , (f + If-fiI)p<Cp(IfIp+ If - filP),

where Cp = max(2p-1,1).

231

EXERCISE 6.106: (a): Let f1,...,fn be measurable functions on] O.

Show that

Ilfl...fnllp0 < IIflIIp1...IIfnllpn

if

0<pi <QO, 0<i<n, 1 1 1

PO p1 pn

(as usual, the convention 1/- = 0 is in effect).

(b): Show that if f,g are measurable and positive

Jfg < IjfIj1 p/rIIgIIqq/r(Jfpgq)1/r

if

14 p<m, 14 q <=, 1< r< co, r=p1

+ 1- 1q

(the convention is made that a0 = 1, including the cases a = 0 or

a=m).

(c): If feLPORm), geLq(um) and

1 p,q<m, p+Q - 130,

232 CHAPTER 6: THE

show that for almost all x the function

t + f(t)g(x - t)

is integrable, and that the function:

h(x) = Jf(t)g(x - t)dt

is such that

- 1. (YOUNG'S INEQUALITY)l i h i l r < 11A p I I g II q if r=

pt

q

00A = vov = 00A = vov = 40A

SOLUTION: (a): If I = {i:pi < co) it is clear that

IIf1... fnllp0 < li IT fillPO iT lifiii.ieI

1 = C 1

PO ieI pi

It may therefore be assumed that pi < m for all i. First consider

the case where n = 2. Then

pOp0p1 p2

Applying Holder's Inequality for p = p1/p0, q = p2/p0 to the func-

tions IflPO,lglPO yields

LP-SPACES 233

which gives the formula in this case. When n > 2 one proceeds by

induction; in fact, if

1 1 1

W p2 pn

then

1 1 1

PO p1 W

whence

IIf1f2...fn11PO E IIf1IIp1IIf2...ffII,E

IIf1IIp1IIf2IIp2...IIffIIPn

SOLUTION: (b): If r = m this is a matter of Holder's formula.

Therefore assume that r < W and consider first the case where

p > 1, q > 1. Then r > p, r > q. Set

_ pr = grp1 r-p , P2 r-q , P3=r.

Then

1+ 1+ 1= 1+ 1- 1= 1,p1 p2 p3 p q r

so that if h1,h2,h3 are three positive measurable functions, by

part (a) above one has

1 p12 2 3 3

Jh1h2h3 (Jhl ) (Jh2 ) (Jh3 ) .

If

h11 = fP, h22 = gqh33

= fpgq,

that is to say, if

234 CHAPTER 6: THE

hl = fl - P/r, h2 = gl - q/r, h3 = fP/rgq/r,

the desired formula is obtained.

When p = q = 1 one has r = 1, and the formula becomes trivial.

Lastly, of p > 1, q = 1, one has r = p, and it is a matter of

proving that:

Jfg 4 ((g)1 -I/P(ffpg)llp,

which is none other than HSlder's formula, because

91 - 1/p(fPg)1/P=

fg.

SOLUTION: (c): Assume for now that p < m, q < -, and that

p + q - 1 > 0,

so that r < -. By noticing that

J g(x - t)Igdt = IIgIIq,

it follows from part (b) that

(J If(t)g(x - t)ldt)' .< IIfIIP PIIgIIq gJlf(t)IPIg(x - t)Igdt,

so that

Jdx(Jff(t)g(x - t)Idt)r

6 IIfIIP-PIIg!Ig gJdxJlf(t)lplg(x - t)Igdt

IIfIIP-p IIgIIq gJlf(t)lpdtJ Ig(x - .t)gldx

LP-SPACES 235

= IIfIIPIIgIIr <

From this it follows that for almost all

J f(t)g(x - t)Idt < -,

which ensures that the function h is defined almost everywhere.

Since

Ih(x)I S JIf(t)(x - t)Idt,

one has

Jh()(rdx , IIfIIPIIgIIq,

which accomplishes the proof in this case.

If q = - one must have 1/p >, 1, hence p = 1. It therefore re-

mains to examine the cases where

p

+

q

= 1, for which r = . By

Holder's Inequality, for all x one then has

Ih(x)I , IIfIIpIIgIIgi

i.e.

Ilhll , Ilfllpllgllq

EXERCISE 6.107: Let 1 < p s - and p + 1 = 1.q

(a): Show that if f e LP

IIfIIP = sup{ I Jfgf:Ilgllq. 1}.

(b): Let f be a positive measurable function.

Show that:

236 CHAPTER 6: THE

IIfIIP = sup{Jfg:g 3 0 and IIgIIq s 1}.

(c): Let f be a measurable function. Assume that there ex-

ists a constant M such that:

JfI M

for every simple function g such that fg is integrable and 11g11q

1.

Show that IIfIIP < M.

(d): Let f be a measurable function such that f g is inte-

grable for every function g e Lq.

Show that IIfIIP <

AVA = VAO = ADA = VAV = DOA

SOLUTION: If f = 0 almost everywhere all these properties are

trivial, therefore it will be assumed that 0 < IIfIIP <

SOLUTION: (a): By Holder's Inequality,

sup{IJfgI ; IIgIIq'< 11 < IIfIIP.

In order to prove the inequality in the opposite direction let us

first of all assume that p < -, and let:

Tx)If(x)Ip-2 if f(x) 0,

g0(x) _to if f(x) = 0,

so that fg0 = IfI.

If p > 1 one has Ig0iq = IfIP, and then if

g(x) = IIfIIP' g0(x) we have IIgII = 1 and Jfg = IIfIIP If p= 1we have IIg0IIm 1 and Jfg0 = IIf1Il Finally, when p = - let-

0 < M < IIfiI, and let EC (If! M) such that 0 < meas(E) <

LP-SPACES 237

Set

1 if x e E,meas E f x

g(x) _0 if x4E.

Then IIgIIl = 1 and:

Jfg = 1 J If(x)Idx 3 M.meas E E

REMARK: In fact we have proved that if 1 4 p <

IIfIIp = max(I Jfgl:IIgIIq = 1).

This is nothing astonishing if one knows that in this case Lq is

the dual of LP. Against this, if 0 < f(x) < 1 for all x, and

IIfIL_ = 1, then for every function g such that IIgIIl < 1 we have

IIJfgl< 1, otherwise there would exist a e a such that Ial = 1 and

Jf(ctg) = 1. If ag = g1 + 2g2, then 11g1111 , 1 and Jfg1 = 1; but

under these conditions one would have 11g+11, <1 and

Jfgi - Jfgl =1,

whence:

Jfgi = 1 > jg1,

that is to say:

0 < Jfgi < 1, Jfgl % 00

J(1 - f)gi 6 0,

which would imply that gi = 0 almost everywhere, which contradicts

Jfgi = 1.

238 CHAPTER 6: THE

SOLUTION: (b) : When I I f I IP

< °° and g e Lq, one has I Jf9 I < Jf'Ii ,

whence the result in this case. If IIfIIP = °° let us set:

f(x) if IxI < n and f(x) < n,

fn(x) _0 otherwise.

By what has preceded, there exists gn > 0 such that II9n11q < 1

and Jfngn >- IIfnIIP - 1. Since Jfgn > Jfgn and IIfnIIP -; IIfIIP

the result is again true in this case.

SOLUTION: (c): Let En = {x:lxl .< n and If(x)I < n}, and let fn =

AE . If g is such that 11g11q < 1 there exists a sequence of sim-n

ple functions (gi) such that IIg - gillq - 0. Then, because fn a LP,

Jfg = limJfngi = limJf(g 1E ).i i n

The functions g nE are simple, II9iIlE Iiq < 1, and f(g IlE ) is in-n n n

tegrable. From this it follows that

IJfn9I < M.

By part (a) this implies that IIfnIIP < M, and therefore that

IIfIIP = 1nmllfnllP < M.

SOLUTION: (d): Let fn be as above, and set

u(g) = Jfg, un(g) = Jfg, 9 e Lq.

The un's are continuous linear forms on Lq. Moreover, for g e Lq,

fng -; fg and Ifn9l E If9I, JIfI < By Lebesgue's Theorem

un(g) -; u(g). By the Banach-Steinhaus Theorem u is a continuous

LP-SPACES 239

linear form on Lq, that is to say there exists a constant M such

that IJfgI s M if IIgIIq < 1. Using part (c), it follows from

this that IIfIIp < M.

EXERCISE 6 108: Denote by En the set of step functions on]Rn.

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

(*)

SOLUTION: By replacing f with zero outside the support of g one

can assume f is integrable. Notice that if f is real and 0 < h

< g, then since f >. f,

Jfh < Jfg.

When f is complex and IhI < g,

I Jfhl , JIfIi

and, furthermore, if f and h are real:

Jfh< IJfhI.

240 CHAPTER 6: THE

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

flhl = If1

1g

(where we have adopted the convention that f1If1I-1 = 0 when fl = 0),

and if f is real

+f1

h=

2 f 91

(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,

Jf1h2

=ffig

-

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 <2e.

Finally, when f is complex,

J IfI9 - I Jfh1I - I JIfI9 - Jfh1I1

and the rest follows as above.

REMARK: En can be replaced by K, the space of compactly supported

continuous functions.

EXERCISE 6.109: Let 1 < p < W and p t

q

= 1.

Show that if f is a locally integrable function and if there

exists a constant M such that:

IJf9l <M

for every step function g such that I9Ilq < 1, then IlfAIp < M.

A0A - 0A0 = A VA = 0t0 = AV

SOLUTION: By the preceding exercise, if g is a real positive step

function such that Ilgllq S 1, then

J fig = sup{IJfhh:h e En and Ihl . g} . M. (1)

To begin, assume that p > 1, and set

If(x)I if Ixl < n and If(x)I < n,(n(x)

0 otherwise.

When g e Lq, g , 0, Ilgllq s 1, there exists a sequence of step

functions gi such that gi 0 and II9 - gillq -' 0 (since q < -).

242 CHAPTER 6: THE

As 9n a LP, one therefore has

Jg = limJ (Pngi < limsupJ If Igi 6 M.i i

By Exercise 6.107, InII IIp 6 M, whence Ilflip < M. When p = 1 note

that if one were to have IIfII1 > M there would exist a rectangle

P such that

J!f I = Jiflap > M,

which would contradict (1), because 1I]11- = I.

EXERCISE 6.110: Let X =]R1, Y = ]R , and let f be a positive meas-

urable function on XxY.

Show that for p a 1:

{J (J f(x,y)dx)pdy) 1/p ,1< J{Jf(x,Y)Pdy}'dx.Y X Y

(GENERALISED MINKOWSKI INEQUALITY)

ADA = VAV = ADA = V AV = AVA

SOLUTION: Set

g(y) = JXf(x,y)dx.

On writing the function y w f(x,y) as it is a matter of

proving that

IIgIILp(Y) <

Let be a positive measurable function on Y. Then

LP-SPACES

J(Y)P(Y)dY = JdxJf(xY)cP(Y)dY,

whence, writing the conjugate of p as q,

Jg(P <

This inequality implies (*) (cf., Exercise 6.107).

243

mEXERCISE 6.111: For every i , 1 < i < n, let X. = i 1, and let fi

be a positive measurable function on X1 x ... x Xi x ... x Xn (wherethe hat indicates the term that must be omitted in the product).

Set

J_i, Îi = fi1...dx...dx

Considering fi as a function on X1x ... x Xn which does not depend

upon the i-th variable, show that

Jf1...fndx1 .dxn 6(I1...n)1/(n-1)

(*)

Deduce from this that if V is a measurable set of R3, and if

31,3 2,S3 denote the areas of the projections of V onto each of

the three coordinate planes, then:

vol(V) c S12 .

LVA - VAV - AVA = VAV = AVA

SOLUTION: For n e 2 inequality (*) is clear (in fact, equality

holds). Assume that this relation has been proved up to order

n - 1 Then

244 CHAPTER 6: THE

I =f= Jf1dx2...dxnJf2...fndx1.

Writing, for 2 .< i 4 no

gi = Jf11_1dxl,

Generalized Holder's Inequality gives (cf. Exercise 6.106)

1/(n-1) 1/(n-1)

Jf1 2 .gn 2.. n'

Applying Holder's Inequality again, with p = n - 1 and q = (n - 1)x

(n - 2)-1, yields

I '< I1/(n-1)(I g1/(n-2) 1/(n-2)dx...dx

)(n-2)/(n-1).

1 2 ...gn n

Now, using the induction hypothesis shows that the integral on the

right side is majorised by

(gidx2. .- 'a1...dxn)1/(n-2)

i=2 J

Finally, by noticing that for 2 < i S n

Ii = Jgidx2...dxi...dxn,

Inequality (*) is obtained.

To obtain (**), it suffices to note that if A1,A2,A3 denote

the projections of V onto the coordinate planes, then

ILV(x,y,z) < 1LA (x,y)ILA (y,z)1LA (x,z).1 2 3

EXERCISE 6.112: Show that if I1f - f 11 - 0, then f -} f in meas-n p n

ure.

LP-SPACES

MMA = VAV = X00 = VAV = MMA

SOLUTION: If a > 0 and En = {If - fn

I 3 a}, then

apmeas(En) <JE

If - fnlp < 'If - fnIIP,n

which shows that meas(E ) - 0.n

245

EXERCISE 6.113: Let (fn) be a sequence of functions in Lp(X),

1 < p < W, that converges in measure to a function f.


(a): f eLP(X) and limllf - fn II P= 0;

n-

(b): The sequence (IfnIP) conserves mass and is uniformly

integrable (for the definitions of these notions see Exercise

3.58).

MMA = VAV = MMA = 040 = AVA

SOLUTION: Assume first that f e LP and fn - f in LP. For every

measurable set E of X, Minkowski's Inequality gives

JEIfnlp)1/p < (IEIf lp)l'p + Ilf - fnllp (1)

Let c > 0. There exists a set B of finite measure, a number s>0,

and an integer N such that

lflp < e2 p,JX_B

JIfIp < e2-p if meas(E) < ,

246

IIf-fntIp< 111p if n > N.

By (1), if n > N and meas(E) < s:

JX-BIfnIp < 6,

JE 'fn' p<C.

CHAPTER 6: THE

Let B1,...,BN be set of finite measure and 01,...,SN > 0 numbers

such that

IfnIp<C if1<n<N,X-B

n

JEIfnIp < E if 1 < n < N and meas(E) < 0n.

Setting A = B U B1 U... U BN and 6 = then for

every integer n,

IfnIp < C,X-A

(2)

J IfnIp < e if meas(E) < 5, (3)

E

which proves that the sequence (If) conserves mass and is uni-

formly integrable.

Now assume that fn - f in measure, and that the sequence (fn)

satisfies Condition (ii). For every c > 0 there exist a set A of

finite measure and a number 6 > 0 such that the conditions (2)

and (3) are satisfied for every integer n. By Fatou's Lemma the

function f also satisfies these conditions. Set

p eEn I'f - fnI > meacA }nA.

LP-SPACES 247

There exists an integer N such that meas( n) < 6 for n > N, and

consequently

11f - fnIIP < (JX-A IfIP)11P + ( JX-A Ifnlp)1/P

+ (J IfI )1/P +JEIfl)n

n

+ (J If - fn1P)1/PA-E

n

< 5c1/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 fn+ f in Lp. 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.

EXERCISE 6.114: Let (fn) be a sequence of functions in Lp(X),

1<p<-.

(a): Show that if fn -> f e Lp(X) almost everywhere, and if

limllfn IIP

= Ilfn- P

then fn -> f in Lp(X).

(b): Show that the conclusion above is still valid if the

convergence almost everywhere of fn

to f is replaced by conver-

gence in measure.

avn - vov - ovo a vov - ovo

248 CHAPTER 6: THE

SOLUTION: (a): Let e > 0. There exists a set B of finite meas-

ure and a number 6 > 0 such that

JXB Iflp < 2

j If IP < 2 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

If Ip.f lflp < e t fA Iflp < e t liminffAX n n

J IfnIp = j Iflp,X X

this yields

j Iflp E + J Iflp - limsupj Ifnlp,X X n- X-A

i.e.

limsupfnIP t C.n- J X-A

Furthermore:

LP-SPACES 249

ffIP)1/P1

IfIP)11P + (LA IIf - ffIIP 4 (

X-A

+ 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 inte-

grable. As Fatou's Lemma is valid for convergence in measure,

the argument used in part (a) above shows that

JX-A IfPI4E

implies that

limsup1 IfnIP 6 e,n'°° X-A

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)

JIf IP > a.

Ek nk

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 partk

(a) above fn would tend to f in LP, and consequently the fnk 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.

(a): Show that if

nil IRe(fn)I = 1,11lX

limmJ I1-I.fnlI=0,X

then

nJ

IIm(fn)I = 0.X

(b): Now assume that

2 = 1.limn

Re(fn) =n 1 If I n J

IfnIi

X X X

Show that

limn JX

Ii - fnl = 0.

LP-SPACES 251

1Vt = VAV = 1VA = VLtV = EVts

SOLUTION: (a): Set fn = un + ivn (un,vn real). And

en = JXI1 - IffII

Now,

IvnI'<

IfnI 4 1 + 11 - IfnII,

and consequently

JXIVn1 4 1 + en.

Therefore the second hypothesis implies

= l YmssupJx IVn I S 1.

ssume that 0 < L 6 1 and choose a A such that 0 < A < £C. ByA

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 $).

252 CHAPTER 6: THE

Then

lira meas(An) = 0.n-

Furthermore

X Ivn'+

J

X-B(1

+B nn n

4 Smeas(Bn) + 1 - meas(Bn) + En.

If 0 < a < 1 this would imply

meas(B) <1 - B

Set Dn

= AnU B

n. On X - D

nwe have simultaneously

Ifnl < 1 + a,

Hence

IvnI > a.

lung < d = (1 + a)2 - S2.

Since lunI < 1 + 11 - Ifnll it would follow that

J Iun1 < 6(1 - meas(Dn)) + meas(Dn) + Enx

6 + (1 - 6)meas(D ) + c .n n

If 0 < 6 < 1, taking into account that

meas(Dn) < meas(An) + meas(Bn),

one would have

1 - A + En

LP-SPACES 253

1=n 1 lunI 6 d+(1-d) 1_S =1- (1-d)1-SX

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

will imply y < 1) and d = (1 + a)2 -S2

< 1. This is always

possible choosing a first and then a so small such that

d < 1.

SOLUTION: (b): With the same notations as in part (a),

lira u= limJ If I = 1.n

The inequalities

IJX nI < JXlunI < JXIfn1

imply

lima lu I = 1.n X n

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,

(1)

(2)

which, as n -> -, tends to 4 or 0 according as e = +1 or -1. From

this it follows that

254 CHAPTER 6: THE

X

Relations (2) and (3) allow us to use part (a), so

(3)

n-oJXIvnI = 0. (4)

But then, from the inequalities:

Il - fnI Il - IffII + (IfnI - fnI

J- - Ifn I I + IvnI + Ifn I - un

and from (1),(3) and (4), it follows that

rl-imm1X I 1 - fn I= 0.

EXERCISE 6.116: Let X be a measurable set of 32m and f a measur-

able function on X. As usual we write

I I f I I P = ( J Iflp)11p if 0 < p < .,X

IIfiIm = ess sup I f IX

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,

thenI I f I 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 . 0. Let A = (f 1).

SOLUTION: (a): If r < p < s and r e If, s e If, when s < m one has

fP < fs on A and fP < fr on x - A, which proves that fP is inte-

grable, and therefore that p e If. If s = -, fP << fr still holds

on X - A and furthermore fP < IIfIIp; since f is integrable we

have meas(A) < W and it again follows that f is integrable.

Assume that X = [0,W[. If f(x) = e-X, then If = ]0,W]. If

f(x) = n when n < x < n + 2-n (n = 0,1,... ) and 0 outside of

these intervals If = ]O,W[. If f = 1, If = (W}, and if f is

continuous, and f(x) v x-1/a

a > 0 when x + W, then If =] - a,W],

but If = [a,-] if f(x) v x-1/a(logx)-b with ab > 1. However, if

f is continuous on ]0,oo[, f(x) = 0 for x >. 1, and f(x) v x-1/a

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 func-tions 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

256 CHAPTER 6: THE

and

p=n+ (ls A) 0< At 1,

then:

llfllp < llfllrllflls-A (1)

If r,s are finite, it suffices to make h = fAp g =f(1-A)p

in

Holder's Inequality

JX

c ( J

X JXX X X

When 0 < r < s = -, one has p = r/A > r, whence

llfllp < IIf(r/A)-rllWI fr = IlfurAllfIIP(1-A) ,JX

which shows that (1) is still valid in this case.

By a classical property of convex functions, from this it fol-

lows that a '+ logllfll1/a is continuous on the interior of its de-

fining 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. Assume first thatr is finite. In this case fp i fr, and does so monotonically on

each of the sets A and X - A. From this it follows that

JAf p- I 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). This proves the result in this

case.

LP-SPACES 257

If r = m and 0 < Y < IIfllm then

meas(f > Y) > 0 and 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

IIfIIp < IIf IIl r-r/Pllfll/p,

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 .< (J f)A(( fs)1-AX X X

It suffices to make h =f1r, g = f(1-a)s

in Holder's Inequality:

rXJfg < ((X ?/A)

X

( fX

fl/(1-a))1-A

J

SOLUTION: (d) : The inequality is trivial if Hf Ilr = m or IIfAIS = °°.

Assume, therefore, that these two numbers are finite. If s is fin-

ite the inequality results from the convexity of logIlfIIp as a func-

tion 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

part (b) above. It is then clear that

LrnLS C Lp.

EXERCISE 6.117: The notations are the same as in the preceding

exercise. It is further assumed that meas(X) = 1.

(a) : Show that IIfIIr IIfil if 0 < r < s < -. Can one

have IIfIIr = IIfIIs < °°?

(b): 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

+

aI. Then

IIfglIr < IIfIISIIgIIa (1)

(cf., Exercise 6.106). On making g = 1 the desired inequality

is obtained. Since (1) is obtained by applying Holder's Inequal-

ity to fr and gr, equality can be had with g = 1 only if f is

equal to a constant almost everywhere.

SOLUTION: (b): By Jensen's Formula,

loglIfIl = 1 logJ fp " Jlogf.P P X X

LP-SPACES 259

On the other hand, since logu 6 u - 1,

1 logJ fP <1

[1 fp - 1] = J up - 1).X P X X

On A = (f > 1) p-1(fp - 1) decreases to logf, and on X - A it in-

creases to the same function. Consequently

limJ P-1 (fp - 1) = I logf,' X 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) =

260 CHAPTER 6: THE

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

ip(cx) = c,y(x) + (1 - e)V(0), 0 4 c 6 1, 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 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,

J 0(f) = 2(a2logu + b) + "(allogy + b).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 =1

(alogf + b)X X

EXERCISE 6.119: For any function * increasing on [l,m[, such

that 4(l) > 0 and limi(p) show that there exists on [0,1] ap-,measurable function f such that limilfilp = and. Il,fllp < *(p)

P-1-for all p.

400 = V AV = AVA = VAV - AVA

SOLUTION: Let L be the set of functions f, measurable on [0,1],

such that:

Ilfll<sup

1f 11°°IIfII,y =

It is clear that L is a vector space and that f - Ilfil is a

norm on this space. Since IIfiL 4 IIfiL. (cf., exercise 6.117)

it is clear thatLm

C L Everything reduces to proving that

L- # LV for then there exists f such that IIf IL = °°, IIfII < 1;by exercise 6.116 one has limllfIIp and on the other hand the

p'°°definition of IIfII implies that IIfIIp s ,y(p) for all p 1.

First, let us show that L provided with the norm f > IIf11 is

complete. For this, consider a sequence (fn) of elements of L

such that:

Y

E Ilfnli <n

262 CHAPTER 6: THE

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. Furthermore, if sN = fl + + fN,

Ilg - SOP < IlfnllP < *(p) nIN Ilfn%,

which proves that g - sN e L., hence that g e L.J. and furthermore

Ilg - SNIIC' E IIffII*,n>N


"M 119 - SN'I* = 0.N-

Assume that Lm = L W) = 1 can be assumed without any loss

of generality. Since

Ilfll = supIlf1IP ¢ Ilfi' p>.1 *'p

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, IIfIIP < IIflIW = A F (p) if p > p0,

and:

Ilfllp< IIfllP =lcVp) when l<p4p0.0

EXERCISE 6.120: (a) : Let p . 1 and f e LP(R). For all x e]R set

F(x) =If(t)dt.

0

Show that F(x + h) - F(x) = o(Ihll - 11p) as h - 0, uniformly

for x e1R.

(b): Show that if f is integrable on It, is continuously

differentiable, and if f' e LPOR) for some p >, 1, then

lim f(x) = 0.

IXI-

SOLUTION: (a): Let e > 0. There exists d > 0 such that meas(E)

< d implies

jE

If 0 < h < S, by Holder's Inequality, if p > 1 one has

x+hlf I p)1/P < eh1 - 1/p.IF(x + h) - F(x) I = If

x+fI h1 -1/P(fXx

264 CHAPTER 6: THE

SOLUTION: (b): By part (a) above, the function f is uniformly

continuous, and consequently f(x) - 0 as jxl -; W (cf., Exercise

3.37).

EXERCISE 6.121: Let X be a measurable set of IlRn such that 0 <

meas(X) < 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 (1

fP)1/P

X1

X

A00 - V AV - eve - VAV = OVA

SOLUTION: One can clearly assume that 1I fP < °D, as otherwise

there would be nothing to prove. Let Fn = min(F,n); it is clear

that (F > A) C (F > A) and that (F > A) _ 0 if A >, n. From

this it follows (cf., Exercise 5.98) that

( (nI Fn - p1 0

nAP-lmeas(F

> a)da

4<p(nAp-2da1

f(x)dxJ 0 (F>A)

( (min(n,F(x))= pJ f(x)dxl

aP-2da =1 f(x)Fn(x)p-ldx.

X 0 X

By Ht5lder's Inequality, this yields, if

p

tq

= 1

LP-SPACES

XFp < -1 )q)1/q.

1

fP)1/P((F np

J p - J J

265

Now (p - 1)q = p,JX n

< - since 0 t Fn 6 n and meas(X) < m, so

(J FP)1/P 6 1 (J

X

f

X n p - 1

P)1/P,

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;

(ii): There exists a constant M such that for every sequence

P1,...,Pn of disjoint rectangles that are of non-zero

measure

n1 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: If f e LP and

p

+

q= 1, then

if fIP 4 meas(P1

.)p/q( IfIp,P. JP.

1 1

and consequently, because p/q = p - 1

266 CHAPTER 6: THE

i 1meas(P.)p-11JP.fIp <

.=iJP.IfIp < IIfIIP (1)

Now assume that condition (*) holds. Let g = c IlP. be ai i

step function, with ci # 0, and the Pi being disjoint non-neglige-

able rectangles. Then

JfI = Ii ciJ flP.i

< (1 1 IJfIp)1/P(G IciI%ieas(Pi)q(p-l)/P)l/q

imeas(P.)p-1

P. i

M1/pq Icil'meas(Pi))1/g = M1/pIIQIIqi

By Exercise 6.109 we therefore have

IIfIIP < M.

Comparing (1) and (2) yields the last part of the problem.

(2)

EXERCISE 6.123: Denote by En the vector space of step functions

on ]R . Let T be a linear mapping from En into Lloc(l ). Assume

that for every f e En

liTfilgo < AollfllPOJ IITfllgl < A111fllp1 ,

where 1 < pi,gi < -, AO > 0, Al > 0.

Prove that if

1 1 - t t 1 1- t t0< t c 1,

P P0 + P1 ' q q0 + q19

LP-SPACES

for all feE

IITfIIq, AO-tA4IIfiIP.

267

(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. Let f e En and

geEn be such that IIfIIP = IIgIIq = 1, where

4

+ q = 1. The

function f may then be written:

U i8f = E r e kIl ,

k=1k

Ak

where rk > 0, 0k eR, and the Ak are disjoint rectangles of Rn;

similarly

v ipg = I PRe nB ,

R=1

with analogous conditions. If a > 0 and a < 1 then set

a(z)/a l0kbfz =

k

rk e Ak,

gP(1-B(z))/(1-B) 1Tp

zk

e IlBR

F(z) =AzAz

RmT(fz)gz.

0 1

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 Ge(z) = eEz F(z) to which

268 CHAPTER 6: THE

one can apply the maximum principle, deduce that IF(z)I S 1 if

0 5 Re(z) S 1. Next examine the case where all - 0) = 0).

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*gIlrIIfIIPpIIgI

q (cf., Exercise 6.106).

AVA = DAD = AVA = VAV = 0VA

SOLUTION: Note first of all that

a(t) = a, 0(t) = 0,

and that if y eat,

Rea(iy) = a0, Re6(iy) = 00,

Rea(1 + iy) = al, Res(1 + iy) = Sl.

(1)

(2)

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). Since

1-S(z))/(1-S)ekkT(IlAk),F(z) =

AlzAzkLk

rka (z)/ap(9

10 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 =

k

rk/ameas(Ak) = IIfAIP = 1,

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 = then Rea(iy) = 0 and

Ilfiyllm =maxlrk(iy)/al = 1).

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.

2If e > 0 and G(z) =

eezF(z), if 0 <, x <, 1, y e ]R, then

IG(x + iy)I =ee(x2-y2)IF(z)I

<e6(1-y2)M,

IG(iy)I : eE,

IG(1 + iy)l <, eE.

The first of these inequalities shows that there exists T such

that:

IG(x + iy)I < eE if 0 <, x _< 1, lyi 3 T.

On the perimeter of the rectangle with vertices ±iT,l ± iT the

modulus of G is bounded by eE, by the Maximum Principle this also

holds over the whole of this rectangle, and consequently over the

whole strip 0 <, Re(z) 1. In particular

IG(t)I = lee' F(t)l <, eE.

Making c -> 0, because ft = f, gt = g by Equation (1), yields

270 CHAPTER 6: THE

IF(t)I = 1AltA

0

JT(f)gl : 1.

By exercise 6.109 this proves that:

IIT(f)IIq <,

Al-tAt

when f e En and IIf1I P = 1. By homogeneity, for f e En

IIT(f)IIq , A1-tA4IIfIIP

Now examine the cases where a(l - B) = 0. In the first place,

if a = 0 one has, for example, a0 = 0; if a1 > 0 then t = 0, and

there is nothing to prove; if a0 = a1 = 0 then

IITfIIq0

< A011fli",. IITfIIq1

-<

As the function a N logIIfII1/ is convex on the interval with end-

points B01B1 (cf., Exercise 6.116),

IITfIIq , AO-tAlIIfIIW.

Finally, when B = 1 one can restrict oneself to examining the

case where B0 = B1 = 1 and a > 0; the function:

F(z) =Al

zAzJRmT(fz)g0 1

is again entire and bounded on the strip 0 <, Re(z) <, 1, and

IF(ib)I < a II T(fiy)II1IIg1Im . 1,0

LP-SPACES

IF(1 + iy)I < A IIT(f1+iy)II11I911 < 1,1

and everything finishes as above.

Let g e Lq, f be a step function, and set

Tg(f) = fig.

If 1 + -r = 1 thenq q

IIT9(f)II , I191IgIIfIIgi

IIT9(f) II q s 119II gIIfII1.

Then if

1 t 1 1- tr=q P 4 +t, O,t 1,

we have

IIT9(f)IIr = IIf*gll . IIflIpII91Iq

271

(3)

(4)

The relations (3) are equivalent to

r

=

p

+

q

- 1 >, 0. Now assume

that f e Lp, g e Lg, f 0, g , 0, 1 : p < There exists a se-

quence (fn) of positive step functions such that

fn- f almost everywhere,

IIf-fnllpi0

As the space Lr is complete one deduces from (4) the existence

of a function h e Lr such that f *g -; h in Lr. Also

Ilhllr ` IIfIIpII9IIq.

272 CHAPTER 6: THE

By considering a subsequence, it can be assumed that

fn*g '+ h almost everywhere.

By Fatou's Lemma, for almost all x,

Jf(y)g(s - y)dy , liminfJfn(y)g(x - y)dy = h(x) <

From this it follows that f and g are convolvable and that f*g < h

almost everywhere, whence

Ilf*gllr< IIhIIr < Ilfllpllgllq

In the general case it is noted that f and g are convolvable if

and only if lfl and Igl are, and in this case if*gl , lfl"lgl.

Finally, if p = m, one necessarily has q = 1, r = m, and in this

case the result is classical.

EXERCISE 6.124: Let X C mn, Y C Iltn be measurable sets such that

meas(X) > 0, meas(Y) > 0. If f is a function on X and A > 0, set

f(x) if lf(x)l < X,

fi(x) _Af(x)lf(x)l-1 if lf(x)l a A,

fA = f - fa

Furthermore, let 1 , p0 < p1 , w and E be a vector subspace of

Lp0(X)C Lp1(X) such that f e E implies that f' e E for A > 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 (1)

for any f e E, g e E.

LP-SPACES 273

(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 Pimeas(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

IITfll0 < f e E. (21)

(c): Assume in addition that meas(Y) < W and that if f e E,

A > 0, and

f(x) if I f(x)I A,fi(x) _

0 otherwise,

then fA e E. Set f = f

Show that if 1 = p0 < p1 t m and (2) is satisfied (and (2') if

p1 = co), then if C A)da, JP_lmeas(ITfI > A)da,

a

274 CHAPTER 6: THE

then make a = 11fII1)

(d): 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 (5)

(Use the decomposition f =VT

+ , then evaluate the inte-

grals:

f

1 rmmeas(ITfI > A)da,

f

meas(ITfI > A)da.0 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), W) = meas(ITfI > A).

Since JTfI 5 ITf + ITffI by (1), it is clear that

(ITfl > A)C (IT?i > 2A)U(ITfXI > jX),

and consequently:

LP-SPACES

W) <, meas(ITfAI > 2A) + meas(ITfAI > IX)

Furthermore, by (2)

2A0IIfAIIP PO

meas(IT? I > a

O,

111faIIPP1

meas(ITfXI > A)1

From (6),(7) and Exercise 5.98 it follows that

IITfIIP =0

p(2A0)PO0aP-PO dal If(x)IPOdxX

(W

+ p(2A1)Pll 1P P11

daj IfA(x)IPIdx.0 X

(IfI > t) if 0 < t < 1,

0 if t 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

275

(6)

(7)

(8)

(III>t)_(IfI>t+A).

276

Thus

meas(I? I > t) = q(t + A),

-meas(IfA I > t)

(p(t) if0<t<A,0 ift'> A.

CHAPTER 6: THE

Using Exercise 5.98 again to evaluate the integrals

andJIfAIP1,

it follows from (8) and (9) that

i p- 1 M P-1IITfIIP 5 PPO(2A0)

OJ 0dAJ t 0 q(t + A)dt

JO o

(9)

J 1' I pO

p1 W p-pl-1 A p1-1+ ppl(2A

J A dAJ t p(t)dt. (10)

0 0

Since p - p0 - 1 > -i and t - A 5 t,

JoAp-pO-1d)LJot1) 0-1c(t

+ A)dt 5

Ap PO1dA= WtPO

1q,(t)dt

tJ

0 Jo

= 1 JtP_1(t)dtp-p00

1) IIfIIP.=p(p-P0

Similarly, because p - p1 - 1 < -1,

JAP-pl-1dA(otPl-1q(t)dt

= J- tpl-1,(t)dtJtAp-pl-idA

0

=

LP-SPACES

1 tP-1cp(t)dt

pl pJO

p pl-Z- l )-llfllP.

(10),(11) and (12) yield

p0 PiP0(2A 0)

IITfIIP . I P - P+ pp,lp ) IIflIP.

0 1

which proves (3).

SOLUTION: (b): When p1 = ' and (2') holds, write

277

(12)

A/2Ameas(ITfI > A) -,-c meas(ITf mI > -)L) + meas(ITfa/2AWI

> #L)'

Now the set (ITf1AAml > A/2) has measure zero, since almost every-

Tfa/2A ..-1 < A. 11fa/2A_I1- . 2


IITfIIP : 'A)dX0

P0W ppO1 m p0-1 A/2A.

pOp(2AO) J a dal t meas(If I > t)dt

0 0

p0p(2A0)poJXp-pO-1 dXJ0tP0

l( P

+ 2Am)dtO=0 o Ilt

(Contd)

278 CHAPTER 6: THE

(Contd) , pop(2A0)p0

1 A

p-p0-1

dxJW t

p0- 1

Q(t)dt

o A/2AW

p0(°° p0-1 (2A t p-p0-1= pop(24 ) J t c(t)dtJ A dA

0 0

p_

po 0

- p0(2A0)P0(2Am)P-P0HA P.

P p0 P

SOLUTION: (c): It can be assumed that a = IIfIIl > 0. It can

also be assumed that 1 = p0 < pl < -; indeed, if p1 and

if p0 < pi < by part (b)

Pmeas(ITfI > A) S lr ITfI

xPl ITfI>A

IITfIIP1P1 Pl APl- S

P;a

This being so,

J0PAP1AdAa 1

meas(Y)J padA = ameas(Y).0

On the other hand

(13)

Vi(a) 4 meas(IT? I > 2A) + meas(IT." I> 'fA) 5

LP-SPACES 279

2A0I II 12A1 Il a II Pi Pi

whence:

jaa

pAP-1iP(A)da 2pA0JaXp-2dXjIf(x)I>aIf(x)Idx

Pi °° P-P1 1 P1+ p(2A1) 1 A dA J If(x) I dx

a

1

In the last integral If(x)Ipi 4 API- If(x)I, and consequently

JpAPhlp(A)dX < 2pA0 J If(x)Idxjap-2da

a X a<A<If(x)I

+ p(2A1)PhIIf II1 FAP-2da

a

1

2pA

pj

If(x)I(aP-1 _ If(x)IP-1)+dx

X

+ ill )P1

Pap-1

II f 111.

Since (aP-1 - If(x)IP-1)+ ¢ aP-1, and since a = IIfIIi, this yields

P

1p[2A0 + (2AI) 1]

pi

(14)j PaP ,y (a)da : 1 - p IIfIIa

and, taking (13),(14) into account,

280

pl

IITfIIp <, {meascY) + - 1(

pl)

}HAP.

SOLUTION: (d): On one hand

1

J *(A)da : meas(Y),0

and on the other

V00 < meas(ITf TI > A) + meas(ITffi > 2

A+

so

[2AiII./rIIJP1A

2AOJW A I f(x)Idx1 If(x)I>,T

+ (2A )P1 daIf(x)Ipldx.

l1 xp1fIf(x)I5y1-x-

-1)/2In the last integral

If(x)IP1

5 A(P1

If(x)I, whence

JP(A)dA : 2A 1 If(x)IdxJ a-1da1 X 1<A<lf(x)I2

+ (2A1 X-(pi +1)/2

dA

CHAPTER 6: THE

(15)

2(2A)p1

_ oj

If(x)Ilog+I f(x)I2dx +p

1 1 11f1I1. (16)x 1

LP-SPACES

If M is such that u 5 M(1 + ulog+u), then

1 1 f 111 < M(meas(X) + JIfhloghfh). (17)

Furthermore, since log+u2 - 21og+u it follows from (15),(16) and

(17) that

I I Tf1I1 5 B +CJX

Ifllog+IfI,

whence:

B = meas(Y) +2M(2A1)

Pi

P - 1 meas(X),1

2M(2A1)p1C=4Ao+ pi - 1

EXERCISE 6.125: Set

u =J 1 i f x < e,

logx if x >, e,

(a): If f and g are two positive measurable functions such

that

Jf2u(f) <

Go,

J 2

u(g)

show that fg is integrable.

<

(b): Assume that f is as above and that ( gn

) is a sequence

of positive measurable functions such that:

282

2

= 0 .l imJ u (9n)n

Show that

limJ fgn = 0.n

CHAPTER 6: THE

(c): Now assume that g is as in part (a), and that

lnmJ fnu(fn) = 0.

Show that

limJfng = 0.n

1VA = VOV = AVA = VLV = OVA

SOLUTION: (a): Let us show that for x > 0, y >. 0,

2

xy <, 2x2u(x) + u(y) (1)

which will show that fg is integrable if f2u(f) and g2/u(g) are.

If x % y/u(y), then xy < 2y /u(y), and (1) holds. Assume, there-

fore, that y < xu(y), If y < e then y < x, and consequently xy <

x2 6 x2u(x); if y 3 e, then noticing that logu 6 u/e, one obtains

xy2"

<`< 2e logy è mi x,

and consequently logy < 2logx, whence:

xy 4 x2logy 5 2x2logx <, 2x2u(x).

LP-SPACES 283

SOLUTION: (b): Let 0 < r < 1. By replacing x by rf and y by gn

in (1), integrating and noting that u(rf) 5 u(f), one obtains

2( ((

Jfgn 5 2+2u(f) +

9

rju(gn

so

limsupJfgn 2rJf2u(f),

and on making r 0

lim1fgn = 0.n

SOLUTION: (c): Now replace x by fn and y by r-1g in (1). This

yields

(

Jfg 2r1fnu(fn) + rl 92

11u(r g)

so

2

limsup gn jf g 1

n r u(r-1g)

If r > 1

((2 22

X22 logg1u( = Jgre

U-`-gu(g) + Jg>re u(g) logg - logr

5 log(re)

(2)

and consequently:

284 CHAPTER 6: THE LP-SPACES

2

lim 1g = 0,r-).. rJ u (r-lg )

which proves, because of (2), that

limJf g = 0.n

CHAPTER 7

The Space L2

EXERCISE 7.126: Let X be a set of ]R such that meas(X) = 1 and

Alp...$An a finite sequence of measurable sets of X such that:

meas(Ai) A c > 0 for all i.

Show that if

nc3e>,C(1-c)

n - 1

there exist two indices i,j (i < j) such that

meas(AiflAi ) 3 c2 - E.

ovo = vov = ovo = VAV = ovo

SOLUTION: Applying the Cauchy-Schwarz inequality to the function

285

286 CHAPTER 7

and setting

nA = meas(Ai)

i=1

yields

A2 < A + 2 E meas(Aini<j

If one had

meas(Aif)Ai ) < c2 - e

for every pair i 4 j, taking into account that A2 - A is increas-

ing for A >, 1, and that A >, no >, 1, one would have

n2c2- no < A2 - A < 2 .n(n 2 1) (c2 - e)

so

< c(1 - c)n - 1 '

contrary to the hypothesis.

EXERCISE 7.127: Let f be a piecewise continuously differentiable

function on the interval [0,a].

Show that if f(O) = 0, then

I tf'(t)f(t)Idt < JIf(t)I2dt.0 0

When does equality hold?

00A=Vt0=MA=VAV-MA

THE SPACE L2

SOLUTION: Let

t

u(t) = J lf'(8)lde.0

Now,

If(t)I 6 u(t) and u'(t) = If'(t)I.

Consequently,

J0'(t)f(t)kt f0u'(t)u(t)dt = u2(a).

Furthermore,

alf'(t)Idt2

u2(a) = IJoJt

< J dtJ If"(t)I2dt = aJalf'(t)I2dt,0 0 0

287

whence the result.

In order to have equality it is necessary first of all that

Ifl = u, that is to say

IJ0f'(s)d8I = J0If'(s)Ids,

which implies that f' = elalf'j, where a is a real constant.

Next, in order that the Cauchy-Schwarz inequality become an equal-

ity, it is necessary that If'I be constant. Taking into account

that f(O) = 0, it follows that f(t) = At, X e (E.

288 CHAPTER 7:

EXERCISE 7.128: Let X be a measurable set of]RP such that

meas(X) < - and (fn) a sequence of measurable functions on X such

that

IffI.< M,

JX Ifnl2= 1 for all n.

(a): Show that if a sequence of complex numbers (an) is

such that the series E anfn(x) converges for almost all x e X,n

then n'n n = 0.

(b): Show that there exists a sequence (bn) of complex num-

bers such that

1 mbn = 0 and I lbnfn(x)!2 =n

for all x's belonging to a subset of X having strictly positive

measure.

ovo - vev = ovo = VtV = AVA

SOLUTION: We shall assume that meas(X) = 1.

SOLUTION (a): If the series E anfn(x) converges almost everywhere,

in particular one has anfn(x) 3 0 almost everywhere. Let d > 0

be such that 1 - M26 > 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

Xlanfnl2

= fAlanfnl2 +

JX_A'fnl2 s (Contd)

289

.< en + M21an126,

2

2 Enlanl 2

1 - M d

which proves that an - 0.

SOLUTION (b): Choose a sequence (bn) such that bn -+ 0 and

I (for example, bn = 1//n). Assume that E Ibnfn(x)I2n0

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) >, X lb12J

If 12 % (1 - M26) E lb i2,n>,n0 n A n n,n0 n1

which is absurd.

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,<i,j,<n'

Prove that:

G(f1,...,fn) = n!JUnIdetll.fi(xj)II1<i,j.<n12dx1dx2...dxn.

AVp = VAV = AVA = VAV = AVA

SOLUTION: We have

nIdetllfi(xj)IIl2 = £(To) f fT(i)(xi)fo(i ),

T,a i=1

where T and a run over the permutation group Of {1,2,...,n) and

£ denotes the signature. On setting

a1. = (filfj) = jUfi.'j,

the right side of the formula to be proved becomes:

n n2n L £(Ta)IT

aT(i) a(i) X e(Ta ) II aT6(1) a(i),

T,a i=1 T,a,

i=1

C Cn

= nt G G e(T)T1

aa T i=1

= det llaij iI

THE SPACE L2 291

EXERCISE 7.130: Let K be a function defined on 3t2 that is meas-

urable and such that

A2=

JJ 21K(x,y)I2dxdy <

For every function f e L2(3t) the function Tf is defined by

Tf(x) = JK(x,Y)f(.)dy. (*)

Show that this defines a linear transformation of L2(Xt) into

itself, and that

IITf112 , A ll!112 (**)

Show next that if T' is defined analogously with kernel K',

then T'T has kernel K", where

K"(x,y) = JK'(x,z)K(zY)dz.

AVL = VAV = AVA = VAV = AVA

SOLUTION: By Fubini's theorem, for almost all x the function

y - K(x,y) is square summable, and consequently Tf(x) is defined.

By the Cauchy-Schwarz inequality

ITf(x)12 I1f11 2f IK(x,y)I2dy.

Integrating with respect to x yields (**). Nevertheless, it re-

mains to prove that Tf is measurable. Let E be a subset of 3t

with finite measure. Then

fdxf'ILE(x)K(x,y)f(y)Idy , 'V'12JE dx(flK(x,y)I2dy)2 <

292

because by Fubini's Theorem the function

X * (IK(x,y)I2dy)i

CHAPTER 7:

is square integrable. The same Theorem implies that the function

x y ILE(x)Tf(x) = J1EK(x.Y)f(Y)dy

is measurable. From this it follows that Tf is measurable.

The last formula results from the following calculations:

T'Tf(x) = JX'(S.z)dzJK(z.v)f(Y)dY

= Jf(Y)dYJX'(xz)K(zY)dz.

and

J

K-(x,y)I2dxdy Jdxd(JIK'(x.z) 12dz)(JIK(z,Y)12d5)

= JIK(x,2)I2dsdzJIX(z.Y)l2dzdY.

EXERCISE 7.131: If f e L2(0,1), for 0 4 x< 1 set

Wf(x) = f(x) +21xex-tf(t)dt.

0(*)

Also let

91(x) = e-x, 92(x) = ex.

Show that (*) defines a linear mapping W of L2(0,1) into it-

self, such that:

THE SPACE L2_ 293

(a): Wc1 = 92;

(b) : If (f l (p 1) = 0, then (W fc'2

) = 0 and DDWf 112= Ilf 112;

(c): If (gIc'2) = 0 there exists f e L2(0,1) such that:

(f1q,1) = 0 and g = Wf.

Deduce from this that W is a bijection.

ovo = vav-= ove = vov - ovo

SOLUTION (a): We have

xe-x + 21 ex-2tdt

0

e-x

-

ex-2t t=x= ex = W.t=o 2

SOLUTION (b): Set W = 1 + U, and determine U*:

X(U*fih) = (fIUh) = 2J0f(x)dxlex-t7htldt

0 0

2J1 TFtTdtf

0 t

Hence

U*f(x) =211et-xf(t)dt.

X

Now calculate U*Uf when

294

1

(flkl) = J e-tf(t)dt = 0:

0

U*Uf(x) =41et-xdtJtet-uf(u)du

Jx 0

X Ixe- (u+x4e f(u)du0

x J2tdtx

+ 411e-(u+x)f(u)duJ1e2tdt

x u

1= 2e2-x a uf(u)du

JO

(1

-2xex-uf(u)du

- 2Jeu-xf(u)du

J0 x

= - Uf(x) - U*f(x).

Thus, whenever (f191) - 0,

W*Wf = (1 + U + U* + U*U) f = f .


(Wfl92) = (Wf,Wq,1) = (W*Wfl(p1) = (fl(vl) = 0,

and

CHAPTER 7:

IIWfI12 = (Wflwf) = (w*wflf) = (flf) = IIfI12

THE SPACE L2

SOLUTION (c): Now assume that (gl(P2) = 0. Then

UU*g(x) =4rxex-tdtJ0

X rleu-tg(u)dut

= 4 rxex+ug(u)du

JO

Te-2tdt

lx+u rx

+ 4 e g(u)duJ

a-2tdtx o

(1 x 1

2exI eug(u)du -2ex-ug(u)du

- 2jeu-xg(u)du

in Jr

= - Ug(x) - U*g(x).

295

Consequently WW*g = g when (glp2) = 0. Setting f = W*g, one has

g = Wf, and furthermore

(flml) =(W*gl(pl)

=(g,Wml) _ (gl(p2) = 0.

Thus W is an isometry of{(p l}L onto

{m2}l. From this it certain-

ly follows that W is bijective. In fact, if

(fI9l) = 0 and W(f + X91) = W f + X(p2 = 0,

then

Wf=0 and a=0,

296

and consequently

f + ail = 0

because

Ilf112= IIWf112=0.

Note that if g e L2(0,1) then

(gk(P2) (gIT2)p2.g = 9

O2 (p 2 + (2I2)

Setting

(g1ro2 (g1W2)f = W (g - T(p 2 ,p 2 W2} +

'p2 (p2W1,

we have Wf = g. Notice also that W*92 = X(p1

and

A(c?11(P1) _ (W*(p 21p1) = (921W91) = (cp2I 2),

so that

f = W"9 - l(W11W1 - P2I42 )J cP2)Tl.

Now,

2 2

(c) llcpl) - e2 , ('21 2) - e 2 1

2e

whence

f=W*g-2(glcp2)cpl.

This relation may be further written as

CHAPTER 7:

THE SPACE L2

f(x) = g(x) -2Jxet-xg(t)dt,

0

which is the "inversion formula" of:

g(x) = f(x) + 2Jxex-tf(t)dt.J0

297

EXERCISE 7.132: Let f be a continuously differentiable function

on 3R such that

Jx2ffxI2dx < Jff(x)12dx <

(a): Show that for x > 0

xlf(x)I2 F Lt2t)l2dt)Lh1(t)I2dt)

..J)I2cx 2(J+ x2lf(x)I2dx)z(J+WIf,(x)I2dx),

(This inequality is Heisenberg's Uncertainty Principle in Quantum

Mechanics).

evo = vov - eve = vev = ovo

SOLUTION (a): If x 3 0 let f0 = f(x) , and

a2 = dt, a2 = JWIf'(t)I2dt.x x

298 CHAPTER 7:

For all t >. 0,

f

x+tf(x + t) - f(x) = f'(u)du,

x

so, by the Cauchy-Schwarz inequality,

I f(x + t) - f(x) I < s,T.

Assume that 0 .< h .<2

0/S so that for 0 .< t < h

If(x+t)I :f0 - 5r:0.

Then it is clear that

a>x/(f0-B1)

Choose the value of V that maximizes the right side of this in-

equality, i.e. i = f0/2S. This yields

xf2 .< 4as,

which is the desired inequality.

SOLUTION (b): From the first part (since an analogous inequality

holds when x < 0) it follows that

lim xIf(x)I2 = 0.IxI 4M

But then, by integrating by parts

tW +If(x)I2dx = - x(f'(x)f(x) + f(x)f'(x))dx,

whence, using the Cauchy-Schwarz inequality,

THE SPACE L2

JIf(x)12dx+W

. 2J+w

x I f I I f' Iw

2(J+

x2If12)'(J+WIf'I2)'.co

299

EXERCISE 7-133: (I): Let X be a measurable set of]Rp such that

0 < meas(X) < and (fn)n>1 an orthonormal sequence of L2(X).

For every integer N > 1 set

N

KN(x,y) = G fn(x)fn(7,n=1

LN(x) = fX IKN(x,y)Idy.

(a): Assume that all the functions fn are real and that

there exists a constant A such that ILN(x)I < A for every integer

N>, 1 and all xeX.

Show that if (an)n31 is a sequence of complex numbers such

that I Ian12 < -, and that if

NsN(x) =

C

G anfn(x),n=1

then for almost all x e X

supIsN(x)I < m.N

(Assume first that the an's are real and introduce a measurable

mapping v of X into {l,2,...,N} such that

300 CHAPTER 7:

sN(x) = maxs(x) = s()(x),ncN

nV

1

sN(x)dxX

is bounded by a constant).

(b): Take the same hypotheses and notations as in part (a).

Show that if

.f L anfnn=1

in the sense of convergence in L2(X), then for almost all x e X

W

f(x) = I anfn(x).n=1

(Introduce an increasing sequence (2n)n31 of numbers greater than

zero, such that wn -> and X mnjanI < m, and a sequence of inte-

gers N1 < N2 < < Nk < such that sN (x) -> f(x) almostk

everywhere; next bound IsN(x) - sN (x)I when Nk <N < Nk+l' using

part (a) applied to wnan).

(c): Assume that X = [0,1], and that for every function f

continuous on X

Nlim G (flfn)fn(x) = f(x)N n=1

uniformly for x e X.

THE SPACE L2 301

Show that there exists a constant A such that ILN(x)I S A

for all N and all x.

(II): We call the HAAR SYSTEM the sequence of functions

(fn)n>0 defined on X = [0,1] in the following way: f0 = 1, and

if n31 and n = 2m + k, 0<k<2m, 04X<1,

2m/2if

2k < 2k + 1<2m+1 2m+1

m/2 2k + 1 2k + 2fn(x) 2 if 2m+1 x < 2m+1

0 otherwise

Assume in addition that

fn(1) = fn(1 - 0).

(a): Show that (fn)n>O is an orthonormal system in L2(X).

(b): Show that for all n > 0 there exists a sequence:

n n0=x0<xl<... <xn+1

which possesses the following property: In order that a function

f belongs to the vector subspace Vn generated by f0,f1,...1fn, it

is necessary and sufficient that f is constant on each of the in-n n

tervals [xi 'xi+1[, 0 < i < n, and that f(l) = f(1 - 0).

From this deduce that if f e L2(X), its orthogonal projection

on nV is, on each interval [xi,X +1[, equal to:

nX

1 J1+1f(x)dx.n n

xi+l - xi xi

302 CHAPTER 7:

(c): From this deduce that if f is continuous on X then

f(x) = E (flfn)fn(x)n=0

(1)

uniformly on X. In particular, (fn)n>o is an orthonormal basis

of L2(X). Using parts (I)(c) and (I)(b), deduce from this that

for every function f e L2(X) the relation (1) is satisfied for

almost all x e X.

(III): The RADEMACHER FUNCTIONS are the functions defined

on X by the formulae

rn(x) =sgn(sin2n+1%x),

n 0.

(a): Show that if 0 4 n1 < n2 < ... < n, then

J

1

0

(x)dx = 0.

p

In particular, (rn)n,0 is an orthonormal system in L2(X).

2

(b): Let (an)n30 be a sequence of real numbers such that

an < - and f = anrn in L 2W.

Show that for almost all x e X

W

f(x) = E anrn(x).

n=0

(Express the Rademacher functions in terms of the Haar functions,

and use part (II)(c)).

(c): With an and f as above, show that for all p > 0

Ap'If112 4 IIf11p < BPIIf112,

THE SPACE L2 303

where

Ap = min(1,2-(2 p)/p), BP = 11 + Ply

(For the second inequality assume first that p = 2k, k an inte-

ger; for the first, if 0 < p < 2 write 2 = Ap + 4(1 - 1),

0 < A < 1).Deduce from this that for every real number p

f1

0

exp(ulf(x)l2)dx <

(d): Show that the functions 1 and r r r (0 6 n1

<nl n2 np

n2 < < np) form an orthonormal basis of L2 (X) (the WALSCH

BASIS).

ADA - VAV = AVd - V AV - eve

SOLUTION (I):(a): One can clearly reduce to the case where the

an's are real. Let f= E anfn in L2(X). Then an = (flfn), so

aN(x) = JXxN(XIY)f(Y)dY.

For 0 4 n 4 N let

An = {x:a (x) < a(x) if 0 t p < n and s(x) 4 s(x)

if n 4 p 4 N}.

The sets An are measurable, mutually disjoint, and have union X.

Setting v(x) = n when x e An, one has

8h(x) = maxs(x) = sv(x)(a,)n6N

n

=JXxv(x)(x,p)f(p)dy.

304 CHAPTER 7:

Note that

N

G IlA (x)Kn(x,y),Kv(x)(x,y) =

C

n=0 n

and that consequently this function is square integrable on X x X.

This remark allows us to justify the various interchanges in the

orders of the integration carried out in the calculations which

follow:

JxsNz"(x)dx = fxf(Y)dyfxKV(x)(x,y)dx

{Jx

dy(1x

xv(x)(x,y)dx)2}Z.

Now,

Jx dy(JXKv(x)(x,y)dx)2

= JJxxxdadRfxKv(a)(a'y)KVW

(S,y)dy,

and furthermore,

v( a) v(R) (

v(a)(a'y)Kv(B)(S'y)dy = I fr(a)fs(R)J fr(y)fs(y)dy.X r=0 s=0 X

Taking into account the orthonormality of the fn's, this yields

jXdy( JIX Kv(x)(x,y)dx)2= JJ v(a)cvKv(a)(a,$)dadR

()

+ ffvKv(S)(a,$)dad6.

(s)<v(a)

The first integral of the right side is bounded in modulus by

.< Ameas(X).J XdaJ

w(a)tw(S)IK v(a)(a,s)Ida c JX L

v (a)(a)

THE SPACE L2 305

Since KV(S)(a,s) = KV(S)(R,a) the second integral is bounded by

the same quantity, so

J8(z)dx 4 11f112 2Ameas x . (2)

The functions sN are integrable, as the sN are, and the maximum

of a finite number of integrable functions is also; further,

aN 4 8* and sN(x) > snpsn(x). The Lebesgue Monotonic Conver-

gence Theorem and inequality (2) show that Wan(x) is integrable,

and consequently that sups (x) < - for almost all X. Replacingn n

f by -f one deduces from this that snp(-sn(x)) < -, that is to

say iRfsn(x) > -m, and therefore that snpIsn(x)l < - almost every-

where.

SOLUTION (I):(b): Here again it may be assumed that the an's are

real. Let n1 < n2 < be a sequence of integers such that

2a- 4 k-4

n>nk

Set n0 = 0, wn = l if 0 << n < nl, and wn = k if nk 6 n < nk+l andk> 1. Then

w2a2

= G a2 + k2 a2n=0 n n n<nl n k=1 nk4n<nk+l

n

CM

a2 +E

k-2 < °°n<n1 n k=1

Moreover, it is clear that the sequence (wn) is increasing and

that wn -> m. Since 8N f in L2(X) there exists a subsequence

sN such thatk

lkm aN (x)= f(x) almost everywhere. (3)

k

306

By part (I)(a), for almost all x

NM = supl I wnanfn(x)I <

N n=0(4)

If x is such that (3) and (4) are satisfied, and if Nk 4 N< Nk+i'

Abel's bound gives:

18N(X) - sN (x)I = I E wnlwnanfn(x)Ik Nk<n4N

b 2M w-1

X Nk'

which shows that sN(x) - f(x).

SOLUTION (I):(c): Let C(X) be the space of continuous functions

on X, with the norm of uniform convergence. If f e C(X) let us

set:

N 1

3N(f;x) = E (flfn)fn(x) =n=1

JKN(x,Y)f(Y)dY.0

For fixed N and x the mapping f - aN(f;x) is linear; denote it by

This map is continuous and its norm is

1sup{ 1 JKN(x,y)f(y)dyI : feC(X), 11f11_ < 1}0

= fIXNXYIdY = LN(x).0

CHAPTER 7:

(This flows from exercise 6.108 on observing that one can replace

the space of step functions by the space of continuous functions).

Assume that

THE SPACE L2 307

supLN(x)N,x

Then there would exist a sequence (xk) of points in X and a se-

quence (Nk) of integers such that

LNk(xk) = IIBNk(.;xk)II -, °°

By the Banach-Steinhaus theorem there would exist f e C(I) such

that

SUP ISNk

(f;xk) I = °D

But this is impossible, because by hypothesis sN(f;x) -+ f(x) uni-

formly in x.

SOLUTION (II):(a): It is clear that

f

1 2f=

1, n30.0 n

Furthermore, if 0 4 n1 < n2 it can be seen without difficulty

that fnlfn

= ofn2, with c = 0,1, or -1, and consequently

J

1f01

fnl fn2 fn2

= 0.0

SOLUTION (II):(b): The result is obvious if n = 0. When n =

2m + k, 0 4 k < 2m, let us consider the end pointsxn

(0 6 i

n + 1) of the intervals obtained by putting end to end, starting

at 0, 2k + 2 intervals of length 2-(m+1)..

then 2m - k - 1 inter-

308 CHAPTER 7:

vals of length 2-m (observe that (2k + 2)2-(m+l) + (2m - k - )2-m

= 1). Let 012'" 0n+1 be the characteristic functions of the

intervals [O,xi[,[x1,x2[,...,[xn,xn+i]' It is clear that each

function fi, 0 - i - n, is a linear combination of the 1 6 j

4 n + 1; since each of these systems of functions is linearly in-

dependent and they have the same cardinality, they generate the

same vector subspace n, which proves the first part of the ques-

tion.

If f e L2(X), let

ni+Ci

hj=i

be its orthogonal projection onto Vn. Then

J

1

(f - h)$n = 0, 1 5 i 4 n+ 1,0

andn,n

= 0 if i 4 j, while

n

iA. f(x)dx.

xi - xi-1 xi+1

SOLUTION (II):(c): If f is continuous on X:

N

sN(x) = E (flfn)fn(W)n=0

is its orthogonal projection onto Vn, and consequently if N(x)

1, 210 4 N <210+1,

and

em = sup{If(x) - f(y)I:Ix - yI .<2-m},

THE SPACE Lz

then

Nrx .

(f(x) - f(y))dyIf (x) - sN(x)I= I N 1 N

NSx - x x.

i-1 i-1

which proves that in this case

f(x) = E (fIfn)fn(x)n=0

309

uniformly in x.

As the continuous functions are dense in L2(X), and as every

continuous function is the uniform limit, and hence also the lim-

it in L2(X), of its orthogonal projections onto Vn, it follows

that the fn form an orthonormal basis of L2(X). Finally, by

parts (I)(c) and (I)(b) formula (1) holds almost everywhere when-

ever f e L2(X)

REMARK: This last result can be proved independently of (I) by

using part (II)(b) and Lebesgue's differentiation theorem (cf.,

Chapter 9). It will be noted that as the linear mapping

nf - I (f l fi )fi

Z=0

is continuous from L1(X) into V , and as L2(X) is dense in L1(X),

n n+1

(flfi)fi =1ai(f)o'

02= 2=

with

n

A.(f) __n

Mnjxn f(x)dx

i

310 CHAPTER 7:

1still holds for f e L W. This and Lebesgue's theorem on differ-

entiation show that in this case (1) is still true almost every-

where.

SOLUTION (III):(a): Observe that rn can be considered as a func-

tion defined on ]R with period 2-n. If nl < n2 < < np the

-n

function rn rn rn has period 2 1, and consequently1 2 p

--n1

1 n1 2Ornlrn2...rn = 2

J

rnlrn2...rn

P 0 p

-(n1+1)

J2_(:l+l)rfl2.

-n1

=2n1 l112

r_ r0 2 np np

2

-n -(n +1) n -n -1 -nSince rn rn has period 2 2, and since 2

1= 2

2 12 2

2 pthe term in parentheses equals zero. In particular, r and r

n n2

are orthogonal if n1 f n2, and as r2 = 1 almost everywhere the

system (rn) is certainly orthonormal.

SOLUTION (III):(b): It is clear that

r = 2n/2n

It follows that

L f..

2n,i<2n+1

N

n0ann

THE SPACE L2 311

is the orthogonal projection of f onto V2N+1_by part (II)(c)

for almost all x

Nlim E an n(x) = f(x).N+°' n=0

SOLUTION (III):(c): If k A 1 is an integer,

N2k (2k)! 0... ON 00... ONC

I n=0anrn l

= 00+ .. +0n_ 2kB0! ... a ! a0 aN r0 rN

Taking into account that rn = 1 and part (III)(a), this yields

f01

N 2k = (2k)!250

2SNI InO an nI

C

00+..+SN_k 200 ,... 20N ! a0

..aN

Now note that

N( E an)k = x

k! 200 2RN

i... ! a0 ...aN

0

and that for every integer n >, 0, 2n 4 (2n)!/n! .< 2nnn, so that

(2k)!00l...0N!

2kkk kTi- (2a 0 !... 2BN)! a0+-+aN = k

2

It follows that

By the preceding part, the function under the summation sign tends

almost everywhere to f2k as N -' . Fatou's Lemma then gives

312

i

IIf112k < k211f112-

CHAPTER 7:

(5)

When p > 0 consider the integer k 3 1 such that 2k - 2 < p <

2k; by exercise 6.117(a)

11f 11P

< IIflI2k < k211f11 2< [1 + P2-]2 I1f112

which proves the first inequality.

If p 3 2 then Ilf112 < IlfIlp (Exercise 6.117(a)); when 0 < p < 2there exists 0 < A < 1 such that 2 = Ap + 4(1 - A), and then (cf.,

Exercise 6.116(c))

1[f 112ilfliPplIfllw(1-A)

But by (5)

11f114 < 2' 11f1121

so

2-2(1-A) IlfIl2A-2 $ IlfIIPp

Since

4A-2=Ap and 1-A=2A 2

this yields

2-(2-p)1pllfll2 < 11Ap

which proves the second inequality.

Finally, to prove that

THE SPACE L2 313

J exp(uf2) <0

we can restrict ourselves to the case where u > 0. Let N be such

that

C

a2 1

n3Nn ` 2eu '

and let

h= I a r, g= I a rn<N

n nn>N

n n

(6)

Since exp(Lf2) 4 exp(2)jh2)exp(2ug2), and since h is bounded, it

suffices to prove that exp(2ug2) is integrable. Now by (5),

fo

exp(2ug2 (2n 1)n lg2n(2u II9I12)n n

n=0 J0 n=0

D'Alembert's Criterion shows that the last series converges if

2epf1gI12 < 1, which is assured by (6).

SOLUTION (III):(d): By part (III)(a) the Walsch system is ortho-

normal. For n a 0 and 0 4 k < 2n let 0n k be the characteristic

function of [k2-n, (k + 1)2-nJ. It is easily verified that

(1 + rn)0n,k= ctn+1,2k'

g(1-

Taking into account that rn = 1, it follows from this that the

vector subspace generated by the Walsch system contains all the

functions 0nk'

and that it is consequently dense in L2(X).

314 CHAPTER 7:

EXERCISE 7.134: Let (an) be a sequence of distinct numbers

strictly greater than -}.

aShow that the function tn form a total sequence in L2(0,1)

if and only if one of the following three conditions is satisfied:

(a): There exists a subsequence of the sequence (an) that

tends towards a finite limit that is also greater than -'-z;

(b): liman =-2 and y lan + zl = W;

( c ) : liman = co andant = W.

From this deduce MUNTZ'S THEOREM: If an > 0 and liman the

asequence of functions formed by 1 and the t n is total in C(0,1)

if and only if Ean-1

= w-

LVA = VAV @ AVA ° VAV = AVA

SOLUTION: First of all we are going to calculate the distance in

L2(0,1) from tp to the vector subspace generated by tal,...,tan

(p > 0 an integer). Recall that if xi,.... xn are elements of a

Hilbert space and if G(xl,...,xn) = detll(xilxi )ll, then xl,...,xn

are linearly independent if and only if G(xl,...,xn) 4 0, and in

this case the distance from a point x to the vector subspace gen-

erated by x1,...,xn is given by

G(x,xl,...,xn)

G ,x...,x1n(The proof of this result is given at the end of the exercise).

In the present case,

G(tal,...,tan) = detlla.+ a. + lP

THE SPACE L2

Now,

if (ai - ai )(bi - bi

detIla-III 2'j(ai + b.)

i j

315

(This is called the CAUCHY DETERMINANT FORMULA, and is proved at

the end of the exercise). Setting ai = bi = ai + 'z, this yields

(a -

a a

ITG(t1,...,tn) =

'<j

(ai + a + 1)

Equation (*) then gives for the distance from tp to the vector

subspace generated by ta1,...,tan.

Ip - 01I...Ip - anId =

2(p + al + 1)

By Weierstrass's Theorem and the property that C(0,1) is dense ina

L2(0,1), the sequence (t n) will be total in L2(0,1) if and only

if for every integer p > 0

IP - all ...IP - anIl in-* p + a + 1 p+ an + 1 - 0'

1

Assume first that (an) has an accumulation point 2, -} < 2 <

By considering, if need be, a subsequence, it can be assumed that

- < a < an -12, shows that there

exists A such that 0 < A < 1 and

IP - anI

P + a + 1 ` A for all n,n

316 CHAPTER 7:

and consequently (**) is satisfied. Next, assume that an = _} +

En, with En > 0 and En - 0. Then whenever n is large enough for

an < p to be true,

IP -n2 c

p+an+1 p+En+i .

Since:

2E 2E

p + En +ry P +

it follows that accordings as I En is finite or not, the limit of

IP - all...IP - anI

(p + al + 1 ...(p + an + 1

is strictly positive or zero (at least if p is different from all

the an's, which is the case when p is large enough). Consequently,

in this case the sequence (tan) is total in L2(0,1) if and only if

En(an+I)=w.

Lastly, if an then for large enough

IP - anI= 1 - 2p + 1

p+an+1 p+an+1

2p + 1 ,L2p+1P t an t 1 a n

From this it follows that the sequence (tan) is total in L2(0,1)

if and only if I and = -. This accomplishes the proof of the

first part of the exercise.

THE SPACE L2 31 7

Assume that an > 0 and an - -. It is clear that the sequencea

(t n) together with 1 can be total in C(0,1) only if it is total

in L2(0,1), hence if Iand

= It remains to prove the converse.

One can assume that an > 1 for all n. Let P be a polynomial.

Applying what has gone before to the sequence (an - 1), for all

e > 0 one can determine numbers en, only a finite number of which

are non-zero, such that

J1IP'(t) - ICntan-l12dt

< e2.

Since for 0 < x < 1

a (x a -1P(x) - P(0) - V Can x

n= I W(t) - Cnt W,n n

the Cauchy-Schwarz inequality gives

a

IP(x) - P(0) - Cnanx nI < e, 0 < x < 1.

As the polynomials are dense in C(0,1), M[lntz's Theorem is proved.

If x1,...,xn are linearly independent elements of a Hilbert

space H, then for all x e H there exist complex numbers

and a vector z e H that are uniquely determined by the conditions

x=z+aixi,

(zlxi) = 0, 1 4 i n.

318 CHAPTER 7:

If d denotes the distance from x to the subspace generated by

x1,...,xn then d2 = (zlz) = (xlz), so that A11...,an,d2 are the

unique solution of the linear system

I ai(xilx.) _ Wx.), 1 < j < n,

i

d2 + I X;(xlxi) = (xjx).

i

That this system has a solution implies that G(xl,...,xn)4 0; (*)

then follows from Cramer's rule.

Let us now prove the Cauchy Determinant Formula. It is clear

that

Rdet ai +-

1<i,j<n (ai + bi)

where R is a polynomial of degree n2 - n homogeneous in the ai

and the bi. If for i< j ai = aj (reap. bi = bj) the determinant

has two identical rows (reap. columns) and consequently is zero.

The polynomial R is therefore divisible by

IT. (ai - a(bi - b

and since this polynomial is of degree n2 - n,

Tf. (ai - aj)(bi - b.)

detIII = C Z<ai + bj 1<i,j4n n TI. (ai + aj)

Z,7

THE SPACE L2 319

where Cn is a constant. Finally, it is verified without diffi-

culty that

lim lim aldet

and that

lim lim al

1a. = det

iT (ai - a.)(bi - b.)1<i<,j<n

b al- (ai t bj )1<Z,J<n

fl (a. - aj)(bi - bj)2<i<j<n

rr (ai + bj)2<i,j<n

1

a 1 2<i, j<n'

so that Cn = Cn-1. Since C1 = 1 the formula is proved.

EXERCISE 7.135: Let f be a holomorphic function in the disc U =

{z:Izl < U.

Show that if f is injective and if

f(z) = E cnzn for jal < 1,

no

the area of f(U) is equal to

n nlcnl2.n=o

000 - vov - AVO - vov - AV1

SOLUTION: the Jacobian of f is equal to If'12, so that:

320 CHAPTER 7:

f1

meas(f(u)) = rdrJ2%

If'(reio)l2d8.0 0

Furthermore, if 0 E r < 1,

f'(rei8) _nrn-1c e

n=1n

Plancherel's Formula gives

J2t0

so

If'(reio)I2d8 = 2n In2r2n-2IdnI2

n=1

CO 1r2n-1dr

meas(f(U)) = 2x n2Ic I2fon=1

n

W

= n I nlcnl2.n=1

EXERCISE 7.136: Let f be a continuous function on the disclzl< l

and holomorphic in its interior. Assume that

f(z) = 1 + anzn for Izl < 1.n=1

Show that if

f2n

2nIf(els)lds < 1 +lall2,

0

then the function f possesses at least one root in the disclzl <1.

THE SPACE L2 321

ovo = vov - AVA = vov = AVA

SOLUTION: If f were never zero in the disc Izi < 1 there would

exist a function g, holomorphic in this disc, such that g(0) = 1

and f = g2. Let

ag(z) = 1 + 2 z + bnzn

n=2

be the expansion of g as an entire series in the disc IzI < 1.

For 0 4 r < 1 Plancherel's Formula yields:

1 + -ja112r2 + GIbn12r2n

_ 127E lg(rel-5)I2d3

n=2 0

1 2n

2aJIf(relNd9.

0

Since

lim f(elor+1-

uniformly in S, it follows that

r2n

2nIf(ei9)IdO > 1 + 41a112.

0

EXERCISE 7.137: Let U be an open set of a and H(U) the set of

holomorphic functions in U.

Prove that H(U)n L2(U) is closed in L2(U).

ovo = vov - ovo - vov = AVA

322 CHAPTER 7:

SOLUTION: Let fn e H(U)f1L2(U) and let fn - f in L2(U). If D is

a closed disc with centre z0 and radius r > 0, and is contained

in U, there exist r1,r2 such that r < r1 < r2, and the closed

disc with centre z0 and of radius r2 is also contained in U.

For all z e D and all p, ri -< p < r2, we have (Cauchy's Formula)

f (z) = 1fn(E)

dn

-rJ2n21E-z01=p

E - z

r2n i8

2nfn(z0 + pelf)

ai0 dO,

0 z0+pe -z

and consequently

r 2n i8

fn(z) = 27r1- r J 2pdp fn(z0 + pelf)e

i8 dO2 1 ri

J0z0+pe -z

-z1 f w 0

dudv,2(r2 - ri Jrill -z01<r2 C - z0 E - z

where E = u + iv.

The function equal to ( - z0)1k - z01-1(g - z)-1 if ri

1E - z01 t r2 and to zero otherwise belongs to L2(U), and thus

lmfn(z) = 277r21- ri 11r1 1-z01<r2f() E - z0 - z dudv.

Since, for z e D and r1 . J E - z01 < r2,

-z0 1-z0 g -z1 6ri-r'

THE SPACE L2

the function

g(z) = limfn(z)n

323

is holomorphic in b, and consequently in U. Clearly g = f almost

everywhere in U, which proves the result.

CHAPTER 8

Convolution Products and FourierTransforms

EXERCISE 8.138: Let A be a measurable set of ]R with positive

measure.

Show that A - A is a neighbourhood of zero. Deduce from

this that if A is an additive subgroup of jR that is measurable

and is distinct from 32, then meas(A) = 0.

ovo = VAV = ovo = vov = ovo

SOLUTION: First assume that A is bounded, and let 9 be its char-

acteristic function. Since cpeL'flL°° the function q,* is con-tinuous. Furthermore,

p*E(0) = meas(A) > 0.

Hence there exists a neighbourhood V of zero on which 9*4 > 0;

but then V C A - A, which proves the property in this case. In

the general case there exists a measurable and bounded set B

such that B C A, meas(B) > 0. Then A - A D B - B is a neighbour-

hood of zero by what has preceded.

If A is an additive measurable subgroup of 3R, and if meas(A)> 0,

then A = A - A is a neighbourhood of zero, which implies that A =3t.

325

326 CHAPTER 8: CONVOLUTION PRODUCTS

EXERCISE 8.139: Let f be a measurable function on ]R such that

for all the elements a of an everywhere dense set A of Bt

f(x + a) = f(x) for almost all x.

Show that there exists a constant c such that f(x) = c for

almost all x.

000 = vov = ovo = vov a ovo

SOLUTION: First assume that f is bounded. Let ((pi) be a compact

approximate identity in L1. The functions fi = cp1f are contin-

uous and fi(x + a) = fi(x) for all i, all a e A, and all x e 3R.

From this it follows, because A is dense in Bt, that each of the

functions fi is equal to a constant ci. By extracting a suitable

subsequence from the sequence (rpi), one can assume that fi(x) -;

f(x) for almost all x. From this it follows that ci -> c and that

f(x) = c for almost all x.

In the general case, consider the functions fn equal to f where

IfI s n and to zero otherwise. It is clear that the functions fn

still satisfy the condition of the problem. Therefore there ex-

ist constants en such that fn(x) = Cn almost everywhere. If for

one n cn 4 0, then f(x) = cn almost everywhere. Otherwise, for

all n, fn = 0 almost everywhere, and consequently f = 0 almost

everywhere.

EXERCISE 8.140: Let f e L°°R) and ft(x) = f(x - t).

Show that if

limIIf - 0,t4 0 t

then there exists a function g that is uniformly continuous on 3R

such that f = g almost everywhere.

AND FOURIER TRANSFORMS 327

ove = vov - ovo - vov = eve

SOLUTION: If (mn) is an approximate identity such that 9n 3 0,

j(Pn = 1, and q) n(x) = 0 when IxI a en (with cn + 0), then

iI *f - f II., < sup IIf - f1100.IJI<En y

In fact, if * e L1 and II* II1 = 1 then

(*)

IJ(f*(Pn -f)V I < JIf(x - y) - f(x)Ign(b)Ik(x)IdxdY

= J gn(y)dyflfy(x) - f(x)IIVi(x)Idx

< sup IIf - f II,,YI« y

n

It follows from (*) that 9tif - f in L'. But the 9 ief are uni-

formly continuous, and since for such functions the "sup" and

the "ess sup" coincide, the (p n*f form a Cauchy sequence in UC,

and consequently Tn*f + g in UC . Clearly f = g almost every-

where.

EXERCISE 8.141: Let I be an open interval of Bt, and let f e

Lloc(I) be such that for some integer n

for every function m e Do*(I) satisfying the conditions

Ji

(p(x)dx xg(x)dx xn(p(x)dx = 0.I I


Show that f is a polynomial of degree at most n.

AVA - Vt0 = AVA - VAT = AVA

SOLUTION: More generally, we shall prove the following result:

If 91""09n are elements of Lloc(I) and if Jf V = 0 for every

function (pe Dm(I) satisfying

fi9i9 =0, 14i4n, (*)

then f is equal (almost everywhere) to a linear combination of

the gi.

Let E = Lloc(I), F = D"(I), and denote by G the subspace of

E generated by the gi. Let Go be the subspace of F formed by the

p e F such that (*) holds (and consequently Jgg = 0 for all g e G).

Let u be the canonical linear mapping of F onto F/G0. There ex-

ists a linear mapping v of F into G*, the dual of G, defined by

(g,v(9,)) = Jg, g e G, 9 eF.

The kernel of V is exactly G0. Hence there exists an injective

linear mapping w of FIG0

into G* such that:

V = wou.

Since G is finite-dimensional, (G*)* is canonically identified

with G, and tw is identified with a surjective linear mapping of

G onto (F/G0)*. Now, the linear form defined on F by

P J' Jf P

is zero on G0 by hypothesis. Hence it defines a linear form 0 on

F/G0, that is to say, an element of (F/G0)*, such that for all


p eF:

ff9 =

Let g e G be such that tw(g) = 0. Then for all 9 e F

Jfa = (tw(g),u((P)) = (g,(w0u)(9))

= (g,v(9)) = Jg9.

Therefore, by the fundamental lemma of the Calculus of Variations,

f = g almost everywhere.

For readers not so familiar with duality and quotient vector

spaces, here is another proof.

Firstly, it can be assumed that the gi are linearly independ-

ent in Lioc(I). Consider the linear mapping of D'(I) into mn

given by

9 µ (X. =J9)11.

This mapping is surjective, otherwise there would exist complex

numbers a1,...,an, not all zero, such that for all g e D"(I)

0 = aiJgi9 = f (I aigi)cp.

By the fundamental lemma of the Calculus of Variations, this

would imply that I aigi = 0 (almost everywhere), contrary to the

hypothesis made earlier about the linear independence of the gi.

From this it follows that for all j = 1,2,...,n one can find

(Pi a such that

330

(

CHAPTER 8: CONVOLUTION PRODUCTS

1g.. = ij, 1 c i c n.

Then let

C. = Ifc., 16,j c n,

and let us show that for all cpe DP(I)

Ji(P = Jg,

which will prove the property. Note that if

cy0(x) = ,(x) - cpj(x) f 9j(p,

then for all i = 1,2,...,n

Jg%p0=

0,

and consequently

Jf0 0.


Jf = CJg.p=

Jg.

EXERCISE 8.142: Let f be a locally integrable complex function

on [0,co[.


(a): Show that for all a > 0 the integral

1fa(x) =

r aj (x -

t)a-lf(t)dt

0

exists for almost all x 3 0, and that fa is locally integrable.

(b): Show that (fa)S= fa+r

(c): Show that if f e LPoc (p > 1) then fa is continuous

whenever a > 1/p.

t00 - V AV = ADA = VAV = Dot

SOLUTION (a): For every real number a > 0

(a x ra ra

dx1 (x - t)(%-'If(t)Jdt = I

lf(t)IdtJ(x -

t)a-1dx

0 0 0 t

= J0t)a - dt <

Jx(x -t)"-llf(t)Idt

<

0

for almost all x, and that fa (which is therefore defined almost

everywhere) belongs to Lloc'

SOLUTION (b):

r t

(fa)s(x) = P a P S J

x(x -

t)0-1dtJ(t -

u)m-lf(u)du

0 0

= 1 f(u)du x(x - t)S-1(t

-u)a-1dt.

r a r S j0 JU

332 CHAPTER 8: CONVOLUTIONS PRODUCTS

On setting t = u + (x - u)r this yields

(f ) (x) = 1 x(x -u)a+0-1f(u)du Ira-1(1

-r)s-1dr

a s - r(«)r(B10 0

r(a +S

x - u)a+-1f(u)du

Jo(

x

=fa+s(x).

The premutation of the orders of integration is valid for almost

all x, because, by part (a), the integral obtained by replacing

f(u) by If(u)I in the penultimate line is finite for almost all x.

SOLUTION (c): Let a > 0. If fa = IL [O,a].f, and

a-1ga(x) = r a , 0 < x .< a,

and ga(x) = 0 if x > a, it is clear that for 0 < x t a

fa(x) = 9a*fl(x).

Now, fa a LP, and if p + 4 = 1 then ga e Lp if q(a - 1) > -1, that

is to say, if a > 1 - Q = p . In this case it is known that 9a*fa

is continuous.

EXERCISE 8.143: Prove that there is a sequence of functions

fn e L20Rp) such that llfn u2 = 1 and fn*fn -; 1 uniformly on everycompact set of ]p.

ovo = vev = evo - vov e ove

AND FOURIER TRANSFORMS

SOLUTION: First assume that p = 1, and let

gnnn[-1/2n,1/2n]'

333

Then IIgnII2 = 1 and gn = gn. If fn = an is the Fourier transform

of gn, then

IIfn II2 = 119n 112 = 1,

and

fndafn = 9n*Sn = 9 nn = gn = ng"n = vnfn.

Now, since gn a LIOR) :

1/2n sinnx

fn(x)=

Vn-1_e-2nixydy

=1 nx n111 1/2n Vn n

and it is clear that ?fn (x) -> 1 uniformly on every compact set

of m. When p > 1 it suffices to consider the functions

fn(x1)fn(x2)...fn(xp)'

REMARK: To determine fn one may draw some insight from the pro-

perty that fn*?n = Ifinl2 must tend, in the sense of distributions,

to the Dirac measure at the origin, whence the idea of taking an

approximate identity for Ifnl2. It is not difficult to verify

that IIfn1I2 = 1, for

t[x 2nrt-sin

n = 12ax = 1.

n

The direct verification of the relation fn*fn = icfn

is a little

more fastidious.


EXERCISE 8.144: Let V be a connected open set of iR and let f e

L10c1


(i): There exists a constant c such that f(x) = c for

almost all x e V;

(ii): For every compact set K C V, as h - 0

fx + h) - f(x)Idx = o(IhI).K

AVA - VAV = AVA = VA0 = AVA

SOLUTION: First assume that f can be extended to an integrable

function on xtp. Let (fin) be an approximate identity formed by

continuous functions such that (p n(x) = 0 if IxI 3 1/n, and let

fn = f*'Pn' If x e V,

0<S<n and K=B(x,S)CV,

then

I fn(x + h) - fn(x) I

If(x + h - y) - f(x - y)1l4pn(y)Idyf

IyI<1/n

< II anII JK If(y + h) - f(y)Idb = o(jhj).

Thus, at every point of V the derivative offn is zero. Hence

fn is equal on V to a constant cn. By choosing a suitable sub-

sequence it can be assumed that fn(x) + f(x) for almost all x.

Consequently an

tends to a limit c, and f(x) = c for almost all

xeV.


In the general case, consider an increasing sequence of rela-

tively compact connected open sets (Vn), the union of which is V,

and which are such that n C V. The function equal to f on n

and to zero elsewhere is integrable and satisfies Condition (ii)

of the problem for every compact set K C V (for K + h C V as

soon as h is small enough). Hence there exists a constant en such

that f(x) = cn almost everywhere on n . It follows immediately

that all the cn's are equal to the same constant c, and that

f(x) = c almost everywhere on V.

REMARK: It may be interesting to indicate how one proves that

every connected open set V of ]R is the union of a sequence of

connected open sets Vn such that n C n+1' with the n being

compact.

Let (Wn) be a sequence of relatively compact open sets with

union V and such that n C ntl' Let us show that for every com-

pact set K such that K C V there exists an integer n such that K

is contained in a connected component of W. There certainly ex-

ists an integer m such that KCm'

One then has K C 01 U U Op,

where the 0i are connected components of m. Let us consider the

connected compact set obtained by taking the union of 01,...,Vp

and of (p - 1) polygonal lines contained in V and joining a point

of Oi to a point of Oi+l (1 . i < p - 1). This compact set is

contained in a W and as it is connected it is in fact contained

in a connected component of W. This proves the assetion made

above. From this it follows that it is possible to construct a

sequence of integers 1 = n0 < n1 < and a sequence of open

sets Vk such that:

Vk is a connected component of n ,k

n C Vk+l'k

It is clear that the Vk satisfy the required conditions.


EXERCISE 8.145: Let V be an open set of i and f eLloc1 (V).

Show that the following two conditions are equivalent:

(i): f is equal almost everywhere to a function g that has

bounded variation Qn every compact interval contained

in V;

(ii): For every compact interval [a,b] contained in V,

b

alf(x + h) - f(x)ldx = 0 lhl as h -r 0.

eve = vAv = AvA = vov= AvA

SOLUTION: It may be assumed that V is an interval. If f is in-

creasing on V, then for 0 < h < 1

J

b (b+h (a+hlf(x + h) - f(x)ldx = J f(x)dx - f(x)dx < 2Mh,

a b a

where M denotes the upper bound of lfl on [a,b + 1]. The case

where h is negative is treated analogously. As every function

with bounded variation is a linear combination of increasing

functions, the implication (i) => (ii) is demonstrated.

In order to prove (ii) => (i) one can clearly assume that f

is real. Let [a,b] C V. There exists 6 > 0 such that

[a - 26,b + 26] C V.

By replacing f by zero outside this interval, it can be assumed

that f is integrable and that there exists a constant M such that

J

b+6_ lf(x + h) - f(x)ldx < Mlhl if lhl < 6.

a 6

Then let (v.) be an approximate identity formed by continuously


differentiable functions whose supports are contained in ]-d,d[

and are such that Il'Pill, = 1. Extracting, if necessary, a subse-

quence, it maybe assumed that

fi(x) = (f*(p i)(x) -; f(x) if x4E,

E a set of measure zero. Lastly, by moving a slightly to the

left it can be assumed that we still have [a - 26,b + 26] C V,

and, further, that a #E. If Ihl < 6 then

frblfi(x + h)- fi(x)I

JI dx

a

l.f(x + h - y) - f(x - y)I Iki(y)Idy<

+fb dxfd

a -6

6 b+a

<

+f-dI,.(y)ldyfa-d lf(x + h) - f(x)ldx

JJ

< M.

The functions fi are continuously differentiable, and consequent-

ly, by Fatou's Lemma

bV(fi;a,b) _ Ifi(x)Idx < M.

a

If a = x0 < xl < < xn .< b, therefore for all i

n-1

I Ifi(xk+l) - fi(xk)l '< M.

k=O

Furthermore, if xk $ E for k = 0,...,n, then passing to the limit

when i - - yields


n--1

kI0 If(xk+1) -f(xk)1 < M.

Now consider the set of sequences A = {a = x0 < x1 < < xn < b}

such that xk 4 E, and let us agree to write A < x if xn < x. Set

n-1

P(A) = E (f(xk+l)- f(xk)+,

k=0

n-1

N(A) = E (f(xk+l) - f(xk)_k=0

By the preceding,

P(A) + N(A) < M.

Finally, when a < x < b let

P(x) = supP(A),A<x

N(x) = supN(A).A<x

The numbers P(x) and N(x) are finite and increase with x. When

x 4E we need only consider the sequences A for which xn = x. In

this case

P(A) - N(A) = f(x) - f(a),

so

P(x) - N(x) = f(x) - f(a).

Thus the function f coincides almost everywhere on [a,b] with

the function P - N + f(a), which has bounded variation.

If [an,bn] is an increasing sequence of compact intervals the


union of which is V. there exist functions gn of bounded varia-

tion which coincide almost everywhere with f on [an,bn]. One can

assume that gn is right continuous on [an,bn]. It immediately

follows from this that gn+1 coincides with gn on [a n,bn]. Setting

g(x) = gn(x) if an < x < bn defines a function on V with bounded

variation on every compact interval contained in V, and which is

equal to f almost everywhere.

EXERCISE 8.146: Let f be an integrable continuous function of

bounded variation on]R, and f its Fourier transform.

Show that

+m 1 +-

If(2nn) =

2nI f(n). (POISSON'S FORMULA)

n= n=-w

(Prove that the function

+-

f4(x) = E f(x + 2nn)n=-w

is continuous, has period 2ir, and has bounded variation on [0,2,r],

and then use exercise 3.42).

1VA = VAV = OVA = VAV = AVA

SOLUTION: Let

V(x) = V(f;-=,x),

limV(x)=V<W.x-).-

Proceeding as in exercise 3.42 it is seen that f' c) is defined

for almost all x, is integrable on x, and that

f

2n4

tm

f (x)dx = j_ f(x)dx. (1)


Let x0 be a point where f(x0) is defined. If x0 4 x < x0 + 27E

then

If(x + 2rn) - f(x0 + 27En)I 4 V(x + 27Cn) - V(x0 + 21En)

4z V(x0 + 2n(n + 1)) - V(x0 + 2xn),

and furthermore

(V(x0 + 2n(n + 1)) - V(x0 + 2nn) = V <

This shows that fa is defined and continuous on [x0,x0 + 2n], and

consequently on the whole of ]R.

If 0 = x0 < xi < ... < xk = 2 n then

k-1

If"(xi+l) - fa(xi)Ii=0

k-1 +-

= II

I f(xi+l + 2xn) - f(xi + 2nn)i=0 n=

k-1 +-

I I If(xi+1 + 2xn) - f(xi + 21En)Ii=0 n=--

4 V.

Applying (1) to the function e-lnxf(x) yields

1

21E

e-inxfl(x)dx =I

+e-inxf(x)dx= 2n

?(n).0

(2)

By the Jordan-Dirichlet Theorem and (2),


+m +W 2n

X f(2un) = fa(0) = G L =

n-mJ0e_1f'(m)

n=--

EXERCISE 8.147: (a): Let I = ]a,b[ and f,g be locally integrable

functions on I.

Show that the following properties are equivalent:

(i): After modification of f on a set of measure zero

(Yf(y) - f(x) = J g(t)dt, a < x < y < b;

x

(ii) : For every function q) eDW(I):

rb b

J

f(x)4)'(x)dx + J g(x)cp(x)dx = 0.

a a

Next, show that if the pair (f,gl) has the same properties,

then g = gl almost everywhere. We then make the convention g = Df.

If g is continuous it will be noted that Df = f'.

(b): Let

H2(I) = {f:feL2(I),DfeL2(I)}.

Prove that the scalar product

(b(fig) =

J(f(x)g Tx7 + Df(x).Dg x )dx

a

defines a Hilbert space structure on H2(I).

(c) : If f e H2(R) show that


Df(y) = iy?(y) almost everywhere.

(d): Let b = - and f e H2(I).

Show that limf(x) = 0.x-W

(e): Assume that a,b are finite. By (i) every function of

H2(I) can be extended by continuity to a and b.

Show that if a 4 c .< b there exists pc e H2(I) such that

f(c) = f(x)* c(x)dx +f e H2(I).b

JaCb

Determine *c explicitly.

(f): Again assume that a,b are finite and set

HD (I) = {f:f a H2(I),f(a) = f(b) = 0}.

Show that D°°(I) is dense in H2(I), and that for a < c < b

there exists 8c a H2(I) such that

bf(c) = 1 Df(x).DBc(x)dx,

af e H2(I).

(g): Let f be an indefinitely differentiable function on

R2 - {0}, zero for jxI 1, and equal to log(log(l/Ixl)) for

0 < 1xI < I.

Show that f,2f/2x1,2f/2x2aL2(z2), and that for (p ee(x2)

J2{faL+ ax. =o, i=1,2.

What conclusion can be drawn from this?

AVA = VAV - AVA = VAV = evn


SOLUTION (a): If f,g satisfy (i) and p e DP(I), let a,$ be such

that a < a < 8 (ii).Now assume that (ii) is satisfied. Let a < x0 < b, and

Jf1(x) = g(t)dt, a < x < b.

x0

By what has just been proved,

f

rb rbI fl9'+J gro=0,a a

9 e D-(-T).

Comparing this with (ii) yields

rbI (f - f1)p' = 0,JJJa

'p e DP(I).

When 9 runs over 9' runs over the set of functions *e D-(I)

such that Ji = 0. 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 substrac-

tion one obtains


b

(g - g1)T = 0, 9 e D(I),a

whence g = g1 almost everywhere.

SOLUTION (b): The Formula defines a positive alternating form

on H2(I), and (fIf) = 0 implies that 1IfII

L2(I

)

, whence f = 0. It

remains to prove that H2(I) is complete in the norm associated

with this scalar product. If (fi) is a Cauchy sequence in H2(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

Jafi.cp' +faDfi.W = 0,

and on passing to the limit:

b rb

J

fp' +1

gcp = 0.

a 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),

+W+0 r+0 Jf(x)c'(x)dxJP?(Y)TTdY_ = 1 _ Df(x)Zxsdx = -

J](Y)iy(y)dJ = 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


L2OR). Hence bf-(y) = iyf(y) almost everywhere.

SOLUTION (d) : First Proof: Note that if A = IIDf II 2 andL (I)

a < x < y, then

If (y ) - f(x) I 4 A.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+

E2 , 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

W x +c 24A

22 2

J If(x)I2dx > I j n If(x)I2dxE E

a n x n 4 4A2n

= W.

Second Proof: We can reduce to the case where I =]R by the fol-

artifice: 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

f

a0(t)dt = f(a).

a

By extending Df to [--,a] by setting Df(t) = 8(t) for t E a, E±nC

setting

xf(x) = f(a) + J Df(t)dt, x e ]R,

a


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); since

(1 + y2) 2 e L2(kt) we have 'e L1(R). But then f is the inverse

Fourier transform of ?, and therefore tends to zero at infinity.

SOLUTION (e): Let us show that f + f(c) is a continuous linear

functional on H2(I). To do that, write

(cf(c) = f(x) + 1 Df(t)dt, a 4 x < b,

x

so

,

If(c)I 4 (b - a)2IIDfII 2 + If(x)I,L (I)

or

If(c)I2 4 2(b - a) IIDfII22 + 21f(x)12. (*)L (I)

Integrating with respect to x from a to b yields

I f(c)12 , 2(b - a) 112

+ b - a112

L2(I)'

which proves our assertion.

The existence and uniqueness of ic

results from this. In par-

ticular, if W e V (I),

c b bcp'(x)dx = J(p(x)*c(x)dx +

a a a

which can again be written, by introducing the characteristic

function Xc of the interval [a,c],


b bJaq,'(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) and D(D*c - Xc) _ *C.

347

Since *c

is continuous Dpc - Xc is continuously differentiable

and its derivative is * c. In other words,

(1)c

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,

a0 = J(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

YIVC(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


(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

cosh(b - c)cosh(x - a)sinh(b - a)

(x) _ccosh(c - a)cosh(b - x)

sinh(b - a

a 4 x 4 c,

c<x4b.

This expression for *c is still valid when c = a or c = b. (If

c = a, for example, then (3) disappears and (4) reduces to 'c(b)

= 0).

SOLUTION (f): It is clear that D -M C HD(I). Furthermore,

Ho(I) is the orthogonal complement of { a''Pb}. To prove that

VI(I) is dense in H2 (I) it suffices to prove that every f e H2(I)

orthogonal to r(I) is a linear combination of Pa and 'b. Now,

the condition:

(b rb1 f(x)p(x)dx + I Df(x)p'(x)dx = 0, T e 'O-(I),

a a

implies that Df e H2(I) and.D(Df) = f; but then as f and Df are

continuous this latter condition can be written f' = f; the or-

thogonal complement of D°°(I) is thus two-dimensional; because it

contains ya and 'Pb the result is proved.


The mapping f - Df is a continuous bijection of HD(I) onto the

orthogonal complement of 1 in L2(I). By a theorem of Banach this

mapping is bicontinuous, which means that f -> IIDf JJ2

is, onLI)

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-cb ba a4x4c,

Xc(x) - b l a JXc(t)dt =a c - a

and consequently

(b - c)(x - a) 4x4c- a) ,a

(c - a)(b - x) <x <b-a . ,c..

REMARK: One can avoid using Banach's theorem by showing that

bf -* (If(a)1 2 + J

a

is a norm on H2(I) equivalent to the one introduced in part (b)

above. One of the two necessary bounds follows from (**) proved

in part (e); the second results from (*), this time integrated

with respect to c, setting x = a.

SOLUTION (g): Set

x = (xl,x2 IxI = r = (x1 + x2)2.

Then

350

1

CHAPTER 8: CONVOLUTION PRODUCTS

r'EI f(x)I2dx = 2n r(loglog1 )2dr <

xl<z o

Moreover, if 0 < r < 'z,

ofaxi

whence

Ix-Lf'(r) '( If'(r)I = 1r logy'

a2

Jlxl ax. (x)I dx

2tfn

( dr

O rllogrl2

Finally, if (p e D%R2 ), and if

w = f(pdx2 on a22 - {o),

then

dw = l ax cp + f x1J

dx1Adx231 1

so (by Stokes' theorem)

(BERTRAND'S INTEGRAL!)

of(P + f 11p'= J f J fwx1 - xI='

JO<CIXkR

If R > 1 this equation can be written as

2L (P

+f a )axf la

1 (2n- eloglog1 cp(ecos ,tsine)cosed8.

0


Making c -> 0, this yields

I 1 (P + f 11c = o.2

An identical relation is obtained by replacing x1 with x2.

This shows that the space H2(IR2), analogous to H2 OR) in two

dimensions, is not made up only of continuous functions. This

property can be linked with the second proof of part (d) and the

fact (1 + Ixl2) 1 does not belong to L2(It2

EXERCISE 8.148: All the functions considered are measurable and2n

have period 27E. Jf means J f(x)dx, and f e L1 if Jlfl <0

(a): Assume that

J Iflog(i + Ifl) < -.

Show that f eL1,

and that if g is a function such that

exp(a-11g1)e L1 for a real number x > 1, then the convolution

product f*g is defined everywhere, and

If*gl < AlogA Jlfl + A Jlfllog(1 + Ifl) + J(eA-1181 - 1).

(Use the inequality

ab < alog(1 + a) + eb - 1,

valid if a >, 0 and b >, 0).

(b):. Set

(*)

G(x) = logl 1cosx '


and fn(x) = f(x) if If(x)I >, 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): Noting that log3 , 1 gives

JIfI = JIfI<2lfl + J IfI>2IfI

< 4n + Jlfllog(l + Ifl ),

which proves that f e L1. 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): If x0 is a root of cosx, then as x -. x0

G(x) ti logIx ix01-


and consequently

eA-1G(x)

ti Ix - x0I-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 - f is bounded. More pre-

cisely, hn is continuous, as LW

n

*L1 C C. For A > 1 and every

integer n

IIf*G - hnil.

< AlogAJlfnl + Xflfnllog(l + Ifnl) + J(ea_1G

- 1).

Since Ifl l >. l f21 >. --- and I fnI - 0, this yields

-1limsupllf.G - hnll 4 J(eXGn--

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 L1then 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.


SECOND SOLUTION: There would exist a > 0 such that

JIf(x)Idx -1 1.

Let p be the characteristic function of the ball with centre zero

and radius a/2. Then if IxI 4 a/2 one would have

f* (P (x) = J

IHI

f(x - y)dy . J

Iy I < aIf(y)Idb < 1,

-< 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-

ively. If fl = F(wl), f2 = F(cp2), then fl eLl(R' ), f2 eL'(Itn),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 = I an T (1)a n n

n=-m


is a continuous operator of L1 (the series (1) converges normally

in L(L1,L1)). Ifn)ne2Z

is another summable sequence, then

Tao Ts = Ty, Yn =

p+Q-n

apq.

Now assume that we have determined a,s so that

TOT0 = 0,

If m e L1, then

(2)

a 4 0, S 4 0. (3)

(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) Sneln n

then:

uv = 0, (4)

and conversely (4) implies (3). Consider the function Isinxl;

its Fourier coefficients are

(n n

en = 1,R1 _ e -ice l sinx ldx = nJ cosnxsinxdx.n 0

It can be shown that


c2n+1=

0,

a-1 = 0-1

The function Isinxl being continuous and piecewise continuously

differentiable, the elementary theory of Fourier series ensures

that

+00

Isinxl = I cneinx

n=-m

But then

+co

(sinx)+= 4 (eix - e-ix) + cneinx

n=--

(5)

+m- (sinx)_ = 42

(e ix - e-ix) -z

C cneinx

n=G-w

so that the sequences a and S defined by

a = s1

1 1 = 1 ,

2 1C2n-n

4 ,

1 1

a2n - s2n n1 - 4n2

a2n+1 = 82n+1 = 0if n + 0 and -1,

satisfy (3). 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,

1 (na2 - - st

- 1)nn n 1 - 4n2 a2n+1 = 02n+1 = 0 if n4 0 and -1


EXERCISE 8.151: 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 >, 1

a enxf(a- x)dxI2 4

JJIf(a - u)f(a - v)Idudv.

a u>-av>-au+v>O

Using Exercise 3.72, deduce from this that f = 0 almost every-

where 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,

J

tom x-WIf(y)g(x - y)Idy = j If(y)g(x - y)Idy = (Contd)

0


(Contd) = Jx0

fM(y)gM(x - y) dy

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

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( enxf(a - x)dx)2 = fJ en(u+v)f(a

- u)f(a - v)dudv

-a (ul'aIV1<a

=JJ e

n(u+v)f(a - u)f(a - v)dudv

u>,-a

v>,-au+v50

+ ff en(u+v)

f(a - u)f(a - v)dudv.

u.<<a

vsau+v>,0


Carrying out the changes of variables u = a - y, v = a - x + y

in the last integral yields

if0sy#xs2aen(2a-x)f(y)f(x

- y)dxdy

r2aen(2a-x)(f*f)(x)dx= 0.

0

Since en(u+v) E 1 if u + V .< 0,

J

a

IJ-a

enxf(a- x)dx12 S

JJ3If(a

- Of (a - v)Idudv.

U -av a-au1VS<0

Now note that

(1)

2atoIJ- ef(a - x)dxl c f If(x)Idx. (2)

a a

(1) and (2) imply

supiJenxf(a

- x)dxl s M <an 0

(3)

By exercise 3.72(b) we therefore have f(a - x) = 0 for almost all

x e [0,a] .

REMARK: For use in the rest of the exercise, note that this shows

that f*f = 0 implies f = 0.

SOLUTION (c): For almost all x >. 0

(f*gl)(x) t (fl,tg)(x) = (Contd)


x x(Contd) = J0 (x - y)f(y)g(x - y)dy + Jyf(y)g(x - y)dy0

(x= xff(y)g(x - y)dy = 0.

But then

(f*gl)*(f*gl) (f*gl)*(fltg)

(f*g)*(fl*gl) = 0,

and by the above remark

f*gl = 0. (4)

SOLUTION (d): Let gn(x) = xng(x), n a 0 an integer. From (4)

one deduces by induction

f'tign = 0, n 3 0.

In other words, for almost all x and all n 3 0

Jyflf(x - y)g(y)dy = 0,

and consequently for almost all x

+m xIf(x - y)g(y)I dy = J If(x - y)g(y)Idy = 0.

0

By Fubini's Theorem

(5)

,J0 =_dxJ_ If(x - y)g(y)Idy = JIf(s)IdxJI9(Y)dY

m


and therefore one of the two functions f or g is zero almost

everywhere, which proves that L+ does not admit a divisor of

zero.

REMARK: (5) implies that for almost all x there exists a set xwith measure zero such that

f(x - b)g(y) = 0, b 4EX.

But since the Ex vary with x, and x runs over a non-denumerable

set, the only way of concluding from this that one of the func-

tions is zero almost everywhere is to use Fubini's Theorem.

However, it is interesting to note that this difficulty does

not occur if f,g are assumed continuous and that the proof of (1)

makes an appeal in this case only to an elementary form of

Fubini's Theorem (for functions continuous on rectangles or tri-

angles). Still, the proof of the fact that for a continuous

function f,

a supiJenxf(x)dxl< m implies f = 0,

n 0

is far from being trivial. In Yoshida's Functional Analysis

there is a proof more elementary than the one we have given in

the solution of Exercise 3.72.

When one has proved that f*g = 0 implies f = 0 or g = 0, for

continuous functions, one can pass to the general case by regular-

isation. In fact if (vi) is a regularising sequence with the

support of Ti contained in [0,1/i], then

(f*rpi)ic(g*(Pi) = 0;

if g # 0 then g*cpi4 0 for i large enough, and consequently

foi = 0. It remains to observe that for all M,


MJ

0If - f*(PiI } 0.

EXERCISE 8.152: Let 1 0 set

F(x) = x1lf(t)dt.0

(a): Show that F e LP(0,°°) and that:

IIFIIP 6 p p I IIfIIP. (HARDY'S INEQUALITY)

(b): Prove that equality holds in the above relation only

if f is zero almost everywhere.

(c): Show that the constant p/(p - 1) in Hardy's Inequality

is the best possible.

(d): Show that if f e L1(0,m), f > 0, then F 4L1(0,-).

ADA = DAD = AVA = DAD = AVA

SOLUTION (a): First Proof: Let G(x,t) = f(xt). Then

I

F(x) =J

G(x,t)dt0

By exercise 6.110,

IIFIIP E IIPdt0

= IIf IIP I t-1/Pdt = p p 1 IIfIIP.0


Second Proof: First assume that f is positive, continuous, and

has compact support in ]0,m[. Then F is zero in a neighbourhood

of zero, and

00F(x) = lrfx 0

for x sufficiently large. In particular,

xF(x)p-> 0 whenx -> 0 orx - w.

Differentiating the original expression for xF(x) gives

xF'(x) + F(x) = f(x),

whence, upon multiplying byF(x)p-1

and then integrating,

JzF(x)F(x)1)1dx + JF(dx = Jf(x)F(x)P1dx.

0 0 0

Integrating the first integral by parts, and taking into account

that xF(x)p vanishes at 0 and -, one obtain

P -1J F(x)pdx = Jf(x)F(x)P1dx.P 0 0

Applying H61der's inequality to the right side yields

IIFIIP " p p l Ilfii(J-F(x)(p-1) 1dx)1/q, where

p

+

q

= 1.0

That is

11F lip , p p l IIf11 JIFIIPlqp

(1)

As IIFIIP < - (because of the value of F(x) in neighbourhoods of


0 and m), it follows that

IIFIIP ,p

p1 Ilfllp.

If f is continuous and compactly supported in ]0,°°[, then

IF(x)I , xjxlf(t)ldt,0

so

(2)

IIFIIP p p 1 IllfI Ilp = p p l Ilfllp.

Finally, if f is any function in LP(0,°°), there exists a sequence

(fi) of continuous functions compactly supported in ]0,°D[ such

that if - fillP

+ 0. If

Fi(x) = L xfj(t)dt0

then

F.(x) + F(x) for all x > 0.i

By Fatou's Lemma,

IIFIIP , limsupl1FillP , p P 1 limllfilhP = 1 Ilf1Ip.i i

SOLUTION (b): First assume that f e Lp(0,co) and f >, 0. Let

be a sequence of compactly supported continuous positive functions

such that f + f in Lp. Then F. -' F in LP, and so by exercisesi 16.105 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


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,

that is to say F = Bf, B >, 0 a constant. Let us note that if

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:

f(x) = Cxa, C > 0, 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) = xf0

lf(t)Idt

yields

-]IF IIp < IIGIIp < p p 1 IllfI Ilp = p p l Ilfllp-

REMARK: The second proof given for part (a) above is more ele-

mentary than the first proof, which uses the generalised Minkow-

ski 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-


ceeding in the following manner, which furnishes a third proof of

Hardy's inequality.

Let

(P(x) = eX/Pf(eX),

I

e-x/qif x 0,

0 ifx<0,

It is not hard to verify that

N 11 If 11

X%tc(x) = ex/PF(ex),

11XII l J1

= (rll/r r 0,

so that

IF 11

Lp(0,°) =iix*c ii

LP(-O,}°)<- 1lx ii L1

(-0,}0)tic 11

LP(--, +W)

= 4iIftiLp(0,co)

by virtue of the classical properties of the convolution product

on:R. Moreover, since X e Lq(-_,+_),

1 (X*c)(x) = 0,1x1-

which implies that

limxl'pF(x) = limxl"PF(x) = 0, (3)

x+O X4-


But then, if f 3 0, we still have

xF'(x) + F(x) = f(x)

for almost all x, by the Lebesgue theory of differentiation, and

furthermore, by (3), the formula

J

xF(x)P-1F'(x)dx = - 11 F(x) dx0 p0

is still valid, which ensures the validity of (1). Note that

the integrability of xF(x)P-1F'(x) is assured, because FP andfFP-1

are integrable .... sinceFP-la

Lq. If

r 3 1,p

+ r >, 1, and 8 = -p tr

- 1,

we have (cf. exercise 6.106)

OX*PlI S . 11X 11 r, II,P II P ,

from which it is easily deduced that

1+1--

r

( (xS/P - 1IF(x) I Sdx)1/s. 4 8 + 1s q

) IIfIIP, S > P,0

(

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 multiplica-

tive group ]R that the preceding is expressed more simply in the

following way: If

9(x) = x1/Pf(x),

and


x1/q for x >, 1,

0 if 0<x<1,then

X*g(x) = xl/PF(x),

where the convolution product this time is defined by

xlkg(x) =J'X(t)g(xt-1) tt = Ot .

0

SOLUTION (c): If f(x) = X -11P when 1 . x 4 A, and 0 otherwise,

then simple calculations show that

0 if 0 < x 4 1,

F(x) = qx 1(x1/q - 1) if 1 4 x 4 A,

q(A1/q - 1)X-1 if x >, A,

IIfIIP = (logA)1/P

(A 1 (A1/q - 1)P 1/PIIFIIP

= q{J x P(xl/q - 1)Pdx +P P'11 -

1A

Since

x P(x1/q - 1)P v xP/q - P = X-1 as x ->

we have

Jx(x1 /q - 1)Pdx ti logA,1


and consequently

IIFII

ITf1 'q asA+oo.P

SOLUTION (d): In this case

(m (J F(x)dx =

xdx

J J

xf(t)dt = J f(t)dt J =

0 O x 0 0 t x

369

EXERCISE 8.153: Let us denote by Lp(t-1dt) the set of measurable

functions on ]0,W[ such that

IIf1ILp(t-1dt) U0lf(t)tP tt)1/p

Also, set

L-(t-1dt) = LW(O,W).

Show that if

f e Lp(t 1dt), g e Lq(t-1dt), p,q >. 1, p+ q 3 1,

then the integral

Jf(t)g(xt') tt0

exists (in the Lebesgue sense) for almost all x > 0.

Denoting by f*g the function thus defined almost everywhere

on ]0,oo[, show next that


IIf'=gII r -1 ` IIf1I p - 11g 11 q -1 if r = p + q - 1,L (t dt) L (t dt) L (t dt)

Lastly, show that if p +

Q

= 1, then f*g is continuous, and

limf*g(x) = lim(ffg)(x) = 0.X-+0 x-

AVA = VAV = AVA = 0E0 = A0t

SOLUTION: This is a matter of developing the last part of the

remark in the preceding exercise. If we set F(x) = f(ex), G(x) _

g(ex), it is not hard to verify that f*g is defined almost every-

where 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: Let p >. 1 and 1 -

p

< a < 2 - p ; for f e Lp(0,co)

set

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 < A(r,p,a) IIf IIp.J0

Show, further, that if p > 1


F(x) =o(xa-1+1/p) asx -> 0 or x

Deduce from this that

suplF(x + h) - F(x)I =o(Ihla-1 + 1/p)

X


SOLUTION: Note that

F(x) = xl/p + a-1( tl/p f(t) sinxt dtJ 0 (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-1J(t)x(rt)

d'.

0

- xl/p +a-1tt .

in

Setting ap(t) = rp(t-1) and using the notations of the previous

exercise, this becomes

F(x) = x1/p+a-1('*X)(x).

Now observe that

371

IN 11 -1 =11w1I =Ilf1ILp(t dt) Lp(t-1dt) p


and that

UXII

Lr

(t-1dt)0<r4 m,

for

p

+ a - 1 > 0, which ensures the result for r and if

r<-

1t-1iX(t)lr = O(tr(l/p+ )a-1)+1

t-1iX(t)Ir _

if

1tr 1/p+a-2 +1 as t; 0 and p+a - 2 < 0.

8 1, p+8al, r=1+1-1,

that is to say, if

rap and s=4+r, where p+q=1;

then

ii hx u Lr(t-1dt)s IIX I

LS(t -1dt)

Ilf IIP,

which is the desired inequality. When r = - then s = q, and con-

sequently if p > 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)cos(x + Zh)tsinj'ht dt,

E to


suplF(x + h) - F(x)I < 2F(2'Ihl) = o(jhla-1 +11P).

x

EXERCISE 8.155: Let p > 1 and f e LP(O,m). Set

F(x) = J0e_Xtf(t)dt, x > 0.

Show that for r >, p

(f - 1I F(x) Irdx)1/r <l-I

1/qr (411/S11f 11p,J

0 Il J

p+q= 1,

Show also that

F(x) = O(x-1/q) as x -* 0 or x -* -.

ove = vev = evo = vev = ovo

373

SOLUTION: Proceed as in the preceding exercise, observing that

if

W(t) = t1/Pf(t), X(t) =tl/qe-t,

then

F(x) =x-1/q(,*f)(x).

Furthermore, if a > 0

IlxIILS(t-tat) =

(jo S/q -1 e- St dt)1/s= I s l 1/qr 1411/S.


EXERCISE 8.156: Show that a subset H of Lp(]R ), 1.4 p < -, is

relatively compact if and only if it satisfies the following con-

ditions:

(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 com-

pact in Lp, H is bounded, which proves (i).

For every e > 0 there exist functions f1,...,fr e Lp such that

) t 6.sup( min IIf - fS IIPfeH l*ssr

There is a compact set K of Ltn such that

J]Rfl-KSIp 4 ep, 1 4 s 4 r.


Then for all f e l l

Iflp < (2e)p, 1<s<r.

which proves (ii).

Similarly one can choose a neighbourhood V of zero in nzn such

that

II(fs)a - fSIIp < E,

and then if f e H

Ilfa - f 1Ip < 3c,

aeV, 1 < s < r,

since

Ilfa - flip < II(f - fs)allp + II(fs)a - fSIIp + Ilfs - flip.

The Conditions are Sufficient: For every compact set K C gtn we

shall set fK = fIlK. Let m be a continuous function with compact

support A. For f e ll the functions fK*y have their support con-

tained in the compact set K + A. and if

p

+

q

= 1 then

Ilf`*"PII_ < IIf"IIphI,P 11q < MII,P IIq,

I (fK*(p )(x1) - (f`*p)(x2) I < 11f`11P

IIkx - ro-x II

1 2q

< MIIcpx 2-x1

(P Ilq.

By (i) the set of functions fKhm,f e H, is therefore bounded and

equicontinuous. It follows from Ascoli's Theorem that for all

a > 0 function fl,...,fi,e H can be found such that


sup( min IIfK*cp - f 5 a.feH 1<s<r

Now let c > 0. Choose K such that (condition (ii)):

(1)

Ilf - fKIIP' e , feH, (2)

and let V be as in (iii). Assume further that m 3 0, lalli = 1,

and that the support of 9 is contained in V. Since

(f*q)(x) - f(x) =fv

(f(x - y) - f(x))9(y)dy,

the generalised Minkowski inequality (cf., Exercise 6.110) furn-

ishes, for f e H, the bound:

IIf*c - fIIP 4 JIIfy - fIIpp(y)dy < e, (3)

and consequently by (2) and (3):

IIfK*a - fKIIP < II(fK - f)*cIIP + Ilf*T - AP +If

- fKIIP

4 3e. (4)

This being so, let us choose fi,..,,fr e H such that (1) is satis-

fied with a = e(meas(K + A))-1/P. Writing

IIf - fs IIP IIf - fKIIP + IIfK - fK*a IIP + IIfK, - fs*9IIP

+ Ilfs`=c - fs IIP + Ilfs - fs IIP

it follows from (1),(2), and (4) that


sup( min hf - fsIIP) 4 se,fell 1Fr4s

which proves that H is relatively compact.

EXERCISE 8.157: (a): Let 0 < a1 < a2 < and an = 0(na).

Prove that

alimsup

n= 1.

an+1

(b): Let f be a measurable positive function on ]R such

that 0 < 11A. < W.Prove that there exists a sequence (9n) of functions such

that

M: q)n a D%RP) and ,(Pn > 0;

(ii): hIf*qnII 1;

(iii): f*(Pn -> 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. Set an = hhf*$n11. and

consider some x e]R such thatn

(f*8n)(xn) I.1 - 1a.

Show that if

hn(x) = a-1t xn


< pone has Iif*hn II = 1 and when x

((f*h ) (x) a I1 -

11 an

n nJJan+1

(c): Deduce from this that if g e Ll(a) is such that f*g 0,

then

Jg(x)dx 3 0.

AVO = VAV = AVA = VAV = avo

SOLUTION (a): If

limsupan

n an+1

one would have, starting from some n0, an+l > )an, and conse-n-n

quently an > an0 x 0 when n 3 n0, which contradicts the hypo-

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


On+1(x + z) = 1 >, 8n(x).

In particular On < $n+1' whence an < anti. Now note that

a1 >. (ffcq) 1)(0) >1 f(x)dx > 0,

I r <2p

and that, denoting the volume of the ball IxI < 1 by Vp,

an < IIOnII1IIf1I. < Vppp(2n + 1)pIIfII..

This being so, if xn and hn are as in the statement of the ques-

tion, then

II f*hn II = an+l IIf*$n+1 II. = 1,

and furthermore, if IzI . I1 - 11 nan+l

n n ` nJJa

a

n+l

By (**) and part (a), for all e > 0 there exists an integer n0

such that

an(1- it

a0 >1-E.

n0 n0+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 = 1and


1 - e .< (f*q,)(z) 4 1

if Izi < p. Denoting by wn

the function corresponding to p = n

and e = 1/n, one obtains the desired sequence.

SOLUTION (c): This follows from

Jg = fg = lim(g,f*p ) = lim(g*f,p )

n n n} n

and the fact that g*f 0 implies that (g*f,wn) 0 if Tn > 0.

EXERCISE 8.158: Let f e Li(R ). Prove that

IJf(x)dxl = 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 A(a)f(x - a).adRp


There is no difficulty in verifying that if al,a2 e P then

Al*(A2*f)=

(al*A2)*.f,

where

1(A1*A2)(a) = I Aba2(a - b)bdRP

(the reader well versed in measure theory will recognise the no-

tion 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


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. To pass to the general case

observe that there exists a e Q, 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 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 con-

vex 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

fa's. In other words there would exist g e L , g non-zero almost

everywhere, and such that

Jf(x - a)g(x)dx > a > (2)

for all a e]R . 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. Let p be as above. Then

(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.


EXERCISE 8.159: If f is integrable on [a,b], and if

fb is xf(x)e n dx = 0a

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

bF(z) = 1 f(x)eizxd

a

This defines an entire function which vanishes at each a . Ifn

the latter have a finite limit point, then F(z) = 0. 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

is xlime nn-9

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

is xlime e nn- n


exists for all the x's belonging to a measurable set A of ]R with

meas(A) > 0.

Show that either

lime = 0,n-,m n

or

lime = c $ 0 and lima = a.n. , n n._,a, n

ADA = DAD = ADA = DAD = ADA

is xSOLUTION

n(a) : The set of the x eat for which lime exists is

an additive subgroup that is measurable and of strictly positive

measure. From this it follows that

is x

g(x) = lime nn-),-

(*)

exists for all x eat (cf. exercise 8.138). Then for every inte-

grable function f on at,

f

+W ( -+Q is x

_f(x)g(x)dx = limJ f(x)e n dx

n-t°° m

(Dominated Convergence Theorem). Assume that for a subsequence

(On) of (an) IsnI -> By the Riemann-Lebesgue lemma,

E 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. The

sequence (an) is therefore bounded. Let a,8 be two accumulation

points of the sequence. By considering subsequences of (an) con-


verging towards a and B, one deduces from (*) that

eiax - eiBx

for all x, whence by differentiation a = B. This proves that the

sequence (an) is convergent.

SOLUTION (b): Since

is x

icne nI = Icn1,

we have that

IimIc I = pn-),w Ti

exists. Assume that p > 0. Then from some point on, cn 4 0. If

is (x-y)

x,y e A, the sequence e n is therefore convergent. Because

meas(A - A) > 0 it follows from part (a) above that a = lima ex-n-).w n

is x is xists. Finally, if x e A the sequences ene n and e n are con-

vergent, which proves that lime exists.n-).w n

EXERCISE 8.161:(a): Let f e L1(T), and

If(x + u)Jdu,f+(x) = suph-1foh>o

f(x) = max(f+(x),f (x)).

Show that for all A > 0,


meas{ (f+ > A) n [-n, n] } < Ln11f111, (*)

and some analogous relations for f and ?. (Use the "Setting

Sun Lemma"). What does the application of Marycinkiewicz's Theo-

rem give in this case? (Cf. exercise 6.124).

(b): Let (Kn) be a sequence of functions of L1(T) and (Hn

)

a sequence of continuously differentiable functions on [-n,n]

such that

(i): IKn(x)I < Hn(x), -n < x < n;

n

(ii): A = sup 2nj Hn(x)dx <-n

(+W(iii): B = sup

1IxH'(x)Idx <

2n 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 (*). If

0 < A < 211fp1, then

sp(a) < 2% < 11fD D.

Furthermore, if IxI 4 1 and h > 0 are such that


211f111 < A < h-1Jhi f(x + u)I du,0

and if N is the integer such that 2n(N - 1) < h < 20, then be-

cause 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 IfI > A for a z such that

x x < z <x+2n}

z< meas(x:-n < x < 3n,

z1 xJ IfI > A for a z such that

xx < z < 3n}.

By introducing the continuous function

xF(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 lfIJ

fn

i

AND.FOURIER TRANSFORMS

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:

lf(x + u)ldul,?(x) =suplh-1

10h

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).

389

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 hsuph 11 gi(x + u)du = h-11 If(x + u)ldu,i o 0

so


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 IIpi 2

(Actually the argument above is valid for every measurable func-

tion 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+Ifln

Ilf°Ilp , Apllf111, 0 < p < 1.

SOLUTION (b): As each function Hn is bounded, so is each Kn, and

the convolution product Kn*f is therefore defined everywhere,

moreover

((n

I(Kn*f)(x)l < ?nJ- Hn(y)lf(x - y)ldy.E

Writing

(1)

Jll'(u)dun(y) = n(n) -y


and changing the orders of integration, this yields

JE(Y)If(x

7c x- y)Idy = Hn(n)JOIf(x - y)Idy

391

(n u- J H'(u)duJ If(x - y)Idy.

0 n 0

Foru>.0

uJ If(x - y)Idy 6 of (x),0

so

JH(Y)If(x - y)Idy 6 00 (n) + JQIuH'(u)Idu)f (x).

Similarly,

0 (JH(y)If(x - y)Idy (iH(-n) +_x n n

so that by (iii) and (1)

I(Kn*f)(x)I < '1(Hn(n) + Hn(-n) + 2B)?(x).

Furthermore, the relation:

YyH (y) = fo H(u )du + Jb'uIc(u)dun

(2)

x n

s J:Hfl(U)dU + f0juU1(u)jdu,


r0 0

-du + I JuH'(u)ldu,n n( - n) 4 I H

m mn _n

and consequently, by (ii) and (iii):

J( n(n) + Hn(-n)) < A + B.

Returning to (2) one obtains

I(Kn*f)(x)l 4 (A + 2B)fA(x).

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 y2 f

K(x,y) = P(x,y) + iQ(x,y) = nz

f

(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:

JP(xPf(z) = - t,y)f(t)dt.

Determine limPf(x + iy) when f is the characteristic func-y;0

tion of an interval [a,b].

1 xQ(x,y) =

2

(c): Drawing inspiration from the preceding exercise, show


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

fact that from every bounded sequence ofLm

one can extract a sub

sequence which converges weakly in this space).

(e): When f e Lp(Ik), 1 6 p < -, set:

Kf(z) = JK(x - t,y)f(t)dt,

+Qf(z) = 1mQ(x - t,y)f(t)dt.

Show that Kf is holomorphic in the half-plane H. Next show

that liiKf(x + iy) exists for almost all x. (Reduce to the casey-*o

where f < 0, set g(z) = expKf(z), and use parts (c) and (d) above.


Hf(x) = limQf(x + iy).y->0

exists for almost all x. (Hf is called the HILBERT TRANSFORM of

f)-


(f): Calculate the Fourier transform Q0,y) of Q(-,y).

From this deduce that 1IHf 112 =If 112 when f e L2 (M).

(g) : Let f e L1Ot), f 3 0. Show that for all e > 0 and ally > 0

logjl t pKf(t + ie)Idt = logll + tiKf(z + ie)I.

f-m (x - t)2 + y


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 0 and

(b

P(x,y)dx = 1 (tan-1b - tan-1 a ). (1)a

SOLUTION (b) : Note that P( ,y) a Lq(,t) for 1 4 q < which as-


sures the existence of Pf(z). Moreover, (1) shows that if f is

the characteristic function of [a,b] then

ifa<x<b,lim?f(x,y) i if x = a or x = b,

0 otherwise.

SOLUTION (c): First note that -PX(u,y) > 0 if u >. 0, and that

I -uP'(u,y)du = 2( u2du -_ 1

0 X 0 (u2 + 1)2 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,


from which the desired inequality follows. This being so, let

(Tk) be a sequence of step functions such that II(Pk111 =OR-4)

and

f = ITk

in LThere exists a set N1 of measure zero such

that

f(x) =

k

(pk(x), x4N1.

Furthermore, for all z e R,

Pf(z) = I Prpk(z).k

(2)

(3)

Finally, by part (b) there exists a set N2 of measure zero, such

that

1imPpk(x + iy) _ (P k(x) for all k and all x 4N2'

y->o

Since

I meas{x:suplPcpk(x + iy)I > k-2} = IOR-2)

<

k y>o k

there exists a set N3 of measure zero such that if x *N3

:

supIPpk(x + iy)I 4 k-2y>o

(4)

starting from some k0

(which may depend on x). In other words,

when x 4N3the 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 (x) = f(x).y->o k y-Yo k k k

When f e LPOR), 1 < p 4 -, for every integer N 3 1 let us set


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. Furthermore,

IPfN(x + iy)l (t)I dt.It 1;.N

(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.

SOLUTION (d): For c > 0 and R > 0 let rc R be the loop formed by

the segment [-R + ic,R + ic] and the semi-circle E * is + Rein

0 < 8 t n. For all z e n, whenever 1/c and R are large enough,

g(z) =2n1

g()iJ

r

-c,R

l0 = 2ni J dre,R

so by subtraction it is deduced that

g(z) = r - z- z d.c,R

As the function g is bounded, the integral along the semi-circle

is 0(1/R), and consequently

_ V- (t + ic)g(z)

nE. t + is - z t + 2c -dt. (5)

Now, let (ck) be a sequence that decreases to zero. Extracting


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

that co < y. Since

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.

SOLUTION (e): The functions 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)

I

1 21 If(t)I when zeC.

((a + t )Z - R)

The function of t which appears in the right side of this inequal-


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, asy-*O

well as limPf(x + iy), moreover. Sincey40

expiQf = gexp(-Pf),

it follows from this that

limQf(x + iy) = Hfy40

exists for almost all x. When f is not negative it is written

as a linear combination of four negative functions.

SOLUTION (f): The function belongs to L2(kt). Its Fourier

transform is, by definition, the limit in L2(1R) of the functions:

i M xsin2nlxr -2nil;E

J -Me Q(x,y)dx - - dx as M 3 m.

M n-M x2+y2

As the integral on the right has a limit as M 3 -, one deduces

from this that

Q(E,y) = - if+C0

xsin2niax ax = - if+" xsin2nyEx d,.

_ ,Xx2 + y2 n1 x2 + 1

It is a classic result that for u >, 0,

1+W

cosux dx = e-u.nJ -m x2 + 1


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}'. Since

Hf almost everywhere,

it follows that Hf a L2OR), Hf = Y7, and therefore that

IIHf1I2 = II?II2 = IIYYII2 = 117112 = IIfI12.

SOLUTION (g): Since

IK(x,y)I s ny

it is clear that


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

Re(z) - 1. By considering the principal value of the logarithm

in this half-plane, the function log(l + pKf(z)), which is holo-

morphic in IT, is defined unambiguously; moreover, its modulus is

bounded by

2+logll+pxf(z)I <2+pIKf(z)I <+ " IIfIIl

(since logll + pKf(z)I > 0). For all e > 0 the function

g(z) = log(1 + pKf(z + ie))

is therefore bounded and holomorphic in H. Since

ling(x + iy) = log(1 + iKf(x + ie))y+0

we have (cf., part (d))

+Wlog(1 + pKf(t + ie)YJ- dt = log(1 + pKf(z + ie)),

n (x-t)2+gand on taking real parts,

+W logll + pKf(t + ie)Idt = logll + pKf(z + ie)I.If_C0

(x - t)2 + y2

We have observed that

nlogll+pKf(z+iE)I (Y+e) IIf1I1,


furthermore, since Kf = Pf + iQf, as Pf,Qf are real and Pf 3 0,

we have

log+IIQf1 < log 11 + 1,KfI.

so that

2

(x-t) 2+b2log+IuQf(t+iE)ldt<uy+E 11f 111.

When y ± oFatou's Lemma gives

(+mlog+lpQf(t + ie)Idt < ullflll.

But then for all A > 0

meas{t:IQf(t + iE)l > A}.log+(ua) < uIIfIl1,

and on making u = e/a,

meas{t:lQf(t + iE)l > A} < a 11f111

If f e L1(R) is complex, one deduces that

meas{t: I Qf(t + iE) I > a} < ae 11f111,

SOLUTION (h): Note that if f e L2(R)

IIf112,

and consequently

(6)

A) 4 a2 11/112.


Coupled with (6) this allows us to use Marcinkiewicz's Theorem

(cf., exercise 6.134). For 1 < p < 2 there exists a constant AP

(independent of E) such that

E) Ilp , Ap Ilfllp

if f e E, the space of step functions on R. To extend this result

to the case where 2 < p < -, let us observe that if f,g a L2(R),

then

(Qf(,E)lg) = ((,e)JI0) = - (M(,E)s) - - (fIQg(,E)).

If

p

tq

= 1, then 1 < q < 2, and if f e E,

IIQf(,E)IIp = sup{ I(Qf(,e)Ig)I: eE,11gllq. 1}

= sup{I(flQg(,t))I:geE,IIg II q. 1}

AgIIfIIp

Now assume that f e LP(R), 1 f in LP(R). For all t eR

IQf(t,E) - Qfk(t,c)I s IIQ(,E)IIgIIf - fkllp,

and Fatou's Lemma allows us to conclude that again in this case

IIQf(,E)IIp c Apilf1Ip

(setting AP = Aq if 2 < p < =). By part (e), for almost all t

limQf(t,c) = Hf(t),E-ro

404 CHAPTER 8

and another application of Fatou's Lemma yields

IIHfIIP < APIIfIIP

CHAPTER 9

Functions with Bounded Variation:Absolutely Continuous Functions:Differentiation and Integration

EXERCISE 9.163: Show that if f is absolutely continuous on [a,b]

then so are the functions If1p for p a 1.

ovo = vov = ovo = vov ®nvo

SOLUTION: This follows immediately from the fact that for p 1

and 1z11 6 M, Iz21 ( M,

1Iz11p - Iz21p1 4pMp-1121 - z21.

EXERCISE 9.164: Let E be a subset of [0,1] with measure zero.

Construct an increasing and absolutely continuous function on

[0,1] such that at every point x e E.

ovo = vov - ovo = vov = ovo

SOLUTION: Let (V 1) be a sequence of open sets such that

E C Vn meas (V 1) < 2-n.

The function

405

406

g : I IVn n

is integrable. Let

xf(x) = Jg(t)dt.0

CHAPTER 9:

The function f is increasing and absolutely continuous. Let x e E.

For every integer N there exists a > 0 such that Iy - xj < a im-

plies y e V1 if 1 4 n<, N. Then

f(y) - f(x)>, N,

y - x

which proves that f'(x) = +-.

EXERCISE 9.165: Let (cn) be a sequence of strictly positive num-

bers such that

OD

n=1 n

(a): Show that one can construct a sequence of intervals

In = ]xn - en,xn + en[ such that xn 4 xm if n 4 m, the sequence xn isdense in ]R and every point of IR belongs to an infinite number of In's.

(b): With the In's the same as in part (a), consider the

function f such that f(xn) = cn and f(x) = 0 if x is distinct from

all the x 's.

Show that if limcn = 0 f is differentiable at no point of iR.

tVA = VAV = AVA = VAV - AVo

SOLUTION (a): One of the two series E c2n' X c2n+1 is divergent.

Assume, for example, that E c2n = -. Let (zn) be a sequence of

BOUNDED VARIATION etc. 407

points everywhere dense in It and such that zn 4 zm if n # M. Let

I2n+1 = ]zn - c2n+1'zn + c2n+1['

Next, consider integers nl < n2 < such that

L c2n > pnpcn<np+1

By induction, for all p one can determine some distinct points yn

(np .< n < np+1) distinct from all the zn and from the yl,y2,...,

yn.1

such that the intervals I2n= ]yn - c2n'y + c2n[ cover [-p,p].

pThe sequence of In's so obtained satisfies the required conditions.

SOLUTION (b): It is clear that

liminff(x) = 0 < f(xn) = cn.x;xn

The function f is therefore not continuous at xn, and in partic-

ular it is not differentiable. If x is distinct from all the x 'sn

then

liminfIf(y) - f(x)y yx = 0.

Furthermore, Ix - xn

I < en for an infinite number of integers n,

and consequently

Cl y pI

f(yy - X(x ) I ln

a measurable set of [0,1] such that for

some a > 0

meas(Ef) [a,b] ) , a(b - a)

408 CHAPTER 9:

for any 0 < a < b t 1.

Show that meas(E) = 1.

AvA - vAv = AvA = vAv = eve

FIRST SOLUTION: On setting f = nE the condition of the problem

becomes

1b

Jf(x)dx ? a.-aaBy using Lebesgue's Differentiation Theorem, one deduces from this

that for almost all x e[0,1] f(x) > a, and consequently f(x) = 1.

This shows that meas(E) = 1.

SECOND SOLUTION: Let

`pn = AIL [0,1/n] I n = 1,2,... .

The functions rpn form an approximate identity in L1(1R). There ex-

ists therefore, a subsequence (ns) such that f*9n -; f almost every-s

where. If 0 < x .< 1, and if n is large enough, then

(xf*cpn(x) = n1 f(y)dy >, a.

x-1/n

From this one concludes, as above, that f(x) = 1 for almost all

the x e [0,1].

EXERCISE 9.167: Let (fn) be a sequence of increasing functions on

[a,b]. Assume that the series

W

f(x) = I fn(x)n=1


converges for all x (a 4 x < b).

Prove that for almost all x e [a,b]

f'(x) _ f'(x). (FUBINI'S THEOREM)n=1

eve = Vev - ove = vov = ovo

SOLUTION: By considering the functions fn - fn(a) one can assume

that fn 0 for every integer n. Let

sn = f1 + ... + fn.

By a theorem by Lebesgue which asserts that an increasing function

is differentiable almost everywhere, one is assured that at almost

all the points of [a,b] all the numbers sn(x) and f'(x) exist.

When a < x <y4 b

8n(y) - 8n(x) -4 sn+1(y) - an+l(x) t f(y) - f(x),

and consequently

sn(x) '< antl(x) 4 f'(x)

almost everywhere. This proves that almost everywhere:

fn(x) <n=1

Now consider a sequence n1 < n2 < such that

f(b) - sn (b) < 2-P.P

Then for all x e [a,b]

410 CHAPTER 9:

E (f(x) - sn (x)) < E (f(b) - sn (b)) 0P

for almost all x. As the sequence sn(x) is increasing almost

everywhere,

8,(x) - f'(x)

which accomplishes the proof of Fubini's Theorem.

EXERCISE 9.168: Let E be a set of 3t that is not necessarily meas-

urable.

Show that at almost all the points x e E

limm*(EnLx -h,x+h]h;0

2

where m*(A) denotes the outer measure of A.

eve = VAV = AVA - VAV = AVA

SOLUTION: Recall that the outer measure of a set A is defined by

m*(A) = inf{meas(V):V an open set, and A C V},

and that


m*(AUB) < m*(A) + m*(B).

First let us show that if A c [a,c] and B c [c,b], then

m*(AUB) = m*(A) + m*(B).

In fact, for all e > 0 there exists an open set V such that A U B

C V and

meas(V) < m*(A U B) + e.

Let

V = Vn]-'.,e + c[,

V+ = vn ]c - c,+oo[.

Then

V = V U V+, meas(V n V+) < 2c, A C V, B C V+,

and consequently

m*(A) + m*(B) < meas(V) + meas(V+) = meas(V) + meas(V n V+)

6 m*(AUB) + 3c,

which proves the result.

Second, note that generality is not lost by assuming that E is

contained in a compact interval [a,b]. Let n be an open set con-

taining E and such that

meas(V ) < m*(E) + 2-nn

For all a 4 x 4 b let

412 CHAPTER 9:

f(x) = m* (En [a,x] ), fn(x) = meas(Virl [a,x] ).

Everything comes down to proving that f'(x) = 1 at almost all the

points of E. If a .< x 4 y 4 b, then by the preceding

(fn(y) - f(y)) - (fn(x) - f(x))

- meas(Vnf) [x,y]) - m*(E1i[x,y]))3 0.

From this it follows that the functions fn - f are positive and

increasing, and that fn - f 4 2-n. Fubini's Theorem (cf., the-,

preceding exercise) applied to the series

I(fn - f)n

implies that for almost all x

fn(a) - f'(x) - 0

It remains to observe that if x e E then fn(x) = 1 for every inte-

ger n.

EXERCISE 9.169: Let f be a continuous function on [a,b] and let

V be its variation on this interval (0 : V 4 W).

Show that for every A < V there exists d > 0 such that

n--iL If(xitl - f(xi)I > A

i=0

if

a = x0 < x1 < < xn = b and max(xitl- xi) < 6.

(* )

ovo - vov - ovo = vov = ovo


SOLUTION: Let a = y0 < < yP = b be such that

p-1

A = E If(yi+l)-f(yi)I

>A.

i=0

There exists 6 > 0 such that ix - yI < 6 implies

If (x) - f(y)I < (AA)

2p

If a = x0 < < xn = b and max(xi+l-x.) < 6 the contribution

to the right side of (*) of the terms corresponding to i's such

that yj E xi < xi+l<

yj+l is greater than

If(yj+1) - f(yj)I -(A

P

X)


n--1

G If(xi+l) - f(xi)I > A - (A - A) = A.i=0

EXERCISE 9.170: Let f be a real continuous function on [a,b].

Show that

+W

V(f;a,b) = J- n(t)dt,

where n(t) denotes the (finite or infinite) number of solutions

of the equation f(x) = t (BANACH'S THEOREM).

A0A = 0i0 - AVt = 0A0 = 40t

SOLUTION: If A = (a = x0 < xl < . < x = b) is a subdivisionP

of [a,b], set as usual

414 CHAPTER 9:

-C1

A(f) =

p

L lf(xi+l) - f(xi)1,i=0

I A I = max(xi+1 - xi:0 4 i s p - 1).

Let fA be the function that coincides with f on A and which is af-

fine linear on each of the intervals [xi+,,xi]. Let no(t) be the

number of solutions of the equation fo(x) = t.

First show that

JnA(t)dt

nA(t) 4 n(t) if t4 f(A).

(1)

(2)

Let us denote the numbers of solutions of the equations f(x) = t

and f,(t) = t in ]xi+,,xi[ by vi(t) and pi(t). When t4 f(4) we

have

p-1 p-1n(t) = I v.(t), nA(t) = I pi(t),

i=0 i=0

and furthermore, pi(t) = 1 or 0 according as t does or does not

belong to the open interval with end points f(xi) and f(xi+1

Therefore

J p.(t)dt = lf(xi+1) - f(xi)I and vi(t) 3 pi(t)

by the Intermediate Value theorem. This proves (1) and (2).

Now note that if A C A' then

fo = (fA,)o and f(o) = fo.(o)

so that (2) gives


nA(t) 4 no. (t) if t 4f(o).

Consider a sequence (AS) of subdivisions of [a,b] such that

AS C ASt1' limIASI = 0.S

(3)

(4)

Let D be the union of the f(A )'s and of the maxima and minima of

f on the subintervals of an arbitrary AS. In other words, if t 4

D, and if a,B are two consecutive points of a subdivision AS, then

t is distinct from f(a) and f(B) and from the maximum and minimum

of f on [a,B]. The set D is denumerable, and we are going to show

that

limn (t) = n(t) if t 4D.s S

Since, by the preceding exercise

limts(f) = V(f;a,b),s

(5)

(6)

the equation to be proved will follows from (1),(3),(5),(6) and

Lebesgue's monotone convergence theorem. To prove (5), consider

t 4D and let k be an integer such that k .< n(t). Let E1 < E2 <

< Ek be solutions of f(x) = t. By choosing s0 large enough,

AScan be assume to separate the E , that is to say in each inter-

val corresponding to AS there is nor more than one i. If a < E10

< B, where a and B are two consecutive points of AS , two cases0

can arise: either f(a) - t and f(B) - t (which are not zero) have

opposite signs, in which case the equation f,6 (x) = t has a so-So

lution in ]a,B[; or these two numbers have the same sign. If they

are positive, for example, then since t is not the minimum of f on

[a,B] there exist 7,6 such that a < y < d s0 so that AS has a point in [y,d],1

416 CHAPTER 9:

and then the equation f. (x) = t has at least one solution ins1

]a,8[. Observing that this property is not lost by increasing

al. one can thus operate step by step and obtain an integer s 3

so such that nA (t) a k. Taking (2) into account, this certainlys

implies (5).

EXERCISE 9.171: Let F be a function defined on [a,b] and 1 < p <

Show that in order for there to exist f e Lp(a,b) such that:

XF(x) = F(a) + J f(t)dt, a 4 x 4 b,

a

it is necessary and sufficient that:

(*)

)IPn-1 IF(x.-

F(x i

< WI (**),_isup

A i=O (xi+1 - xi)p

where the supremum is taken over all partitions A = (a = x0 < xi

< < x = b) of the interval [a,b].

Show also that under these conditions the left side of (**)

equals 1IfIIP.


SOLUTION: In what follows, with A being given, set

L. = xi+l - xi, F. = F(xi+l) - F(xi).

Now assume that (*) holds, with f e Lp(a,b). If

p

+

a

= 1 then

F1I = xitlflC of/q((xi+llfiP)1/p'

x. x.


and since p/q = p - 1 this yields

nIl JAFilpA-(p-1) nel xi+ll.flp= jblflp (1)

i=0 i=0 JXi a

Next assume that (**) is satisfied and let M be the value of the

left side of this relation. Let us first show that F is absolute-

ly continuous. For that, consider a finite sequence (ai,bi) of

intervals contained in [a,b] and mutually disjoint. Using Hol-

der's inequality for finite sequences this yields

CIF(bi) - F(ai)Ipli/p

a1 /qIF(bi) - F(ai) 11

( (bi - ai)) /qi i (bi - ai)p i

< M1/pq (bi - ai))1/q

so that

lim E IF(bi) - F(ai)I = 0.E(b.-a.)-'0 i

Having thus proved that F is absolutely continuous, it can be

asserted that (*) holds for some function f e L1(a,b). It remains

to show that f e Lp(a,b). If A is a decomposition of [a,b], let

fA be the function equal to AFi/Di on the (i + l)-th subinterval.

As F'(t) = f(t) for almost all t, it is clear that if As is a

sequence of partitions such that maxAi - 0 then -} f almost

severywhere. Furthermore

jblfolp =

nIlInFilpn(p-1)

6 M.a i=o

Therefore by Fatou's Lemma

418 CHAPTER 9:

bP< M,JIfI

which, taking (1) into account, completes the proof.

EXERCISE 9.172: Let 1 < p < m and

p

!-+.L = 1.

Show, using the preceding exercise, that for every contin-

uous linear function u on LP(R) there exists g e Lq(R) such that

u(f) = Jfg, f e LP(1t).

AVA = vov = AVA = vov = AVA

SOLUTION: Let

{u1

(O,xl ) if x 0,

F(x) _- u(Il(X,01 ) if x 0,

so that

F(B) - F(a) = u(1(a,01 ), a < B. (1)

Let A > 0 and A = (-A = x0 < xi < < xn = A). Observe that if

n-1

g =i0

e11(Xlpxl+il

then using the notations of the previous exercise,

Inii c.AF 1.I = Iu(g)I <

IIuII(n11

Ic1.IPA

1

)1/P

i=01 i=0


In this inequality substitute

r 1 IAF,"Iq-2Ur-

Aq-1i

if F1 $ 0,

c i =

0 otherwise

This yields

n-l IiF'.1q In-1 IAF.1p(q-1) 1/p

2=0 Q _IIu Il

CC

j=0 &p(q-1 -1I i

= IIu II

n-l J A Fi l

q 1/p

i=0I

so

n-1 I AF .i1 q

111

4 Ilullg.i=0 Aq

i

By the preceding exercise g(t) = F'(t) exists for almost all

t e [-A,A], and

xF(x) = J g(t)dt, IxI 4 A,

0

+A

J-AIg(t)Igdt < IIu1Iq.

In view of the arbitrariness of A, g(t) = F'(t) exists for almost

all t eR, IIg IIq < IIu II, and

XF(x) = J g(t)dt, x e1Z. (2)0

420 CHAPTER 9:

Let

Ug(f) = Jf(t)g(t)dt, f eLY(it).+

This defines a continuous linear function on Lp(12), and taking

(2) and (3) into account (1) can be written

u(Il[a sl) = ug(n[a $] ), a c a.

As the functions IL[a,sl

form a total set in LP(R) we therefore

have u = ug.

EXERCISE 9.173: Two functions f,g that are positive and measur-

able on ]0,1[ are called EQUIMEASURABLE FUNCTIONS if

meas(f > y) = meas(g > y) for all y > 0.

(a): Show that if f,g are equimeasurable, then

J

1 1

0

f(x)dx =J0g(x)dx.

(b): Let a be a positive Borel function on [0,'[.

Show that if f,g are equimeasurable, then so are sof and

Bog.

(c): With f positive and measurable on ]0,1[ show that

there exists an unique function f* that is positive, decreasing,

right-continuous on ]0,1[, and equimeasurable with f. (Show that

{y:y > 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 JE To


and that for every positive and decreasing function w on ]0,1[

1 1

1

f(x)cp(x)dx 4J

f*(x)q(x)dx.0 0

(e): If f is positive and integrable on ]0,1[, then for

0<x<1 set

(( (u x

Of (X) = sup! 1 uJ f(t)dt, 0 1 then

II$fllp 0 there exists a number A depend-

ing only upon c such that

422 CHAPTER 9:

(1 1 1

J Of(x)dx 4 2 f(x)log+f(x)dx + AJ f(x)dx + E.0 0 0

AVA = 4A0 = 00A = 000 = 0VA

SOLUTION (a): This property follows from the equation (cf. exer-

cise 5.98)

f

1 W

f(x)dx = J meas(f > y)dy.0 0

SOLUTION (b): It is easily verified that if f,g are positive on

] 0 , 1 [ then the set of sets A C [0,W[ such that f 1(A) and g-1(A),

are measurable, and

meas(f 1(A)) = meas(g-1(A))

is a a-algebra. When f,g are equimeasurable this a-algebra con-

tains all the intervals ]y,'[, y > 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. In other words there exists T*(x) >. 0 such

that


(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>x ),nn

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 suf-

ficient that for all y > 0

cp(y) = meas(f > y) = meas(g > y) = meas(gl> y) = g*(y),

that is to say, by (2), that gl = T*.

SOLUTION (d):

31JE = J0E.f =

JQE*.

424

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

J

eas(E)

Ef)otJ

fineas(E)

J f - fE 0 0

In particular, if 0 < c < 1, then

fO E JO rt

From this it is immediately deduced that

J1 1

0fW< J0f*,

CHAPTER 9:

(3)

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): Let

XF(x) = Jf(t)dt, 0 < x < 1.

0

425

The set (9f > y) is formed of the x's for which there exists u

such that 0 y) = E (bn

- an

)

n

1by an =

yJ(8f>y)f

1 meas(8f>y)I

0f*.

y

Therefore when meas(Of > y) > 0

1 rmeas(8f>y)

meas 8f > y J0(4)

aNote that since f* is decreasing, a-1J decreases as a increas-

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

(8f)*(x) sxrx

f*0

SOLUTION (f): As the functions soBf and so(8f)* are equimeasur-

able,

JosoBf = Joso(8f)* S Js{.Jf*}dxoo .

SOLUTION (g): In particular, if s(x) = xP, p > 0, then

PAP -< JOjoJPdx.

After extending f* by zero on [1,W[ and setting

JxG(x) = x-1 f*,

0

(5) and Hardy's inequality gives

IIof1Ip 4 IIGUP s p Ilf llp = p IlfllP

since f and (f*)P are equimeasurable.

(5)

SOLUTION (h): If 0 < p < 1 then Holder's inequality for the pair1 T-1p

p1 applied to (5) yields

BOUNDED VARIATION etc.

110f 11P F P-- p)O xP 0 x

rr10 -11-PLJ0

xPJp J

< IIf IIiJO xP

so

II$f IIP <1 1 pl 1/P

IIf Ill.

SOLUTION (i): Finally, if p = 1, then starting from (5), one

obtains

J

OOf 61 0 J o = I of*(t)dt1 t = 10f*(t)log tdt.

Let e > 0 and choose a such that 0 < a < 1 and

f

a

log1 s0

< .

Then

f

1 a 1 1 1

of1

f*(t)log t dt + log a 1 f(t)dt.0 0 0

i

Let E be the set of t 's such that 0 < t < a and f*(t) 4 t

Then

427

Jf*(t)log..dt < Jiog t + 21 f*(t)log+f*(t)dt < (Contd)VT

o

428 CHAPTER 9

1

(Contd) e + 2Jf(t)log+f(t)dt,0

whence, at last,

J1 1 1

Of 2 flog+f + log

1-JOf +

0 Jo a

CHAPTER 10

Summation Processes:Trigonometric Polynomials

EXERCISE 10.174: (a): If the series

W

Iun=0

n

converges, and if

0

n0 Ixn - An+ll <I

show that the series

00

L Anunn=o

converges.

(b): Conversely, prove that if the series (3) converges

whenever the series (1) converges, then (2) is satisfied.

ovp = vov - nvo - vev = ovn

(1)

(2)

(3)

429

430 CHAPTER 10: SUMMATION PROCESSES

SOLUTION (a): (2) implies that An = A t en, withcn

3 0. As

L len - entll < °°and suplu0 + + unI <

the Abel-Dirichlet Theorem shows that the series E enun is con-

vergent. 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-1fN(s) = I ()n - An+1)sn t ANsN.

n=0

This linear functional is continuous, and it is a classical re-

sult that its norm is

N-1

IlfN ll = E l An - Xntl I + I AN1'n=o

Since

NfN(s) = A0s0 t =an(sn - sn-1),

n1

and as the series I (sn - 8 n-1) is convergent, limfN(s) existsN_

for every s e E. By the Banach-Steinhaus Theorem one therefore

has

Su "fNII<

which implies that:

TRIGONOMETRIC POLYNOMIALS

G Ian - xn+1I<

n=0

Second Proof: Let

An - an+1 = rJn, rn > 0, IEnI = 1,

Ensn+rO+r1+ +rn

U 0 = 3 0 , un = sn - sn-1 if n >. 1.

If it is assumed that

G Ixn - xn+1I G rnn=0 n=0

431

then sn - 0, and the series E un is therefore convergent. Fur-

thermore,

N N-1C

0

Anun = aNsN +n 0I

(an - an+1)snnI.

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 rC

n-01+r0+ n


and tends to infinity with N by a classical property of positive

divergent series. It follows that the series Anun is divergent:

EXERCISE 10.175: Let (un)na0 be a decreasing sequence of positive

real numbers.

Show that if

then

un = of n).

ovo = V1V = tV1 = VAV = ovo

SOLUTION: Let

ENn=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: Let (an)n30 be a sequence of real numbers. Set:

_ 2 _fan

= an - an+1,1 a

n == a 4a n+1'

TRIGONOMETRIC POLYNOMIALS 433

The sequence is called a CONVEx SEQUENCE if

02an > 0 for all n > 0.

(a): Let c be a convex function on [0,-[. Show that the

sequence an = 9(n) is convex.

(b): Show that if the sequence (an )n>0 is convex and bound-

ed, then

(i): It is decreasing;

(ii): L1an = o(1/n);

(iii): L (n + 1)t2a. = a0 - lima..n=0 n

(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): This follows from

2an = an - 2a n+1 + a

n+2

= 2['9(n) + 3(p(n + 2) - p('n + L(n + 2))].

SOLUTION (b): The condition A2a 0 for all n means that the

sequence (dan) is decreasing. If, for some integer r Aar < 0,

then for every integer n > r

n-1an = ar - E Aas ar + (n - r)( Aar),

s=r


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 2 N-1

I

(n + 1)A an =I

Aan - NAan = a0 - aN - NAaN,n =O n =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+lif k 0, q)1 = c0,

lpk-+ 0. (1)

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)


enk< (P

k+l'

(Pk-1 (Pk

<(Pk-2 (Pk-1

nk - nk-1nk-i nk-2

(4)

(5)

This is possible because the sequence (cn) tends to zero. Note

that (4) is valid for all k > 0. This being so, let m be such

that cp(nk) = k and that is linear on each of the intervals

[nk,nk+l].This function is decre sing and tends to zero at

infinity by (1), and it is convex by (5). If nK< n < nK+1,

then by virtue of Relation (4)

en < en 5 9(nk+1) < (P(n).k

EXERCISE 10.177: Let (un,p

)n30,p

be a double sequence of com-

plex numbers, such that

W

(i): X lu 1 4 M for all p >. 0;n=0 n,P

W

(ii): l}i.)m I uP+00

nP

= 1;00 n=0 ,

(iii): un,P = 0 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


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 In

and let us consider on e0 the linear functionals

cc

fp(e) _ I un Pen.n=0 '

It is clear that these latter are continuous and that

cc

IIfP II S E I un, p I . M.n=o

(1)

It is a question of proving that fp(c) -> 0 for all c ec0. Let

en be the sequence such that em = 6nm. From the relation

N

I1c - Ccnen II = sup I en I

n=0 n>N

one deduces that the eats form a total set in c0. By (1) it

therefore suffices to prove that for all n

limfp(en) = 0.

But this is nothing other than (iii), because

f (en) = uP n,p

TRIGONOMETRIC POLYNGd'4IALS 437

In the following six exercises the following notations are

used:

If (cn)ne2Z is a sequence of complex numbers, set:

CC

c

80+81+. . .+8N

sN Ini<N n' GN N + 1

tN InI`N Inlcn

It will be recalled that the seriesL

cn is called a CONVER-

GENT SERIES if limsN exists. It is called a CESARO CONVERGENTN

SERIES if limaN exists. By the preceding exercise convergenceN

1implies Cesaro convergence. (Let uN,p + 1 for N s.

AVA = VAV - AVA = VAV = AVA

SOLUTION: Let s' = limsN. If a'< then s = a', by what wasrecalled in the note above. If s' _ -, then for p < N

s0 + ... + apN

N' N+1 +Nsp'which implies that s , s

p, and consequently s = s' =

EXERCISE 10.17,: (a): Show that limaN exists if and only if theN

series


W

I N 2tNN=1

converges. (Use exercise 10.174).

(b): Assume that aN -> s. Show that sN - s if and only if

tN = o(N).

(c): Again assume that aN -)- s. Show that if for some real

number p , 1

GInlp-llcnIP

<

then sN -), s.

(d): Prove the following result (HARDY'S LEMMA): If cn =

0(1/n) then aN -> a implies sN -> s. (Assume that for a sequence

N1 < N2 < tN , BNS, B > 0, for example, and show thats

tN +v >' 2 BNS if 0 4 v 42 A

NsS

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

tN = n cn = I n(c + c_n) = E n(sn sn-1InI,N n=1 n=1


tN = N(sN - aN-1), (1)


which proves (b). Now

sN = (N + 1)GN - NoN-1'

and substituting this expression in (1) yields

tN = AN + 1)(aN- 'N-1

).

Therefore, aN converges if and only if the series

(a - a ) N=i AN + 1)N N-1NL

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): If p = 1 the condition becomes

E ICnI < 00,

439

(2)

in which case sN converges to a value that can only be s. There-


fore assume that p > 1, and let q be such that

p

+ q = 1; it is

now a matter of proving that N 1tN -+ 0 (cf., the part (b)). Now,

if r < N then

It

N

Since

< NI I nlcnl + NIInll/P InI1 - 1/pcnl

Inl-4r r<lnl.<N

Nq/P l/q p -l pl/p.I cI + 1(2 n ) ( Inl I I )

N InI<rn N

n=1 lnl>rn

NC

G nq/P .< 1 (N + 1)q,n=1

it follows that

tNN 4

N L cnI + I

?11/q N + 1( C lnlp-11 nlp)1/P

I ni4r Ini>r

and consequently

tN [2]1/q

N InIP-1IC lp)1/plimssupii < 4N

InI>rn

By making r + - the desired result is obtained.

SOLUTION (d): Assume that aN -+ s and that

Icn + c-nI E n , n

Let us show that N1tN

- 0 under these conditions, which by part

(b) will prove that sN - s. Otherwise there would exist a B > 0

and an increasing sequence of integers Ns

such that


tN > BNs.S

If VS is the integral part of BNs/2A, then for 0 < v < vs

tN +v= tN + I n(cn + c-n

s s N <n.<N +vs S

>. BN - vAS

BN BN>BN

S2 -2ss

441

Furthermore, by part (a), the series I N 2tN must be convergent,

and in particular

L

7 2 4

NN ' N<N +vs S S

must tend to zero as s But this quantity is bounded from

below by

sBN N+v -2 B vs B2> 0'2 j

s

Nx 2 Ns +'vs 2B + 4A

S

a contradiction.

REMARK: We have implicitly assumed that the cn's were real. If

this were not the case, decompose them into their real and imagin-

ary parts. Moreover, in assuming that N-1 tN

did not tend to zero

we supposed that limsupN1tN

> 0. If this supremum were zero then

necessarily liminfN 1tN < 0, and it would suffice to change the

signs of all the cn's.

442

EXERCISE 10.180: Set

CHAPTER 10: SUMMATION PROCESSES

=8P + 8

p+1+ +

8p+Nap,N N + 1

(a): Express ap,N as a function of the aN's.

Deduce from this that if aN-> s then a

p-* s for every

,NP

sequence of integers (Np) such that NP > ap, a > 0.

(b): Prove that if ian! < Alni for n >, 1, then

.Iap,N - spI < A N

(c): Deduce from the preceding a new proof of Hardy's Lemma

(cf. the preceding exercise).

AVA - VAV = AVA = VAV = AVA

SOLUTION (a):

ap,N N + 1

(p + N + 1)ap+Npop-1

ap+N + NN p+N - ap-1

If the sequence (Np) is such that Np > ap, a > 0, then

QN< a< 1

(1)

so that if aN-> s then a -> B.

p,Np


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 G (N + p + 1 - n)(cn + c-nn=p+1

N

p N2+ 1r =

pr=1

SOLUTION (c): If Icnl . AIn(-1 for n a 1 and aN - s, let

E p sup IaN - 0 1I,P N>p-1 P

so that ep -> 0. Let N be the integer such that

N 4pp<NP+1.

Then by equation (1) proved in part (a)

IoP,Np - op+Npl p

On the other hand, by the bound obtained in part (b),

P,IaN - 81 < AP

p p

This proves that sP - s.


EXERCISE 10.181: Assume that cn -> 0 and that en = 0 except when

n = 0,±n1,±n2,... with

p+l>, q > 1.p

n

Show that if aN 4 s then sN -> B. (Let ap = max(Icn l,ic-n 1)

p pand show that for np .< N <

np+l

N 1tN < 2 ¶ argr p;r=1

then use exercises 10.177 and 10.179).

ADA = VAV - AVA - DAD = ADA

SOLUTION:

n-E >, qp-r if p >, r,nr

so that if np < N < np+1' then

IN 1tNl = IN 1 ¶ nr,(cn + c_n )Ir=1 r r

<2 a nr 2 ¶ a qr-p.

r=1 r np r=1 r

Now note that ar + 0, and that

l im I qr-p = qPer=l q-1'


limgr-P = 0.

P4°°

By exercise 10.177 it follows that

lim L argr P = 0,P-'°° r=1

and therefore that N1tN

-> 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,

then the series

+00

alnlgn

is convergent.

SN = 0[a N)

SOLUTION: Using the notations Aan,42an of exercise 10.176, we

have

N N

I alnlcn = a0c0 + I an(cn + c-nn=-N n=1

= a0s0 + an(sn - sn-1)n=1

(Contd)

446

(Contd)

CHAPTER 10: SUMMATION PROCESSES

N-i= a

N8N

+ I Aa.8n=0

N-i

= aN8N

+ E Aan((n + 1)an - no n-1)n=0

N-2= a

NaN

+ NAaN-1aN-1 + E (n + 1)A2anann=0

By hypothesis aNaN

-r 0. Moreover, as the sequence (an) is convex

and bounded and further (cf. exercise 10.176)

00

E (n + 1)IA2an.an1 s A (n + 1)A2an = a0A.n=0 n=0


lien a c = (n + i)A2a a .

N-n n

EXERCISE 10.183: 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},

N

M(f) = sup{If(x)I:0 < x 4 2n}.


Prove that

I

- 21 1 fmf (1 11 1i n n+ 1 inf M : f e E'J < l2 + J 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.

J71/2n f(x)dx = 2 cp - c1 + 3 - 5 + <

2m(f).

By noticing that

it CO - c1 ;., (2 - iJcp 3 12 - 1)n + 1

and that :

we obtain:

m(f) Ii 21 1l nJn+1

On the other hand, if:

447

1 + cosx + + cosnxP x) = n + 1


an elementary calculation shows that:

1 1 sin(n + Z)xf(x) =n+12+ 2s in x j -

It follows from this that if n/2 < x < ir:

If(x)I 2n+1 (ltsi-n n) _

EXERCISE 10.184: 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

A = IlfII = f(x0).

Show that if Ixl < n/N then

f(x0 + x) 3 AcosNx.

(c): f is no longer to be assumed to be real. Show that

Ilf'II. < NRfIIW. (BERNSTEIN'S INEQUALITY)

AVA = VAO = AVA = DAD = AVA

SOLUTION (a): If

f(x) _ cne1lnlkN

consider the polynomial


P(z) = zN C 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

then gE(xr) >, e if r is odd and gE(xr) < -e if r is even. From

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 inter-

val [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 (Contd)

211f 11 > f (x0 + 2N) - f (x0 2N, = J- f' (x0 + x )dx >.n/2N

450 CHAPTER 10

(Contd)rn/2N 2m

mJ_ cosNxdx =N ,

n/ 2N

which proves the result in this case.

In the general case let x be a real number. There exists an

a such that lal = 1 and If'(x)I = af'(x). Let us set of = u + iv,

where u and v are real. Then

If'(x)I = u'(x) 6 Nlu I

' NIIafIIm = NIIfil..

CHAPTER 11

Trigonometric Series

EXERCISE 11.185: Let z e a, z $ . Prove the following formulas

ezx=e2nz - 1(2z 1 +

Xzcosnx - nsinnxl

0 < x < 21;n

n=1 z2 + n2 ) I

zx _ eTIz - 1 2e +

[(-1)nenz - 1] z2osn2nz n

, 0 < x < n;n=1 z +n

W

ezX = n E [1 -(-1)nenz] nsinnx 0 < x < R.

n=1 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). Therefore

451

452

e2nz - 1 1(n) = 2n z n

Since

2tcz(0)

= 2nz

n $ 0.

CHAPTER 11:

the first formula is proved (Jordan-Dirichlet Theorem). For

x = 0 the sum of the series is

z(f(x + 0) + f(x - 0) = (e2nz + 1)


W

ncothnz = 1 + V2z

z n=1 z2 + n2

Now consider the functions g and h, with period 2n, respectively

even and odd, and which coincide with f on ]0,n[. Then

[g'] = zh, [h'] = zg,

so

ing(n) = A(n),

inTi(n) = zg(n) + 2n (2 - 2e1Ze-nn).

Therefore if n 4 0 then

g(n) _(-1)nenz - 1 z

n z2 + n2,

di(n) = 1-(1-(T nZit 2 2z + n

TRIGONOMETRIC SERIES

Moreover

nz

9(0) =e

nz 1 ,7Z(0) = 0,

453

which proves the two other formulas. In fact the second is valid

for x = 0. From this it is easily deduced that

m

sinhnz z +C

n=1

(_1) n

z

+

n2

EXERCISE 11.186: Find the sums of the following series:

I acosnxc

nsinnx2,2 2n=1 n+ a n=1 n2 + a

SOLUTION: Let

W

g(x) _ Iacosnx

n=1 n2 + a(1)

inz

G(z)2e 2 z = x + iy, y >, 0. (2)

n=1 n + 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

iz

G" - a2G =ae

Re(z) > 0.e - 1

(a real, a + 0).

(3)

But the G can be analytically continued to a minus the two real

454 CHAPTER 11:

half-lines ]--,0] and ]2n,+m[. From this it follows that g is

infinitely differentiable (and even analytic) on ]0,2n[ and that

on this interval satisfies the differential equation:

g,, - a2g = aRe

eix a

le ix 2

Hence

g(x) = acosha(x - n) + Ssinha(x - n) - 12a

for 0 < x < 2n, and hence also, by continuity, for 0 < x < 2n.

Since g(0) = g(20, 0 = 0.

Now,

r212asinhanit

0

so that

acosnx = ncosha(n - x) - 1

n=1 a2 + n22sinhna 2a '

0<x< 2n.

The derivative of the right side is continuous on [0,2n], hence

one can differentiate term by term, giving

00

C nsinnx = nsinha(n - x)

n-1 a2 + n22sinhna

0<x<2n.

(The formula is false at x = 0, for at this point the function

with period 21 that coincides with the right side for 0 < x < 2n

possesses right and left limits equal, respectively, to n/2 Ad

-n/2).

TRIGONOMETRIC SERIES 455

EXERCISE 11.187: Find an expansion of x2 as a trigonometric ser-

ies that is valid for 0 < x < 2n. Also, determine expansions of

x2 as a series in cosines, then in sines, valid fo 0 < x < n.

ovo = vov - ovo - vov = ovo

SOLUTION: If f has period 27 and f(x) = x2 for 0 < x < 21, at

zero this functions has a step equal to -4n2. We have [f'](x) _

2x, 0 < x < 2n, so therefore

[f ](n) = - -r , n f 0,

and

in71(n) in - 27,

so

?(n) = 2-

2n

2 znn

4723 '

2 = 47 + 4 c cosnx - nnsinnx 0 < x < 2n.3 n=1 n2

If x = 0 the right side is equal to

l(f(x + 0) + f(x - 0)) = 272,

2

which recovers the well known formula

456

C 1 i2n==1 n2 - 6 .

If g is even and g(x) = x2 for 0 < x .< it, then

[g'](x) = 2x, -n < x < it,

hence if n # 0

[g'](n) _(_1)n+i 2

in '

and consequently

Furthermore,

Thus

x2 = n2 + 4 c (-1)3n cosnx

n=i n2,

For x = 0 we recover the formula

CHAPTER 11:

(1)n+in-2 = n2

n=i12


Lastly, if h is odd and h(x) = x2 for 0 < x < it, then

[h'](x) = 21sT,

and h has a step equal to -2n2 at the point it. Since

2 1 - (-1)n[h'] (n) _ -it 2

n + 0,n

and

inle(n) = [h ](n) -ne-inn

we have

lt(n) _ -2 1 - (_1)n

+ (-1)n+lit

It . 3 inzn

and consequently

457

2 n+l sinnx 8 sin(2n + 1)xx =2n E(-1)n -n 3 0<x<1.

n=1 n=1 (2n + 1)

REMARK: It is known that the first series has 2'x as its sum;

therefore

sin(2n+1)x=1 (=_x2),n=1 (2n + 1) 3 8

04x< it.

EXERCISE 11.188: Prove the following formulas:

I sinnx (_) lcot'zx, 0 < x < 2n,n=1

+mC einx (- 0, 0 < x < 21.

n=-m

458 CHAPTER 11:

(The (c) over the equals sign indicates that the equation holds

when the sum is taken in the sense of Cesaro).

AV4 = VOV = AVA = VAV = IVA

SOLUTION: If 0 < x < 2%,

N _ cosjx - cos(N + ')xsN =

Il sinnx 2sin2xn

and consequently

N

aN = cot'x -2(N + 1 sin

T- Ixcos(n + 2)xn=

=0

_ cot ix -sin(N + 1)x

4(N + 1)sin2ix

which proves that

am -> -fcotix.

Similarly, for the second sum

N inx sin(N + ')xsN = 1 e = DN(x) =

sin ixn=-N

and

_ 1 sin(N + 1);x 2oN =

FN(x)N -+I ( sine,

so aN -* 0. Not that if x = 0, then

sN= 2N + 1,

so


aN=N+1+,,.

EXERCISE 11.189: Let f be a continuous function.

Show that the following two conditions are equivalent:

F(i): There exists a function continuous on the closed

disc lzi 4 1 and holomorphic in the open disc lzl < 1, such that

f(x) = F(eix), x em;

(ii): f is periodic with period 2n, and ?(n) = 0 if n < 0.

AvA = vev = ova - vAv - eve

SOLUTION: (i) =a (ii): For all 0 < r < 1 and all n > 0,

0

-znF(z)dz = irn+l

2nei(n+1)xF(reix)dx.

1lzl=r 10

As the function F is continuous on lzl E 1 the limit r -* 1 may

be taken, yielding

?( - n - 1) = 0, n 0.

(ii) => (i) : If:

N

PN(z) = E I 1 - N + 1)?(n)zn,n=0

then by Fejer's Theorem

PN(e ix ) + f(x) uniformly on I .

By the Maximum Principle, it follows that

lim ( sup IPM(z) - PN(z)l) = lim ( sup IPM(z) - PN(z)l) = 0.N,M+- l z l c1 N,M+- l z l =1

460 CHAPTER 11:

Therefore

PN(z) -* F(z) uniformly on z < 1,

is continuous on IzI .< 1, holomorphic on IzI < 1, and

f(x) = F(eix) if x e]R.

EXERCISE 11.190: Let f e L1(T). Show that the following condi-

tions are equivalent:

(i): If(n)I = O(e-cInl), e > 0;

(ii): f is the restriction to]R of a function holomorphic

on the strip IIm(z)I < 6, 6 > 0;

(iii): The series

+00?(n)einz

n=-oo(*)

converges on a strip IIm(z)I < 6, 6 > 0;

(iv): The series (*) converges at two points, z1 = x1 t i61

and z2 = x2 - i62, with 61,62 > 0.

4V4 = 040 = 4MA = VAV = t0A

SOLUTION: Let

an(z) = ?(n)einz +?(-n)e-inz

Then

2inz1

2inz inz1

inz

(e - e 2)j(n) = an(z1)e - an (z2 )e2


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

n62 -n61

1?(n)l < Me + e - 0(e-elnl),

le2n62 - e-2n611

461

if c = min(61,62). 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< J k I < e6e ,

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. Let

anEn

462 CHAPTER 11:

be its Laurent expansion. Since the circle IzI = 1 is contained

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): Let f e C00(T). Show that if

IIf(S)II0 = O(Rsr(as + 1)), R>0, a>0,

?(n) =0(InI1/20'e-(InI/R)1/a).

(b): Now assume that (*) is satisfied. Show that

11f(s)II = O(ssRsr(as + 1)), 0 = (a - )+.

(c): Deduce from this that f is analytic if and only if

If(n)I =O(e-elnl)

for some e > 0.

AVA = VAV - OVA = VAV = AVA

SOLUTION: (a): If A is such that

IIf(S)II < ARSr(as + 1), s a o,

then for all n 4r 0


1 (n)I = I(in)-sf^ (n)I < AI IRIJsr(as + 1).n

Let us take for s the integral part of (1/a)(Inl/R)1/a. Then

ISIl r(as + 1) < (as)-asr(as + 1) ti

e-as 2.lnl

Since

I I

J

1/a

- a < as < [J..L]/a

(*) certainly holds.

SOLUTION: (b):

+00

f(s)(x) = E (in)sf(n)e11"`, s > 0,-00

and consequently

IIf(s) II00 = 0( 1 ns +1/2a e- (n/R)1/a

).n=1

Let a = a + 1/2a and

(x) =xae-(x/R)1/a

ax 0.

9 attains its maximum at

xo = R(aa)a.

463

Let Na be the integral part of xa. Then

464 CHAPTER 11:

Nspa(n) F Ja()dx + q) a(NQ) + 9

a ( Na+ 1) +

1N+1Ta(x)dx

n=1a

S (Pa(x)dx + 2(p a(xa).

Now,

J(x)dx = aRo+lr(aa + a),

0

cpa(xa) =Ra(aa)aae-aa

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)),n=1

y = max(a,2),

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


IIf(s)II Co(1)= O(RSal), R > 0.

By part (a) above, (1) implies that

I?(n)I = 0(InI2e-InI/R) =0(e-£InI)

if 0 < e < R

On the other hand, if

IJ(n)I = O(e-EInI),

then by part (b),

IIf(s)II0o

= 0(s2E-SS!) = O(Rss!) when R > 1 .g

EXERCISE 11.192: Show that if f e L1(T) and:

Co

g(x) _f(n)einx

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 every-

where.

EXERCISE 11.193: Let f e L1(T). Show that if

Ifcx + t) + f(x - t) - 2sI tt <I0

466 CHAPTER 11:

then

?(n)elnx = s.

n=-

nvn = vnv - nvn = vnv = nvn

SOLUTION: In fact

(DINT'S TEST)

r

a (f;x) - s =1J

n

(f(x + t) + f(x - t) - 2s)sin(Nt+ J)t

dt0

+ eN(x),

with eN(x) - 0. Since the function

t 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. Why?).

Show that then

?(n) = 0(n-a).

nvn = vov - nvn - vnv = nvn

SOLUTION: In fact


(2n= 4nj OX) - flx + n, )e-inxdx,

0 l J

so

?(n)I 2(rc)a.

EXERCISE 11.195: Let f e L. (T).

467

(a): If at a point x where f(x + 0) and f(x - 0) exist, one

has for0<t<6,

f(x + t) + f(x - t) - f(x + 0) - f(x - 0)I s Mta, a > 0,

then the Fourier series of f converges at this point to

1(f(x + 0) + f(x - 0)).

(b): If f e Lip(a) the Fourier series of f converges uni-

formly to f.

nve = vov = eve o vov = nvn

SOLUTION: (a): Let

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

It is known that

a (f;x) - J(f(x + 0) + f(x - 0)) =1 6((t) sin(Nt+ j)t dt

N0x

+ eN(x),

468 CHAPTER 11:

where

lime (x) = 0.N

By the hypothesis made, t-lcx(t) is integrable, so the result

follows from the Riemann-Lebesgue Theorem.

SOLUTION: (b): If f e Lip(a), f is continuous and it is known

that in this case eN(x) - 0 uniformly in x. Moreover,

ITx(t)I < Mta,

which proves the result.

EXERCISE 11.196: If

f(x) =IxI-a0(x), 0 < a < 1,

where 0 has bounded variation on show that

I?(n)I =0(na-1).

400 = vpv = AVA = VAV = AVA

SOLUTION: One can assume that is increasing and positive on

Then

I Ipnlj

11

elxla axlnl

Now,

n -: nx

J

e dx = na-l(G(nn)- G(in)),

C Ix la


where

rx -itG(x) =

I

e dt.0 0 Itla

It is known that G(x) tends to finite limits as x + ±m; IGI is

therefore bounded by a constant M, and consequently

If(n)I .<

Mo(n) na-1

EXERCISE 11.197: Let f e LP(T), 1 $< p 4< -. Set

fl=.f, fk - fkfk-1 for k 3 2.

(a): Show that:

P(f) _II.fkII/kP

exists.

(b): Show that if p(f) = 0 then f = 0 almost everywhere.

ovo = VAV = ovo = vov = MA

SOLUTION: Note that if f,g e LP(T) then

II.f*gIIP < IIfll1IIaIIP S IIfIIPIIgIIp

SOLUTION: (a): It follows from what was noted above, that, set-

ting

ak=IIfk11pI

470

one has

ak+k 4akak' ak4 IIfIIP1 2 1 2

It is then classical that

p(f) = limak/kk

exists.

SOLUTION: (b): For n ea and k = 1

1?(n)Ik = I?k(n)I , IIfkIIl s IIfkIIp'

which implies that

I f(n) I < P(f), n e 2Z.

CHAPTER 11:

Therefore, when p(f) = 0 one has = 0, and consequently f = 0

almost everywhere.

EXERCISE 11.198: Let f e L1(T). Assume that for some a > D

27E

If(x t h) - f(x)Idx =0(Ihla).

0

(a): Show that if

f1 = f and fk = f"fk-1'k 3 2,

then

fk e L2(T) when k >2a


(b): Show that fk coincides almost everywhere with a con-

tinuous function when k > 1/a.

AV4 = VtV = 4VA = VAV - 4VA

SOLUTION: Note that

IXf(n) = 1

2n

{f(x) - f+ n)}e-inxdx,

0 J

so

If(n)I =0(Inl-a).

SOLUTION: (a): It follows from (1) that

Ifk(n)I =0(Inl-ak),


E Ifk(n)I2 < W if k > 21an

But then fk e L2(T).

SOLUTION: (b): If k > 1/a, then

G I?k(n)I <n

(1)

in which case fk coincides almost everywhere with the sum of its

Fourier series, which is absolutely convergent, and hence contin-

uous.

472 CHAPTER 11:

EXERCISE 11.199: Let f eL1(T) be such that f(n) = 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

Elnlp-ilf(n)I

< -,

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,

rxg(t)dt = (f(n)el"x - f(n)).

0

Since

Y If(n)I < -,


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' eL2,

then

I I?(n) 14 IL I11 + " IV' II2 .n


SOLUTION: In fact, 11(0)1 : If II1, and

E I1(n)I 4 (I n-2)2(1 Inf(n)I2n40 n40 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): 11f111<1+E.

OVA - VOV = AVA = VAV = AVA

SOLUTION: If A C [-r,r] let

f(x) = (2N + 1)-1( elnx )( L elm),

where N is an integer to be determined later. It is easy to see

that

474 CHAPTER 11:

1 if Inj 4 r,

f(n) = 1 - 2N + 1 if r < Inj < r + 2N + 1,

0 if Inj >r+ 2N + 1.

Using the Cauchy-Schwarz inequality and Bessel's equation, it

follows that

211

1If(x)I dx < ON + 1)-1(2N + 1)2 UN + 2r + 1)2

27E0

2N+2r+1( 2N+ 1

If N is chosen sufficiently large, then

ILrII <1+e.

EXERCISE 11.202: Let ((pi) be an orthonormal basis of L2(T).

Prove that

LOA - VAV = AVA = VAV = AVA

SOLUTION: If f e L2(T) then

I

2= I

I(fIki)I2.if112

2

In particular, letting f(x) = einx ,

-Ti(n) 12 = 1, n ea.


Now consider a sequence (cn)nea such that

c > 0, E c =n nnW c2 <

onn

475

and let g eL2(T) be such that g(n) = cn. If I I1pi111 < °° were tohold, then we could write

E cn = I In n i

_

i n

_ (g*pilgi)

= G (g I Ai,ec"pi)Z

I19112 l1q'i111,

which is absurd.

EXERCISE 11.203: Let 1 < n1 < n2 < < np

< be a strictly

increasing sequence of integers. Set

N inxfN(x) =

N

ep

P=1

Show that for almost all x

limfN(x) = 0.N

AVA - V AV - t0A = VAV - t0A

476

SOLUTION: By Plancherel's Formula

CHAPTER 11:

Cm 21[

2n L J If2(x)I2dx = 2

m=1 0 m m=1 m

and consequently for almost all x:

< °°,

limf 2(x) = 0. (1)M m

If m2 c N < (m + 1)2, then

2

fN(x) - N fN

2(x) =NI

I e

in x

pm

2p=m+1

(m+1)2-1- m2` 2N rN-

Since m2/N -' 1 as N -* -, it follows that

limfN(x) = 0N

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: For every integer k > 1 let Vk be an open set such

that


E C Vk C ]0,2n[, meas(Vk) E 2-k

Let9k

be the characteristic function of Vk and f the function,

with period 2n, equal to G k9k on [0,2n]. If 0 kcpk

implies that

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) 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

1

pi(t)dt_

pi(t)f(t)dt = 2n(9i*f)(0)JatlxI 1t n

2nf(0) = 0.

This shows that (pi) is an approximate identity, because, fur-

thermore, Ti > 0 and

IIcpiII1 = ?i(o) ->- I.

ExERCISE 11.206: Let (an)n0 be such that an > 0 for all n, andA } W. in

Show that there exists a continuous function f such that

liml upxInl I f(n)I I al", I f(n)I2 = ._W


in x

(Consider a series I An'Ie s , where the integers ns increases

quickly enough).

nvn = vov = ovo = vov = ovo

SOLUTION: Let n1 < n2 < be a sequence of integers such that

The series

1 in x

f(x) _ antes

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,

E a1n1If(n)12 = I X"I?(ns)I2 = E 1 = .

n s S s

EXERCISE 11.207: Let (an)ne2Z be a sequence of complex numbers

such that

E 1Xnf(n)I < for all f e L1(T).n

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 - N + 1, lanl < A,lnl<N l

and by making N--> W this yields

L Ian1 < A.n

EXERCISE 11.208: Let (), n)ne2Z be a sequence of real numbers such

that


X a 0 , E X=- E X 2

n n

Show that there exists a function f e L2(T) such that

(i): If(n)I = o(an);

(ii): 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 .

co is a Banach space. If u e c0, there exists u e L2(T) such that

u (x) '\ Xanune",

n

and

g1122 = E anlunl2 < IIuI12 E an

n n

The mapping u -; u is therefore continuous from c0 into L2(T). If

I is a compact interval of length greater than zero, let

pI(u) = ess sup I u(x) I .

xeI

482 CHAPTER 11:

It is clear that pI is a semi-norm on co; this semi-norm, fur-

thermore, is lower semi-continuous. In fact, for every function

g, continuous on I, the mapping

is continuous on c0 and

pI(u) = sup{IJIugl:g 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'sO

Theorem). If one had AS +O one would deduce, taking the convex-

ity of AS into account, that for some p > 0,

huh 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]. If

is n

hull < 1, and u(r) _ (une r nea'

then also

11U(r)11 F 1,

and by (1)

ess sup Iu(x)I = ess sup I c(x + ar) I =pl(u(r))

4

PI+a I

r


(since u(x+ar)=u(r)(x)). Hence s/p would hold if Ihu1141.For every function f e L1(T) the mapping

2n

u 2nI z7(x)f(-x)dx0

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). By the preceding exercise

this would imply -

IXn <

n

contrary to the hypothesis.

Let (1k) be a sequence of compact intervals with lengths great-

er 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. The setk

484 CHAPTER 11:

E = n Ekk

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 Ik

EXERCISE 11.209: Assume that

cn ' cn+1 3 0, en < n , n > 1.

(a): Show that

N

I

I c sinnxl < 2Av.n=1

n

(b): Deduce from this that the functions

W

f(x) = E cnsinnxn=1

belongs to L1(T), and that the right side is its Fourier series.


SOLUTION: (a): One can assume that 0 < x < it. Let M be an ipte-

ger between 1 and N - 1. Noting that

sinnxI < nx and Isinpx + + singxl <si- x < x 9

1


one obtains

N M NAn

1 csinnx1.1 111+1 E 15AMx+ M+1x.n=1 n=1 n=M+1

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.

REMARK 1: f is continuous on ]0,n], since

I ensinnxi 4x aN+l'

n=N+l

REMARK 2: Plancherel's Theorem shows directly that there exists

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): The series (*) is uniformly convergent;

(ii): The series (*) is the Fourier series of a continuous

function;

(iii): an = o(1/n).

(c): Show that the following conditions are equivalent:

(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): an = 0(1/n).

(d): Show that the following conditions are equivalent:

(i): The series (*) is the Fourier series of a function

of L1(T);

(ii): The sum of the series (*) is a function in L1(T);

(iii) : E n-1 a < -.

Show that in this case the partial sums of the series (*)

converge in L1(T).

000 = Vov - AVO = vav = ovo


SOLUTION: First recall that the formula:

2sinnx.sin-x = cos(n - J)x - cos(n t J)x

shows that for N < M and 0 < IxI 4 it

D (x) =N

sinnx =cos(N - ')x - cos(M + })x 1

N,M n=N -

2sin ( )


IDN,M(x)I< since

iT , 0 < IxI 5 n.

On the other hand,

M M-1I a sinnx = aMDN M(x) + E Aan.DN n(x),n=N

nn=N '

where

4an= an - anti

3 0.

From this one deduces that

Mp naN

InLNansinnxl < 'T , 0 < IxI < it.

2)

(3)

(4)

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 I ansinnxl} = 0.N-' x gN+1,<n4N

In particular,

lim a sin nn = 0.N- 2N+14n4N n 2N

Now,

Na

I asin 2N NZN+itn<N

n2V2-

which proves that NaN - 0, and thus that (i) => (iii).

Let

eN = supnan.n>,N

If 0 < IxI < it, there exists an integer p such that

it

FT-1

(5)

Using (4) with N replaced by N + p and letting M one then

has

I I ansinnxl 4 1 1 ansinnxl + I : ansinnxIn>.N N4n<N+p n>'N+p

IxI I na +naN+

Non<N+p n ix

.pIXICN+(p+1)xN+P

so that, because


pIxl < it and (p + 1)aN+p .< (N + p)aN+p S EN,

we have at last

a sinnxl < (n + 1)eI

nN_n3N

(6)

which proves (iii) =>;(i).

If the series (*) is uniformly convergent, its sum is a con-

tinuous function f whose Fourier coefficients are obtained by in-

tegrating 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,

hence sinnx a2nx

= N, when n < N. Thus

n1a2N(77E

N) ;' n1N (1 2N + 11 N 2N n

2N I 1 ) a

n<N

We have thus proved that (ii) => (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

proved that (i) => (ii). On the other hand, if (ii) is satisfied,

the aN are uniformly bounded, and the inequality

a2N (2'N, > 4 (N + 1)aN

proved in part (b) shows that (ii) => (iii).

SOLUTION: (d): Letting N = 1 and M - in (3) yields

f(x) _ ansinnx =L

Aanb1 n(x).n=1 n=1 '

Note that by (1)

cosix - cos(n + )x sin2 x ,D1,nx_-

2sinx tangy+ "2sinnx,

sin2'nxg(x) =n=1

Dan tanx

h(x) = 2 1 ansinnx.n=1

Since

Aan > 0 and E Aan = a0,

the function h is continuous (note, furthermore, that as the ser-

ies which gives h is absolutely convergent, this series is cer-


tainly the Fourier series of h). Furthermore, because

sin 22nx ' 0 for 0 < x 4 n,tanix

g >. 0 and

(ncWc ' n

Og(x)dx =n=1

°anj0 tan xax

491

(7)

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

J71

f(x)sinnxdx = Da

lrn(x)sinnxdx = Aa = a

n.

nk=1

k nJIn

1,k k=n k 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-1an

<

Now, since

1-

2tan x X

is integrable on (0,n), we have

JO sinntx dx = 2J1 s x dx + 0(1) =ji7c

+ 0(1)z

n logn,

492

and because

lognti1+2+ to,

(ii) is therefore equivalent to

I1 t2

+ t .1 Aa <

11

nn=1

n

Since

(8)

NI n = I1 +

2+ ... + N)aNtl + 1 I1 +

2

+ ... + nlpan

n=1 n=1 l 11

it is clear that (iii) implies (8). The converse is also true,

because

I1 t

2

+ +NJaN+l =

[i++ ... +

n:Ntltan

1 {1 +

2

+ + nldan.n>.N+1

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 N '

But by the preceding

CHAPTER 11:

II51 nIll ti 1 logn,


hence by (8)

IIf-SNII130.

493

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 + ancosnxn=1

converges for 0 < x < 2n, and that

Co

f(x) = 2 1 (n + 1)A2anFn(x),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 eL1,

and that if:

an + 0, n&an -> 0, 1 (n + 1)IA2anI < m,n=0

then the right side of (1) is the Fourier series of f.

(1)

(2)

evo - vev - 4aL - vov = pv4

494 CHAPTER 11:

SOLUTION: (a): We have (where Dn is the n-th Dirichlet kernel):

a N N8N(x) = 2 + E ancosnx =

2 =an(

n(x) - Dn_1(x))

n=1 n=0

N-Ci

2 aNDN(x) +

2

G AanDn(x).n=o

Since

D0 + + Dn = (n + i)Fn,

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:

ID 1 , IF (x)l < 1N(x) sin x n(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:

2f(k) _ (n + 1)62anFn(k)n=0

m

= I (n + i - k)t2an.n=k


Now,

N

E (n + 1 - k)a2an = ak- aN+l

+ (N + 1 - k)DaN+1'n=k

Consequently, if, in addition,

NAa -> 0n

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

-

n a

1

f(t)dt <aJ-f(t)dt.

1 it a

(b): Let (cn)ne2Z be a sequence of positive real numbers.

Suppose that there exists d > 0 and he L2(-S,S) such that

+m a

E ccp(-n) = 71-h(x)(p(x)dxn=-m

nS

(*)

for every function cp that is infinitely differentiable and has

compact support contained in ]-S,S[.

Show that

aC2

<

aJ

Ih(x)I2dx.n=-w 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. Then

0<W< aZn and W(x)=0 ifa< Ixl <n.

Furthermore,

?(n) = 18(n)12 > 0.

Therefore

2nfa f(t)dt > 1

it +W

f(t)c(t)dt = 2n = f(n)s(- n)a n n

2nf(0)$(0) =I8(0)I2n

f(t)dtfn

2 n

(2 n) f _nf(t)dt,

and the result follows

SOLUTION: (b): Set, as usual

(f19) = J_1r.

For s = 1,2 ... let 85 a C'(T), 85 0, 11-5sll l = 1, and 85(x) = 0if 1/s << 1x1 < it. Then

lim85(n) = 1, (1)S

for all n ea. Denote by Cam(-6,5) the set of infinitely differen-

tiable functions with compact support contained in In

the usual manner Cam(-d,d) is identified with a subspace of CS(T).


s*0s e C (-d,d) whenever s is large enough. Applying (*) to this

function yields

I en1;S(n)I2 <n

Hence one can define a continuous function

gs(x) = Ian1SS(n)I2einx.

n

Let 0 < y < 6 and p e C (-y,y). Whenever s is large enough

p68s°e4s e C and consequently

(gSIW) _ a 1iS

(n)I2 ThT= (hlpda8se8s)n

= (h* *9sk,P)

Therefore

gs = h*e JS on ]-Y,Y].S

Taking the complex conjugates of both sides of (2) and then

multiplying term by term, one sees that the Fourier coeffi-

cients of Igs I2 are all positive. By part (a) one therefore

has

IIgs112

Y

2 YJ-YIh*Os*8sI2 Y Ilh*ss*;s 112 <2 Y Ilhll2,

and on making y - 6:

IIgsII22 6 IIhlI2.

(2)

(3)

Taking (1) and (2) into account, we have limgs(n) = cn8

498

and, using (3),

CHAPTER 11:

IIcn12

= sup EIcn12

= sup(lim X (0 s(n))2)

n N InkkN N s Inl4N

< anIIh112.

EXERCISE 11.213: With every function f e L1(T) there is associ-

ated its ADJOINT FOURIER SERIES:

I -isign(n)f(n)e1RX,

where

-1 if n s -1,

sign(n) = 0 if n = 0,

1 ifn1.The ADJOINT FOURIER SUMS of f are the numbers

SN(x) = I -isign(n)f(n)eln".

InI,<N

(a): Show that

,s* = f*D*N N'

where DN is a function to be determined.

(b): Show that

ti 2logN,J D(u)du0

(c): If c is an integrable function such that

TRIGONOMETRIC FUNCTIONS

lime(u) = 0,u+0

show that

J e(u)DN(u)du = o(logN).0

(Set

AN(u) = J(u) - sinNu,

and note that

AN(u) 0 if 0 '< u 4 n).

(d): Deduce from the preceding that if

t = lim(f(x + u) - f(x - u))U-0

then

3N(x)lim

NtogN

499

(e): Then prove the following result: If f has left and

right hand limits at all points, and if

If(n)I = ° n(1)

then f is continuous.

AVA m VAV m AVA - VDV ° AVA

SOLUTION: (a): It is clear that

500 CHAPTER 11:

NDN(u) = 2 1 sinnu.

n=1

Multiplying both sides by sinNu we find

(1)

D*(u) = cosju - cos(N + ')u0 U (2)

N sinNu

SOLUTION: (b): By (1),

it ["(N-1)] 1-n-.7JD(u)du=2

0N n=02

ti 2logN.

SOLUTION: (C): It is easily seen that for 0 < lui 4 it,

ON(u) = DN(u) - sinNu - (1 - cosNu)cotju.

By the Riemann-Lebesgue Theorem:

Let a > 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

iJ eA*1 < an

J

A* +n

J e(u)cotiudu + o(1).0

N oN d


Since

JOAN = JODN + 0(1)

ti 2logN,

from this one concludes that

limsup(logN)-11E J E 2a,fo

n

N

and, since a is arbitrary, that

J

EAN = 0(logN).0

By (3) we also have:

J0CDn

= o (logN).

501

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 (4)N

502 CHAPTER 11:

SOLUTION: (e): If f has left and right hand limits at all points

then £ exists at every point x and is equal to

f(x + 0) - f(x - 0).

Further, if

then

f(n) = a n11 J '

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

AU cn n

Set:

u1 + 2u2 + + nun

an = n ,

bn = n ,

W

yn = n L ussin22n

s=1

u2 + 4u2 + + n2u2

Show that the conditions an -> 0, bn - 0, yn -* 0 are equiv-

TRIGOROMETRIC SERIES

alent. (Prove that

bn Aan, a 2n < bn yn

and that for every integer v

n2vb +A

2

Yn 4 nv v

(b): Let f be a real function with left and right hand

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)I2r=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.

503

Show also that if f is continuous and has bounded variation,

then (pn -> 0 uniformly on R.

(c): Calculate:

r2n

J pn(u)du,0

using Plancherel's Formula.

(d): With the aid of what has gone before, prove the fol-

504 CHAPTER 11:

lowing theorem (owed to Wiener): A necessary and sufficient con-

dition for a function f with bounded variation to be continuous

is that:

lim f(1) + 2f(2) + ... + nf(n)= 0.

n n

AVA = V/V = AVA = VAV = AVt

SOLUTION:2 2

.<(a): Since s us Asus, it is clear that b .<n Aan.

The Cauchy-Schwarz Inequality shows that

aIu2 + 4u2 + + n2u 2 2 = b12.

l Jn n

Furthermore, noting that sinu . 2u/n if 0 . n I us(;8 )2 = bn.s=1

yn < n X us (2n) 2+ A2n L 2s=1 l s>nv s

2 r 2 2A"

4bnv + A2nl a2 _

n4vbnv +

v1nv x

All this shows that an - 0 is equivalent to bn - 0, then that

yn -)- 0 implies bn + 0, and finally that if bn -* 0 then for every

integer v > 0

2l i nsupyn - A

and consequently yn + 0.


SOLUTION: (b): 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 <u+2n,

and then

E + 2kn # u + xr for all r.

Hence there exists r such that

u+xz,< +2kn<u+xr+1, 04r<2n.

By (1) and the periodicity of f, this implies that

If(u + xr+1) - f(u + xr)I > d,

and we have thus proved that

9 (u) > d2 if n > n0 and u $ E + nip.

If it is now assumed that f is continuous and has bounded var-

iation, 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-1qpn(u) < e I If(u + xr+1)

-f(u + xr)J e EV(f;0,2n).

r=0

506 CHAPTER 11:

This shows that n i 0 uniformly on1R.

SOLUTION: (c):

f(u + xr+l) - f(u+ xr) ti

(eisxr+1 - eisxr)f(s)eisu

s

is(x + n/2n)(( ll

ti 2ie rsinl2nJf(s)eisu.

s ll

Plancherel's Formula gives

m(

gn(u)du = 8n Y IJ'(s)I2sin212nJ

2

2nl 0 s= l

and as f is real, ?(-s) = f(a), so that

(27[ M

2j0

pn()du = 16n = Ij(s)I2sin2[22n-

s=1

SOLUTION: (d): By the results proved in parts (b) and (c),

n If(s)I2sin2[In- 0

s=1 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

_ I ascosst.S=O

(b): Denote by yp the number of indices s such that:

0<s< 3 (4p- 1), as40.

Show that

= 3P.yptl - yp

(1)

(c): Deduce from this that there exists a continuous func-

tion f with period 21, such that for all x

xf(x) x + limJ (1 + cost)(1 + t cos4p-1t)dt.

p-0 0(4)

(d): Show that f has bounded variation and that its Four-

ier coefficients are not o(1/n). Calculate f(4n)).

ovo = vov = ovo = VAV = ovo

SOLUTION: (a): First note that

(1 + cost)(1 t cos4p-1t)

is a linear combination of functions cosst for

0 < s < 1 + 4 + + 4P-1 = 1 (4P - 1).

Multiplying this combination of functions by cos4Pt, one obtains,

508 CHAPTER 11:

for s 4 0, the two terms

cos(4p + s)t,cos(4p - s)t.

The coefficients 4p + s and 4p - s lie respectively between 4p t1

and 3 (4p+1 - 1) and between 3 4p +

3

and 4p - 1. Now

(4p - 1) <3

4p + 3

which shows that there exists a sequence (as) satisfying (1) for

all p > 1. By what has gone before it is clear that as = 2-u

with u eJi, for all s such that as # 0; in particular, 0 < as < 1.

SOLUTION: (b): -yP+1- yp is equal to the number of terms which

appear upon multiplying the right side of (1) by cos4pt and re-

placing the products of cosines by sums; by the analysis carried

out in part (a)

yptlyp = 2(yp - 1) + 1 = 2yp - 1,

and consequently:

(yp+l yp) = 3(yp - yp-1

One can show that y2 - yl = 3, so

3p.yptl - yp

SOLUTION: (c): Let us denote by Fp(x) the value of the integral

which appears in the right side of (2). We have

aIFp+1(x) - Fp(x)I .< .. < (Contd)

(4p-1)/3<s4(4p+1-1)/3


(Contd)

Yp+1 - yp = 3p+1

3(4p-1) 4p-1

Therefore -x + Fp(x) converges uniformly on ]R to a continuous

function f. Furthermore, the integral over [0,21] of

- 1 + (1 + cost)(1 + cos4p-1t)

509

is zero. Hence -x + Fp(x) has period 21. Therefore so does f.

SOLUTION: (d): For all p,FF is increasing; thus the same is true

for its limit. But then f has bounded variation because it is

the difference of two increasing functions. The Fourier coeffic-

ients of f are the limit of those of -x + F (x), so

a sinsxf(x) S

s=1s

It is clear that:

2if(4s) =4-s

which proves that si(s) does not tend to zero.

EXERCISE 11.216: The object of the exercise is to construct a

function f e Lip(a) whose Fourier coefficients are not o(n-a) (cf.

exercise 11.194). Let a > 1 and 0 < a < 1. Set

f(x) _ a nacosanx, -n < x 4 R.n=0

(a): Let u a 0. Show that

(If(x + a-u) - f(x - a-u)I 4Aa1-a) [u] -u + Ba-c'

510 CHAPTER 11:

where A,B are constants depending only upon a,a, and where [u]

denotes the integral part of u.

Deduce from this that f e Lip(a).

(b) : Set

F(x) =sinTrx

if x # 0, F(0) = 1.

Prove that

(1-a)n

f(s) _ a nFlan - s).

n=1 a + s(*)

(c): Let 0 < A < loga, and let n0 be an integer such that

n0 1a >2loga-A

For every integer v > n0 denote by sv the integer such that

v- 1-zsav<sv+

Show that

IF(an - s )I < 1 ifn>v,V t(avloga - J)

and

IF(an - sv)I < 1n if n0 s n < v.naa

(d): Deduce from this that

Vf(sv6 > 0


whenever v is large enough, and that, in particular, ,f(s) is not

O(s-a).

REMARK: When a = 1 every function f e Lip(1) has bounded variation.

In this case then see exercise 11.215.

ovo - VtV - ove - V1V - ovo

SOLUTION: (a):

f(x + a-u) - f(x - a-u) = - 2 1 a-nasinan-using x,n=1

so

lf(x + a-u)-u [u]-1 (1-a)n-u -na- f(x - a ) <2 E a +2 a

n=1 n= [u]

2a(1-a)[u]-u + 2a-a[u]

a1-a - 1 1 - n-a

Setting:

A 2 B = 2 ,a 1-a - 1 1- a_a

the desired inequality is obtained. Then

If(x t a-u) - f(x - a-u)i < '[Aa(1-a)([u]-u) + Baa(u-[u])1

2a au

< -(A + Baa),

which shows that f e Lip(a).

SOLUTION: (b): The series which defines f is absolutely conver-

512

gent, and consequently

a-na lI (cosanx)(cossx)dx.n=0

n0

On observing that

(an + s)F(an + s) = (an - s)F(an - s),

and that consequently

nF(an + s) + Flan - s) = 2a F(an - s),

(an + s)

this yields

(1-a)n

(s) _ a F(an - s).

n=0 an + s

SOLUTION: (c): Note that

if x # 0,F(x)I <n x17

and that if n > v

CHAPTER 11:

v a an av > (n - v)aloga > avioga - 2.a ' - 8

Since

n navioga - 1 > a ologa - z > as 0 > 0,

it follows that

IF(an-8V

)I <1

n(avloga - 1)if n > v.

TRIGONOMETJ?IC SERIES 513

If it is now assumed that n0 E n < v then

sv - an > av - an - '7 > (v - n)anloga - ' >. anloga - 'z > aan,

because

n n0a >, a > 2 logo - A

Thus

IF(an - sv)I < 1n

if n0 4< n < V.

naa

SOLUTION: (d): Taking into account part (c) and the inequalities

IF(x)l < 1 for all x and IF(x)l >. 2/n if jxj < 'zf we obtain

In0 1 and-a)nF(an

- sv)I . _ n r 1 a(l-a)n- 0(a-v),

n=0 a+ s v n=0v

v-1

n s

a-an = 0(a-v),

n=n0 v n=0

s 1 a-an = 0(a-v),

n=v+1 n(avloga - Z) n=0

and

(1-a)v (1-a)va F(av - s )>,2aav+sv v 71 2av+

ti1 -av7Ca

It follows that if 0 < 6 < 1/rz then

514 CHAPTER 11:

av1f(s)I> a

when v is large enough. Since sv ti aV we also have

SC, l?(8V) >

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 inXPr(x) = I rlle

1 - r21 - 2rcosx + r2

(b): For every function f e L1(T) set

+00

Ar(f;x) _ Lrlnl?(n)e"x.

Prove that if f(x + 0) and f(x - 0) exist at a point x, then

lim A(f;x) = '2(f(x + 0) + f(x - 0)).r-*1-

r

(c) : Set

(xF(x) =

Jf(t)dt.

n

Prove that if F(n) = 0, then


A f'( x) = 1n F(x + t) - F(x - t) Q (t)dt

r 2nj_n 2sint r '

where Qr is a function to be determined.

(d): Prove that as r -> 1 (r < 1) the Qr's form an approx-

imate identity in L1(T).

(e): Prove that if at some point x the "symmetric deriva-

tive of F", Dsf(x) = lim[F(x + t) - F(x - t)]/2t, exists and ist->0

finite, then

limA (f;x) = D F(x).r+1_ r s

(f): Compare this result with the Lebesgue-F6jer theorem.

AVl = V AV = p0A = OAV = A0A

SOLUTION: (a):

ix -ix 2P(x)=1+ re + re - 1-rr 1 - relx 1 - relx 1 - 2rcosx + r2

SOLUTION: (b): 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

1n

Pr > 0,2n

J- P (x)dx = 1,rn

and for all 6 > 0

516 CHAPTER 11:

P (x)dx < P (d), lime (d) = 0.2n dslxl<n r r r-*1 r

Set

s = 1(f(x + 0) - f(x - 0)),

Q(t) = f(x + t) - f(x - t) - 2s.

Taking into account that P (-x) = Pr(x), one easily sees that

(nA(f;x)

- s = J QtP tdt.r2n

0

Let e > 0; if d is chosen so that

IQ(t)I < e if 0 < t < d,

then

IAr(f;a) - sI . e + Pr(d)IIQIII,

which proves that

limA (f;x) = s.rr+1-

SOLUTIONS: (c);(d): Integrating by parts and taking into account

that Pr'(-t) = -Pr(t) yields

it

1n

Ar(f;x) =12J- f(t)Pr(x - t)dt - 2nJ- F(t)Pr(x - t)dt

n n

n n= 2nJ- F(x - t)P'r (t)dt = - 2nJ- F(x + t)P'(t)dt,r

n n

so


Ar(f;x) = - 2nn

j(F(x + t) - F(x - t))P'(t)dtn

setting

Qr(t) sint.P (t),

we have

Ar(f;x) = 2IfF(x +

2sintF(x- t) Qr(t)dt.

(1)

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

2nIQr(t)dt = r.

n

Finally an elementary calculation shows that

(

Qr(t) = 2r(1 - r2 )Isint

2)

2

l1 - 2rcost + r )

attains its maximum when

cost =2r

1 + r2

Therefore, if 0 < d < it then

sup Qr(t) 4 Qr(d),66t67[

518 CHAPTER 11:

as long as

cosd < 2r(1 +r2)-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

to 0. Setting

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 fol-

lows. (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 + u) - f(x))du = 0.h-*O 0

Therefore this result is better than that of Lebesgue-Fejer,

which says aN(f;x) - f(x) if


hlimb-11 If(x + U) - f(x)ldu = 0.h->0 J 0

519

EXERCISE 11.218: The series en converges to s in the sense of

Borel (written

(B)c s)n_m

if

ns r

limer

ni = sr- n=0

where

sn = IpI e

<n

(a): Show that if

in the ordinary sense, it also converges to s in the Borel sense

(use Exercise 10.177).

(b): Show that

n (B) 1z 1 z if Re(z) < 1.n=0

(c): For every function f e L1(T) set:

520 CHAPTER 11:

nBr(f;x) = e-r 8 (f;x) x,

n30n n'

Show that

Br(f;x) = (f*Br)(x),

where

B (x) =e-2rsinÌx sin(lx + rsinx)

r sinix

(d): Prove that

J(x) =rx

J I su

u du tin logx as x0

(e): Prove that

IIBr11, = 7c/2e-2ru2 Isin(2r 1)ul du + 0(1) as rJO


116111ti 21ogr, rrR

(f): Using (e), show that there exist continuous functions

whose Fourier series does not converge in the Borel sense at cer-

tain points.

(g): Prove that if f e L1(T), and if

1= 0lim(f(x + t) - f(x))log

t-r0 I I


at a point x, then

?(n)einx(B) f(x).

AV0 = V AV - AVA = VAV = DOA

521

SOLUTION: (a): If Cr.) is a sequence of real numbers that tends

to +-, and if:

n-r. r.

un,i = e n! '

then:

(i): Un,i

>. 0;

w

(ii) : E u = 1;n=0 n,i

(iii): limuni = 0.ti '

By exercise 10.177 this implies

ns rlimer

ni s ifsn->s.X1->- n=0

SOLUTION: (b): In this case

1-zn+1sn 1-z

and consequently

522 CHAPTER 11:

n-r

sL n

re

n=0 n! = 1 1z

(1 - ze

from which the result follows.

SOLUTION: (c): Because f*Dn it is clear that

Br(f; ) = f*Br,

where

Drn-r C n

= e.Br n=O nt

Recall that

i(n+ )x

Dn(x) = ImIesi-fix 0 < lxIz

and consequently

i )Br(x) = se.x 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 n

J(nn) = Jsinx

dx.sn

8=0 0X

The integral that appears in this expression is equivalent to

2/ns are s -> -, which proves that


J(ma) ti 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 sinuIsin(u + rsin2u)Idu.

0

As the function 1.1 -U

that as

1 I is bounded on [0,1/2], it is clears inu

2 n/2 e-2rsin2u _

IIBrIIl = nju

Isin(u + rsin2u)Idu + o(1).0

Using the mean value theorem and the inequalities

sinu 32n , 0<u< 2

4u2 - sin2u c 9 ,

523

this yields

524 CHAPTER 11:

lx/2(e-2rsin2u_ e-2ru2) du

< 23-frn/2u3e-8ru2/n2du = 01

0u O r

Consequently

22 n/2 a-2ru

IIBr1I1 = 'If UIsin(u + rsin2u)ldu + o(i).

0

Applying the mean value theorem once again, and using the inequal-

ity

3

u- sinu F 6, u 3 0,

it follows that

3

Isin(u + 2ru) - sin(u + rsin2u)l t4r3u

and since

43 f o/2u2e-2ru2du= ClrJ

one deduces that

22((

0n/2 a-2ru

IIBrlll = u Isin(2r + 1)uldu + o(1).,if

To evaluate the integral which appears in the preceding formula,.

choose a d > 0 and write that

-2de J((2r + 1)r-'6) t 4 J((2r + 1)r-16) logr,

foit


n/2 e-2u

611F

fm

SU

du.

Then

2 IB ill JIB rillr2 e - 2 < liminf logr limsup loy 1<2

1E-

10r II1 _ 2

nlogy 2

SOLUTION: (f): The linear functional on C(T) defined by

f - Br(f;x)

525

have JIBr11, as norm. By the principle of condensation of singul-

arities, for all x there exists a GS everywhere dense in C(T)

such that

SuPIBr(f;x) I = 00

if feG6.

SOLUTION: (g): Note that

n n n

71 1Br =

e-r

n>,0 n! 21J-1Dn

n=

e-rIL = 1.

n>.On.

2

If, for 0 < Iti 4 it, we set

526

w(t) = (f(x + t) - f(x))log1--1,t

Qr(t) =

then

Br(t)

log l t l

U

cp(t)Qr(t)dt.Br(f;x) - f(x) = 2nf-X

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

2n/2 a-2ru

IIQr1I1 = jcJ Isin(2r + 1)uldu + o(1).0 ulogu

(1)

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-') + 1

T u

a2u2

du.logn log2

IIQx,111 4< A, A = constant.

CHAPTER 11:

But then if E > 0 and if 6 > 0 is such that ItP(t)I < e for 1t1.< 6.


by (1) we shall have

2rsin2dIB(f;x) - f(x)I < eA

-

r +l0 2 sin d.l Icp(t)Idt,

d4Itkit

which proves that

limB(f;x) = f(x).r-+W

r

EXERCISE 11.219: Let (an) be a sequence of numbers an >. 0, and

An = a0 + + a .

Suppose that

aAn An-> 0.

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

ans0

+ ... + a08n

vn = An

where

528 CHAPTER 11:

8 = X C .

IP14nP

(I): (a): Show that if a series converges in the ordinary

sense it converges in the Narlund sense. (Use exercise 10.177).

(I): (b): Prove that if 0 < a0 .< a1 < Cesaro converg-

ence implies Narlund convergence.

(II): In the whole of this Part (II) assume that:

a0>a1>,.., >,an>,**, 9 an-*0.

For f e L1 set

nNn(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) = Asinsn x X acos(p + )x,

n p=0

p

1 cos(n + 1)xn

Psin(p + 7)x.V

= An

sin ixp=0

a

Show that

(*)

(1)

(2)

1/n

Jn(x)Idx


is bounded for n >. 1.

529

(II): (b): Show that there exist constants B,C,D such that:

Isin(p t 1)xJ

n

1/n 2sin2ixdx 4 B(p + 1)log p + C(p + 1), 14 p < n,

nslnxI

dx < Dlogn, n > 2.1/n 2sin2Ix

(II): (c): By carrying out some suitable Abel transforma-

tions on (1) and (2), prove that

IIVn Ill < 1,

and:

fIUn(x)Idx < C + (Elogn + B Aplogll +

1/n n p=1 `

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 n-lJO(sin2(n + 1)W){cossnntx )x

+ AnnL

Asin(p + 1)xfdxl p= 2 JJJ

530 CHAPTER 11:

stays bounded, then by studying the behaviour of the integrals:

f

7E sin2(n + 1)x.cos(n + z)

0

xs infix dx'

lTE

0

cos(2n + 2)x.sin(p + 1)xdx,

deduce that

1 n Asup

A I -- < .n n p=1 P

(II): (e): Conclude from the preceding that when (*) is

satisfied the following conditions are equivalent:

(i) : For all f e L1(T)

inx (N)f(n)e = f(x)

at every point x where f is continuous;

n A

sup AP < .

n p=1 P

V,

(III): For every real number4 set

na = (a + n)

n n! -

and assume that

a_

Aa-1

n n


(III): (a): Show that

A = Aan n

531

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 fin-

ally, if a = 1 Cesaro's method is recovered.

(III): (b): It is said that:

An1 + ... + Sn Aa g0a =n 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 con-

tinuous.

ovo = VAV - ovo = vov = ovo

SOLUTION: (I): (a):

vn P10 un,pap,=

where

U =n,p

a

A-pif 0 n.

532

It is clear that

u = 1.

p>0 n,P

Moreover,

an_p

un,P { An-p

and consequently

limu = 0.n n,P

By exercise 10.177 sn -> s implies vn -> s.

CHAPTER 11:

SOLUTION: (I): (b): Writing an for the n-th Cesaro sum of the

series X Cn,

nvn an-p((p + 1)ap -

PCP-1n p0

(n + 1)a0

n-1A an + A (p + 1)(an-

- an- 1)an n p=0 P P p

p=O an,paP,

where

n,p

(P (an- - a 1) if 0 4 p < n - 1,P n-Pn

(n + 1)a0An

if p = n,

0 ifp>n.


As the sequence (an

) is increasing, all the An,p >. 0. It is

clear that

lima = 0.n n,P

Furthermore, when all the sn's are equal to one, the same holds

for all the a 's and v 's. son n

A = 1.p=0 n,P

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 1n

sin(n - + )xI aDn-p(x) = A E ap

sin xn p=0

pn p=0

1=

A

sin(n ± 1)xn

apcos(p + )x

n p=0

-7-1 i IXAn

cossn 1)xn0 apsin(p + J)x,

p

which implies the first result desired.

Note that

sin(n + 1)x2(n + 1) ,

in xs

and consequently

IUnI < 2(n + 1),

so

534 CHAPTER 11:

f1/n ( l

J U 4 211 + l 4.0

SOLUTIONS: (II): (b);(c): Denoting by Fn the n-th Fejer kernel

and setting

A = ap - ap+i'

we obtain

n-11

V (x) = A cos(n + 1)x{(n + 1)anFn(x) + E (p + 1)h F (x)}.n p=0 P P

(1)

Since

AP;0, 0 and IIFPIII=1,

it follows that

n-111711 A {(n + 1)an + E p + i)AP} = 1.

n p=0

Similarly, noting that

n

I cos(p + ')x =p=o

we obtain

sin(n + 1)x2sinx

Un(x) = A sin(n + 1)xn

(2)

x Ja sin(n + 1)x n,l A sin(p t 1)x)n

2sin21x + p=0p

2sin2.,x J


Now note that

2fitsinx dx 4 (n xdx = it

log(nn) ` Dlogn.1/n 2sin tx 1/n (2x )

for all n >. 2 if D is chosen large enough.

When 14p<n,

fn Isin(p2

1)xldx 4 (p + 1)J 1/p xdx +

f'dx

2J1/n 2sin Ix 1/n (2x 1 /p (2x

n2

< B(p + 1)log a + C(p + 1),

with B = C = 12/2. Therefore

11I U ( C [(n + 1)a +n n p=1 P

[D0ognn-1

+ + B (p + 1)log P1n p=1

p

Notice that the first bracket-is equal to An - Ap, and that

n-1 n

E (p + 1)A log 2allogn + X alogp

p=1P

p p=2

P

n((

pl

+ E palogl1 - .

P=2

PI J)

The last term of the right side is negative, and

n n-1(( l

la logn

= I A logl1 +1- Alogn.

p=2 P p p=1 P t P1

536 CHAPTER 11:

Therefore, writing

E = D00 + 2Ba1,

we have

J7E

IUnI < C + A (Elogn + BnIl

Alog1i + pJ).1/n n p=1

p

SOLUTION: (II): (d): With the help of an Abel transformation

one obtains

U (x) =1 sin(n + 1)x

n An

s inmx

r( n-1x 1Ancos(n + )x + 2sinZx A sin(p + 1)x}

P=0 P J

= sin(n + 1)x.Gn(x),

where

n-1Gn(x)

=cossinxT- + A2- 1 A sin(p + 1)x.

n p=o p

Since

sin2(n + 1)xlGn(x)I < Isin(n + 1)x.Gn(x)I = IUn(x)I,

we have

rn

supJ sin2(n + 1)xlGn(x)Idx <0

Let


M = sup I cot''x - 2xOtxtn

Since

it sin2(n + 1)xcos(n + )x

Jdx

0 s ink1)xdx + Jsin2(n + 1)xcos(n + 1)xcotixdx,

0 0

this integral remains bounded in absolute value by

jn(n+l )

it(M + 1) + 2sin2u.cosu du

0

Now, the generalized integral:

I

sin2u.cosu du

0u

is convergent, from which it follows that

n-1 rn

1 E A1

s i n

A(n + 1)x.sin(p + 1)xdx

n p=0 p 0

remains bounded. Also, if 0 < p t n - 1, then

f cos(2n + 2)x.sin(p + 1)xdxl2n + p + 3 + 2n - p + 1

0

537

Since

538

Since

n-1

E A 6 n,An p=0 P

it follows that:

-n1 it ApA

A 1 sin(p + i)xdx =2

(2p + 1)Ti p=0 p 0 n 062p6n-1

remains bounded. Writing

n A A1 g am= 1 C r1 + 1 l 2p

An p=1 p An 162psn pJ 2p + 1

CHAPTER 11:

1G

2 A2p-1

+ TiA262pcn+1 l1+ 2p - 11 2p + 1)

because A2p-1 6 A2p one obtains

An A A1 nt2-k6 sAn p=1 p A An

Since

A a t a an+

An

at last we obtain

2

2p + 1n+2 062pEn+1

41+2 n+1A ,An n

n Asup A E

- < m.n An p=1 p

1 tn+1 n+2

SOLUTION: (II): (e): Let


1 nOn = Un - n = A I aDn_p

n p=0P

It is clear that

1n

_2itj On

Since the n-th F6jer kernel is

Fn (x) _1 [sin(n t 1)x12,

n t 1 sinx J

by using the expression (1) for n found in (II)(b)(c), one ob-

tains for all 6 > 0

n-1 asu IV 1(x) I < 1 2i lan + APl An 2 .

An sin 16 p0 n sin 16

Similarly, the expression (2) for Un provides the bound

su Un(x) < An 126<IxT<n n 2sin Zd

Thus

M(6)= su IO(x)I-*0n 6<IXOn

If

n Asup A E <n n p=1 p

asn -}w.

(3)

then

540 CHAPTER 11:

nsup A <

n n p=1

i.e.

1sup A loge <n

By the inequalities

rl/nllnII1<1, I

0

JUnI <4,

n-lJ171

/nl nI <C + An (Elogn + B

pI1Aplog1 + p)

log11 + pl `p

it is then seen that (3) implies that

sup lln ll l-(4)'

In particular, in this case (fin) is an approximate identity in

L1(T). If f e L1(T) is continuous at x, the bound

INn(f;x) - f(x) I c 114nl11Isup If(x + t) - f(t)I + M(6)llfllltI<6

shows that

limNn(f;x) = f(x). (5)

If it is now assumed that (5) is satisfied for every continuous

function f and all x, then in particular

suPiNn(f;o)I < feC(T).


By the Banach-Steinhaus Theorem, (4) then holds, and therefore

because

IIVnII1 4 1.

But the (II)(d) this implies (3).

SOLUTION: (III): (a):

W

Anxn = (1 -x)-a,l

n=0-1<x<1,

from which it is immediately deduced that

nC Aa-i = Aa.LP=0 P n

Thus by setting an = Aa-1 we haven

An = An, limAnn

a alim-=lnma+n=0.An

Furthermore,

an+1 (a - i)a = 1 + n + 1 'n

which proves that an increases with n if a > 1, and decreases

with n if 0 < a < 1; when a = 1 we have an = 1 for all n.

SOLUTION: (III): (b): If a 1 the (CA)-convergence of

E ?(n)elnx to f(x) at every point where f is continuous results

542 CHAPTER 11:

from Fejer's Theorem and from part (I)(b) above. If 0 < a < 1

note that

aAa nn'Pa+1

so

ac A na aL p aMa+1 n

p=1

Thus, (3) is satisfied, and by part (II)(e) this implies (C,a)-

convergence at every point of continuity.

EXERCISE 11.220: Let 1 4 p 4 2 and p+ 4= 1.

(a): Show that if f e LP(T), then

+W

E I f(n) I q IIfIIp.n=

(b): Let (en)neM be a sequence of complex numbers such that

+00

E Icn IP<n=-w

Show that the series

+00I einx

n=-m

converges in Lq(T) to a function f, and that


+W

Ilfllq < 1 IC Ip.q nn=-W

(Use the Riesz-Thorin Theorem, cf. exercise 6.123).

avn a VAV Q nvn ° vnv - nvn

543

SOLUTION: (a): Let T be the mapping which associates with f eW

L (T) the sequence of its Fourier coefficients. Then

IIT(f) IIW < Ilflll,

as

I f(n) I r Ilflll for all n,

and by Plancherel's Theorem

IIT(f) II2 = IIf112

If0<t<1and

1p= 1 1 t+

2tt

1 1- t t t2 = ,q - + 2W L 2

then by the Riesz-Thorin Theorem

IIT(f)11q

< IIf1IP

(1)

(2)

(3)

Note that p runs over the segment [1,2] when t varies from 0

to 1, and that

p

+

4

= 1. Formula (3), proved for f e LW(T), is

still valied if f e Lp(T), since LW(T) is dense in Lp(T).

544 CHAPTER 11

SOLUTION: (b): Consider the space E of sequences c = (cn)nea,

all the terms of which are zero except for a finite number, and

let

+W

$(c)(x) =V. cneln"A x e3R.

n=- ,

Then

110(c) IIW < 11c 111,

k(c)112 = Iic 112.

so

IIa(c) IIq < 16 11P.

We finish by observing that E0 is dense in .2p(2z).

(4)

(5)

Erratum to Exercise 3.45

THE PROOF THAT (iii) => (ii) IS VALID ONLY IF f(O) = f(l)

If this condition is not fulfilled, consider a continuous

function 8 such thatE

0 < $E < 1, 8c (0) _ $E(1) = 0,

and write

n - rln 1 .f(x.) 1 f

ti=1 1 0

e (x) = 0 if E<x<1-EE

l(1 - 8 )f= n l (x1.)f(x 1.) - f

o fo2=1 E E

-1

n+ n (1 - e (xi))f(xi)

2=1

Assuming that Ifl < M, we have

In 1n

f(xi) - J'A < In-1 I -, E(xi)f(xi) - Jl0Efl +ti=1 0 i=1 0

(Contd)

545

546 ERRATUM TO EXERCISE 3.45

(1 -1 n(Contd) + M1 (1 - 9 ) + Mn (1 - 8e(xi)),

0e

i=1

and therefore

1- fl

.) fl <2Mf1(1

- 8 ) < 4eM.limsupn-1n

f(x J

2=1 0 0

E

Bibliography

N.A. BARY: A Treatise on Trigonometric Series, VoZ.s I, II,(Pergamon Press, Inc., New York), (1964).

J. DIEUDONNE: EZxments d'Analyse, Tome II, (Gauthier-Viliars,Paris), (1969).

R.E. EDWARDS: Fourier Series, VoZ.s I, II, (Holt, Rinehart andWinston, Inc., New York), (1967).

C. GEORGE: Cours d'Int9gration (Ecole des Mines, Nancy), (1977).P.R. HALMOS: Measure Theory, (Van Nostrand, New York), (1950).Y. KATZNELSON: An Introduction to Harmonic Analysis, (John Wiley

and Sons, Inc., New York), (1968).A. KOLMOGOROV and S. FOMIN: Elements of the Theory of Functions

and Functional Analysis, (Mir, Moscow), (1974).L.H. LOOMIS: An Introduction to Abstract Harmonic Analysis, (Van

Nostrand, New York), (1953).F. RIESZ and B. NAGY: Lecons d'Analyse Fonctionelle, (4th edn.),

(Budapest-Paris), (1965).W. RUDIN: Real and Complex Analysis, (McGraw-Hill, New York),

(1966).K. YOSIDA: Functional Analysis, (Springer-Verlag, Berlin), (1965).A. ZYGMUND: Trigonometric Series, Vol.s I, II, (Cambridge Univer-

sity Press), (1959).

Name Index

The notions introduced in Chapter 0 do not appear in this

Index. The numbers below refer to the exercises.

Abel-Dirichlet Theorem 11.209- sums 11.217-'s Theorem 3.33

Airy function 4.78

Cauchy determinant 7.134Convergence in measure 3.54,

3.56, 3.57, 6.112,6.114

Baire's Theorem 1.17, 11.208Banach-'s Formula 9.170-Steinhaus Theorem 6.107,

7.133, 11.218, 11.219-'s Theorem 6.119, 8.147Bernstein's Inequality 10.184Bessel function 5-95Bienaym6-Chebycheff Inequality

. 3.67Borel-Cantelli Lemma 1.3- sums 11.218

Dini Test 11.193Dirichlet integral 5.89Duality Theorem 9.172

Equi-measurable function 9.173

Fatougeneralisation of -'s Lemma

3.36F6j6r's Formula 3.47Fubini's Differentiation Theo-

rem 9.167, 9.168

550

Gauss 3.67Gram determinant 7.129, 7.134GraphClosed - Theorem 11.207

Haar functions 7.133Hahn-Banach Theorem 3.65,

8.158Hardy-'s Inequality 8.152, 9.173-'s Lemma 10.179, 10.180Heisenberg's Uncertainty

Relations 7.132Hilbert transformation 8.162

Integrabilityuniform - 3.58, 6.113

Jensen's Inequality 3.69, 3.70,3.71, 6.100, 6.101, 6.102,6.117

Lindelt3f's Theorem 3.63Liouville's Theorem 3.72Lyapunov's Theorem 3.60

Marcinkiewicz's Theorem 6.124,8.161, 8.162

Maximum Principle 3.72, 6.123,11.189

MinkowskiGeneralised - Inequality

6.110Mtlntz's Theorem 7.134

NAME INDEX

Poincare's Formula 1.1Poisson's Formula 8.146

Rademacher functions 7.133Richter's Theorem 3.61Riesz-Thorin Theorem 6.123

SandPile of - Formula 5.94Seriesadjoint Fourier - 11.213

Simpson's Formula 5.94Stone-Weierstrass Theorem

3.46, 7.134Stokes' Formula 8.147

Titchmarsh's Theorem 8.151

Walsch basis 7.133Wiener's Theorem 11.214

Young's Inequality 6.106,6.123, 8.148

Ndrlund sums 11.219

Claude George Exercises in Integration

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