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Problem Books in Mathematics Series Editor: P.R. Halmos
Unsolved Problems in Intuitive Mathematics, Volume I: Unsolved Problems in Number Theory by Richard K. Guy 1981. xviii, 161 pages. 17 illus.
Theorems and Problems in Functional Analysis by A.A. Kirillov and A.D. Gvishiani 1982. ix, 347 pages. 6 illus. Problems in Analysis by Bernard Gelbaum 1982. vii, 228 pages. 9 illus.
A Problem Seminar by Donald J. Newman 1982. viii, 113 pages.
ProblemSolving Through Problems by Loren C. Larson 1983. xi, 344 pages. 104 illus.
Demography Through Problems by N. Keyfitz and J.A. Beekman 1984. viii, 141 pages. 22 illus.
Problem Book for First Year Calculus by George W. B/uman 1984. xvi. 384 pages. 384 illus. Exercises in Integration by Claude George 1984. x. 550 pages. 6 illus. Exercises in Number Theory by D.P. Parent 1984. x. 541 pages.
Problems in Geometry by Marcel Berger et al. 1984. viii. 266 pages. 244 illus.
Claude George
Exercises in Integration
With 6 Illustrations
SpringerVerlag New York Berlin Heidelberg Tokyo
Translator J.M. Cole
Claude George University de Nancy I UER Sciences Mathematiques Boite Postale 239 54506 Vandoeuvre les Nancy Cedex France
17 St. Mary's Mount Leybum, North Yorkshire DL8 5JB U.K.
Editor Paul R. Halmos Department of Mathematics Indiana University Bloomington, IN 47405 U.S.A.
AMS Classifications: OOA07, 2601, 2801
Library of Congress Cataloging in Publication Data George, Claude. Exercises in integration. (Problem books in mathematics) Translation of: Exercices et problemes d`int6gration. Bibliography: p. Includes indexes.
1. Integrals, GeneralizedProblems, exercises, etc. I. Title. II. Series. QA312.G39513
1984
515.4
8414036
Title of the original French edition: Exercices et problemes d'integration, © BORDAS, Paris, 1980. © 1984 by SpringerVerlag New York Inc. All rights reserved. No part of this book may be translated or reproduced in any form without written permission from SpringerVerlag, 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 0387960600 SpringerVerlag New York Berlin Heidelberg Tokyo ISBN 3540960600 SpringerVerlag Berlin Heidelberg New York Tokyo
Introduction
Having taught the theory of integration for several years at the University of Nancy I, then at the Ecole des Mines of the same city, I had followed the custom of the times of writing up detailed solutions of exercises and problems, which I used to distribute to the students every week.
Some colleagues who had had
occasion to use these solutions have persuaded me that this work would be interesting to many students, teachers and researchers. The majority of these exercises are at the master's level; to them I have added a number directed to those who would wish to tackle greater difficulties or complete their knowledge on various points of the theory (third year students, diploma of education students, researchers, etc.).
This book, I hope, will render to students the services that this kind of book brings them in general, with the reservation that can always be made in this case: that certain of them will be tempted to look at the solution to the exercises which are put to.them without any personal effort.
There is hardly any need to
emphasize that such a use of this book would be no benefit. the other hand, the student who
On
after having worked seriously
upon a problem, seeks some pointers from the solution, or compares it with his own, will be using this work in the optimal way.
V
INTRODUCTION
vi
Teachers will find this book to be an important, if not exhaustive, list of exercises, certain of which are more or less standard, and others which may seem new. I have also noted (and this is what led me to edit these sheets)
that from one year to another one sometimes forgets the solution of an exercise and that one has to lose precious time in redisThis is particularly true for those solutions of
covering it.
which one remembers the heuristic form but of which the writing up is delicate if one wishes to be clear and precise at the same time.
Now, if one requires, quite rightly, that students
write their homework up correctly, then it is befitting to submit impeccable corrections to them, where the notations are judiciously chosen, phrases of the kind "it is clear that ... " used wittingly, and where the telegraphic style gives way to conciseness.
It is often the incorporation of these corrections which
demands the most work; I have therefore striven to take pains with the preparation of the proposed solutions, always remaining persuaded that perfection in this domain is never attained.
If
this book encourages those who have to present (either orally or in writing) correct versions of problems to improve the version they submit, the object I have set myself will be partly realised. In this book researchers will find some results that are not always treated in courses on integration; they are either properties whose use is not as universal as those which usually appear and which are therefore found scattered about in appendices in various works, or are results that correspond to some technical lemmas which I have picked up in recent articles on a variety of subjects: group theory, differential games, control theory, probability, etc.,
...
.
In presenting such a work it is just as well to make explicit those points of the theory that are assumed to be known.
This is
the object of the brief outline which precedes the eleven chapters of exercises.
INTRODUCTION
vii
In view of the origin of this book, it is evident that I took as a reference point the course that I gave at the time.
After
having taught abstract measure theory one year, I opted the next for a course expounding only the Lebesgue integral.
This is not
the place to discuss the advantages and inconveniences of each of the two points of view for the first year of a master's programme. I will say only that I have always considered the course that I gave to be more a course in analysis in which it is decided to use the Lebesgue integral than as a dogmatic exposition of a particular theory of integration.
The choice of exercise reflects
this attitude, especially in the emphasis given to trigonometric series, thereby paying the hommage due to the theory which is the starting point of the works of Cantor, Jordan, Peano, Borel, and Lebesgue.
From this it results that, except for the seven exer
cises of Chapter 2 concerning aalgebras, all the others deal with Lebesgue measure on]Rr.
The advantage that has to be conceded to
this point of view is that it avoids the vocabulary of abstract measure theory, which constitutes an artificial obstacle for those readers who might not yet be well versed in this theory.
As for
students who might have followed a more sophisticated course, I can assure them that by substituting du for dx and u(E) fore meas(E) they will essentially rediscover the problems as they are commonly put to them, except for pathological examples about measures that are not afinite and the applications of the RadonNikodym Theorem.
Furthermore, on this latter point the more per
spicacious amongst them will not fail to see that the chapter treating the relationships between differentiation and integration is not foreign to this theorem.
Truthfully, there is another
point that is not tackled in this book, namely the matter of Fourier transforms of finite positive measures and Stone's Theorem, which to my mind is better suited to a course on probability. As was mentioned above, numerous exercises are devoted to trigonometric series, which provides an important set of applications
Viii
INTRODUCTION
of Lebesgue's theory.
This has led me to include some exercises
on series, summation processes, and trigonometric polynomials. Other exercises use the theory of holomorphic functions.
In
particular, some results of the PhrUgmenLindeltff type arise on
two occasions; in each instance I have given its proof under the hypotheses that appear in the exercise.
Quite generally, I have
included in the solutions, or in an appendix to them, the proofs of certain points of analyis, topology, or algebra which students may not know.
I have chosen to make each solution follow immediately after the corresponding problem.
The other method, which consists of
regrouping the former in a second part of the work, seemed to me (from memories I have retained from my student days) much less
manageable, especially when the problem is long, for it then becomes necessary to return often to the back of the book in order to follow the solution. I find it difficult to cite the origin of these exercises.
Many are part of a common pool of knowledge, handed down, one might say, in the public domain.
Others are drawn from different
classic works where they are proposed without proof or followed by more or less summary indications (in this respect it is interesting to note that in forcing oneself to write down the solutions one discovers a certain number of errors just as many in the questions as in the suggestions offered).
Certain of the exer
cises in this book were communicated to me orally by colleagues; I would thank them for their help here.
Lastly, others are, as
I have already said, lemmas found here and there, and which I have sometimes adapted.
Table of Contents
INTRODUCTION
...
...
...
...
...
V
CHAPTER 0: OUTLINE OF THE COURSE ...
...
...
...
...
1
CHAPTER 1: MEASURABLE SETS ... ... 1.21) (Exercises 1.1
...
...
...
...
37
CHAPTER 2: aALGEBRAS AND POSITIVE MEASURES Exercises 2.22  2.28)
...
...
...
79
CHAPTER 3: THE FUNDAMENTAL THEOREMS. (Exercises 3.29  3.72)
...
...
...
89
CHAPTER 4: ASYMPTOTIC EVALUATION OF INTEGRALS... (Exercises 4.73  4.78)
...
... 177
CHAPTER 5: FUBINI'S THEOREM... ... (Exercises 5.79  5.99)
...
...
...
... 199
... CHAPTER 6: THE LP SPACES ... (Exercises 6.100  6.125)
...
...
...
... 225
CHAPTER 7: THE SPACE L2. ... ... (Exercises 7.126  7.137)
...
...
...
... 285
...
...
...
...
CHAPTER 8: CONVOLUTION PRODUCTS AND FOURIER TRANSFORMS (Exercises 8.138  8.162) CHAPTER 9: FUNCTIONS WITH BOUNDED VARIATION: ABSOLUTELY CONTINUOUS FUNCTIONS: DIFFERENTIATION AND INTEGRATION (Exercises 9.163  9.173)...
ix
... 325
... 405
x
TABLE OF CONTENTS
CHAPTER 10: SUMMATION PROCESSES: TRIGONOMETRIC POLYNOMIALS.. 429 (Exercises 10.174  10.184) CHAPTER 11: TRIGONOMETRIC SERIES ... ... (Exercises 11.185  11.230)
...
ERRATUM TO EXERCISE 3.45
...
...
...
...
BIBLIOGRAPHY
...
...
...
...
...
...
...
... 547
NAME INDEX..
...
...
...
...
...
...
...
... 549
...
... 451
... 545
CHAPTER 0
Outline of the Course
aALGEBRAS AND MEASURES
0.1
DEFINITION: A family A of subsets of a set X is called a aALGEBRA ("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 aalgebra A, and that the aalgebra A is closed under countable intersecFor two sets A,B e A let us denote A  B = {x:x a A,x $ B};
tion.
then we have (A  B)e A. The smallest aalgebra containing the open sets of ]R aalgebra of BOREL SETS of ]R
;
is the
this aalgebra is also the small
est aalgebra which contains the closed (resp. open) rectangles
of]R
.
DEFINITION: A (positive) MEASURE on a a algebra A is a mapping u of A into [0,oo] such that if E is the disjoint union of a sequence of sets En e A, then u(E) = I u(En).
It follows that u(o) = 0, and then, if E is the union (not necessarily disjoint) of the sets En, U(E) 5 L u(En).
An equi
valent definition is the following: If E is the union of a fin
1
CHAPTER 0: OUTLINE
2
ite number of Ei's, each of which is in A and which are pairwise disjoint, then p(E) = p(E1) +
+ p(Ep); and furthermore p(A) _
limu(An) when A is the union of an increasing sequence of sets An of A.
If p is a measure and A is the intersection of a decreas
ing sequence of sets An e A and if u(A1) < , then p(A) = limp(An) There exists one and only one positive measure v on the aalgebra 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 submanifold of iRp that is of dimension less that p is negligeable. DEFINITION: A set of ]R
is called a LEBESGUE MEASURABLE SET (or
simply a MEASURABLE SET) if it belong to the smallest aalgebra containing the Borel sets and negligeable sets of]R1. In order that E C ]R
be measurable it is necessary and suffic
ient that there exist the Borel sets A and B such that A C E C B and v(B  A) = 0; upon then setting meas(E).= v(A) one unambiguously defines a positive measure on the aalgebra of Lebesgue measurable sets of ]Rp.
SURE ON I(.
This measure is called the LEBESGUE MEA
A set is negligeable if it is measurable and of
(Lebesgue) measure zero.
This is why one also uses the expres
sion SET OF MEASURE ZERO to denote a negligeable set.
The Lebesgue measure is invariant under translation as well as under unimodular linear transformations (i.e., those with determinant equal to ±1).
A homothety of ratio A multiplies the
Lebesgue measure by JAJp (where p is the dimension of the space).
OF THE COURSE
3
DEFINITION: If A is a oalgebra of subsets of X and B is a Q algebra of subsets of Y, a mapping f:X + Y is said to be an A BMEASURABLE MAPPING if f 1(B) e A for a Z When Z B e Y= B.
Ilzp and B is
the aalgebra of Borel sets, one says, simply, that f is an AMEASURABLE 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 aalgebra of Borel sets or the Lebesguemeasurable sets of X. If f:X ;Ill, in order that f be Ameasurable 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 Ameasurability of an ARITHMETIC FUNCTION, that is to say of a mapping of X into [co,+m] =3R.
If f is an Ameasurable mapping of X into Iltp and g
a Borel mapping of Ilzp into zzq, then gof is Ameasurable. note that every continuous mapping of Ilzp into Ilzq is Borel.
Let us If
(fn) is a sequence of Ameasurable arithmetic functions, the functions supfn,inffn,limsupfn,liminffn are also Ameasurable. DEFINITION: A function is called a SIMPLE FUNCTION (with respect to the aalgebra A) if it is a linear combination of characteristic functions of sets of the aalgebra A.
For every Ameasurable positive arithmetic function f there exists an increasing sequence of positive simple functions which converges to f at every point of X.
DEFINITION: A property holding on the points of a set A of7Rp is said to be true ALMOST EVERYWHERE ON A SET A if the set of points
of A for which this property is not satisfied has measure zero.
If f and g are two mappings from z
into zzq (or m) such that
f is measurable and f = g almost everywhere, then g is measurable.
CHAPTER 0: OUTLINE
4
This allows the notion of measurability to be extended to functions that are defined only almost everywhere.
DEFINITION: A function defined on]RP is called a STEP FUNCTION if it is a linear combination of characteristic functions of rect
angles of]RP. Every measurable arithmetic function on]R
is the limit almost
everywhere of a sequence of step functions.
THEOREM: (Regularity of the Lebesgue Measure): For every measurable set E ofiRP one has: sup{meas(K):K compact K C E}; meas(E) = inf{meas(V):V open V D E}.
THEOREM: (Egoroff): Let X be a measurable set of]R
such that
meas(X) < co and (fn) a sequence of measurable functions such that
fn  f almost everywhere on X.
For every e > 0 there exists a
measurable set A C X such that: (i): meas(X  A) < ci (ii): fn > f uniformly on A.
0.2
INTEGRATION OF MEASURABLE POSITIVE FUNCTIONS
NOTATION: If cp is a simple function on Min that takes the positive
values a1,...,ap on the (disjoint) measurable sets Al....)Ap, we set
I
n
q,(x)dx =
aimeas(Ai),
9 _
i=1
OF THE COURSE
5
with the convention that a.(+)
or 0 according as a > 0 or
a = 0. DEFINITION. With the above notation, if f is a positive measurable arithmetic function on ]Rn there exists an increasing sequence ((pi) of positive simple functions which tends towards f One then sets:
at every point.
f(x)dx = if = limf
J
. .
This element of [0,+]=]K +, which does not depend upon the sequence
(Ti) selected, is called the (LEBESGUE) INTEGRAL of f on7Rn.
This (Lebesgue) integral possesses the following properties (where f and g denote measurable positive arithmetic functions): PROPERTY (1): If f = g almost everywhere, then if = Jg;
PROPERTY (2): Jf = 0 if and only if f = 0 almost everywhere;
PROPERTY (3): if <  implies f < m almost everywhere;
PROPERTY (4): f 4 g almost everywhere implies if
PROPERTY (5):
PROPERTY (6)
1
b
10
CHAPTER 0: OUTLINE
we write
when f is integrable over [a,b], and a "Chasles' Formula" can then If  < a < b < +, then f is Lebesgueintegrable
be written.
over [a,b] whenever it is Riemannintegrable 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 SEMICONVERGENT) 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
f J_
M M}oj0
f
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)
a
sup IJxgl. 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 jjXxf (x,y)dxdy
then the formula
fXdxjf (x,y)dy =
=
dyfXf(x,y)dx fy
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 (according 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
JJ1If(x,y)ldy
L1 3 L1 (continuously), the convolution alge
bras D and L1 operate continuously in all the function spaces which have previously been defined.
NOTATIONPROPOSITION: For every function f defined on1Rn we set: f(x) = f(x),
.'(x) = f(x).
When f and g are convoZvable, so are fand I (resp. f and g), and
(f*g)" = f*.,
(f g) = f"g.
Furthermore : = 1, we set
NOTATIONPROPOSITION: If g e LP, h e Lq, p + 4
(g,h) = Jgh = (g*)(O) (g1h) = Jgh = (g*h)(O).
When f eL1, geLP, heLq, p + q = 1,
(f*g,h)  (g,f*h),
(f,eg1h) = (gl?*h).
OF THE COURSE
23
NOTATION: We denote by LPQ1r) the set of measurable functions on iR
that have period 2nand are such that
If 11p =
[
p
J2nIfIP)1/P 0
IIfII,, = ess sup I f(x) I
< Co,
05x62n
The set of kfold continuously differentiable functions on ]R with period 2n is denoted Ek(W) (here 0 < k 4 Co, and E0(i') is
also denoted C(ur)). DEFINITION: The CONVOLUTION PRODUCT OF TWO MEASURABLE FUNCTIONS f and g WITH PERIOD 21E is defined by the formula: 2n
f(x  y)g(y)dy.
(f*g)(x) = 2nJ 0
Defining fi > f in Ek(IF) to mean that f(S) f(S) uniformly for every integer s with 0 4 s < k, the following hold: LP(a)&rLq(a) C C(a)
(continuously), if
L1(a)*LP(a) C LP(Ir)
(continuously),
L'(T.')*E (a) C Ek(a)
(continuously),
= 1,
+
p
4
as do the inequalities:
IIf* IIC 5 IIfIIpIIgIIq, if feLP(a), geLcl(r), p +
= 1, q
IIf*IIp 1 IIfIIlIIgIIp,
if feL'(l), geLP(a).
By defining J`,J as above, and
24
CHAPTER 0: OUTLINE 21
(f,9) = 2nJ
2n f9,
f9
(.fig) = f1J 0
0
one obtains the same formulae as above.
Let us note that in the
formulae defining f*g,(f,g) and (fig), one can replace the range of integration (0,2n) by any interval of length 2n.
Clearly one can consider functions having an arbitrary period T > 0, it is then just a matter of replacing it by T/2 everywhere.
0.8
REGULARISATION OF FUNCTIONS
DEFINITION: One calls an approximate identity in L1 every sequence (rpi) of integrable functions that satisfies the following conditions:
(i): There exists a constant M such that 11,Pi11l < M for all i; (ii): lim J'Pi = 1; i
(iii): For every a > 0,
limJ i
m = 0, JxJ>,a i
An approximate identity (ml) is said to be compact if all the functions cp. vanish outside the same compact set of]Rn. i
In L1 there exist compact approximate identities consisting of
functions belonging to V ; these are called REGULARISING SEQUENCES. THEOREM: Let (rp
be an approximate identity in L1.
If E is one
of the spaces LP (1 < k < .o) or UC , then f o r every function f e E
one has 9 i *f > f in E. If the approximate identity (W.) is compact this property extends to the spaces LPoc,Lp (1 < p < o), Ek,Vk (0 < k From this one deduces the following corollaries:
THEOREM: (Density); D
!1 < p < c ), Ek, Vk.
is dense in each of the spaces Lp;LPoc, LP c
OF THE COURSE
25
LEMMA: (Calculus of Variations): If f e L10c is such that Jf9 = 0 for any rpe D , then f = 0 almost everywhere.
DEFINITION: An approximate identity in L1(1r) is a sequence (p.) of integrable functions with period 27t, such that
(i):II(piII1,M; (
(ii) : lim 2l 71 i = 1; (iii): For all a, 0 < a < n, liml
i
p.(x)dx = 0.
a 0, there exists 6 > 0 such that for every finite sequence of mutually disjoint subintervals ]ai,si[ of [a,b] one has
i
If(si)  f(ai)I < e
whenever
i
(R.  ai) < 6.
THEOREM: (Lebesgue): If f is absolutely continuous on [a,b] f is differentiable almost everywhere, f' is integrable and
b
f(b)  f(a) = J.f'(x)dx, a
b V(f;a,b) = J If'(x)Idx. a
Conversely, if F is integrable on [a,b], and if
f(x) = JF(t)dt, x a
a < x < b,
CHAPTER 0: OUTLINE
30
then f is absolutely continuous and f' = F almost everywhere. The theory of differentiation makes use of the two following Lemmas, which are interesting in their own right. LEMMA (1): Let K be a compact set of atn covered by a family of open cubes.
From this family there can be chosen a finite sequence
C1.$.$Cp of mutually disjoint cubes such that
meas(K) < 3n
meas(Cp). k=1
LEMMA (2): (The Setting Sun Lemma): Let f be a real continuous function on [a,b]., E the set of points x of this interval for
which there exists a y such that x < y < b and f(x) < f(y).
Then
the set E is the disjoint union of a sequence of intervals with end points an < bn such that f(an)
0.11
f(bn)
TRIGONOMETRIC SERIES If (un)nea is a sequence of complex numbers indexed by some
positive or negative integers, one sets
W
+W

un = u0 +
I
N (un + u_n) = lim
E
un,
N n=N
n=1
when this last limit exists.
luni <  one also has
When I00
u
n
=lim N
M E
n=N
u.n
M
A trigonometric series is a formal series of the type
OF THE COURSE
tm c e n
31
m
a
inx
=
2
(ancosnx + bnsinnx),
+
n=1
where, upon agreeing to set b0 = 0,
an = cn + cn, b
n
C
= i(c n  c n ),
 ib ), n = Y(a n n
n > 0.
cn = '(an + ibn),
In what follows it is assumed that f e L1(7r) (L1(w) has been
defined in Section 0.7).
For every n e 2z we set
(2n einxf(x)dx.
?(n) = I
0
The ?(n) are called the FOURIER COEFFICIENTS of f, and the formal series
inx
is the FOURIER SERIES of f.
We use the notation
L oneinx
f(x) ti
W
to indicate that the second member is the Fourier series of f, that is to say that cn = ?(n) for every n ez.
It will be noted
that:
12n r
an
f(x)cosnxdx,
n 0
bn =
1 n
2n
f(x)sinnxdx.
On setting en(x) = einx we have the following FORMULAS
32
CHAPTER 0: OUTLINE
.'(n) _ (f l en), fsaen = f(n)en,
f(n) = ?(n)
and f(n) = ?(n),
f*g(n) = f(n)g(n),
P (n) = in(n)
if f is absolutely continuous.
For the explicit calculation of the Fourier series of a function f the following result, which generalises (5) in the Formulas above, is useful:
PROPOSITION: If f is a pieeewise continuously differentiable function, that is to say, if there exist points.
u 4 al < a2 < ... < a p
it
, 0 and for all
x e]R.
OF THE COURSE
35
THEOREM: If one of the functions f or g is of bounded variation
on [0,27E], then:
1r2n 2n
f(x)g(x)dx =
1
f(n)g(n)
n=W
0
.(n)(2neinxg(x)dx.
=
E
n=0
0
In other words: In order to integrate f with respect to g(x)dx one can integrate its Fourier series term by term. THEOREM: (Fejer): If f(x + 0) and f(x  0) exist at a point x, then lima N(f;x) = j(f(x + 0) + f(x  0)).
N Whenever f is continuous on a compact interval, one has aN(f;x)> f(x) uniformly on this interval.
The Fejer kernels form an approximate identity, so that if
f e LP(T), 1 < p
f in LP(a).
THEOREM: (FejerLebesgue): 1imoN(f;x) = f(x)
(0.11.7)
at every point x such that
lim LJhlf(x + t)  f(x)ldt = 0. 0 h+0
In particular, Equation (0.11.7) is true for almost all x. THEOREM: (Plancherel): In order that f e L2('T) it is necessary and
sufficient that
CHAPTER 0: OUTLINE OF THE COURSE
36
E
If(n)I2
0 construct an open set U everywhere
EXERCISE 1.2:
dense in R, and such that meas(U) < e.
ovo = vov  ovo = vov = AVA
SOLUTION:
Let (rn) be the sequence of rational numbers, and for
all n let us denote by In the open interval with centre rn and length e2n.
The union U of the In is an everywhere dense open
set of 3R (for it contains all the rational numbers) and furthermore,
meas(U) S
e2n
1
= e.
n=1
EXERCISE 1.3:
Let (En) be a sequence of measurable sets such
that
E meas(E n
,1
meas(E.). i=1
1
In the general case one considers, for n > s, the sets Hs n formed by the points which belong to at least s of the sets E1,.. ,nIt is clear that Hs
n C Hs n+l
and that
Hs = U Hs,n.
n=s Consequently
meas(H ) = Jim meas(H ) n s,n S
n Jim
. 1 the number
ap(x) = u(2p1 x)
is the pth term in the binary development of x e [0,1[.
Whatever may be the numbers el,...,ep (ei = 0 or 1), the set of x's such that 0 4 x < 1 and al(x) = ell
has measure 2P.
ap(x) = cp,
From this it follows that the set Bp of the x's
such that 0 4 X < 1 and
MEASURABLE SETS
49
a1(x) = 0,
a3(x) = 0,
. ,
a 2p1 (x) = 0,
has as its measure 22p+1 x
2p1
= 2p.
Since B is the intersection of the Bp one certainly has meas(B) =0.
SOLUTION: (b):
The set A' = 2B is also negligeable.
Now A' is
the set of numbers of the type
Y
a 2(2p+1)
p
p=0
a p = 0 or 1.
As every real number may be written
n+
ap=0or1,
ap2p,
p
it follows that ]R = A + A'.
EXERCISE 1.10:
Let f be a complex measurable function on ]R such
that f(x + 1) = f(x) for almost all x.
Show that there exists a
function g such that f = g almost everywhere and g(x + 1) = g(x) for all x.
AVA = DA4 = ADA = DAD = ADA
Let E = {x,f(x) + f(x + 1)} and let us set
SOLUTION:
+co
(E t n),
F= n=co
as well as
CHAPTER 1:
50
 f(x) if xeF. It is clear that g has all the properties desired.
EXERCISE 1.11:
Let A be a bounded set of ]R
Sip = meas{x: I Ix II
and let
r 1},
where lixll denotes the Euclidian norm of x. Let (xi)i>.1 be a sequence of points of A; we set
do = inf{Ilxi  xi II1 1 4 i < ,j 4 n}. Prove that
liminfndP n n
y for n n sufficiently large, as well as the balls B. with center xi and radius ri, where
if 1i y1/Pn  1/P whenever n is large enough, and
yl/pn1/P.
ri t rj =
n, one has
When 1 4 i < j E 2pn and j
 1/p, IIxi  xi II > dj > yl/Pj and we also have
ri + r.
yl/Pn 1/P
zyl/P(2j1/p
+
 n1/P)
=
yl/Pj 1/p.
From this it follows that the balls Bi are mutually disjoint. If one sets A(e) = {x:d(x,A) .< e), one will then have
urn  1 2
LL
n
Pn (2i1/P + 2C
 n 1/P)Pl
< meas(A(e )), n
J
i==n
where
e
n
=
2yl/Pn  1/p
We have 2Pn 1 (2i'/P 
lnm
i=n
n1/p)P
Pn [2rnl 1/p  1]p = 2C = lim n i==n
Il
J
(Contd)
CHAPTER 1:
52
r2P (2,l /P  1)Pdx
(Contd) JI
1
=
(1  t)P
1tt
p1
dt,
(the last integral is obtained by setting x = (1 t t)P).
On the
other hand, the sets A(En) are decreasing and have A as intersection; furthermore, as A is bounded they are of finite measure. From this it follows that
meas(A) 1
c
St
p
P
and consequently that
liminfndp 4
n
n
EXERCISE 1.12: = 1.
1 meas(A)
c
P
SZ
P
Let X be a measurable set of ]RS such that meas(X)
Let u be a bijection of X onto itself such,that for every
subset E of X, E is measurable if and only if u(E) is measurable, and then meas(E) = meas(u(E)).
Furthermore, assume that if N is
a measurable subset of X and if u(x)e N for almost all the points of N, then N or X  N is negligeable. Let E be a measurable subset of X such that meas(E) > 0, and
if x e X if uP(x)$ E for all p > 1, inf(p:p >, 1,uP(x)e E}
(one sets u1 = u, uPt1 = u0up).
Show that
otherwise
MEASURABLE SETS
53
1 f n(x)dx=1. E
A00 = 0A0 = A0A = VAV = A00
SOLUTION:
E
n
For 1 .< n 6 W let us set
= {x:x e E and n(x) = n}.
Let us also set GO = E,
n
u_P(E),
Gn = E  U
n . 1,
p=1
where u p = (U1 )P).
It is clear that
14n<W,
En=Gn1Gn, E
= o GG.
For n > 1 one has n1 un(Gn)
=
un(E)
 U
up(E).
P=O From this it results that the sets un(G ), n 3 0, are mutually n
disjoint, and that
y = l_J un(G ) n=0
= lJ un(E). n=0
Since u(y) C y and meas(y) > meas(E) > 0, the second hypothesis
54
CHAPTER 1:
made on u implies that meas(y) = 1; in other words, since u preserves the measure, m
Go
1 =
meas(G.). E E measun(G ) = n=O n=0 n n
Let us note that the Gn are decreasing; consequently meas(E ) = lim meas(G ) = 0.
n
n
Thus one has
N n(x)dx =
E
n=1
nmeas(E ) = lim n
N
X
n=1
N
(N1
= liml I
N In=O
(n + 1)meas(G )
n
rN1

I
n=O
l
nmeas(G )J
l
= liml I meas(Gn)  Nmeas(GN)J N n=O 1
It remains to be observed that
meas(Gn) = 1
L
and
meas(Gn) 3 meas(Gn+1)
implies Nmeas(GN) } 0.
Thus
I n(x)dx = 1. J
meas(G )} n
n{meas(G n1
E
REMARK: We have used the following classical result:
n
MEASURABLE SETS If un
55
> un+l >
0 and G Un < m, then nun > 0.
We shall briefly recall the proof: let c > 0 and let p be such that
(n  p)un S up+1 + ... + Un S E
for all n
p; then
liminfnun s E. QED n
EXERCISE 1.13:
One says that a set A of]Rp is ALMOST OPEN if
almost all the points of A are interior points of A. Let f be a real function defined on an open set U of Iltp.
Prove that the following conditions are equivalent: (a): f is continuous at almost all the points of U;
(b) :
For all a e]R the sets (f > a) and (f < a) are almost open sets. AVA = VAV = AVA = V0V = AVA
SOLUTION: (a):
Let E be a negligeable set in U such that f is
continuous at every point of U  E.
If x e (f > a)  E one' will
have f(y) > a for all the points y of a neighbourhood of x, therefore x is interior to (f > a).
This set is therefore almost open.
One argues similarly for (f < a). (b):
f is continuous at x if for every rational number
r < f(x), x is interior to (f > r), and if, for every rational number s > f(x) x is interior to (f > s).
If r is rational, let
us denote by Ar (reap. Br) the set of points of (f > r) (reap. (f < r)) not interior to this set.
If these sets are neglige
able then so is their union, which, by the preceding, contains the set of points of discontinuity of f.
56
CHAPTER 1:
EXERCISE 1.14:
One says that a bounded real function f defined
on ]R is ALMOST EVERYWHERE CONTINUOUS if the set of its points of discontinuity is negligeable. (a):
Give an example of a function that is almost everywhere
continuous and such that there exists no continuous function coinciding with it almost everywhere. (b):
Show that in order for a bounded real function f to be
almost everywhere equal to an almost everywhere continuous function, it is necessary and sufficient that there exists a set A of R such that ]R  A is negligeable, and that the restriction of f to A is continuous. (c):
Deduce from (b) that f is measurable and that there ex
ists a sequence of continuous functions fn which is convergent at every point of R and whose limit is almost everywhere equal to f. (d):
Show that a rightcontinuous function is continuous
except at the points of a set that is at most denumerable, and therefore is almost everywhere continuous. A0A = V AV = A0A = 0M4 = AVI
SOLUTION: (a):
If f(x) = 0 for x < 0 and f(x) = 1 for x > 0, f
cannot coincide almost everywhere with a continuous function g, for with the complement of a negligeable set being everywhere dense in R, one would be able to find two sequences xi < 0 < yi tending to zero, and such that g(xi) = f(xi) = 0,
g(yi) = f(yi) = 1.
On passing to the limit one would have g(0) = 0 and g(O) = 1 at the same time, which is absurd. SOLUTION: (b):
The condition is evidently necessary.
show that it is sufficient.
To do that let us set:
Let us
MEASURABLE SETS
57
g(x) = Iim{sup(f(y):y e A,Iy  xI < a)}. a*0 a>0
This definition has.a meaning, for A is everywhere dense. one has f = g.
On A
Furthermore, if x e A and if c > 0 there exists
a > 0 such that f(x)  c .< f(y) . f(x) + c if yeA and I y xI < a. Then if ix  x '
l y  x' I
l
< a one has
l y  x l
< a' = a  Ix  x' I
< a for all y e ll such that
.
From this it follows that g(x)  e .< g(x') .< g(x) + c,
which proves that g is continuous at each point of A. SOLUTION: (c):
For every x and all c > 0 there exists a > 0 such
that f(y) .< g(x) + c if y e A, ly  xI < a.
As above, from this one deduces that
g(x') . g(x) + e if Ix'  xl < a.
The function g is therefore bounded and upper
semicontinuous.
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 rightcontinuous.
If A
is a nonempty 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 rightcontinuity of f,
for all x e A there exists a
x
> 0 such that
a(f(]x,x + ax[)) < a. But then:
]x,x+ax[(nA=0. Let us choose a rational number rx in ]x,x + ax [ .
If y e A and
x < y, one therefore has
rx < y < r y , which proves that x e A  rx is an injection of A into Q, and con sequently that A is denumerable.
EXERCISE 1.15: < .
Let A be a measurable set of ]R such that meas(A)
Show that the function x y meas(A(1]co,X]) is continuous.
AVA = V AV = AVA = V AV  AVl
SOLUTION:
If xn is a decreasing sequence, and tends to x, one
has
A(1]co,x] = n {An].,x n]}, n
and consequently
MEASURABLE SETS
59
meas(Afl],x]) = 1im meas(Afl]m,xn]) n If xn is a strictly increasing sequence that converges towards x then
Ar)].,x[ = U {Afl]W,xn]}, n and consequently, because meas({x}) = 0, meas(A r)]W,x]) = meas(Af)]o,x[) = lim meas(A r)]m,x ]),
n
n
which proves that the function x 1+ meas(Afl]oo,x]) is continuous.
EXERCISE 1.16:
Let 0 < A < 1.
For any measurable sets A,B C[0,1]
of positive measure, do there exist 0 < x < y < 1 such that
meas(Ar)[x,y]) = Ameas(A), meas(Br)[x,y]) = Ameas(B)?
AVA = V AV = AVA = VAV = A VA
The answer is affirmative if and only if A = 1/n, where
SOLUTION:
n = 2,3,...
First of all let us assume that A = 1/n.
.
Since the
function f(x) = meas(A r) [O,x])
is continuous and increasing on [0,1], there exist points 0 =t t1
0 be two numbers such that
(n+1)a+ns= 1, and let us decompose [0,1] into 2n + 1 contiguous intervals, al
ternatively of length a and B, the two extreme intervals having length a.
Let A be the union of intervals of length a, and B
that of intervals of length S.
If
meas(Afl [x,y]) = Ameas(A), then [x,y] contains at least one of the intervals of length S, as otherwise one would have
meas(Afl[x,y]) < a =
n + 1
meas(A) < Ameas(A).
But then,
meas(Bf)[x,y]) >, B =
meas(B) > Ameas(B). n
EXERCISE 1.17:
Let I be a compact interval of 3R such that meas(I)
> 0 and 0 < S < 1.
We shall say that the operation T(s) is carried
62
CHAPTER 1:
out on I if one subtracts from I the open interval having the same centre as I and of length Smeas(I).
More generally, if I is
a disjoint union of a finite number of compact intervals of nonzero lengths, to apply T(S) to I consists in carrying out this operation on each of the intervals forming I.
Now let (0n) be a sequence of real numbers 0 < Sn < 1; we shall denote by In the compact set obtained by successively carrying out the operations T(S1),T(S2),...,T(Sn)
starting from the interval
[0,1].
Show that
(a):
In+1 C In,
meas(In)
2)...(1 
1)(1 
n
and that every interval contained in In has a length less than 2n
From this deduce that
(b):
K
R In n
is compact, nonempty, nowhere dense, has no isolated point, and that
meas(K) = lim(l n (c):
1
S) n
)(1  a
Assume that Sn = 1/3 for all n.
Show that meas(K) = 0
and that X e[0,1] belongs to K if and only if it can be written in base three uniquely using only the digits 0 and 2.
From this
deduce that K is not countable.
(d):
If Bn = 1 
al/n(n+l),
0 < a < 1, show that meas(X) = a
(which proves the existence in [0,1] of nowhere dense compact sets whose measure is arbitrarily close to 1).
MEASURABLE SETS (e):
63
Deduce from part (d) the existence in [0,1] of a se
quence An of sets of the first category that are mutually disjoint and such that: (i): meas(An) = 2n n
(ii): Kn = U A
is a nowhere dense compact set;
i=1
(iii): Every interval contiguous with n (that is to say, every connected component of the complement of Kn in [0,1]) contains a set in An+1 of measure greater than zero. From this dedua
A
that
U An n=1
is of first category, has measure one, and that its complement in [0,1] is a set of second category of measure zero.
(f) :
Let
E nU0 A2nt1'
Show that for every interval I contained in [0,1] and of nonzero length there holds 0 < meas(E()I) < meas(I).
(g):
From part (f) above deduce the existence of Borel sets
E C IR such that for every interval I of nonzero length one has 0 < meas(E(lI) < meas(I). (h):
Can one have meas(E) < m for such sets?
Deduce from the preceding that there exist positive
functions that are Lebesgue integrable, but that are not limits almost everywhere of increasing sequences of positive step functions.
CHAPTER 1:
64
SOLUTION: (a):
It is clear that
In+l C In
and
meas(In) _ (1  S1)...(1  On).
Moreover, n is formed by 2n mutually disjoint intervals of equal lengths; when the latter property. SOLUTION: (b):
The set K is compact and nonempty 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) < 2n and consequently meas(I) = 0.
In other words the interior
of K is empty, which is the definition of a nowhere dense compact set.
If x e K and e > 0, for large enough n one of the intervals
forming In will be contained in ]x  e,x + e[; for this it is sufficient that 2n < e.
Now, the two endpoints of this interval
belong to K, which shows that x is not an isolated point of K. Lastly,
meas(K)= lim meas(In) = lim (1 
Sn).
n
SOLUTION: (c):
In this case, one has: ( ln
meas(K) = liml3J
n
= 0.
Furthermore, I1 is equal to the set of x's which are written in base three as
x = O.ala2...an...'
with a1 = 0 or 2 (it will be noted that 1/3 = 0.0222 and that
1 = 0.222 ).
Similarly it is seen that x e In if and only if
ai = 0 or 2 for 1 i i < n.
From this one deduces that x e K if for
all i ai = 0 or 2; the expansion of x in this form is then unique. If with every set A C iN one associates xA =
0
MEASURABLE SETS
65
if i eA and ai = 2 if i4 A, a bijection between P(v) and K is realised; K is therefore not denumerable. a1/n(n+l),
SOLUTION: (d):
If Bn = 1 
0 < a < 1, one has
a
meas(K) = lima n, n
with
1 _
1
n+1
(n
whence meas(K) = a. SOLUTION: (e):
By the preceeding it is seen that in every non
empty open interval I there exists a nowhere dense compact set whose measure is imeas(I).
The let Al be a nowhere dense compact
set of [0,1] such that meas(A1) = z.
In each interval contiguous
to Al let us choose a nowhere dense compact set the measure of which is half that of this interval, and let us denote by A2 the union of these compact sets; A2 is of first category (for the set of intervals contiguous to a compact set is denumerable) and its measure is 1.
Furthermore, K2 = Al U A2 is closed; in fact,
if x is a limit point of K2 and does not belong to Al it belongs to an interval I contiguous to A
1
and is therefore a limit point
of IflA2, which is compact, whence x e A2.
On the other hand, it
is clear that [0,1]  K2 is dense in [0,1]  K1, which itself is dense in [0,1], which proves that K2 is nowhere dense.
Quite
generally, An+1 will be constructed by choosing in every interval contiguous to the compact set K
n
a nowhere dense compact set of
CHAPTER 1:
66
measure equal to half the length of this interval, and by taking the union of these compact sets. is of first category and that Kn+l
As above, it is seen that An+1 = Kn U An+1 is
Further
closed.
more,
n
2i
meas(Kn) _
= 1  2n, i=1
whence:
z[1  (1  2n)] = 2(n+1)
meas(An+l)
Finally, Condition (iii) is satisfied by construction.
A
If
U An n=1
this set is of first category, for it is the countable union of sets of first category, and

W
meas(A) =
2n = 1.
1
n=1
Its complement is not of first category, by a theor m of Baire, and its measure is zero. SOLUTION: (f):
E
n
Let
A2n+1'
F
i A2n' l
If I is an interval of length greater than zerocontained in [0,1], its intersection with E or F is of measure greater than zero, since meas(E U F) = 1.
meas(If1E) > 0.
Let us assume, for example, that
Then ICE contains at least two points x < y.
Let n be such that these two points belong to Al U A2 U
U A2n+1'
MEASURABLE SETS
67
They therefore belong to K2n+1, and as this set is compact and nowhere dense there exists an interval J contained in [x,y] which is contiguous to it; one then has meas(If)F) > meas(JflA2n+2) > 0.
Since
meas(I) = meas(IfE) + meas(If)F), it follows from this.that meas(Ef)I)
< meas(I).
If E is the set studied in Question (f) above,
SOLUTION: (g): then
E= U (E + n) nea
answers the question.
By going back to the proof of Question (f)
again one sees that for E one can take the set
EN
N
n
A2n+1'
If
E = U (EINI + N), ne2Z
then for every interval I of positive length one has: 0 < meas(If1E) < meas(I), and furthermore, W
m
m
2(2n+1) = 10
2(2n+1) + 2
meas(E) _ n=0
N=1 n=N
9
CHAPTER 1:
68
SOLUTION: (h):
Let f be the characteristic function of the set
E defined above, and let cp be a positive step function such that rp 4 f almost everywhere.
If I is an interval of length greater
than zero on which 9 is equal to a constant, since meas(I  E) > 0 this constant is zero.
Thus cp = 0 almost everywhere.
From this
it follows in particular that f cannot be the limit almost everywhere of an increasing sequence of step functions. REMARK:
To prove the existence of a nowhere dense compact set of
[0,1], the measure of which is arbitrarily close to unity, one may also consider the set E = [0,1]  Q.
There holds
1 = meas(E) = sup{meas(K):K compact, K C E}, and every compact set contained in E is evidently nowhere dense.
EXERCISE 1.18:
Consider a double sequence (fm n) of measurable
complex functions on X = [0,1] such that for all m the sequence converges almost everywhere towards a function gm and
(fm
that the sequence (gm) converges almost everywhere towards a function h.
Show that there exist two sequences of strictly increasing inconverges n s' S Generalise this result to the case where
tegers (ms) and (ns) such that the sequence (fm almost everywhere to h.
X =P. (orEP).
wA=vw=w0=0AV =MA SOLUTION: set E C X.
First assume that every convergence is uniform on a Then there exists ml < m2
1.
= gm
On setting
E=nm m=O
the desired result is obtained.
Let us now use an argument known as the "diagonal process". By applying the preceding result to e = 1 one first determines a measurable set E1 and two strictly increasing mappings T1,5 1 of
N* into itself, such that meas(X  E1) < 1 and
f91(n),O1(n) ' h
on E1.
Now applying the same result to e = Z and the double sequence
f
(m) 0 (n) one obtains a measurable set E2 and two strictly in1
creasing mappings 92102 of 1V* into itself such that meas(X  E2)
h
on Er.
Let us then set
E = U r, r mS = 910...op (a),
ns = 010...o0s(a).
S
The set E is measurable and meas(X  E) = 0. every r 3 1 the sequence (fm
Furthermore, for
)sr is a subsequence of the se
s,n
s
From this it follows
quence that
fm in s s
> h
on E.
It will be noted that in order to use Egoroff's Theorem it suffices to assume that X C 1R
and meas(X) < .
By writing ]R
as
the union of a sequence of such sets, a new application of the diagonal process allows this result to be generalised to the case
where X = ]R
.
EXERCISE 1.19:
sets of]R set K of ]R
A mapping t  F(t) ofIR into the set FP of closed
is called a MEASURABLE MAPPING if for every compact the set {t:F(t)f1K 4 01 is measurable.
MEASURABLE SETS
71
(a):
Show that {t:F(t) = O} is measurable.
(b):
Show that t ; F(t) is measurable if and only if
{t:F(t)f1 B * 0} is measurable for every open ball B of ]RP, or
again if t  F(t)1K is measurable for every compact set K of ]R Show that if t 
(c):
.
F1(t) and t  F2(t) are measurable
mappings of]R into the closed sets of)R
and]Rq respectively,
then t  F1(t)x F2(t) is measurable. Show that if t  K(t) is a measurable mapping of ]R into
(d):
the compact sets of ]R , and if f is a continuous mapping of IlRP into 3R q, then t  f(K(t)) is measurable.
Show that if t > Fn(t) are measurable, then
(e):
t  n Fn(t) n is also measurable. (f):
Show that t  F(t) and t 
of Et into the sets of ]R
K(t) are measurable mappings
which are respectively closed and compact,
then t ; F(t) + K(t) is measurable.
1V = VV =Ova=V V =AVA SOLUTION: (a): n.
Let Bn be the closed ball with center 0 and radius
We have: Co
{t:F(t) = O} =]R  U {t:F(t)f1Bn * 0}. n=1
SOLUTION: (b):
If t ; F(t) is measurable and if V is an open set
of]RP, there exists a sequence of compact sets Kn of which V is the union, and:
CHAPTER 1:
72
{t:F(t) n V 4 0) = U {t:F(t) n n # 0}. n=1 If now {t:F(t)n B 4 0} is measurable for every open ball B, as every open set V is the countable union of such balls, the set {t:F(t)n V 4 0) is measurable.
If K is compact, let us consider
the open sets Vn = {x:d(x,X) < 1/n} where d(x,K) denotes the distance from x to K for a norm on]R .
One has:
{t:F(t)f1K 4 0) = U {t:F(t)f1Vn 4 o}. n=1
In fact it is clear that the first set is contained in the second; moreover, if for all n there exists xn a F(t) n Vn one can find a
subsequence x
n
such that x .
It is clear that x e K (as
> X. .
K) = 0), and that x e F(t) also, for F(t) is
d(x,K) = linrl(x Z
n
1
closed.
If t > F(t) is measurable and K is compact, t  F(t)n K is
evidently measurable (this is true, moreover, if K is closed). If t + F(t)n K is now measurable for every compact set K, by considering afresh the balls Bn (cf., Question (a) above), we have
{t:F(t) n K # 0} = U {t:(F(t)f1K)n Bn # 0}, n=1 which proves that t SOLUTION: (c):
F(t) is measurable.
Note that we have not specified the norm on ]R P.
If we choose the norm II(xl,...,x P of ]Rp ]R
)II
= Maxlxil, every open ball
is of the form B1 xB2, where B1 and B2 are open balls of
and itq respectively.
Then:
{t:(F1(t)xF2(t))n(B1xB2) 4 0} = n {t:Fi(t)nB. 4 O}. i=1,2
MEASURABLE SETS SOLUTION: (d):
73
For every open set V of Rp we have:
{t:f(K(t))nv # f } = {t:K(t)nf 1(V) # 0}. SOLUTION: (e):
Consider first the case of two measurable mappings
t  F1(t) and t  F2(t).
Let us denote the 'diagonal' of RpxRq
For every compact set K of Rp,
by A.
{t:F1(t)nF2(t)nK = 0} = {t:(F1(t) x (F2(t)nK))nt + f}. By what has gone before, t + F1(t)x (F2(t)n K) is measurable, and on the other hand 0 is a countable union of compact sets.
that t } F1(t)n F2(t) is measurable.
the result for a finite intersection.
This shows
By recurrence one obtains In the general case one
writes n
{t: ( n F (t))nK + y} = n {t:(n Fr(t))nK + 0}, n=1
n=1
r=1
an equality that results because the decreasing compact sets
n
n (Fr (t)nK) r=1
have a nonempty intersectionif and only if each of them is nonempty.
SOLUTION: (f):
It is known (see a course on Topology) that
F(t) + K(t) is closed if F(t) is closed and K(t) is compact. Moreover, if u denotes the mapping (x,y) ; x + y of Rp x Rq into
Rp, we have u(F(t)x K(t)) = F(t) + K(t) By the preceding, t r F(t) x K(t) is measurable, and it is seen,
as in Question (d) above, that u(F(t)x K(t)) is also.
CHAPTER 1:
74
Let Fp be the set of closed sets of R
EXERCISE 1.20:
set of nonempty compact sets of i2P, x >
l
ix1
1
,
KP the
a norm on ]R
and
,
for every nonempty set F of Fp d(F) = Min{ IjxII:x a F}. For F E Fp write:
(a):
cp0(F) _
{x:x eF, jjxjj = d(F)}
if F # 0,
{0}
if F = 0.
Show that t u cp0(F(t)) is measurable if t  F(t) is measurable
(for the definition of the measurability of a oneparameter family of closed sets see the preceding Exercise).
Let ei(x) =X. if x = (x1,. .. ,xp) e:IR
(b) :
.
For K e K0 set
e.(K) = Min{e.(x):x a K} and cpi(K) = {x:x a K,ei(x) = ei(K)}.
Show that t 
(K(t)) is measurable if t  K(t)E K0 is measurP
able.
From this deduce that there exists a mapping 9:F
(c):
>iRp
P
such that:
(i) : ip(F) e F if F e Fp and F $ 0; (ii): t  q(F(t)) is measurable if t y F(t) is.
Let f:jRp +R be continuous.
(d):
a 2Rq x F
>
P
Show that there exists
PP such that :
(i) : If F e F and x e f(F), then a(x,F) e F and f(a(x,F)) = x; P
(ii): If t H x(t)e]i
,
t i+ F(t)e Fp are measurable, then
t * a(x(t),F(t)) is.
(e):
Assume that the mapping t f* Fi(t)e FP (i = 1,2) and
MEASURABLE SETS
75
t > x(t) eF1(t) + F2(t) are measurable. Show that there exist t > xi(t)e Fi(t) that are measurable and
such that x(t) = x1(t) + x2(t) for all t. A0A = V AV = A0A = 0A0 = A0A
SOLUTION: (a):
Let K be a nonempty compact set of ItP.
The set
{t:g0(F(t))nK # 0} is equal to
{t:F(t)f1K # 0,d(F(t)) = d(F(t)f1K)}
(*)
when 0 4K, otherwise it would be necessary to add {t:F(t) = O}. The latter set is measurable (cf., Exercise 1.19(a)).
Since the
set {t:F(t)n K + 0} is measurable by definition, and as
K
is measurable it suffices to prove that t } d(F(t)) is measurable on its defining set.
Now, for all a 3 0
{t:d(P(t)) < a) = {t:F(t)n Ba # 0},
where B
a
SOLUTION:
denotes the closed ball with centre 0 and radius a.
(b):
Similarly, on the measurable set {t:K(t)n K # 0}
one has:
{t:ei(K(t))n K # 0} = {t:ei(K(t)) = ei(K(t)f1K)},
and one sees, as above, that t  e .(K(t)) is measurable (by note
ing that {t:K(t)n F # y} is measurable for F closed). SOLUTION: (c):
1op0)(F) is reFor every F e F the set (cp P P It is clear that F > q(F) has all the re
duced to a point p(F). quired properties.
76
CHAPTER 1:
SOLUTION: (d):
Let us, for x ERq and F e Fp, write:
a(x,F) = p(f 1(x)t F). It is clear that Condition (i) is satisfied.
To prove that (ii)
is also satisfied it suffices to show that t > f 1(x(t))r)F(t) is
Now, for every compact set K of Rp
measurable.
{t:f 1(x(t))r)K 4 O} = {t:x(t)e f(K)}, and f(K) is compact.
SOLUTION: (e) :
Let f 2R x]R > Rp be defined by (x,y) > x + y,
and let a be defined as above. Let us set a(x(t),F1(t)x F2(t)) = (x1(t),x2(t)).
Then x1(t) and x2(t) satisfy the properties required.
EXERCISE 1.21:
In this Exercise it is proposed, by consideration
of the Axiom of Choice, to prove the existence of nonmeasurable The Axiom of Choice appears in the following form:
sets of R.
Given a nonempty family (B ) of mutually disjoint nonempty sets i
of ]R there exists a set E of 3R which contains one and only one
point of each Bi. (a):
Show that there exists a set E C [0,1] such that for
every x eR there exists an unique y e E such that x  y is rational. (b):
Let S be the union of the sets E + r, where r runs over
the set of rational numbers lying between 1 and 1. Show that [0,1] C S C [1,2],
and that if r,s are two distinct rational numbers, then E + r and
MEASURABLE SETS
77
E + s are disjoint. (c):
Deduce from this that E is not measurable.
ovo = vov = ovo = vov = ovo
SOLUTION: (a):
Let Q be the set of rational numbers, and if x,y
are real numbers let us express that x  y eQ by writing x ti y. This defines an equivalence relation on]R which all the equivalence classes intersect [0,1].
The existence of E follows from
this and the axiom of choice. SOLUTION: (b):
It is clear that
E + r C [0,1] + [1,1] = [1,2], and therefore S C [1,2].
On the other hand, if 0 _< x s 1 there
exists y e E and re Q such that x = y + r.
1 which proves that [0,1] C S.
One has Iri = Ix  yJ
Lastly, if r # s one has (E + r)
fl (E + s) _ 0, otherwise there would exist z eIIZ and y1 a E, Y2 e E such that
z =y1+r=y2+s, and consequently y1 4 y2, which contradicts z being equivalent to a single element of E. SOLUTION: (c):
If E were measurable S would also be measurable,
and
meas(S) =
I
meas(E + r).
reQfl [1,1] Now, all the numbers meas(E + r) are equal to meas(E), so that meas(S) = 0 if meas(E) = 0, and meas(S) =  if meas(E) > 0.
This
is absurd, because by virtue of the inclusions proved in (b) one would have to have 1 4 meas(S) 4 3.
CHAPTER 2
vAlgebras and Positive Measures
EXERCISE 2.22:
Let (X,C) be a measurable space, (xi) a sequence
of points of X, and (mi) a sequence of real numbers mi > 0.
For
every set E e C set: V(E) =
I
m..
x.eE i
(a):
Show that u is a measure on C.
(b):
Show that if {xi}e C for all i then one has C,1 = P(X),
and conversely.
AVA = VAV = AVA = VAV = AVA
SOLUTION: (a):
It is clear that u(O) = 0 (as usual, the conven
m = 0). Let (E ) be asequence of mutuE i n ieo For every inally disjoint sets of C, and let E be their union. tion is adopted that
teger N one evidently has: N
11 (E1U...UEN) =
1 n=1
(E),
ir(En)
79
80
CHAPTER 2: aALGEBRAS
whence
u(En)
E
u(E).
n=1 Moreover, for every finite set A of 1N one has
E
m
X eE
= 1
i
m. 4
I
G
n=1 x.eE
n
i
1
1 n=1
u(En
ieA
ieA
whence M
u(E) = sup
X
m.
1 one of the two integers 2n,2n + 1 belongs to one of the A. and that the other belongs to none of them.
This shows that the
union of the Ails also possesses the property. SOLUTION: (b):
It is clear that f is a bijection.
Furthermore,
if A has the property and if n 3 1 one has 2n e fl(A) if and only if 2(n + 1)e A, hence if and only if 2(n + 1) + 1 e A, that is to say, if and only if 2n + 1 e f 1(A). urable.
This proves that f is meas
Finally note that A = {0} has the property, but that
f(A) = {2} does not, since 2e f(A) and 3 * f(A), so f_1 is not measurable.
EXERCISE 2.24:
Let C be a family of subsets of a set X.
If
M C X, set: CM = (Mf)E;E a C}. (a):
Show that if C is a aalgebra on X, CM is a aalgebra
on M (CM is called the aalgebra INDUCED on M by C). (b):
If Me C give a simple characterisation of CM.
(c):
If C is generated by a family A of subsets of X, show
that CM is generated by AM. (d):
Deduce from part (c) that if M is a subset of a topo
CHAPTER 2: aALGEBRAS
82
logical space X, the Borel aalgebra associated with the topology induced by X on M is equal to the aalgebra induced on M by the Borel aalgebra 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 aalgebra on M.
SOLUTION: (b) :
If M e C, then
CM = {E:E e C and E C M}.
SOLUTION: (c):
Since AM C CM, C(AM) C CM.
subsets E C X such that MflEe C(AM).
Let C
be the set of 0
Evidently one has 0 e Coy
and equalities (*) and (**) show that C0 is a aalgebra on X. Since A C CO, one has C C CO, which proves that CM C C(AM).
SOLUTION: (d):
This follows from TM being the topology induced
on M by the topology T of X.
If M is a Borel set, the Borel sets of M can be interpreted either as the Borel sets of X contained in M or as the Borel sets of the topological
EXERCISE 2.25:
ability on C.
subspace M of X.
Let C be a aalgebra on X and let p be a probLet M C X be such that E e C and E D M implies
p(E) = 1. Show that a probability 11
M
is defined on CM by setting
83
AND POSITIVE MEASURES
PM(MnE) = u(E) for all E e C (the induced aalgebra CM was defined in the preceding Exercise).
Let us first prove that if E e C, F e C, and Mr) E = Mf1 F, then u(E) = u(F). Now, in this case one has SOLUTION:
(X  E) U (EnF) > M,
(X  F)U(EfF) > M, whence
1u(E)+u(EnF)=1, 1  u(F) + u(E n F) = 1,
and consequently u(E) = u(F). thus defined unambiguously.
The mapping
uM It is clear that:
of CM into [0,1] is
PM(0) = uM(Mflf6) = i(f) = 0, um(M) = PM(Mf1X) = u(X) = 1.
Moreover, if E e C, F e C, and (MUE) fl (MUF) = Mn(EUF) _ 0,
one has:
u(EnF) = 0, whence
uM{(MnE)U(MnF)} = uM{Mn(EUF)} = U(EUF) _
(Contd)
84
CHAPTER 2: aALGEBRAS = p(E) + p(F)
(Contd)
= 1M(MfE) + uM(Mf)F).
Hence
is additive. It remains to prove that it is continuous. uM To do this let us consider a sequence (En) of elements of C such that
Mfl En C Mf1En+1
Let us set Fn = El U M(lEn = MnFn,
UEn, so that
Fn Fn+1'
If F is the union of the F 's, then F e C and n
Iimll (MfFn) = 1imu(Fn) = u(F) = uM(MnF)
= M(U (MnFn)). n
EXERCISE 2.26:
Let X be a nonempty set.
Show that the aalge
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 aalgebra 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 aalgebra generated by the {x},
x e X contains all the sets indicated.
these sets form a aalgebra.
It remains to prove that
The family of them is closed under
complementation and contains the empty set.
If (En) is a sequence
of such sets their union is countable if all of them are; if the
AND POSITIVE MEASURES
85
complement of one of them is countable the complement of their union will be, a fortiori.
This proves the first part of the
Exercise.
Let us now note that if the En are mutually disjoint there can be at most one of them that is not countable.
The countable
additivity of it follows from this.
EXERCISE 2.27:
Let N be the set of natural number, P(N) the a
algebra of all subsets of N.
For every natural number n denote
by nN the set of multiples of n. Show that there cannot exist a probability p on P(ri) such that for every integer n , 1 there holds:
u(nO = n1 . d0A
SOLUTION:
0L0
AVA
V AV
LVL
Since {0} C n N for all n , 1 one would have 1j({0})4 1/n,
and consequently p({0}) = 0.
Furthermore, since 0 is the unique
integer that is divisible by an infinite number of prime numbers, one would have:
{o}
r=1 i=r
{piNJ
where (pi) denotes the sequence of prime numbers.
From this it
would result that
0 = limp{ U pi3N r i=r 1
(*)
By virtue of an elementary property of arithmetic one has:
pi w n ... npi 3N = pi ...pi N 1
a
1
a
< ia)
CHAPTER 2: aALGEBRAS
86
Therefore one would have
u(p. Nn ... np. i) = 11
1
pi , ..pi
la
a
1
r 4 s Poincare's Formula (cf., Exercise 1.1) would give
If 1
l
s
u(U
pi1VJ
i=r
p1a r4i1 1 one would have
ll
tir
ll
((
i1VJ =
o[
s>
i=r
piJ = 1,
which would contradict (*). Now for the proof of (**).
Let As be the set of nonzero in
tegers that do not have prime factors greater than ps. s
iT
1
i=1 1 
1
Pi
k,
=IT{E i=1
n=0 pi
keAs
Then:
AND POSITIVE MEASURES
87
whence:
s 1im
s
1 1
i=1 1 
EXERCISE 2.28:
= lim
I
s keA S
pi
_
I
k= .
k=1
Show that there does not exist a aalgebra having
a countably infinite number of elements.
AVA = VOV = AV1 = VAV = AVA
Let X be a set and let C be a countable family of sets
SOLUTION:
of X which is a aalgebra on X.
For all x e X the set of the E e C
such that x e E is countable; therefore the intersection Ex of
these sets belongs to C, and this is the smallest set of C that Since for all E e C the point x c (Ex  E) or x e ExflE
contains x.
one has either Ex  E = E{ or ExflE = Ex, that is to say ExflE = O In particular, for two arbitrary points x,y e X one
or Ex C E.
has ExflEy = 0 or Ex = Ey.
Let I be a countable set such that
{Ex}xeX  {EiIieI'
with E. # E
.
if i + j.
For every subset A C I,
7
EA = U B. e C, ieA
and EA $ EB if A # B. P(I) onto C.
It is clear that A ' EA is a bijection of
Then if I is finite C is finite, and if I is infin
ite C is not countable.
CHAPTER 3
The Fundamental Theorems
Calculate
EXERCISE 3.29:
(1
xnlogxdx 0
for every integer n 3 0, and deduce from it the value of
1 10
x dx,
1
given that 1
1 n=1 n2
12 6
Av4 s vov ° AVA  vov  VA
SOLUTION:
By setting x = et one has:
Jixnlogxdx =  Jte(nt1)tdt = 0 0
89
(Contd)
CHAPTER 2: THE
90 W
_
(Contd)
1 2J (n + 1) 0
tetdt = 
1 2 (n + 1)
Moreover, if 0 < x < 1, OD
logx
1x
_
 xnlogx.
I n=0
As the functions x + xnlogx are positive, one can integrate term
by term, which gives
J
l logx dx = O 1 x
C
(n +
n==O
EXERCISE 3.30: (a):
=  n2
1

1)2
Let a > 0.
6
For what values of s em is the
function x + xseax integrable on gt+?
Also calculate the value
of its integral with the aid of the r function. Show that for Re(s) > 1,
(b):
n n=1
s
=rs
x1 s x
J
0e 1
dx.
ADA = VAO = ADA = DAD = AVA
SOLUTION: (a):
Since
Ix seaxl = xRe(s)eax
it is clear that this function is integrable only when Re(s) >  1, and that then:
rxseaxdx
a(s+1)( xsexdx = a(s+l)r(s =
0
0
+ 1).
FUNDAMENTAL THEOREMS SOLUTION: (b):
xs1
91
For x > 0
xs1
= e x
ex  1
ex =
1 
xs1e nx
L
(*)
n=1
Since C
xRe(s)lenxdx
I
n=1
n=1 J0
0
F(Re(s))
00
E
nRe(s) < w
n=1 if Re(s) > 1, equation (*) can be integrated term by term, giving
r x s1
ds =
EXERCISE 3.31:
C
n=1
0 es  1
r xslenxdx = r(s)
I ns. n=1
0
For every integer n 3 0 calculate:
0 2
and from this deduce that for all z e c the function t
et coszt
is integrable on [0,Co], and calculate its integral. AVA
SOLUTION:
V AV
AVA
X04
By making the change of variable t  ti one has:
t2net2dt =
0
J'o
000
otnetdt
Jo
=
r(n +
) _
(Contd)
CHAPTER 3: THE
92
(Contd)
= z(n 
... ''r(Z)
'z)
1.3.
(2n  1) 2n+1

(2n)!
22n+1n
since r(z) = u. Therefore 2n
2
2 t2net
e t coszt =
n=0
(1)n (2n
,
!
and 2n G
(1)n
J
n0
0
T
2
Cw
2
t2net
(2n)!
I
dt = n G n=O 2
Thus, 2n
°°
!t2net
n (2n
et cosztdt n=O
2
dt
0
2 n
(z l
(1)n
= 2 E
n4
'
n0 e
z2/4
2
EXERCISE 3.32:
snax
J Oex1
Establish the relation:
dx
12n
2n+1
a
n=1n+a 2 2
AVA = OAV = AVA = V1V = t1V4
n!
/Tr
FUNDAMENTAL THEOREMS SOLUTION:
93
If x > 0,
s in=
Co
=
ex  1
e I's inax .
E
(*)
n=1
Furthermore, by using the inequality Isinul < lul, one has for
n
1
le nxsinaxldx < aJ xe
J
0
0
rix
dx = a n2
so that:
J0ienx sinaxldx
0 X
dx
p
>. 0,
and this holds for all e em.
el$Jxf(x)dx = 
By choosing 0 so that
Jf(x)dxl X
one obtains
f f(x)dx = 0.
JX
EXERCISE 3.42:
Let f be an integrable function on at and let a'>0.
Show that for almost all x eat the series
FUNDAMENTAL THEOREMS
107
+W
n= f[.+nI
(*)
is absolutely convergent, and that its sum F(x) is periodic with period a and is integrable on (O,a).
tVA = VOV = AVA = VAT  OVA
SOLUTION:
Setting u =
+ n on obtains
a
I J:If l
+
n]
=a
I
dx
n=w
C'fxI
=a
JI
.f (x) I dx
so that the series in (*) converges absolutely at almost all points of (O,a).
As this series does not change when x is replaced by
x + a, it follows from this that it is absolutely convergent at almost all points of ]R, and that its sum coincides with a function
F of period a (setting, for example, F(x) = 0 at the points x where the series is not absolutely convergent) that is integrable on
(0,a). Since one can integrate term by term, one obtains by proceeding as above: fa
i
F(x)dx =
J(x)dx.
0
EXERCISE 3.43:
Let f be an integrable function on Ir, and let a > 0.
Show that for almost all x e 1R:
limn af(nx) = 0.
n*
CHAPTER 3: THE
108
DOA  VAV = X00 = VAV = 400
SOLUTION:
t
r(
(x)Idx,
so
inCL
n=1
f(nx) I dx
(ii) of part (a).
We may assume that f is
Let us note that if (xn) is equidistributed then (*) holds
real.
for every characteristic function of an interval [a,b], 0 4a 4b4 1.
This formula therefore also holds, by linearity, for every step function.
Now for every e > 0 there exists a step function 9 such
that: 1
J TSe+
f5 cp,
J0
J0
f
From this one deduces that:
1
N
l im supN
E
n=1
N
f (x n )
.
(iii) is trivial upon setting f(x) =
e2nipx
in
N. Finally, let us show that (iii) => (i).
We first show that
CHAPTER 3: THE
112 (iii) => (ii)t.
Now, if (iii) holds, then (*) is true for every
trigonometric polynomial.
If f is continuous on [0,1], and if
e > 0, there exists such a trigonometric polynomial 9 for which If  91
To simplify the
E (the StoneWeierstrass Theorem).
ensuing calculation, set
uN(f) = N
((1 CN
f(xn)  J f
=1
0
Then IUN(f)I < IuN(9)I + IuN(f  q))I < IuN((P)I + 2E,
and consequently, since uN(P) > 0,
lim supluN(f)I .< 2E, N
which proves that (ii) is satisfied. let
Now let 0 < a < b 4 1 and
be the characteristic function of [a,b].
For every E > 0
there exist two continuous functions gE and fE such that
0< f
E
. 1 is equidistributed and if the boundary of E has measure zero, then
limN 1v(N;E) = meas(E),
N
where v(N;E) denotes the number of integers n such that 1 < n < N
and xn a E.
EXERCISE 3.46:
Let f be a continuous function on ]R2 such that:,
f(x,y) = f(x + 1,y) = f(x,y + 1)
FUNDAMENTAL THEOREMS
115
for any x,y. Show that for every irrational number
f(x,y)dxdy = lim
J1
T
0'<x,y 0
there exists a step function h such that
E I f  hI < E. Setting
TI g(x)dx, 0
one has (taking. account of lyl < M)
((+00
II
+m
f(x)[g(nx + an)  y]dxl < Jh(x)[g(nx+a)Y]dxI+2Mc ,
and consequently,
117
FUNDAMENTAL THEOREMS tW
lnmssupl l
f(x)[g(nx + an)  Y]dxI 4 We, _W
which proves (*).
When g(x) = eix and an = 0 for all n, one obtains the
REMARK:
RiemannLebesgue Lemma: +m f(x)e1nx
lima
n. _,
dx = 0
if f is integrable.
EXERCISE 3.48:
Let (pn) and (an) be two sequences of real numbers
such that
I Ipncos(nx + an)I < n for all the x's of a set A of measure greater than zero.
Show that L
Ipnl
0.
Hence there exists an N such that meas(AN) > 0.
Let E C AN, 0 < meas(E)
0 when p meas(E(1
it follows from the preceding that
The set of x's from [0,2i] such that
= 0.
limsupjcos(n x + s 34w
)I 4 1
S
being the union of the E1
(n 3 1), the proposition is proved.
 1/n
Let f be a positive integrable function on an open
EXERCISE 3.50:
such that meas(X) < w.
set X of ]R
Show that there exists a function g, lower semicontinuous Qn X, such that g >. 1/f and J fg < W.
X 401 = 000 = 404 = V1V = 001
Let us set:
SOLUTION:
An
n) + 1 < f 4
fn
(n = 0,1,2,... ),
and
o). There exist open sets VD and n contained in X such that
Vn D A_,
JV n
f(x)dx
0 and the series E tnfn(x) converges for almost all x;
(iii): There exists a sequence ItnI of real numbers such that
LltnI
and the series I tnfn(x) is absolutely
convergent for almost all x.
122
CHAPTER 3: THE ova = vav = eve = vev = ave
SOLUTION: (i) => (ii),(iii):
Without loss of generality fn > 0
By Egoroff's Theorem there exists a sequence
almost everywhere.
of measurable sets A
C A2 C
1
such that meas(X  As) < 1/s and
f  0 uniformly on each of the A n S of integers n1 < n2
a for all s.
If for al
s
most all x the series E tnfn(x) converges, then in particular
to fn (x) > 0, and consequently fn (x) > 0. s
s
s
(iii) => (i):
Letting g(x) = I Itfn(x)I, by hypothesis one
has g(x) <  almost everywhere.
Then if A is the union of the
As = {g < s} one has meas(X  A) = 0.
I Itnl JA IfnI = JA 9
0 there exists S > 0 such that if E is measurable and meas(E) < d then
jE
Ifl < E. AVO = VAV = AVA = VAV = AV!
CHAPTER 3: THE
124 SOLUTION:
0 (since only 1A appears in
We may assume that f
the statement of the exercise). SOLUTION: (a):
Let:
A0= (f=0),
An=
II
f uniformly on A.
Hence there exists no such that
for n 3 n0 Ifn  fI
and
on A,
< meas(A)
 f)I < E. IJX(fn
By writing
JXA
if n
fn=J
f +JX (fn  fJAn
BA f+JXB
f),
n0 one obtains:
JX_Afnl
< Mmeas(B  A) + E + E +
mess A
meas(A) < 4E.
Furthermore, if
g = IfI + meas A)
on A
and
g = 0
on X  A,
it is clear that g is integrable and that Ifnl < g on A for n 3 n0. SOLUTION: (b):
By Fatou's Lemma,
JX_AIfI < E,
and by the Lebesgue Dominated Convergence Theorem,
JA If  fnI } 0.
FUNDAMENTAL THEOREMS
127
Then by the inequality
fx If  fnl < JXA Ifl + JXA Ifnl + Jlf  fl one concludes
li pJX If
 fnj c 2e,
which proves that
J/n J[ SOLUTION: (c):
Here are three counterexamples where all the
functions considered are zero outside [0,1], which comes down to taking this interval as X. (i): Let
fn(x)=(n 22) if1 x< for n ? 2.
1
and fn(x)=1 if
Then fn > 1 almost everywhere.
If A =
n<x41 one has
IfnI < 1 on A and
fn = 0. XA Nevertheless,
J/n
= 2
(ii): Let
fn(x) = x sin x
if
n< x c 1
and
fn(x) = 0
if 0 4 x 4
n
128
CHAPTER 3: THE
Since
1 sin 1 dx
lint
X
a>O ax a>0
exists, for every e > 0 there is an a > 0 such that if A = [a,l] one has
IJXAfnl < E.
On A the Ifnl are uniformly bounded.
Now, fn(x)  (l/x)sin(l/x)
almost everywhere and this function is not Lebesgue integrable. (iii): Finally, let
fn(x) = x sin x
if 2 < x s
and fn(x) = 0 otherwise. n
Then fn  0 and
 0.
JX f
n
If the conditions of part (b) were satisfied by the f 's they n would also be satisfied by the Ifnl, and one would have: 2n
lin
IsixnxI
lim= 0. fXlfnl
dx =
n
Now, by Exercise 3.48,
1im12n Isinnxl dx = limJ2 Isinnxl ax
n.=i n
x
n
=
1
x
1 sinxdxl Xk 0
1
= 21o g2
FUNDAMENTAL THEOREMS EXERCISE 3.54:
129
Let (fn) be a sequence of measurable functions
on a measurable set X of ]R
.
One says that this SEQUENCE CON
VERGES IN MEASURE towards a measurable function f if for all a >0 lim meas(If  f
n
n).
I
>
a) =
0.
Show that if fn  f in measure there exists a subse
(a):
quence (fn ) which converges to f almost everywhere.
Give an ex
S
ample showing that the sequence (fn), itself, need not converge to f almost everywhere. (b):
Show that if the fn's are positive, converge in meas
ure to f, and if
(Fatou's Lemma for convergence in measure). (c):
Show that if the fri converge in measure to f, and if
there exists a positive integrable function g such that IffI < g for all n, then f is integrable and
n'° X if l imJ
 fn I= 0
(Dominated Convergence Theorem for convergence in measure). (d):
Assume that meas(X) < .
Show that if the fn's con
verge towards f almost everywhere, they also converge towards f in measure.
Give an example showing that the condition meas(X)
<  is indispensible.
CHAPTER 3: THE
130
tVA = VAV = AVA = VAV = AVA
SOLUTION: (a):
Choose a sequence of integers n1 < n2
measlIf  fn
I
s)
S
Then the set of points that belong to an infinite number of sets (If  fn
I
> 11s) has measure zero (cf., Exercise 1.3).
Now,
s
this set contains the one formed of the points where (fn ) does s
For every integer n
not converge to f.
1 let fn be the charac
teristic function of the interval [r2s,(r + 1)2s], where n and s are the integers such that n = 2s + r, 0 < r < 2s.
The sequence
(fn) converges in measure on [0,1] but does not converge at any point of this interval. SOLUTION: (b):
By part (a) there exists a subsequence which con
verges to f almost everywhere, and the classic Fatou's Lemma immediately furnishes the result. SOLUTION: (c):
Let e > 0.
There exists an integrable set A C X
such that
g<E. XA
Set An = (If  f > E/meas(A)) and note that by part (a) one has n IfI < g almost everywhere; then, by decomposing the integral on X I
into a sum of integrals on X  A, A  A n, and Af1An, one obtains:
If  f I < 2e + e +
limsupJ
n
X
n
limsupJA
2g.
n7
Now, meas(An)  0 by hypothesis, so (cf., Exercise 3.5M ,
FUNDAMENTAL THEOREMS
131
2g = 0,
1imJ
n'°° A
n
SOLUTION: (d):
By Egoroff's Theorem, for all c > 0 there exists
a measurable set E such that meas(X  E) < c and fn > f uniformly
Then, for all a > 0 one has (If  fnl > a) C X  E when
on E.
ever n is large enough, which proves that fn > f in measure.
The
functions fn = n[n n+1] converge to zero almost everywhere, but not in measure.
EXERCISE 3.55:
Prove that if f and the fi's are pos
Let a > 0.
itive measurable functions and if fi ; f in measure, then fi  fa in measure.
AVD = VAV = AVA = VIxV = AV6
SOLUTION:
If 0 < a s 1 this results immediately from the inequal
ity: Ifa
 fil , If  files.
When a > 1 one has (by the mean value theorem):
lfa fil
aI f  fil (f V
fi)a
Let e > 0 and let
where f V fi denotes the maximum of f and fi. M > 0.
Then
meas(Ifa  fll > c)
mead If  fil > l
meas(fV fi > M). > M, then
Note that if f v f i
E
1)
CHAPTER 3: THE
132
If  fiI >
or
f > zM
M,
for otherwise one would have
fi = f + (fi  f) < zM + ,M = M. Thus:
meas(lf°` l > E) < mead if  fil >
E1
"Ma
+ meas(If  fil > ZM) + meas(f > 2M), so that
limsup meas(lf°`  fal > e) < meas(f > 15M). Z
Since lim meas(f > M/2) = 0,
M+=
lim meas(Ifa
> e) = 0.
Z
EXERCISE 3.56:
0 < meas(X) < m.
Let X be a measurable set of Iltp such that
Denote by M the vector space of measurable comIf f e M, set
plex functions on X.
p(f) =
j
+
Ifl)1
(a):
Show that fn  0 in measure on X if and only if p(fn)
(b):
Show that (f,g) y p(f  g) is a metric on M if one
+ 0.
agrees to identify two functions that are equal almost everywhere
133
FUNDAMENTAL THEOREMS
Show that p is not a norm on M, but that the sum and
(c):
the product of two functions of M are continuous operations in the metric defined by p.
Show that M is complete in the metric defined by p.
(d):
AVA = 0A0 = A00 = VAT = MMA
This follows from the fact that if Ac = (IfI > c)
SOLUTION: (a):
1 + e
meas(Ac) < p(f) < meas(Ac) + 1 +
meas(X).
If f,g a M, from the inequality
SOLUTION: (b):
1 + If
c
1+ f+ 1+ g
g
one deduces that p(f + g) < p(f) + p(g).
From this it results
that p(f  g) satisfies the Triangle Inequality.
Furthermore,
if p(f  g) = 0 then f  g = 0 almost everywhere, that is to say that f = g in M. SOLUTION: (c):
p is not a norm, for in general p(Af) + IXlp(f)
Nevertheless, if fn  f and gn > g in measure, then
P(f + g  fn  gn) < p(f  fn) + p(g  gn) + 0,
so f + g > fn + gn in measure. Assume now that fn  0 in measure and that for all c > 0 there exists a > 0 such that
m meas(Ig limsup
I
n
> a) < c
(which is the case if gn + 0 in measure, or if gn = g for. all n).
Since, for all a > 0,
CHAPTER 3: THE
134
(IfngnI > B) C (IgnI > 0)U(IffI
>
one will have limsup meas (I fngn I
> S) < e,
which shows that fngn  0 in measure.
From this it follows imme
diately that if fn  f and gn  g in measure, then fngn  fg in measure.
SOLUTION: (d):
Let (fn) be a Cauchy sequence in M.
There exists
a subsequence (fn ) such that s
p(fn
 fn ) < 4s. s+1
S
By the inequality from the answer to part (a), for all s one has:
 fn
meas(Ifn s+1
I
> 2s) s (1 +
2s)4s.
S
By exercise 1.3, for almost all x one has
Ifn
s+1
(x)  fn W1
0 there exists a measurable set
A C X
such that
meas(A)
c)
For every E > 0 one can choose c such that the integral of f on (f > c) is less than e/2 for all f e H.
Then, if meas(E) < £/2c,
f < e for any feH. E (ii) => (i): First let us prove that
lim meas(f > c) = 0 uniformly for f e H.
(*)
c
Otherwise, there would exist a > 0, a sequence ek
and a se
quence fk e H, such that
meas(fk > ck) > a
for all k.
Let 0 > 0 be such that meas(E) 6 0 implies for all f e H:
f < 1. JE
CHAPTER 3: THE
138
For every k one can find Ek C (fk > ek) such that:
meas(Ek) = min(a,s).
One would then have:
1 >
1
fk > ekmin(a,s), Ek
which contradicts ek >
Now let e > 0 and 6 > 0 be such that
meas(E) < 6 implies
f<E jE for all f e H.
There exists e0 such that e a c
0
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
0 such that
meas(En) = a2n,
f
feHJ En
But this manifestly contradicts Condition (ii). SOLUTION: (c):
Assume first that there exists such a function G.
Let e > 0 and set
M = sup J G(f).
feH X There exists c0 such that t1G(t) > M/E if t > c0.
Then for all
c > c0 and all fEH J
f s L G(f) s E. Mx
(f>c)
Now assume that H is uniformly integrable and look for G in the form
G(t) =
rt
g(u)du,
J 0
where g(0) = gn = constant on [n,n + 1[, with 0 = go 4 gl 6 and g w. Then G willbe increasing (and hence measurable), n convex (since g is increasing), and will satisfy
,
140
CHAPTER 3: THE
limt1G(t) t
for, g(u) >  as u
and this implies that the mean of g on
[O,t] tends to infinity when t i
It remains to choose the g
n
suitably in order that
supJ {G(f) < feH X
(1)
Define
an(f) = meas(f > n); by noting that G(f) = 0 if 0 .< f < 1 (since g0 = 0) one obtains:
G(f) _
X
(as G is increasing).
G(n + 1)(an(f)
G(f)
c) = 0 Cya, n
Nevertherless,
CHAPTER 3: THE
142
uniformly in n, because if e > 0 and c > 1/e one has meas(fn > c)
= 0 if n . c
meas(fn>c)=n1
p
We are going to show that the A(t) realise some sort of "homotopy"
between A10 and Al
1,1 ,
,
in the sense that they satisfy the follow
ing properties: (1):
A(0) = A10 and ,
(2):
,
meas(A(t)) = 2meas(A);
(3):
f1 .
= 0,
JAW (4):
A(1) = A11;
1 4 i F n;
For every function f integrable on A the function e+
t
A(t)f
is continuous.
CHAPTER 3: THE
144
This will imply the property P+1, for by virtue of the Intermediate Value Theorem there exists a t (0 4 t S 1) such that
A(t)fntl =
J
2( JA(O)f'+l
=
(/
+
fn+1 +
JA(l)fn+l)
f
Al,l
A1,0
fn+1)
JAfn+l = 0.
It remains to prove properties (l)(4).
A1(0) = Al
First of all,
0, ,
and if p >. 1,
A
p+l
(0) = U A p+3,q,q
=
U
05q 0 for all a e A, then for every integrable mapping f of X into A
Z(J f(x)dx) = 1 t(f(x))dx : 0, X
X
which proves that:
1AdxCTA. X
Now assume that, with f being as above,
y=1 f(x)dx6FA)P(A).
CHAPTER 3: THE
150
Since the interior of 1T A is equal to that of r(A) (a classical result about convex sets), y must be a boundary point of r A
.
There would then exist a nonconstant 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.21(0)n A, so then y 4 r(.21(0)nA).
Since it can always be assumed that A affinely
generates R , and consequently that the affine dimension of .11(o)n A is strictly less than that of A, one is led to a contradiction.
REMARK: More generally, one can show analogously that if g is measurable and positive on IItn, and
Jg(x)ds = 1,
the set of points
Jf(s )g(s )dx,
where f:3tn N. A is integrable, is equal to F(A). EXERCISE 3.63: (a):
Let f be a mapping of X =
into Y = IItn
Show that there exists a smallest closed set Af such
that f(x) a Af for almost all x e X, and that Af is nonempty.
Next show that if f is measurable, then y e Af if and only if meas(f 1(V)) > 0 for every open neighbourhood V of y. (b):
Now assume that f is integrable (or measurable and
bounded) and consider the set If formed of points of Y of the type:
FUNDAMENTAL THEOREMS
151
1
mess E
f(x)dx,
E
where E is an arbitrary measurable part of X such that 0 < meas(E) < .
Show that If is convex.
What are the relations between Af and
If?
AVA = VIV = AVA = VAV = AVA
SOLUTION: (a):
In order to prove the existence of Af it suffices
to prove that of a largest open set Vf of Y such that f 1(Vf) has measure zero; Af will then be the complement of Vf.
To do that
it suffices to prove that if (Va)aei is a family of open sets of
such that f 1(V
has measure zero for all a e I, then the union
Y of the Va's possesses the same property.
By virtue of LindelSf's
Theorem (cf., the end of the Exercise) there exists a sequence (an )n
1
of elements of I such that W
V= U V n=1
n
and consequently,
f 1(V) = U f 1(Va n=1
n
Certainly has measure zero.
Furthermore, as X = f1(Y), one has
1'f + Y, and consequently Af $ 0.
Now assume that f is measurable.
If y e Af and if V is an open
neighbourhood of y, Vf u V is an open set which strictly contains Vf, so that:
f 1(Vfuv) = f 1(Vf)uf 1(V)
CHAPTER 3: THE
152
has positive measure; since f 1(Vf) has measure zero, meas(f 1(V)) > 0.
Conversely, if y4 Af and V = Vf one has meas(f 1(V)) = 0.
SOLUTION: (b):
Let E. (i = 0,1) be two measurable sets of X such
that 0 < meas(Ei)
0.
By the remark at the end of
Exercise 3.59 there exists a family (Ft)0Ct41 of measurable sets such that
Ft C E,
F1=E,
F0 = O,
t * meas(Eif1Ft)
is continuous,
and JF IlE1(f  yi) = 0,
which we may write as:
f = meas(EifFt)y,.
J Ei f1 Ft Set:
Gt = (E0  Ft) U(E1f1Ft);
then
FUNDAMENTAL THEOREMS
153
f = meas(E0  Ft)y0 + meas(E1 n Ft)y1. J/1
t
Now note that Gt D EOflE1, so that meas(Gt) > 0 for all t; the function
meas(E0  Ft) t .
meas( t)
is therefore continuous and varies from 1 to 0.
For all a, 0 < a
,< 1, there therefore exists a t for which it takes the value a, and then
mess Gt
f = ay0 + (1  a)y1. G
When meas(E0f)E1) = 0 one can assume that E0f1E1 = 0.
It is poss
ible to determine two families (Fi,t)0, 0 for y e Af, and
UP=o, JE
one would be able to assume that Af C £1(0), and one would end by arguing by induction on the affine dimension of Af.
Let us
note that if f is bounded, then Af is compact, and consequently
If = r(Af).
LINDEZAF'S THEOREM: In a topological space possessing a countable
basis of open sets (U).which is the case for
Rm
the union of
an arbitrary family of open sets (V ) is the union of a countable a
subfamily of them.
This is proved by considering initially the subset J of integers n for which Un is contained in at least one of the Va. with every n e J one associates an index an such that Un C Va n Then,
U Va =U V a
neJ an
Then,
155
FUNDAMENTAL THEOREMS Indeed, if x e Va there exists n such that x e Un C V
of a basis of open sets).
(definition
But then n e J and x e Va , which proves n
the Theorem.
Let f be an integrable function on at.
EXERCISE 3.64:
Show that if b
f(x)dx = 0 for all real numbers a and b, then f = 0 almost everywhere. AVA = VAV = AVA = VAV = OVA
Every open set V of ]R being the countable union of in
{SOLUTION:
ttervals, for all such V one has:
Lf E is measurable there exists a decreasing sequence of open Sets (Vn) such that
E C n V n
meas(E) = lim
and
n
ffo that :
jE
=limfV n
f=0.
n
Lbus the set If of all the numbers
1 meas E
Jf E
Mtluces to {0}.
0 < meas(E) < 0, '
Now if contains the smallest closed set Af such
CHAPTER 3: THE
156
that f(x)e Af for almost all x (cf. the preceding exercise). Hence one has Af = {0}, or in other terms f = 0 almost everywhere.
EXERCISE 3.65: (a):
Let p be a norm on 3kn.
Show that if E is a subset of measure zero of IIz+, then
{x:p(x) a E} is a subset of measure zero of ]Rn. (b):
Let V = meas(p c 1) and f a measurable function on at+.
Show that:
nVJtnlf(t)dt
= 12 f1f(p(x))dx
0
if f 3 0, or if J0t1t)dt < AVA = VAV  OVA = ViV = AVA
SOLUTION:
Note first that:
meas(p = r) = lim meas(r 4 p < r + 6+0 = meas(p < 1)lim((r +
0n
 rn) = 0.
ey0
This allows us to write
rb
meas(a 6 p . b) = V(bn  an) = nVJ toldt, a and consequently the formula nVJtnlf(t)dt
= 3t f'f(p(x))dx
0
(1)
FUNDAMENTAL THEOREMS
157
is valid for every step function. Let E be a subset of measure zero of 3R
contained in [0,R].
There exists a seqeunce (fi) of step functions, zero on [2R,m[, such that
04fi4fi+l,
and fi * DonE.
sUpJ0fi<M,
i
Then fiop . f i+1op, supJf.op = sup(nVTO 1
i
tn1f(t)dt)
i
1
nV(2R)n1 supJ
, 0 is locally integrable, if f is measurable, and
if
J ifIg < , there exists a sequence of step functions (f.) such that
(f  fi)g 
0
(a.e.),
jIf  filg 
0.
This is a particular case of a general proposition in measure theory.
It can be proved directly by observing that if h e L
and
Jgh = 0
for every rectangle P, then gh = 0 almost everywhere.
In other
words the annihilator in L°' of the set of functions IlPg (which
belongs to L1 because g is locally integrable) is the vector subspace {h:h a
this subspace.
0 a.e.}.
Now fg a L1 and is annihilated by
Hence fg is the limit in L1 of functions fig,
where the fi's are step functions (by the HahnBanach Theorem), and by taking a subsequence one can assume that fig 
fg almost
everywhere.
If one wants to avoid using the HahnBanach Theorem (as well,
as the property (L1)' = L) one can argue in the Hilbert space
FUNDAMENTAL THEOREMS
159
L2, and note that if f >. 0 then g e L2.
If g were not to be
long to the closed vector subspace generated by the functions
IlP/, the projection theorem would assure us of the existence of
a function he L2 such that
JhV>
Jh1
0,
=0
P a rectangle of ]R' .
P This is absurd, for the second condition implies that hV = 0 alHence there exists a sequence (fi) of step func
most everywhere.
tions such that fiV g in L2; but then the functions f2 are also step functions, and
If  filg
. 0 there exists a sequence
(fi) of positive step functions (cf., the second proof of the case p = 1) such that fig > fpg in L1
,
and consequently fi pg
liP
1'
fg1/p in Lp (cf., Exercise 6.105). Instead of considering sequences of step functions one can, for example, consider infinitely differentiable functions with compact support.
EXERCISE 3.66:
Let V be a convex bounded set of tn.
Show that:
meas(V  V) .< I nnJmeas(V). (If p(x) = inf(X > O:x e X(V  V)) is the gauge of the symmetric
}
160
CHAPTER 3: THE
convex set V  V, p is a norm on in. there exists y e]R
Show that for all x e V  V
such that:
(1  p(x))V + y C Vn (V + x) and use the preceding exercise).
OVA = VAV = AVA  VAV = AVA
SOLUTION:
If x e V  V then p(x) < 1 and x = p(x)z, where z beLet (zi) and (z2) be two sequences
longs to the boundary of V  V.
of points of V such that zi  zi * z.
one can assume that z1
X=
p(x)(z1  z2 ),
*
As the open set V is bounded
z1, z2 ; z2, so that:
z1 a V, z2 a V.
Let y = p(x)z1 and observe that as V is open and convex (1  t)a +
tb e V if 04 t < 1, a e V, bet. Consequently, for a e V (1  p(x))a + y = (1  p(x))a + p(x)z1 a V,
(1  p(x))a + y = (1  p(x))a + p(x)z2 + x e V + x,
which proves that
(1  p(x))v + y c vn (v + x). Denoting the characteristic function of V by cp one obtains:
(1  p(x))nmeas(V) = meas{(1  p(x))V + y)
s meas(Vr(V + x)) = whence meas(V)I 11 VV
(1  p(x))ndx c
t f
n(q)*)(x) = meas(V) 2,
161
FUNDAMENTAL THEOREMS and consequently, because meas(V) > 0,
(1  p(x))ndx < meas(V).
VV By the preceding exercise (1
(1  p(x))ndx = nmeas(V  V)J to1(1  t)ndt
f
VV
0
_
nr(n)r(n + 1) meas(V  V) r(2n + 1
(n!)2
(2n)!
meas(V  V),
rhence, at last, meas(V  V)
.
1
1 +A`+AA 1
,
+ 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
1
+ f` < 1 + f, one has
f2 = 1 + f,
and so
f = 0.
EXERCISE 3.71:
Let X be a measurable set of ]R
such that meas(X)
1, and let f,g be two positive measurable functions on X. Show that if fg , 1, then:
Jx f.1 g >. 1. X
AVA = VAV = AVA = VAV = AVA
SOLUTION:
The inequality fg >. 1 in fact implies that f(x) > 0
and g(x) > 0 for all x e X, so that
f > 0, Jx
g > 0. Jx
We therefore need only consider the case when
FUNDAMENTAL THEOREMS
173
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.
Show that if g is integrable on E, and
(a):
If
T gI s M,
n = 0,1,2,...
E then:
n = 0,1,2,...
= 0,
.
JE
From this deduce that if g is integrable on [0,a] and
(b):
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 therefore has, uniformly on E: eXf
=
xn n0 n
.
0, one
CHAPTER 3: THE
174
and consequently n fexf91
= IX n
niJf
I
¢
x.
Setting F = f  1, one therefore has 0 < F . B = A  1, and
I J exFgl  1,
x >. 0.
(1)
For all z = x + iye Q set JeZFg.
(2)
4)(z) =
Then
IeZFgl
< eBx IgI, which proves that ¢ is an entire function
and that IIgII1eBx+
(3)
By Inequality (1) it is also true that
I0(x)I .< 1,
x
0.
(4)
(3) and (4) imply, on setting C = max(IIgII1,l), that C
if
Re(z) < 0 or z eR+.
(5)
We will show that in fact:
zea.
Io(z)I 4 C,
(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 1 ) n
J F
n >, 0.
FUNDAMENTAL THEOREMS
175
But
fn
Cn(f  1)S s
s
which implies the desired result. We now next prove (6). Theorem.
This follows from the PhrXgmenLindeldf
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(  eeian/4a )4(a)
where c > 0 and 1 < a < 2.
8 . n/2, then
If a = pe18, p >, 0, 0
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+ < p one has
IG(z)I < Cexp(  epacos(8  40 + Bp).
(8)
L$ince an/4 < a(8  n/4) < an/4 < n/2, one deduces from (8) that:
IG(z)I < Cexp(epacosl4) t Bp).
(9)
Because a > 1 the right side of (9) tends to zero as p
there
'pre there exists p0 such that IG(z)I c C;
Ipw by (7) and (10)
Re(z) >, 0,
Im(z) >, 0,
Izl
>, p0.
IGI 6 C on the boundary of the domain
(10)
CHAPTER 3: THE FUNDAMENTAL THEOREMS
176
Re(z) > 0, Im(z) > 0,
IzI
< p0, by the Maximum Principle one also Finally, if z = pelf, p  0, 0 < 0
has IGI .< C in this domain.
< 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,
SOLUTION: (b):
J0tgt)dt
Re(z)
0,
Im(z) y 0.
By the preceding,
= 0,
n  0.
It follows that for every polynomial P,
e
a
P(u)g(logu ) uu = 0.
J
1
If p is a continuous function on [l,ea] and if (P.) is a sequence of polynomials which tends uniformly to (p on 3this inter
val, then since u1g(logu) is integrable, one obtains
e
a
c(u)g(logu) du = 0. 1
But then u1g(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 a
dx F(t) = ot+W(x) f
Show that if 0 4 r < 1,
(a):
ra
(t) t>0
0
dx
q(x)
(b): Show that if r > 1, then as t } 0
F(t) ti
(c):
IT
1
rAl rsin(n/r) tl  1 r
Show that if r = 1, then as t > 0
F(t) ti
log 1
A
CHAPTER 4: ASYMPTOTIC
178
(For part (c), show that this can be reduced to the study of the integral a
at
(t + (p(x))1dx,
and make the change of variable x = aty).
000 = 0A0 = A0A = voo = ova As t decreases to zero, (t + T(X))1 increases to
SOLUTION: (a):
1/y(x) for all x e]O,a], whence the result. SOLUTION: (b):
Make the change of variables x = (ty)1/r.
F(t) = rtl  1/rJ0
rt
Then
y1/r  1
1 + t1cP(t1/ry1/r)
dy.
Set:
y
1/r1 if 0 < y 4 ar/t,
1 + tqt1/ryl/r)
ft(y)
if y > ar/t.
0
Since, for y > 0 fixed and t  0, t 1p(t1/ry1/r) 
Ay,
1/r  1 lTO 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.
ft(y
Therefore
1/r  1
1+By .
EVALUATION OF INTEGRALS
179
The Dominated Convergence Theorem can therefore be applied, yielding
(
1
lY
tr0 +o
ft(y)db =
1 +
0
J
SOLUTION: (c):
dy =
0
A
l r
it
sin(77r)
Note that the relation proved in part (a) is true
0, so that lim F(t) = W if r >, 1. t+0
for all r
Now,
10t
j0
t +dp(x
0 one has 9(aty) v aAty and t = o(ty). Consequently
ty t>0 t + p(aty) lim
Furthermore, for 0 4 y s 1 and t > 0,
t + P(aty)
t + aBty
from this it follows that
< aB
CHAPTER 4: ASYMPTOTIC
180
1
t}'
lim f
t>0 0t+ip(ats')
dy =  .
Let p,q,r be three positive real numbers.
EXERCISE 4.74:
Find
the necessary and sufficient condition that
fxP
dx < 0.
+0 1 + xqjsinxxjr
What type of counterexample does this furnish? AVA  V AV = A0A = ono = AVA
SOLUTION:
The integral is equal to
n=0 where
I
=
fn+1
n
n
xp
_
(x + n)p
J1
o 1 + (x + n)q(sinnx)r'
1 + xq, sin= Ir
For t > 0, set
(p(t) _
dx 0 t + (sinnx) r
then
7 (2np
n + 1)q
m((n + 1)q)
0,
if04r 1.
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 integrand 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
(iii):
l9(x)le
J
0
f(x) dx

I
g(x)etf(x)dx
J0
(2):
'
i a
+ 1l (S+1)/a rP1)(At)
Prove that
B(p,q) =
r(p)r(g) r(p+q)
(Begin by showing that
B(p,q) =
SOLUTION:
p q+ 4 B(p + 1,q).)
Let 0 < B < A, y > 1; there exists b such that 0 < b
< a and a
f(x))3 Bx ,
0 4 g(x) 4 yx
if 0 < x 4 b.
Since f(x) > f(b) > 0 for b 4 x < a, when t > 1 one has
t(S+1)/alag(x)etf(x)dx 1b
4
t(B+1)/ae(t1)f(b) a lg(x) lef(x)d., J0
which proves that the left hand side tends to zero as t 3 . Moreover, the change of variable x .+ xt1/a gives
t(0+1)/a j bg(x)etf(x)dx 0
where
=
JF(x)dx
(*)
'EVALUATION OF INTEGRALS
183
is/°g(xt1/a)etf(xt1/a)
bt1/a
if 0 < x
Ft(x) =
if x >
0
btl/a.
By (i) , a xseAx
t
limFt() =
and by (*) , a 0 < Ft(x) < yxseB"
Applying Lebesgue's Theorem, and recalling that the integral from b to a tends to zero
limt(S+1)/a(a
g(x)etf(x)
10
dx = JX'e Axadx
t
0
1 A(Otl)/ar(B + 11
a
a
FIRST APPLICATION: Setting x = a(u + 1) one obtains
r(a + 1) = J xaexdx =
aa+1ea(W ea(ulog(u+l))du.
1
0
0
The above result can be applied to each of the integrals with a = 2, A =
,
B = 0, giving
0
F
2xr()(2)
ea(ulog(u+1))du ti
1
l
=
an
and
CHAPTER 4: ASYMPTOTIC
184
r(a + 1) ,
aa+gi
eaV For p > O,q > 0,
SECOND APPLICATION:
(1
p(l 
B(p + l,q) = J 17XX)
o
(1
=
=
p
x)p+q1dx
+ q 0 (1 
x)p+q
p1
x 1
x
dx
(1  x)2
B(p,q).
p
From this it follows that for every integer n 3 1, B(p,q) _ (p + q)(p + q + 1)...(p + q + n) B(p + n + 1,q). n) p(p +
Now it is known that as n
a(a +
n)
nl
na
 r(a)
,
So
(p +
q + n) p...(p + n)
,,
r(p) nq r(p + q)
On the other hand,
B(p + n + l,q) =
x)penlog(1x)
11xq1(1 
dx.
J0
Applying the first part with a = A = 1 and S = q  1,
B(p + n + l,q) % r(q)n q, So
EVALUATIONS OF INTEGRALS
185
B(p,q) = r(p)r(q) r(p + q)
Prove that for n
EXERCISE 4.76:
, n x`` sinx4dx
x e
0,
= 0,
0
and then that as t > ,
1
f eitxx4sinx"dx
r()ein/8t5/4
ti 4
J0
004  040 = 400  V AV = A4A
SOLUTION:
The function
x4n+3ezxdx
F(z) = J
0
is holomorphic for Re(z) > 0.
F(z) =
When z is real one has:
(4n + 3)! 4n+4 z
By analytic continuation this formula is valid for every z with Re(z) > 0.
In particular, if z = 1 + i,
x4n+3e(i+i)xdx
=
(1)n+i On + 3)!
T
22n+2
Taking the imaginary parts of both sides yields:
J x4n+3exsinxdx 0
= 0,
CHAPTER 4: ASYMPTOTIC
186 1
and carrying the change of variable x i x" gives
(WnxIT J x e
, sinx'dx = 0.
0
Now set: 1
1
f(z,t) = exp(itz  z')sinz",
which for fixed t is holomorphic for Re(z) > 0 and continuous 1
(One chooses the principal branch of z4 in this
for Re(z) > 0. halfplane).
Taking into account the majorisation
Isin(x + iy)I < ey,
which is valid for y > 0, then for R > 0 and t > 0 one has:
I
If(Re l$, t)
I
I
< exp[ tRsina  R4(cos 4  sin!)), 4
< exp
2t7 8

,2R"sin i

(1)
Moreover, for z = x + iy, y 3 0:
Iz "et"f(z,t)I < min(IzI 4,Iz "sinz"I), and there therefore exists a constant M such that
(2)
Y A 0.
If(z,t)I < MIzI"ety,
Therefore, using (1), the integral of f(z,t) along the circular quadrant 8  z = Re'
(0 < 8 < 7t/2) is majorised in modulus by
I
Rexp( 0
l
2t exp(  v
sin 8) ,
LUATIONS OF INTEGRALS
187
consequently tends to zero as R 
By Cauchy's Theorem
therefore has
J f(x,t)dx = if f(iy,t)dy 0 0
=
it5/4ein/8 OtIeit[/B flit1y,t)dy. J
0
is easily verified that
limt ein/8f(it1y,t) = y e y
by (2) It4ein/8f(it1y,t)I < My4e y.
refore by Lebesgue's Theorem
limt5/4JWf(x,t)dx t3
ieln/BrWy4eydy
=
0
0
iein 8r [T5) = i_ r(4)ein B. 4
RCISE 4.77:
For every integer n
1 denote by do the number
partitions of a set with n elements. (a):
Set d0 = 1.
Show that for all complex numbers z:
n n=0 n n, =
(b):
exp(ez  1).
Deduce from this that for every real number u > 0:
CHAPTER 4: ASYMPTOTIC
188 u+2co
i
do
= 2nieJu_i z(ntl)exp(eZ)dz.
Let u
(c):
n
be the unique real root of the equation
zeZ=n+1. Show that
d , n
u U exp(e n  u e nlogu  lu n n n
n'
eJ
and deduce from this that logdn '' nlogn.
Ava = vev = ove = vov  ovo
SOLUTION: (a):
To determine a partition of En+1= {1,2,...,n + 1}
one may first fix the part of En+1 which contains n + 1; if this
part contains p + 1 elements (0 < p < n) there are (p) possible choices.
Next it remains to choose a partition of the n  p re
maining elements, which gives, taking account of the convention do = 1, n
do+1
p0
n lP ,dnp
(1)
This relation can be written as do+l (n + 1)

(n + 1)!
n
1
d p
p)! p_0 p! (n 
.
If the series n
f(z) _
do n, n=0 L
(2)
EVALUATIONS OF INTEGRALS
189
has a radius of convergence R > 0, then taking the derivative and using the above relation shows that for jzj < R
f'(z) = ezf(z), and consequently, since f(0) = 1,
f(z) = exp(ez  1).
This function is entire, and if its Taylor expansion is written in the form of Equation (2) the coefficients do have to satisfy Equation (1), so, since d0 = 1, the proposition follows. SOLUTION: (b):
By Cauchy's Theorem:
z(n+1)exp(ez)dz
IN
C
n = 2nieI
 M
where r denotes the rectangle indicated by the figure.
Note that, on
B
O
D IN
A
this rectangle, if z = x + iy, lexp(eZ)I = exp(excosy) .< exp(eu),
and consequently
.< exp(e
JBCDA
u
2u + 4M )
0+1
On the other hand: iy)(n+l)exp(eu+iy)l
j(u +
(U2 =
+
y2)(n+l)/2exp(eu)
From this one deduces that for n 3 1: =
do
n1
2x_f+ (u + iy)(n+l)exp(eu+iy)dy,
(3)
190
CHAPTER 4: ASYMPTOTIC
the integral being absolutely convergent. SOLUTION: (c):
Introducing the principal branch of log z in the
halfplane Re(z) > 0, and if u is chosen so that ueu = n + 1,
+W do
n!
exp(eu  ueulogu)J g(y,u)dy,
(4)
where
g(y,u) = exp{eu[ely  1  ulogI 1 + mil]}. Note that
l g(y,u) I = exp{ 
2
((
eu[2sin2 + 2 log I 1 +
2 ] }. U,
111111
Therefore
if ly l >7E
+ u 2y2)jue
9(y,u)dy1 < 2J (1
u
dy
n
2
dy Jn 1 + u Ie u y 2
The last integral has the value eu/242utani( au/2 fu)
whence
fly,>ng(y,u)dy
= 0(ueu).
If lyl < it then Isin(y/2)I
lg(y,u)I < explI
o
lyl/n, so that:
2 euy2) It
(5)
191
EVALUATIONS OF INTEGRALS Setting y = tou/2 yields:
+x
xeu/2 eu/2Jxeu/2
g(y,u)dy fx
g(te,u)dt, u/2
=
and for t > 0, 2
limg(teu/2, u) = et /2
Since for 0 < ItI 4 xeu/2 2
a 2t /x
2
Ig(teu/2,u)I .
0, carrying out the change of variable z + t2z in Equation (1) and setting R = t la,a = t3/2 yields (
2
4(t) =
expixlz +
J
LS
l
3 ) dz.
CHAPTER 4: ASYMPTOTIC
194
Since the derivative of z + z3 /3 vanishes for z = i, let us take S = 1 in the formula above.
This yields
+m ( 3l 'r O(t) = e2X/3+ expal u2 + 23 Idu 1t
2
l
1 e 2a/3
(+mexp
111
I
u2 + 3 Idu.
The latter integral can be written
V 1,
+
 1)du.
J+WeCO
The absolute value of integral above is majorised by 2
JIuI3e_u
du.
m
From this, one deduces that as t *
¢(t) = 't "exp( 3 +3/21(1 t O(t3/ )).
Setting A = t3/2 as before, 3
(
rit"
(t) = J
)dx. 0
As the derivative of x3/3  x vanishes for x = 1, one is led to
the change of variable x = 1 + u//, which gives
(
(t) = 1
Re{e2ia/3T
3
(
xpi 1u2
+ 3
/T) d.)
EVALUATIONS OF INTEGRALS
195
Set ((
f(z) = expila2 + 3x) = Rei*, where, as z = pale, 3
R = exp(  p2sin28   _ sin30), 3
= p2cos20 +
p COS33.
Assume that 0 < 0 4 n/4 and p
.  v.
Then sin20 3 0 and sin38 3 0.
From this it follows that for every number F such that 0 4 E F 1:
sin20 + .
n8  8 = a8,
sin30
Where a = 4/n  1 > 0.
Therefore
2 e'p
IRI
0 1 + xa + ys + zy y>O z>O and then calculate this integral.
AVA = VAV = AVA = VAV = eve
First of all it is clear that one must have a > 1,
SOLUTION:
0 > 1, and y > 1, for if, for example, a s 1:
dx
J O1+xa+ys+zy for all y > 0 and all z > 0.
x = u
2/a
Y = v
,
2/$ ,
Set
z=w 2/y
The integral may be written
_
I
u2/a  lv2/S  lw2/y  1
S
aOy 111u>0
l+u +v +w 2
2
dudvdw. 2
v>o w>o
Now change to spherical coordinates u = rsinecoscp,
which yields
V = rsinesincp,
w = rcose,
CHAPTER 5:
212 8
rm r2(1/a + 1/0 + 1/Y)1
1= aSO
r2+1
0
Ia/2sin2(1/a
drJ ic/2cos2/a  19sin2/5  1(pdW 0
+ 1/B)l6cos2/Y
X
 13d8
0
1 aRY
r
ars ratsPY r(q + 2 +
r(a +
rm ul/a 1+U
0 Y)
Thus if
a + a +
1
< 1
y
then
1 =
n
rlalr[oil r[1]
aay
r {cx + B +
sinn l a+ yj
0+ Y)
which can also be written
aBy r (a) r (S) r Y) I
EXERCISE 5.91:
a
Y)
Calculate:
fff 0<x,y,z «
dxdydz 1  cosxcosycosz
OVA = VAV = MVA = VAV = OVA
SOLUTION:
Set
du
FUBINI'S THEOREM
213
V =
u = tanix,
z = taniz,
so that the integral becomes:
I
4
dudydw
JJJ>0 u2 + v2+ W2 +
u2v2'W2
v>o w>0
Changing to spherical coordinates yields
/2d3f/2dTf
0
I = J0
1 + r4sin4Ocos28cos2
0
2
Finally, setting
r = t4(sin8)cos
3cos cpsin cp,
we have _
(n/2
I
_1
_1
(n/2
cos 1sdolo
(W
cos 'gsin 2gdgj0
= f0
i r(i)r(i) r(1)r(1) 4
r(
)4,
r() sin it
7E /2
=
4
,41
1 + u
du
r(3 r(4)
Noticing that
r(;)r(4) =
sin 4n =
nom,
the answer can be written
I = 4x(4)44
EXERCISE 5.92: ence:
Use a double integral to represent the differ
214
CHAPTER 5: J+_
f(x) 2 dxfg(x)2dx  lIJ f(x)g(x)dxl2
00
ava = vav = ova = vov = ave
SOLUTION:
It will be found that
2JJ2(f(x)g(y)  f(y)g(x))2dxdy
satisfies the requirement.
EXERCISE 5.93:
Show that the centre of gravity G of a homo
geneous cone satisfies:
= loo, where 0 denotes the vertex of the cone, and G
0
the centre of
gravity of its base.
AVA = VAV = 1v6 = Vtv  pvt
FIRST SOLUTION:
On placing the coordinate origin at 0 and
making the plane xOy parallel to the base of the cone C, the coordinates E,n,r of G are given by:
JJJ'ydz
dxdydz
'SIC and two analogous formulae.
If
are the coordinates of
G0,S0 the area of the base, and CZ the crosssection of the cone cut by the plane parallel to xOy with height z, then:
215
FUBINI'S THEOREM 1
r
dxdy =
JJC
1
z
JJ xdxdy =
ydxdy
V0,
3 n0S0,
1
2 = zdxdy = zICoJS0
11
JJ
s0,
3
z
Jcz
0J
2
z
0Jc6S0, 3
1
ll
so that
J 00
NZ013 C0S0dz
3
=E , 4
0
etc.....
2 S0dz J00NEO)
SECOND SOLUTION:
G is the centre of mass of the segment OG0
weighted with density ku2, where u = OM and k is a constant. It follows immediately that
EXERCISE 5.94:
Show that the volume bounded by a ruled surface
and two parallel planes is equal to
V = 6 Sl + S2 + 4S3 where h denotes the distance between the two planes, S1 and S2 are the areas of the crosssections cut out by these two planes, and S3 is the area of the crosssection cut out by the plane parallel to the other two and located at a distance h/2 from
216
CHAPTER 5: (The Pile of Sand Formula).
each of them.
t0E = V AV = M1L  010 = AVA
SOLUTION:
As the zaxis is perpendicular to the planes under
consideration, the crosssection cut out by the horizontal plane with height z is bounded by a curve given parametrically by equations of the type x = a(t) + zb(t),
rz
{l
y = c(t) + zd(t),
where a,b,c,d are periodic functions that we shall assume to be piecewise continuously differentiable.
The area S(z) of the
corresponding section, given by
xdb  bdx
S(z) = 2J r
Z
is therefore a seconddegree polynomial in z.
From this it fol
lows that (Simpson's Formula) h V =
I
11
S(z)dz = 6 (S1 + S2 + 4 S3). 0
Show that if
EXERCISE 5.95:
then the order of the integration can be inverted in
0
0
217
FUBINI'S THEOREM From this deduce that J0(y)
afm
I
sinax
dx =
0 a2+y2
+x`
1
d y, y,
where JO is the Bessel function:
2
rn/2 cos(ycose)d8.
J0(y) = nJ 0
AVL = VAV = LVA = V1V = OVA
If 0 < E < X < w then
SOLUTION:
Jsinaxdxjf(y)ex3'dy = JW f(y)dyjex3'sinaxdx, 0
E
0
E
since
x r Isinaxf(y)exyldy < (X  E) Je Eylf(y)
dxJ fE
0
dys .
0
The Second Mean Value Theorem gives
r
x exysinaxdx
J
e
2
a
and the inequality Isinaxl
.
ax implies
X
J_XYid
E
Y
Consequently,
X f(y)f exysinaxdx E
4 a Il(0,1)(y)If(y)I + an(11)(y)lf(y)Iy2
CHAPTER 5:
218
The Dominated Convergence Theorem leads to the formula
J sinaxdxJ f(y)exydb = 0 0
J
f(y)dyJ exysinaxdx. 0 0
Now,
exysinaxdx =
a
,
y 2 + a2
0
so
f (y)
of
sinaxdxJf(y)exydy.
dy =
o a2 + y2
0
0
Since J0 is continuous and IJ01 < 1, it is clear that the conMoreover:
ditions of the problem are satisfied if f = J0.
JJ0(y)e XYdY =
Icos(ycos9)d8
feexydy
J0
0
0
n/2 =
W
nd8J exycos(ycose)dy fo
=
2
0
(n/2 x2
0
2

xd8 +
cos28
tan
dt 0 1 + x2 + x212
2x rm n
t
1
W
11
f
`n 1
7,77j=: xt + x t=0
(Switching order of integration is legitimate, since
W
n/2
exyjcos(ycos8)Id8 < x < ").
2J dyJ n
n
1
+ x
FUBINI'S THEOREM EXERCISE 5.96:
219
Let H be a continuously differentiable function
For all r > 0 set
on [0,°[.
m(r) = sup(xlogr  H(x)).
show that if
JoeH'(x)
0 one sets
(x
u(r) = supJ log(rcp(t))dt, x>.0 0
then
I' (r) dr < f0
r2
Notice that u(r) > 0 for all r > 0.
On the other hand,
log(rp(t))dt < J1og+(rq(t))dt, x
I
0
0
CHAPTER 5:
220 so
Jlog+(r(t))dt.
0 < u(r)
0 by using the formula 2n
Jo logll + zeitldt = log+lzl.
AVA = VM0 = t01 = VLV  AVA
SOLUTION:
It may be assumed that meas(X) = 1.
Let p > 0.
Then:
2n
dtJ log 11 + Peitf(x)Idx = 0. f0,
X
If Fubini's Theorem can be applied then 2n 0
dx
X
log 11 + peitf(x)ldt = 21
log+lpfl.
X
JO
From this it follows that IfI < p1 almost everywhere, hence that f = 0 almost everywhere.
It remains to be seen that Fu
bini's Theorem actually can be applied.
JX log+11 + zfl =
By hypothesis
log 11 + zfI JX
for every complex number E.
Therefore
J0dtJ1H1 + Peitf(x)Ildx = 2J I
dtJ lo11 + pitf(x)ldx 0
X
4 4n(1 t p JX If 1) < m
CHAPTER 5:
222
(We have used the inequality log+Il + z1 s 1 + IZ1).
EXERCISE 5.98:
With every function f that is positive on E = 3Rn
associate the set Df C E x3R formed by the points (x,t) such that
0 c t '< f(x). (a):
Show that f is measurable if and only if Df is measur
(b):
Show that if f is measurable and p > 0:
able.
Jf(x)Pdx = pJWtPlmeas(f > t)dt.
E
0
Show that if f is measurable its graph is a set of
(c):
measure zero in E x]R.
000  VAV = 000  VtV = 000
SOLUTION: (a): is also.
If f is measurable, the function p(x,t)= f(x) t
Since Df = (p
0) it follows that Df is
0)fl(t
measurable.
Now assume that Df is measurable.
Then x Ir meas((Df)x)= f(x)
is measurable (Fubini's Theorem; here, for A C E x]R and x e E,
AX denotes the set of is such that (x,t) e A).
SOLUTION: (b):
J fP = E
Fubini's Theorem also shows that
pJEdx(0(x)tpldt = pJOtp1dtJ(flt)dx
M =
pJtP1meas(f > t)dt. 0
SOLUTION: (c);
One can show, as in part (a), that the set D'f
FUBINI'S THEOREM
223
of (x,t)'s such that 0 < t < f is measurable, and the calculation carried out in part (b) proves that
f f = meas(Df) = meas(D'f).
J
E
Consequently Df  D'f, which is the graph of f, has measure zero
EXERCISE 5.99:
Let (Dn) be a sequence of closed discs, contain
ing in the unit disc D, of radii rn > 0, and mutually disjoint.
show that if meas(D  U D ) = 0 then G r = . n n
n
AVO = VAV = AVO = VLV = AVo
SOLUTION:
Let In be the orthogonal projection of Dn on the meas(In) = 2 1 rn, and consequently if E rn
0 for n n.
case one would have D
all n and because the Dn's are mutually disjoint and contained It is then clear that:
in D. k
meas(Lxf1Dn ) < meas(Lxf)D) i=1
(1)
1
for almost every x e]1,1[.
If i
is the characteristic func
tion of D  U D , then (1) means that for almost all x (IxI < 1) n
JP(x)dy > 0,
CHAPTER 5: FUBINI'S THEOREM
224
and consequently:
meas(D  U Dn) = JdxJP(xiy)dy n
> 0.
CHAPTER 6
The LPSpaces
EXERCISE 6.100:
Prove Holder's Inequality by using Jensen's In
equality with the function P(x) = xP, x >. 0, p > 1.
MMA = 0M = MMA = VAT = AVA
SOLUTION:
Since cp is convex, for a positive measurable f and a
positive g with integral equal to one,
(Jfg)p < JfPg
(*)
Now assume that f 3 0, g , 0,.and: jgq JfP

1, =
.where q is such that
1
p
+ 1 = q
.
Replacing f in (*) by fgl q and
g by gq yields:
Jfg' 1. EXERCISE 6.101:
Prove Minkowski's Inequality using Jensen's In
equality and the function 'p(x) = (1  xl'P)P, 0 6 x s 1, p , 1.
225
CHAPTER 6: THE
226
AVO = VAV = AVO = VAV = OVA
The function p is convex, for it is continuous on
SOLUTION:
[0,1], and on ]0,1[ its derivative is equal to
1/PlP1
(1 l
x l/p
)
which is increasing in x.
Consequently, if 0 < f E 1, g 3 0, and
Jg = 1, then
1 < (Jfg)1/P + (J(1  ?/p)pg)1/p Now assume that f
0, g
(*)
0, and
J (f + op = 1. Replacing g in (*) by (f + g)p and f by 0 where f + g = 0 and by fP(f + g)P otherwise, yields
1
. 0, g > 0, and
Jg = 1, then
1 + (Jfg)1/p < (J(1 + fl/P)Pg)l/P.
(*)
If now f >. 0, g > 0, and
JgP
1, =
then replacing g by gp in (*) and f by 0 when g = 0 and by
fPgP
otherwise, yields
1+
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< p41,
CHAPTER 6: THE
228
as can be seen by studying the function
x>xp+1 (x+l)p. From this it follows that if f,g,h e L
d(f,g) = Jif  gIP < J(If  hl +
f in LP.
Since d(Af,0) = Ialpd(f,0), the mapping f H d(f,O) is not a norm
onLpif0 R, and of that defined
by IxI < R, Ix  al > R.
These two sets are mapped into each
CHAPTER 6: THE
230
other by the symmetry x  a  x, and the first is contained in the 'annulus' R
0 denote by LP the set of positive func
tions whose pth powers are integrable.
Show that for all a > 0 the mapping f * fa is a topological isomorphism of LP onto LP/a
AVA = V AV = AVA = VAV = AVA
SOLUTION:
This is a matter of proving that
J If  file >
0
Jjfa  file/a
implies
> 0.
When 0 < a 4 1 we use the inequality
Ixa  ya'I ' Ix  yla,
x > 0, y a 0,
which gives
Jlfa  falPla < JIf  filP.
Now let us consider the case a > 1.
IJ
 Til
If  fiI (f V
fi)a1,
By the Mean Value Theorem,
f V fi = max(f,fi),
and by HUlder's Inequality for the pair a,a(a 
1)1
LPSPACES
f I fa
231
iIp/a

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
SOLUTION: (b):
IIf1IIp1IIf2...ffII,
IIf1IIp1IIf2IIp2...IIffIIPn
E
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 p1
_ rp pr ,
P2
=
rq , gr
P3=r.
Then
1+ 1+ 1= 1+ 1 1= p q r p2 p3 p1
1,
so that if h1,h2,h3 are three positive measurable functions, by part (a) above one has p1
1
Jh1h2h3
(Jhl
)
2
(Jh2
2 )
3
(Jh3
If
h11 = fP,
h22 = gq
that is to say, if
h33
= fpgq,
3 )
.
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
SOLUTION: (c):
fg.
=
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 IIfIIPPIIg!Ig gJdxJlf(t)lplg(x  t)Igdt
IIfIIPp IIgIIq gJlf(t)lpdtJ Ig(x  .t)gldx
LPSPACES
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
J
h()(rdx , IIfIIPIIgIIq,
which accomplishes the proof in this case. If q =  one must have 1/p >, 1, hence p = 1.
mains to examine the cases where
+
It therefore re
= 1, for which r = . By
Holder's Inequality, for all x one p then q has
Ih(x)I , IIfIIpIIgIIgi i.e.
Ilhll , Ilfllpllgllq EXERCISE 6.107: (a):
Let 1 < p s  and p + 1 = 1.
q
Show that if f e LP
IIfIIP = sup{ I Jfgf:Ilgllq. 1}. (b):
Let f be a positive measurable function.
Show that:
CHAPTER 6: THE
236
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 < By Holder's Inequality,
SOLUTION: (a):
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)Ip2
if f(x)
0,
g0(x) _ if f(x) = 0,
to so that fg0 = IfI
g(x) = IIfIIP'
.
If p > 1 one has Ig0iq = IfIP, and then if
g0(x) we have IIgII
= 1 and Jfg = IIfIIP
If p= 1
we have IIg0IIm 1 and Jfg0 = IIf1Il Finally, when p =  letM) such that 0 < meas(E) < 0 < M < IIfiI, and let EC (If!
LPSPACES
237
Set
1 meas E
f x
if x e E,
g(x) _
if x4E.
0 Then IIgIIl = 1 and:
Jfg = meas 1 E JEIf(x)Idx 3 M.
REMARK: In fact we have proved that if 1 4 p
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)
Y X
1/p ,1< J{Jf(x,Y)Pdy}'dx.
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)
0,
and an integer N such that
lflp < e2 p,
JX_B
JIfIp < e2 p
if meas(E) < ,
CHAPTER 6: THE
246
IIffntIp< 111p if n > N. By (1), if n > N and meas(E) < s:
JXB
IfnIp < 6,
JE 'fn'
p 0 numbers such that
XB n
IfnIp N, and consequently
11f  fnIIP < (JXA IfIP)11P + ( JXA Ifnlp)1/P
+ (J
IfI
)1/P +
JE
n
+
(J AE
Ifl)n
If  fn1P)1/P
n
< 5c 1/p which proves that f e Lp and fn , f in LP.
REMARK: If fn e Lp(X), fn  f almost everywhere, and Condition
(ii) is satisfied, one again has f + f in Lp. n
It suffices to
replace the sets En in the proof above by a set E (Z A such that meas(E) < 6 and fn > f uniformly on A  En, which is possible by Egoroff's Theorem.
Let (fn) be a sequence of functions in Lp(X),
EXERCISE 6.114:
1 f in Lp(X). (b):
Show that the conclusion above is still valid if the
convergence almost everywhere of f
n
to f is replaced by conver
gence in measure.
avn  vov  ovo a vov  ovo
CHAPTER 6: THE
248 SOLUTION:
Let e > 0.
(a):
There exists a set B of finite meas
ure and a number 6 > 0 such that
JXB Iflp < 2
< 2
j If IP
if meas(E) < 6.
E
By Egoroff's Theorem there exists a set A C B such that meas(B  A) < 6 and fn + f uniformly on A.
Therefore, by Fatou's
Lemma
f
X
lflp < e t fA Iflp < e t liminffA If Ip.
J
n
X
n
IfnIp = j Iflp, X
this yields
j Iflp
E+ J
X
Iflp  limsupj
X
i.e.
limsupfnIP t C. n
J
Furthermore:
XA
n XA Ifnlp,
249
LPSPACES
If  ffIIP 4
IfIP)11P + (
( 1
XA
LA ffIP)1/P I
+ suplf  fnl.meas(A)1/P, A
whence
l ymsupllf  ffIIP
2e1/p
which proves that fn * f in LP(X).
SOLUTION: (b):
By the preceding exercise it suffices to show
that the sequence (IfnIP) conserves mass and is uniformly inteAs Fatou's Lemma is valid for convergence in measure,
grable.
the argument used in part (a) above shows that
JXA IfPI4E implies that
limsup1 IfnIP n'°° XA
6 e,
which proves that the sequence (If
P) conserves mass.
If this
sequence were not uniformly integrable there would exist a > 0,
a sequence (Ek) of sets of finite measure and some integers n1 < n2
1. Lastly, if f(x) =x, If = 0. Considering the sums of two functions of the above type, any integral in ] 0,oo] can be obtained for if. SOLUTION: (b):
This is a question of proving that if r e If, s e If
CHAPTER 6: THE
256
and
p=n+ (ls A)
0< At 1,
then:
llfllp < llfllrllfllsA
(1) f(1A)p
If r,s are finite, it suffices to make h = fAp g =
in
Holder's Inequality c ( J JX
JX
X
When 0 < r < s = , one has p = r/A > r, whence
llfllp
0.
Show that:
limjjfjjP
=
exp(JXlogIfl ),
where by convention exp(m) = 0.
ovo = VAV  ovo = vov = ovo
SOLUTION: (a):
Let a be such that
+
=
r
s
I. Then a
IIfglIr < IIfIISIIgIIa
(1)
On making g = 1 the desired inequality
(cf., Exercise 6.106).
Since (1) is obtained by applying Holder's Inequal
is obtained.
ity to fr and gr, equality can be had with g = 1 only if f is equal to a constant almost everywhere. By Jensen's Formula,
SOLUTION: (b):
loglIfIl
P
=
1 logJ fp " Jlogf. P
X
X
LPSPACES
259
On the other hand, since logu 6 u  1,
1 logJ fP
1) p1(fp  1) decreases to logf, and on X  A it increases to the same function.
limJ P1 (fp  1) =
' X
Consequently
logf,
I
X
which proves that
lim
logIIfIIp =
Jlogf.
NOTE: As in the preceding exercise, we have assumed that f >. 0.
EXERCISE 6.118:
The notations are the same as in the preceding
exercise (in particular, meas(X) = 1).
Find all the functions 0 on ]0,[ such that
a(l ollflip) = J (f ) for every bounded measurable function f > 0. 00A = V AV _ AVA = 000  MMA
SOLUTION:
Let 0 < c < 1, and let A C X be such that meas(A) = c.
The function f, equal to x > 0 on A and to 1 on X  A, is such
that : l
llfllp = expJXlogf = exp(clogx) = xc
(cf., the preceding exercise).
As (f) = fi(x) on A and ¢(f) =
CHAPTER 6: THE
260
4(1) on X  A, one must therefore have: O(xc) = co(x) + (1  c)O(i) for x > 0 and 0 4 c < 1 (the formula is in fact trivial if c =0 or c = 1).
If we set ct(x) _ iy(logG), where p is defined on at, the
preceding relation becomes
0 4 c 6 1,
ip(cx) = c,y(x) + (1  e)V(0),
x eat.
From this it follows that there exist three constants al,a2,b such that:
ip(x) =alx+b
ifx>.0,
4,(x) =a2x+b
ifxt 0.
Let us then consider u,v, 0 < u < 1 < v such that uv > 1 and B C X such that meas(B) = Z.
For the function f equal to u on B
and tov on X  B
lIIfIIP = exp(ilogu + Zlogv) = u > 1. Consequently:
4(,IIfIIp) = jal(logu + logy) + b. On the other hand,
0(f) = 2(a2logu + b) + "(allogy + b).
J
X From this it follows that al = a2.
Hence 4 must be of the form
O(x) = aloge + b.
Conversely, if 0 is so defined, then
LPSPACES
261
w( mII.fIIp) = alog(limll flip) + b p;G (
= a1 logf + b = X
EXERCISE 6.119:
(alogf + b) 1
X
For any function * increasing on [l,m[, such
that 4(l) > 0 and limi(p)
show that there exists on [0,1] a
p,
measurable function f such that limilfilp =
and. Il,fllp < *(p)
P1for all p. 400 = V AV = AVA = VAV  AVA
SOLUTION:
Let L
be the set of functions f, measurable on [0,1],
such that:
IIfII,y = sup
Ilfll 1f 11
N
IIffII*,
and consequently that
"M 119  SN'I* = 0. N
Assume that Lm = L of generality.
Ilfll
'
W) = 1 can be assumed without any loss
Since
= sup p>.1
Ilf1IP
*'p ¢ Ilfi
the norms 11.11, and 11.11 would be equivalent by virtue of a
theorem of Banach; that is to say, there would exist a constant
M such that for every f e L
Ilfli', < MlIfII*' Now this is absurd, as for every A > max(M,l) there exists p0 such that iy(p) >, A for p > p0; then if
LPSPACES
263
f=AIL
P [O,A
0l
one has
IIfil = A,
if p > p0,
IIfIIP < IIflIW = A F (p)
and:
when l 1 one has
IF(x + h)  F(x) I
=
x+fI
If x
h1 1/P( x+hl fX
1/p. f I p)1/P < eh1 
CHAPTER 6: THE
264
By part (a) above, the function f is uniformly
SOLUTION: (b):
0 as jxl ; W (cf., Exercise
continuous, and consequently f(x) 3.37).
EXERCISE 6.121: meas(X)
A) .
X)f,
then for all p (1 < p < 00)
(1 FP)l/P `
p
X
(1
1
fP)1/P X
A00  V AV  eve  VAV = OVA
One can clearly assume that 1I fP < °D, as otherwise
SOLUTION:
there would be nothing to prove.
that (F
Let Fn = min(F,n); it is clear
> A) C (F > A) and that (F
> A) _ 0 if A
>, n.
From
this it follows (cf., Exercise 5.98) that
( I
Fn  p1(n0 APlmeas(F n > a)da
p(nAp2da1
f(x)dx
4
A)
(min(n,F(x)) ( = pJ f(x)dxl
X
aP2da
=1
0
By Ht5lder's Inequality, this yields, if
t
p
q
f(x)Fn(x)pldx.
X
= 1
LPSPACES
265
fP)1/P((F np1 )q)1/q.
XFp
and using the
Monotonic Convergence Theorem.
EXERCISE 6.122:
Let 1 < p < .
Show that if f is locally inte
grable the following conditions are equivalent:
(i) :
f e Lp; There exists a constant M such that for every sequence
(ii):
P1,...,Pn of disjoint rectangles that are of nonzero measure
n 1
meas(P.)p_111PfIP E M.
(*)
JP.
i=1
If these conditions hold, the smallest constant M that can be
taken in (*) is equal to
II f lip.
ove  vov  ovo = VAV = ovo
SOLUTION:
= 1, then
+
If f e LP and
q p if
fIP 4 meas(P .)p/q( P. 1
1
IfIp,
JP. 1
and consequently, because p/q = p  1
CHAPTER 6: THE
266
meas(P.)p11JP.fIp
i 1
0, 0k eR, and the Ak are disjoint rectangles of Rn;
similarly
ip
v
g= I
PRe
nB ,
R=1
with analogous conditions.
If a > 0 and a < 1 then set
a(z)/a l0kb rk e
fz =
Ak,
k
gz
P(1B(z))/(1B) 1Tp e
k
F(z) = 0
AzAz 1
IlBR
RmT(fz)gz.
Show that F is an entire function, bounded on the strip 0 4 Re(z) E 1, and that for all y eR IF(iy)I 4 1,
IF(1 + iy)I < 1. 2
By considering for e > 0 the functions G (z) = eEz F(z) to which e
CHAPTER 6: THE
268
one can apply the maximum principle, deduce that IF(z)I S 1 if Next examine the case where all  0) = 0).
0 5 Re(z) S 1.
Use the RieszThorin Theorem to prove that if f e Lp, g e Lq,
r
=
1 + 1  1
0, then f and g are convolvable, and IIf*gIlr
IIfIIPpIIgI
q (cf., Exercise 6.106). AVA = DAD = AVA = VAV = 0VA
SOLUTION:
Note first of all that
a(t) = a,
0(t) = 0,
(1)
and that if y eat, Rea(iy) = a0,
Re6(iy) = 00, (2)
Res(1 + iy) = Sl.
Rea(1 + iy) = al,
Also notice that if z belongs to the strip 0 4 Re(z) < 1 then so do a(z),R(z),l  z, and that if a > 0 then min(l,a) 4 IaZI s max(l,a).
F(z) =
Since
Al zAz kLk 1 ' 0
rka (z)/ap(9
1S(z))/(1S)ekk
T(IlAk), 1
it follows that F is an entire function bounded by a constant M for 0 4 Re(z) 6 1.
Hence by (2), upon setting qi = 1/(1
q' = 1/0  0), one has
Ilflyllpo = Ilfl+iyllpl
rk/ameas(Ak) = IIfAIP = 1,
=
k
Ilgiyllgo
= II gl+iyllg1 = R ameas(BQ) = Ilgllq = 1,
that is to say:
LPSPACES
269
Ilfiyllpo = IIf1+iyllpi = Ilgiyllgo = IIg1+iyllgi = 1 (if one of the numbers p0,p1,q
or qi is equal to =, these form
ulae are still true; for example, if p0 =
Ilfiyllm =
then Rea(iy) = 0 and
= 1).
maxlrk(iy)/al
HSlder's Inequality then gives
IF(iy)I
A0IIT(fiy)IIg0llgiyllgo
1
F(1 + iy) I . AiI I T(f1+iy) I I q1 l I g1+iy ll ql . 1.
If e > 0 and G(z) =
IG(x + iy)I =
eez
2 F(z), if 0 0.
Fin
ally, let T be a mapping of E into the set of measurable functions on Y such that
JT(f + g)l < lT(f)l + IT(g)I for any f e E, g e E.
(1)
273
LPSPACES (a):
Assume that p1 < m and that there exist numbers AO > 0,
Al > 0 such that for all f e E and all A > 0
AiIIfIIP Pi meas(ITfI > A)
t),meas(IfAI > t)).
(b):
Show that (3) still holds when p1 = W, if the inequal
ity in (2) for i = 1 is replaced by
f e E.
IITfll0 < (c):
(21)
Assume in addition that meas(Y) < W and that if f e E,
A > 0, and
fi(x) _
f(x) if I f(x)I 0
then fA e E.
A,
otherwise,
Set f = f
Show that if 1 = p0 < p1 t m and (2) is satisfied (and (2') if p1 = co), then if C < p < 1 there exists a constant Ap such that
IITfIIp . APIIfII1,
f e E.
(4)
(Evaluate the integrals:
J0pxPlmeas(TfI > A)da,
JP_lmeas(ITfI > A)da,
a
274
CHAPTER 6: THE
then make a = (d):
11fII1)
The hypotheses are as for Part (c), and assume further
than meas(X) < oo.
Show that there exist constants B,C such that
I I TfI I14 B + CJX IfIlog+lfI
(Use the decomposition f =
VT
+
(5)
, then evaluate the inte
grals: 1
rm
meas(ITfI > A)da, f0
meas(ITfI > A)da. f
1
Also note that there exists a constant M such that u 6 M(1 + ulog+u) for all real u).
NOTE: Formula (3) is a weakened form of a more general theorem of Marcinkiewicz (cf., for example, R.E. Edwards: Fourier Series, Vol. II, (Holt, Rinehart and Winston, Inc.), pp. 157 et seq.).
/VA = VAV = AVA = vtv = A4t
SOLUTION: (a):
Let
p(a) = meas(IfI > A),
Since JTfI 5 ITf
W) = meas(ITfI > A).
+ ITffI by (1), it is clear that
(ITfl > A)C (IT?i > 2A)U(ITfXI > jX), and consequently:
275
LPSPACES
W) 2A) + meas(ITfAI > IX)
(6)
Furthermore, by (2)
2A0IIfAIIP
PO O
meas(IT? I >
,
a
P1
(7)
111faIIP meas(ITfXI >
1
A)
From (6),(7) and Exercise 5.98 it follows that
IITfIIP = 0
p(2A0)PO0aPPO
dal If(x)IPOdx X
(W
1
+ p(2A1)Pll 1P P1
daj IfA(x)IPIdx.
X
0
(IfI > t) if 0 < t < 1,
if t
0
a,
and that If"(x)I > t > 0 if and only if If(x)I > A, and If(x)I(1AIf(x)I1)>t,
or in other words
(III>t)_(IfI>t+A).
(8)
276
CHAPTER 6: THE
Thus
meas(I? I > t) = q(t + A),
meas(IfA I
> t)

(p(t)
if0 i and t  A 5 t, JoAppO1d)LJot1) 01c(t
+ A)dt 5
t
1 =
J
WtPO
dA
Jo
0
=
1
Ap PO
q,(t)dt
1
JtP_1(t)dt
pp00 1
=p(pP 0 )
IIfIIP.
Similarly, because p  p1  1 < 1, pl1,(t)dtJtApplidA
APpl1dA(otPl1q(t)dt
J
= 0
J t
=
LPSPACES
277
1
tP1cp(t)dt
pJO
pl
p plZ
l )llfllP.
(12)
(10),(11) and (12) yield
Pi
p0
IITfIIP .
P0(2A 0) I
pp,lp ) IIflIP.
+
P  P0
1
which proves (3). SOLUTION:
(b):
When p1 =
' and (2') holds, write
A/2A
meas(ITfI > A)
mI > )L) + meas(ITfa/2AWI
,c meas(ITf
> #L)'
Now the set (ITf1AAml > A/2) has measure zero, since almost every
Tfa/2A
1 < A.
2..
11fa/2A_I1 .
From this it follows that
'A)dX
IITfIIP : 0
W ppO1 pOp(2AO) P0 J a 0
m p01 A/2A. meas(If dal t
I
> t)dt
0
p0p(2A0)poJXppO1 dXJ0tP0
=
O 0
l
o
+
( P
2Am)dt
Ilt
(Contd)
CHAPTER 6: THE
278
pp01
p0 , pop(2A0) 1
(Contd)
o
p0(°° p01
= pop(24 )
t
J
p0 1
Q(t)dt
t dxJW A/2AW
A
(2A t pp01
c(t)dtJ
A
dA
0
0
_
p

po
p0(2A0)P0(2Am)PP0
HA P.
p0
P
SOLUTION: (c):
0
P
It can be assumed that a = IIfIIl > 0.
also be assumed that 1 = p0 < pl < ; indeed, if p1 if p0 < pi
A) S lr xPl
ITfI ITfI>A
IITfIIP1
 S P1
Pl
APl
P;
a
This being so,
a J0PAP1AdA
1
meas(Y)J padA = ameas(Y). 0
On the other hand
Vi(a) 4 meas(IT? I
> 2A) + meas(IT."
I
>
'fA) 5
(13)
279
LPSPACES
2A0I
II
1
2A
1 Il a II
Pi
Pi
whence:
ja
2pA0JaXp2dXjIf(x)I>aIf(x)Idx
pAP1iP(A)da
a
Pi °° PP1 1 + p(2A1)
P1 dA J
1 A
If(x) I
dx
a
1
In the last integral If(x)Ipi 4
JpAPhlp(A)dX
< 2pA0 J
a
+
If(x)I, and consequently
API
If(x)Idxj X a,T
1
If(x)Ipldx.
+ (2A )P1 l
da 1 xp1fIf(x)I5y1x(P1 1)/2
If(x)IP1
In the last integral
JP(A)dA : 2A 1 If(x)IdxJ 1
If(x)I, whence
5 A
X
a1da 1 0.
Let d > 0
By Egoroff's theorem there exists a
set A of X such that
meas(X  A) < 6
and
En = suplanfn(x)l > 0. A
Then lan12 = J
lanfnl2
X
= fA
lanfnl2 +
fnl2 s
JX_A'
(Contd)
289
.< en + M21an126,
2
lanl
En
2
2
1Md
which proves that an  0. SOLUTION (b): I
b
Choose a sequence (bn) such that bn + 0 and
(for example, bn = 1//n).
almost everywhere.
Assume that E Ibnfn(x)I2
n0
Ib f (x)I2 < 1 n n
for all x e A whenever n0 is large enough (Egoroff's Theorem). Now, for every n.
1=JAIfnl2+JXAIfn 12 < JAIfn12 + M26,
so
1 3 meas(A) >,
which is absurd.
X n>,n0
lb
12J
n
lbn1i2, E A Ifn 12 % (1  M26) n,n0
290
CHAPTER 7:
EXERCISE 7.129:
Let U be an open set of ]Rp and H = L2(U).
Re
call that the Gram determinant associated with elements f1,..., fn of the Hilbert space L2(U) is defined by
G(f1,...,fn) = detll(.fil.fj)II1, m, and therefore that snpIsn(x)l <  almost everywhere.
SOLUTION (I):(b):
Here again it may be assumed that the an's are
Let n1 < n2
nk
Set n0 = 0, wn = l if 0 1.
Then 2
2
w a = G a2 + k2 a2 nk4n (ii). Now assume that (ii) is satisfied.
f1(x) =
g(t)dt,
Let a < x0 < b, and
a < x < b.
Jx0
By what has just been proved,
rb
rb
a
a
I fl9'+J gro=0, f
9 e D(T).
Comparing this with (ii) yields
rb I
(f  f1)p' = 0,
'p
e DP(I).
JJJa
When 9 runs over such that Ji = 0.
9' runs over the set of functions *e D(I)
By exercise 8.141 there exists a constant C
such that f  f1 = C almost everywhere, which proves (ii) _> (i). Lastly, if the pairs (f,g),(f,gl) satisfy (ii), by substraction one obtains
CHAPTER 8: CONVOLUTION PRODUCTS
344
b
9 e D(I),
(g  g1)T = 0, a
whence g = g1 almost everywhere.
The Formula defines a positive alternating form
SOLUTION (b):
on H2(I), and (fIf) = 0 implies that
2(I , whence f = 0.
1IfII
L
It
)
2
remains to prove that H (I) is complete in the norm associated 2
with this scalar product.
If (fi) is a Cauchy sequence in H (I),
(f ) and (Df ) are Cauchy sequences in L2(I).
Hence there exist
f,g e L2(I) such that fi  f, Dfi  g in L2(I). If p e DP(I), then b
b
Dfi.W = 0,
Jafi.cp' +
fa
and on passing to the limit: b
rb
fp' + J
a
gcp = 0. 1
a
Consequently g = Df and fi  f in H2(I).
SOLUTION (c):
If cpe D00(JR), then by Plancherel's Formula and the
relation ,'(y) = iyV(y),
+0
_ JP?(Y)TTdY
r+0 = 1 _ Df(x)Zxsdx
+W
Jf(x)c'(x)dx
=
J](Y)iy(y)d J
=
J00Y(Y)EYYdY. _
As the y e DW(G) are dense in L2(It), and as the Fourier transform
ation is an isometry of L2(R) onto itself, the $ are dense in
AND FOURIER TRANSFORMS
345
Hence bf(y) = iyf(y) almost everywhere.
L2OR).
First Proof: Note that if A =
SOLUTION (d) :
IIDf II 2
L (I)
and
a < x < y, then
4 A.
If (y )  f(x) I
If f(x) were not to tend to zero as x >
there would exist E > 0
and a sequence of points xn such that 2
xn+l > xII
+
E
2 ,
If(xn)I 3 E.
4A
Consequently, for xn < x < xn + E2/4A2
I f(x) I a
I f(xn) I If(X)  f(xn) I
E A 2A =
2
But then
x +c 2
W
J If(x)I2dx > I j n
a
n xn
4A
2 E
If(x)I2dx
2
2 E
= W.
n 4 4A2
Second Proof: We can reduce to the case where I =]R by the folartifice: if a > m, f is extended by continuity at a (cf.
(i)
of part (a)); let a < a, and let S be a continuous function with support contained in [a,a] and such that
a 0(t)dt = f(a). fa
By extending Df to [,a] by setting Df(t) = 8(t) for t E a, E±nC setting
x f(x) = f(a) + J Df(t)dt, a
x e ]R,
CHAPTER 8: CONVOLUTION PRODUCTS
346
an extension of f is obtained that belongs to H2(]R).
Now, for
f eH2(JR) the functions ?(y) and iy7(y) belong to L2(B) by part (c).
From this it follows that (1 + y2)27(y) e L2(]R);
(1 + y2) 2 e L2(kt) we have
since
But then f is the inverse
'e L1(R).
Fourier transform of ?, and therefore tends to zero at infinity.
SOLUTION (e):
Let us show that f + f(c) is a continuous linear
To do that, write
functional on H2(I).
(c
f(c) = f(x) + 1 Df(t)dt,
a 4 x < b,
x so ,
If(c)I 4 (b  a)2IIDfII
2
L (I)
+ If(x)I,
or
If(c)I2 4 2(b  a) IIDfII22 L (I)
21f(x)12.
+
(*)
Integrating with respect to x from a to b yields
I f(c)
12
, 2(b  a)
112
112
+
ba
L2(I)'
which proves our assertion.
The existence and uniqueness of i
c
results from this.
In par
ticular, if W e V (I),
c
b
cp'(x)dx = a
b
J(p(x)*c(x)dx + a
a
which can again be written, by introducing the characteristic function Xc of the interval [a,c],
AND FOURIER TRANSFORMS
347
b
b
Jaq,'(x){Dhc  Xc(x)}dx + JaV(x)*c(x)dx = 0.
Comparison with (ii) of part (a) shows that this implies
DOC  Xc e H2(I)
Since *
c
and
D(D*c  Xc) _ *C.
is continuous Dpc  Xc is continuously differentiable
and its derivative is * c.
(1)
c
In other words,
is continuous;
(2)
p" = i on [a,b]  {c};
(3)
y' (c  0)  *C, (c + 0) = 1.
Furthermore, if a < y < c and if f(x)
x)+, then Df = X
,
and consequently,
0 = aJ(Y  x)*c(x)dx  j
Y
 x)*(x)dx.
c(Y)  c(a) = Ja
If M is the maximum of I'Pci on I, it follows from this that Y
IVC(Y)  c(a)I < MJ (Y  x)dx < 'M(Y  a)2, a
and consequently *'(a) = 0.
Proceeding analogously for the other
end point, to conditions (1),(2),(3) can be added
(4)
*c(a) = *c(b) = 0
CHAPTER 8: CONVOLUTION PRODUCTS
348
(this is in the case where a < c < b).
From (2) and (4) one ob
tains
Acosh(x  a),
a < x < c,
Bcosh(b  x),
c < x C b.
Conditions (1) and (2) imply the system Acosh(c  a)  Bcosh(b  c) = 0, Asinh(c  a) + Bsinh(b  c) = 1. From this it is not difficult to deduce that
c (x)
cosh(b  c)cosh(x  a) sinh(b  a)
a 4 x 4 c,
cosh(c  a)cosh(b  x) sinh(b  a
c<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).
Ho(I) is the orthogonal complement of {
a''Pb}.
Furthermore,
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
1 f(x)p(x)dx +
a
rb I
Df(x)p'(x)dx = 0,
T e 'O(I),
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 orthogonal complement of D°°(I) is thus twodimensional; because it contains ya and 'Pb the result is proved.
AND FOURIER TRANSFORMS The mapping f 
349
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 JJ
2 LI)
is, on
H2(I), a norm equivalent to that induced by the norm of H2(I).
The existence (and uniqueness) of 8c follows from this. The orthogonal projection of Xc onto the orthogonal of 1 in L2(I) is:
bc b Xc(x)  b l a JXc(t)dt = a
ba
a4x4c,
c  a
and consequently
(b  c)(x  a) 
a)
(c  a)(b  x)
a REMARK:
a
4x4c ,
c..<x ., n, fn(x) = 0 otherwise.
Show that f*G is defined everywhere and that there exists a continuous function hn such that
f*G = hn + fn*G.
Conclude from this that f*G is continuous.
AVA = vov = AVO = VAV = AVo
SOLUTION (a):
JIfI
= J
Noting that log3 , 1 gives
IfI2IfI
< 4n +
Jlfllog(l + Ifl ),
Furthermore, if a,b , 0 then
ab < alog(1 + a) + eb  1.
(YOUNG'S INEQUALITY)
Consequently, if A > 1
J If(x)g(y  x)ldx < xJIfIlog(1 + AI.fi) +
JceA'I
Now log(1 + alfl) < loga + log(1 + IfI), so
If*gI < AlogAJIfl + AJIfIlog(1 + Ifl) + J(eA1Igl  1).
SOLUTION (b):
G(x) ti log
If x0 is a root of cosx, then as x . x0
Ix ix01
AND FOURIER TRANSFORMS
353
and consequently eA1G(x) ti
Ix
 x
0I1/A
This shows that G e L1 and exp(A1G)e L1 if A > 1.
The function
fxg is therefore defined everywhere, as is fn*G, moreover, since fn also satisfies (*).
As for hn = (f  fn)*G, this is also a
function defined everywhere, since f  fn is bounded. cisely, hn is continuous, as LW *L1 C C.
More pre
For A > 1 and every
integer n
IIf*G  hnil. 1
< AlogAJlfnl
Since Ifl l
>.
+ Xflfnllog(l
+ Ifnl) + J(ea_
G
 1).
>.  and I fnI  0, this yields
l f21
1
limsupllf.G  hnll 4 J(eXG nMaking A } W shows that hn > f*G uniformly, and therefore that f*G is continuous.
EXERCISE 8.149:
Prove that the algebra L1(Rn) does not have a
unit element.
AVA = V1V = AVM = VtV = OVo
FIRST SOLUTION:
If f e L1 were such that f*g = g for all g e L1
then taking the Fourier transform gives
=
Now, for all
x e32n there exists g e L1 such that g(x) + 0, and consequently
?,(x') = 1, which is impossible, because f must tend to zero at infinity.
CHAPTER 8: CONVOLUTION PRODUCTS
354
There would exist a > 0 such that
SECOND SOLUTION:
JIf(x)Idx 1
1.
Let p be the characteristic function of the ball with centre zero
and radius a/2.
f* (P (x) =
Then if IxI 4 a/2 one would have
f(x  y)dy .
J
If(y)Idb < 1,
J
Iy I < a
IHI < a
a contradiction.
EXERCISE 8.150:
Prove that the algebra L1(,Rn) possesses divisors
of zero.
AVO = vov = ove = VAV = ovo
SOLUTION:
First Proof: Let V1 and V2 be two nonempty and dis
joint bounded open sets of Mn.
Let p1,(P2 be nonzero indefinite
ly differentiable functions compactly supported in V1,V2 respect
If fl = F(wl), f2 = F(cp2), then fl eLl(R' ), f2 eL'(Itn),
ively.
fl 4 0, f2 4 0, and f1*f2 = 0, since F(f1''f2) _ 9192 = 0.
Second Proof: We shall assume that n = 1.
If a e]R, denote by Ta
the operator of translation by a, that is to say, that Taf(x) = f(x  a).
It is known that Ta is an operator on L1 with norm 1,
and that
TaOTb = Ta+b'
Ta(f*g) = Taf*g.
For every summable sequence a = (an)nex' +M T
a
=
I n=m
an T
nn
(1)
AND FOURIER TRANSFORMS
355
is a continuous operator of L1 (the series (1) converges normally in L(L1,L1)).
If
n)ne2Z
Tao Ts = Ty,
is another summable sequence, then
apq.
Yn =
(2)
p+Qn Now assume that we have determined a,s so that
TOT0 = 0,
a 4 0,
S 4 0.
(3)
If m e L1, then (TQ*Ta9) = (ToT$)((p*c) = 0.
It will be possible to ensure that Tag # 0, T9 # 0 by taking for 9 a function zero outside ]0,1[ and such that 11N111 > 0; then the functions Tncp will have disjoint supports, so that, for example,
11Tam 111 =
11m 111
G 1%1 > 0. n
In order to determine a,s so that (3) is satisfied, note that if
u(x) _
anell"{,
v(x)
Snel n
n then:
uv = 0,
(4)
and conversely (4) implies (3).
Consider the function Isinxl;
its Fourier coefficients are
en =
n
(n 1,R1
_ e ice n
It can be shown that
l sinx ldx
=
nJ cosnxsinxdx. 0
CHAPTER 8: CONVOLUTION PRODUCTS
356
c2n+1
2
C2nn
0, =
1
The function Isinxl being continuous and piecewise continuously differentiable, the elementary theory of Fourier series ensures that +00
cneinx
I
Isinxl =
n=m But then
(sinx)+
4
=
+co
(eix  eix) +
cneinx
n=(5)
 (sinx)_ = 42
+m
(e ix  eix) 
C
cneinx
z
n=Gw
so that the sequences a and S defined by
a1
= s1
1
=1,
a1 = 01
4 , 1
a2n
 s2n
1
n
1  4n2
a2n+1 = 82n+1 = 0 satisfy (3).
if n + 0 and 1,
One can, if one wants, obtain real sequences on re
placing x by x + 1/2 in (5), which gives a1=S1=a1=S1,
a2 n
  stn
1 n
(
 1)n
1  4n2
a2n+1 = 02n+1 =
0
if n4 0 and 1
AND FOURIER TRANSFORMS EXERCISE 8.151:
357
Let L+ be the set of functions locally integrable
on ]R and zero on ],0[.
Two functions of L+ that are equal al
most everywhere are identified. (a):
Show that if f,g a L+, then they are convolvable, and
fiegeL+. (b):
Show that if f e L+ and if f*f = 0 almost everywhere on
[0,2a], a > 0, then for every integer n >,
a enxf(a

x)dxI2 4
1
If(a  u)f(a  v)Idudv. JJ
a
u>a v>a u+v>O
Using Exercise 3.72, deduce from this that f = 0 almost everywhere on [O,a]. (c):
Show that if f,g e L+ and f*g = 0, then, setting f1(x)
= xf(x), g1(x) = xg(x),
f*gl + f1*g = 0, and consequently f*g1 = 0.
(d):
Conclude from the preceding that the algebra L+ does
not possess divisors of zero (TITCAMARSH'S THEOREM). tVt = VAV ° AVA s VtV = AV1
SOLUTION (a):
Let M > 0 and let fM(x) = f(x) if x < M, fM(x) = 0
if x > M, gM being define analogously. Then if x < M,
x
tom
J
WIf(y)g(x
If(y)g(x  y)Idy =
 y)Idy = j 0
(Contd)
358
CHAPTER 8: CONVOLUTION PRODUCTS
(Contd)
= Jx
fM(y)gM(x  y) dy
0
= f
fM(y)gM(x  y) dy.
Since fM,gM are integrable, it follows first that (f?,g)(x) is de
fined for almost all x < M, and consequently for almost all the
x eat, and then that x f(y)g(x  y)dy.
(f*g)(x) = 0
This shows that (fig)(x) = 0 if x < 0.
Furthermore, for almost
all x e [0,M] (frcg)(x) = (fM*gM)(x),
which proves that f*g is locally integrable, and hence in L.
SOLUTION (b):
a
fJ en(u+v)f(a
enxf(a  x)dx)2 =
(
a
 u)f(a  v)dudv
(ul'a IV1,a
v>,a u+v50
+
ff u., 0.
Let
be a sequence of compactly supported continuous positive functions
i
such that f + f in Lp.
Then F. ' F in LP, and so by exercises 1
6.105 FP1 + FPtin Lp/(p1) = Lq.
Replacing F by Fi in (1) and
passing to the limit, shows that this formula is still valid for f.
As (2) results from Holder's Inequality applied to the right
AND FOURIER TRANSFORMS
365
side of (1), equality can hold in (2) only if it holds in this Holder inequality, that is to say, if there exists a constant A >, 0 such that
FP = F(p1)q = A/p, Let us note that if
that is to say F = Bf, B >, 0 a constant.
f $ 0, we necessarily have F 4 0, whence B > 0.
Since F is to
be continuous f would be also, and by differentiation one would obtain
Bxf'(x) = (1  B)f(x), that is to say:
C > 0,
f(x) = Cxa,
a =
1  B B '
This is absurd, because whatever a may be such a function does not belong to Lp(O,m). If f e LP(O,W) without being positive, but if f 4 0 almost
everywhere, then setting
G(x) = xf
lf(t)Idt 0
yields
]IF IIp < IIGIIp
, 8n(x).
In particular On < $n+1' whence an < anti.
Now note that
f(x)dx > 0,
a1 >. (ffcq) 1)(0) >1
I r . I1  11 n ` nJJa n n n+l an+l
By (**) and part (a), for all e > 0 there exists an integer n0 such that
(1 it
a n
0 >1E.
n0 an0+1 Setting 9 = hn
one has therefore determined for every p large
enough and all c > 0 a function p e D", c 3 0, such that Ilf*c L = 1 and
380
CHAPTER 8: CONVOLUTION PRODUCTS 1  e .
0.
Prove that
infJIA1f(x  a1) + ... + Anf(x  an)Idx,
where the infimum is taken over all the systems of elements a1, ...,an e]R
and positive numbers ai,...,A
such that Al + tan = 1.
(Use the result proved in the preceding exercise).
AvA = vAv  AvA = vov = AvA
SOLUTION:
First let us fix some notations.
As usual set f(x)=
f(x  a); in addition, if f e L1 we shall set
I(f) = Jf(x)dx.
Let P be the set of positive functions on]Rp that are zero except at a finite number of points and are such that:
E
A(a)=1.
aeatp If A e P, set:
(A*f)(x) =
I
adRp
A(a)f(x  a).
AND FOURIER TRANSFORMS
381
There is no difficulty in verifying that if al,a2 e P then
Al*(A2*f)
=
(al*A2)*.f,
where
(A1*A2)(a) =
I
1 Aba2(a  b)
bdRP (the reader well versed in measure theory will recognise the notion of the convolution product of f with discrete probability measures).
If A e P, f e L1, it is clear that
Ila*fIll
0 there exist X1,a2 e P such that
Ila1*fll1 < p(f) + I1a2*9111
'c p(9) +
so that p(f + g) < II(xl*x2)*(f + g)111 11X2*(a1*f) II1 + llal*(a2*9) 111 11X1*fI11 + Ila2*9111 p(f) + p(g) + 2e.
Now assume that one had proved that f e L1, f real, and I(f)= 0 implies p(f) = 0.
Then if f e L1 is real and if cp >, 0, I((p) = 1,
one would have II(f)I s p(f) 4 p(f  I(f)T) + II(f)l = II(f)I, which would prove (1) in this case. observe that there exists a e Q,
To pass to the general case
lad = 1, such that II(f)l = I(af),
if of = u + iv, where u,v are real, then p(u) = lI(u)I = 1(u),
p(v) = II(V) I = 0,
AND FOURIER TRANSFORMS
383
and consequently:
p(f) > II(f)I = I(u) = p(u)
>. p(u + iv) = p(af) = p(f).
Therefore let us assume that f e L1, f real, and I(f) = 0.
Note that p(f) is the distance in L1 from zero to the closed convex envelope of the fa's, a e ]R
.
If one had p(f) > 0 the Hahn
Banach Theorem would assure us of the existence of a number a > 0 and of a nonzero continuous linear form on L
bounded from be
low by a on this convex envelope, and in particular on the set of In other words there would exist g e L , g nonzero almost
fa's.
everywhere, and such that
Jf(x  a)g(x)dx > a > for all a e]R
(2)
Since I(f) = 0 the left side of (2) is not changed
.
by adding a constant to g, which allows us to assume that g > 0. Let h(x) = g(x); then h > 0, h e Lm, and (2) can be written (3)
Let M =
IIhII..
Note that by (3) M > 0.
(f  M q,)*h > a  M M = 0. By the preceding exercise one would have
I(fM9)=I(f)M> 0, which is absurd.
Let p be as above.
Then
384
CHAPTER 8: CONVOLUTION PRODUCTS
EXERCISE 8.159:
is x
b
f
If f is integrable on [a,b], and if
n dx = 0
f(x)e a
for a sequence (an) of complex numbers having at least one finite limit point, then f = 0 almost everywhere on [a,b].
AVn = VAV = AVA = VAV = evn
SOLUTION:
For all z e a: let
b
F(z) = 1 f(x)eizxd a
This defines an entire function which vanishes at each a the latter have a finite limit point, then F(z) = 0.
.
If
n In partic
ular the Fourier transform of the function equal to f on [a,b] and zero elsewhere is zero.
Consequently f = 0 almost everywhere
on [a,b]. EXERCISE 8.160: (a):
Let (an) be a sequence of real numbers such
that
lime
n9
is x n
exists for all x's belonging to a measurable set A of 3R with meas(A) > 0.
Show that the sequence (an) is convergent. (b):
Let (cn) be a sequence of complex numbers and (an) a
sequence of real nubmers such that
lime e
n n
is x n
AND FOURIER TRANSFORMS
385
exists for all the x's belonging to a measurable set A of ]R with meas(A) > 0.
Show that either
n,m n
lime
= 0,
lime
= c $ 0
or
n. ,
n
lima
and
n._,a, n
= a.
ADA = DAD = ADA = DAD = ADA
SOLUTION
(a) :
The set of the x eat for which lime
is nx
exists is
an additive subgroup that is measurable and of strictly positive measure.
From this it follows that
g(x) = lime n),
is x n
(*)
exists for all x eat (cf. exercise 8.138).
Then for every inte
grable function f on at, +W
f_
( +Q
is x
f(x)g(x)dx = limJ f(x)e n dx nt°°
m
(Dominated Convergence Theorem).
E
Assume that for a subsequence
By the RiemannLebesgue lemma,
(On) of (an) IsnI >
0
for every integrable function f, and consequently g(x) = 0 for
almost all x, which is absurd because Ig(x)I = 1 for all x. sequence (an) is therefore bounded. points of the sequence.
The
Let a,8 be two accumulation
By considering subsequences of (an) con
386
CHAPTER 8: CONVOLUTION PRODUCTS
verging towards a and B, one deduces from (*) that eiax
eiBx

This proves that the
for all x, whence by differentiation a = B. sequence (an) is convergent.
Since
SOLUTION (b): is x I
icne
n
= Icn1,
we have that IimIc
exists.
I
= p
Ti
n),w
Assume that p > 0.
Then from some point on, cn 4 0.
is (xy) is therefore convergent. x,y e A, the sequence e n
If
Because
meas(A  A) > 0 it follows from part (a) above that a = lima exn).w n
is x
is x
ists.
Finally, if x e A the sequences ene
vergent, which proves that lime
n).w n
EXERCISE 8.161:(a):
f+(x) =
Let f e L1(T), and
suph1fo If(x
+ u)Jdu,
h>o
f(x)
exists.
= max(f+(x),f (x)).
Show that for all A > 0,
n
and e
n
are con
AND FOURIER TRANSFORMS
387
meas{ (f+ > A) n [n, n] }
A for a z such that
IfI
x < z <x+2n}
x z < meas(x:n < x < 3n,
By introducing the
z
1 xJ IfI > A for a z such that
x
x < z < 3n}.
continuous function
x F(x) = J lf(u)ldu  Ax 0
the last inequality may be written p(a) < meas(V), where
V = {x:n < x < 3n,F(z) > F(x) for a z such that x < z < 30 By the "Setting Sun Lemma" V is the disjoint union of a sequence of intervals (a.,b.) such that F(a.) < F(b.), and consequently
b lfI i
J
n
f
AND.FOURIER TRANSFORMS
389
which accomplishes the proof.
Proceeding similarly for f and observing that
(f° > A) _ (f+ > A)U(f > A), one obtains
meas{(f> a)fl< !
11f111}
Let us note that: 10h
?(x)
=
suplh1
lf(x + u)ldul,
so
Furthermore, if f e L(T) then
Ilf 11. S IIfll.. By Marcinkiewicz's Theorem (cf. exercise 6.124) for all p, 1 < p 4 m, there exists a constant AP such that
Ilf Ilp < APIIf11P,
f eLW(T).
The relation above is still valid if f e Lp(T).
In fact, if 9i
is an increasing sequence of simple functions that tends to Ifl at every point, then for all x and all h > 0
h
h
suph 11 gi(x + u)du = h11 If(x + u)ldu, i
so
o
0
390
CHAPTER 8: CONVOLUTION PRODUCTS
supgi(x) = f(x). 2
From this it is deduced that
supgi() = ?(X), i
and therefore
Ilf lip = Sup Ilg" IIp < Apsup 1191 lip = AP Ilf IIp
i
2
(Actually the argument above is valid for every measurable function with period 2%).
In particular, fA e Lp if f e Lp, 1 < p 4 .
Using the last part of Exercise 6.124, one can show similarly that there exist constants A,B,Ap (0 < p < 1) such that
II? ll t A + Bj, Ifi log+Ifl n
0 < p < 1.
Ilf°Ilp , Apllf111,
SOLUTION (b):
As each function Hn is bounded, so is each Kn, and
the convolution product K *f is therefore defined everywhere, n moreover
((n
I(Kn*f)(x)l < ?nJ Hn(y)lf(x  y)ldy. E
Writing
n(y) = n(n)  Jll'(u)du y
(1)
AND FOURIER TRANSFORMS
391
and changing the orders of integration, this yields x
7c
JE(Y)If(x
 y)Idy = Hn(n)JOIf(x  y)Idy
(n
u
 J H'(u)duJ If(x  y)Idy. 0n
0
Foru>.0 u If(x  y)Idy 6 of (x),
J
0
so
JH(Y)If(x  y)Idy 6 00 (n) + JQIuH'(u)Idu)f (x).
Similarly, 0
_x n
f(x  y)Idy
(iH(n) + n
(JH(y)I
so that by (iii) and (1)
I(Kn*f)(x)I < '1(Hn(n) + Hn(n) + 2B)?(x).
Furthermore, the relation:
Y n )du + Jb'uIc(u)du yH (y) = fo H(u
x
n
s J:Hfl(U)dU + f0juU1(u)jdu,
(2)
CHAPTER 8: CONVOLUTION PRODUCTS
392 0
r0
n n(  n) 4 I n
H m
du + I
JuH'(u)ldu,
_n
m
and consequently, by (ii) and (iii): J( n(n) + Hn(n)) < A + B.
Returning to (2) one obtains
I(K *f)(x)l 4 (A + 2B)fA(x).
n
One can then deduce for the operator f + K*f the same properties
as for f
f".
EXERCISE 8.162:
In what follows set z = x + iy, where x eat and
y > 0, and
P(x,y) = n
2
x
1
y f
Q(x,y) =
2
f
K(x,y) = P(x,y) + iQ(x,y) = nz
(a):
Show that the functions
form, as y + 0, an ap
proximate identity in L1(gt).
(b) :
If f e LP(It), 1 < p < m, set:
Pf(z) =
JP(x
 t,y)f(t)dt.
Determine limPf(x + iy) when f is the characteristic funcy;0 tion of an interval [a,b]. (c):
Drawing inspiration from the preceding exercise, show
AND FOURIER TRANSFORMS
393
that if f e L1(R) and A > 0, then
meas{x:supIPf(x + iy)l > Al c 2a1iIfII
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
halfplane II
= {z:Im(z) > 0}.
Show that there exists g e L %R) such that g = P.
(Use the
Lm
fact that from every bounded sequence of
one can extract a sub
sequence which converges weakly in this space).
(e):
When f e Lp(Ik), 1 6 p < , set:
Kf(z) =
Qf(z) =
JK(x
 t,y)f(t)dt,
+ 1mQ(x  t,y)f(t)dt.
Show that Kf is holomorphic in the halfplane H.
Next show
(Reduce to the case that liiKf(x + iy) exists for almost all x. y*o where f < 0, set g(z) = expKf(z), and use parts (c) and (d) above.
Deduce from this that Hf(x) = limQf(x + iy). y>0
exists for almost all x. f)
(Hf is called the HILBERT TRANSFORM of
CHAPTER 8: CONVOLUTION PRODUCTS
394 (f):
Calculate the Fourier transform Q0,y) of Q(,y).
From this deduce that 1IHf 112 (g) :
If 112 when f e L2 (M).
=
Let f e L1Ot), f 3 0. Show that for all e > 0 and all
y > 0
logjl t pKf(t + ie)I
fm
dt = logll + tiKf(z + ie)I.
(x  t)2 + y
Deduce from this that
JlogIQfct
+ ie)Idt < uIIfIll.
Next show that for every f e L1cR)
meas{t:IQf(t + ie)I > a} < ae IIfII'. (h):
Using parts (f) and (g) and Marcinkiewicz's Theorem
(cf. exercise 6.124) show that if 1 < p < 2 there exists a con
stant Ap such that (IHfIIp < ApIIfIIp
for all feLp(gt).
Extend this result to the case where 2 < p
0 and
P(x,y)dx = 1 a
SOLUTION (b) :
(tan1b
 tan1 a ).
Note that P( ,y) a Lq(,t) for 1 4 q
. 0, and that
SOLUTION (c):
I uP'(u,y)du = 2( 0 X 0
u2du
(u2 + 1)2
_
1
2
.
Using the notations of the preceding exercise, this yields
JOP(t,y)lf(x ± t)ldt = JIf(x ± t)IdtJtPX(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) < A1IIf ilI,
CHAPTER 8: CONVOLUTION PRODUCTS
396
from which the desired inequality follows.
This being so, let
(Tk) be a sequence of step functions such that
f = I
Tk
OR4)
and
II(Pk111 =
in LThere exists a set N1 of measure zero such
that
f(x) =
x4N1.
(pk(x),
(2)
k Furthermore, for all z e R,
Pf(z) = I Prpk(z).
(3)
k Finally, by part (b) there exists a set N2 of measure zero, such that 1imPpk(x + iy) _ (P k(x) y>o
for all k and all x 4N
(4)
2'
Since
OR2)
I meas{x:suplPcpk(x + iy)I > k2} = I
y>o
k
o starting from some k
when x 4N
3
0
(which may depend on x).
In other words,
the series (3) is uniformly convergent with respect
to y. It then follows from (2),(3) that if x+N1UN2UN3, limPf(x + iy) = X limPq) (x + iy) = X k k yYo k y>o
k
(x) = f(x).
When f e LPOR), 1 < p 4 , for every integer N 3 1 let us set
AND FOURIER TRANSFORMS
397
fN(x) = f(x) or 0 according as IxI < N or not, and fN = f  fN.
Since fN a L1(IR),
limPfN(x + iy) = f(x) Y+0 for almost all x such that IxI < N.
IPfN(x
+ iy)l
Furthermore,
It 1;.N
(t)I dt. (x  t) 2
From this one concludes that limPf(x + ig) = f(x) Y0 for almost all x e]N,N[, and consequently for almost all x e]R.
For c > 0 and R > 0 let rc R be the loop formed by
SOLUTION (d):
the segment [R + ic,R + ic] and the semicircle E * is + Rein For all z e n, whenever 1/c and R are large enough,
0 < 8 t n.
g(z) =
g() 
2n1iJ
d
l Jr 0 = 2ni e,R
r c,R
so by subtraction it is deduced that
g(z) =
r c,R
 z z
d.
As the function g is bounded, the integral along the semicircle is 0(1/R), and consequently
g(z)
E
_ V
n
.
(t + ic) t + is  z t + 2c 
dt.
Now, let (ck) be a sequence that decreases to zero.
(5)
Extracting
CHAPTER 8: CONVOLUTION PRODUCTS
398
a subsequence, one can assume that the functions t ' g(t + ick) converge weakly in L OR) to a function g.
One can also assume
Since
that co < y.
It + iEk  zI a It + iE0  zI
It + iEk  21
,
> It  51,
Lebesgue's Theorem shows that the functions
t '* (t + iEk  Z)1(t + ick 
8)1
tend in norm in L1OR) to the function
t y ((x  t)2 +
y2)1,
and consequently one deduces from (5) that
g(z) = Pg(z).
Note that by the preceding part of the exercise, for the sequence under consideration, limg(x + iy) = g(x) y+0 for almost all x.
The functions
SOLUTION (e):
belong to Lq(nt) for 1 < q
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)) +W
YJn
log(1 + pKf(t + ie)
(xt)2+g
dt = log(1 + pKf(z + ie)),
and on taking real parts,
+W If_C0
logll + pKf(t + ie)I dt = logll + pKf(z + ie)I. 2 + y2 (x  t)
We have observed that
nlogll+pKf(z+iE)I
(Y+e)
IIf1I1,
CHAPTER 8: CONVOLUTION PRODUCTS
402
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
(xt)
+b2log+IuQf(t+iE)ldt 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 < =).
limQf(t,c) = Hf(t), Ero
By part (e), for almost all t
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
and 1z11 6 M,
Iz21 ( M,
1Iz11p  Iz21p1 4
EXERCISE 9.164:
pMp1121  z21.
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) < 2n.
The function
405
1
406
CHAPTER 9:
g : In IVn is integrable.
Let
x Jg(t)dt.
f(x) =
0
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 implies y e V1 if 1 4 n, N,
which proves that f'(x) = +.
EXERCISE 9.165:
Let (cn) be a sequence of strictly positive num
bers such that OD
n=1
(a):
n
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 is dense 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
p
L
npcn a, and consequently f(x) = 1. This shows that meas(E) = 1. SECOND SOLUTION:
Let
n = 1,2,...
`pn = AIL [0,1/n] I
.
The functions rpn form an approximate identity in L1(1R).
ists therefore, a subsequence (ns) such that f*9n
There ex
; f almost every
s
where.
If 0 < x .< 1, and if n is large enough, then
(x
f*cpn(x) = n1
f(y)dy >, a.
x1/n From this one concludes, as above, that f(x) = 1 for almost all
the x e [0,1]. EXERCISE 9.167: [a,b].
Let (f ) be a sequence of increasing functions on n Assume that the series
W f(x) =
I
n=1
fn(x)
BOUNDED VARIATION etc.
409
converges for all x (a 4 x < b). Prove that for almost all x e [a,b]
f'(x) _
(FUBINI'S THEOREM)
f'(x).
n=1
eve = Vev  ove = vov = ovo
SOLUTION:
that fn
By considering the functions fn  fn(a) one can assume 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 0 and A = (A = x0 < xi
. 1.
If it is assumed that
G
n=0 then sn 
G
Ixn  xn+1I
n=0
rn
0, and the series E un is therefore convergent.
Fur
thermore,
NC
N1
(an  an+1)sn
Anun = aNsN +
nI.
n I0
0
The first term of the right side is bounded by IA0I + r0 + ... + rN1
1 + r0 +
'
+ rN
a quantity that remains bounded as N  ; the second term is equal to
N1 C
r
n01+r0+
n
CHAPTER 10: SUMMATION PROCESSES
432
and tends to infinity with N by a classical property of positive It follows that the series
divergent series.
Anun is divergent:
Let (un)na0 be a decreasing sequence of positive
EXERCISE 10.175: real numbers.
Show that if
then
un = of
n). ovo = V1V = tV1 = VAV = ovo
SOLUTION:
Let
EN
n=N+1
u
If p > N then (p  N)uP < uN+1 + ... + up 4 EN,
and consequently
li ppuP
EN,
which implies the desired result upon making N
EXERCISE 10.176:
fan
Let (an)n30 be a sequence of real numbers.
_
= an  an+1,
2
1 a
_ =
n =
a
4a n+1'
Set:
TRIGONOMETRIC POLYNOMIALS
433
The sequence is called a CONVEx SEQUENCE if
02an > 0
for all n > 0.
Let c be a convex function on [0,[.
(a):
Show that the
sequence an = 9(n) is convex.
Show that if the sequence (an )n>0 is convex and bound
(b):
ed, then It is decreasing;
(i):
(ii):
L1an = o(1/n);
(iii):
(n + 1)t2a. = a0  lima..
L
n
n=0
(c):
Show that for every sequence (cn)n>0 which tends to
zero there exists a convex sequence (an)n>0 which tends to zero, and which is such that Ic.I < an for all n a 0. ADA  VAV = A0A = V AV = DOA
SOLUTION (a): 2
This follows from
an = an  2a n+1 + a
n+2
= 2['9(n) + 3(p(n + 2)  p('n + L(n + 2))]. SOLUTION (b):
The condition A2a
sequence (da ) is decreasing.
n
0 for all n means that the
If, for some integer r Aa
then for every integer n > r
an = ar 
n1 E
s=r
Aas
ar + (n  r)(
Aar),
r
< 0,
434
CHAPTER 10: SUMMATION PROCESSES
and so the sequence (an) would tend to +W, which is absurd. Therefore Dan 3 0 for all n, which proves (i). But then the sequence (an) is decreasing and bounded, and thus has a limit.
Since:
N1 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,
N1
N1
2
Aan  NAan = a0  aN  NAaN,
(n + 1)A an = n I=O
n I=O
which implies (iii), since by (ii) the last term tends to zero. SOLUTION (c):
On replacing cn by sup(IcSl:s i n) it can be as
sumed that the sequence (cn) is positive and decreases to zero. By part (a) it suffices to construct a convex function q on [0,W[ that tends to zero at infinity, and that is such that 9(n) > cn for all n.
To do that, consider a sequence (q)k) such that
9k ' qk+l
if k
0,
q)
1 = c0,
+ 0.
(1)
lpk
Let no = 0, and let n1 be such that
ni > n0,
cn1
nk1'
(3)
TRIGONOMETRIC POLYNOMIALS
435
< (P
(4)
k+l'
enk (Pk1
(Pk
(Pk2
(Pk1
nki
nk2
0. that cp(nk) =
Note
This being so, let m be such
and that is linear on each of the intervals
k
This function is decre sing and tends to zero at
[nk,nk+l].
infinity by (1), and it is convex by (5).
If nK
< n < nK+1,
then by virtue of Relation (4)
en < enk 5 9(nk+1) < (P(n). EXERCISE 10.177:
Let (u
)
n,p n30,p
be a double sequence of com
plex numbers, such that W (i):
lu
X
n=0
4 M for all p >. 0;
1
n,P
W (ii):
l}i.)m
00 P+00
I
n=0
un
,P
(iii): un,P = 0
= 1; for all n > 0.
Show that if sn + s, then
lim
I
un p8n = s.
P" n=0
'
1Vt  V AV  AVA a VAV = ovo
SOLUTION:
If an > a, then on setting an = s + cn it is seen that
436
CHAPTER 10: SUMMATION PROCESSES
by (ii) one is reduced to the case where s = 0.
Let c0 be the
Banach space formed by the sequences c = (en) that tend to zero, with the norm 11C11 = sup f cn I
n
and let us consider on e0 the linear functionals
cc
fp(e) _
un Pen.
I
n=0
'
It is clear that these latter are continuous and that cc
IIfP II S
I un, p
E
I
.
M.
(1)
n=o
It is a question of proving that fp(c) > 0 for all c ec0.
en be the sequence such that em = 6nm.
Let
From the relation
N
I1c 
C
cnen II = sup I en I
n>N
n=0
one deduces that the eats form a total set in c0. therefore suffices to prove that for all n limfp(en) = 0.
But this is nothing other than (iii), because
f (en) = u P
n,p
By (1) it
437
TRIGONOMETRIC POLYNGd'4IALS
In the following six exercises the following notations are used:
is a sequence of complex numbers, set:
If (cn)ne2Z
80+81+. . .+8N CC
sN
Ini s.
Show that sN 
s if and only if
tN = o(N). (c):
Show that if for some real
Again assume that aN ) s.
number p , 1
InlpllcnIP G
, 1).
ta0 = VAV = MMA = VAT = 00A
SOLUTIONS (a);(b): N
n cn =
tN = InI,N
I
N
n(c
+ c_n) =
n=1
E
n(sn
sn1
n=1
from which it is easily deduced that tN = N(sN  aN1),
(1)
TRIGONOMETRIC POLYNOMIALS which proves (b).
439
Now
sN = (N + 1)GN  NoN1'
and substituting this expression in (1) yields
tN = AN + 1)(aN
 'N1
(2)
).
Therefore, aN converges if and only if the series
NL
N=i AN + 1)
(a N  a N1 )
converges.
Since the sequences
aN = N(N + 1) 1
and
aN =
N1
(N + 1)
satisfy
N=1 IAN 
1, and let q be such that
+ q = 1; it is
now a matter of proving that N 1tN + 0 (cf., the p part (b)). if r < N then
Now,
It
Inll/P InI1  1/pcnl
N
< NI
nlcnl + NI
I
r 0 t
and an increasing sequence of integers N
s
such that
TRIGONOMETRIC POLYNOMIALS tN
441
> BNs. S
If VS is the integral part of BNs/2A, then for 0 < v < vs
tN +v
= tN
s
+
>. BN
>
s
S
 vA
S
BN BN
n(cn + cn
I
N 0'
a contradiction. REMARK:
We have implicitly assumed that the c 's were real. n
If
this were not the case, decompose them into their real and imaginary parts.
Moreover, in assuming that N1 t 1
N
did not tend to zero
If this supremum were zero then > 0. N necessarily liminfN 1tN < 0, and it would suffice to change the we supposed that limsupN
signs of all the cn's.
t
442
CHAPTER 10: SUMMATION PROCESSES
EXERCISE 10.180:
Set
8P + 8 =
ap,N
(a):
+ p+1 N + 1
+
8p+N
Express ap,N as a function of the aN's.
Deduce from this that if a
N
> s then a
p ,NP
* s for every
sequence of integers (Np) such that NP > ap, a > 0.
(b):
Prove that if ian! < Alni
Iap,N  spI < A
(c):
N
for n >, 1, then
.
Deduce from the preceding a new proof of Hardy's Lemma
(cf. the preceding exercise).
AVA  VAV = AVA = VAV = AVA
SOLUTION (a):
(p + N + 1)ap+N
pop1
N + 1
ap,N
ap+N + NN p+N  ap1 If the sequence (Np) is such that Np > ap, a > 0, then
N< a< so that if a
N
1Q
> s then a
p,Np
> B.
(1)
TRIGONOMETRIC POLYNOMIALS
443
SOLUTION (b):
p+N
(N + I)ap,N = (N + 1)8
+
G
(N + p + 1  n)(cn + cn),
n=p+l so that
p+N Iop,N  spI
N +
11
(N + p + 1  n)(cn + cn
G
n=p+1 N
p N2+ 1 r=1
SOLUTION (c):
E p
P
If Icnl
.
r=
p
AIn(1 for n a 1 and aN  s, let
sup IaN  0 1I, P N>p1
so that ep > 0.
Let N be the integer such that
N 4pp 0 and that en = 0 except when
EXERCISE 10.181:
n = 0,±n1,±n2,... with
p+l>, q > 1. n p
Show that if aN 4 s then sN > B. and show that for np .< N
, r,
>, qpr
r
so that if np < N < np+1' then
IN 1tNl = IN 1 ¶
nr,(cn
r=1
0.
Then by exercise 10.179 sN > s if
aN >s.
EXERCISE 10.182:
Let (an)n>0 be a sequence of strictly positive
real numbers that is convex and tends to zero. Show that if
'ON' < A,
SN = 0 [a N)
then the series +00
alnlgn
is convergent.
SOLUTION:
Using the notations Aan,42an of exercise 10.176, we
have N I
N
alnlcn = a0c0 +
n=N
I
an(cn + cn
n=1
an(sn  sn1)
= a0s0 +
n=1
(Contd)
446
CHAPTER 10: SUMMATION PROCESSES Ni = a 8
(Contd)
N N
+
Aa.8
I
n=0 Ni
= a 8 + N N
Aan((n + 1)an  no n1)
E
n=0 N2
= a a + NAaN1aN1 + N N
By hypothesis a a N N and bounded
E
(n + 1)A2anan
n=0
Moreover, as the sequence (an) is convex
r 0.
and further (cf. exercise 10.176)
00
E
(n + 1)IA2an.an1 s A
(n + 1)A2an = a0A.
n=0
n=0 From this it follows that
N lien
a
N
EXERCISE 10.183:
c
n
=
(n + i)A2a
a
n
.
Denote by En the set of trigonometric polynom
ials of the form
f(x) = e0 + e1cosx +
+ cncosnx,
where
c0 a c1 > ... > cn > 0. Set m(f) = sup{ If(x)I:Zn s x C n},
M(f) = sup{If(x)I:0 < x 4 2n}.
TRIGONOMETRIC POLYNOMIALS
447
Prove that
I
fmf
1 i  21 n n+ 1
inf M
:fe
(1 E'J < l2 +
11 J
1
n+ 1
tVt  VtV  AV = VAV = t4A
SOLUTION:
We can restrict ourselves to the case where
cp + c1 +
+ cn = 1.
We then have M(f) = 1
and
cp
(n + 1)i.
f(x)dx = 2 cp  c1 + 3  5 +
n
J71/2
B y noticing that
it
CO  c1
;.,
(2  iJcp 3 12  1)n + 1
and that :
we obtain:
m(f)
Ii l
21
1
nJn+1
On the other hand, if:
P x) = 1 + cosx n+ + 1 + cosnx
. 0 let ge(x) = f(x)  (A + E)cosNx.
If a is small enough then gE(y) < 0.
Furthermore, if xr = rn/N From
then gE(xr) >, e if r is odd and gE(xr) < e if r is even.
this it follows that for e > 0 gE has at least one zero in each of the intervals ]y,xl[,]xl,x2[,...,]x2N_2,x2N_1[.
By making
e > 0 one concludes from this that g0 has at least 2N  1 zeros in the interval [y,2n[.
Moreover, g0(0) = g;(0) = 0; g0 would
therefore be a trigonometric polynomial of degree less than or equal to 2N which would have at least 2N + 1 zeros in the interval [0,2n[, which is absurd (because g0 # 0, since g0(y) < 0).
SOLUTION (c):
First assume that f is real.
Replacing f by f,
if necessary, it may be assumed that for some x
m = 11f,11. = f'(x0). By part (b)
n/2N 211f 11
> f (x0 +
2N)
 f (x0
2N, = J
(Contd)
f' (x0 + x )dx >. n/2N
450
CHAPTER 10 rn/2N
(Contd)
mJ_
cosNxdx =
2m
n/ 2N
N
,
which proves the result in this case. In the general case let x be a real number. a such that lal = 1 and If'(x)I = af'(x). where u and v are real.
Then
If'(x)I = u'(x) 6 Nlu I ' NIIafIIm = NIIfil..
There exists an
Let us set of = u + iv,
CHAPTER 11
Trigonometric Series
EXERCISE 11.185:
ezx
=
Let z e a, z $
e2nz  1(2z 1 + n
e
zx _ eTIz  1 nz
+
X n=1
zcosnx  nsinnxl z2 + n2
)
0 < x < 21; I
[(1)nenz  1] z2osn2
2 n
Prove the following formulas
.
z +n
n=1
,
0 < x < n;
W
ezX = n E
n=1
[1 
(1)nenz] nsinnx
0 < x < R.
z +n
What are the values of these series when x = 0? AVO = VMV = AVA = VOV = AVo
SOLUTION:
Let f be a function of period 27E such that f(x) = ezx
if 0 < x < 21.
Then [f'] = zf, and consequently
in f at 0 is 1  e21z).
451
Therefore
452
CHAPTER 11: e2nz  1
(n)
1
n $ 0.
z n
2n
=
Since
2tcz
(0)
=
2nz
the first formula is proved (JordanDirichlet Theorem).
For
x = 0 the sum of the series is
(e2nz + 1)
=
z(f(x + 0) + f(x  0)
from which it is easily deduced that W V
ncothnz = 1 +
2z
n=1 z2 + n2
z
Now consider the functions g and h, with period 2n, respectively even and odd, and which coincide with f on ]0,n[. [g'] = zh,
[h'] = zg,
so
ing(n) = A(n), inTi(n) = zg(n) + 2n (2  2e1Ze
Therefore if n 4 0 then
g(n) _
di(n) =
(1)nenz  1 n
z
z2
+
z2
+ n2
1(1(T Zit
n2
n
,
nn).
Then
TRIGONOMETRIC SERIES
453
Moreover
e
nz
9(0) =
7Z(0) = 0,
1
nz
,
which proves the two other formulas. for x = 0.
In fact the second is valid
From this it is easily deduced that m
(_1) n
C
sinhnz
z +
n=1
n2
z
+ EXERCISE 11.186:
I
Find the sums of the following series:
acosnx 2
n=1 n+ a
SOLUTION:
nsinnx c
2
n=1 n2 + a
2,
(a real, a + 0).
Let
W
g(x) _
acosnx
(1)
I
n=1 n2 + a inz
G(z)
n=1 n
2e
2
z = x + iy,
y >, 0.
(2)
+ a
The series in (2) converges absolutely for y > 0 and defines on the halfplane y > 0 a holomorphic function that satisfies the differential equation
ae
G"  a2G = e
iz
Re(z) > 0.
(3)
 1
But the G can be analytically continued to a minus the two real
454
CHAPTER 11:
From this it follows that g is
halflines ],0] and ]2n,+m[.
infinitely differentiable (and even analytic) on ]0,2n[ and that on this interval satisfies the differential equation: 2
g,,  a g = aRe
eix
le ix
a
2
Hence
g(x) = acosha(x  n) + Ssinha(x  n) 
1
2a
for 0 < x < 2n, and hence also, by continuity, for 0 < x < 2n.
Since g(0) = g(20, 0 = 0. Now,
r212asinhan it
0
so that
acosnx = ncosha(n  x) 2sinhna n=1 a2 + n2

0<x< 2n.
1
2a '
The derivative of the right side is continuous on [0,2n], hence one can differentiate term by term, giving 00
nsinnx = nsinha(n  x) 2sinhna n1 a2 + n2 C
0<x 0,
znF(z)dz = irn+l
0
 1lzl=r
2nei(n+1)xF(reix)dx. 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
n=0
I 1 
N + 1)?(n)zn,
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) =
N,M+ l z l c1
lim ( sup IPM(z)  PN(z)l) = 0.
N,M+ l z l =1
460
CHAPTER 11:
Therefore PN(z) * F(z) uniformly on
is continuous on IzI
.
0; The series
(iii):
+00 ?(n)einz
(*)
n=oo converges on a strip IIm(z)I < 6, 6 > 0; The series (*) converges at two points, z1 = x1 t i61
(iv):
and z2 = x2  i62, with 61,62 > 0.
4V4 = 040 = 4MA = VAV = t0A
SOLUTION:
Let
an(z) = ?(n)einz +
?(n)einz
Then 2inz (e
1
2inz
 e
inz
2)j(n) = an(z1)e
inz
1
 an (z2 )e
2
TRIGONOMETRIC SERIES
461
If (iv) is satisfied there exists a constant M such that lan(z1)I .< M,
Ian (z2)I < M,
n >. 0,
so that if n j 0 n61 + e 1?(n)l < M le2n62  e2n611 n62
e
if c = min(61,62).

0(eelnl),
This proves that (iv) _> (i).
If (i) is satisfied, the series (*) converges in norm on every strip IIm(z)I < 6, 6 < e
Therefore (i) => (ii).
Since it is trivial that (iii) => (iv), it remains to prove that (ii) => (iii).
Let F be a function holomorphic on the strip Im(z) < 6, 6 > 0, and such that f(x) = F(x) if x is real.
As the relation
F(x + 2n) = F(z) is true on at, it holds on the entire strip IIm(z)I < 6 (principle of analytic continuation).
Hence there exists a function G de
fined on the annulus _6 e
0, a>0,
0(InI1/20'e(InI/R)1/a).
=
Now assume that (*) is satisfied.
11f(s)II
(c):
=
Let f e C00(T).
=
O(ssRsr(as + 1)),
0 = (a 
Show that
)+.
Deduce from this that f is analytic if and only if O(eelnl)
for some e > 0.
If(n)I =
AVA = VAV  OVA = VAV = AVA
SOLUTION: (a):
If A is such that
IIf(S)II < ARSr(as + 1), then for all n 4r 0
s a o,
TRIGONOMETRIC SERIES
463
I(in)sf^ (n)I < AI IRIJsr(as + 1).
1 (n)I =
n
Let us take for s the integral part of (1/a)(Inl/R)1/a. IS
Il lnlr(as + 1) < (as)asr(as + 1) ti
eas
Since
1/a I
 a < as < [J..L]
I
/a
J
(*) certainly holds.
SOLUTION: (b): +00
f(s)(x)
=
s > 0,
(in)sf(n)e11"`,
E
00
and consequently
IIf(s) II00
= 0( 1 ns +1/2a e (n/R)
1/a
).
n=1
Let a = a + 1/2a and
xae(x/R)1/a
a
9
(x) =
x
0.
attains its maximum at
xo = R(aa)a.
Let Na be the integral part of xa.
Then
2.
Then
464
CHAPTER 11:
Ja()dx + q) a(NQ) + 9 N
spa(n) F n=1
a
(N
a
+ 1) +
+1Ta(x)dx
1N a
S
(Pa(x)dx + 2(p a(xa).
Now,
(x)dx = aRo+lr(aa + a), J0
Ra(aa)aaeaa
cpa(xa) =
Since r(x + h) ti xhr(x) as x } 00,
aRa+1r(aa + a) = 0(aaRar(aa)). Furthermore,
Ra(aa)aaeaa
= O(a'Rar(aa))
Therefore 00
ma(n) = 0(aYRar(aa)),
y = max(a,2),
n=1 so
11f(s)JI = 0(syRsr(as + 2)) = O(sy ZRsr(as + 1),
which is precisely (**).
SOLUTION (c):
It is known that f is analytic if and only if
TRIGONOMETRIC SERIES
IIf(s)II
Co
465
= O(RSal),
R > 0.
(1)
By part (a) above, (1) implies that 0(e£InI)
I?(n)I = 0(InI2eInI/R) =
if 0 < e < R
On the other hand, if
IJ(n)I = O(eEInI),
then by part (b),
0(s2ESS!) = O(Rss!)
= IIf(s)II0o
EXERCISE 11.192:
when R > 1g .
Show that if f e L1(T) and:
Co f(n)einx
g(x) _
almost everywhere,
n=then f = g almost everywhere.
A0A = VAV = AVA = VAV = A0A
SOLUTION:
By the LebesgueFejer theorem
aN(x) + f(x)
almost everywhere,
But SN(x)  g(x) implies that aN(x)  g(x), so f = g almost everywhere.
EXERCISE 11.193:
I Ifcx 0
Let f e L1(T).
Show that if
+ t) + f(x  t)  2sI tt
2a
k 3 2,
TRIGONOMETRIC SERIES (b):
471
Show that fk coincides almost everywhere with a con
tinuous function when k > 1/a.
AV4 = VtV = 4VA = VAV  4VA
SOLUTION:
Note that 2n
f(n)
= 1
{f(x)  f+ 0
IX
n)}einxdx, J
so
0(Inla).
If(n)I =
SOLUTION: (a):
(1)
It follows from (1) that
0(Inlak),
Ifk(n)I =
and consequently that
E Ifk(n)I2 < W n
if k > 21 a
But then fk e L2(T).
SOLUTION: (b):
If k > 1/a, then
G I?k(n)I
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 lnlpilf(n)I E
< ,
which ensures that f is (p  1) times continuously differentiable, and that
If(P1)(n)I
0(InI(0'p+1))
=
with (a  p + 1)  1 >
Assume, therefore, that a > 3/2, and let us prove that f' exists almost everywhere and belongs to L2(T).
We have
Y Inf(n)12 < M.
Hence there exists g e L2(T) such that
g(n) = in?(n),
n ea.
Since it is possible to integrate a Fourier series term by term,
rx
g(t)dt = 0
Since
Y If(n)I < ,
(f(n)el"x  f(n)).
TRIGONOMETRIC SERIES
473
the right side is equal to f(x)  f(0), which proves that f'(x)= g(x) almost everywhere.
EXERCISE 11.200:
If f is absolutely continuous and f' e
L2,
then
I I?(n) 14 IL I11 + " IV' II2 .
n
ovo = vov = ovo = vov = ovo
In fact, 11(0)1 : If II1, and
SOLUTION:
E
I1(n)I 4 (I n2)2(1
n40
n40
Inf(n)I2
n40
t Ilf' 112
ExERCISE 11.201:
For every finite set AC 2Z and all e > 0 show
that there exists f e L1 such that
(i) :
0 4 f(n) 4 1 for all n ea;
(ii): f(n) = 1 if n e A;
(iii):
11f111r+ 2N + 1.
Using the CauchySchwarz inequality and Bessel's equation, it follows that
211
If(x)I dx < ON + 1)1(2N + 1)2 UN + 2r + 1)2
27E 1
0
2N+2r+1 (
2N+ 1
If N is chosen sufficiently large, then
0,
E cn
=
c2
0 such that:
E IAnf(n)1 < Allf111, n
f e L1(T).
If in the preceding inequality we take for f the Fejer kernel FN, one obtains
I1 lnl W this yields
L Ian1 < A. n
EXERCISE 11.208: that
Let (), n)ne2Z be a sequence of real numbers such
TRIGONOMETRIC SERIES
X
481
EX2
E X=
a0,
n
n
Show that there exists a function f e L2(T) such that (i):
(ii):
If(n)I = o(an);
For any a < b, ess suplf(x)I = . asxxb
ovo = VtV = ova = VAV = AVA
SOLUTION:
Let c0 be the vector space of sequences u = (un)nea
of complex numbers such that Inlm IunI = 0,
provided with the norm
Hull = S1PIunl .
If u e c0, there exists u e L2(T) such that
co is a Banach space.
u (x) '\
anune",
X
n and
g112
2 = E anlunl2 < IIuI12 E an
n
n
The mapping u ; u is therefore continuous from c0 into L2(T).
I is a compact interval of length greater than zero, let
pI(u) = ess sup I u(x) I .
xeI
If
482
CHAPTER 11:
It is clear that pI is a seminorm on co; this seminorm, furthermore, is lower semicontinuous.
In fact, for every function
g, continuous on I, the mapping
is continuous on c0 and
sup{IJIugl:g
pI(u) =
continuous, J1ii $ 1}.
For every integer s a 1 let us consider the closed set AS = {u:pl(u) < s}.
We are going to show that AS = 0, which will prove that pl(u) for all the u belonging to an everywhere dense Gs set (Baire's O
Theorem).
If one had AS +O one would deduce, taking the convex
ity of AS into account, that for some p > 0, huh < p => pI(u) '< s.
(1)
Let (ar), 1 < r 4 k, be a sequence of real numbers such that the intervals I + ar cover [0,2n].
hull < 1,
and
If
u(r) _ (une is n r nea'
then also 11U(r) 11
F 1,
and by (1)
ess sup Iu(x)I = ess sup I c(x + ar) I I+a r
I
=
pl(u(r))
4 P
TRIGONOMETRIC SERIES
483
(since u(x+ar)=u(r)(x)).
s/p would hold if Ihu1141.
Hence
For every function f e L1(T) the mapping
2n
u
2nI
z7(x)f(x)dx 0
would be a continuous linear functional on c0.
Thus there would
exist a sequence (an)ne7G such that
((2n
E lanI
(ii).
If (*) is the Fourier series of a continuous function, then the Fejer sums aN(x) of (*) converge uniformly.
Since aN(0) = 0, it
follows that
N02N(2'N) = 0. If x = 7t/2N then nx < it, hence sinnx 3 0, when n < 2N and nx < 7E12, 2nx Thus when n < N. hence sinnx a = N,
n1
a2N(77E
N)
;' n1N (1
2N n (iii). SOLUTION: (c):
Inequality (5) proved above again shows that (i)
=> (iii), and in the same way (iii) => (i) follows from (6). Furthermore, (i) implies that the sum of the series (*) belongs to LW(T), and Lebesgue's Theorem justifies obtaining the Fourier
490
CHAPTER 11:
coefficients of this sum by integrating term by term, so that (*) is certainly the Fourier series of its sum.
It has thus been
On the other hand, if (ii) is satisfied,
proved that (i) => (ii).
the aN are uniformly bounded, and the inequality
> 4 (N + 1)aN
a2N (2'N,
proved in part (b) shows that (ii) => (iii). Letting N = 1 and M 
SOLUTION: (d):
f(x) _
ansinnx = n=1
in (3) yields
Aanb1 n(x). n=1 L
'
Note that by (1)
cosix  cos(n + D1,nx_
g(x) =
2sinx
n=1
h(x) = 2
)x
sin2 x , + "2sinnx, tangy
Dansin2'nx tanx 1
ansinnx.
n=1
Since Aan > 0
and
E Aan = a0,
the function h is continuous (note, furthermore, that as the series which gives h is absolutely convergent, this series is cer
491
TRIGONOMETRIC SERIES Furthermore, because
tainly the Fourier series of h).
sin 22nx ' 0
for 0 < x 4 n,
tanix g >. 0 and
(n
'
cWc
Og(x)dx =
n=1
n
°anj0
(7)
ax
tan x
It follows that f e L1(T) if and only if the series which appears in the right side of (7) converges.
Furthermore, in this case
the Fourier series of f certainly is (*), for the integrals of g(x)sinnx and h(x)sinnx are obtained by integrating term by term, and therefore lrn
J71
n
n
f(x)sinnxdx =
Da
k=1
k
nJI
n
1,k
(x)sinnxdx =
k=n
Aa
k
= a
.
n'
Thus (ii) => (i).
As the fact that (i) => (ii) follows from exercise 11.192, in order to prove that (ii) (iii) it remains to prove that the convergence of the right side of (7) is equivalent to I n1a Now, since
1
tan x 
2 X
is integrable on (0,n), we have
ji7c
JO sinntx dx = 2J1 s x z n logn,
dx + 0(1) =
+ 0(1)
n
N (cf.
'
(3)), and consequently
Ilf  SN II1
N
But by the preceding
II51 nIll ti 1 logn,
'n
Ill.
TRIGONOMETRIC SERIES
493
hence by (8)
IIfSNII130. EXERCISE 11.211: (a):
If (an)n30 is a sequence of complex num
bers such that 00
an ' 0,
E
IA2anI < 00,
n=0
prove that the series Co
f(x) = la0 +
ancosnx
(1)
n=1 converges for 0 < x < 2n, and that Co
f(x) = 2
1
(n + 1)A2anFn(x),
(2)
n=0 where Fn denotes the nth Fejer kernel. (b):
Show that if
an > 0,
1
(n + 1)Io2anI
0,
1
(n + 1)IA2anI < m,
n=0 then the right side of (1) is the Fourier series of f. evo  vev  4aL  vov = pv4
CHAPTER 11:
494
We have (where Dn is the nth Dirichlet kernel):
SOLUTION: (a):
N
N
a
8N(x) = 2 +
ancosnx =
E
n=1
2 n=0 =
an(
n
(x)  Dn_1(x))
NCi
2 aNDN(x) +
AanDn(x).
G
2
n=o
Since
+ Dn = (n + i)Fn,
D0 +
this yields N1
sN(x) = 'aNDN(x) + jN aN_1FN1(x) +
E
(n + i)A2an.Fn(x).
n=0 For 0 < x < 2n:
1 sin x
IDN(x)
,
IF (x)l
0 so that Ie(u)I < a
if
0 E u < d.
The RiemannLebesgue Theorem and the fact that
AN(u)>.0
if 0su4 R.
show that n
r
n
n
iJ eA*1 < a A* + J e(u)cotiudu + o(1). N J d 0 oN
lui 4 it,
TRIGONOMETRIC SERIES
501
Since
JOAN = JODN + 0(1)
ti 2logN,
from this one concludes that n
limsup(logN)11E
J
fo
N
E 2a,
an d, since a is arbitrary, that
EAN = 0(logN). J
0
By (3) we also have: n
J0CD
= o (logN).
SOLUTION: (d):
As the function DN is odd, it is easily seen that n
N(x) _  2nJ (f(x + u)  f(x  u))DN(u)du, 0
If
f(x + u)  f(x  u) = C + e(u)
with lirE(u), u*0
from parts (b) and (c) above one deduces that
lim(logN)1sN(x) _  n N
(4)
CHAPTER 11:
502
If f has left and right hand limits at all points
SOLUTION: (e):
then £ exists at every point x and is equal to f(x + 0)  f(x  0).
Further, if
f(n) = a
11n
J
'
then
IsN()I 6
I
(11(n) + f(n)) = of I I = o(logN).
(n=1
n=1
From (4) it is then deduced that £ = 0, which proves that f is continuous.
EXERCISE 11.214: (a):
Let (un)n)1 be a sequence of positive num
bers such that
U
n
c
A
n
Set:
u1 + 2u2 +
an =
+ nun
n
,
2 2
u2 + 4u2 +
bn =
+ n u
n
,
W
yn = n
L
s=1
ussin2
2n
Show that the conditions an > 0, bn  0, yn * 0 are equiv
TRIGOROMETRIC SERIES
503
(Prove that
alent.
an2 < bn
Aan,
bn
yn
and that for every integer v
n
2
vb
2
+A
Yn
4
(b):
Let f be a real function with left and right hand
nv
v
limits at all points, and with period 2n.
For every integer
n > 1 set 2n1
9n(u) _
If(u + xr+l)
f(u + xr)I2
r=0 where xr = nr/n.
Show that if
If(C + 0)  fU  0)j > d > 0
at some point E.
then whenever n is large enough
rp n(u) > d
for almost all u.
Show also that if f is continuous and has bounded variation, then (pn > 0 uniformly on R.
(c):
Calculate:
r2n
pn(u)du,
J
0
using Plancherel's Formula. (d):
With the aid of what has gone before, prove the fol
CHAPTER 11:
504
A necessary and sufficient con
lowing theorem (owed to Wiener):
dition for a function f with bounded variation to be continuous
is that: 2f(2) + ... + nf(n) = 0. lim f(1) +
n
n
AVA = V/V = AVA = VAV = AVt
SOLUTION:
2 2
Since s us
(a):
.
0 finite, then limA (f;x) = D F(x).
r+1_ (f):
r
s
Compare this result with the LebesgueF6jer theorem. AVl = V AV = p0A = OAV = A0A
SOLUTION: (a):
re ix
P(x)=1+
1  relx
r
SOLUTION: (b):
+
re ix 1  relx

1r
2
1  2rcosx + r2
Replacing the f(n) by their defining expressions
in the formula for Ar(f;x) shows that Ar(f;x) = (f*P )(x).
Note that the r form an approximate identity in L1(T) as r } 1, since
Pr > 0,
and for all 6 > 0
1
2n
n
J P (x)dx = 1, n
r
516
CHAPTER 11:
2n
P (x)dx < P (d),
dslxl 0; if d is chosen so that if 0 < t < d,
IQ(t)I < e then
IAr(f;a)  sI . e + Pr(d)IIQIII, which proves that
limAr (f;x) = s.
r+1
Integrating by parts and taking into account
SOLUTIONS: (c);(d):
that Pr'(t) = Pr(t) yields
it
1
1
n
Ar(f;x) = 2J f(t)Pr(x  t)dt  2nJ F(t)Pr(x  t)dt n
n
n = 2nJ
F(x n
so
t)P'r (t)dt
n =  2nJ F(x + t)P'(t)dt, n
r
TRIGONOMETRIC SERIES
517 n
j(F(x + t)  F(x  t))P'(t)dt
Ar(f;x) =  2n n
setting Qr(t)
sint.P (t),
we have
F(x +
Ar(f;x) = 2If
 t) Qr(t)dt. (1)
2sintF(x
n
It is clear that
Qr(t) = Qr(t)
and
Qr >. 0.
Also, if f(x) = cosx then F(x) = sinx; therefore taking f(x) _ cosx and y = 0 in (1) yields n
2nI
Qr(t)dt = r. n
Finally an elementary calculation shows that
(
l1
 2rcost + r
attains its maximum when
cost =
2r
1 + r2 Therefore, if 0 < d < it then sup Qr(t) 4 Qr(d),
66t67[
2
sint
Qr(t) = 2r(1  r2 )I
2) )
CHAPTER 11:
518
as long as
r2)1
cosd < 2r(1 +
Since Qr(d) > 0 as r > 1, the Qr form an approximate identity in L1(T).
SOLUTION: (e):
By adding a constant to f, F(n) may be assumed
Setting
to 0.
fi(t) = F(x + t)  F(x  t)  D F(x), 2sint S we have n
Ar(f;x)  D(x) = 2nJ (t)Qr(t)dt + (r  1)DF(x), n
since i(t) > 0 as t  0, one can show as in (b) that the integral
on the right side tends to 0 as r  1, from which the result follows.
(Note that there is no problem at ±n since at these points
Qr(t) " (constant)sin2t).
SOLUTION: (f):
In particular
Ar(f;x) + f(x)
at all points where f(x) = F'(x), i.e. almost everywhere.
Now,
the condition f(x) = F'(x) can be written rh
limh_1 h*O
0
+ u)  f(x))du = 0.
Therefore this result is better than that of LebesgueFejer, which says aN(f;x) 
f(x) if
519
TRIGONOMETRIC SERIES
h limb11 If(x + U)  f(x)ldu = 0. h>0 J0
The series
EXERCISE 11.218:
en converges to s in the sense of
Borel (written
(B)
c _m
n
s)
if
s r
r
r
lime
n
ni
= s
n=0
where e
sn = IpI
. 0;
n,i
w
(ii) :
u
E
n=0 (iii):
limu
n,i = 1;
ni
ti
= 0.
'
By exercise 10.177 this implies
s rn
limer
X1>
ni
SOLUTION: (b):
ifsn>s.
In this case
1zn+1 sn
s
n=0
1z
and consequently
522
CHAPTER 11:
e r
s rn n n=0 n! L
= 1 1
z
(1  ze
from which the result follows. Because
SOLUTION: (c):
f*Dn it is clear that
Br(f; ) = f*Br, where
Br
= e
r
C
n Dr n
n=O
nt
.
Recall that i(n+ )x
sifix
Dn(x) = ImIe
0 < lxI
z
and consequently
x
i
Br(x) = se. =
Im(ejx+reix
)
er(1cosx) sin(x + rsinx) sinix
SOLUTION: (d):
First of all, if n > 0 is an integer, then
n1 J(nn) =
n
J
8=0
0
sinx X
sn
dx.
The integral that appears in this expression is equivalent to 2/ns are s > , which proves that
TRIGONOMETRIC SERIES
J(ma) ti
523
loge.
Therefore, when x > m:
0 < J(x)  J(1[x/71])
.O
n
21J1Dn
= 1.
If, for 0 < Iti 4 it, we set
526
CHAPTER 11:
w(t) = (f(x + t)  f(x))log11 t
,
Br(t)
Qr(t) =
log l t l then U
Br(f;x)  f(x) = 2nfX cp(t)Qr(t)dt.
(1)
Note that for 0 < 6 < ItI E n:
2rsin2'z6
IQr(t)1
log2 sin 6
On the other hand, some calculations similar in every way to those carried out in (e) show that
n/2 a2ru
2
Isin(2r + 1)uldu + o(1).
IIQr1I1 = jcJ
0
ulogu
Decomposing the integral above into an integral taken between 0
and 1/1, and another taken between l/V and n/2, it is seen to be bounded by:
J((2r + 1)r') +
logn
IIQx,111 4< A,
a2u2
1
log2
T
u
du.
A = constant.
But then if E > 0 and if 6 > 0 is such that ItP(t)I < e for 1t1.< 6.
TRIGONOMETRIC SERIES
527
by (1) we shall have
 2rsin2d
IB(f;x) r  f(x)I < eA +
l0 2 sin d.l
d4Itkit
Icp(t)Idt,
which proves that
limB(f;x) r = f(x). r+W
Let (an) be a sequence of numbers an >. 0, and
EXERCISE 11.219:
An = a0 +
+a
.
Suppose that
a
n> 0.
An
A
n The series +W L
en
is said to converge to s in the sense of NBrlund, written +w (N)
E Cn  8,
_m
if
vn =
where
a s + ... + a 8 n 0 0 n
A n
528
CHAPTER 11: 8
=
C
X
IP14n
(I):
.
P
Show that if a series converges in the ordinary
(a):
sense it converges in the Narlund sense. (I):
(Use exercise 10.177).
Prove that if 0 < a0 .< a1
a1>,.., >,an>,**,
an*0.
9
(*)
For f e L1 set n Nn(f) = A 1 apsep(f), n p=0 where s(f) denotes the nth Fourier sum of f. (II):
(a):
Show that
Nn(f) = f,t(Un  n), where
Un(x) = A
sinsn x X
V
1 cos(n + 1)x = A
n
Show that 1/n
n(x)Idx J
p acos(p +
)x,
(1)
a sin(p + 7)x.
(2)
p=0
n
sin ix
n p=0
P
TRIGONOMETRIC SERIES
529
is bounded for n >. 1. (II):
n
Isin(p t 1)x dx 4 B(p + 1)log p + C(p + 1),
J1/n n
1/n
Show that there exist constants B,C,D such that:
(b):
14 p < n,
2sin2ix
slnxI
dx < Dlogn,
n > 2.
2sin2Ix
(II): (c):
By carrying out some suitable Abel transforma
tions on (1) and (2), prove that
IIVn Ill < 1, and:
IUn(x)Idx < C + f 1/n
Aplogll +
(Elogn + B
p=1
n
`
where E is a constant. (II): (d):
Now assume that
SnPIIUnII, By carrying out another Abel transformation on (1) for Un and observing that
sin2(n + 1)x < Isin(n + 1)x I,
show first of all that the quantity
71
JO(sin2(n
nl
x )x
+ 1)W){cossnnt l
+ An
Asin(p + 1)xfdx p= nL
2
JJJ
CHAPTER 11:
530
stays bounded, then by studying the behaviour of the integrals:
s infix
+ 1)x.cos(n + z) x
7E sin2(n f0
dx'
lTE
cos(2n + 2)x.sin(p + 1)xdx, 0
deduce that
nA
1
sup n
< I n p=1 P
A
(II):
.
Conclude from the preceding that when (*) is
(e):
satisfied the following conditions are equivalent:
For all f e L1(T)
(i) :
f(n)e
inx (N)
= f(x)
at every point x where f is continuous;
n
sup A n
A P
p=1 P
< . V,
For every real number4 set
(III):
na = (a +
and assume that
a
_
n
n)
n! 
n
Aa1
n
TRIGONOMETRIC SERIES (III): (a):
A
n
531
Show that
= Aa
n
and that if a > 0 one obtains a summation method in the style of NSrlund.
Next show that if 0 < a < 1 one is in the case studied in
part (II) above, if a > 1, in that of part (I)(b); and that finally, if a = 1 Cesaro's method is recovered. (III): (b):
g0 An1 + ... + S n
a a
n
It is said that:
=
A
Aa n
is the CESARO SUM OF ORDER a of the series I en, and that the latter is (C,a)CONVERGENT if 6n is (a > 0).
Prove that for every a > 0 the Fourier series of a function f e L1(T) is (C,a)convergent to f at every point where f is continuous.
ovo = VAV  ovo = vov = ovo
SOLUTION: (I): (a):
vn
P10 un,pap, =
where a
Ap n U
n,p
if 0 < p 6 n,
=
0
if p > n.
CHAPTER 11:
532
It is clear that
= 1.
u p>0
n,P
Moreover,
a n_p un,P { Anp
and consequently = 0. limu n n,P
By exercise 10.177 sn > s implies vn > s. SOLUTION: (I): (b):
Writing an for the nth Cesaro sum of the
series X Cn,
n
anp((p + 1)ap 
vn
PCP1
n p0
n1 (n + 1)a A 0 an + A (p + 1)(anP  an P 1)a p n n p=0
p=O
an,paP,
where
(P (ann
(n + 1)a0 n,p
A
0
P
 anP
1)
if 0 4 p < n  1, if p = n,
n
ifp>n.
TRIGONOMETRIC SERIES As the sequence (a
n
)
533
is increasing, all the An,p >. 0.
It is
clear that = 0.
lima
n
n,P
Furthermore, when all the sn's are equal to one, the same holds for all the a 's and v 's. so n n
= 1.
A
n,P
p=0
By Exercise 10.177 an * s implies vn  s.
SOLUTION: (II):
(a):
Nn(f) is equal to the convolution product
of f with 1
n I
n p=0
n
1
p aDnp(x) = A
=
+ sin(n sin x
ap
E
n p=0
sin(n ± 1)x
1
n
apcos(p +
A
cossn 1
7
i IX
1)x
n 0
p which implies the first result desired. Note that
sin(n + 1)x s in x
and consequently IUnI < 2(n + 1),
so
)x
p=0
n
An
)x
2(n + 1) ,
apsin(p + J)x,
534
CHAPTER 11:
1/n
f0
l
(
4 211 + l
JU
SOLUTIONS: (II): (b);(c):
4.
Denoting by Fn the nth Fejer kernel
and setting A
= ap  ap+i'
we obtain
V (x) =
1
An
n1
cos(n + 1)x{(n + 1)anFn(x) +
(p + 1)h F (x)}. P P
E
p=0
(1)
Since
and
AP;0, 0
IIFPIII=1,
it follows that
11711
A {(n + 1)an +
n1
p + i)AP} = 1.
E
p=0
n
Similarly, noting that
n I
cos(p + ')x =
sin(n + 1)x
2sinx
p=o we obtain
Un(x) = A sin(n + 1)x
(2)
n
x
Ja
n
sin(n + 1)x 2sin21x
n,l A + p=0
p
sin(p t 1)x) 2sin2.,x
J
TRIGONOMETRIC SERIES
535
Now note that fit
2
sinx
xdx
(n
dx 4
1/n 2sin tx
=
it
log(nn) ` Dlogn.
1/n (2x )
for all n >. 2 if D is chosen large enough.
When 14p