TRANSLATIONS OFMATHEMATICALMONOGRAPHSVOLUME 107

TRANSLATIONS

MATHEMATICALMONOGRAPHS

.107

B. M. Makarov, M. G. Goluzina,A.A. Lodkin, and A. N. Podkorytov

Selected

Problems in

Real Analysis

Mathematical Society

Translated from the Russian by H. H. McFaden

1991 Mathematics Subject Classification. Primary 26-01, 28-01.

Library of Congress Cataloging-in-Publication Data

lzbrannye zadachi p0 matematicheskomu analizu. English.Selected problems in real analysis/B. M. Makarov...[et all.p. cm.—(Translations of mathematical monographs, ISSN 0065-9282; V. 107)Includes bibliographical references.ISBN 0-8218-4559-4 (alk. paper)1. Functions of real variables. 2. Mathematical analysis. I. Makarov, B. M. II. Title.

III. Series.QA331.5 19313 1992 92-15594515'.8—dc2O CIP

Copying and reprinting. Individual readers of this publir.ation, and nonprofit libraries actingfor them, are permitted to make fair use of the material, such as to copy a chapter for usein teaching or research. Permission is granted to quote brief passages from this publication inreviews, provided the customary acknowledgment of the source is given.

Republication, systematic copying, or multiple reproduction of any material in this publication(including abstracts) is permitted only under license from the American Mathematical Society.Requests for such permission should be addressed to the Assistant to the Publisher, AmericanMathematical Society, P.O. Box 6248, Providence, Rhode Island 02940-6248. Requests can alsobe made by e-mail to reprint-permission@ams.org.

Copyright ®1992 by the American Mathematical Society. All rights reserved.Translation authorized by the

All.Union Agency for Authors' Rights, Moscow.The Amencan Mathematical Society retains all rights

except those granled to the United States Government.Pnnsed in the United States of Amenca

The paper used in this book is acid.free and falls within the guidelinesestablished to ensure permanence and durability. ®

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1098765432 0201009998

Contents

Foreword

Notation ix

Problems Solutions

CHAPTER I. Introduction 3 143§1. Sets 3 143§2. Inequalities 8 147§3. Irrationality 13 152

CHAPTER II. Sequences 17 157§1. Computation of limits 17 157§2. Averaging of sequences 19 162§3. Recursive sequences 21 164

CHAPTER III. Functions 25 167§1. Continuity and discontinuities of functions 25 167§2. Semicontinuous functions 28§3. Continuous and differentiable functions 28 171

§4. Continuous mappings 31 173

§5. Functional equations 33 175

CHAPTER IV. Series 35 179§1. Convergence 35 179§2. Properties of numerical series connected with

monotonicity 36 180§3. Various assertions about series 38 185

§4. Computation of sums of series 40 190

§5. Function series 41 192

§6. Trigonometric series 43 194

CHAPTER V. Integrals 47 201§1. Improper integrals of functions of a single variable 47 201

§2. Computation of multiple integrals 49 204

CONTENTS

Problems Solutions

CHAPTER VI. Asymptotics 53 213

§1. Asymptotics of integrals 53 213

§2. The Laplace method 55 217

§3. Asymptotics of sums 59 221

§4. Asymptotics of implicit functions and recursivesequences 63 227

CHAPTER VII. Functions (continuation) 67 231

§1. Convexity 67 231

§2. Smooth functions 73 237§3. Bernstein polynomials 77 244§4. Almost periodic functions and sequences 80 253

CHAPTER VIII. Lebesgue Measure and the LebesgueIntegral 87 263

§1. Lebesgue measure 87 263§2. Measurable functions 89 267§3. Integrable functions 91 268§4. The Stieltjes integral 98 280§5. The e-entropy and Hausdorif measures 99 282§6. Asymptotics of integrals of higher multiplicity 104 290

CHAPTER IX. Sequences of Measurable Functions 109 301

§1. Convergence in measure and almost everywhere 109 301§2. Convergence in the mean. The law of large numbers 111 302§3. The Rademacher functions. Khintchine's inequality 114 305§4. Fourier series and the Fourier transform 120 311

CHAPTER X. Iterates of Transformations of an Interval 125 317§1. Topological dynamics 125 317§2. Transformations with an invariant measure 132 333

Answers 347

Appendix I 357

Appendix II 361

Appendix III 363

Bibliography 367

Subject Index 369

Foreword

This problem book is intended first and foremost for students wishing todeepen their knowledge of mathematical analysis, and for lecturers conduct-ing seminars in university mathematics departments. It is somewhat differentfrom the usual problem books in the greater difficulty of the problems, whichinclude a number of well-known theorems in analysis. Despite this, no specialpreparation is required to solve the problems in Chapters 1—Vu and in §1 ofChapter X, and many of them are accessible even to first-year students in thesecond semester. All the facts needed to solve these problems are containedin the standard university analysis texts, in particular, in the books of Zorich[7], Kudryavtsev [16], Rudin [23], and Fikhtengol'ts [29]. The problems inChapters VIII and IX and §2 of Chapter X require a somewhat higher levelof preparation of the reader and presuppose that he is familiar with the fun-damental concepts of measure theory. The corresponding facts can be foundin the last chapter of the cited text of Rudin and, in more complete form, inthe books [14] of Kolmogorov and Fomin and [4] of Vulikh.

The contents of the first seven chapters, which include about two thirdsof all the problems, do not go outside the framework of the classical topicsof analysis (functions, derivatives, integrals, asymptotics). Both here andin the subsequent chapters we make no attempt at maximal generality and,when we have to choose between a more general and less general formula-tion of a problem, often show preference to the latter. Chapters VIII—X areless traditional for a problem book in analysis. Besides the tastes of the au-thors, the program of the mathematical analysis course adopted at LeningradUniversity served as a criterion in choosing the material for these chapters.Problems going outside its framework and relating to the "theory of func-tions of a real variable" (in spite of the arbitrariness of this division) are notincluded in the book. For example, we do not use many attractive problemswhose solution is based on the Lebesgue theorem on differentiation of anintegral with respect to a variable upper limit. Also;almost no reflections ofproblems connected with complex analysis are found. We refer the readersinterested in this circle of questions to the widely known collection of Pãlyaand Szegô [21] and to the book of Titchmarsh [27].

FOREWORD

We tried to combine problems dealing with separate topics or methodsinto cycles within whose confines it would be possible to exhaust, step bystep, various circles of questions with sufficient thoroughness. Partly for thisreason we were not able to avoid a certain lack of homogeneity in the degreeof difficulty of neighboring problems, which can grow markedly in the courseof a single cycle. Therefore, it is not rare that more difficult problems giveway to relatively simple problems, and the reader who has not solved someproblem must not feel disheartened and can hope in full for success in thesolution of subsequent problems.

Brief but often detailed solutions of most of the problems are given in thesecond part of the book. However, we advise the reader not to be in a hurryto use this part of the book and thus miss the chance of devising a bettersolution than the one presented there.

The literature in analysis and, in particular, textbooks and problem col-lections, contains an enormous amount of material, and we think that onlya few of the problems presented can pretend to be original. We saw it asour goal first and foremost to try to systematize and introduce into everydaypractice problems contained (sometimes in implicit form) in almost inac-cessible sources (especially for students) and in the mathematical folklore.The experienced reader will notice that along with the traditional materialare borrowings from "Matematicheskoe Prosveshchenie," Mathe-matical Monthly," and the collections [19], [22], [24]—[26], [38], and others.Problems are accompanied by references to the literature only in exceptionalcases.

The general editing of this problem book was done by B. M. Makarov.We express sincere gratitude to our friends and colleagues A. B. Alek-

sandrov, D. A. Vladimirov, E. D. Gluskin, Yu. G. Dutkevich, V. V. Zhuk,K. P. Kokhas', M. Yu. Lyubich, G. I. Natanson, A. V. Osipov, A. I. Plotkin,0. I. Reinov, B. A. Samokish, S. V. Khrushchev, and D. V. Yakubovich,whose frequent advice and numerous critical remarks were of great help tous. We are also obligated to them for a number of elegant problems.

The following system was used for numbering problems and for references.Within a single chapter the problems are labelled by two numbers, the firstdenoting the section and the second the problem in that section. In refer-ring to a problem in another chapter we first indicate the chapter number (aRoman numeral). For example, problem Vll.2.5 is problem 2.5 in ChapterVII.

We will be grateful to all readers for their opinions and comments.

The authors

Notation

N is the set of natural numbers;Z is the set of integers;

is the set of rational numbers;R is the set of real numbers;

is the extended set of real numbers;C is the set of complex numbers;R" is the additive n-dimensional space;0 is the empty set;A x B is the direct (Cartesian) product of sets A and B;card(A) is the cardinality of a set A;fog is the composition of mappings (functions) f and g:(fo g)

(x) = f(g(x)).f(A) is the image of a set A under a mapping f;r '(A) is the complete inverse image of a set A under a mapping

1;is the closure of a subset A of the space

B(x, r) is the open ball of radius r about a point x;is the ball B(O, r) in

is the ballis the unit sphere about zero in

(a, b) stands for any one of the four kinds of intervals ([a, b], (a, b),[a,b),or (a,b]);

11, (11) denote the character of monotonicity of a function f(nonincreasing, nondecreasing);

11 A, (fT A) denote the equalities limx..,a f(x) = A(A E for anonincreasing (nondecreasing) function f;

f(x) = O(g(x)) for x E A (or on the set A) means that If(x)I �C is some positive number;

1(x) = O(g(x)) as x a (or f(x) xa O(g(x))) means that

f(x) = O(g(x)) on some neighborhood of the point a;

f(x) g(x) as x —, a (or 1(x) g(x)) means that f(x) =

9,(x)g(x), where —, 1 as x —, a;

NOTATION

1(x) = o(g(x)) as x — a (or f(x) o(g(x))) means that

1(x) = ço(x)g(x) . where 0 as x —, a;a sequence (a mapping defined on the set of integers �

C ER);denotes a series and its sum;

1(x) on A denotes uniform convergence of a sequenceof functions to a function f on a set A;

In the notation connected with sequences, namely,as n —, 00, etc., we often omit the expression n x and

write , etc., let mean and letmean

C(X) is the set of functions continuous on a set X;is the set of functions r times continuously differentiable on

aninterval AcR (0�r�-i-oo);is the set of functions f satisfying the condition 1(x)—

1(Y) = —pr) on the square Ax Ac R2 (a Lipschitz condi-tion on the interval c R with exponent a> 0);

is the set Lipa(R);2°(A) is the set of Lebesgue-measurable functions defined on a

(measurable) subset A of the space R" and finite almost every-where;

gr(A) is the set of functions 1 in 2°(A) such that the functionfir is integrable on the set A;

2°°(A) is the set of functions f in 2°(A) such that the essentialsupremum of fl is finite;

ess supAf is the essential supremum of a function f inis Lebesgue measure on the space R";

2. is the measureis the volume (n-dimensional Lebesgue measure) of the ball B';

[x] is the integer part of a real number x;x mod y stands for the number x — [xfy]v (x. r E R, r > 0);

a.e. is an abbreviation for "almost everywhere";

Problems

CHAPTER I

Introduction

§1. Sets

A binary sequence is defined to be a sequence "consisting of 0's and l's,"that is, a sequence = {ej, where ek = 0 or 1 for any k E N. The set ofall possible binary sequences will be denoted by

In speaking of sets whose elements, in turn, are sets, we use the expression"system of sets."

1.1. Let 9(N) be the system of all possible subsets of N. Prove that:a) the sets and €9(N) have the same cardinality;b) the sets and E x E have the same cardinality.1.2. Prove that the set has the cardinality of the continuum.1.3. Prove that the sets R2 and R3 have the cardinality of the continuum.1.4. Prove that the set of all sequences of real numbers has the carclinality

of the continuum.1.5. Prove that the set of continuous functions on an interval [a, b] has

the cardinality of the continuum.1.6. Determine whether there is a system Qt of subsets of N satisfying

the following conditions:a) 21 has the cardinality of the continuum;b) card(AflB)<+oo forany A,BE21.1.7. Determine whether there is a system Qt of subsets of N satisfying

the following conditions:a) Qt has the cardinality of the continuum;b) for any number I and any sets A, B E 2t the inequality a — bI <t is

valid only for finitely many of the points a E A and b E B.1.8. Determine whether there is a system 21 of subsets of N satisfying

the following conditions:a) 2t has the cardinality of the continuum;b) the system 2t is linearly ordered by inclusion, that is, of any two sets

in Qt, one is contained in the other.1.9. Let 9(N) = {A C NI card(A) < +oo} be the system of all finite

subsets of N. Prove that:a) the system 9(N) is countable;

I INTRODUCTION

K

FIGURE 1

b) there exists a function : —, N such that ço(A) � if A C B(A, B

1.10. a) Define an equivalence relation in the set = (0, +oo) bytaking x y if x/y E Prove that the intersection of each equivalenceclass with any (nonempty) open interval contained in is nonempty.

b) Define an equivalence relation on the circle S' = {z E Izi = 1} by

taking z if = e2lnio , where 0 E Prove that the set of limit pointsof any equivalence class coincides with S1.

1.11. Let S' = {z ECIIzI = 1}. Sets A, BC S1 aresaidtobecongruentif there exists a number a E R such that B = E A}, that is, B "isobtained from A by a rotation through the angle

Prove that there exists a sequence of pairwise disjoint and congruent

sets such that S' =1.12. Prove that the plane it is possible to arrange a continuum of

disjoint figures five, but at most countable many figures eight.1.13. A bird track is defined to be a set on the plane that is the union of

three segments with a common endpoint (the vertex of the track) and lyingon distinct rays (Figure 1). Prove that on the plane it is possible to arrangean at most countable set of disjoint bird tracks.

1.14. By a T-shaped figure we understand a union of two mutually per-pendicular segments with the midpoint of one an endpoint of the other. Ob-tain an upper estimate for the number N of disjoint T-shaped figures formedby segments of unit length and contained in a square with sides of length a.

1.15. Prove that in space it is possible to arrange at most countably manydisjoint "hoops" (cylindrical rings) of fixed radius (the thickness of the hoopsis equal to zero).

1.16. After losing all competitions with Balcia, the devils (there are in-finitely many of them) decided to develop sports skills and organized sportclasses. In each class there were only finitely many devils, but there were somany classes that in any infinite company of demons it was possible to find

SETS 5

at least two registered in the same class. Prove that with the exception offinitely many lazy demons each devil is registered in infinitely many classes.*

1.17. Suppose that a sequence of finite subsets of N thickly cov-ers N, that is, for any infinite set B C N there is an index m such thatcard(B fl Am) � 2. Prove that:

a) the natural numbers belonging only to finitely many of the sets forma finite set;

b) there exists an infinite set E C N such that C {1, 2

1.18. If and are two sequences of finite sets, each thicklycovering N (see problem 1.17), then there exist indices p and q such thatcard(ApflBq) �2.

1.19. Suppose that E CR and (x+y)/2 E E for any x, YE E.a) Is it true that E j [500, 1000] if E j [0, 1] and 1992 E E?b) Prove that if IntE 0 then E is an interval.1.20. Find all limit points of the following sets:

a){n'+m'In,mEN},

1.21. Let E C (0, +oo), E 0. Prove that if x/2 E E and E

E for any x, y E E, then = [0, +oo).1.22. Let E C Prove that:a) if the family {Ba}QEA of open balls is such that E C BQ, then

there exists an at most countable set A0 C A such that E Cb) there exists an at most countable subset of E whose closure contains

E.1.23. a) A set is said to be discrete if any point in it is isolated. Prove

that every discrete set on the plane is at most countable and its closure cannothave interior points.

b) A point a in a set E C R is said to be semi-isolated if there exists anumber e > 0 such that at least one of the intervals (a—e, a) and (a, a+e)does not contain points of E.

Prove that the set of semi-isolated points of any set E C R is at mostcountable.

1.24. Let ECN, cardE=+oo. Provetheexistenceofanumber a>!such that infinitely many of the numbers [a'l (k E N) are contained in E.

1.25. Let G be an open subset of R that is unbounded above. Doesthere exist a positive number x0 such that the set G contains infinitely

many points of the form nx0 (n E N)?

*This refers to a poem of Pushkin well known in Russia: "The tale of the priest and hisworker Bakia (Cxa3Ka o none pa6oTHHKe cr0

I. INTRODUCTION

1.26. Let be a sequence of open subset of R that are unboundedabove. Prove that there is a number x0 > 0 such that each of the setscontains infinitely many points of the form mx0 (m E N).

The Cantor set K is defined to be the intersection of the sets C R(n E N) defined as follows. The set K1 is obtained by removing the middlethird (1/3, 2/3) of [0, 1]. In other words, K1 is the union of the twoclosed intervals

= [0, 1/3] and = [2/3, 1],

which we call the intervals of first rank.The set K2 is obtained by removing the middle thirds of the intervals of

the first rank, that is, the union of the four closed intervals

= [0, 1/9], = [2/9, 1/3], = [2/3, 7/91, = [8/9, 1],

which we call the intervals of second rank.The subsequent construction continues by induction: the set is ob-

tained by removing the middle thirds of the (closed) intervals of nth rank.It is convenient to "enumerate" the intervals of nth rank with the help of theindices es,..., es,, where takes the values 0 and 1. The indices of theintervals of first and second ranks have already been indicated, and the restof the indexing is done by induction. Suppose that the intervals of nth rankhave already been supplied with indices, and / is one of them. Thetwo intervals of (n + 1)st rank obtained after removing the middle third of

are as follows: the left-hand one is denoted by£

and the right-hand one by 1.The set is the union of all possibleclosed intervals of (n + 1)st rank.

1.27. Prove thata) K has the cardinality of the continuum;b) K is closed and does not have isolated points;c) the sum of the lengths of the open intervals making up the set [0, 1]\K

is equal to 1.1.28. Let K be the Cantor set.a) Prove that a number t belongs to K if and only if it is representable

in the form t = where is equal to 0 or 1.b) Describe the sets

K+K={s+tlt,sEK} and K—K={s—tls,tEK}.1.29. Suppose that a set E C R has the property that for any points

x,yEE, x<y,thereexistsapoint ZEE suchthat x<z<v Doestheclosure of E necessarily contain an interior point?

1.30. Construct a discrete set on the plane whose closure has the cardi-nality of the continuum. Does there exist a discrete set on the line with theproperty? (See problem 1.23 for the definition of a discrete set.)

SETS 7

1.31. Let K be the intersection of the sets C IR (n E N) defined asfollows. The set K1 is obtained by removing from the closed interval K =[a, b] a nonempty open interval 5 = (p. q) whose endpoints are differentfrom a and b. In other words, K1 is the union of the two closed intervals

= [a, p] and A1 = [q, b], which we call the intervals of first rank.The set K2 is obtained after removing from and open intervals

and whose endpoints are different from those of and , respectively.The set Ac\St (e = 0, 1) consists of two closed intervals, the left-hand oneof which we denote by Ac0, and the right-hand one by Ac1. Thus, K2is the union of four closed intervals A10, A11, which we callthe intervals of second rank. The subsequent construction is continued byinduction. Suppose that we have constructed the set made up of theclosed intervals of nth rank. It is convenient to "enumerate" the intervals ofnth rank with the help of the indices er..., , where can take the valueo or 1. The indices of the intervals of first and second ranks have alreadybeen indicated, and the subsequent indexing is carried out as follows. Inconstructing the (closed) intervals of (n + 1)st rank we remove from eachclosed interval of nth rank a nonempty open interval whose

endpoints are different from those of Acc

The difference Ac £

consists of two intervals of (n + 1)st rank, the left-hand one of which wedenote by Ac

c 0' and the right-hand one by / The set is the

union of all the intervals of (n + 1)st rank. If the set K = fl1 does nothave interior points, then it is called a generalized Cantor set.

Prove that:a) K has the cardinality of the continuum;b) K is closed and does not have isolated points;c) K is a generalized Cantor set if and only if I,, 0, where I?, is the

maximal length of the intervals of nth rank.1.32. Let {Ac c }' where n E N and can take the values 0 or 1, be

a family of nonempty bounded intervals satisfying the following conditions:

1) Acc2) for n E N the intervals Ac...c and are disjoint if

Let = £Prove that the set A = fl1 has the

cardinality of the continuum.1.33. Prove that a nonempty open interval cannot be represented as the

union of a sequence of disjoint closed sets.1.34. Prove that the plane cannot be covered by a sequence of closed

disks without common interior points.1.35. Suppose that —oo a � +oo and (a, b) C U1 Prove

that the closure of at least one of the sets has an interior point.

1.36. Prove that the set of irrational numbers is not the union of a se-quence of closed sets.

8 1. INTRODUCTION

1.37. Let ii = be a strictly increasing sequence of positive integers,and let = = 0 or 1}.

a) Prove that is closed and does not have isolated points.b) Prove that the following assertions are equivalent:

1) does not contain interior points;2) > + infinitely many times;

3) the expansion i = (where 0 or 1) is unique.

§2. Inequalities

2.1. Prove that for any finite sequence of real numbers there isan index m E {0 n} such that

E ak

m = 0 the first sum is taken to be zero, and for m = n the second).2.2. Let be a finite sequence of positive numbers, and let M =

maxl<k<fl ak and m = minl<k<fl ak. Prove thata)

2n&�1<k<n 1<k<n k

b)

2 1 2(m+M)24mM

1<k<n 1<k<n k

2.3. Let be a finite sequence of nonnegative numbers with S =a1 + a2 + ••. +; < 1. Prove the inequalities:

a) 1/(1 —S) � +ak) � 1 +S;b) 1/(1 +S) —ak) � 1—S.The left-hand inequalities are strict if 5> 0. The right-hand inequalities

are strict if at least two of the numbersa finite sequence of real numbers with > —1 for

k = 1 n. Prove that:a) if S = a1 + + � 0, then

fi1<k<n

with equality possible only in the cases when n = 1 or a1 = .. =

a a

1+a1

INEQUALITIES 9

thena a'7 [J(i+a)1! n!

1<k<n

with equality possible only in the case when a1 = =; = 0.2.5. Let be the sequence of prime numbers, in increasing order

(p1 = 2). Prove the inequalities:a' 1

/ 2/ 3) •

b) 1+ E1<n�m 1/pa> lnlnpm.2.6. For finite sequences and of real numbers let the

expressions and (respectively, and de-

note the nondecreasing (respectively, nonincreasing) rearrangements of thosesequences. Prove the inequalities:

E dzkbk � E akbk � E1<k<n

2.7. Let and be finite sequences of real numbers. Provethe inequalities:

� E � akbk.

In the solution of the following problems the Abe/transformation can turnout to be useful:

E akbk = + (ak — ak+l)Bk,

where Bk = b1 + ... + bk (k = 1 n). It is especially convenient to usethis equality in those cases when the sequence is monotone.

2.8. Let and be finite sequences of real numbers, with

the sequence nonincreasing and nonnegative. Prove the inequalities:

a) a1 mkrn(bl +."+bk) � � a1 max(b1 +... +bk);

b)I

akbkl � a1 maXl<k<fl lb1 + + bkl;

c) m>1<k<fl ak � akbk � M>1<k<fl ak,where m =minl<k<fl +. + bk) and M = maxl<k<fl +... + bk).Here M cannot by a smaller factor, nor m by a larger factor.

2.9. Let and be finite nonnegative sequences, withnonincreasing. Consider an index m E {1, 2 n} such that

m maxl<k<fl bk � bk. Prove that:

E akbk � bk) E1<k�n

L INTRODUCTION

If m maxl<k<fl bk = bk, then

(max bk) � akbk.— — n--m<k<n

2.10. Let be a finite sequence of real numbers. Prove that:a)

2

ak)

where = maxl<k<fl ak — ak+lI,

b) if is monotone, then

52 1 2 1

where 5 = minl<k<fl ak — ak+lI.In what these inequalities become equalities?2.11. Let be a nondecreasing convex (or nonincreasing concave)

sequence of real numbers. Prove the inequalities

ak)

l<k<n

If is nonincreasing and convex (or nondecreasing and concave), thenthe inequalities should be reversed.

2.12. Suppose that and are finite sequences of nonneg-ative numbers. Prove that:

_______

a) if minb<k<fl(bk—ak)=A�O,then

b)

2.13. Prove that the functions x (1 + 1/x)x and x i-' (1 +are strictly monotone on each of the intervals (—oo, —1) and (0, +oo).

2.14. Prove that: a) <n! for n E N; b) n! for n � 11.These inequalities imply that n! = 0 < < 1. More preciserepresentations of factorials are considered in problems 11.2.10 and 11.2.11.

2.15. Prove the following inequalities for any .v E 1]:a) (p�1);b) (1 + (1 — <2P1(1 (p � 2);c) (1 +(1 <2(1 (p � 2);(Which of the inequalities b) and c) is stronger than the other?)d) (1 +x)2+(p— 1)(1 —x)2 <4lhIP(1 (1 �p � 2).

INEQUALITIES

2.16. Let n E N and 1 [0, n]. Prove the inequalities:a) 0 � e' —(1 — � for n � 1;b) t2e'/n2 � e' —(1 — for n � 2;c) e' —(1 for n � 36.d) Prove that

/ —n 2

o�(i+-i _e'<Le'\ n/ —n

for n � 1 and 1 E [0,2.17. Suppose that n E N and 1 E [0, n]. Prove the inequalities

0 — e' (1 — <_L\ ni

The following problem considers relations supplementing the classical in-equalities

sinx <x <tanx, ln(1+x) <x <eX_ 1,

arctanx <x <arcsinx.

2.18. Prove the inequalities:a) (1+x)1n2(1+x)<x2 for x>—1,

x > 0;c) x2 <In(1 +tan2x) <(sinx)tanx for x E (0, ir/2);d) 3x—x3<2sin(irx/2) for xE(0, 1);e) x3 < (sin2x)tanx for x E (0, ir/2).f) Is the inequality <(sin tan x true on (0, ir/2) for some fixed

positive number e?2.19. Prove the inequalities:a) (ex — 1)ln(1 +x) > x2 for x > 0;b) (tanx)arctanx>x2 for xE(0,1r/2);

(What happens for p <0?)d)((1—x)"—1)(1—(1+x)'1")>x2fOrXE(0,1)andP<—1.(What happens for p E (—1, 0)?)2.20. Prove the inequalities

(sin arcsinx <x2 <(sinx) arcsinx

for xE(0,1).

Problems 2.21—2.27 concern integral inequalities which are analogues ofthe summation inequalities already considered. They can be proved by twopaths: either by taking a limit in a summation inequality, or by modifying

12 L INTRODUCTION

the proof used in the discrete case. We remark also that by passing to thelimit, the inequalities in problems 2.23 and 2.24 can be carried over to thecase of improper integrals over an interval of the form [a, +oo).

2.21. Suppose that the function f is integrable on [a, bl and m =inff> 0, and let M = supf. Then the following inequalities hold:

a) ljjdx�

b)

(b - a)2� jbjb

� (b -

2.22. Suppose that the functions I and g are monotone on [a, b]. Ifthe monotonicity is of a single type, then

f(x)g(a + b — x)dx�

f(x)dxj g(x)dx

� j f(x)g(x)dx.

If the monotonicity is of different types, then the inequalities are reversed.2.23. Suppose that the functions I and g are integrable on [a, b],

fj, andf� 0. For x E [a, b] let G(x)= f'g(t)dt. Then the followinginequalities hold:

a) f(a)infa<x<b G(x) � f f(x)g(x) dx � 1(a) supa<y<b G(x);

b)I f f(x)g(x) dxl � 1(a) supa<x<b

c)

(inf1b

f(x) dx<

f(x)g(x) dxa<x�bX—a a a

I G(x)\ 1b� I sup —i f(x)dx.

\a<x�b' —a1 Ja

2.24. Suppose that the functions f and g are integrable on [a, b], Iand 0 g 1. Let c = f g(x)dx Then

/ / f(x)g(x)dx� / f(x)dx.Jb--c Ja Ja

IRRATIONALITY 13

2.25. Suppose that f E C'([a,b]), 5 = min(abjlf'I, and A =maxiabjifi. Then

— a)2ô21 j 12 (x)dx —

jb)�

Equality holds only for linear functions.2.26. If a continuously differentiable function f is nondecreasing and

convex (or nonincreasing and concave) on [0, 1], then

-x)3(f(x))2dx � jf2(x)dx_ (ji)2

� x3(f'(x)f dx.

For nondecreasing concave (or nonincreasing convex) functions the inequal-ities are reversed. Equality is possible only for a linear function.

2.27. Suppose that the functions f and g are positive and integrable on[a, b]. Then the following inequalities hold:

a)

exp1b

lnf(x)dx) f(x)dx

(an analogue of the Cauchy inequality on the arithmetic mean and the geo-metric mean);

b)

exp Inf(x)dx) + exp In

� exp1b

ln(f(x) + g(x))dx)

(an analogue of Minkowski's inequality; see problem 2.12b)).2.28. Let K be the set of all nonnegative and nonincreasing functions f

on [a, b] such that af(a) = bf(b) and f f(x)dx = 1. Prove that for anyfunctions f and g in K

2v'E

Iamax(f(x), g(x))dx�

For what functions is equality attained?

§3. Irrationality

3.1. Let x E IlL Prove the equivalence of the following assertions:

a) x

14 1. INTRODUCTION

b) there exist infinitely many irreducible fractions p/q (q E N, p E Z)such that Ix — < 1/q2.

c) there exists a sequence of irreducible fractions (qk E N,E Z) such that —, +oo and x — = as k —, +00.

In the following problems we use the notation ö(x) = x — [x] for thefractional part of a real number x.

3.2. Suppose that x E Prove that the sequence {ö(nx)} is densein (0, 1).

3.3. a) Prove that the sequences {sin n2} and {sin 4fl} do not have limits.b) Prove that if the sequence has a limit, then it is equal to

zeroand qEN and p EZ.3.4. Suppose that x E R. Prove the equivalence of the following state-

ments: a) x E b) the set {ö(n'992x)In E N} is finite.3.5. Show by examples that the sequence can be dense in

(0, 1), and that it can be nowhere dense.3.6. Find the set of limits of all convergent subsequences of the sequence

2/3 . . 3/2a) {sin n }; b) c) {sin irn };

d) {sinlnn}; e) {sin(irnlnn)}.

3.7. Find the set of limits of all convergent subsequences of the sequence:a) {ö(nx)}, where x E Q; b) {ö(nx)}, where x Ec) {ö(na)},where 1);d) {ö(n512)}.3.8. Does there exist a real number x such that 6(x'1) E [1/3, 2/3] for

all n?3.9. Suppose that f E C([0, 1]) and x E Prove that

f(ö(kx))-4Jf(t)dt1<k<n 0

as n—'-i-oo.3.10. Suppose that x = (x1,..., x,,,) E Rm\Qm. Prove that there exist

integers p1..., and a positive integer q such that

1x1 — p1/qI<q_(l+l/m) (i = 1. 2 in),

and the denominator q can be taken as large as desired.3.11. Let = minflEZ — in/ni, n E N. Prove that:a) (2n)2 <

b) if a strictly increasing sequence of indices is such that cxn1 � C/nj(j E N), where C is a fixed number, then increases at least like ageometric progression: � 1 + 1/(8C) (thus, the lower estimate inthe inequality in a) is coarse for most indices).

§3. IRRATIONAUTY 15

In problems 3.12—3.18 {ek} is an arbitrary sequence of +1's and —l's,and is a strictly increasing sequence of positive integers.

3.12. Prove that the sum of the series is irrational if

lim(nk+l — = +00.3.13. Let lim(n1n2 .. = 0. Prove thata) the sum of the series >(_l)k/nk is irrational;b) if k E N, then the sum of the series is

irrational.3.14. Suppose that = +00. Prove that:

a) the sum of the series is irrational;b) k E N, then the sum of the series is

irrational.3.15. Suppose that is an algebraic number of degree at most n (that is,is a root of an algebraic polynomial of degree n with integer coefficients).

Prove that if is irrational, then there exists a number ca > 0 such thatforall PEZ and qEN.

3.16. Suppose that lim = +00. Prove that:

a) the number >(_l)k/dc is not a quadratic irrational;b) if k E N, then the number is not a

quadratic irrational.3.17. Prove that the numbers (the Liouville numbers) are tran-

scendental.3.18. Suppose that lim 1. Prove that:

a) the sum of the series is a transcendental number;b) if k E N, then the sum of the series is

a transcendental number.3.19. Using the expansions e = 1/n! and e' =

prove that:a) the number e is irrational;b) if A,B,and C areintegersnotallequaltozero.3.20. Irrationality on the number it.a) Let

1cn/2 2"1=—! I——t costdt (n=0,1,2,...).

/

Prove that = where is an algebraic polynomial of degree at

most n with integer coefficients.b) Derive from this that it2 (and hence also it) is an irrational number.

3.21. Irrationality of the values of the exponential at rational points.

16 1. INTRODUCTION

a) Let

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

Prove that = + where and are algebraicpolynomials of degree n with integer coefficients.

b) Derive from this that er Q if r E Q, r 0.3.22. Irrationality of the values of the tangent at rational points.a) Let

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

Prove that = sinx where and are algebraicpolynomials of degree at most n with integer coefficients.

b) Derive from this that tanr if r E Q, r 0.

CHAPTER II

Sequences

§1. Computation of limits

1.1. Let = Find the limit1.2. Let = — l)!!/n!!. Find

lim lim

the limit the sequence where

= [J sinl<k<n

1.5. Let a ER and = >1<k<fl((a + Find the limit1.6. Leta) Find the limitb) Determine the character of monotonicity of the sequencec) Is it true that � 2 for any n E N?1.7. Compute the limit lim + 1)(n + 2).. (n + n).1.8. Suppose that the sequence of positive integers is such that

kr/n —, p, � n. Compute the limit lim1.9. Find the limits:

a) lim b) lim —

k . .rc) lim d) lim sin

O�k<2n

e)

1.10. Find the limits:

a) limsin(2ren!); b) limnsin(2ren!);2. p. ,—

c) limn sin(2ren!); d) limn sin(r(v2+1) ).

II. SEQUENCES

1.11. Suppose that the nondecreasing positive sequence has finitelimit a. Prove that the series converges if and only if x1x2

for some C>0.1.12. Let a > 0 and let = -i--i- (n square roots).

Prove that: a) —, 'a = + b) — Ca/(21a)'1 for some

Ca>0. Whatisthevalueof C2?

1.13. Let = + + .. + (n cubic roots). Prove that

C/(12)'1 for some C >0.

1.14. Let p> 1 and = + (n roots). Prove thatthe sequence converges to the positive root of the equation 1 =0.

1.15. Suppose that p > 1 and is a nonnegative sequence. Let

= + .. . + for n E N. Prove that the sequence

converges if and only if the sequence In is bounded above.

1.16. Prove that:

a)

b)

c) cosx = +• for lxi � ir/2;

d) 9 = + + where is the se-

quence of Fibonacci numbers: u1 = u2 = 1, and = + forn>2;

e) 2x2 = + + +•• for x � 1, whereis the sequence of values of the Chebyshev polynomials at the point x,

namely, = 1, = x, = — for n � 1.

1.17. Let = + + ... + Prove that ; —, a E R, and

1.18. Prove that —, 0 if — (1/2); —, 0.1.19. a) Prove that + � e for any positive sequence

b) For any number a � e determine a positive sequence such that((a1 —, a.

1.20. Let be a sequence of real numbers such that an+m � +am.Prove that the limit E R exists and equals

1.21. Suppose that the sequence of real numbers is such that —

—, 0. Prove that the set of limits of its convergent subsequences is theinterval with endpoints and

§2. AVERAGING OF SEQUENCES 19

1.22. a) Suppose that the sequence of real numbers is such thatxfl.F1 + —, 0. Prove that the set of limits of convergent subsequences ofthis sequence is either infinite or contains at most two points.

b) Suppose that the sequence + xflH +"• + is convergent.Prove that the set of limits of convergent subsequences of the sequencecan be represented as a union u U U , where the i\ are closedintervals in IL

1.23. Suppose that the sequence of real numbers is such that —

Prove that1.24. a) Give an example of a bounded divergent sequence such

thatb) Does there exist a bounded sequence such that xflF1 — —, 0,

but the sequence {(x1 + + does not have a limit?1.25. Let be a sequence of real numbers such that the limit

exists for any C> 1. Prove that has a limit.

§2. Averaging of sequences

2.1. Let be a sequence of real numbers. Prove that +•—, L if —, L. The converse is true only if k(xk — Xkl) = o(n).

2.2. Suppose that � 0 and = ak +00. For an arbitrary

sequence of real numbers let = akxk. Prove that

� � �In particular, —, a if —' a (for = 1 this implies the first assertionin problem 2.1).

2.3. Suppose that A is an infinite subset of N, A = {n1, n2, .. . } and

let = card{m E Aim � n}. If the limit 0(A) = lim exists, then it iscalled the density of the set A (or the density of the sequence {nk}). Provethat:

a) 0(A) = lim(k/nk);b) foranynumber E [0, 11 thereexistsaset Ac N suchthat 0(A) =2.4. Let be a sequence of real numbers. Prove that:a)if then 21<k<fllxk—aI -4 0;

b) if iXk —al —, 0 for some a ER, then there exists a sequence{nk} of density! (see problem 2.3) such that —, a;

c) if is bounded, and —, a for some sequence {nk} of density

1, then2.5. Let <1 for nEN.a) Show by examples that the existence of one of the limits

lim !(a1 + ... + or lim + ... +

does not imply the existence of the other.

20 IL SEQUENCES

b) Suppose that (a1 + + a and + + b. Prove

that a2 � b � a and that b can take any values between a2 and a.2.6. Suppose that and = +00. If the sequence

of real numbers is such that the limitx—x —

limyn

exists, then —, 1 (a theorem of Stoltz, a discrete analogue of UHôpital'srule). Formulate and prove a similar assertion for indeterminants of the type0/0.

2.7. Prove that:

a) inn; b) ka fl—1);

i<k<n

d)Ink Inn i—a

k>n i<k<n

Ink Inn i—a

k>n

2.8. Prove that:

a) >0); b) <0);

1<k<n k�n

c) (a>0); d)

l<k<n i<k�n

e)

k>n

The results in problems 2.9 and 2.10 play an important role in mathemat-ical analysis. They will be used in the solution of many of the subsequentproblem.

2.9. Prove that >i<k<fl = Inn + y + o(1) for some number y. Thenumber y(y = 0.5772. .)is called the Euler constant.

2.10. Prove that ln(n!) = = Withthe help of the Wallis formula show that C = which implies theStirling formula: n!

2.11. Prove that for all n E N

<n! <

In particular,

RECURSIVE SEQUENCES

2.12. Prove that as n—'+oo:

a) kP 1n1+P++(l) ifpE(—1,O];1<k<n

b) ifpE(O,1];l<k<n

c) 1n1+P+1P+P (1) ifpE(1,2].

2.13. Provethatas n—'-j-oo:1+p

a) ifp>—1;l<k<n

b)l<k<n

c)

i<k<n

2.14. Prove that:

a) Ei<k<n

b) ink! = + 2n + Inn — 3n2

i<k<n

+ 1)n+C2+o(1).

§3. Recursive sequences

3.1. Let x0 = a, x1 = b. and = + (n E N). Provethat the limit exists, and find it.

3.2. Let x0 = a and x1 = b. Find the limit in the followingcases:

b) = (1 — + (n E N).3.3. Let x1 E R and = (2+ — 1.

a) For what values of x1 does the sequence diverge? What is theasymptotic behavior of

b) For what values of x1 does converge? What is3.4. Suppose that b E R, a E R, and xflF1 = Prove that:a) if bk 1;

b) if IbI> 1 and

c) SbThi if IbI> 1 and S = X1 + >ak/b 0.

22 IL SEQUENCES

3.5. Suppose that the sequence and the number p E R are suchthat —p);1 for any n = 2,3 Prove that if p >0,then the limit E exists, but for p 0 this assertion is false forsome sequences.

3.6. Suppose that the positive sequence and the number p E R are

such that � for n = 2, 3 Prove that if p E (0, 2), thenconverges, while for any other values of p this assertion is false for

some sequences.3.7. Suppose that the sequence of nonnegative numbers is such that

� + for n = 3, 4 Prove that = 0(1/n!).3.8. Suppose that k E N, 0 < /3 <cr/k, and is a nonnegative

sequence. Prove that:

a) if; = 0(Xn_i

then =

0 , then =

(n!)a )3.9. Suppose that

kEN,

and is a nonnegative sequence such that = +•

Prove that = if 0< /3 < 1/(k(1 + 0)).3.10. Suppose that k E N, 1 +00, A1 > 1 , and is a nonnegative

sequence such that � + ... + for n > k. Prove that= 0(e3'(A1 .. . where = In particular,

= 0((1 for any e >0; but if the seriesconverges, then = 0((A1 .

3.11. Suppose that k E N, a> 0, and is a nonnegative sequence.Prove that:

a) if � + . + xflk)/(ln n)°, then

((10

forany e>0;

§3. RECURSIVE SEQUENCES 23

b) if; � +••. + Xfl_k)/fl , then

o exp(

k

,,) forOczo<k,

for > k;c) if � + ... + xfl_k)/e , then =d) if � +"+Xflk)/(fl!) , then

an/2k

).

CHAPTER III

Functions

§1. Continuity and discontinuities of functions

1.1. Describe the sets of functions f: R -+ R having one of the followingproperties (e,ö,x1,x2ER):

a) Ye 3ö>0b) Ye>0 3öc) Ye>0 3ö>0d) Ve>0 Vö>0e) Ye>0 3ö>0f) Ye>0 3ö>0g) Ye>0 3ö>0h) 3e > 0 Yö >0 (1x1 — x21 <ö If(x1)—f(x2)I <e);i) Ye>0 3ö>0

1.2. Let the function f be defined on R. Prove that the following prop-erties of f are equivalent:

a) I is continuous on R;b) the set f'(G) = {x E RIf(x) E G} is open for any open set G Cc) the set f'(F) = {x E RIf(x) E F} is closed for any closed set

FcR;d) for any c ER the sets c)) and r'uc, +oo)) are (men;e) for any c E R the sets c]) and r'uc, +oo)) are closed.f) Is the condition "for any c E R the set r'({c}) is closed" sufficient

for the continuity of f?

1.3. Prove that any function defined on a set E C R has an at mostcountable set of points of clisconunity of the first kind.

1.4. Let EL and ER be countable subsets of [0,1] with EL fl ER = 0.Define a function on [0,1] that is continuous only from the left at points ofEL, continuous only from the right at points of ER, and continuous at theremaining points of [0,1].

1.5. Let be the set of points at which a function f onR is discontinuous from the left (respectively, from the right). Is it true

26 III. FUNCTIONS

that if one of these sets is at most countable, then the other is also at mostcountable?

1.6. Let 10 be the Dirichlet function:

11 for xEQ,f0(x)

= 1 o for x E R\Q.

a) Prove that f0(x) = for any x E R.b) Does there exist a sequence {fj of continuous functions on R such

that f0(x) = for any x ER?1.7. Prove that the set of points of discontinuity of an arbitrary function

on an interval C R is an P-set, that is, the union of a sequence of closedsets. Is this true if the function is defined on an arbitrary metric space?

1.8. Define on R a function with a given set E of points of discontinuity

a) E is a closed set;b) isaclosedsetforany nEN.1.9. Define on the interval (0, 1) a function such that any nonempty

open interval C (0, 1) contains a continuum of its points of continuityand a continuum of its points of discontinuity.

1.10. Prove that a function f defined on an interval (a, b) is continuousif and only if:

a) f has the Cauchy property (that is, the image of each interval (p, q) C(a, b) is an interval);

b) the set f'({y}) is closed for any y E R.1.11. Prove that a function on an interval [a, b] is continuous if and

only if its graph is connected and closed.1.12. Let E C R, and let I be a bounded function on E that has a

local extremum at each nonisolated point of E. Prove that the set 1(E) is

then at most countable. Is this true if E is a separable metric space? Anarbitrary metric space?

1.13. Describe all the continuous functions on an interval [a, b] havinga local extremum at each point of (a, b).

1.14. Suppose that a function 1 is defined and one-to-one on (a, b) andcontinuous at a point c E (a, b).

a) Is r' necessarily continuous at the point 1(c)? Does alwayshave one-sided limits at 1(c)?

b) Is r' continuous at the point 1(c) if I is strictly monotone?1.15. Let f E C((a, p6)). Prove that f is one-to-one if and only if it is

strictly monotone.1.16. Suppose that a positive function I is defined and differentiable on

[a, +oo). Prove that if infv>a f'(x)/f(x) > 0, then:a) f(x)=o(f((1+ö)x))is x—'-i-oo forany ö>0;b) f'(z) f'(ez) as —' +oo for any e >0.

CONTINUITY AND DISCONTINUITIES OF FUNCTIONS 27

1.17. Prove that if f E C([0, +oo)) and the limit f(nx) existsfor any x � 0, then the limit f(x) exists. Prove this if the limit

f(nx) exists only for points x in some nonempty closed set withoutisolated points.

1.18. Define on [0, +oo) a function f satisfying the following condi-tions:

a) = 0 for any x E [0, +oo);b) f is bounded on any finite interval and has discontinuities of the first

kind only;c) the set of limit points of f at infinity fills R (compare with problem

1.17).1.19. Define on [0, +oo) a function f satisfying the following condi-

tions:a) f(nx) = 0 for any x E [0, +oo);b) the set of values of I on any nonempty open interval C [0, +oo)

fills R (compare with problem 1.17).1.20. Prove that if f E C([0, +oo)) and f(x+h)—f(x) —, 0 as x —, +00

for any h E R, then f(x + h)— f(x) 0 as x —, +00 on any finite interval,and hence f is uniformly continuous on [0, +oo).

1.21. Let 5 > 0 and f E C([0, 1]). The graph of f is said to havea horizontal chord of length ö if there is a point x E [0, 1 — ö] such that1(x) = f(x + ö). Prove that if f(0) = 1(1), then for any n E N there is ahorizontal chord of length 1/n. Show that for chords of a different lengththis is not true in general.

1.22. Prove that if fE C(R) and f(x+h)— 2f(x)+f(x—h) —, 0 as

h —, +oo for any x E R, then f is a linear function.1.23. A function 1: R —, R is said to be Cësaro-continuous if the condi-

tion +. . . —, x0 implies —(f(xg)+• . . —'1(x0). Describeall functions that are Cësaro-contiuous.

1.24. Let f(x) = 11k>O(1 + x < 1. Prove thata) there exist C1 > 0 and C2 > 0 such that c1 � � for all

xE[0, 1);b) the function is not monotone on the interval [0, 1);c) does nothavealimit as x—' 1—0.1.25. Does there exist a continuous function on [0, 1] such that all sets

of constancy are countable (have the cardinality of N)?1.26. Suppose that the function I is defined and separately continuous

on the square Q = [0, 1] x [0, 1]. Prove that there is a sequence {fj offunctions continuous on Q such that f(x, y) y E[0, 1] (that is, I is a function of the first Baire class on Q).

1.27. Let f be defined and separately continuous on R2. Prove that if

1 vanishes on a dense subset of R2 (that is, a set whose closure coincides

with R2),then fEO.

28 III. FUNCTIONS

§2. Semicontinuous functions

Let C R be an arbitrary intervaL A function f defined on and pos-sibly taking the value —00 (+oo) is said to be lower (upper) sernicontinuousif 1(a) � 1(x) (1(a) � for any point a E

2.1. Suppose that a function f is defined on R. Prove that the followingproperties of f are equivalent (compare with problem 1.2):

a) f is lower semicontinuous;b) the set f'((c, +oo)) is open for any c E R;c) the set f'((—oo, c]) is closed for any c E R;d) if —, a, then 1(a) � supflENe) for any point a E R and any number e > 0 there is a neighborhood

V of a suchthat f(x)�f(a)—e for xEV.2.2. Prove that:

a) the characteristic function of a set E C R is lower semicontinuousif and only if E is open;

b) the function f: R R defined by

0 if x is an irrational number,

f(x)= 1/n ifx=rn/nwherenEN,rnEZ,and rn/n is irreducible,

is upper semicontinuous.

2.3. Prove that if a function is lower semicontinuous on [a, b] and takesonly finite values, then it attains the smallest value on [a, b] (and hence isbounded below). Does it have to be bounded above?

2.4. Let g be an arbitrary function defined on an interval C R, andlet 1(x) = g(y) (x E - Prove that f is lower semicontinuous.

2.5. Let be a family of lower semicontinuous functions definedon an interval C R, and let 1(x) = supQEA fQ(X) < +oo for any x E

Prove that f is lower semicontinuous. Is this true for the functionh = infQEA ia? Is h lower semicontinuous if the family {fa}iE4 is finite?Countable?

2.6. Prove that a function f: [a, b] —, R is lower semicontinuous if andonly if I = E C([a, b]). g � f}.

§3. Continuous and differentiable functions

3.1. Let = = f(x+h)—f(x)a f a polynomial of degree at most rn if and

only if = 0 for any x, h E ILL

3.2. Let f E C'([O, +oo)). Prove that f'(x) + 1(x) —, L E R as x+00 if and only if f(x) —, L as x —' +00 and 1' is uniformly continuouson

CONTINUOUS AND DIFFERENTIABLE FUNCTIONS 29

3.3. Suppose that f is differentiable on [a, b]. Prove that if f'(a).

f'(fl) < 0, then there exists a point c E (a, b) such that f(c) = 0.3.4. Suppose that a function f is differentiable on [a, b]. Prove that

if the set (f)'(x) = {y E [a, b]If'(y) = x} is closed for any x ER, thenIEC'([a, b]).

3.5. Suppose that a function f is defined on [a, b]. Prove that I E

C'([a, b]) if and only if the ratio + h) — 1(x)) tends to a finite limituniformly on [a, b] as h 0.

3.6. Let f E C((a, b)), and suppose that for any x E (a, b) the limitf(x — h)) = g(x) exists and is finite.

a) Provethatif g�0 on (a,b),then f isincreasing,andifthen f is constant.

b) Provethatif gEC((a,b)),then IEC'((a,b)).3.7. Suppose that functions f and g are defined on (a, b) and satisfy

the condition that for any x E (a, b) there exists a positive number > 0

such that f(x+h)—f(x—h)=2hg(x) for 0<h Provethatif I is

differentiable, then it is a polynomial of degree at most two. Is this true if Iis continuous? If both 1 and g are continuous on (a, b)?

3.8. Suppose that F C R is a closed set without interior points. Provethat there exists a strictly increasing function f E C'(R) such that f'(x) = 0

if and only if x E F.3.9. Suppose that f E C([0, 1]). Determine whether the following state-

ments are equivalent:

a) I is constant on [0, 1];b)

—x

x

Suppose that I is defined on the interval (a, b) C R - The variation

of a function f on (a, b) is defined to be the supremuin of allpossible sums where xk <xk+l, xk E (a,b),k = 1 n. The function I is called a function of bounded variation

on (a, b) if <+00. The symbol V((a, b)) denotes the set of allfunctions having bounded variation on (a, b).

3.10. Let fE V([a, b]), and let g(x)= Var(f) (x E (a, b]), g(a)=0. Prove that if I is continuous at a point c E [a, b], then so is g.

3.11. Let I E C'([a, b]). Prove that = f dx.

30 III. FUNCTIONS

3.12. Suppose that 1 < 2, let 1(x) = x2 for 0 < x � 1,and let 1(0) = 0. Prove that f V([0, 1]) and f is differentiable on[0, 1], but the function g(x) = Var(f) is not differentiable at zero. Doesg'(O) exist for = 1?

3.13. Supposethat for 0<x�1and f(0) = 0. Prove that:

a) f E V([0, 1]) if and only if /3;b) I E Lip1([0, 1]) if and only if � /3 + 1;c) if < /3 + 1, then I E Lip7([0, 1]), where y = cx/(/3 + 1).

3.14. Let 0 < y < 1. Construct a function f E 1]) not havingbounded variation on any nondegenerate interval C [0, 1].

3.15. Find functions f, g E C([0, 1]) such that:a) f E V([0, 1]) and I 1]) for any 0;b) g V([0, 1]) and g E 1]) for any < 1.3.16. Suppose that a function f: R —, R is continuous and 2ir-periodic,

and letf(x — t)co(x)=j

s/iTdt (xER).

Prove that E Lip112; more precisely, prove that 19,(u) — �For xE[0, 1] let

k�12

The graph of I is pictured in Figure 2. The function 1 is called the Cantorfunction (and its graph the "Cantor staircase").

3.17. Prove that:a) if x = Ek�1 2ek/3k, where ek = 0 or 1, then 1(x) = ek/2,

'1I

4 I

I I I

I .__i I I I

4 I I I

I IIIO 278i993

FIGURE 2

CONTINUOUS MAPPINGS 31

b) JEC([0,1]);c) on the middle third of each interval defined in the construction

of the Cantor set (see §1.1) the function f is constant and takes the value1

3.18. Find the length of the graph of the Cantor function.3.19. For what a> 0 is the Cantor function in the class 1])?3.20. Let K be a generalized Cantor set, and the closed intervals

of nth rank appearing in the definition of K (see the of a gener-alized Cantor set in problem 1.1.31). Further, suppose that a = inf K, b =sup K, and are the open intervals removed from the closed intervals

in constructing the closed intervals of (n + 1)st rank. Define the func-

tion on [a, b] as follows: = 0, = 1/2 if XE [a,

1

if XEK, a<X.Thefunction willbe called the Cantor function corresponding to the set K.

a) Prove that is increasing and continuous on [a, b].b) Prove that

for any point X0 E K and any number h > 0 is increasing at any pointof K).

3.21. Suppose that F C R is a closed set without isolated points. Provethat there exists a function f E C(R) such that:

a) f'(X) = 0 for any X E R\F;b) f(X + h) — f(X — h) > 0 for any X E F and any h > 0.

A function f on an interval containing a point c is said to belong to theclass Lip(a, c) (cr> 0) if there exist positive numbers L and ö such thatIf(x)—f(c)I � for any point x in the intersection , c+ö).

3.22. Supposethat is

the set of critical points of 1. Prove that:a) if 1' E c

f then

§4. Continuous mappings

4.1. Suppose that P(z) = + a1 z + am? (z E C). Prove that theimage P(F) of any closed set F C C is again a closed set. Is this true foran arbitrary polynomial in two variables?

32 IlL FUNCTIONS

4.2. Let F0 = {z E Clizi = 1, z —1} and ço(z) = Imz/(1 +Rez).Prove that is a one-to-one mapping of r'0 onto R. Find

4.3. Does there exist a partition of [0, 1] into two disjoint sets A andB such that each can be mapped onto the other in a one-to-one bicontinuousmanner (homeomorphically)?

4.4. Define functions and on [0, 1] as follows. Let t E [0, 1],

fl2 = a3 if a2 is even, and fl2 = 2 — a3 if is odd; fly, = a2n_la2 + a4 + a2n2 is even, and = 2— a2n_l if a2 + a4 + a2n_2

is odd (n � 2). Similarly, = E where: = a2 if a1 is even,and y1 =2—a2 if a1 isodd; —a2, if iseven,and

= 2— if + a3 + ... + a2nl is odd (n � 1).a) Prove that and w are well defined and continuous.b) Let f: [0, 1] -' R2 be the mapping with coordinate functions and

vi (f(t) = vi(t)) for t E [0, 1]). Prove that f maps [0, 1] onto thesquare [0, 1] x [0, 1]. The mapping f is described schematically in Figure3. Each of the intervals k = 1 9. is mapped onto the square withthe same index (Figure 3(a)). Figure 3(b) shows the circuit of the unit squarewhen the interval is partitioned into 92 parts.

c) Prove that the system

J =

1. vi(t) =

of equations has at most four solutions for any E [0, 1]. Find andsuch that this system has four solutions.d) Prove that the sets of constancy of and vi (that is, the sets

and c E R) do not have isolated points.e) Prove that E Lip112([0, 1]) ,but 1]) if 1/2.4.5. A continuous mapping of a closed interval onto a square is called

a Peano curve. Let u and v be the coordinate functions of an arbitraryPeano curve defined on [0, 1]. Prove that if u, v E 1]), thena � 1/2. (Thus, the smoothness of the coordinate functions of the Peano

(a)

A

(b)

iii

t

o 119 2/9 3/9 419 519 619 719 8/9 I

FIGURE 3

§5. FUNCTIONAL EQUATIONS 33

curve constructed in the preceding problem is the best possible.)4.6. Prove that any two generalized Cantor sets are homeomorphic, that

is, a one-to-one bicontinuous correspondence can be established betweenthem.

§5. Functional equations

5.1. Describe all the functions I E C(X) satisfying the equation f(2x)= 1(x) for all x E X in the following cases:

a) X=R;b) X=(0,-i-oo).5.2. Suppose that the function 1: R —. R satisfies the equation f(x +y)

= 1(x) + 1(y) (x, y ER). Prove that:a) f(rx)=rf(x) xER);b) if f is bounded above on some nonempty open interval, then it is

linear, that is, 1(x) = ax (x ER), where a =c) if f is discontinuous at at least one point, then its graph is dense in

R2.5.3. Describe all the monotone functions 1: (0, +oo) — R satisfying the

equation f(xy) = 1(x) + 1(y) (x, y > 0).5.4. Describe all the functions f E C(R) satisfying the equation f(xy)

= xf(y) +yf(x) for any x, y ER.5.5. Find all functions 1: R — R satisfying the system of equations

f(x+y)=f(x)+f(y), f(xy)=f(x)f(y) for x, yER.5.6. Let n E N, n> 1. Describe all the functions f: R —, R such that

forany x, yER.

5.7. Suppose that for some a > 0 the function I E C(R) satisfies the

condition 11(x) + 1(y) — f(x +y)I � a for any x, y ER. Prove that f isrepresentable as the sum of a linear function and a function not exceeding ain absolute value.

5.8. Suppose that the function I E C(R) satisfies the condition f(x) +

f(y) = for any x, y E R. Prove that 1(x) = ax2 for anyxER,where a =1(1).

5.9. Suppose that the function f E C(R) satisfies the conditions f 0

and f(x)f(y) = for any x, y E R. Prove that f(x) = e°'2

for any x E R (a is a fixed number).5.10. Let a > b > 0. Describe all the functions f E C((0, +oo)) such

that the difference [(ax) — f(bx) does not depend on x E (0, +oo).5.11. a) Describe all the functions E C((0, 1]) satisfying the condition

for 0<x�1.b) Describe all the nondecreasing functions f E C((0, 1)) satisfying the

condition 1(x2) = (f(x))2 for 0 <x < 15.12. Find a continuous strictly increasing function f: R —, R such that

f(x2)=(f(x))4 forany xER.

34 III. FUNCTIONS

5.13. Let = {z E Clizi � 1}. Find all the continuous one-to-one map-pings f of the disk onto itself satisfying the condition f(z2) = (f(z))2forall ZEB.

5.14. Prove that there exists a unique monotone function 1: [0, 1] —' R

satisfying the conditions 1(x) = 2f(xf3) and 1(x) + f(1 — x) = 1 for anyXE[O,1].

5.15. Find all the functions 1 E C2(R2) satisfyingtheconditions f(x, y)and Z) foranyx,

y, ZER.5.16. For which continuous functions 1: R R does there exist a func-

tion g: R2 —, R such that f(xy) = g(x, 1(y)) (x, y ER)?

Suppose that G and G' are arbitrary groups (with the multiplicative no-tation for the group operation). A mapping G — G' is called a homo-morphism if it preserves the group operation, that is, if =for any x, y E G. If S' is the group of complex numbers with modulus 1,then a homomorphism of G into S' is called a character of the group G.

5.17. Find the general form of a character for the following groups:

a) Z; b) Z x Z; c) Zm(the group of residues modulo m).

5.18. Find the general form of a continuous character on the following (ad-ditive) groups: a) R; b) R x Z; d) C. The same for the following(multiplicative) groups: e) S'; f) = (0, +oo); g) = C\{0}.

5.19. Describe all the continuous homomorphisms of a group G intoa group G' in the following cases (the notation introduced in the precedingproblem isused): a) G=R, G'=R;b) G=R,

G=R, G'=S';f) G=C,

CHAPTER IV

Series

§1. Convergence

1.1. Does the series(') ejn converge, where = 0, if the decimalexpression for n contains the digit 9. and = 1 otherwise?

1.2. Do the following three series converge:

a) b) c)

1.3. Prove that the series In converges only for a> 1.1.4. Determine the values of p > 0 for which the following series con-

verge:

n>2fl +PiMu)

,,c) i'(n + 1)'

where i'(n) is the number of digits in the decimal expression for n.1.5. Determine whether the following series converge:

a) b)

c) E d) E — (n—

Here is the number of divisors of n, is the nth prime number, andço(n) is the number of positive integers � n that are relatively prime to n.

1.6. Prove that for a> 1 the series1

I

converges.

(1) Recall that stands for the series

36 IV. SERIES

1.7. Determine whether the following double series converge:

a)2

1

2' b)min(n,m)

n,m�1 +m

n +m +m

+m n +mLCM(n m) 1

g)n4+m4

h)(LCM(n,

m) is the least common multiple of the numbers n and m,GCD(n, m) is their greatest common divisor, and e(n, m) = 1 when n andm are relatively prime and 0 otherwise.

1.8. Verify that the sum of the series

11 111111111111111111

(the group of terms + — is repeated times) is equal to 1. Finda rearrangement after which the sum becomes equal to —1.

§2. Properties of numerical series connected with monotonicity

2.1. Suppose that > 0, 1. Prove that the series E arccos2

converges if and only if the sequence is bounded.2.2. Let j 0. Prove that:a) the series E converge or diverge simultaneously;b) if = +oo, then 1/ n) = +oo;c)if lflnn)=+oo.

The Abel transformation for series is used in the solution of many ofthe subsequent problems: if sequences and are such that the se-quence converges (where = b1 +. • then the seriesand — converge or diverge simultaneously. In the case whenthe series E converges, another variant of this assertion is useful: if thesequence converges (where = + then the series

and — converge or diverge simultaneously. Theseassertions are easy to prove by applying the Abel transformation (see Chap-ter I, §2) to the sums Ei<n<jv The Abel transformation is especiallyconvenient in the case of I iionotone sequence

2.3. Let 0. Prove that <+00 if and only if = o(lfn) and— <+00.

§2. SERIES CONNECTED WITH MONOTONICITY 37

2.4. Let j 0. Prove that >an/n <+00 if and only if = o(1/lnn)and E(an —an+i)lnn <+00.

2.5. Let p> 1 and an j 0. Prove that the series converges if the

series E converges. Is the monotonicity essential?2.6. Let an j 0. Prove that:a) if the series E converges, then an Xk = o(1);b) if an > 0 for all n E N, then there exists a iioiinegative sequence {xn}

such that an >.1<k<n Xk = o(1) and >anxn = +00. Is the monotonicity of{an} essential in a)and b)?

2.7. Suppose that an j 0 and the series converges. Prove that theseries converges, and >k>n akxk = o(an).

2.8. Suppose that p > 0, q, E N, and Q> 1. Let {an} be a non-negative sequence, and let An = a1 + ... + an. Prove the equivalence of thefollowing assertions:

a) E an/nt <+oo; b) <+00; c) <+00.

2.9. Prove that the series converges if e > 0 and the sequence{xn} is such that Xk —. 0. Is the convergence always absolute?Does the series xjn converge?

2.10. Suppose that an j 0, j 0, and = = +00. Does theseries min(an, bn) always diverge?

2.11. Suppose that an � 0 and —. oo, where = a1 + an

Prove that: a) Ean/Sp = +00; b) <+00 for any e >0.2.12. Suppose that an � 0 and <+00, and let = ak.

Prove that: a) an/cup = +00; b) < +oo for e > 0;c) <1. d) Is it always true that

2.13. Suppose that p � 1, an � 0, and = >k>n <+oo. Prove

that: a) � b) if = for some > 0, then= for any q > + 1).

2.14. Let {an} be a positive monotone sequence with 1 for all nand with an a <+00. Prove that the series

1 (an_i/an)—!

diverges if a = 0 or a = 1, and converges otherwise. What happens if

a = +00?2.15. Let f be a positive strictly increasing function on [1, +00), with

f(x) +oo as x +00. Prove that the series 1/1(n) and (n)

converge only simultaneously is the function inverse to f).2.16. Suppose that f decreases to zero on the interval [0, +00), the

sequence {an} is nonnegative and bounded, and an = +00. Prove that the

series f(n) and E + .. + an) converge or diverge simultaneously.

38 IV. SERIES

2.17. Suppose that j 0 and let = A. For an arbitrary set i/ C N

the sum of the partial series corresponding to ii is defined to be A(v) =(by definition, A(ø) = 0). Prove that the following assertions are

equivalent:a) the sums of all partial series fill the interval [0, A];b) � + +... for all indices n.2.18. Does there exist a series such that j 0 and the sums of

all its partial series (see 2.17) fill the Cantor set?2.19. Suppose that j 0 and = +00. Prove that there exists a

sequence of indices such that <flk+l, < 1/k, and =+00. Can the monotonicity of be dropped?

2.20. a) Prove that if a series converges, then there exists a se-quence I +oo such that the series converges.

b) Suppose that the series diverges, and —, +00. Does the seriesalways diverge? What happens if I +00?

2.21. Let be a sequence of positive numbers.a) Prove that the series

'—'n t

diverges.

b) Is it true that the series + diverges if 0 and= +oo?

§3. Various assertions about series

3.1. Does there exist a sequence 0, such that the series

and converge? Can be selected to be positive?3.2. Suppose that the series converges. Can the series di-

verge? Can diverge?3.3. Suppose that >0 and >a, <+00. Prove that

a) b)

3.4. Suppose that 0 < < 1. Prove that if the seriesconverges, then so does the series + n). Is the converse true if

0? Can the monotonicity be dropped?3.5. Suppose that � 0 and <+00. Prove that there exists

a sequence of numbers with density 1 (see problem 11.2.3) such that—, 0.

3.6. Prove that the following assertions are equivalent:a)

§3. ASSERTIONS ABOUT SERIES 39

b) for any sequence of numbers with density 0 (See problem 11.2.3)

the series converges.3.7. Suppose that the series converges absolutely and = 0

forany kEN. Provethat forall n. —

3.8. Prove that if the series E converges for any sequencetending to zero, then E < +00.

3.9. Prove that the following two properties of the sequence areequivalent:

a)

b) if the series converges, then so does the series3.10. Suppose that the function is monotone and bounded on [0, +oo).

Prove thatc—'+O

for any convergent series Ear.

3.11. Construct a positive sequence with the following properties:a) —, 0; b) = +00; c) if

a the following properties:a) —, 0; b) = +00; c) if an increasing sequence of indices issuch that supk�J <+00, then the series E converges.

3.13. Let be a sequence of positive numbers, and let = a1 ++ and = + a.... Prove that the following six assertions are

equivalent:

a) = b) =

c) � q1 < 1 for all n EN;

d) <+00 and � < 1 for all n E N;

ifp>0;

ifp>O.

If a)—f) hold, then so does the assertion g) for any Q > 1 there ex-ists an index L = LQ such that afl+Lfafl � Q for n E N, that is, the

sequence can be decomposed into L subsequences = {aflL+1}

(1= 0, 1 L — 1), each increasing no more slowly than a geometric pro-

gression: � Q. Prove that if the sequence is monotone, then

g) a). Is this assertion true for nonmonotone sequences?3.14. Let be a sequence of positive numbers, and let = a1 +

+ and = + Prove that the following seven assertions

are equivalent:

c) = d) a1 <+00 and =

IV. SERIES

1 1 Ii\.ifp>O;n+2

g) =

If a)—g) hold, then: h) —, +00. Show by examples that h) a), noteven for monotone sequences; but if 1 +00, then a)—g) hold.

3.15. Suppose that p> 1 and � 0 and let = (a1 +

a) � (pf(p — 1))

b) � (pf(p — 1))P3.16. Suppose that > 0 for all n E N, and the series con-

verges. Prove that the series a2• • converges, while the series>(a1 + diverges.

3.17. Let f be a function defined on R. Prove that the following prop-erties of f are equivalent:

a) f(x)=O(x) as x—'O;b) for any absolutely convergent series the series converges

absolutely;c) for any absolutely convergent series the series E converges.3.18. Let f be a function defined on R. If the series converges

for any convergent series then 1(x) = Cx in some neighborhood ofzero.

3.19. Suppose that Q is a countable subset of (0, 1) with infQ = 0 andsup Q = 1. Prove that for any number x E (0, 1) it is possible to numbertheset Q={q1,q2, ...} insuch awaythat

§4. Computation of sums of series

Compute the sums of the series in problems 4.1—4.11 (the sums in 4.7—4.9can be computed in terms of the Euler constant—see problem 11.2.9).

4.1.

4.2.(rEN).

4.4. + 1)! (IzI <4).where in E N, in > 1, = 1 if n is not

divisible by in, and = 1 — in if n is divisible by m.

21

4.6. 1)e'. 4.7. ( —n�2 m�2

FUNCTION SERIES 41

4.8. inn. 4.9. nJ.

4.10. >iislnxd 4.11. slnxd

4.12. Prove that

1 4 (2N)!! çn/22 2N

I xcos xdxfl>Nn ir(2N—1)!!J0

In particular, 1+ ...+ ...+= 4.4.13. Let s> 1. Prove that:

a) = lim(1 — 3_S)_1

b) = lim(1

where p1 =2, p2 = 3, ... are the prime numbers, numbered in increasingorder, = 1 if the number of prime divisors of n (counting multiplicity)is even, and = —1 otherwise = 1).

4.14. Prove that

4.15. Prove that for ti < 1

a) b)

where is the number of divisors and a(n) the sum of the divisors ofn.

§5. Function series

5.1. For what p. q >0 and x, y E R does the series

x .

Econverge?

5.2. Suppose that —' 0. Does the series converge uni-

formly on R? Estimate Iunder the assumption that <

I forany nEN.5.3. Does the series

xn

42 IV. SERIES

converge uniformly on [0, +oo)? Estimate

xnsup>

2x?o

5.4. Does the series min(xn!, l/xn!) converge for x > 0? Does itconverge uniformly on (0, +oo)? Estimate >min(xn!, 1/xn!).

5.5. Suppose that ER, >0, = and = +00, andassume that the series B(t) = >fl>O converges for ti < 1. Prove that

the series A(t) = converges for ti < 1, and

lim <lim— — B(t) — B(t) —

(compare with problem 11.2.2). In particular, if c 0, thenA(t) cB(t) as t —, 1—0; if = A(t) = o(B(t)) as t —, 1—0.

5.6. Prove that if the series >fl>O converges, then —4

>.n>O as t —, 1 —0 (Abel's theorem).5:7. If the partial sums of the series are such that the limit

lim((S1 + . =1 exists and is finite, then the series A(t) =converges for ti < 1, and A(t) as t —, 1 —0. Thus, the Abel methodfor summation of series (computation of the limit a/) isstronger than the arithmetic mean method (the Cesãro method). —

5.8. Suppose that � 0, = +00, and the series B(t) =converges for ti < 1. If the sequence is such that thi limitlim(a0+ • )/(b0+.. = I exists and is then the series A(t) =

converges for fri < 1 , and A(t)/B(t) -4 I as t —, 1 — 0.59. Suppose that the sequence is such that > —1 for n E

N, and 11(1 + = A E (0, +00). Prove that for fri < 1 the productfl(1 + = A(t) converges, and A(t) —, A as t —' 1 — 0 if <+00.Can the condition <+00 be dropped?

5.10. Let 1/2. Prove that:a) the series converges for any x E R;b) for 1/2 < < 1 the convergence is uniform on [a, AJ, 0 <a <A,

but not uniform on [0, AJ;c) for = 1 the convergence is uniform on [0, AJ, A > 0.

5.11. Let 1(x) = Prove that

0< if(x)i/lnlnx<+oo, if(x)i=0.x—.+oo

5.12. Let 1(x) = sin nxi, where � 0 and < +00. Provethat f E Lip1 if and only if <+00.

5.13. Suppose that the sequence of polynomials satisfies the con-ditions:

a)thedegreeof isatmost mEN forall nEN;

§6. TRIGONOMETRIC SERIES 43

b) the limit P(x) = exists and is finite for all x E [cr, flu,a</3.

Prove that P is a polynomial of degree at most m • and P on anyfinite interval.

5.14. Suppose that —00 � a < b � +oo, E C((a, b)), and � 0(n E N), and let f = f,. Prove that if I is continuous on (a, b), then

b b

j 1(x)dx = j dx.

§6. Trigonometric series

6.1. Let {An} and be numerical sequences, with An —, +00.Prove that in any nonempty open interval there is a point x such that

cos(Anx + cOn) = 1.6.2. Provethat an,bp-4O if

a c [0, 2irJ with the cardinality of the continuum and asequence —, +oo such that sin Bnx 0 on E.

6.4. Prove that +IbnI) <+00 if � constOn

6.5. Prove that for x E (0, it)

Sn(x)= sinx+"•+sinnx= O(min(n, 1/x));

Cn(x) cosx+"•+cosnx=O(min(n, lfx)).

6.6. Compute the limits

1 sinAxi 1flt/2 sin2 Axlim —, . dx and lim — i

A—'+oo in A j0 sin x A—'+oo in A Jo sinx

What is the asymptotic behavior of the integrals Ln = and

= dx?6.7. Prove that the series * exp(inir(x — diverges for any

x ER.6.8. Compute the sums of the following series:

sinnx cosnxa) s—; b) L1y—jnsinflx

c) —; d)

6.9. Prove that:

In problems 6.10—6.21 the sequence decreases to zero. The Abeltransformation (see §2) can prove to be useful in solving these problems.

44 IV. SERIES

6.10. Prove that the series

and

converge uniformly on any closed interval not containing points 2irk (k EZ).

6.11. Prove that for XE (0, it)

n�N

6.12. Prove that the series

E sin nx sin mx and cos nx sin mxn�l

(m E Z) converge uniformly on ILL6.13. Prove that if g(x) = sinnx for x ER, then

2P= —j

g(x)sinmxdx.

6.14. Prove that the series sin flX converges uniformly on R if andonly if = o(1/n).

6.15. Prove that the partial sums of the series sin nx are uniformlybounded if and only if = 0(1/n).

6.16. Let = 0(1/n) and suppose that the function is absolutelyintegrable on [a, bJ. Prove that the series sin nx can be integratedtermwise on [a, bJ. Using the result of problem 6.8a), compute the sum ofthe series

6.17. Let f(x) = cosnx. Prove that If(x)Idx < +oo if<+00.

6.18. Let g(x) = sinnx. Prove that < +oo if andonly if <+00.

6.19. Suppose that < +oo f(x) = cosnx g(x)= sinnx, and E C([0, irj). Prove that the series cos nxand sin nx can be integrated termwise on [0, irj. In particular,

1(x) cos mx dx = (ni E N).

6.20. Prove that for XE (0. it):a) > 0;b) � —1.

6.21. a)Supposethat Provethat g�0 onb) Let 1(x) = + cos iz.v. Prove that if the sequence is

convexand —'0, then f�0 on Rand —

§6. TRIGONOMEThIC SERIES 45

6.22. Let be the nth partial sum of the series 1(x) =E(sin nx)fn, and let M, = Prove that

= I G= j dU> = f(+0)

(the Gibbs phenomenon).6.23. Let 1(x) = EIkI<N cke end m > 0. Prove that

Mfir a

fork=±1 (x,y€R).6.24. Let 1(x) = E2' Prove that f E Lip112, but I

for6.25. Let 1(x) = sin where A E (1, 3). Prove that I E

where = log3 A, but f Lips for /3 >6.26. Let 1(x) = Sin Prove that I E for any E

(0, 1). Is it true that f E Lip1?6.27. Let f(x) = E2' sin(2irn!x). Prove that:a) I for any 1J;b) f is nondifferentiable at every point.6.28. Prove that the following three properties of a trigonometric series

cos(2'x) +n>O

are equivalent:a) + <+oo;b) the partial sums SN(x) = + are uni-

formly bounded on some interval fri ,bJ, ac) the partial sums SN(x) are uniformly bounded on ILL6.29. Consider the two sequences and {Qk}k>o of algebraic

polynomials (Rudin-Shapiro polynomials) recursively as follows:

= Q0 = 1, = + = —

(n = 0, 1, 2,...). Prove that:a) the polynomials and have degree — 1;

b) all the coefficients of the polynomials and are equal to ±1;

c) if Izi = 1, then + = 2'".6.30. Let RN (N = 0, 1, 2, ...) be the natural projection of the set of

all algebraic polynomials onto the set of polynomials of degree at most N(RN annihilates the monomials of degree greater than N and does not changethe polynomials of degree at most N). Prove the following inequalities for

IzI=1 andforany N, neN;

IRN(Pfl)(z)I � IRN(Qfl)(z)I �

46 IV. SERIES

6.31. Prove that there exists a series eke with €k = ±1 whosepartial sums satisfy the inequalities: —

a) ISm(O)I� b)j2

6.32. Using the sequence {ek}k>o in problem 6.31, we see that there

exists a continuous function f of the form f(O) = ake such that

= +00 for any p <2.6.33. Suppose that the sequence c C satisfies the condition (cf.

problem 6.31)

=1<k<n

Prove that for > 0:a)

ek ikO amax =0(n);

1<k<n

b)

max > = 0(na);k>n

c) the functionikOf(O)=

k>1

is in the class for E (0, 1).6.34. Prove that there exists a function I E Lip112 of the form f(O) =

cke such that ICkI = +oo (cf. problem IX.4.13).

CHAPTER V

Integrals

§1. Improper integrals of functions of a single variable

In problems 1.1—Li! investigate the convergence of the corresponding in-tegrals.

1.12. Determine whether the convergence of the integral f(x) dximplies that of the following integrals:

a) j f3(x)dx; b) j1.13. Let 1(x) = (2lnIl _t21 dt (x > 0). For what e > 0 does

the integral converge?

1.14. Compute the Frullani integral f°°(f(ax) — f(bx))/xdx, where

I is continuous on [0, +oo) and satisfies one of the following conditions:a) the integral f(x)/x dx converges;b) f(x+T)=f(x) forsome T>0 andall x�0;c) the limit I = 1(x) exists and is finite.

1.15. Suppose that the integral f°° 1(x) dx converges. Prove that:

a) the integral f°° dx converges for any e> 0;

+00

1.1. [ IsinxldxJO eX2Smn2X

13 r+00

1.5. [ Isinx"IdxJ1

+00 dx1.7. [

1.9.[00

cos(x3 — x) dx.JO

+00

1.11. [ sin(xlnx)dx.JO

i.2. /dx

Jo

1.4. /11

1.6.[+00

Isin(x" +1dxx

1.8. / xPlsinxlXdxI'

1.10. / sin(x" + ax + b)dx, p> 1.JO

48 V. INTEGRALS

b) —, f(x)dx as e —, 0 (cf. Abel's theorem—problem P1.5.6).

In problems 1.16—1.38 compute the integrals. Some of them are expressedin terms of the Euler constant y (see problem 11.2.9).

1.16. 1.17.Jo x

118lnx—dx. 1.19.Li-x

+00 x2dx1.20. 1 1.21.

Jo co&x+00 sinx

1.22. [ —dx. 1.23.x

'+00— cosx

1.24. / dx.x

1 25 J— 1)(eibX — 1)

dx (a, b ER).—co

+00 mx1.26. 1

x x

x

q > 0).x

1.31. Jo (1—x)1n2(1—x)

J ln(sinx) dx.

1' dx.Jo 1+x

x[+00 x— sinx dx.

Jo x3

1.27.

1.32.

1.33. Using the result of problem I.2.16a). prove that

+o(1).

With the help of the Stirling formula find the value of the Euler-Poissonintegral dx.

§2. COMPUTATION OF MULTIPLE INTEGRALS 49

1.34. Je_e dx k=O,1,2,3,4,5.

1.35. 1 edJo

1.36.

j+OO (2 2/2)

1.37. a) I —dx; b) IJo J0

1.38. f°° ir(x)/(x3 — x) dx, where ir(x) is the number of prime num-bers not exceeding x.

§2. Computation of multiple integrals

In problems 2.1—2.7 it is required to compute the given integrals.

2.1. f°° j;FOOJInx —

2.2.

a)

J x)dx1 . . . dxv;

c) 1 1 (e>O).J(1.+oo) x1 xv))

2.3. fR dx.

2.4. a) ffR2 lax +

b) fR l(x, 0"2dx (a E R'1, > —1).

2.5. lix — yll1 dy1 dy2 dy3 (x E R3).

2.6. dx dy du dv.

2.7. f(Axx)<1 dx (x E R4, A a positive-definite matrix).

2.8. Let A = {x = (x1, x2,x3,x4) EI

x1}. For

what t E is the integral K(t) = dx finite?What is itsvalue?

In problems 2.9—2.13 we understand a random choice of points to meanthat the probability of choosing a point belonging to some set is proportionalto its measure (length, area, volume).

SO V. INTEGRALS

2.9. Two points are chosen randomly in the interval [a, bJ. Find themean value M of the distance between them. What is the probability Pthat this distance is greater than M?

2.10. Three numbers are chosen randomly on the interval [0, aJ. Whatis the probability that they are the lengths of the sides of some triangle?

2.11. Two points u and v are chosen randomly on the interval [—a, aJ.a) What is more probable: that the roots of the equation z2 + uz + v =0

lie on the real axis (probability P1(a)), or that the roots of this equation donot lie on the real axis (probability P2(a))? What do the probabilities P1(a)and P2(a) tend to as a —, +oo?

b) What is the probability that the biquadratic equation z4 + uz2 + v = 0

has both real and complex roots?2.12. Two points are chosen randomly in the unit disk. What is the mean

value S of the area of the disk constructed with the segment joining themas a diameter?

2.13. Two points are chosen randomly on a circle of radius R. Find:a) the mean value L of the length of the chord joining them;b) the mean value of the angle (� it) formed by the radii drawn to

these points.2.14. Let p < 1 and let

=11 ((x —v)2 +(y —

u2+v �1

11 2 2'1'(x, = II ln((x — u) + (y — v) ) dudv.JJu2+v <1

Prove that 'I' E C'(R2) and find 'I'.2.15. Let = {x = (x1,..., E

I El<k<flxk/k 1, x1,...,� 0}. Prove that for any t E R

1... 1 dx dx =J t /

2.16. Prove that for m, n E N with m � n and for a > 0

I dx (_1)m_1n—rn

I Ina)JLO,Ir (a+x1 (in— 1)!(n —m)!

where = = f(a + 1) — 1(a)and =2.17. Let ak (k = 0, 1 n) be positive numbers.a) Prove that for any nonnegative continuous function f on [0, +oo)

f...jf(a1+(a2_ai)x1+(a3...a2)xx2+...+(a_a)x...x)

x dx1 . . d;j...jf(a+(a

§2. COMPUTATION OF MULTIPLE INTEGRALS 51

where = {(x1,..., ER IEI<k<flxk � 1, X1 � 0. � 0}.b) Compute the integrals

0 (a1 + (a2 — a1)x1 + (a3 —a2)x1x2 +• + (a0 —

dx1 . . . d;I (a0 + (a1 — a0)x1 + (a2 — a0)x2 + —

2.18. Prove that:a) if f is continuous and nonnegative on R, then

! f(ax+by)dxdy +00 du

=I (a,bER);(l+x2)(l+y2) —00 l+u

n)

'p(a,

dx = (cos ( laki)/

2.19. Let I E C2([O, 1] x [0, 1]). Find the limit

tim fl2(Jhfhf(xxdyiJ_1/2)'\

0 0 l�j�n )

CHAPTER VI

Asymptotics

§1. Asymptotics of integrals

1.1. Suppose that —oo <a <b � +00, f, g E C([a, b)), and g > 0.Prove that:

a) if and as thenfg(t)dt as

b) if fg(t)dt= +00 and 1(t) =o(g(t)) as b—O, then ff(t)dt =o(f: g(t) dt) as x —, b —0;

c)if fg(t)dt<+00 and as t—'b—O,thenas x—'b—O;

d) if fg(t)dt <+00 and 1(t) =o(g(t)) as t—' b—0,theno(f g(t)dt) as x —' b —0.

1.2. Find the asymptotics as t —, 1 —0 of the following integrals:

b)

c) 2'(t) = — 2tcosx +t2)' dx.1.3. Find the asymptotics as A —, +00 of the following integrals:a)

b)

c) (a ER);

d) f°° dx;e) (a ER).1.4. Find the asymptotics as A —, +00 of the following integrals:

a) (p>0);b)

c) I Mdx;

d) (p ER).1.5. Find the asymptotics as n •—' +00 of the following integrals (n E N):

a) = dx (p 0);

53

VI. ASYMPTOTICS

b) (p>0);c) = (p > 0).1.6. Prove that as e —, +0:a) 1/2e;

b) 1/2€;

c) —, 1;

d) f00(cosx)/xC dx ire/2.1.7. Prove that as p —, +oo:a) sin(x")dx ir/2p;b) f°°cos(x")dx—'l.1.8. Suppose that the function is continuous on R and has period

T> 0, and that = + ço(t)dt 0. Prove that f11°° c,/eas e—'+O.

1.9. Prove that as A —, +oo:

n/2 ita)] dx=—A+0(1);o sinx 2

f'cosAx IT 11

1n/2 cos2xdx itc)jn/21 Ad) [ xl

J0 \sinx/ 3

1.10. Find the asymptotics as A —, +00 of the following integrals:

a) fsin2x dx (p � 1); b)

1A Isinxllcosxl dx (p > —1);

c) j (p � 1); d) j (p> 1).

1.11. Suppose that the function I decreases to zero on [a, +oo). thefunction is continuous on R and has period T> 0, and fj dt =0.Prove that

I

fFOO � 1(A) for A � a.1.12. Suppose that I decreases to zero on [a, +oo) , and is continuous

on R and has period T> 0. Prove thatJA

dt =JA

f(t)dt + I + 0(1(A)),

where = fj dt and I =f is nonnegative and monotone on [a, +oo), is

continuous on R and has period T> 0, and C, = + f' ço(t)dt 0. Provethat:

§2. THE LAPLACE METHOD 55

a) if f°° 1(t) dt = +00 and f7' f(t) dt = dt), thendt 1(t) dt as A —, +00;

b) if f°° 1(t) dt < +oo and f(t)dt = 1(t) dt), then

f E C([a, bJ), and is continuous on R and hasperiod T> 0, and let = f' co(t)dt. Prove that

f(t)ço(At) dt = dt.

1.15. Let f E C([a, bJ x [—1, 1J). Prove that

1(A) = jbf(x sinAx)dx11b

(sint)dt) dx.

1.16. Find the asymptotics as e —, +0 of the following integrals:

a) b)

1.17. Prove that as e —, +0:

3a) jxb) j = 1— 2€— e2lne + 0(e2);

c) 1' dx = 1 — e — + o (.-L).J0 Inc Inc

1.18. Suppose that I is positive and integrable on (0, 1) and that theintegral In dx is finite for some number p in (1, 2J. Prove that

j f(x)dx = 1+ Inf(x)dx + O(e"), e —, +0.

1.19. Prove that:

sin nxj 2a) J x

dxb) I max Isinkxl— Inn;

Jo 1<k<n xIsinkxl dx

c) I maxJo 2<k<n Ink x

§2. The Laplace method

2.1. Let f E C([0, 2E/2J). Find the limit

lim n I xdx.n—,00 J0

VI. ASYMPTOTICS

2.2. Suppose that 0 > 0 and c9 = dt. Prove that as A —, +oo:

2 A C0 dx C0a)j (1-x) b)j

2.3. Let 0 > 0. Find the asymptotics as A +00 of the integral

fO/'.TAA

Jcos

2.4. Verify that the asymptotics as A —, +00 of the following integralsdo not depend on the parameter 0 > 0:

2A dxa) j (1 —x ) dx (0 � 1); b)

J Jdx

2 A'2 o (1+x+x)

e) j°(ln(1+x)y1d o (0�ir);

g) j lnA(e — x)dx (0 � e — 1).

2.5. Find the asymptotics as A —, +oo of the following integrals:a) (p>O);

b) f (p>O);

c) 10fl/2xP cosA xdx (p> —1).

The characteristic feature of problems 2.2—2.5 consists in the phenomenonof localization—the asymptotic behavior of the integral depends on the be-havior of the integrand only in a neighborhood of a single point. This effectalso appears clearly in a more general situation in the study of importantintegrals of the form

=A

where is nonnegative and piecewise monotone. For large values of theparameter A the graph of the function ç0A has sharply expressed "humps"in neighborhoods of those points at which has a strict local maximum.By breaking up [a, bJ into several intervals if necessary it can be assumedthat is monotone on [a, bJ. Here it suffices to consider only the casewhen is decreasing. Then the values of ç0A at points far from a arenegligibly small in comparison with its values at points near to a, which givethe main contribution to the integral c1(A). To find the principal part of

it remains to approximate in a neighborhood of a by a simplerfunction and to compute the integral obtained. The Laplace method for

§2. THE LAPLACE METHOD 57

investigating integrals of the form (J?(A) and modifications of them amountsto the realization of the scheme presented.

Most often encountered is the case when the difference — is aninfinitesimally small quantity of power type, that is,

C.(x—a)° (C,a>O).x—'a+O

It is investigated in problem 2.6. The result obtained there is called theLaplace asymptotic formula (for brevity of formulation it is assumed inproblem 2.6 that = 1). The reader can formulate an analogous as-sertion without difficulty if is increasing on [a, bJ and

C(b — x)°. This implies at once the Laplace asymptotic formula also forpiecewise monotone functions.

2.6. Suppose that —00 <a < b � +00, the function is positive and

decreasing on [a, b), andx—.a+O

(C, p > 0). Prove that

A f(1+p)= Ia

ço (x)dx(AC)"

In particular, 1/AC for p = 1 ,and for p = 1/2.2.7. Suppose that —00< a < b � +00, the function is positive and de-

creasing on [a, b), fco(x)dx <+00, and =+00 for some p > 0. Prove that f ço4(x) dx = as A —, +00.

2.8. Suppose that —00 <a < b � +00, the functions f and g areabsolutely integrable on (a, b), f> 0, and 1(x) g(x) as x —, a + 0.Assume further that is nonnegative and strictly decreasing on [a, b).Prove that

Jg(x)co'4(x)dx

Jf(x)ço(x)dx.

a A'+OOa

2.9. Find the asymptotics as A +00 of the following integrals:

A cos(ax)a)

Jsin x dx; b)

(1 + x2)Adx (a E R);

çAA sinx dx

c)J

e (A — x) dx; d) (p <2);

e) 1' dxf)

Jo Jo

g) j dx.

2.10. Find the asymptotics as n —, +00 (n E N) of the following inte-

grals:a)

VI. ASYMPTOTICS

b) dx;c) f(1 — 4x + 2x2)hZ

In the following problems we consider integrals of the form c1(A) =

f f4(x)dx, where the function 14, while not now a function of the formconsidered in problem 2.6, still preserves its characteristic feature: as A in-creases, the graph of 14 has a sharper and sharper "hump" in a neighborhoodof the point x4 at which 14 attains its largest value. Although the result inproblem 2.6 is not applicable here, the idea of the solution is preserved: rep-resenting 14 in the form , and replacing lnf4 in some neighborhood ofx4 by its Taylor expansion and the integral over (a, b) by the integral overthe neighborhood of x4, we find the principal part of c1(A). The choice ofthe neighborhood is the basic difficulty here. On the one hand, it must notbe too large, since otherwise the error due to the use of Taylor's formula be-comes influential. On the other hand, in order to neutralize the second errordue to the replacement of the integral over (a, b) by the integral over theneighborhood, this neighborhood cannot be too smalL The main content ofthe solution involves a successful choice of the neighborhood, allowing goodestimates of both these errors.

We mention also that if x4 —, x0 as A —, +00, then it can be more con-venient to consider the Taylor expansion of the function lnf4 in a neigh-borhood of the point x0.

2.11. Find the asymptotics as A —, +oo of the following integrals:

f+OOf 2x \A dx pxa) I —i J —; b) (Ax) dx (p > 0);

Jo \1+x/ J0

c) j eXP(A _x)A dx (0 <p < 1).

2.12. Prove that for any real number p

IJo A—.+x A

2.13. Provethatas A—'+oo:

1' dx 1l/2

1a) j b) j x dx

In the following problems we study the asymptotic properties of the Eulerr-funczion r(x) = IX_le_f di. x > 0.

2.14. Prove Stirling's formula: + x)

2.15. Prove that r(x + xcr(x) for any c E ILL

2.16. Prove that di 4r'(l + x).2.17. Suppose that is a positive function defined on (0, +oo).

§3. ASYMPTOTICS OF SUMS 59

Prove the following assertions:

•

_______

ia) lim = 0 urn I ?e 'dt = —.

x—'+oo f(1 + x) Jo 2

• — x .b) hrn = +00 lim I t e dt = 1;

x—+oo x—.+oof(1+x)j0

• — x 1c) urn urn I t e dt=0.

x—'+oo x-'÷oor(1+x)j0

§3. Asymptotics of sums

3.1. Suppose that an > 0 and —' 0. Prove that:a) if = +00 and bn an , then bk E1<k<n ak;b) if Ean = +00 and bn = o(an), then bk = ak);c) if >an <+00 and bnd) if Ean <+00 and bn = o(an), then Ek>n bk = O(>k>n ak).3.2. Suppose that j 0, y � 0, and E ILL Prove thaff

a) if >1<k<n kaak = then an =

b) kaak = o(nY), then an =

c) if Ek>n kaak = then an =d) if >.k>n k°ak = then =3.3. Suppose that y > 0 and E1<k<n kaXk Prove that:

a) if y > then = x1 + +b) if y = then Sn = x1+

y <cr, then the series EXn converges, and

an = Xk —fl.k>n

3.4. Suppose that y > 0, E R, the series E converges, and

>.k>n kcxxk Prove that:

a) if y <—cr, then = x1 + ••• + + yI;b) if y = —cr, then n;

y > —cr, then the series Exk converges, and an = Ek>n Xk

+ y).3.5.

where 0 <y < 1.a) Prove thatb) Can it be asserted that a, 1/n if inn?

3.6. Suppose that an 1 0, E an < +00, and >.k�n ak n', where

y > 0. Prove that an

vi. ASYMPTOTICS

3.7. Suppose that > 0 and let

i(i)k1<k<n

Prove that min(l,3.8. Let be the number of divisors of a natural number k. Prove

that T(n) = >1<k<fl = n(ln n + 2y — 1) + Here y is the Eulerconstant (see 11.2.9).

3.9. Suppose that j 0, Ecu,, <+00, and f(t) = t +for t � 0. Prove that the behavior of the function f as t —' +00 is closelyconnected with that of the sequences {njç}; more precisely, there exists anumber C > 0 such that the following inequalities hold:

a) sup,>0f(t) �b) �c)

3.10. Let 1: [0, +oo) —' R be a nonnegative nonincreasing function, andlet = hEf(kh), h > 0. Prove that = J°°f(t)dt. Is

this true if I is continuous but not monotone?3.11. Determine the asymptotic behavior of the following sums as t —'

+oo:

a) b)k�1

3.12. Suppose that 1(p) = as p > 0. Prove thatf(p)= 1/2+0(p).

3.13. Suppose that I is a nonnegative decreasing function on [1, +x).Prove that:

a) if = +00, then El<k<flf(k) = f(z)dt+ C + o(l).in particular, El<k<flf(k)

b) if f°°f(t)dt + 0(1(n)),in particular, if 1(x) = o(J°°f(t)dz) as x —, +oc, then f(k)f°°f(t)dt

3.14. If f is a monotone function on [in, n] (in, n E Z), then

1(k)_f f(t)di �max(If(n)I, If('n)I).,n<k<n I??

In particular, if f is monotone on [1. +oo), f(t) dt = +00, andf(x) = as x —' +00, then El<k<flf(k) J11zf(t)dt.

3.15. Suppose that M, N E Z. M � N. and the function I (complex-valued in general) is integrable on the interval [M — N + Let =

)JM—1/2

§3. ASYMPTOTICS OF SUMS 61

a) Prove that

1 N-i-1/2Var(f).

In particular, if f is monotone, then 41f(N + 4)— f(M—b) If fE C2([M—+, then

and hence � If"(t)Idt. In particular, if f is convex or concave,then

3.16. Prove that

fl (i + ._L) As— and iii (i +l<k<n 1<k<n

for some positive numbers A and B.3.17. Supposethat YER\N, nEN, and

n n!

Prove that there exists a number Ca 0 such that

C0

n )

3.18. Prove that:a)

b)

c) .22.33..3.19. Let 0. Prove that as n —, +oo (n E N):

a) b)

c) d)

3.20. Prove that flk/k!

3.21. Prove that

1<k<n

62 VI. ASYMPTOTICS

3.22. Supposethat NEN, zEC,and Re:>0. Provethat

where the constant on the 0-term is independent of z and N.3.23. Supposethat NEN, ZEC,and Rez>0. Provethat

= (V'N_ 1/2+ (1 +

where the constant in the 0-term is independent of: and N.3.24. Let E (0, 1). Prove that

and

as x —, +0, where A and B are positive numbers dependent on3.25. Prove that there exists a number E (0, 1) such that for any

the partial sums of the series E(cos nx)/ncx are uniformly boundedbelow, but for any E (0, this is false.

What can be said about boundedness below for the partial sums of theseries , XE [0, irJ?

3.26. Suppose that yl E C'([a, +oo)), y,, > 0, and = on[a, +oo). Prove that = yr'(l/e)+ 0(1) as e — +0.

3.27. Find the asymptotic behavior as t —, 1 — 0 of the sums of thefollowing series(2):

a) (p >0); b) (p > 1);

c) (p > 0); d) it? (a> 1);

e) (a > 0); f) In.

3.28. Find the asymptotic behavior as t —, 1 — 0 of the sums of thefollowing series:

a) (p> —1); b) (p E R).n>2

3.29. Find the asymptotic behavior as t —' I — 0 of the sums of the

(2)Recall that stands for the series

§4. IMPLICIT FUNCTIONS AND RECURSIVE SEQUENCES 63

following series:

t nt

C

(1 (1

.

/

k) 1_1n2' 'Ld(11fl)23.30. Find the asymptotic behavior as t —. 1 — 0 of the sums of the

following series:

a) b) E(1 +12fl)2;

e)

3.31. Find the asymptotic behavior as t —' 1 — 0 of the sums of thefollowing series:

a) T(t) = b) S(t) =

where is the number of divisors and a(n) the sum of the divisors of anumber n.

§4. Asymptotics of implicit functions and recursive sequences

4.1. Suppose that z(t) = (t E R), and let t(z) be thefunction inverse to z(t). Prove that t(z) as z —, +0.

4.2. Let x = tanx lying in interval(irn, ir(n + 1)), n E N. Prove that

4.3. Prove that each of the following equations determines an infinitelydifferentiable implicit function y defined in some neighborhood of the point(a, b), and find the coefficients in the expansion

64 VI. ASYMPTOTICS

a)ye'—x=O. a=b=O, n=3;2 2b)y +lny—x=O, a=b=1. n=3;

a=1. b=O, n=3:d)ylny—x=O, a=O, b=1, n=3;e)e"+x2+y—1=O, a=b=O, n=6;f)arctan(x+y)—x—2y=O, a=b=O, n=6.

4.4. Prove that Kepler's equation y = M + x siny, where M is a fixedpositive parameter, determines an infinitely differentiable function y definedin a neighborhood of the point (0, M), and

y(x) = M+ xsinM+ + 0(x3), .v —, 0.

4.5. Verify that the domains of the implicit functions considered in prob-lem 4.3 contain the half-line [a, +oo) , and prove the following as x —, +oo:

lnlnx flnlnxa) y(x) = lnx —lnlnx + + 0 1mx \ lnx

ln2x ln4x I ln'1xb)

lnx lnx 1 ln2x 11n2xc)

x x 2x x

x xlnlnx x(lnlnx)2 fxlnlnxd) y(x)=—+2 + +01mx lnx lnx \ mx

2 •—x3+x 2(x--x3)e) y(x)=—x +1—e +0(xe );

x ir 1 111)

4.6. Let 1),where for x�0 (a andp are positive constants). Prove that

(.v0>0).

4.7. Suppose that x0 = and ; = — (ii E N). Provethat with accuracy 106.

4.8. Find the asymptotics of the recursive sequences ; = x0>0, in the following cases:

a) f(x)=x(1—x), x0<1; b) f(x)=sinx. x0<ir; c)1 +x

d)f(x)=arctanx; e) f(x)=ln(1+x); f) f(x)=1_e_X;

g) 1(x) = h) 1(x) =

§4. IMPLICIT FUNCTIONS AND RECURSIVE SEQUENCES 65

4.9. Suppose that C, p, x0 >0, then = x,1 + Prove that:

a) b) = (Cpn)lhJ (i + + 0 (i))

4.10. Construct a positive function f on (0, +oo) such that the recur-sive sequence = x0 > 0, satisfies the relation 1/(ln n).

4.11. Suppose that x0 E (—1, 2), = — 1) (n E N) Findthe asymptotics of the sequence

4.12. Suppose that a0 = 1, and = + (>.0<k<n ak) (n E N).

Prove that4.13. Suppose that < 1, a0 = 1, and = + ak)

(n E N). Find the asymptotics of the sequence

4.14. Suppose that p E R, a0 = 1, and = +(n E N). Find the asymptotics of the sequence

CHAPTER VII

Functions (Continuation)

§1. Convexity

A function 1: (a, b) —+ R is said to be convex (strictly convex) if for allpoints x1 , x2 E (a, b) and all numbers > 0 with + = 1 theinequality

f(t1x1 + t2x2) � t1f(x1) + t2f(x) (f(t1x1 + t2x2) <A1f(x1)

holds for x1 x2. If —f is convex, then f is said to be concave.

1.1. Prove that a convex function f satisfies Jensen's inequality

I ( �J l<j<n

for any x1,..., E (a, b), E [0, 1J, with = 1.1.2. Suppose that a function f is defined on the interval (a, b), and let

x1 <x2 <x3 be points in (a, b). Consider the chords joining the points ofthe graph of f with abscissae x1 and x2, x1 and x3, and x2 and x3. Theslopes of these chords are

f(x2) — f(x3) — f(x3) — f(x2)

respectively. Prove that f is convex if and only if

f(x2)—f(xl) f(x3)—f(x2)— x3—x1 —

for any triple of points x1 <x2 <x3 in (a, b) (the three chords lemma).Verify that this pair of inequalities is equivalent to

1 x1 f(x1)1 x2 1(x2) �0.1 x3 1(x3)

1.3. Suppose that a function 1 on an interval satisfies the inequality

<1(f(x) +f(x2))

68 VII. FUNCTIONS (CONTINUATION)

for all x1, x2 E Prove that:a) f satisfies Jensen's inequality

f ( � xj E (J = 1 n)I<j<n

for any rational E [0, 11, = 1;b) if I is continuous, then it is convex.1.4. Prove that the condition of continuity in problem 1.3b) can be re-

placed by the following condition: f is bounded above on some nonemptyopen interval q) C = (a, b) (or even on (p,

1.5. Prove that the following three statements are equivalent for a func-tion 1: (a, b) —,

a) f is convex;b) the supergraph = {(x, y) E R2 I x E (a, b), y � f(x)} of f is a

convex set;c) through each point of the graph of f it is possible to draw a straight

line supporting the supergraph (i.e., a line such that all the points of thesupergraph lie above it).

1.6. Prove that if f is a convex function on [0, 1J, then the function= 1(x) + 1(1 — x) is decreasing on [0, 1/21.

1.7. Prove that if a function on an interval is locally convex, then it isconvex.

1.8. Prove that a convex function I on (a, b) is continuous on (a, b),and has there finite increasing one-sided derivatives and and

1' (x) exist everywhere except on an at most countable set of points. Provethat I E for any closed interval C (a, b).

1.9. Prove that a function f E C((a, b)) is convex if and only if for allx E (a, b)

�0.In particular, if I1(x) = 0 for all x E (a, b),then I is a linear function;

if f is twice differentiable on (a, b),then it is convex if and only if f' � 0on (a,b).

1.10. Prove that a function I E C((a, b)) is convex if and only if

1

x all x E (a, b).1.11. Every convex continuous function on [a, bJ is the limit of a uni-

formly convergent decreasing sequence of:a) piecewise linear convex functions;b) twice continuously differentiable convex functions.

§1. CONVEXITY 69

1.12. Prove that if a function f is convex and strictly monotone, thenr' is either convex or concave.

1.13. Suppose that a function f is convex on [a, bJ, and a functiong is convex and increasing on [c, dJ. Prove that if the composition g o Imakes sense, then it is convex.

1.14. Let f E C((0, +oo)). Prove that the functions xf(x) and f(lfx)are then convex or nonconvex simultaneously.

1.15. Suppose that a function f is convex on R and =f(x)fx = 0. Prove that f is constant.

1.16. Suppose that a function f is convex on [a, +oo). Prove that thelimit f(x)fx =1 E exists, and / > —00.

1.17. Suppose that a function f is increasing and concave on [0, +oo),and f(0)=0. Provethat f(x+y)�f(x)+f(y) for all x,y>0.

1.18. Prove that if 1: [0, 00) R is differentiable, concave, and non-negative, then xf'(x) � 1(x) for all x � 0.

1.19. Prove that if f is a convex function on [a, oo), then the function= f(b + x) — 1(x) is increasing (b >0).

1.20. Suppose that a function f is convex and increasing on [0, 00),and f(0) = 0. Prove that

E (_1)kf(ak)�f(

for any numbers a0 � a1 0.1.21. A sequence C R is said to be convex down if = —

� 0, i.e., � for any fl E N. Prove that if asequence is convex, then:

a) � + q EN);b) the sequence is monotone or there is a number m> 1 such that

a bounded convex sequence (see problem 1.21).Prove that:

a)

b) =o(lfn);c) + = a0 — a.

1.23. Suppose that f E C([0, 1J), f � 0, P = f(x)dx, and M =maxf. Prove the following inequalities:

a) if f is concave, then

a') j xf(x)dx � a") j x2f(x)dx �b) if f is concave and monotone, then

b') j ñx)dx <b") j f3(x)dx � 2P3;

70 VII. FUNCTIONS (CONTINUATION)

c) if f is convex and minf = 0, then

, 2P2 Ii 2 2P3c) xf(x)dx � c) x 1(x) dx �

d) if 1 is convex and monotone, and mini =0, then

d') j t (x)dx � iMP; d") j f3(x)dx �All eight inequalities have a simple mechanical meaning. For example,

the inequalities b) mean that among equal-area subgraphs of functions of thegiven class it is the triangles that have the greatest static moment and thegreatest moment of inertia with respect to the x-axis.

1.24. Suppose that the function f is nondecreasing on [0, 1/21, and1(1 — x) = 1(x) on [0, 11. Prove the inequality

fI fl flI 1(x)ço(x) dx �! f(x)dx! ço(x) dx

JO JO JO

for any convex function on [0, 11.1.25. Prove that the integral f(x)ço(x) dx is nonnegative for any con-

vex function on [0, 11 if and only if the function f satisfies the followingconditions:

a)

b)c) — x)f(x) dx � 0 for any a E (0, 1).1.26. Suppose that the function I is nonnegative on (0, +oo) and

1(x) 0 as x +00. Prove the following inequalities:a) if f is decreasing and a >0, then 1(x) sin ax dx � 0;b) if f is convex, then � 0.In both cases equality is possible only for f 0.1.27. a) Suppose that � 0 (n E N), and 1(x) =

Prove that lnf is a convex function on IR.b) A positive function I is said to be logarithmically convex if the function

lnf is convex. Prove that a logarithmically convex function is convex, andthat a sum of logarithmically convex functions is logarithmically convex.

1.28. Suppose that I E C((0, +oo)), f(1) = 1.a) Prove that the following assertions 1) and 2) are equivalent:

1)1(x) f(y)

f(x)f(y) � f(xy) (x, y E (0, oo));

2) there exists a p E [0, 11 such that 1(x) = x" (x E (0, oo)).

Verify that condition 1) cannot be replaced by the condition

1') f(x)f(y)�f(xy)(x,yE(0,oo)).

§1. CONVEXITY 71

b) Prove that the following assertions 3) and 4) are equivalent:

3) f <1(x) f(x)f(y) � f(xy) (x, y E (0, oo));

4) there exists a p E (—oo, 0J U [1, +oo) such that f(x) =(xE(0,oo)).

1.29. Suppose that a1, b1 � 0, / = 1 n, p, q> 1, + = 1.Prove the Holder inequality

/ \'/P/ \1/q

>ajbj�( >afl (

1<j<n

Prove that:a) K is monotonically increasing;b)

limic(p) = (x1 ..

c) = max{x1,... ,

d) ic(p) = min{x1,... ,

e) the function = (ic(p))" is logarithmically convex on (0, +oc).Formulate analogues of these assertions for nonnegative functions in

C([0, 1J).1.31. Suppose that f is a piecewise continuous function, 1: [a, bJ —,

[m, MJ, is a convex function on [m, MJ, p is a continuous and non-negative function on [a, bJ, and fp(x)dx = 1.

Prove that in this case:

:f(x) dx) � dx;

(dx) ço(f(x))p(x)dx;

c) if f> 0, then

(Lbp lnf(x) dx)

� J p(x)f(x) dx.

72 VII. FUNCTIONS (CONTINUATION)

1.32. Suppose that A0 are the consecutive vertices of a convexpolyhedron inscribed in a circle, with A0 and fixed. How should thepoints A1,..., be selected in order to maximize the perimeter and thearea of the polyhedron (for a given n)?

1.33. A point source of light located at a point (0, b), b > 0, illumi-nates an area that is the subgraph of a nonnegative convex function f E

C'([O, oo)), f(0) > b, and the light rays are reflected from the graph off and from the x-axis according to the well-known law. Is the whole areailluminated if 1(x) = 0?

Let f be a convex function, f: R —, L The Legendre transform of I isdefined to be the function

ISo= sup(xt — f(t)) (x ER).

(ER

The quantity f*(x) shows how much it is necessary to lower a straight linethrough the origin with slope x in order to make it a supporting line of thesupergraph of f (Figure 4).

1.34. Prove that if a function f: R —, (—oo, ooj is convex, then:a) f is convex;b) the set 91(f) = {x E R I fOx) <+oo} is an interval, and if a point

is not an endpoint of then

1(t) = sup{g(t) I g is a linear function, g � 1);

c) (f)*(t) = 1(t) if t is not an endpoint ofd) Young's inequality f(t) + f(x) � xt holds for all x, t ER;

FIGURE 4

§2. SMOOTH FUNCTIONS 73

e) the function f* is finite on the interval = (inff'(t), supf'(t)) ,wherethe infimum and the supremum are over those points t E at whichf'(t) exists, and for x E the supremum in the definition of f*(x) isattained;

f) if s = f'(t) < ff is linear on f is not differentiable at a point p = f'(t),t E (a, b);

g) if = (a, b) and f is strictly convex and differentiable on (a, b),then 1' is continuous on (a, b), f is differentiable on the open interval

= {f'(t) I t E (a, b)}, and the functions I' and (f*)1are mutually inverse;

h) if 1 � g, then1.35. Prove that if the function is strictly increasing and continuous on

[0, +oo), = O,and = ,then for a E [0, +oo) and b E [0,the inequality holds:

ab< I IJO JO

1.36. Find the Legendre transform and write Young's inequality (seeproblem 1.34d)) for the following functions:

a)

b) (p�l);c) the graph of f is a convex polygonal curve;

d f —t"fp ift�0,+00 t<0 (0<p<l);

( ifltl�a,( +00 ifItI>a (a>0);( ift>0,

+00 ift�0;g) f(t)=e'.1.37. Consider the collection W of pairs (f, g) of convex functions onsatisfying the condition xy � 1(x) + g(y) for all x, y E ILL Call a pair

g0) E W extremal if the relations f � g � g0, and (1' g) E Wimply that I = 10 and g = g0. Prove that the pair g0) is extremal if

and only if = g0 and g0* = 10.1.38. Give examples of functions satisfying the condition f(x) = f(—x)

(x E R). Prove that the only function coinciding with its own Legendretransform is the function 1(t) = t21 2.

§2. Smooth functions

2.1. Prove that if f is differentiable m times on the interval (a, b) and

x, x+mh E(a, b), then hmf(m)(x+0h),where 0<0 <m (see

74 VII. FUNCTIONS (CONTINUATION)

problem 111.3.1 for the definition of2.2. Supposethat ak,bk,ckER, for k=1 n. Provethat

if f E C(R),

f(x)= E akf(bkx+cky)1<k<n

for all x,yER,then f€C°°(R).2.3. Prove that for any function f E C°°([a, bJ) and any numbers e> 0

and n E N there is a polynomial P such that — p(k)(x)I <e for allxE[a,bJ,k=0,1 n.

2.4. Suppose that f E C°°(R), f 0. Prove that iff vanishes on some

nonempty interval, then = +00.

2.5. Let 1(x) = where Qk (k = 1 n) arereal polynomials, x E R trove that I has finitely many zeros.

2.6. Suppose that f E C°°([0, +oo)), [(0) = 1(x). Prove that

for each n E N the function has a zero.2.7. Let [E C2((0, 1J), and suppose that f(x) = o(1) and f"(x) =

O(x2) as x —' +0. Prove that f(x) = as x —, +0.2.8. Let f and g be smooth even 2T-periodic functions that are de-

creasing on [0, TJ. Prove that their convolution, that is, the function h(t) =— x)dx, has the same properties.

2.9. Let f be n times differentiable on [0, 1J, and suppose that= J(k)(1) = 0, k = 1 n — 1. Prove that for some x the inequalityholds:

� — 1(1)1.

2.10. Suppose that is an interval, f E and Mk =k=0. 1,2.

a) Prove that M1 � in the case = R, and that the constantcannot be decreased.

b) Obtain an analogous inequality in the case when =c) Prove that if = [0, 2J and M0, M2 � 1. then M1 � 2 (the estimate

is sharp).d) Study the question for an arbitrary interval = [a, bJ.2.11. Prove that if f E C2([0, +oo)). then the following inequalities

hold:a) 11(0)12 � lf(x)l dx lf"(x)l dx;b) � + lf"(x)l) dx.

Are these estimates sharp?2.12. Find minf If"(x)I2dx under the condition that f E C2([0, 1J),

1(0) = [(1) = 0, and 11(0) = a.2.13. Let [E C°°(R). Prove that if for each x E R there is a number

§2. SMOOTH FUNCTIONS 75

fl E N with f is a polynomial.2.14. Let be an arbitrary sequence of real numbers. Prove that

there is a function I E C°°(R) of the form 1(x) = +such that =

a E thatif 1(a) = 0, then

f(x)=

where x = (x1,..., xv), E ,and = (j = 1 n).

2.16. Let a E and suppose that F is a mapping of the setinto R satisfying the following conditions g E

a) F(f+g)=F(f)+F(g);b) (a ER);c) F(fg) = F(f)g(a) + f(a)F(g).Prove that

F(f)=J

forsomec1,...,2.17. Let f and g be nonnegative continuous functions defined on

[0, oo). Prove that if for some number C � 0

f(t) � c-Ej f(x)g(x)dx, t �0,

then

f(t) � Cexp (j g(x) dx)

for all t � 0 (Gronwall's inequality).Deduce from this that if h E C'([O, oo)), h(0) = 0, M � 0 and Ih'(t)I �

MIh(t)I for t � 0, then h 0.

A function f: C is said to be positive-definite if

(1)

j,k=1

foranyt1,...,A function I is said to be conditionally positive-definite if(1) holds under

the additional condition that z1 + + Zm =0.

2.18. Prove that:a) the function f(t) = (t E Rl?) , where a E R'1, is positive-definite;

b)thefunction f(t)=Acos(a, t)+B A,BER,and A > 0, is conditionally positive-definite.

76 VIL FUNCTIONS (CONTINUATION)

2.19. Prove that if is a finite measure on then the function

1(t)= j djz(x) (t Er)

(the Fourier transform of the measure u) is positive-definite.2.20. Prove that the functions 1(t) = e" and g(t) = 1/(1 +t2) (t ER)

are positive-definite.2.21. Prove that:a) the function 1(t) = (t E is conditionally positive-definite;b) for 0 <p <2 the functions f(t) = (t E are conditionally

positive-definite.2.22. Prove that if a function f: C is positive-definite, then:a) f is positive-definite;b) f is bounded;c) continuity at zero implies uniform continuity of 1.2.23. Prove that if the function 1(t) = is positive-definite for all

s > 0, then the function C is conditionally positive-definite.

Suppose that = [a,b], f E and f(x) 0 (x E is). TheSchwarzian derivative Sf of f is defined by

Sf(x) — f"(x) 3

— f'(x) 2

— 6d ( f(x)

—e

2.24. Verify that the following assertions are valid for real functions sat-isfying the above conditions:

a) if the composition fog is defined, then

S(fo g)(x) = Sf(g(x))(g'(x))2 + Sg(x):b) if h(x) = /J)/(yx +ö) and the composition g = h of is defined,

then Sg(x) = Sf(x);c) if Sf(x) <0 for all x E then also S(f)(x) <0 (x E for any

n EN;d) if Sf(x) <0, then S(f')(y) >0 at the point y =e) the inequality Sf(x) <0 (x E holds if and only if the function

g = IfhIhhI2 is convex;0 if Sf(x) <0 for all x E then the function il does not have a

strictly positive local minimum in (a b) (the modulus princi-ple").

It can be seen from problems X.1.37—40 why it is negativity (and notpositivity) of the Schwarzian that is important.

2.25. a) Verify that the functions f(x) = sinx, f(x) = 1(x) =xe_X, f(x) = e' ,and 1(x) = Ax2 +Bx+C satisfy the condition Sf(x) <

§3. BERNSTEIN POLYNOMIALS 77

0 if f(x) 0, while the functions 1(x) = x + x3, f(x) = tanx, and

f(x)=cotanx donot (x,A,B,C€R).b) Verify that any polynomial f of degree n � 2 with all zeros real

satisfies the condition Sf(x) <0 if f(x) 0.2.26. Let

(z — w)(y — x)R(w,x y,z)=

(z—y)(x —w)

(w. x, y. z be pairwise distinct numbers). Prove that if f E f ispiecewise monotone, and

R(f(x1), f(x2), f(X3), f(x4)) > R(x1, x2, x3, x4)

for any points x1 <x2 <x3 <x4 belonging to a single interval of mono-tonicity of f, then the function il does not have a strictly positive localminimum at interior points.

§3. Bernstein polynomials

3.1. Let n E N and r = 0, 1 For x ER let

Snr(X) = — — flXy

O<k<n

Prove that:a) Snr+i(x) =x(1 --1(X)), rEN;b) = = = nx(1 — =

nx(1 —x)(1 —2x), = nx(1 —x)(1 +3(n—2)x(1 —x)).3.2. Let n E N and ö > 0. For x E [0, 1] let

O<k<nIk/n--xI�o

Prove that:1 1 17 —2n5

a) b) c)

Let n E N. The polynomial

x) = —

O<k<n

is called the nth Bernstein polynomial of the function f: [0, 1] ILL

3.3. Find the Bernstein polynomials of the following functions:

a) f(x)=xm, m=0, 1,2; b)f(x)=aX, a>0.3.4. Suppose that f is defined on [0, 1], and let n EN. Prove that:a) and 1)=f(1);b) x) � x) for all x E [0, 1].

78 VII. FUNCTIONS (CONTINUATION)

3.5. Prove that:a) if f is increasing (decreasing) on [0, 1], then the Bernstein polynomial

is increasing (decreasing) on [0, 1];b) if f is convex (concave) on [0, 1], then the polynomial is

convex (concave) on [0, 1].3.6. Suppose that I is defined and bounded above on [0, 1]. Prove

that:a) x) � sup10 for all n E N and x E [0, 1];b) if the interval is contained in [0, 1], then x) � f

for all xE/s;c) if the intervals and are such that C C [0, 1] , and I

then there is an index N = N(f, large enough that x)f for all x E and n > N.

3.7. Suppose that the functions f and g are bounded on [0, 1] andcoincide in a neighborhood of a point x0 E [0, 1]. Prove that the sequences

x0)} and x0)} are convergent or divergent simultaneously,and in the case of convergence they have a common limit.

3.8. Suppose that the function f is bounded on [0, 1]. Prove that:a) if f is continuous at x0 E [0, 1], then x0) f(x0);b) if f is continuous at each point of a closed interval C [0, 1], then

onc) if x0 E (0, 1) is a point of discontinuity of f of the first kind, then

—, + 0) + f(x0 — 0)).3.9. Suppose that f E C([0, 1]). Prove that:a) if dx = 0 for n = 0, 1 then f 0;b) if = 0 for n E I is a linear function.

3.10. Suppose that f E C([0, 1]) and f(x)g"(x)dx = 0 for anyfunction g E C2([0, 1]) equal to zero in neighborhoods of the points 0 and1. Prove that f is a linear function.

3.11. Suppose that I E C([0, it]) and e > 0. Prove that there exists atrigonometric polynomial T of the form T(t) = >0<fl<N cos nt such thatIf(t) — T(t)I <e for all t E [0, it]. —

3.12. a) Let f E C([0, 1] x [0, 1]). For n E N and x, y ER let

x, y) = L) —

Prove that I on [0, 1] x [0, 1] as n —, +oo.b) Suppose that f E C([0, it] x [0, it]) and e > 0. Prove that there exists

a trigonometric polynomial T of the form

T(u, v) = ak) cosku cosjv

such that If(u, v)—T(u,v)I <e forall u,v E[0, it].

§3. BERNSTEIN POLYNOMIALS 79

3.13. Suppose that f E Cr([0, 1]). Prove that fi) on [0, 1].3.14. a)Let IELiP [0, 1], Provethat

M(x(1 — x a positiveconstant depending only on f.

b) Let E [0, 1J and let fa(x) = Prove that �n

f is convex on [0, 11, then x) � 1(x) for all n ENand xE[0, 1].

b) If a continuous function f on [0, 1J is such that x) � 1(x) forall n E N and x E [0, 1], then I is convex on [0, 1].

3.16. Let f E C2([0, 1]). Prove that:x(1—x)

x) — 1(x) = 2n(11(x) +

where the sequence {ej of functions converges uniformly on [0, 1] to zero(the Voronovskaja formula).

3.17. Suppose that f E C([0, 1J), g E C2([0, 1]), and g is equal tozero outside (a, b), where 0 <a <b < 1. Prove that:

O<k<n

g —x)g(x))"dx.

3.18. Let the function f: [0, 1] —. R be such that

sup x) — f(x)I = o(1/n).O<x<I

Prove that I is linear.3.19. Let f E C([0, 1]). Prove that the condition 1(0), 1(1) E Z is

necessary and sufficient for I to be the limit of a uniformly convergentsequence of algebraic polynomials with integer coefficients on [0, 1].

3.20. Let f be defined on [0, 1]. Prove that:

x) = 1(0) + E xk

1<k<n

where = f(y + — 1(Y)' = k EN.3.21. Let 1(x) = xm (x ER), m = 0, 1, 2 Prove that:a) 41 on any finite interval;b) x)I � max(1, IxItm) for any x ER.3.22. Let 1(x) = cmxm for x E (—R, R) where R> 1. Prove

that:a) 4f on any interval [—r, r] if 0< r <R;b)ifCm�0 forany m,then on [0,1],and on

[1,R).3.23. a) Prove that for a fixed n E N the polynomials —

k = 0, 1 n, form a basis in the space of polynomials of degree at mostn.

80 VII. FUNCTIONS (CONTINUATION)

b) Prove that the coefficients in the expansion of an mth-degree poly-nomial P with respect to such a base with n � in, which form the table

a0 1 ,n mi-i

am+1 am+i am+1 am÷i

... ::: ::: ::: ::: ::: ::: ...

satisfy the equalities

0 0 n+i n k k—i k (k= 1,2 n)

(cf. the equalities for the binomial coefficients (i)).c) Prove that P(x) > 0 for x E (0, 1) if and only if for sufficiently large

n the numbers (k =0 n) are not all zero and are nonnegative.

§4. Almost periodic functions and sequences

We write x y if Ix — <e, and x y (mod 1) if there exists aninteger k such that Ix — y — kI <e.

A set X C R is said to be relatively dense if there exists an L > 0 suchthat each interval of length L contains at least one number in X.

4.1. Prove that if is an irrational number, then for any a E [0, 1) andany e > 0 there is an n E Z such that a (mod 1). Verify that the setof such n is relatively dense on the line (cf. problem 1.3.2).

4.2. a) Prove that if 0 and the ratio is irrational, thenfor any a1, a2 E [0, 1) and any e > 0 there is a t E R such that

(mod 1), (mod 1). (1)

Conversely, if the system (1) is solvable with respect to t for any a1, a2,and e, then is ilTational.

b) Prove that if and are irrational, then the set of integersolutions of the system (1) is relatively dense on the line.

4.3. What condition must the numbers E R satisfy in orderthat the system of equations

(mod 1), i=1,2 k,

have for any e > 0 a solution t with > for some 1?4.4. Find a condition which the numbers E R must satisfy in

order that the system of equations

(mod!), i=1 k,

have for any afl..., ak E [0, 1] and any e > 0:a) a real solution t;b) a relatively dense set of solutions.

ALMOST PERIODIC FUNCTIONS AND SEQUENCES 81

4.5. Find a condition which the numbers E R must satisfy inorder that the system of equations

(mod!), i=! k,have for any a i,..., ak E [0,!) and any e > 0:

a) an integer Solution m;b) a relatively dense set of integer Solutions.4.6. Let f(x) = sin(ax + b) + sin(cx + d), x E R, where ac 0. Prove

that if f is not identically equal to zero, then it takes values of differentsigns.

A complex-valued function f on R is said to be uniformly almost periodic(UAP) if for any e > 0 there is a number L = L(e) > 0 such that eachinterval of length L contains at least one number r (an e-almost period)for which

If(x+z)—f(x)I<e forallx€R.

4.7. Prove that each trigonometric quasipolynomial

f(x) = E R, ak E C)1<k<n

has the property of uniform almost periodicity. Verify that f is periodic ifand only if the factors .., are pairwise commensurable (that is, theirratios are rational).

4.8. Let f E be a function having period 1 with respect to eachargument, let E R be rationally independent numbers, and let

E R. Prove that the function

tER

is uniformly almost periodic.4.9. Prove that if f is a UAP function, and the numbers L(e) are

bounded for small e, then f is periodic.4.!0. Prove that a continuous function on R that is UAP is uniformly

continuous on the line and is bounded. Does there exist an unbounded non-periodic UAP function? Does continuity follow from boundedness?

4.11. Prove that the limit of a uniformly convergent sequence of UAPfunctions on R is a UAP function.

4.!2. Prove that if a UAP function has a derivative that is uniformlycontinuous, then the derivative is also a UAP function.

4.!3. Prove that if a primitive F of a continuous UAP function f isbounded, then it is also a UAP function.

4.14. Suppose that f E C(R), h E R, and fh(x) = f(x + h). Provethat f is UAP if and only if for each numerical sequence the sequence

{fh } has a subsequence uniformly convergent on R (compactness of the

of translates).

82 VII. FUNCTIONS (CONTINUATION)

4.15. Prove that a sum or a product of continuous UAP functions is againa UAP function.

4.16. Prove that the mean value

1çT

M(f) = lim —! f(x) dxT—'+oo Tjr

exists for each continuous UAP function.4.17. Let (f, g) = where f and g are continuous UAP func-

tions, and the functional M is defined in the preceding problem. Provethat:

a) (f, g) has the properties of an inner product (in particular, 1) = 0

if and only if I 0).b) the functions = t E R, form an uncountable orthogonal

family with respect to this inner product;c) the quantities = E R, k = 1 n) satisfy the

inequality

1<k<n

(Bessel's inequality for the system Conclude from this that 0for at most countably many values A. Can this set be an arbitrary countablesubset of R?

4.18. Let a nonperioclic sequence. Define thesequences and as follows:

b J0 a—flifa ifa

Prove that at least one of the sequences and is nonperiodic.A binary sequence e = {ek}kEN is said to be a!?nost periodic if for any

"word" (ek+ ek+1) (we also write €k+1 in e there exists a num-ber LEN such that the set {n EN I = j = 1 1} intersectsevery segment of the natural numbers of length L. Almost periodicity fortwo-sided sequences {ek}kEZ is defined similarly.

4.19. Let {ek}kEz be an almost periodic sequence, and let 1(x) =(x E R, [x] the integer part of a number x). Is I a UAP function?

4.20. Is it true that for every almost periodic sequence {ek}kEz theCesàro means have a limit

lim—n<k<n

4.21. Let S = {(x, y) I kx � y � k(x + 1) + 1}, a strip in R2 withk an irrational number, and let Z = S n Z2 be the set of integer points inS. Prove that the orthogonal projections of the points in Z on a side of thestrip (see Figure 5) partition it into segments with lengths taking exactly two

§4. ALMOST PERIODIC FUNCTIONS AND SEQUENCES 83

values a and b. After numbering the Segments by the integers in order oftheir location on the line, let = 0 if the ith Segment has length a, and

= 1 otherwise. Prove that the sequence {ej}kEZ is almost periodic.4.22. Consider the sequence = = 0110100110010110... (the

Morse sequence) formally defined by the equalities = 0, €2 = 1, €k =1 — for <k � (n E N). Prove that it is almost periodic.

4.23. For a finite sequence ("word") A = €l€2 of 0's and l's let A°denote A, and A' = where = 1

— €k. Define a sequence eby determining its initial segments A,, recursively: A0 is an arbitrary word,

A0 = e, A, = 4 ,where is some word of 0's andl's, A2 = E {0, 1}, and so on. Thus, A0 serves asthe initial part of A,, A, as the initial part of A2, and so on. The infinitesequence determines some sequence e. Prove that e is periodic oralmost periodic.

4.24. A sequence offolds is defined to be a sequence s = of 0'sand l's whose initial segments of length 2k — 1 (k = n E N) havethe form . . •. (where = 1 — si). The name "sequenceof folds" is due to the fact that with the word it is possible to associatea sequence of folds of a paper strip that are formed on the strip as a resultof bending it n + 1 times (first in halves, then once more in halves, and so

FIGURE 5

a

84 VII. FUNCTIONS (CONTINUATION)

on). If upon moving along the strip a 1 is associated with each fold leadingto a bend to the left, while a 0 is associated with a "right bend" fold, then

is obtained (see Figure 6). The sequence s is uniquely determined byspecifying the numbers 2

Prove that every sequence of folds is almost periodic, but not periodic,not even if we confine ourselves to terms with sufficiently large indices.

4.25. Prove that sequences of folds (see problem 4.24) have the follow-ing property (Besicovitch almost periodicity): for each e > 0 there exist anumber L >0 and numbers E N (m E N) such that:

a) if is the number of falling in the interval C R4, thenfor all intervals of length L;

b) for each m E N

c)

— <e;

iim1im—L

4.26. Prove that if = n E N, where is a sequence offolds, then for any t E R the limit

1 --2nktzC =lim— xke1<k�n

(Fourier coefficient) exists; moreover,(m+2)= if t = (2! + 1)2 in 0.

and C, = 0 otherwise. Verify "Parseval's equality"

lim! IXkI1(k(n O�td

4.27. Consider the binary sequence

r = = 00010010000111010001001011100010...

FIGURE 6

§4. ALMOST PERIODIC FUNCTIONS AND SEQUENCES 85

(the Rudin-Shapiro sequence), whose initial segments of length have

the form = where and are words of length andA1 = C1D1 = 00, while and are defined recursively for n> 1:

= = = is obtained from bythe substitution 0 1).

Prove the almost periodicity of this sequence. We note that the sequenceis the sequence of coefficients of the Rudin-Shapiro polynomials

(see problem IV.6.29).

CHAPTER VIII

Lebesgue Measure and the Lebesgue Integral

In this chapter the word "measure" means Lebesgue measure on oron the sphere 5m1 = {x E Rm

I lxii = 1}. Lebesgue measure on Rm isdenoted by the symbol 'tm' and for m = 1 by the letter

Lebesgue measure

1.1. Suppose that Ek C (0, 1), (k = 1, 2 N), and >1<k<Nt(Ek)> N — 1. Prove that Ek) > 0.

1.2. Suppose that E C S1, z ZN E S1 , and ji is Lebesgue measure

on S1 ("arclength"). Prove that if jz(E) > — 1/N). then the set E,suitably rotated, contains all the points zi,..., ZN, that is, there is a point

ZOES'suchthatZOzkEEfork=l N.1.3. a) Prove that if E C Rm and A,,(E)> 1, then there are two (dis-

tinct) points x, x1 E E such that the differences between the correspondingcoordinates are integers.

b) Let V c Rm be a convex set centrally symmetric with respect to zero,and let > ?. Prove that V contains a point a 0 with integercoordinates.

c) Let V C Rm be a convex set centrally symmetric with respect to zero,and let > where N is some positive integer. Prove that Vcontains at least 2N nonzero points with integer coordinates.

1.4. Find the measure of the set of points in (0, 1) whose decimal frac-tion expansion:

a) contains a 4 at a given place;b) contains a prime number at a given place;c) contains given numbers at two given places;d) contains numbers of different parity at two given places.1.5. What is the measure of the set of numbers in (0, 1) whose decimal

expansions contain a 0?1.6. Let Ek C (0, 1) (k E N). Determine whether there is a subsequence

{Ek} such that > 0 if one of the following conditions holds:

a) = 1; b)

88 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

1.7. Suppose that > 0, and <00. Consider the set

A = {x E (0, 1)1 the inequality

Ix — (p E Z, q E N) has infinitely many solutions}.

Prove that = 0.1.8. Supposethat 0<0<1, EcRm,and 0<A(E)<oo. Prove that

there is a cube such that <A(E fl1.9. Let E, E0 c R be arbitrary measurable sets of positive measure.

Prove that:a) 0 is an interior point of the set E — E = {x — y

Ix, y E E};

b) E contains two (distinct) points separated by a rational distance;c) the sets E+E0 = {x+y XE E, y E E0} and E.E0 = {xy xE

E • y E E0} have interior points.1.10. Suppose that E c R and > 0. Prove that if (x + y)/2 E E

for any points x, y E E, then E is an interval (cf. problem 1.1.19).1.11. Suppose that ii = {nk} is a strictly increasing sequence of positive

integers, and let

E,,= ek2k I = 0 or 1 }

(see problem 1.1.37).a) When isb) Prove that if Iim(nk+l — nk) � 2 + log2 m (m E N), then the set

E(m) = + E,, + + (m terms) has measure zero.n}.

Prove that if — Ilk) = oo, then = 0.1.12. Let E = {>k>1 eklk! I = 0 or 1}. Prove that E is a closed set

without interior points, and = 0. Is it true that the measure of the setE + E + ... + E (m terms) is equal to zero for any m E N?

Let be a closed interval of length Let {1n}n>O be a sequence of pos-itive numbers satisfying the condition (n E N). Consider a setE1 C containing the endpoints of and consisting of two closed intervals

and of length each. The intervals and are called the inter-vals of the first rank. Replacing by 12 and repeating the same operationwith and instead of we get four intervals ofsecond rank U C U C whose union is denoted byE2. A similar construction applies for the intervals of the third, fourth, etc.,ranks. The union of the intervals of rank n is denoted by E,. The setE = is called the set of Cantor type constructed with the help of thedefining sequence {1,} on the interval (cf. the construction of the Cantorset and of generalized Cantor sets (problems 1.1.27 and 1.1.31)).

1.13. a) What is the defining sequence for the Cantor set?b) What is the defining sequence for the set in problem 1.12?

§2. MEASURABLE FUNCTIONS 89

c) When is the set in problem 1.11 a set of Cantor type? In this case whatis the defining sequence for it?

1.14. Construct a set E c [0, 1] of Cantor type having positive measure.Show that the measure of such a set can be arbitrarily close to 1.

1.15. Prove that there exists a set A c [0, 1] such that for any (non-empty) interval C (0, 1)

and

1.16. Let A c R2 be a compact convex set, and let be its s-neighbor-hood, that is, the set UXEA B(x, e). Prove that = a + be + Ce2, andfind the coefficients a, b. and c.

1.17. Prove that every (nonempty) open Subset of Rh can be representedas the union of a sequence of disjoint balls and a set of measure zero.

1.18. Let be a sequence of pairwise congruent subsets of S' satis-fying the following conditions (see problem 1.1.1 1):

S1 = n Em 0 for n m (n, m E N). Prove that the setsare not measurable.

1.19. The mapping x '—' {x + O} where {y} is the fractional part of anumber y and x E [0, 1), is called translation by 0 modulo 1. Replac-ing rotation by the angle by translation by 0 modulo 1, construct anonmeasurable subset of [0, 1) by analogy with problems 1.1.11 and 1.18.

1.20. Using the result of problem 1.19, show that every set E Chaving positive measure contains a nonmeasurable subset.

1.21. Let 2t be a system of subsets of N satisfying the following condi-tions:

1) if card(A) = card(N\A) = 00, then A E 2t if and only if N\A 2t;2) if A E 2t, C C N, and card(C) <00, then AU C E 2t and A\C E 2L.

(The existence of such a system of sets can be proved with the help of Zorn'slemma (see, for example, [13], p. 33)).

Prove that the set E = {EkEA2—k

I A E 2t} is nonmeasurable.1.22. A set E C Rm is said to generate a tesselation if its translates by all

possible vectors with integer coordinates cover Rm without overlapping, thatis, Rm = U,Ez'"(l + E), and (I + E) n (1' + E) = 0 for I 1' (1, 1' E Zm).

Prove that the measure of a measurable set generating a tesselation is equal

1.23. Suppose that E C Rm, > 0, and A is a countable densesubset of Rm. Prove that + E)) = 0.

Measurable functions

The symbol denotes the set of all Lebesgue-measurable functionson a (measurable) set E C that are finite almost everywhere. Two func-tions in are said to be equivalent on a set E0 C E if they coincide

90 VIIL LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

almost everywhere on E0. The symbol x4 denotes the characteristic func-tion of a set A C

2.1. Suppose that f 5! °(E), and fl � 1 almost everywhere (a.e.) onE c Prove that if Ill < 1 on a set of positive measure, then there existmutually nonequivalent functions 11 , 12 E 5!°(E) such that

1111 � 1 and 1121 � 1 a.e. on E, and I = + 12)/2.

2.2. Give an example of a monotone function on R that is not equivalentto a continuous function, not on any (nonempty) open interval.

2.3. The Dirichlet function (equal to 1 at rational points and to 0 atirrational points of the real line) is discontinuous at each point, but equivalentto a continuous function. Does there exist a function in 5!0([0 1]) suchthat any function equivalent to it on [0, 1] is discontinuous at each point ofthis interval?

2.4. Let K be the Cantor set, f the Cantor function (see problem111.3.17)), and g(x) = (x E [0, 1]). Prove that:

a) g is strictly increasing;b) A(g(K)) > 0;c) there is a (measurable!) set H c K such that g(H) is nonmeasurable.

Let E c and f E 5!°(E), and suppose that for any t E R the measureof the set E(f < t) = {x E E I 1(x) <t} is finite. The function

F(t) = < t) (t E R)

is called the distribution function of f (on the set E). Functions having thesame distribution functions are said to be equimeasurable.

2.5. Prove that a distribution function is left-continuous. When is itcontinuous at a point t0 E R?

2.6. Find the distribution functions on [0, 22E] of the following func-tions:

a) sinx, b) cosx, c) sin d) sin(2x —

2.7. Determine whether the functions sin x and (n E N, E

R) are equimeasurable on the following sets: a) [0, 22EJ; b) [0, 2E].2.8. Let F be the distribution function of a function f E T]).a) Find the distribution functions of the functions f+ c, cf (c > 0), and

- b) Let I be the extension of I to R with period T, and let g(x) =f(x — c) (x E (0, T)). Find the distribution function of g.

2.9. Suppose that E c = 1, and I E 2°(E).a) Prove that there exists a unique number t0 such that � t0) �

and forany e>0.b) Prove that � t0) �

§3. INTEGR.ABLE FUNCTIONS 91

c) A number M satisfying the inequalities

� M) � � M) �is called a median of f. Is a median unique?

Let f€Sf°(E).

The function f* is called the nonincreasing rearrangement of f.

2.10. Prove that the functions I and are equimeasurable. Assumingthat the distribution function F of a function f is continuous and strictlyincreasing, determine how F and f are connected.

2.11. Find the nonincreasing rearrangements of the following functions:a) sinx on [0, 22E]; b) sinx/2 on [0, 22E]; c) tanx on (0,

2.12. Find the nonincreasing rearrangement of the function2 2 2 2f(x,y)=x +y (x +y �1).

2.13. Let f(x) = >1<k<flak/(x—ck), where a1 >0 >0, andC1,..., çER. Provethatforany t>0

where A = ak.

§3. Integrable functions

In this section all the sets and functions under consideration are assumedto be measurable. The set of functions integrable with respect to Lebesguemeasure on a set E C Rm is denoted by as usual, fE 1(x) dx denotes

the integral of a function I over a set E with respect to Lebesgue measure.The characteristic function of a set A is denoted by X4.

3.1. Suppose that the union of any three sets in the family E1,..., ENcoincides with [0, 1]. Let

S1 = and S2 = fl Ek).

Express the integral

dx0

I

in terms of S1 and3.2. Suppose that each point of [0, 1] belongs to at least k of the sets

in the family E1, ... , EN. Prove that � k/N for some n.3.3. Suppose that E C Rm, <00, and f E 2°(E). Prove that

the following conditions are equivalent:

92 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

a) f€Sf(E),b) >.k>1 � k)) <00,c) >.k�1 � Ill <k + 1)) <00.3.4. Suppose that the sets Ek C R'1 (k E N) are distinct and that

<00, and let

= {x E R'1 I x E Ek for precisely n values of k},

= {x E Rm I x E Ek for at least n values of k}.

Prove that the sets and are measurable, and that

�k>n

= =n�1

3.5. a) Prove that the series where is an arbitrarynumerical sequence, converges almost everywhere on ILL

b) Construct a function in that is nonintegrable on every (non-empty) interval.

3.6. Can an everywhere differentiable function on [0, 1] have a deriva-tive that is not integrable on [0, 1]?

3.7. Determine for what p E R the function f is integrable on the givenset in the following cases:

a) xE(+1,+oo);b) xE(0,+oo);c) xE(0, 1);d) xE(0,+oo);e) f(x)=x"/(1+x6sin2x), xE(0,+oo).3.8. Are the functions f below integrable on the set [—1, 1] x [—1, 1]?

Find their iterated integrals

J (J f(x, Y)dY) dx and J (1 f(x, y) dx)

for:

a) b)

c)f(x,y)= XY ,, d)f(x,y)=(x +

(in all these cases the value f(0, 0) can be defined arbitrarily).3.9. Determine for what p E R the function f(x. y) = 1 —

((x y) E R2) is integrable on the set E in the following cases:

§3. INTEGRABLE FUNCTIONS 93

a) E=[0,1]x[0,1],b) E=[0,2]x[0,2],c) E={(x,y)€[0, 1]x[0, 1]I(x—2)2+y2>2, x2+(y—2)2�2}.3.10. a) Let p > 0 and E c Rm. Prove that if' = r)),

then fEdy/lix — y11P � for any ERtm

b) Prove thatI fE

eiX dxl � for any set E C [0, 22E].c) Prove that

for any set E (E c R2) of finite measure.3.11. Let p, t >0, E c Rm, and f E Prove that

� t) � (Pj dx.

In particular, for p = 2 we get inequality:

� t) � t2j If(x)I2dx.

3.12. Prove that if E 2(E) (E C Rm), then

� t) = (t —, +oo).

Given an example showing that the converse is false.3.13. Suppose that p > 0, E C Rm, I E and AmE(If1 � t) =

0(C") (t — +oo). Prove that if <00, then E 2(E) for anyE (0, p).3.14. Let E C Rm and f E Prove that

JEdx

p-.+O (Ill 0).

3.15. Let f, g E f, g � 0. Prove that if fE g(x)dx = 1 and< oo then

I/p

(JEf"(x)g(x) dx)

p—.+OefE

Can the condition fE g(x)f(x)dx <00 be dropped?

3.16. Let x E R, and let x = [x] + where = ek(x) = 0 or1, be the binary representation of x.

a) Find the integrals and ek(x)em(x)dx (k, m EN, k

94 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

b) Find the integral — dx, where =

c) Prove that — dx = O(n2).3.17. Let x E R, and let x = [x] + eklO, where ek = =

0, 1 9, be the decimal representation of x. Let I be one of the num-bers 0,1

(1 ife1(x)=j,10 if

Find the integrals (k, m E N, k m)

a) b)

3.18. Suppose that a E R, a 0, 1, and x =10k

is thedecimal representation of x E (0, 1). Let

a

if the first 0 has even index;

0 otherwise.

Prove that f is measurable, and find 1(x) dx.3.19. Open intervals of length 1/3, 1/9, etc., were removed successively

from [0, 1] in the construction of the Cantor set (see the discussion preced-ing problem 1.1.27). Let 1(x) = n on the open intervals of length and1(x) = 0 on the Cantor set.

a) Prove that I is measurable, and find 1(x) dx.b) Find the distribution function F of I and the nonincreasing rear-

rangement f* of f (see definitions before problems 2.5 and 2.10).3.20. Let K be a closed subset of [a, b], and let p(x) =

inf{Ix — tI It E K} be the distance from a point x E R to the set K. Prove

that for any s > 0 the function = — dx (y E K) isintegrable on K.

3.21. Prove the equalities (0 <a oo)

a) f(maxl<k<mlxkl) dx = m. f 1(t) dt,

b) f(x)dx = 1(y))dt, where f isassumed to be nonnegative, and is (m— 1)-dimensional Lebesgue mea-sure on the boundary of the cube [—1,

Below, the symbol denotes Lebesgue measure (surface area) on the

sphere SmI = {x ERtm I lixil = 1}. The area of the sphere, i.e., the quantityis denoted by

§3. INTEGRABLE FUNCTIONS 95

3.22. Prove by induction that

a) j f(x1,...,=

o {j_2 f(u1 sinO sinO, cosO) 2(u)} dO

(n � 3),

b) j f(x)dx= j {J f(tw) dt (n � 2),

where f is assumed to be integrable on or respectively. Note thatthe formula a) remains true also for n = 2 if S° is understood to be theboundary of [—1, 1], that is, the two-point set {—1, 1}, and the measure

is assumed to consist of a unit load at the points +1 and —1.3.23. a) Let r be a nonnegative continuous function on and let

T = {x EJ0 < lxii � r(x/ilxll)}. Prove that

= ! j rh(w)

b) Let A be a bounded neighborhood of zero in R", and let p(x) =sup{l(x, I y E A} and V = {x E p(x) � 1}. Prove that

=

In the solution of problems 3.24—3.26 it is useful to employ the formula(see problem 3.22)

J f(IIxII) dx =Jb

a

a (measurable) nonnegative function on the interval (a, b) (0 �a < b � oo), and = 1)).

3.24. Let E be a (measurable) subset of (0, +oo), and let

n—iProve that = dt.

In problems 3.25, 3.26, and others the symbol denotes the standard

Gaussian measure on , that is, the measure with density

3.25. Compute (r>0).3.26. For a> 0 let Ka be the cone in given by

4 2 2 2 22Ka{(xi,x2,x3,x4)EIR 1x2+x3+x4�ax1}.

Find y4(K0).

96 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

3.27. Compute the volume of the generalized torus T c given by

where B is the disk of radius R <a about the point (0, a).3.28. Suppose that A is a compact subset in the upper half-plane of the

space R2. If is the body obtained by rotating A about the x1-axis in R3,then by Guiclin's theorem, = where is the ordinate ofthe center of gravity of A. Regarding R2 as canonically imbedded in R4,consider the body T "obtained by rotating A about the x1-axis in R4 ," that

Prove the following modification of Guldin's theorem: =where p is the radius of inertia of A with respect to the x1-axis, that is, thepositive number determined by the equality

= Jj y2 dxdy.

3.29. Let A be a compact set in the half-plane x2 � 0 in R2. Regard-ing R2 as canonically imbedded in consider the "body T obtained byrotating A about the x1-axis in ," that is,

Prove that

a) = 2Jj dx1 dx2,

b) = Jf dx1 dx,,

where is the "Gaussian volume of T" is the measure on with

density3.30. Find the measure of the set

(p>O).1<k<n

3.31. a) Compute the integral = I(x,b) Prove that

+ ...+

— r((m+q)/2) r(n/2) (— r(n + q)/2) r(m/2)

for m,nEN, 1�m<n,and rn+q>0.

§3. INTEGRABLE FUNCTIONS 97

3.32. Let a, b E and let 0 be the angle between the vectors aand b (0 � 0 � it). Prove that

J sgn(a, x)sgn(b, = (1 —

3.33. Let = r))(1 r)))dr. Prove that:a) = Ilixil —

b)

3.34. Let A c be an open set star-like with respect to zero, let q4(x) =

inf{t > 0 I ('x E A} (x E be the gauge function of the set A, and letji be a probability measure on such that I(x, y)Idu(x) <+00 forany y E Prove that

j ji(tA)(1 — jz(tA))dt=

q4(x) — <+00.

3.35. Suppose that f is an integrable function on the cube Q = [0, l)m

and has period 1 in each variable. Prove that:a) the integral f(x)dx does not change under a translation of Q by any

vector, that is, 1(x) dx = fa+Q f(x)dx for any a E Rm, where a + Q ={a+y yE Q};

b) if A is an arbitrary m x m integer matrix with determinant 1, then

J f(x)dx=JQ 4(Q)

Is this true if the determinant of A is equal to 1, but its elements are notnecessarily integers?

3.36. Let T be an arbitrary compact convex body in Prove that ifthe (n — 1)-dimensional volume of the projection of T on any hyperplane isnot less than S. then diam T � nV/S. where V =

3.37. Suppose that T C is a compact convex centrally symmetricbody with respect to zero, and let 3 be an ellipsoid of maximal volumecontained in T:

3 C T, =I

C T, an ellipsoid}.Prove that T C3.38. Let K C C be a closed bounded set of positive (planar) measure.

Define the function by

=JJProve that:

a)

b) is holomorphic outside K.

3.39. Identifying C2 in the natural way with R4, consider on the sphere

S3 the functions coma (m, n = 0, 1, 2,...) defined by the equalities

C2) =

98 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

a) Compute the integrals

C2) (m, n = 0, 1,2,...);

b) Prove that the system {com n} n-O is orthogonal. —

c) Prove that if the series >mn>O amncomn is uniformly convergent on B4and its sum is equal to f, then following generalization of the Cauchyintegral formula is valid:

C2) = 2,E21s3 (1 —C1z1—C2z2)dji3(z1, Z2),

where C2)EB4, B4= C2)EC2 I 1C112+1C212 < 1}.

§4. The Stieltjes integral

In problems 1—4 the function h is defined on the interval over which theintegral on the right-hand side is taken, and it is measurable and nonnegative.

4.1. a) For 0 � a < b � 00 let E = {(x1,..., Xm) E Rm a �maxl<k<m IXkI � b}. Prove that

I h ( max dx = m2mJE Ja

b) Prove that

J h(maxi<k<mlxkl)dx = mJ (t— 1)m_ih(t)dt.[i,+oo)'" 1

4.2. a) Let 0mfr) = {(x1 Xm) E Rm I >.1<k�m IXkI � r}. Prove that

h IxkI) dx= (m

jrtm_lh(t) dt.

b)Forak>0 (k=1,2 m),let

E= (x1,..., xm)ERmI x1�01% 1<k<m

Compute the integral fE e dx.4.3. a)For 0�a<b<oo let E={xERmIa<IIxII<b}. Provethat

JEll = mcxmjtm_1h(t)dt.

b)Forp>0 let A(p)={(x1 �1}. Provethat

i/p

jh)

dx = mCm(p)j tm_I h(t)dt,

§5. c-ENTROPY AND HAUSDORFF MEASURES 99

where Cm(p) (see problem 3.30 about the value of Cm(P)).4.4. For E C lRtm let = n B(0, t)) (t> 0). Prove that:a) fE h(IIxII) dx = 1000

b) if E C'[O, +oo), then fEh(IIxII)dx =4.5. Suppose that A c 1R2 is a convex bounded set, and let p(x) =

inf{IIx —yll I y E A} be the distance from a point x E 1R2 to A. Find theintegral fR2 dx.

4.6. Let a be the Cantor function (see problem 111.3.17). Compute thefollowing integrals:

a) J xda(x), b) J x2da(x), c) J a(x)dx,

d) Ja(x)da(x), e) Ja(1—x)da(x).

0

4.7. Let K be the Cantor set, and a the Cantor function. Determinethe values of p E R for which the following integrals are finite:

a11 ('da(x)da(y)

JO JO (x2 + y2)P/2

I I 2 2 2' where(u,v)EKxK;Jo Jo ((x—u) +(y—v) )P/

4.8. Let K be the Cantor set, and a the Cantor function. Define thefunction by

9,(z)= I (:€C).Jo J0

Prove that (cf. problem 3.38):a) E C(C),

'-'001;

b) is holomorphic outside K x K.

e-entropy and Hausdorif measures

Let A be a subset of R'1 and e a positive number. A set C C R'1 is calledan e-net for A if A C UXEC B(x, e). Let

N(A, e) = min{card(C) C is an e-net for A}

be the minimal cardinality of an e-net for A. If the set A is bounded, thenN(A, e) <00 for any e >0. The function H(A, e)=log2N(A, e) is calledthe e-entropy of A.

5.1. A set E C RI? is said to be e-distinguishab!e if the distance betweenany two points in it is at least e. Let

M(A, e) = max{card(E)IE C A, and E is e-distinguishable}

100 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

be the maximal cardinality of an e-distinguishable subset of A. Prove thatN(A, e) � M(A, e) � N(A, e/2).

5.2. Let A be a bounded subset ofa) Prove that if A has an interior point, then

(e—'+O).

b) Recall that the outer measure of a set A C is defined to be the

quantity = I A C G, G an open subset of Prove that

if > 0, then N(A, e) � Ce'1, where C is some positive constant.5.3. Find the asymptotic behavior (as e —, +0) of the e-entropy of the

following sets:a)

b)

c) InEN};d) {n° I n E N} > 0);e) {1/ln(n+1)InEN}.

The metric dimension (metric order) of a bounded subset A of is

defined to be

H(A,e) I H(A,e)M- dim A = lim I respectively, r(A) = lim

5.4. Find the metric dimensions and metric orders of the graphs of thefollowing functions:

a) 1(x) = sin 2Ex, 0 � x � 1;b) 1(x) = 0< x < 1.5.5. Find the metric dimensions and metric orders of the following sets:a) Ax[—1,1]cR2,whereb) the graph of the function 1(x) = sin(2Efx2), 0 <x < 1;c) where S(r) is the circle about zero with radius r.5.6. Find the metric dimensions and metric orders of the following sets;a) Ax[—1, 1]cR2,where A={1/ln(n+1)InEN};b) the graph of the function f(x) = 0< x 1;c) S( 1/ ln(n + 1)).5.7. Find the metric dimension and metric order of the Cantor set.5.8. Find the asymptotic behavior of the e-entropy of the set

5.9. Define an increasing sequence {nk} of positive integers by setting= k + m! if m! � k < (m + 1)!, and consider the set A =

§5. c-ENTROPY AND HAUSDORFF MEASURES 101

=0 or 1}. Further, let A, = A + A +•.• + A (the arithmetic sum, with /terms). Prove that:

a) M-dim(A) = 1, r(A) = 1/2;b) = 0 for any / EN.5.10. Let [0, +oo) —, [0, +oo) be a strictly increasing function such

that C N. Consider the set A =I

= 0 or 1}. Provethat

H(A, e) (e —, +0).5.11. Find the metric dimension and metric order of the set

IY—>.'1k3''lk—O, 1,2,

L €k and17k

have the same parity for any k,(the Hironaka curve).

5.12. Let A C R' and let A, = A + + A (1 terms).Prove that:a) if r(A) = 0, then = 0 for any / E N;b) if r(A) then ).(A,) = 0.c) Is it true that if r(A)> 1/2, then )(A + A) > 0? (Consider the set in

problem 5.3e).)

Hausdorif measures. L. et A C and let p e >0. Define

UB(xk,rk), rk<ek�1

The infimum in the definition of e) is over all possible countablecoverings of A by balls, including balls with centers not in A. The function

defined on the system of all subsets of by

= sup e) I e > o} = e),

is called the p-dimensional (outer) Hausdorif measure.

5.13. Let p > 0. Prove that:a) if A is countable, then = 0;

b) if A cA', then

c) if A C Uk>1 Ak, then �d) if A and B are separated by a positive distance, that is, if d =

inf{IIx—yIIIxEA,fore<d/2.

5.14. Let A C Prove that:a)if0<p<n andb) = 0 if and only if = 0.

5.15. Let A C and let be Lebesgue measure on ("sur-

face area"). Prove that:

102 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

a) if 0 <p < n — 1 and = 0, then = 0;b) = 0 if and only if = 0.5.16. Prove that:a) if 0 <p <q, then � iiq(A);b) if 0 < <oo, then jiq(A) = 0 for q > p. and /Lq(A) 00 for

q <p.The number p = inf{q

IJLq(A) = 0} = sup{q

I= oo} is called the

Hausdonff dimension of the set A. We denote it by the symbol climH(A).5.17. a) Prove that

dim (A)<r(A)= limH— £:Th log2 1/e

for a bounded set A Cb) Let Ak C (k = 1, 2,...). Prove that

dimH (u Ak) = SupdimH(Ak).k

5.18. The diameter of a set C C is defined to be the number diam(C)=Sup{IIx—yIIIx,YEC}. Let p>O and Definethe modifiedp-dimensional Hausdorff measure by

= sup inf >(diam(Ck))" I A C U Ck, diam(Ck) <ec>0 k�1

Prove that � � and hence can be used insteadof in the definition of the Hausdorif dimension.

5.19. Find the HauSdorif dimension p and compute for the fol-lowing sets:

a) A=[a,b];b) A=S';c) A=[0, 1]x[0, 1];

d) A=[O,

5.20. Find the Hausdorif dimension p and compute for the Can-

tor set K. Prove that the function defined by = n [0, x])(0 <x < 1) coincides with the Cantor function.

5.21. Find dimH(A) and M-dim(A) for the following sets:a) A = K, K the Cantor set;b) A = A0, where A0 = {1 + 1/ln(n + n E N};c) A=KUA0.

Let us recall now the concept of a set of Cantor type, which is needed in thefollowing problems (cf. the definitions of the Cantor set and of a generalizedCantor Set in of Chapter I). Let {ln}n>0 be a sequence of positive numbers

§5. c-ENTROPY AND HAUSDORFF MEASURES L03

Such that < (n E N). A set of Cantor type with defining sequence{'n}n>O is defined to be an intersection of sets constructed as follows.Let be an arbitrary closed interval of length 'O• Remove from anopen interval 5 symmetric with respect to its midpoint in such a way thateach of the closed intervals making up has length The left-handone of these closed intervals is denoted by and the right-hand one by

The closed intervals and are called the intervals of first rank.Let K1 = U Suppose that the set which is the union of theintervals of nth rank with length in, has already been constructed. From eachclosed interval of nth rank remove an open interval symmetric

with respect to its midpoint in such a way that each of the closed intervalsmaking up the set \5c...c has length 11141. The left-hand one of theseclosed intervals is denoted by and the right-hand one byThe closed intervals are called the intervals of (n + l)st rank; let

K11+1 = Uc1 ,...,The set of Cantor type generated on [0, 1] by the defining sequence

{211P}11>0, where p> 1 , will be denoted by K(p). Note that for p = log2 3

the set R(p) is the classical Cantor set.

5.22. Suppose that E C [0, 1] is a set of Cantor type, 0 < <00,and 0E is the Cantor function corresponding to E. Prove that

PP(Efl[O,x])=UP(E)aE(x) (0�x� 1).

5.23. Suppose that 0 <p � 1, E C R is a set of Cantor type, and 1,, is

the length of the intervals of nth rank. Prove that:a) if 1im211/ = 0, then = 0;b) if >0, then >0;c) = oo if and only if —, oo.

d) dim11(E x E) = 2dim11(E).5.24. Suppose that A is the set in problem 5.9.a) Verify that A is a set of Cantor type, and find its defining sequence

the Hausdorff dimension p of A.c)Find5.26. Let A and B be sets of Cantor type with defining sequences

and respectively, and let p = dim11(A) and

q = dim11(B). Prove that:a) p = q = log3 2 (recall that log3 2 is the Hausdorif dimension of the

Cantor set);b) = 0, = +00, and hence for any t > 0 the quantities

ji,(A) and ji,(B) can take only the values 0 and +00.5.26. Suppose that n E N, E C {0 n — 1}, m = card(E), and

104 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

1 � m < n. Find the Hausdorif dimension of the set

A= >akn lakEEforanykEN1%k>1

5.27. Let A = {>.k>1 ak!O I ak E {O, 1, 2, 3, 4} for any k E N}.Find dimH(A) and + A) and verify that dim11(A) > 1/2, but

dirn11(A+A)< 1.5.28. Let K be a set of Cantor type, with dim11(K) > 1/2. Prove that

there exists a continuous function C —, C such that zço(z) = 1

and is holomorphic outside K x K (cf. problem 4.8).

§6. Asymptotics of integrals of higher multiplicity

In this section the symbols and denote the n-dimensional(open) ball and the (n — 1)-dimensional sphere about zero with radius r. Ifthe radius is equal to 1, then we write simply and The Lebesgue

measures in R' and are denoted by (volume) and (surface

area), respectively. The volume of the ball is denoted by and the

area of the sphere by The symbol denotes the standard

Gaussian measure on , that is, the measure with densityIf x = (xi,..., is an n-dimensional vector and 0 < p < +00, then

lxii,, is defined to be (>.1<k<n kkl) while lixiL = maxl<k<fl lXkl. As

before, the notation lixil used without an index to denote ihiEuclideannorm of a vector x, that is, the quantity llxll2. Finally, as usual, the meanvalue of a function f integrable with respect to a measure ji on a set E(0< < +00) is understood to be the quantity (1//IE)fE I djz.

In the solution of a number of problems in this section it is convenient touse the formula (see problem VIII.3.22b))

j 1(x) dx= j (j f(rw) dr, f E (*)

1 rflF2a) —I dx=—n+2

b) _!_ I IxIdx

c) _!_IB'(r) 1x1 I" dx

f((1 +p)/2) (2)P/2(p> —1).

6.1. Prove that:6.2. The thickness of the rind of a watermelon is 1/20 of the radius

of the watermelon (assumed to be a ball). What percentage of the volumeof the watermelon is filled by its rind: a) in 3-dimensional space? b) in100-dimensional space?

§6. INTEGRALS OF HIGHER MULTIPLICITY 105

6.3. After falling into 1000-dimensional space, Alice was asked how tomake thin glass hoops (spherical zones) in such a way that their width isequal to 1/10 of the diameter. "It couldn't be simpler," replied Alice. "Youneed to blow up spheres, and then cut off the excess" (see Figure 7). Whatpercentage of glass is wasted with this technique?

6.4. A point x = (xi,..., is chosen at random in the ball What

is more probable for large n: 1x1 I < or Ix1 I > (It is assumedthat the probability of the point falling into a given set is proportional to themeasure of the set.)

6.5. Two points are chosen at random on the sphere S'1'. Let bethe mean value of the distance between them. Find

6.6. Let be normalized Lebesgue measure on the sphere S'1'Prove that the coordinates on this sphere "are asymptotically distributed ac-cording to a normal law," that is, that for any a, b E R with a < b

6.7. Suppose that the function I is continuous on the closed ball B'1,M = and If(r, w) — f(w)I � h(1 — r) for any w E S'1' andr E [0, 1], where h is an increasing continuous function on [0, 1] thatvanishes at zero. Prove that the mean values of f on the ball B'1 and onthe sphere S'1' are close in the following sense: for any E (0, 1/2)

cc' j 1(x) dx — j f(w) � 2(M(1 — e)'1 + h(e)).

6.8. Prove that for any q E R the integrals Ix1 . . havea finite positive limit as n —' 00.

6.9. Prove that for any q E R and 0 <p < oo

f6.10. Let be the measure on R" with density (n/2ir)'1'2 . e'1H2h12.

a) Verify that = 1.

b) Find the limit

c) Prove that IIHxH—1I>e < for 0< e < 1.

FIGURE 7

106 VIII. LEBE5GUE MEASURE AND THE LEBESGUE INTEGRAL

6.11. Prove that for any q ER and 0 <p � 00

a;:L

(see problem 6.10 for the definition of the measure

6.12. Prove that (q ER).6.13. Prove that for any q ER

If and are positive numenal sequences (used in problems 6.14and 6.15 and some other problems), then the expression x means that

<oo.nbWe use this notation in an analogous sense not only for sequences but alsofor arbitrary families of positive numbers.

6.14. Suppose that q E R, m, n E N, 2 � m � n, in + q> 0, andx=(x1

Prove that:a) b)

6.15. Prove that for any q E R and 0 <p < oo

i)=

Consider successively the following cases:a) q>O, 0<p�2;b) p�2;c)

d) q<0.6.16. Prove that for any q E R and 0 <p <00

--I I q q(1/p--l/2)xii

6.17. Prove that the radius of the ball with volume 1 is ex-pressed by the formula

fl fl \fland find a and b.

6.18. a) The ball B'(r) is tangent to all the edges of the n-dimensionalcube [—1, Compare for large n the volume of the cube and the volumeof the balL

§6. INTEGRALS OF HIGHER MULTIPLICITY 107

b) The cube [—1, lJ'1 is partitioned by the coordinate planes into 2'1

cubes. In each of them we inscribe a ball, and then we consider the ballB'1(p) tangent from the outside to all the balls inscribed in the cubes. Forwhat n is B'1(p) c [—1, 1J'1? For large n which is larger: the volume ofthe cube [—1, iJ'1 or the volume of the ball B'1(p)?

6.19. Find the asymptotics of the radius of the ball whoseGaussian volume is equal to 1/2.

6.20. Let and be the mean values of the function l/11x111 on

the respective sets B'1, and [—1, 1J'1, where is the n-dimensionaloctahedron: = {(x1,...,

I >.1<k<n IXkI � 1}. Prove that =n/(n— 1), and 1/n.

6.21. Let be the mean values of the function 11x112/Hx111

on the respective sets and [—1, 1J'1. Prove that:

6.22. Let + eB'1 be the e-neighborhood of the octahedrona) Find the limit

b) Find the asymptotics of the ratio + as n —, 00.

6.23. Prove that 2(2/1r)'1Fl , where

A—'1(xx)6.24. Let be a convex set centrally symmetric with respect to zero

that is contained in the cube (—1, 1)'1. Prove that if > (3/2)'1, thenfor sufficiently large n the set contains a point a = (a1,..., with

integer coordinates such that >1<k<fl akl > n/10.6.25. Prove that:

a) J lxii dx —,

b)f(X1

dx (1 C([0, 1J));

c) J f (qx1 . . dx —, f(e') (1€ C([0, 1J));

d) j f (kd+ " + lXnl)—, I

(1€ C(R"), suplfl <sc).

6.26. Suppose that I � 0, = 1, and 0 <p <oo. Prove

108 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

that:

a)

for any c> 0;

b)

foranyOE(O, 1/p).

c) Can n9 in b) be replaced by where > 0, —, 0?

6.27. Prove that for any n and any vector a E

1/p

� J I(x, a)I" dx) �where = [—1, and A,, and B,, are positive constants dependent

only on p (cf. problem 3.31a)).

CHAPTER IX

Sequences of Measurable Functions

In this chapter, as in the preceding one, the symbol stands for Lebesgue

measure on the space Rm = and the symbol 5! °(E) stands for theset of all measurable functions defined on a set E and almost everywherefinite.

We assume that the reader is familiar with the basic facts in the theory offunctions (the theorems of Lebesgue and Riesz about the connection betweenconvergence in measure and convergence almost everywhere, Lebesgue's the-orem on taking the limit under the integral sign, the theorem on termwiseintegration of positive function series, Bessel's inequality, the Riesz-Fischertheorem, the completeness of the trigonometric system of functions). Someof these theorems are constantly used in the solutions (for example, the the-orem on termwise integration of positive series), and it seems to us thatwithout others (for example, the theorems of Lebesgue and Riesz) the veryformulation of a number of problems cannot be sufficiently understood, evenif these theorems are not formally used in the solutions.

§1. Convergence in measure and almost everywhere

1.1. Determine whether the function sequences below convergein measure and almost everywhere on the intervals they haveintegrable majorants:

a) = nx/(1 + n2x2), = (0, 1);b)

c) = = (0, 1);

d)

e) = n2x2/(n4+x2), (1,

g) 1).

1.2. For a E R let {a} = a—[a] be the fractional part of a, and let bethe characteristic function of the interval with endpoints {ln n}, {ln(n + 1)}

110 IX. SEQUENCES OF MEASURABLE FUNCTIONS

(n E N) (if {ln(n + l)} < {lnn}, then we take to be identically zero).Prove that:

a) = 1 for any x E (0, 1);b) —' 0 almost everywhere on (0, 1) for any c> 1. Compare

this result with the result of problem 11.1.25.1.3. Construct a sequence C 1) such that —, 0 in mea-

sure as n —, 00, = +oo a.e. on (0, 1), and =—oo a.e. on (0, 1).

1.4. Suppose that (n E N) is monotone on (0, 1), and —, Iin measure as n —, 00. Prove that f coincides almost everywhere with amonotone function 10, and

—+f0(x) at the points of continuity of

1.5. Let = I + where x E (0, 2ir), andand are real numerical sequences such that > 0 and

—, +00. Prove that —, 0 in measure on (0, 2ir) as n —' It isnecessary that —' 0 almost everywhere on (0, 2ir)? Consider the

example = — where m = and k = n — m2.

1.6. Let and be real numerical sequences with —, 00 as

fl —, oo. Prove that + = 1 a.e. on R anda.e. on lit

1.7. Prove that n1"3 + dx —i 0 for any sequence

CR.1.8. Suppose that c a > 0. Prove that if

E > a} <00, then � a ac. on lit1.9. Suppose that c and —' 0. Prove that if

E � e,j <00, then:

a) —' 0 a.e. on R;b) for any number e > 0 there is a set e C R such that <e and

0 on R\e as n —' 00.1.10. Prove that convergence almost everywhere is "stable" in the sense

that if C 1) and —' 0 a.e. on (0, 1), then there isa sequence C R such that —' oc and —' 0 a.e. on

(0, 1). Does this assertion remain true if the interval (0, 1) is replaced by

1.11. Suppose that C 2°(0, 1), and 1(x) a.e. on

(0, 1). Prove that there exist a function g E p0(0 1) (aregulator") and a sequence C R such that —, 0 as n —, 00, and

— f(x)I � ac. on (0, 1) for any n E N. Does this assertionremain true if the interval (0, 1) is replaced by R?

1.12. Using the result of the preceding problem, prove Egorov's theorem:

§2. CONVERGENCE IN THE MEAN 111

c 20(0, 1), and —' 1(x) a.e. on (0, 1), then for anye > 0 there is a set e c (0, 1) such that <e and f on (0, 1)\eas n —, 00. Is this theorem still valid if the interval (0, 1) is replaced by

1.13. Let C 20(0, 1). Prove that —, 0 in measure as n —, oo

if and only if subsequence has, in turn, a subsequence

converging to zero almost everywhere on (0, 1).1.14. Suppose that C 2°(R) g E 2°(R) and � g(x)

a.e. on R for any n E N. that if g satisfies the condition

foranye>0, (1)

then convergence of the sequence to zero almost everywhere on Rimplies that 0 in measure as n Is this true if the condition (1)fails?

1.15. Let C 2°(0, 1), C 2°(0, 1), and E

2°(0, 1), and suppose that 1nkk

in measure, and in

measure. Prove that there is a strictly increasing sequence of in-dices such that in measure ("diagonal sequence theorem ").

Determine whether the analogous assertion is true if convergence in measureis replaced by:

a) uniform convergence on (0, 1);b) convergence almost everywhere on (0, 1);c) pointwise convergence on (0, 1)?

§2. Convergence in the mean. The law of large numbers

In this section the letter E denotes a measurable subset of Rm with fi-nite (Lebesgue) measure. The symbol 2r(E) (2°°(E)), where r is a posi-tive number, denotes the set of functions in 2°(E) satisfying the condition

fE If(x)(dx <00 (HIlL esssupXEElf(x)l <00). The symbol lIflir de-

notes (fE We say that —, 10 in the space 2r(E) (conver-gence in the mean with exponent r) if — f0lL —4 0.

2.1. Let f E 2"(E). Prove that:a) if 0 <s <r, then f Eb) if tm(E) = 1 , then the function s lulL is increasing on (0, rJ.2.2. Suppose that C 2r(E), and g0 E 2r(E).

Prove that:a)if inmeasureas fl—'OO

b) if llf,i10llr —i 0 and lIgn—golIr —' 0,then f0g011r12 —i 0.

2.3. Suppose that C 2'(E). Prove that if —, 0 in measure

as n —, oo and <00, then JE —' 0 for any

function gE21(E).

112 IX. SEQUENCES OF MEASURABLE FUNCTIONS

2.4. Suppose that C 2'2(E) and 10 E Prove that if

f —, f0 in measure as n —, and llfnll2 1110112' then 0•

2.5. Suppose that f, g E and 0< Let

fa(f) = EEl lf(x)1 > t}

The quantity )(f) is called the quantile (compare the definition ofwith the definition of the nonincreasing rearrangement of f before prob-1cm VIII.2.10; the quantile //2(f) is the median of the function Ill—seeproblem VIII.2.9). Prove that:

a) forb) fa(af) = for any a ER;c)

d) la,;— inmeasure

0 for any with 0<2.6. Let M C and suppose that 111112 � Clllll1 for any function

f E M. Prove that:a) if 0< r < 1 ,then there is a number 0 = 0(r) such that � C°lllllr

for any function f E M;b) there is a number = > 0 such that for

any fEM;c) if CM, then

—,0 if and only if

—,0 in measure.

2.7. Let C and suppose that 111fl112 � C and � öfor some C, ö > 0 and any n E N. Prove that if the seriesconverges absolutely almost everywhere on E, then <ci.

2.8. Prove that if sin(nx + convergesin a set of positivemeasure, then

2.9. Let be an orthogonal system in p2(E), and let =

lk• Prove that if —' 0 then —, 0 in mea-

sure. (Diverse variants of this assertion bear the name of the law of largenumbers.)

2.10. Prove that the following sequences converge in measureon the interval (—it, it):

a) = sin2 kx.

b) =

c) = + 1/k)ksin2kx.

d) = — k/n)sin2kx.2.11. Find the ilnilt

lim I arctan ! k sin2 kx dx.k

§2. CONVERGENCE IN THE MEAN 113

2.12. Let be an orthogonal system in 22(E) with � Malmost everywhere on E for any n E N, and let ; = fk• Provethat:

a) Hd7k2112 <so;

b) —' 0 almost everywhere on E.(The last statement is often formulated by asserting that the strong law oflarge numbers holds for the sequence

2.13. Let be an orthogonal system in 22(E), and let =

>.1�k<flfk• Prove that if <00, then —, 0 a.e. onE, that is, the strong law of numbers holds for the sequence

Prove that, moreover, the function h = belongs to2.14. Let be the nth digit in the binary expansion of a number

x E (0, 11, and leta —' 1/2 in measure;

b) 1/2 a.e. on (0, 1).The latter means that, for almost all numbers in (0, 1), 0's and l's occurequally often in their binary expansions.

2.15. Fix one of the digits 0 9, say m ,and let Cm(X) = 1 if the dec-imal expansion of the number x—[xJ has the form 0.m..., while Cm(X) = 0

otherwise. Let = Cm(1OkX). Using the same method as inthe preceding problem, prove that:

a) —' 1/10 in measure;

b) —' 1/10 a.e. on (0, 1).

A number x E (0, 1) is said to be normal if for any p E N its p-aryexpansion contains all (the numbers 0, 1, ... p — 1) equally often.It is known, for example (see [12]) that the number

x

is normal (the decimal expansion of x consists of all the natural numberswritten one after the other (in decimal notation)).

2.16. Generalizing the problems 2.14 and 2.15, prove that almost all thenumbers in the interval (0, 1) are normal.

2.17. Suppose that f E p2(0 1) and f(x)dx = 0. For arbitraryn E N and e > 0 let

f(xk)

1<x<n

Prove that —' 0.

2.18. Suppose that C 22(0, 1), and = 0 for any

114 IX. SEQUENCES OF MEASURABLE FUNCTIONS

n E N. For arbitrary n E N and e >0 let

(x1,..., 1)" >€I.

1<k<n

Prove that —' 0 if IIfnhI2 0.

2.19. Suppose that E C(R) and <00. Find the limit

dx.

2.20. Suppose that c R and is an orthogonal sequence

in and let = akfk. Prove that if <oo, then:

a) Is — S = akik,b) the series converges almost everywhere on E;

c) E

§3. The Rademacher functions. Khintchine's inequality

Let n E N, k E Z, and = (k + Define the function

on R by = (_1)k for x E = 0. The functionsare called the Rademacher functions.3.1. a) Prove that for any n E N and any collection {ek}1<k<fl, where

the numbers €k take the values +1 and —1 , there is a point t 1) suchthat rk(t) = €k for k = 1 n. What is the measure of the set of suchpoints?

b) Prove that r1(t + 1) = r1(t), and = r1(2''t) (t ER, n EN).c) Let be the nth digit in the binary expansion of a number x E

(0, 1). Prove that = (1 — almost everywhere on 1).d) Find the sum of the series (x E (0, 1)).3.2. Prove that:a)

J =

b)

. .. = fi J1<k<n

3.3. Let m1 be distinct positive integers. Prove that:

a) f0' rm(t) rm(t))dt = r2(t)

b) frm(t)rm(t).rm(t)dt=0,in particular, is an orthonormal system in 1).

§3. THE RADEMACHER FUNCTIONS 115

3.4. Compute the following integrals:

a) j dx,

b) jc) (z,ckEC).

d)Provethatif ER,

J1e1' dx — 0 for t 0.

3.5. Prove that for any numbers C1,...,a)

I ckrk(x)I dx <A(>1<k<fl lckfl,b) � l>.1<k<flckrk(x)ldx, where A and B are

absolute constants.3.6. Prove that for any numbers Ckj (k, / E N)a)

b)

f \1/2

(�BJJ ckfrk(t)rJ(s) dtds,

J

where A and B are absolute constants.3.7. Prove that:a) {rjrk}l<j<k is an orthonormal system in 1);

b) for any numbers Cik (1 � j <k)

J dx � A (0

1/2

( VIkI) �BJ cfkrJ(x)rk(x) dx,0

where A and B are absolute constants.

116 IX. SEQUENCES OF MEASURABLE FUNCTIONS

3.8. Prove that for any numbers c1,..,

a) dx �A (1�k<n

f \1/22

b)(

� BJ cksin2lcx dx.

J 1<k�n

where A and B are absolute constants.3.9. Using the results of problems 3.5—3.8, prove that for any t> 0 and

any numbers Ck and

a) _<A1(4

(kkI)

b) 4xE(0, 1) clkrl(x)rk(x) >t

c) E (0, 22E) >4 (1%

where A1, A2, and A3 are absolute constants.3.10. Let (xE(0, 1)). Provethat:a) E (0, � < oo;

b) a.e.on (0, 1).

Each of the assertions a) and b) implies the equality = 0 al-most everywhere on (0, 1) , which is known under the name of the strong lawof large numbers. Using the connection between the Rademacher functionsand the digits of the binary expansion (see problem 3.lc)), deduce from thisthat the binary expansion of almost everywhere number contains "equallymany" 0's and l's (see problems 2.14 and X.2.30).

3.11. Let C be an arbitrary real n x n matrix, and A the set of allpossible vectors in with coordinates equal to ±1. Prove that

a) y)4 � K1 . VE4(Cx. y)2)2.

b) � K2 where K1 and K2are absolute constants.

3.12. Using the uniqueness theorem for Fourier transforms of finite Borelmeasures, find the distribution functions of the following functions:

a) 1(x) = + (x E (0, 1));

§3. THE RADEMACHER FUNCTIONS 117

b) (xE(O, 1)).

where {mk}k>1 is an arbitrary (not necessarily increasing) sequence of dis-tinct positive integers.

3.13. Let be one of the intervals Am,j (see the definition of theRademacher functions). Prove that for m � n (in, n E N, t> 0, ak E R)

the following inequalities hold:

a)J E akrk(x) dx�J E akrk(x) dx,

A A

b) j e' ak%(x) dx < j d akrk(x) dx.

Moreover, if E ={x E(0, t}, then:

c) J dx�J dx;E E

d) Lak%(x) dx � j e' akrk(x) dx.

3.14. Using the inequality cosh(s) � (s E R), prove that the fol-lowing inequalities hold for a1, a2, ... E R and c1, c2, ... E C:

a) E (0, l)II>21<k<flakrk(x)I > t} � 2 where A2 =

b) E (0, 1)1 I ckrk(x)I > t} � where C2 =

ICkI2.

c) if

E = x E (0, 1) sup akrk(x) >

then � Alt and � , where A2 =and

g(x)= sup akrk(x)iz�1

then dx < 00 for any number s> 0.

3.15. Prove that if >.k�1 IakI <oo, then the series akrk(x) con-

verges almost everywhere.3.16. Prove that if the series akrk converges in measure on some

set E of positive measure, then akl <00.

118 IX. SEQUENCES OF MEASURABLE FUNCTIONS

3.17. Prove that the following statements are equivalent:a) >.k>lIakI<00;b) the series converges in 2P(0, 1) for some p E (0, 2);c) the series akrk converge in measure on the set (0, 1);d) the series converges almost everywhere on (0, 1).

3.18. Prove that if the series1

sin 2kx converges in measure on

some set of positive measure, then akl <oo.3.19. Let <00. Prove that:a) if

esssup <ooxEA k�1

for some (nonempty) interval then lakI <oo;b) if

esssup sin 2kx <00k�1

for some (nonempty) interval, then lakI <00.

In problems 3.22 and 3.21 we consider the Haar system {hm}n,>0. Thissystem is defined as follows. Each positive integer in can beuniquely in the form m = + k, where n and k are nonnegative integers,and 0 � k By definition, let h0(x) = 1 for any x E (0, 1) and

for <x <

for mEN.

3.20. Prove that:a) is an orthonormal system in 22(0, 1):b) if E = {x E (0, 1)11 akhk(x)I > t} then for n � in

J akhk(x) dx � J akhk(x) dx;£ O<k<ni E

c) if

E = x E (0. 1) akhk(x) > t— — O<k<m

then ).(E) = AC' ,where A = IakI),d) if IakI < 00, then the series akhk(x) converges almost

everywhere on (0, 1). —

§3. THE RADEMACHER FUNCTIONS 119

3.21. Using the fact that for any function f integrable on (0, 1) theFourier series in the Haar system converges to f in 21(0, 1), prove that:

a) the series converges almost everywhere on (0, 1);b) if E = {x E (0, 1)1 > t}, where is the nth partial sum

of the Fourier series for I in the Haar system, then t(E) �3.22. Using the result of problem 3.14, prove that Khintchine's inequality

holds forany p>0 andany a1,...,

1<k<n 1<k<n P

where A,, and B,, are positive constants dependent only on p, ande112. The precise values of the quantities A,, and B,,

are found in [37J.3.23. Prove that the assertions a)—d) of problem 3.17 are equivalent to

the assertione) the series akrk converges in 5f2P(0 1) for any p in (0, +oo).

3.24. Prove the following for any trigonometric polynomial eke

(n � 2):a) there exist numbers ek = ±1 (k = 0, ±1, ±2 ±n), such that

2nI ikx : 2

JICkI

IkI�n

where c> 0 is an absolute constant;b) there exist numbers €k = ±1 (k = 0, ±1, ±2 ±n) such that

/ 1/2

sup ekcke � IckI)xER

IkL<n IkI�n

3.25. Prove that for any a> 2 there are numbers > 0 such that theseries

+ .. +

converges almost everywhere. Deduce from this that

-almost everywhere.

3.26. Let be an arbitrary increasing sequence of positive inte-

gers. Prove that —

1r1(x) + r2(x) + ...+ (x)I

j++OO

120 IX. SEQUENCES OF MEASURABLE FUNCTIONS

almost everywhere on (0, 1). If / = 0(ln ni), then we get from this a veryweak variant of the law of the iterated logarithm:

—lim '

almost everywhere on (0, 1).

§4. Fourier series and the Fourier transform

4.1. Compute the convolution j * J in the following cases (c> 0):

1 222a) f(x) = C (x ER);

cy2it

b) f(x) = ER");

c) (xER).

4.2. Suppose that 0 < 1 and f and are 2it-periodic (measur-able) functions with esssupRlf(x)l <oo, = for lxi � it.

Prove that the function = — y)dy (x E R) is in theclass Lip1_0.

In problems 4.3-4.15 the symbol J(n) denotes the Fourier coefficient ofa function f integrable in (—it, it) with respect to the orthogonal systemrInxl1e

1(n) = j f dx (n E Z).

4.3. a) Let f, g E r), h = fg. Prove that if f(n) = =0for n<0,then h(n)=0 for n<0.

b) Let f E it) and let c1 = e'. Prove that if 1(n) = 0 forn<0,then also for n<0.

4.4. Let f E it). Prove that <00 if and only iff(x) = g(y)h(x — y)dy almost everywhere on (—ir. it), where g andh are 2it-periodic functions in the class it).

4.5. Let f E it). Prove that the numbers (n E Z,n 0) are the Fourier coefficients of a function in the class Lip112.

4.6. Prove that if 0 � 1 and f is a 2it-periodic function in theclass

f is a function of bounded variation on [—it, itJ , then1(n) = 0(1/n).

4.8. Let f(x) = x . cos 1/x for 0 < lxi � it. and f(0) = 0. Prove that,although f is not a function of bounded variation on [—it, itJ (see problem111.3.13 a)), f(n) = 0(1/n).

§4. FOURIER SERIES AND THE FOURIER TRANSFORM 121

4.9. Let 0 < < 1 and = Prove that the function1(x) = (x E R) is in the class Is this true for a = 1?

4.10. Let 1(x) = (x E R). Prove that the Fourierseries of f diverges at the point x = 0.

Suppose that f E 2"(—ir, ir), and = f(k)e"° (lxi E it.n = 0, 1, 2, ...) is the corresponding partial sum of the Fourier series off. Let

n

The sums are called the Cesàro-FejEr sums. It is easy to see that

a (x) = J_ [ —

2irn J—jr sin2((x — y)12)

4.11. Prove that the sequence of Cesäro-FejEr sums corresponding to a2ir-periodic continuous function converges to the function uniformly on R.

4.12. Suppose that f is a 2ir-periodic function in 0 <a <1. Prove that:

a) � C/nU for any x E R, where C is a constant dependingon a and f, but not on x;

b) if— = O(nU), where is the nth partial sum of the Fourierseries of f.

4.13. Let f be a 2ir-periodic function in Prove that if a> 1/2,then if(n)i <00.

For a = 1/2 the assertion becomes false (see problem IV.6.34).4.14. Let f E 2"(—ir, ir). Prove that if ill — = o(lfn), then

f const almost everywhere.4.15. Let f E 22(—ir, it). Consider the function f defined by

f(x) = (x E (—it, ir)).nEZ

Prove that I is real if f is real.

The function I is called the conjugate of 1. The reader can becomeacquainted with one of the numerous applications of this important conceptin the following two problems, borrowed from [41].

4.16. Let 1 = n0 < < be positive integers, let a0 am bearbitrary complex numbers, and let fk(x) = (for j > m

the numbers are taken to be zero). Further, let = i1ki' hk = —

and = Ike >J�k , where e is an arbitrary positive number, andgk is the function conjugate to (see the definition in problem 4.15). Prove

that:

122 IX. SEQUENCES OF MEASURABLE FUNCTIONS

a) forany xE(—lr,lr);b) if � q < then

— aqi � IIfkII2 IIfjfl2j�k

4.17. Let 1 � < (n1 E N). Prove that for any complexnumbers c1 Cm

dx � y

where y is a positive absolute constant (y � 102).

The Fourier transform of a function f E 2"(R'1) is defined to be thefunction f given by

f(s)= j dt (s E

where (s, t) is the inner product of vectors s, t E R".

4.18. Compute the Fourier transform of the functions in problem 4.1.4.19. Find the Fourier transform of the Hermite functions =

ex212(e_x2)('1) (x E R, n = 0, 1, 2,...).The Fourier transform of a finite measure ji defined at least on the Borel

subsets of R'1 is defined to be the function ft with

ft(s)= j d4u(t) (s ER'1),

where (s, t) is the inner product of the vectors s, t EThe Fourier transform of the measure on R' generated by a nondecreasing

bounded function g is denoted by the symbol

4.20. Let F be the distribution function of a function I Ewhere E C and <oo. Prove that

P(s)= j dx (s ER).

4.21. Let ô(s) = (s ER), where a is the Cantor function(see problem 111.3.17). Prove that ô(s) 74 0 as s —, 00.

4.22. Let a1(x) = a(x — y)da(y) (x E R), where a is the Cantorfunction (a(x) = 0 for x <0, and a(x) = 1 for x> 1). Prove that:

a) a1 is strictly increasing on [0, 2];b) ô1(s)740 as 5—*oo.

The solutions of problems 4.23—4.27 are based on the following uniquenesstheorems:

§4. FOURIER SERIES AND THE FOURIER TRANSFORM 123

If f and f 0, then 1(x) =0 a.e. on RI?If p is a finite Borel measure on and 0, then p = 0 (see, for

example, [20], p. 302).

4.23. Prove that if the convolution of an integrable function f onwith the function is identically equal to zero, then 1(x) = 0 a.e.

4.24. Let g(t) = sup{>2k>l €k2I €k2 < t, = 0 or 1}for t > 0, and let g(t) = 0 t <0. that:

a) gEC(R);b) g'(t) = 0 almost everywhere on R;c) = fld) —

0 forx<0,

J g(x — t)dg(2t) = x for 0 x 1,

1 forx>1.4.25. Let = be an arbitrary strictly increasing sequence of

positive integers, and let = and = Further, let

g(t) = sup � = 0 or 1

for t > 0 and g(t) = 0 for t <0, and let the functions g1 and g2 be con-structed analogously with the help of the sequences and 1/2, respectively.

Prove that:a) g€C(R);b) g(t) = 0 almost everywhere on R, and = 0 almost everywhere

on R.c) Verify that the function g is "divisible" in the sense that g(x) =

for any xER.4.26. Prove that if two finite Borel measures on take the same values

on all possible half-spaces (that is, on sets of the form {x E a) � c},where a E and c E R, then they coincide.

4.27. Find the general form of a Borel probability measure on R" that isinvariant with respect to orthogonal transformations and is at the same timea product of two measures concentrated on nontrivial orthogonal subspaces.

CHAPTER X

Iterates of Transformations of an Interval

§1. Topological dynamics

1.1. Suppose that the function f: N —. N satisfies the inequalityf(n+1)>f(f(n)) forany nEN. Provethat f(n)=n forall nEN.

1.2. Determine whether there exists a continuous function f: R —,

satisfying the equation Jo 1= F in the following cases:

a) F(x)=—x; b)F(x)=eX; c) F(x)=x2—2.

Denote by (or simply Id) the identity mapping of X into itself (i.e.,Id(x) = x for any x in X). A mapping h: X —' X is called an involutionif ho h =

1.3. Let 1(x) = x+ 1 and g(x) = x— 1 (x E R). Choose two involutionsh and k such that ho k = f and k oh = g.

1.4. Prove that each bijection is a composition of two involutions.

A fixed point of a mapping f: X X is defined to be a point x E Xsuch that x =1(x).

1.5. Prove that a continuous mapping f: —, R of an interval has a

fixed point in each of the following cases:a) = [a, b], C

b) = [a, b], jc) = (a, b), I is an involution.1.6. Suppose that a function I E C([a, b]) is monotonically increasing,

and f([a, b]) C [a, b]. Prove that for each x0 E [a, b] the sequence =(n E N) converges to a fixed point. Is this true if f is a decreasing

function?1.7. Let f E C(R). For x0 E R let x1 = 1(x0) and xk = f(xk_l)

(k E N). Prove that if the sequence f(xk)} is bounded for somex0 E R, then 1 has a fixed point. Is this true for a continuous mappingf: C-'C?

1.8. Prove that a finite group of affine transformations of the plane hasa fixed point (common for all transformations in the group).

126 X. ITERATES OFTRANSFORMATIONS OF AN INTERVAL

Mappings f: and g: are open intervals onthe line) are said to be topologically conjugate if there exists a continuousbijection (homeomorphism) h: such that f = h o go h' (that is,h o g = fo h). In this case we use the notation f g, and the mapping hwill be called a conjugating mapping.

In problems 1.9—1.12 it is assumed that the mappings are continuous, aredefined on open intervals, and map them into themselves.

1.9. Prove that the relation of topological conjugacy is an equivalencerelation, i.e.: a) f;b) g if and only if f; c) if and

12 then 13.1.10. Prove that topologically conjugate mappings have or do not have

the following properties simultaneously:a) being one-to-one;b) being monotone of a definite type;c) being bounded above or below (without specification of which);d) boundedness.1.11. Prove that topologically conjugate mappings have:a) the same number of intervals of monotonicity;b) the same number of fixed points.1.12. Prove that if f g, then intervals on which f(x) > x are carried

by a conjugating mapping h into intervals on which the same or the oppositestrict inequality holds for g (depending on the character of monotonicity ofh). Prove that if f and g are defined on R and one of the mappings fand g is odd, then the bijection h can be assumed to be increasing.

1.13. Determine whether the mappings f and g on R are topologicalconjugates in the following cases:

a) f(x)=x2, (n=2,3,...);b) f(x)=cosx, g(x)=sinx;c) f(x)=cosx, g(x)=—cosx;d) f(x) = sinx, g(x) = — sinx;e) f(x)=x+sinx, g(x)=x+cosx.1.14. a) Let 1(x) = x2 and g(x) = x2 + ax + b (x ER). Describe the

set of pairs (a, b) such that I g.b) For which a E R is there a b E R such that the mappings 1(x) =

1—ax2 and g(x) = bx(1 —x) (x ER) are topological conjugates? Describethe set of all numbers b arising in this way.

c) Prove that the mappings g and [0, 1] —, [0, 1], g(x) = =3x(1 — x). are not topological conjugates.

d) For which a E R and which intervals = [p, q] C R is the mapping1: 1(x) = 1—ax2, topologically conjugate to a mapping g: [0, 1J —,[0,1] of the form g(x)=bx(1—x)?

1.15. Prove the topological conjugacy of the continuous mappings 1. g:

§1. TOPOLOGICAL DYNAMICS

(a)

(b)

FIGURE 8

[—1, 1] —, [—1, 1] given as follows:a) f(x)= 1—21x1, g(x)= 1—2x2 (xE[—1, lJ);b) f = (n E N) is a "saw-tooth" mapping that takes the values

at the points —1 + 2k/n, 0 � k � n, and is linear in the intervals betweenthem, and g = is the corresponding Chebyshev polynomial determinedby the relation t) = cosnt (see Figure 8);

c) Prove that the functions in a) are conjugate as mappings from R to R.1.16. Suppose that 1: [a1, b1] —, [a1, b1] and g: [a2, b2J —' [a2, b2] are

homeomorphisms of intervals that do not have interior points as fixed points.Prove that I g. Verify that the assertion remains in force if the closedintervals are replaced by open intervals.

1.17. Classify, up to topological conjugacy, the homeomorphisms of aclosed interval having a finite set of fixed points that includes the endpointsof the interval.

Suppose that I is a mapping of an interval C R into itself, and letf°(x)=x, f'(x)=f(x),and for n>1. Wesaythatthe method of successive approximations with initial point x0 E converges

if the limit exists, where =

f that

128 X. ITERATES OFTRANSFORMATIONS OF AN INTERVAL

the method of successive approximations converges for any initial point x0 E[a,b] if:

a) f(x)<x for x>a, f(a)=a;b) f(x) > x for x <b 1(b) = b.1.19. Prove convergence of the method of successive approximations for

any initial point x0 E and find for the following mappings f:a) f(x)=arctanx,b)c) f(x)=(x—2)/2,d) f(x)=x(x2+3)/(3x2+1),e) f(x)=(x+1/x)/2,f) 1(x) = arcsin(2x/(x2 + 1)), = (0, +oo);g)

1.20. In the following cases determine for which initial points x0 in thedomain of a mapping f the method of successive approximations converges,and find

a) f(x)=x+sinx;b) f(x)=x(x—1);c) 1(x) = 1 + ax2 (0 <a � 1/4);d) f(x) = 1 — ax2 (0 <a � 3/4);e) 1(x) = x2 + c (id � 1/4);f) f(x)=x2+(1—2c)x+c2 (c€R);g)

I 1/(4—3x) for

1/3 forx=4/3;h)

— f — 1) if lxi 1/v's.

A fixed point x* of a mapping I is said to be attracting if for x0 in

some neighborhood of x* the sequence = f'(x0) converges to f. Iffor some neighborhood U of the fixed point and an arbitrary pointx0 E U, x0 x*, there is an index n0 such that U, then x* iscalled a repelling fixed point.

1.21. Let x* be a fixed point of a mapping f, and suppose that f is

differentiable at x*. Prove that:a) if lf'(x')l < 1 then x* is attracting;b) if lfl(x*)l < I then is repelling;c) show by examples that a fixed point can be neither attracting nor re-

pelling.1.22. For what values of the parameter a and for what x0 E R does the

sequence of iterates = f(x0) converge if:a) f(x)=1+ax;

§1. TOPOLOGICAL DYNAMICS 129

b) 1(x) = 1 — aixi (a > 0);c) f(x) = 1 — ax2 (a > 0);d) f(x)=a' (a>O).

Let be a differentiable function on an interval and let N, be themapping defined at the points x E with 0 by the formula

N,(x) = x —

It is easy to see that the fixed points of N, coincide with the zeros of thefunction We say that the Newton iteration process with initial point x0converges to c if xk = —, c.

1.23. Suppose that E c E = 0, and > m > 0for x E Prove that if x0 is sufficiently close to c, then:

a) the Newton iteration process starting at x0 converges to c;

b) "superconvergence" holds, namely, lxk — ci � Mq2, where M > 0,0<q< 1,and kEN.

1.24. Prove that if E c E = = = =0, and 0, then for initial points x0 sufficiently close to c theNewton iteration process converges to c, and lxk — Cl � Mqk (M > 0,0<q< 1).

1.25. Let E C2([a, bJ) be a function such that >0 in (a, b) and= = 0 (or = = 0). Verify that for any initial point in

[a, bJ the Newton iteration process converges to a zero of1.26. Let P be a polynomial having only simple real roots. Prove that

if x0 is an inflection point of P (that is, P"(x0) = 0), then the Newtoniteration process starting at x0 converges to a root of P, and all roots exceptfor the minimal and maximal roots can be found in this way.

The set = ft'(x0)ln E N} is called the trajectory of a point x0. Ifthe trajectory is finite, then there is obviously an n E N such that the pointsx0 are distinct, and x x0

is called a periodic point, the number n is called its period, and the tuple(x0 is called an n-cycle. We say that an n-cycle is attracting ifx0 is an attracting fixed point for I", and repelling if x0 is a repelling fixedpoint of f'.

1.27. Let x x_1) be an n-cycle of a mapping f. Prove that:a) if < 1 , then for x0 in some neighborhood of

fnn+k(x)___..._,x; k=0 n—i;

b) if 1,thenforsome e >0 andany, there is an m such that lf"(x0)—x1 � e for all k = 0 n—i.

130 X. ITERATES OF TRANSFORMATIONS OF AN INTERVAL

1.28. Let f be a continuous mapping of a closed interval into itself,and let I be a finite or infinite sequence of closed intervals suchthat c and c for any n � 0. Prove that:

a) there exists a sequence of closed intervals K0 3 K1 3 ... , K0 C

such that = (n � 0);b) there exists a point x E such that = ftZ(x) E for any n.1.29. Prove that if a continuous mapping of a closed interval into itself

has a 4-cycle, then it also has a 2-cycle.1.30. Let 1: —, be a continuous mapping of a closed interval into

itself. Prove that the following statements are equivalent:a) there exists a point a E such that the points b = 1(a), C = f2(a),

and d = f3(a) satisfy the inequalities d � a <b <c or c < b <a � d;b) contains a point having period three;c) for any m E N, contains a point having period m.

The implication b) —, c) and (problem 1.29) is a special case of the fol-lowing theorem of Sharkovskii [301. Let f: —' be a continuous mapping

= [a, bJ). We consider the series of positive integers, ordered as follows:

1,2,4,8 2m.7,2m.5,2m.3

2.7,2.5,2.3 7,5,3.

If p is to the left of q in this series and f has a q-cycle, then f also hasa p-cycle.

1.31. Prove that for n � 2 the Chebyshev polynomial defined on[—1, 1J (see problem 1.15), has cycles of arbitrary order.

1.32. Find the values of the parameter a such that the mapping 1(x) =1 — aixi (x E R) has: a) a 2-cycle; b) a 4-cycle; c) a 3-cycle. Determinewhether these cycles are attracting.

1.33. Let 1(x) = 1 — ax2 (x E R). Find the infimum of the values ofthe parameter a such that f has: a) an attracting 2-cycle; b) a repelling2-cycle; c) a 4-cycle; d) a 3-cycle.

1.34. Let 1(x) = bx(1 — x) (x E [0, 1J), 0< b � 4. Find the infimumof the values of the parameter b such that I has: a) an attracting 2-cycle;b) a repelling 2-cycle; c) a 4-cycle; d) a 3-cycle.

1.35. Let x0 E (0, 1) be a fixed point of the continuous mapping f:[0, 1J —, [0, 11. Prove that if the one-sided derivatives f'±(x0) at x0 arenegative and 1 , then 1 has a 2-cycle. In particular, a 2-cycleexists if f'(x0) < —1.

1.36. Let g be a piecewise smooth mapping of [0, 1J into itself. Provethat if Ig'(x)l � a > 1 at all points where the derivative exists, then ghas a for any n E N. Show by an example that this is false ifdifferentiability of the mapping fails at countably many point.

§1. TOPOLOGICAL DYNAMICS 131

In problems 1.37—1.39 we consider mappings of = [a, bJ into itselfthat belong to the class and are such that the Schwarzian derivativeSf (see problems VII.2.24 and VII.2.25 and the definition preceding them)is negative at the points x that are not critical points of I (that is, wheref(x) 0). The class of such mappings is denoted by

1.37. Let f and suppose that the set of critical points of I isfinite. Prove that for each n E N the transformation I can have only finitelymany n-cycles.

1.38. Let g=ft (n� 1),andlet x<y<z besuccessivefixed points of g. Prove that if [x, yJ does not contain critical points ofg,then g'(y)>l.

1.39. Suppose that f E has a unique critical point c E (a, b).Prove that each attracting cycle of I attracts the trajectory of one of thepoints a, c, or b. and hence f can have at most three attracting cycles.

1.40. Prove that for a � 2 the mapping 1(x) = 1 — ax2 (x E R) cannothave more than one attracting cycle, and for a =2 it has an n-cycle for anyn E N, but none of them are attracting.

1.41. Let be a continuous mapping of the circle into itself having afixed point and a 3-cycle.

a) Prove that also has n-cycles for all n > 3.b) Does necessarily have a 2-cycle?1.42. Let f be a continuous increasing function on the line that satisfies

the conditionf(x + 1) = f(x) + 1.

Prove that:a) the mapping z = = gives a homeomorphism of

the circle S' = {zI Izi = 1} that preserves the order of succession of anythree points, and an arbitrary orientation-preserving homeomorphism of thecircle can be given in this way;

b) if the limit

lim =

exists for some x0 E R, then it exists also for any x E R;c) if the mapping ç0m has a fixed point for some m E N, then the limit

exists and is a rational number;d) if no power of has a fixed point, then the limit exists and is an

irrational number;e) if two distinct functions J and 12 determine a single homeomorphismthen the corresponding limits and differ by an integer.

The number mod 1 determined in this way for an arbitrary orientation-preserving homeomorphism 5' is called the rotation number of thishomeomorphism. This concept was introduced by Poincaré.

132 X. ITERATES OF TRANSFORMATIONS OF AN INTERVAL

§2. Transformations with an invariant measure

Let X be a measurable subset of R'1, and p a a-finite measure on the(Lebesgue-) measurable subsets of X. A nonnegative measurable functionf is said to be a density of p if p(A) = IA 1(x) for an arbitrarymeasurable set A C X (in this case the notation used is dp = = Idx).Two densities give rise to the same measure if and only if they coincide almosteverywhere on X. By the well-known theorem, p has adensity if and only if it is absolutely continuous (with respect to Lebesguemeasure), that is, if and only if p(A) = 0 when = 0 (see [14J). Ameasure p is said to be equivalent to Lebesgue measure if p(A) = 0 if andonly if = 0. It will be said that a mapping T: X —, X preservesthe measure p, or that the measure p is invariant with respect to T, ifp(A) = p(T'(A)) for any measurable set A C X. If ii is a measure on themeasurable subsets of a set Y C Rm, then ii is said to be the image of punder a mapping T: X —, Y if i'(B) = p(T'(B)) for any measurable setB C Y (in this case we write ii = Tp).

2.1. Verify that the following transformations preserve Lebesgue mea-sure:

a) T(x)=(2x)modl (xE[O, 1J);b) T(z)=? (zEC, IzI=l), mEN;c) T(x)=x—1/x (xER\{O}).2.2. Verify that the transformation

T(x, y) = ((2x) mod!, + [2xJ)), (x, y) E [0, 1) x [0, 1)

(the "baker's transformation") preserves the Lebesgue measure2.3. Supposethat X=[O, 1)x[O, 1), A= isanintegermatrix,

and detA = ±1. Define a transformation 7'4: X —, X by the formulas

= (a11x1 + a12x2) mod!, y2 = (a21x1 + a,1x1) mod 1,

where (x1 , x2) E X. Prove that:a) TA is one-to-one;b) TA preserves Lebesgue measure.

The mapping (x1 , x2) '— allows us to identify X with thetorus T2 = S' x S1 and TA with an automorphism of T2 as a group. Thislets us call the transformation 7'4 an automorphism of the torus.

We recall a fundamental result: on every compact group G there is anontrivial finite measure p that is translation invariant: p(gA) = p(Ag) =p(A), where A C G, g E G, gA = E A}, and Ag = {xgIx EA}. This measure is uniquely determined (up to a constant factor) and is

§2. TRANSFORMATIONS WITH AN INVARIANT MEASURE 133

called Haar measure. Its uniqueness can be employed in the solution of thefollowing problem.

2.4. Let the mapping TA: X X be the same as in the preceding prob-lem, but let det A E Z,

Idet 1. Such a mapping (now not one-to-one)

is called an endomorphism of the torus. Prove that it preserves Lebesguemeasure.

2.5. Verify that the mapping T(x) = f mod! (x E (0, 1J) preserves themeasure dp = (C/(! + x))dx.

2.6. Find an absolutely continuous invariant measure:a) for the mapping [—!, !J —. [—!, !J,where is the corresponding

Chebyshev polynomial (see problem !.!5b));b) for the mapping f: R —, R given by f(x) = 1 — 2x2.2.7. Prove that a mapping T has an invariant (respectively, finite in-

variant, absolutely continuous invariant) measure if and only if some powerTIZ (n EN) of it has this property.

2.8. Let T: [a, bJ —, [a, bJ be a StflCtly monotone continuous transfor-mation. Prove that it has an absolutely continuous invariant measure.

2.9. Suppose that the mapping T: [a, bJ —, [a, bJ is piecewise monotoneand smooth. Prove that if it has an attracting cycle (see §!), then T doesnot preserve any finite measure equivalent to Lebesgue measure.

2.!O. Determine whether there is a nontrivial measure on R that is in-variant with respect to all:

a) homotheties Ha: x ax (a 0);b) affine transformations Aab: x '-. ax+b (a, b ER, a 0).

If the answer is positive, give an example.2.!!. Determine whether there is a nontrivial measure on (0, +oo) that

is invariant with respect to all transformations of the form:a)

b) Sb,P:x—4bxt' b>O).If the answer is positive, give an example.

A measure p on is said to be quasi-invariant with respect to translationsif for each a E the condition p(A) = 0 is equivalent to the conditionp(A + a) = 0.

2.!2. Prove that:a) if p is equivalent to Lebesgue measure on then p is quasi-

invariant with respect to translations;b) if ji is quasi-invariant with respect to translations, and the measure of

any bounded set is finite, then p is equivalent to2.!3. Let T be a transformation of a set X C into itself such that

the image of is absolutely continuous with respect to and let fbe an arbitrary integrable function on X.

134 X. ITERATES OFTRANSFORMATIONS OF AN INTERVAL

a) Prove that there exists a function 9(X) such that

J f(x)dx =J f'(x)dx (Ac X) (1)T'(A) A

and f is uniquely determined up to values on a set of measure zero.The correspondence P: f f is called the Perron-Frobenius operator

associated with T.b) If X c is a domain, T is a smooth one-to-one mapping, and

then

Pf(Tx)= I

(x E X).

c) If X = [0, 11, T is piecewise differentiable, and the set T'(x) isfinite (x E X), then

P1(y) =xET'(y)

at all points for which the right-hand side is defined.d) If X = [0, 11, f E C([O, 1J), and T is piecewise monotone, then

f(t)dt.dx

e) The measure dp = I dx is invariant with respect to T if and only ifPf(x) = 1(x) for almost all x E X.

Let nEN,and

= (i = 1 n).

Consider the nonnegative functions on [0, 1) that are constant on the inter-vals Each such function is determined by the vector c = (c1,..., ofvalues it takes on Let us denote it by The set of nonnegativevectors in R'1 is denoted by

2.14. Let T: [0, 1) —. [0, 1) be an arbitrary transformation.a) Prove that for each function of the form (c E there is a unique

function 1d (d E satisfying the equalities

J =J fd(x)dx, i = 1 ii,A,

and the vector d has the form d = ire, where ir = is some matrixdetermined by T.

The mapping c irc is a discrete (finite-dimensional) variant of a Perron-Frobenius operator.

b) Find the matrix ir and prove that = 1 (j = 1 n).

§2. TRANSFORMATIONS WITH AN INVARIANT MEASURE 135

c) Prove that there exists a vector v E v 0, such that irv = V

(with this goal consider the sequence of vectors = (1/rn)-

2.15. a) Let T: [0, 11 [0, 11 be the mapping defined by the formula

(2x1

Find the general form of an absolutely continuous measure invariant withrespect to T.

b) Let T: [0, 1J be the mapping defined by

(x+1L.2—2x forx>4.

Find an absolutely continuous measure that is invariant with respect to T.c) Let T: [0, 11 —, [0, 1J be the mapping given by T(x) = x/2. Find a

fixed point ; for the diScrete Perron-Frobenius operator (for even n) andprove that there is no invariant finite absolutely continuous measure for T.What is the limit of measures = dx as n —, 00 if is normalizedby the condition 1)) = 1?

2.16. Let Q = {z E dIm z > O}. Find an absolutely continuous mea-sures that is invariant with respect to all the conformal automorphisms ofQ, that is, the mappings T(z) = (az+b)/(cz +d), where a, b, c, d ERand

2.17. Let P = {(x, y) E R21x > O} and suppose that for each point(a, b) E P the transformationS Lab and Rab of the half-plane P intoitself are defined by

La,b(x, y)= (ax, ay+b), Rab(x, y) =(ax, bx+y).

Find absolutely continuous measures and on P that are invariantwith respect to all the transformations Lab and Ra ,b' respectively.

We remark that the transformations under discussion correspond to leftand right shift transformations on the group (with respect to multiplication)of matrices of the form

((x,y)EP),

that is, to the transformations

MCbMX,Y and MxyMa,b•

The measures and are called Haar measures (left and right) for thisgroup.

2.18. Prove that if a transformation T: [a, b] [a, b] preserves somefinite absolutely continuous measure, then T is topologically conjugate tosome transformation of the same interval that preserves Lebesgue measure.

136 X. ITERATES OFTRANSFORMATIONS OFAN INTERVAL

The existence of an invariant measure (not necessarily absolutely continu-ous) is often established with the help of the following theorem of Bogolyubovand Krylov:

For every continuous mapping of a compact set into itself there exists afinite invariant measure.

2.19. Prove that if T is a homeomorphism of the circle S' with anirrational rotation number (see problem 1.42), then T is conjugate to therotation of the circle through the angle = in the

sense that h o T = o h for some continuous mapping h: S' —, S'.

In what follows, it is assumed that the measure ji is finite and satisfiesthe normalization condition p(X) = 1 (such measures are called probabilitymeasures).

Let T be a transformation of the set X. A set A C X is said to beinvariant if T'(A) = A. A transformation T preserving a measure pis said to be ergodic if for each invariant set A C X either p(A) = 0 orp(X\A) =0.

2.20. A measurable function f defined on a set X with measure p issaid to be invariant with respect to a transformation T if f(Tx) = 1(x)for almost all x. Prove that T is ergodic if and only if every invariantmeasurable function coincides with a constant function almost everywhere.

2.21. Determine whether the following transformation are ergodic:a) T(x) = (x + a)mod 1, X = [0, 1), =b) T(x) = (2x)mod 1, X = [0, 1), p =c) T(z) = z E X = S1, p is Lebesgue measure on the circle;d) T is an automorphism of the torus (see problem 2.3);e) T is an endomorphism of the torus (see problem 2.4).2.22. Suppose that X is a subset of that is the closure of the set

of its interior points, and T is a transformation of X preserving Lebesguemeasure. Prove that if T is ergodic, then for almost all x E X the trajectoryof x is dense.

2.23. Let X = [0, 1)1? be the n-dimensional torus (n E N), and let Sbe the shift x y = S(x) defined by the formula = (xk + mod 1,k = 1 n. For what values of is the shift ergodic?

2.24. Prove that the mappings of [0, 1] into itself given by the formulas

T(x) = (2x)mod 1, S(x) = 1 — 12x —

are metrically conjugate. (Transformations T: X —, X and 5: Y —, Y pre-serving the respective measures p and ii are said to be metrically conjugateif there exists a measurable mapping h: X —, Y such that hjz = ii andSoh(x)=hoT(x) for almost all xEX.)

§2. TRANSFORMATIONS WITH AN INVARIANT MEASURE 137

2.25. Prove that a mapping T of [—1, 1] into itSelf preserving a measureji is ergodic if:

a) T(x) = 1 — 21x1, d/1 = dx/2;b) T(x)=1—2x2,c) T = is a Chebyshev polynomial (see problem 1.15b)), d,i =

2.26. Prove that if T: X X is a transformation preserving a proba-bility measure ji, and f is a positive function, then = +00 foralmost all x E X (with respect to ji).

A fundamental theorem in the theory of transformations with an invariantmeasure (or "ergoclic theory") is the Birkhoff-Khintchine ergodic theorem (see[3], [15]):

Let T be a transformation of a set X preserving a probability measureand let f be a measurable function with <+00. Then

for almost every x E X (in the sense of ii) the limit

lim! >2f(Tk(x))=7(x) (*)

O<k<n

exists. Furthermore, the function 7 is invariant, and

j7(x)dP(x) =

If T is ergodic, then 7 turns out to be constant (see problem 2.20) andequal to f(x)dji(x) (that is, to the mean value of f on X). The limiton the left-hand side of (*) is called the mean value of I along the trajectoryof the point x.

2.27. Let T: X —, X be an ergodic transformation preserving the prob-ability measure and let A be a measurable subset of X. Let = 0

if A, and = 1 if E A (n � 0). Prove that the limitlim >0<k<fl €k exists (the mean number of times the trajectory hits the setA) for almost all x E X, and find it.

Probability measures ii1 and on X are said to be mutually singularif there exists a set A C X such that ii1(A) = 1 and ii2(A) = 0.

2.28. Let ii1 and ii2 be two probability measures on X, and let T bea transformation that is ergodic with respect to both ii1 and ii2. Prove thatthe measures ii1 and either coincide or are mutually singular.

2.29. Consider an irrational shift

S(x) = (x + a)mod 1(x E [0, 1) = X, a E

138 X. ITERATES OF TRANSFORMATIONS OF AN INTERVAL

It is ergodic, according to 2.21a). By the ergodic theorem, the equality

lim! f(Sk(x))=J 1(x) dx (1)O<k<n 0

holds for any integrable function I on X and for almost all x E X.a) Prove that if f E C([0, 1]), then for the shift S it is possible to

assert more, namely, that (1) is valid for any x E [0, 1) (assume first thatf(0) = f(1), and then consider the general case).

b) Prove that if = (a, b) C [0, 1), = 1 for x E and = 0

for x then

lim! XA(Sk(X))= (2)

forall xE[0, 1).If is replaced by an arbitrary sequence {xk}. then the property

in b) is referred to by saying that the sequence is uniformly distributed, andthe equality (1) is called Weyl's criterion for uniform distnbutivity.

2.30. Let x = Xk E {0, 1}. be the binary expansion of anumber x E [0, 1). Prove that the limit lim Xk (the frequency ofl's) exists for almost all x E [0, 1), and find it. Sofve the analogous problemfor p-ary expansions.

2.31. Prove that if a transformation T: X —' X preserving a probabilitymeasure ji has the property

Iimji(A fl = ji(A)ji(B) (A, B C X) (*)

(the mixing property with respect to ji), then T is ergodic.2.32. Prove that the transformation T(x) = (2x) mod! (x E [0, 1)) is

mixing (with respect to Lebesgue measure).2.33. Suppose that p E (0, 1), q = 1 p, and is the measure on

[0, 1] given on closed intervals with dyadic rational endpoints as follows: ife1,..., 1} and

n)}

(here x = Exk/2k is the binary expansion of a number x). then

= fJl<k<n

and on the remaining closed intervals the measure is extended by additivity.Verify that:

a) the transformation T(x) = (2x) mod 1 is mixing with respect to themeasure

b)c) the measures and are mutually singular for p 1/2;d) /2k, and are mutually singular when p1 p2.

§2. TRANSFORMATIONS WITH AN INVARIANT MEASURE 139

2.34. Let be the first digit in the decimal expansion of the number2k (k=O, 1,...),thatis, a0= 1, a1 =2, a2=4, a3=8, a4= 1What is the frequency of the digit p E {1 9} in this sequence?

As known, every number x in (0, 1] can be represented uniquely as acontinued fraction

x=a +—a2+•

which is also written in the form x = [a1, a2, ...], where a1 = a1(x), a2 =a2(x), ... are positive numbers. Such a continued fraction has finitely manyterms if and only if x is rational. A finite continued fraction [a i,...,can be written uniquely in the form of an irreducible fraction Aninfinite continued fraction is a limit of finite continued fractions:

[a1, a2, . -.1 = lim[a1,... , =

and the number = is called a convergent. It is not hardto verify that

—

The Gauss transformation

T(x) = mod 1 (x E (0, 1])

is connected with expansion of numbers in continued fractions. Indeed, letx = [a1,a2, ...] and y = [a2,a3,...]. Then x = l/(a1 +y), whichimplies that y = T(x). With the help of T the numbers can be written

as follows:

= [l/x], a2(x) = a1(T(x)) =

The Gauss transformation preserves the Gauss measure d/1 = A dxf(l + x)(problem 2.5). The measure ji is a probability measure when A = lf(1n2).

2.35. a) For the Gauss transformation T prove that for some C> 0 theinequalities

� n B) � Cji(A)ji(B)

are valid for A, B C (0, 1] and n E N. Deduce from this the ergodicity of

T.b) Prove the following for almost all x = [a1, a2,...]:1) the relative frequency of a number k E N among the numbers a1,

a2, ... is equal to1

1

(k+1)2

140 X. ITERATES OF TRANSFORMATIONS OF AN INTERVAL

2)a +••-+aurn—

n

3)Ink/1n2

4)

where is the denominator of the nth convergent.

Solutions

CHAPTER 1

Introduction

Sets

1.1. b) Consider the mapping e (a', a"), where a = {ak}, ={e2k1}, and a" = {a2k}, and show that it is bijective.

1.2. Use the binary expansions of numbers in [0, 1).1.3. Use the results of problems 1.2 and 1.lb).1.4. Replace the set R by the set S of binary sequences, and prove that it

has the same cardinality as the set EN = xE xE x... For this consider themapping a = {a,,} —, E EN where = {,1(k) k)

1k) = a2k—(211) (k, J E N), and show that it is bijective (compare with thesolution of problem 1.lb)).

1.5. Use the result of problem 1.4, with the fact that a continuous func-tion on [a, b] is determined by its values at the rational points.

1.6. Replace the set N be the set and consider a system of convergentsequences of rational numbers with distinct limits.

1.7. Use the result of problem 1.6, replacing the set N by the set{2,22,...}.

1.8. Replace the set N by the set and consider the system of sets ofthe form aER.

1.9. b) Consider the set = {A E .9(N)I maxA = n} (n E N), andprove that = Setting = 0, define by induction with

the help of the equality (A E n E N) andshow that the function + 1 has all the required properties.

1.11. Using the Axiom of Choice, form a set E1 by choosing one pointfrom each equivalence class described in problem l.lOb). Consider all pos-sible rotations of E1 through angles 2ith where 0 E and 0 < 0 < 1.

1.12. Associate with each figure eight a pair of rational coordinates, tak-ing one point inside each loop.

1.13. Consider first the bird tracks with lengths of segments bounded offzero. Use the same idea as in problem 1.12, taking points with rationalcoordinates in all three angles corresponding to a bird track and sufficientlyclose to its vertex.

1.14. Associate with each T-shaped figure a half-disk of radius 1/4

143

144 L INTRODUCTION

constructed as shown in Figure 9. Show that these half-disks are disjoint.Adding their areas, we get that � a2, that is, N � < 1 1a2.

1.15. Let P be the plane of symmetry of a hoop 0 perpendicular toits axis. On this axis fix points M and M' symmetric with respect to Pand separated by a distance equal to the diameter of the hoop. Considerthe balls about M and M' with radius ö (see Figure 10). We say thatthese balls accompany the hoop 0. Consider hoops of width at least e > 0.Prove that for any e > 0 there is a number ö > 0 such that if both ballsaccompanying a hoop 0 intersect the balls accompanying a hoop c/, thenO and 0' themselves intersect.

1.16. See the solution of problem 1.17a).1.17. a) Let = {n E rn}. Arguing by contradiction, consider

an infinite set M = {m1, m,, ... } C N such that ) <+oo for any

j E N. Show that in some infinite subset of M there be two pointsin the same set

b) Let = {nlk E By a), there is an index in1 such that the setE1 = is infinite. Replacing N by E1 and by = n E1

FIGURE 9

M'

FIGURE 10

§1. SETS 145

I I I I I

p q p+a,j C a0 a2

FIGURE 11

and again using a), we see that there is an index m2 > m1 such that theset E2 = is infinite. Using induction, we construct a set E ={m1, m2, .. . } which is the desired object.

1.18. Consider the set E in b) of problem 1.17 and show that ifcard(E fl � 2, then there is an index p such that fl � 2.

1.19. b) First prove by induction that if x, y E E and k = 0then forany nEN.

Let [p, q] C E and a E E, and assume for definiteness that a> q. Weprove that [p, q] C E. Let c = sup{x > x] C E}. It is clear that

Assumethat c<a. Considerapoint a0EE suchthat c<a0and (p + a0)/2 <c (see Figure 11). Then

(x+a0 fp+a0 c+a0

and hence E j [p. (c + a0)f 2), which contradicts the definition of c, since(c+a0)/2>c.

1.21. Show that along with each of its points a the set E contains allpoints of the form (n E N), and E also contains arbitrarily smallnumbers.

1.22. a) Let be the system of all possible disks with rational centerand radius. It is clear that if x E then there is a disk B' E such thatx E B' c For each disk in fix arbitrarily a disk Ba containing it (ifsuch a disk exists). Show that the system of disks chosen in this way is thedesired object.

b) For each n E N find an at most countable set E such thatE C UXEE B(x, 1/n). Consider U1

1.23. a) Prove the countability of the set

1/n)\{x}=ø}.b) Prove the countability of each of the sets

E = {x EEIEfl(x, ={xTo do this verify that for distinct points x, x' E E

x x' (x and(x' — 1/n, x')) are disjoint.

1.24. Apply the result of problem 1.25 to the set

G=nEE

146 I. INTRODUCTION

1.25. Note first that if 0 <p <q. then there is a number C such that[C, +oo) C nq).

Let 0 <p <q and m E N. Using the remark above, we see that thereexist numbers x1 E (0, q) and n1 E N such that n1 > m and n1x1 E G.Consequently, there is a nondegenerate closed interval [p1, q1] C [p. q] suchthat n1x E G for any x E [p1, q1]. Replacing [p, q] by [p1. q1] and m by

n1 and repeating the arguments, we construct a nondegenerate closed interval

[p2, C [p1. q1] and a positive integer n2 > n1 such that n2x E G for anyx E q12]. Continuing this process, we get a sequence of nested intervalswhose common point is the desired one.

1.26. Argue as in the solution of problem 1.25, replacing the set G at eachstep by a set Gk in such a way that each of them is used infinitely many times(forexamples,inthesequence G1,G2,G1,G2,G3,G1G2,G3,G4,...).

1.27. a) Show that the Cantor set has the same cardinality as the set ofbinary sequences (see problem 1.2).

1.28. b) Use the result in part a).1.29. Consider, for example, the set E = K \ where K is the Cantor

set.1.32. Show that A contains a set with the cardinality of the continuum.

To do this consider the sets

= U (n E N),ce,..., cE(O,l}

where denotes the closure of and prove as in problem 1.27a) that theset Q = has the cardinality of the set of binary sequences. Verifythat Q \A is at most countable.

1.33. Assume that (a, b) is a bounded interval, and is a sequenceof disjoint nonempty closed sets contained in (a, b). Prove that (a, b) \U1F,, 0. Let p = infF1 and q = supF1. Then the intervals =(a, p) and = (q, b) are nonempty, and the sets = F,, n and

F, = F,, n are closed. Replacing (a, b) by and the sets by thenonempty sets we construct the intervals and Carrying outthis construction for we get the intervals and The intervals

and are disjoint from the set F1 U F2 and from eachother. Continuing this process by induction, we construct a family ....of open intervals satisfying the conditions of problem 1.32, and also tcieequality n any e1,..., It now remainsto refer to problem 1.32.

1.34. Find a straight line not containing points of tangency of the givendisks. Use the result of problem 1.33.

1.35. Let be the closure of Arguing by contradiction, constructa sequence of nested closed intervals C (a, b) such that = 0 forany nEN.

1.36. Use the result of problem 1.35.

§2. INEQUALITIES 147

Inequalities

2.1. Let = E1<k<m ak—Em<k<n ak. It is clear that S0 = Using

the fact that the sequence changes sign, choose an index p such that5,, and have different signs, and show that at least one of the numbers

IS,,I and '5p1 I does not exceed maxl<k<fl lakI.2.2. a) Use the inequalities

� + � 1+

b) The right-hand inequality follows at once from the right-hand inequalityof a). The left-hand inequality is proved by induction. Note also that the left-hand inequality in b) is a special case of the inequality in 2.7 for bk = i/ak.

2.3. The right-hand inequalities are easy to prove by induction. Toprove the left-hand inequalities it suffices to observe that + ak) x

— ak) � 1 and to use the right-hand inequalities.a) Use the Cauchy inequality about the arithmetic mean and the

geometric mean, along with the binomial formula.b) Use the inequalities 1 + a � and ? � 1 + + ... + for

a>—i and a�O.2.5. a) Represent each factor on the left-hand side as the sum of a geo-

metric progression, and multiply the series obtained.b) Using the inequalities a) and ln(i +1) 1, we have that

E ln(14)� E � Hk

= fi1<k<m k

Taking logarithms, we get that

I / i\\ / 1

lnln(i+Pm)=ln(/

k

2.6. It suffices to prove only the right-hand equation (the left-hand in-equality follows from the right-hand inequality, applied to the sequences

{ak}7 and {—bk}7). Here it can be assumed that the sequence is

nondecreasing—this can be achieved by changing the numbering. Assuming

that > a1 for some indices / < 1, prove that transposing the numbers

and in the sequence {ak}7 increases the sum >1<k<n akbk.

148 1. INTRODUCTION

These inequalities have a simple mechanical interpretation. Suppose thatn weights of magnitude a1..., a lever rotating in thevertical plane above a fixed point 0 at distances bi,..., to one side of

0. Then >1<k<fl akbk is the static moment of the system. It is maximal ifthe the weights increase in the direction away from 0. But ifthe magnitudes of the weights decrease, then the static moment is minimal.

2.7. To prove the right-hand inequality consider the double sum

1<k<n

and use the fact that its terms are nonnegative.The left-hand inequality can be proved similarly or can be derived from

the right-hand inequality, applied to the sequences {ak}7 and2.8. To prove the inequalities a) and c) use the Abel transformation. The

inequality b) follows from a).The factor M cannot be decreased: fix / E {1 n} and consider the

sequence ak = 1 for 1 � k � j ak = 0 for / <k n. Then

akbk=bl+...+bJ�Mj=M ak.1<k<n

Consequently, M � (b1 + ... + for any j E {1 n}.2.9. To prove the first inequality use the Abel transformation and the in-

equality b1 + . + bk � min(k, m). The second inequality fol-lows from the first, applied to the sequences {a1 — aflk+1}7 and {bflk+l}7.

2.10. Use the identity

1 2 1 1 2—

— — j = (ak — as).1<k<n J

To get an upper estimate of the double sum use the inequality ak —a lower estimate use the inequality ak — � ölk — Il

(it follows from the monotonicity of the sequence Note that theinequality b) is valid also for a nonomonotone sequence if we set

a. —a.ö=min k—j

2.11. The solution is analogous to that of the preceding problem, but theinequalities (j—k)(ak+l—ak) � � for 1 � k <J � nare used instead of the inequalities ölk — fI � ak — � INk — ii.

2.12. a) Using the Cauchy inequality about the arithmetic mean and thegeometric mean, prove that the derivative of the function =

+x) (a +x) is not less than 1.b) Using the inequality a), prove first the particular case when ak = 1

for k = 1 n. The general case reduces to the particular one if we setbk = ckak for k = 1 n.

§2. INEQUAL!TKES 149

2.13. To determine the sign of the derivatives use the inequalities

2.14. The inequalities can be proved by induction. To prove the induc-tion step use the inequalities (1 + <e <(1 + which followfrom the result of the preceding problem. To prove b) for n = 11 use thefactthat e<11f4.

2.15. The inequality c) is stronger than the inequality b), because

0(x) = 2P2(1 +x") —(1 +x � 0.This follows from the easily verified relations 0(1) = 0 and 0'(x) � 0.

c) For p = 2 the inequality is obvious. For p> 2 let q = pf(p — 1),q E (1, 2), and prove that the function

1(x) = (1+ + (1 —

(1

takes its largest value on [0, 1] at the endpoints of this interval. To dothis, show that f first decreases and then increases, that is, f'(x) <0 forx E (0, c) and f(x) > 0 for x E (c, 1), where c is some point in (0, 1).It is clear that

— p(1 (1 — 1(x

— (1 (1

Since we are interested in the sign of f(x), we study the function

g(x)= (1 — (1

Let r = p — 1 and 1 = lfx, and note that q — 1 = 1/r and r> 1. Theng(x) = + 1)P1 where

1>1.

Itisclearthat as t—' 1+0,and as t—'+oo. Itnowsuffices for us to verify that is increasing. Uncomplicated transformationsgive us that

=- v(t)),

where a = 1— lfr,t+1 I

u(t) = 2r-j—j_ and v(t) = —,—-—

Note that u(t) = v(t) = 2. Therefore, to prove the in-equality (t) > 0 it suffices to see that both the functions u and v arestrictly increasing (in this case v(t) <2 <u(t) for t> 1). Since

u'(t) =+

—

150 1. INTRODUCTION

and the function = hr + a/i — is positive for I > 1 = 0

and c3!'(i) = — C2) > 0), it follows that u'(t) > 0 for 1> 1. Nowconsider the function v. Since

I 2r \fi_1\r

it remains for us to see that the function

/3(t)=

is positive for 1> 1. This follows from the obvious relations /3(t) —' 0 as

I -, +00 and

/3'(t) = (11)2 (1—

d) Prove that the derivative of the function = + —

(1 + x)2 — (p — 1)(1 — x)2 is negative on (0, 1). To do this, it is useful torepresent in the form and investigate the sign of the functionip on (1,+oo).

2.16. The left-hand inequality follows from the relation ln(1 + x) � x,x> —1. To prove the right-hand inequality, verify that the function =12/n + e'(l — is increasing on [0, n]. To do this, represent inthe form = iW(t) and show that � > 0.

b) It must be verified that the function = 12/n2 + et(1 — 1/n)'1 doesnot exceed 1 on [0, n]. Since the inequality holds at the endpoints ofthe interval, is suffices to establish that � 1 if = 0, that is, ife'(l — i/n)'1' = 2/n. But in this case

IE[0,n], n�2.n ii1

c) It can be assumed that I E [0, Let t = where 0 E[0, 2). Then the inequality to be proved is equivalent to the inequality 1 —

— 0 � = + nln(1 — —

ln(1 — 0/2). Differentiating with respect to n, show that �0 1 0\ 0 1 10

where = u+21n(1 —u)+u/(1 —u). Since = 0 and w'(u) =(u/(1 — u))2 � 0, it follows that � 0, that is, � It remains toprove that p36(0) � 0 on [0, 2). Since

= (6— 0)(2 —0)

it follows that

= = +1n4 = 0.0297....

§2. INEQUALITIES 151

Note that the inequality c) holds for n � 30, but does not hold for n � 29.2.17. The left-hand inequality follows from the relation ln(1 —x) � —x—

x2/2 for 0 <x < 1. To prove the right-hand inequality we consider thefunction = — e'(l —

Since � and ço(n) �it suffices to prove that � if = 0, that is, =

e'(l — But in this case = �2.18. a) In proving the inequality = x2/(1 + x) — 1n2(1 + x) � 0,

represent the function ço' in the form = (1 + x)w(x) and verify thatsgn = sgn x.

b) Study the character of monotonicity of the function = arctanx —

x/(1 + 2x/ir) and use the fact that 0 = = 9,(x).

d) Consider the function = 2sin(irx/2) + x3 — 3x for x E [0, 1].Starting from the graph of ç0ffl, construct successively the graphs of the func-tions ço", ço', and

e) Consider the function = sin2 x tanx — x3, and prove successively

that ço'">0, ço">0, ço'>0,f) Compare the Taylor expansions of the functions cosx and

2.19. These inequalities are special cases of the inequality

� xy, (*)

where x, y > 0, y � f(x), and f is such that f(x)/x 1. The inequality(*) follows from the inequalities f(x)/x � f(:)/z for z = x.The sign should be changed in the inequality c) for p <0. For the proof itsuffices to observe that for p <0

and

For p E (—1, 0) the inequality d) is valid for x close to 1, but it doesnot hold in a neighborhood of zero—for a proof it suffices to consider theTaylor expansion of the function on the left-hand side of the inequality.

2.20. The right-hand inequality is a special case of the inequality (*).The left-hand inequality is a special case of the inequality opposite to (*):

x > 0, y � 1(x), and the positive function f is such that f(x)/x 1.The inequality (**) follows from the inequality f(x)fx f(z)fz for =

2.28. See [34]. It suffices to prove the inequality for step functions in K.If f E K for any number in (0, 1).It is easy to see that for any function g

max(f, g) g)+(1 g).

152 I. iNTRODUCTION

Therefore,

jbmax(f(x),g(x))dx

b

max (1 max(f1(x), g(x))dx, j max(f2(x),

Ifthe number of "steps" corresponding to the function f E K is greater thantwo, then the function is easy to represent as half the sum of two differentfunctions in K. Indeed, let c1 � c2 � � c, > 0 be all the values of

n � 3. Then there are numbers e, ö > 0 such that the functionsf± having the same intervals of constancy and taking on them the valuesc1(1±e), c3 c,1, c,(l±e) arein K. Thisleadstothe integral

f max(f(x), g(x))dx attaining its maximal value on functions consistingof two steps. Each such function has the form f = C1X1a s) + C2X15,bI, where

S E (a, b), and the coefficients c1 and c2 are uniquely determined from thecondition f5 E K: c1 = bf(s(b—a)) and c2 = af(s(b—a)). If a <1 <s <bthen

b

jThe right-hand side is maximal for s/i = and is equal to

v's) in this case.

Irrationality

3.1. a) b). It suffices to prove that for any number Q E N there existintegers p and q such that 1 � q � Q and Ix — < 1/(qQ) (sincethe right-hand side of the inequality tends to zero as Q +00, while theleft-hand side is positive by condition a), this implies that there are solutionsof the inequality Ix < 1/q2 with arbitrarily large denominators q).

Fixing a number Q E N, we partition the interval [0, 1) into Q congruentintervals [0, 1/Q), [1/Q. 2/Q) [1 — 1/Q, 1) and consider the points

= nx—[nxj for n = 0 Q. It is clear that E [0, 1). Since thenumber of these points is greater than the number of intervals, at least oneof them contains two points, say 0k and 0 � k <j � Consequently,1/Q>

— = Ijx — [jxJ — kx + [kxJI. Taking q = j — k and p =[jxJ—[kxj, we get that 1/Q> qx —pa.

c) a). Assuming that x = p/q. where p E Z and q E N, we getthat — —' 0 as k —, +00. Since — E Z, this implies that

— = 0, that is, p/q = for all sufficiently large indices k. andthis contradicts the assumption.

3.2. Use the result of problem 3.1.3.3. a) It suffices to prove that the sequence does not have a

limit. Assume that this is not so: sin —, 1. If F = 0, then4 = + where k,, E Z and —, 0.

§3. IRRATIONALITY 153

Consequently, 0 = — + —Therefore, for sufficiently

large n we have = , that is, = A E But then theequality 4fl = + ç, implies that 1 = irA. This is impossible, because thenumber it is irrational (see problem 3.13).

If F 0, then we get from the equality

sin4n+1 = 4 sin 1 — 2 sin2 41?)

that the sequence {cos 41?} has a limit. Consequently, = 2irk, + +where 0 < < it, E Z, and —. 0. Since 4fl+1 = + + itfollows that 0 = 2ir(4k1? — + + —

, and this is possible onlyfor = ±2ir/3. Therefore, for sufficiently large n the difference —

is constant and equal either to 1 (if = —2ir/3) or to —1 (if = 2ir/3).In the first case it is easy to show by induction that k1? = A

. + 1/3, whereA E But then it follows from the equality 4' = 2irk1? — 2ir/3 + thatit E The case = 2ir/3 is handled analogously.

b) Represent the number 0 in the form 0 = irm + it wherem = [0/in and = 0 or 1.

3.5. Consider the numbers 0.101001000100001000001 ... (the nth 1 isfollowedby n 0's) and 0.1234567891011121314151617181920... (allthenatural numbers are written out in the decimal expansion).

3.6. In the case a), b), and d) it is possible to use the result in problem11.1.21. In the case c) consider the numbers n = (2k)4 +1, / = 1 4k.Assuming k to be large, show by means of the Taylor formula that

sin(1rn312) = sin 4) +

3.7. b) Use the result of problem 3.2.c) Use the result of problem 11.1.21.d) Use the same device as in the solution of problem 3.6c).3.8. Prove that for any interval [p. qJ, 0 � p < q � 1, there exists a

number x E R such that 5(x1?) E [p. qJ for n E N. To do this choose anumber m E N such that the numbers a = m +p and b = m + q satisfy theinequality

a(q—p)�2. (*)

It is clear that 5(y) E [p, q] for all y E [a, b]. It follows from the inequality(*) that b2 — a2 � 2. Therefore, the interval [a2, b2] contains an intervalof the form [m1 + p, m1 + qJ, where m1 E N. Let [a1, b1J C [a, bJ be an

interval such that = m1 +p and = in1 +q It is clear that 5(y2) E [p. qJfor all y E [a1 , b1J. It follows from the inequality (*) that the length of

[at, is at least two:

— � — = a1(q — p) � a(q — p) � 2.

154 1. INTRODUCTION

Therefore, [at, contains an interval of the form [m2+p, m2+qJ, whereE N. Suppose that [a2, b2J C [a1, b1J is an interval such that = m2+p

and = m2 +q. It is clear that 5(y3) E [p. qJ for all y E [a2, b2J. By (*),the length of the interval is no less than two: � �a2(q — p) � a(p — q) � 2. Continuing the construction by induction, we geta sequence {[ak, bkj} of nested intervals. The desired point is taken fromtheir intersection.

In conclusion note that the numbers x satisfying the conditions of theproblem form a set having the cardinality of the continuum. For the proofit suffices to make a slight change in the solution, replacing the inequality(*) by the inequality a(q — p) � 3. This enables us to double at each stepthe number of intervals having the required properties. As a result we get ageneralized Cantor set with all points satisfying the condition of the problem.

3.9. Direct computations prove the assertion for functions of the formf(t) = cos(2irmt) (m = 0, 1, 2,...). This implies that the assertion is validfor all functions I of the form f(t) = >0<m<M am cos(2nmt). It remainsto use the Weierstrass theorem (see VII.3.1 1).

3.10. Use the same idea as in the proof of the assertion a) b) inproblem 3.1: for any number Q E N there exist integers q, p1,.., suchthat 1 � q � Qm and I 1 m. Fora proof,partition the cube [0, 1)m into Qm congruent disjoint cubes and considerthe points = (O(nx1) O(flXm)) E [0, lIm for n = 0 Qm.

3.11. a) The right-hand inequality is obvious. Assuming that —

m/nI� 1/4n2 forsome in, nEN,wegetthat

_____

1

— —

Consequently, 12n2 — m21 < 1 , which is impossible.b) It follows from the inequalities

= � C/nj and = �that — — — � that is.

2C

——

n E N, it follows that

— ——

which implies the inequality to be proved.3.12—3.14. After verifying that the remainder of the series is majorized

by the first term of this remainder, show that the assertion c) of problem 3.1holds.

§3. IRRATIONALITY 155

3.15. Let P be an algebraic polynomial of degree n with integer coeffi-cients such that P(cx) = 0. It suffices to consider fractions p/q close enoughto that — < 1 and P(p/q) 0. Since E Z\ {0}, itfollows that � On the other hand, expanding the polynomialP(t) in powers of 1— we get that

1<k<n

Thus, � — The assertion proved is calledLiouville's theorem.

3.16. b) Let = E ek'nk and prove that for any e > 0 there existsolutions of the inequality — < with arbitrarily large q E N.Since (see problem 3.14), this gives us by Liouville's theorem thatthe number is not a quadratic irrational.

Since — E1<k<m(ek/nk)I � it suffices to prove that

.= 0. Assuming the opposite, we get that �

Let Lm = Then the last inequality takes the form

Lm+i and hence

� Lm+i � C(CLm1)3 <... � � (CL1)3.Consequently, � CL1 , which contradicts the condition.

3.17. See the solution of the next problem.3.18. e) Arguments analogous to the solution of problem 3.16 show that

the equality limk..,X = +00 for some N E N, N> 1, implies thatthe sum of the series is not an algebraic number of degree N. Itremains to see that the condition çiiç> 1 implies that 1 =+00 for all N E N.

3.19. a) It is clear that e—El<k<fl(l/k!) = 1)!, where 0 <3. Assuming that e = p/q. p, q E N, and multiplying by qn!, weget that

pn!—qn!O<k<n

The left-hand side of this inequality is a positive integer, while the right-handside tends to zero with increasing n, which leads to a contradiction.

b) If Ae2 +Be + C = 0, then Ae + Ce' E Z. This is possible only forA = C = 0; the proof is analogous to the solution of problem a).

3.20. a) Integrating by parts, verify that

'n+l (n EN).

b) Assume the opposite: ir2 = p/q. where p. q E N. Then

q

156 I. INTRODUCTION

This equality leads to a contradiction, since its left-hand side is a positiveinteger, while the right-hand side tends to zero as n increases.

The result obtained, like the results in the next two problems, was estab-lished (by another method) by Lambert. In the book [32] the reader can finda history and expanded exposition of this theme.

3.21. The solution is analogous to that of the preceding problem.3.22. a) The solution is analogous to that of problem 3.20a).b) Assume the opposite: tan r = p/q, where p E Z and q E N. Let

r = a/b where a, b E N. Then

c,

From this,

qb'C, =

The left-hand side of this equality is an integer, while the right-hand sidetends to zero as n increases. To reach a contradiction it remains to see that

0 for sufficiently large n. If r � ir/2, then this is obvious. Supposethat r> ir/2. Then

= 2j (r2 — t2)' costdt + o ((p2 — n2)fl)

2�j

CHAPTER II

Sequences

§1. Computation of limits

1.2. Verify that the sequences {x2k} and {x2kl} increase. The Wallisformula can be used to compute their limits.

1.3. Show that the sums under consideration differ little from a geometricprogression.

1.4. Write in the form = and study using

Taylor's formula.1.5. The first half of the terms gives an infinitesimally small contribution.

Therefore,

>n/2<k�n O<j<n/2

=O<j<n/2 i�O

1.6. a) The main contribution is given by the last terms:

(J)flJ

=j�O

The sum of the remaining terms is infinitesimally small, since

= 4/n2.b) Represent the difference in the form of a sum where

/1 4 27\ 1 1 4 27 256U 1++Il+ 3+ 4" \n n2 n31 (n+1) (n+1) (n+1)

(fk\kV,,

4<k<n

158 II. SEQUENCES

Obtain the inequality � 0 by using the fact that the function f(x) =(x/(a + x))x is decreasing on (0, +oo) for any a > 0. Verify that > 0

for n � 8. Establish directly the inequalities between the for 1 n 8.1.7. To investigate the quantities (n + 1)(n + 2). (n + n) = (2n)!/n!

use the result in problem 1.2.14. Another solution is obtained by noting that+ 1).. . (n + n)/n) is an integral sum corresponding to the integral+x)dx.

1.8. Use the result of problem I. 2.14.1.9. In problems a)—d) use the fact that the sums under consideration

differ little from integral sums. In problem e) show that the sum lies between

the integrals — 1))112 di and —

1.10. a)—c) Use the equality e = Ek>o 1/k!.d) Compare 1)1? with the integer 1)1? + (1 —1.11. The convergence of the sequence to a positive

limit is equivalent to the convergence of the series

1.12. a) The convergence of the sequence follows from its monotonicityand the inequality 1+ which is proved by induction. To computethe limit use the identity = +

b) Since = a + it follows that

i —xa 1?

and hence — = — x1) XkY'. Therefore,

and it is possible to use the result of problem 1.11:

fl1<k<n

For a = 2 it is not hard to see that = 2 and hence 2— x,,

1.13. Since

2—x = =

4+2; + 4+ +

= ...(2..x1) fl1<k<n

it follows that 2— = Using the result of problem 1.11, we getthat +2; +

§1. COMPUTATION OF LIMITS 159

1.14. The monotonicity of the sequence {xn} is obvious. Boundedness

follows from the inequality � � that is,1.15. If —' c < +00, then = + •.• + � c, and hence

c, that is, � c" . To prove the converse assertion it suffices

to observe that the inequality ç � C" implies that

Using the result of the preceding problem, we get that {Xn} is bounded, andthis suffices for convergence, since the sequence is increasing.

1.16. It follows from the result of the preceding problem that the conver-gence of the sequence

(*)

where ak, bk � 0, is equivalent to boundedness above of the sequence+ >.1<k<n Therefore, the expressions on the right-hand

sides of equalities a)—e) are the limits of convergent sequences (use the in-equality � in example d), and the inequality � for x � 1in example e); these inequalities are easy to prove by induction). To find thelimit of the sequence defined by (*) consider a sequence {On} with

(nEN). (**)

Replacing an by in (*), we get an upper estimate for

Xn

(the last equality follows from (**)). To get a lower estimate for x, use the

inequality (for a, b � 0 and c � 1) n — 1 times.

This gives us that Yn � Xn Thus, 01 � Xn � and

hence Xn —' if —, 0.Verify that in examples a)—e) the relation (**) is satisfied by the following

sequences {0n}

a) n + 2; b) n + 2; c) cos(x/21Z_l);

d) 3U2n+2 e)

In the solution of examples d) and e) use the following representations of the

160 II. SEQUENCES

Fibonacci numbers and the Chebyshev polynomials:

— i

These equalities can be proved by induction.1.17. The convergence of the sequence follows from the result

of problem 1.15. To estimate the difference a — XN, write it in the form(a — xN) = — xv), and consider the quantity — x,:

TI 1(k) (k)

where = + J(k + 1) + . + Using the recursion formula

= + prove successively that:

a)

b) for 1�k<n;c) = 0(1/k312)) for 1 � k <n.

(Here the constants in the 0-terms are independent not only of k but alsoof n.) With the help of the equality c) and the result of problem 2.12a) showthat

that is, — const — 1)!). This implies immediately that

a—xN —xN — 1)!). Therefore, using the resultof problem 1.2.14, we get that

1.18. Let = — x,11/2. Then

1<k<n 1<k<m

§1. COMPUTATION OF LIMITS 161

Since 2k 0, for large in the second sum is small for all n � in. For fixedin the first sum is arbitrarily small if n is sufficiently large.

1.19. a) Assuming the opposite, we get for sufficiently large n that

\ / \ n /which implies that

a1

n+1 n n+1This is impossible, because the partial sums of the series

n+ldo not exceed a1.

b) Consider the sequences = nC (for c> 1) and = n ln(n + 1).1.20. Let I = E [—oo, +oo) and L > 1. Fix an index in

such that <L. Representing an arbitrary index n > m in the formn = km + j, where k E {0, 1, ...} and / E {1 m}. we get that

= akfl+J � + a1 � karn+ and hence

Thus, / � � � L for any number L> 1, which impliesthat = =1, that is, ar/n —i!.

1.22. Apply the result of the preceding problem to the sequencesand {x2kl}.

1.23. Apply the theorem of Stoltz (see problem 2.6) to the sequences{x2k/k} and

1.24. b) To construct such a sequence proceed as follows. Consider asequence of closed intervals such that is located to the left of

Define the function f E C(R) by the equality 1(t) = if I E ,and let

f be linear between the intervals and Let = 1(n). It is clearthat if the lengths of the open intervals adjacent to the increase withoutbound, then — —* 0. On the other hand, by choosing successively thelengths of the to be sufficiently large it is possible to ensure that thereare numbers arbitrarily close to +1 and arbitrarily close to —1 among thenumbers + x2 + + xv).

We get another example by considering the sequence {sin(ln n)}.

1.25. Let <a <b <limx,,, and consider the sets E1 = {n E

NI; <a} and E2 = {n E NI; > b} As established in problem 1.1.24—1.1.26, there is a c > 1 such that each of the sets E1 and E2 containinfinitely many numbers of the form [ck]. but this is incompatible with theassumption that the limit limk..,+,O exists.

162 IL SEQUENCES

§2. Averaging of sequences

2.2. The left-hand inequality follows from the right-hand inequality, ap-plied to the sequence To prove the right-hand inequality it sufficesto consider the case when = 1 <+00. Let L >1 and let m = mL bean index such that <L for n > m. Then

— 1 1= E akxk + E akxk

l<k<m m<k<n

m<k�n

Consequently, � L for any L >1, that is,

2.3. a) If � then k/nk+l � pa/n � kink.b) If 0 < � 1, then, for example, A = {n E — 1)]}. In

thiscase2.4. a) Apply the result of problem 2.1 to the sequence — al}.b) Let > 0, and let Ae = {n E Nil; — ai � Verify that 0(4) = 0.

Let pk(n) = card{Al,k n [1, n]}, and let the numbers 1 = n0 < n1 be

such that < for n > Define A = Uk>l Al/k andB = N\A = {m1, m2, ... }. Then 0(B) = 1 and Xm -4 a.

2.5. To construct a sequence having the limit lim((a1 +••

but not a limit + + it suffices to change somewhat thesolution of problem 1.24b): let {bk} form an alternating sequence of +1and —1 on each closed interval and on the adjacent open intervalslet bk = 0. For a suitable choice of the lengths of the closed intervalsand the lengths of the adjacent open intervals the sequence {(2 +is the required sequence. To construct a sequence having the limit

+• • not having a limit lim((a1 + . . . consider

the sequence +Another example is obtained by considering the sequence with

iseven,

1 if[log2n] is odd.

Of the three limits lim + + lim + ... + andlim + + only the second exists.

b) The inequality a2 � b follows from the Cauchy inequality. To constructa sequence with given limits a = lim((a1 + + and b =

+... + where 0 � a2 � b � a 1, consider the sequencetaking the value on a set A C N of density a (see problem 2.3) and thevalue /3 on the set N\A. Prove that the required equalities can be ensuredby suitably choosing the parameters and /3.

§2. AVERAGING OF SEQUENCES 163

2.6. It can be assumed that / � 0. Here it suffices to consider onlythe case / = 0: if 0 < / < +00, then the sequence — should beconsidered instead of and if / + 00, then the sequences and

should be interchanged. (The conditions > and +00hold because — > — for sufficiently large n.) Assuming that

ep

prove that —. 0. Since

(xk—xkl)=xl+1<k<n 1<k�n

it follows that

= + — Yk1 + — Yk_1

1<k�m

m the second sum is small for all n > m, because €k 0. Forfixed m the first sum is small, because —' +00.

We get another solution if we apply the result of problem 2.2 to the se-quences = and = (instead of

2.7. Use the theorem of Stoltz.2.8. Use the theorem of Stoltz. In examples d) and e) use the result of

problem 1.2.14.2.9. To prove convergence of the sequence (Inn — 1/k} use the

fact that convergence of the sequence is equivalent loconvergence ofthe series — The series arising here converges, as follows from

the easily verified inequality 0< 1/k — ln(1 + 1/k) < 1/2k2.Another solution is obtained by using the inequality

1 1 1 1 10<———<--—---——< 2[x] x x—1 X (x—1)

for x> 1 and observing that the improper integral1\

I t———idxJ2 xj

converges; therefore, the sequence

J =J2 xj 2 3 n—i

has a finite limit.2.10. The solution is analogous to the solution of problem 2.9.2.11. It follows from Stirling's formula that to prove the right-hand in-

equality it suffices to see that the sequence = n!/(V n(n/eyzehl'12fl) is

increasing. Since1 1 1\ n-i-i

=

164 II. SEQUENCES

where

6(1 +:2+t —ln(1+t),it suffices to verify that > 0 for z > 0, but this follows from the equal-ities = 0 and 9,'(i) = + 1)2(2 + 1)2). The left-hand inequalitycan be proved analogously.

2.12—2.14. The solution is analogous to that of problem 2.9.

§3. Recursive sequences

3.1. Prove by induction that the sequence has the form {Aq'1+B},and determine the parameters A, B, and q.

3.2. a) Since

1 xo—x1— = — = =

it follows that

= x0 + —n>1

= x0 + (x0 — x1)n>1

=xo+(xo_x1)Q-_1).

3.3. Since at each step the term in the sequence almost doubles, itis natural to consider the sequence = It is obvious that

n+1 1

yn+1 = —

that is,

yn+1 Yn

n+1 n n+1

Consequently, = n(y1—

2_k/k). From this,

k>n

Therefore, — ln(e/4)) for x1 ln(e/4), and otherwise

'-' 1x =n2 -p1."

3.4. Assuming that b 0, we consider the sequence = Then

= where = It is clear that = Y1

§3. RECURSIVE SEQUENCES 165

If IbI> 1 and thenS

S akYn+l = Y1 + — = —

1<k<n k>n

— a a1 — bk>l

This yields b). It remains to consider the case when 0 < Ibi < 1. We havethat

— a a — a + o(1)Yn+i — Li +

—1<k<n 1<k<n —

Therefore, = —' a/(1 — b).3.5. If p � 0, then the sequence = (p — 1)1? does not have a limit,

although = p � 1 , then either � for alln E N, or � for n � n0 (since the inequality � impliesthat � for n � n0).

Let 0 <p < 1, and define / = and L = Since thesequence is bounded above (by the number max{x1 , L < +oo.Assuming that / <L, fix two numbers 1' and L' with / <1' <L < L'; theywill be made more precise later. Then <L' for all sufficiently large n.Further, n can be chosen so that <!',and hence <p!'+(l —p)L' =

Since it follows that <A, Therefore,L = � that is, L � pl'+ (1 —p)L' But this is not true if thenumber 1' is chosen sufficiently close to 1, and L' sufficiently close to L.

3.6. Since the sequence (In has a limit for p E (0, 2) (see thepreceding problem), it suffices to prove that for such values of p the se-quence is bounded. This is obvious if p E (0, 1], since in thiscase � max{x1 , If p E (1, 2), by multiplying the inequalities

� for n = 2 N we get that XN+l � x =Consequently,

XN+l

This implies that the sequence is bounded for p E (1, 2).If p (0, 2], then the sequence {exp(p— 1)1z} diverges, although =

For p = 2 it is possible to take the sequence {2h1}.3.7. Prove that the inequalities � c/(n — 1)! and � c/(n —2)!

imply the inequality � c/n!. The constant c is chosen so that the lastinequality holds for n = 1, 2.

3.8. The proofs of these assertions are similar. We confine ourselves tothe cases c). Fix a fi E (0, cr/k). By assumption, there exists a numberc = > 0 such that

166 II. SEQUENCES

xe

for n=k+1,k+2 . Let M>0 beanumbersuchthat

for j = 1 n — 1. Then < 12/eon Therefore, �if kC — /3k) + /3k2/2). Now fix an index np large

enough that kC � — /3k) + /3k2/2) for n � np. It can be assumed

that np � k. The inequality � holds for n = 1 np andfor sufficiently large M = Its validity for all subsequent indices n isensured by the choice of np.

3.9. Arguing just as in the solution of the preceding problem, we get thatit suffices to prove that fi(n — k) lnAfl k — (fin — —. +oo. For brevitylet a number y > 0; its choice will be made more precisebelow. For sufficiently large indices j we have that j(1 — a11/a1) <y, thatis, 1 — y/j. Consequently,

aflkH

J n--k<j�nThus,

I k 11

and hence

/3(n — k — (fin —

if y is chosen sufficiently close to the number 0.

3.10. Let C > 0 be a number such that Xm � Am for1 � m <n, n > k. Then

C Ix�—I +...+

+ + + + 1).

Therefore,

C? 1

<CeS

—

(the last inequality follows from the inequality_e' � 1 + t). Choosing thefactor C large enough that � CeS/yA for n = 1 k, we getfrom this that the inequality holds for all indices n.

3.11. Use the result of the preceding problem. In example d) use theresult of problem 2.14b).

CHAPTER III

Functions

§1. Continuity and discontinuities of functions

1.2. g) Consider, for example, a discontinuous strictly increasing func-tion.

1.3. Let w1(x) be the oscillation of f at a point x, that is,

co1(x) = lim( sup 1(y) — inf f(y)),ö—'O Iy—xko Iy—xko

and let A be the set of points of discontinuity of the first kind for 1. Itis clear that A = where = {x E A1w1(x) � 1/ n}. If A isassumed to be uncoui able, then at least one of the sets is uncountable,say Am• Let x0 be a nonisolated point of Am (see problem I.1.23a)), andlet {xk} C Xk —, x0, Xk x0 (k E N). It can be assumed withoutloss of generality that Xk > x0 for any k E N. We prove that f does nothave a limit from the right at the point x0. Indeed, since Xk E thereare points and such that <IXk — x01, — XkI <IXk — x01,

and — f(x)I> 1/2m. It is clear that —, x0 and —' x0. Thus,

1(x) — 1(x) � — � >0.x—'x0+O x—'x0+O m

Another solution can be obtained by proving countability of the sets

A1 = {x E E I there exists the finite right-side

limit f(x + 0) andf(x + 0) > f(x)},A2 {x E E I there exists the finite right-side

limit f(x + 0) and f(x + 0) <f(x)}.

The sets A3, A4 are defined in the similar way using left-side limits. Toprove countability of A1, associate with each point x E A1 a rectangle

(x, x +e) x (1, L) f(x) f(x +0) , and

is small enough that 1(t) > L for all t E (x, x + e). It is easy to see thatif x,yEA1 and Choosingineachrectangle apoint

168 In. FUNCTIONS

with rational coordinates, we get a one-to-one mapping of into x

The proofs that A2, A3, and A4 are countable are analogous.1.4. Consider the following functions and st':

(0 for x <0, (0 for x <0,= 1. 1 for x >0; = 1. 1 for x � 0.

The desired function can be obtained as the sum of a series with terms oftheform (EEL and SEER.

1.5. Assume that the set DR(f) is at most countable, and prove that theset DL(f) is at most countable. Let w1(x) be the oscillation of I at a

point x (see the solution of problem 1.3). It is clear that = Ak,where Ak = {x E I

� 1/k}. If we assume that the set DL(f) isuncountable, then at least one of the sets Ak. say must be uncountable,and with it also the set A = Afl\DR(f). Let x0 be a point in .4 suchthat (x0, x0 + e) n A 0 for any e > 0 (see problem I.1.23b)). Fix asequence CA such that —. x0, > x0 (ii EN). Since ; E A,there are points and such that x0 <xv, x0 <x <;,and

� 1/2m. Using the continuity of f from the right at x0 andpassing to the limit in the last inequality, we arrive at a contradiction.

1.6 b) Assume that a sequence {fj with the indicated properties exists.We find a point c E R such that the sequence does not converge,and thereby come to a contradiction.

Let be an arbitrary closed interval, and x1 an arbitrary rational num-ber interior to Take an index n1 such that 1f0(x1)— < 1/3, thatis, 1 — < 1/3. Using the continuity of , fix a closed interval

centered at x1 such that C and Ii — (x)I < 1/3 for any x ELet x2 be an irrational number interior to Take an index n2 > n1 such

that 1f0(x2) — < 1/3, that is, < 1/3. Fix a closed intervalcentered at x2 such that C and < 1/3 for any x E

Now take an arbitrary rational number x3 interior to and repeat the con-structions above, replacing by and x1 by x3, etc. By induction weconstruct a sequence of indices and a sequence of closed nestedintervals such that Ii < 1/3 for x E '-'21—1 and < 1/3

for Itisclearthat 1/3 atapointand hence the sequence cannot have a limit.

1.8. a) Consider the function equal to 1 on an at most countable subsetof E with closure E (see problem I.1.22b)) and equal to 0 at the remainingpoints.

b) Consider the series where is the function constructed asdescribed in the hint to part a), for the set E =

§1. CONTINUITY AND DISCONTINUITIES 169

1.9. Study the function

1f(n + 1) if the ternary expansion of x E (0, 1)

hasnl's(n=O,l,...),0 if the ternary expansion of x E (0, 1)

has infinitely many l's.

1.10. Prove that if 1(x) < c 1(x) and c f(x0),

then the set r'({c}) is not closed.1.1 1. Use the result of the preceding problem.1.12. Use the fact that it can be assumed without loss of generality that

f has a maximum at each point of E. Consider the set E EIf(y) �f(x) for Ix — <e, y E E} (e is a fixed positive number) and prove thatf is locally constant on Use the result in problem I.1.22a). Anothersolution can be obtained by modifying the arguments in the second solutionof problem 1.3.

The arguments remain in force for a separable metric space. The resultis not true in the case of a nonseparable space. To get a counterexampleconsider the space XxS without isolated points, where X is a closed intervalwith the discrete metric and S = {z E C I Izi = 1}, along with the functionf(x, z) = x.

1.13. Use the result of problem 1.12.1.14. a) Consider, for example, the function

l—x for—I<x�0,x1 +[x']

for 0 <x < 1f(x)= l+x'+[x']

1 + (2— x)' + [(2— x)'] for 1 <x < 22+(2—x)' +[(2—x)']

and the point c = 0.1.15. Arguing by contradiction, use the Bolzano-Cauchy theorem to prove

the absence of bijectivity.1.16. a) Use the mean value theorem on the interval [x, (1 + ö)x].b) Assuming that e < 1 and arguing by contradiction, find a sequence

such that —. +oo and � where 5 is somepositive number. Use a).

1.17. Assume that the limit 1(x) does not exist. Then thereare numbers a and b such that a <b and the sets Ga = {x > OJf(x) <a}and Gb = {x > 011(x) > b} are not bounded. Let x0 be a point such thateach of the sets Ga and Gb contains infinitely many points of the form nx0(n E N) (see problem 1.1.26). Since the sequence {f(nx0)} cannot have a

limit, we arrive at a contradiction.

170 111. FUNCTIONS

1.18. Consider the function

f(x)—11.0 foranynEN,

where is the sequence of prime numbers, and is a sequence whoseset of limits of convergent subsequences fills R (for example, enumerate theset arbitrarily).

1.19. Consider a partition of = (0, +oo) into equivalence classes,with x y if xfy E After establishing arbitrarily a one-to-one corre-spondence between the set of equivalence classes constructed in this way andthe set R, define the function I on (0. +oo) by 1(x) = tf(1 +[x]), where

E R corresponds to the equivalence class containing x.1.20. Prove that the assertion f(x + h) — 1(x) 0 as x — +00 on

any finite interval is equivalent to the assertion that for all e > 0 there existand such that and supa if(x + h) — f(x)i

e. Prove the last assertion by arguing by contradiction and constructing asequence of nested closed intervals and a numerical sequencesuch that —, +00 and + h) — � e for h E

1.21. Prove that the function = f(x + 1/ n) — 1(x) takes a zerovalue on [0, 1 — 1/n]. If 5 1/ n and S > 0, then consider the functionf(x) = — Ax, where E C(R). = 0. = A 0, and has

period ö.1.22. Prove that if 1(x) has the property indicated in the condition of

the problem, then so does f(—x), and reduce the problem to the cases when

f is either even or odd. In the first case show that 1(x) —, f(0) as x —, +00,and conclude from this that f const. In the case when I is odd prove that

—, 1(x) for any x E R, and using this verify that f(Ax) =for any ER.

1.23. As in the preceding problem, reduce the question to the case whenf is either even or odd. If I is even, then assuming that f(x0) [(0),consider the sequence If I is odd, then prove that f(Ax) =

considering successively the cases E N, E Z. 2 E and A E ILL

1.24. a) It follows from the equality (1 + x)( 1 + x2) (1 + x2 ).. =(1 — x)' (lxi < 1) that f(x)f(x) = (1 — .v)'. Therefore, f2(x2) �(1 — � f2(x), that is, 1 � < ffl <

b) Computing the logarithmic derivative of the function

F(x) = (1 - x)f2(x) =1

k>0 1 + X

we get that

= = S(x) +

n�0 n�7

§3. CONTINUOUS AND DIFFERENTIABLE FUNCTIONS

Since 11(1 + 1) <2t2/(1 + 12) for t E (0, 1), it follows that the series

/(1+x2 ) satisfies the Leibnitz test when x2 E (0, 1),and its sum is negative. Direct computations show that S(x) <0 for 0.93 <x <0.945.

c) By what was proved in b), there are points s and t in 1, 1) suchthat s <I and F(s) > F(t). Let s,, S4, = (n = 0, 1, 2,...).Using the identity

F(x) =1+x

and the fact that the function (1 +x)/(1 +x2) is decreasing on 1, 1),we get the inequality

1+s 1+1F(Sn) — = —

1+5?, 1+1,?

� — F(tni)) � — F(tni)•

Thus, F(s) — f(t) >0, although —, 1.

1.25. Consider the set Q = •f'(K) C [0, 1], where f = isa Peano curve (see problem 4.5), and K = {(x, y)I0 � x � 1, y =0, 1, f, .. . }. Verify that the restriction has the required property(use the result of problem 4.4c)). Extend the function in a suitable wayfrom Q to [0, 1].

1.26. For each n EN and x E [0, 1] define fn(x, y) by linear interpo-lation, with values f(x, k/n) at the nodes Yk = k/n (k = 0, 1, 2 n).

1.27. Arguing by contradiction, we see that If(a, � c for Li' — bI <e

forsomea,bER,c>0,ande>0.LetEk={yE(b—e,b+e)If(x,y)�c/2 for Ix—aI� 1/k} (kEN). Showthat anduse the result of problem 1.1.33. Prove that if the closure of Em contains aninterval on (a—

§3. Continuous and differentiable functions

3.1. Show that the mth-degree interpolation polynomial P for f con-structed for the nodes 0, h, 2h mh coincides with I at all points kh(k E Z). Conclude from this that P(t) = 1(t) for t = (k E Z, n EN).

3.2. Show that f(x) —. 0 as x —. +00, assuming that thelimit 1(x) exists. Prove the existence of the last limit by arguing by

172 III. FUNCTIONS

contradiction: if = 1(x) < P = 1(x), then there is asequence +00 such that =0, — a, and —, /3.

Another proof of the equality 1(x) = 1, extremely short but notas instructive, can be obtianed by applying L'Hospital's rule to the pair offunctions exf(x) and eX.

b) To prove the converse, show that if f(x) does not tend to 0 as x —+00, then (since f' is uniformly continuous) the function does not satisfythe Bolzano-Cauchy criterion at infinity.

3.3. Assuming that f'(a) <0 < f(b), prove that the smallest value off is attained in the interior of [a, b].

3.4. Use the results of problems 3.3 and 1.10.3.5. In checking the necessity use the uniform continuity of 1' on [a, b].3.6. a) Assume first that g(x) > 0 for x E (a, b), and suppose that there

are points x1, x2 E (a, b) such that x1 <x2 and f(x1) > 1(x2). It canbe assumed without loss of generality that f(x1) = 1 and 1(x2) = —1 Let

= sup{x E (x1 , x2)lf(y) � 0 for y E (x1 , x)}. There exists a sequencesuch that > 0, — 0, and + <0. Therefore,

g(x) = —

f is increasing. In the general caseconsider the function f(x) = 1(x) + ex, where e is an arbitrary positivenumber. If g 0, then the result already obtained should be applied to thefunctions f and —1.

b) Apply part a) to the function f— G, where G is a primitive of g.3.7. If I is differentiable, then g = 1" and hence f E C°°((a, b)).

Differentiating the identity f(x+ h) — f(x— h) = 2hg(x) twice with respectto h , show by using 3.6a) that 1" const. Using 3.6b), we see that this resultis preserved if f and g are continuous on (a, b). If only the continuity ofI is assumed, then the result becomes false (1(x) = lxi. g(x) = sgn .v).

3.8. Consider a primitive of the function defined by =inf{Ix—yIIyEF}(xER).

3.9. To prove the assertion b) a), study the set E [0, 1]l1(x) — 1(0)1 � ex}, where e is an arbitrary positive number. Prove that

n (0, c)) = c for any c E (0, 1). The statement c) does not implya). A corresponding counterexample can be obtained by considering one ofthe coordinate functions of the Peano curve constructed in problem 4.4 andusing the fact that a set of constancy of this function does not have isolatedpoints (see assertion d) of this problem).

3.12. Estimate the increment of the function I on a region of mono-tonicity. For = 1 it is useful to employ the result in the preceding problem.For wehave g'(0)=2f,E.

3.13. a) See the hint for the preceding problem.b) Estimate f'(x).

§4. CONTINUOUS MAPPINGS 173

c) Show that If(x + h) — f(x)I = O(If(x0 + h) — f(x0)I), where x0 is anendpoint of the smallest region of monotonicity of I whose length is notless than h.

3.14. Consider the series with terms of the form — ak), where fis the function in the preceding problem with suitably chosen parameters.

3.15. Consider the functions f(x) = (ln(2fx))' and g(x) =for 0 < x � 1, 1(0) = g(0) = 0.

3.16. Show that

19,(x + h) — � J w(z + h) — z1h121 dz,

where w is a 2ir-periodic function coinciding with on the interval[—it, it], and estimate the integral on the right-hand side of the inequality.

3.17. To prove the continuity of f use its monotonicity.3.18. Fixing an n E N, get a lower estimate of the length of the polygonal

curve with vertices at the points = (j = 0, 13fl)

by considering the sums and

where K is the Cantor set and p is the metric in R2.3.19. Compute the increment of the Cantor function on [0,

§4. Continuous mappings

4.1. Prove that the boundedness of the sequence is equivalent tothe boundedness of the sequence This is not true for arbitrarypolynomials in two variables (for example, for Q(x, y) = x). Verify thatQ(F) = R\{0}, where F = {(x, y)Ixy � 1} is a closed set.

4.3. To construct one of the possible examples, consider two-sided nu-merical sequences {xk}kEz and {Yk}kEz satisfying the conditions

...<xk<yk<xk+l<...,infxk_O,supxk_1,

and form the sets A = {0} U UkEz[xk, Yk) and B ={1} U Uk€z[Yk, Xk+l).The desired homeomorphism can be obtained by mapping each interval[Xk, Yk) C A onto the interval [yk' xk+l) C B and 0 into 1.

4.4. a) The unambiguity of the definition can be verified by a direct com-putation of and ip at ternary-rational points. The continuity of andip follows from the coincidence of any previously specified number of firstdigits obtained in the representation of sufficiently close points t, ? E [0, 1]as ternary fractions (without the digit 2 in a period).

b) Using the expansion of numbers in [0, 1] as ternary fractions

= E where = 0, 1, or 2, show that if ,j E [0, 1],= , and = 0.y1y2y3..• , then = and = where

= , and the digits are determined successively as

174 III. FUNCTIONS

follows:

=(v1

if is odd;

if + + + is odd;

— J+ + +

2n— 1 is odd.

c) Different solutions of the system = = ,j can be obtainedonly due to nonuniqueness of the representation of the numbers ij E [0, 1]

as ternary fractions. Verify that for = 2/9 = 0.02000... = 0.01222... and= 1/3 = 0.1000... = 0.0222... the system has four solutions: 5/108 =

0.001020202..., 11/108 = 0.00220202..., 13/108 = 0.010020202...,and 19/108 = 0.011202020....

d) Let t = E ço'({c}). If infinitely many of the digits areeven, then, replacing a2k by 2— for sufficiently large k, we get a pointin arbitrarily close to t. If a2k = 1 for k � k0, then consider thepoint (k � k0) . • — obtained from

by replacing the digits = 1 and = 1 by 0's, and by thenumber 2 — The arguments are similar for the consideration of theset

e) Prove that E Lip112([0, 1]). For 2 < in EN let p = [in/2], and take

t,? E[0, 1], t = = ,such that 3m <3—m+1 If = = , then

= =

and hence 19,(t) — 9,(t')I � � — Assume now that there isan index k � in — 1 such that Let k is the smallest indexwith this property. Then in view of the inequalities 3m — <the expansions for t and ? must have the form t = .. .

and ? = — 1)22... ... (if at leastone of the digits i,.... is not 0 or at least one of the digits

is less than 2, then t —? � 3.m+l) The following fourcases are possible:

1) k = 21 is even, and + + • +k is odd;

3) k = 2/ — 1 is odd, and + +... +k is odd, and + + + is odd.

In the first case we see that = 0.161/32... /3,0... and =

§5. FUNCTIONAL EQUATIONS 175

0.131132... /3,... Consequently, 19,(t) — � 3.3k— . The remaining cases are handled similarly.

4.5. Let e > 0 and N = 1 + [1/a]. The points tk = k/N (k = 1 N)form an e_net* for [0, 1]. Verify that the points (u(tk), v(tk)) form a 5-net for the square [0, 1] x [0, 1], where S � Cea. Prove that the numberof points in a 5-net for the square is not less than C0ö2.

4.6. Let E be the set of all binary sequences, and let E N,

E {0, l}} and E N, es,..., E {0, l}} be the fami-

lies used to define the generalized Cantor sets Kand K (see problem 1.1.31).Define the mappings —' K and —, K by the equalities =and 0(e) = i, where t E and I E (a = E E).

Prove that and are bijections, and I = o is a homeomorphismof K onto K.

§5. Functional equations

5.2. c) It is evident that f is not linear, and hence f(x0) x0f(1)for some x0 E ILL This inequality is equivalent to the linear independenceof the vectors e' = (1 , f(1)) and e" = (x0, 1(x0)). Therefore, the set{se' + te"Is, t E R} coincides with R2 and the set {se' + te"Is, t E is

dense in 1R2 It remains to see that the points Se' + te" belong to the graphof I for any rational s and t (this follows from a)).

5.3. Apply the result of the preceding problem to the function g(x) =f(eX).

5.4. Show that f(1) = f(—1) = 0, and prove that I is odd. Considerthe function g(x) = +1(x) (x > 0) and use the arguments in the solutionof problem 5.3.

5.5. Use the result of problem 5.2.5.6. Prove successively that 1(0) = 0, 1(x'1) = (1(x))'1, and f(x+y) =

1(x) + f(y), for x, yE IR. Using the equality f((rx +y)'1) = (rf(x)+y)'1,where r E show that f(x2) = ±(f(x))2 for any x E IR; use the result ofproblem 5.2.

5.7. Prove by induction that

E f(xk)—f i(P—i) (pEN). (*)

Derive from this that the ratio (1/n)f(±n) (n E N) is bounded. Fix strictlyincreasing sequences (nk) C N and {mk} c N such that the limits

f(n) f(-m)and B=lim k

—mk

The definition of an e-net is given in VIII.5

176 III. FUNCTIONS

exist. We prove that

= A and = B.

For any e > 0 fix an index such that O'flk <e and — Al <e.Let x > 0 and p = [x/nk]. Then x = +S, 0 � s < It follows from(*) that + f(s) — f(x)l � ap, and hence

—?——f(nPflk+S k x x

x the last inequality implies that lf(x)/x — Al <It is proved similarly that f(x)/x —, B as x —, —00. Passing to the limit asX —, +oo in the inequality

we see that A = B. Consider now the function f(x) = 1(x) — Ax and ver-ify that lf(x)l � a for any x E ILL Indeed, it is clear that f(x)/x —,0 as x —, 00, and f, like f, satisfies the inequality (*). Therefore,11(x) — f(nx)/nl � a for any x E R and n E N. Assuming that x 0and passing to the limit in the last inequality, we get what is required.

We remark that the estimate for f cannot be improved, as we can seefrom the example of the function 1(x) = max(1 — xl, 0).

5.8. Show that 1(0) = 0 and that f is even. Prove that if f 0,then 1(x) 0 for x 0. To do this verify that if [(a) 0 (a >0),then 1(x) 0 on (0, a). Indeed, suppose that this is not so, and let c bethe largest zero of I on (0, a). Then = 0 for any n E N, andhence f(cV'l + 22n) = 1(c) + = 0. Since cV'l + 2_2n E (c, a)for sufficiently large n, we arrive at a contradiction to the choice of thenumber c. Note further that a can be chosen arbitrarily large, because

= 4hf(a) 0. Accordingly, 1(x) 0 for x 0. Assume thatf(1) = 1 and consider the set A = {x > 011(x) = x}. Verify that itsclosure coincides with [0, +oo) (see problem 1.1.21).

5.9. Prove that 1(0) = 1 and that f is even. Show that if f(x0) = 0,then 1(t) = 0 for ti � 1x01. Derive from this that 1(x) > 0 for any x E R.Verify that the function g(x) = lnf(x) (x E R) satisfies the condition ofthe preceding problem.

5.10. Prove that the function g(x) = f(eX) is the sum of a linear func-tion and a periodic function.

5.11. a) Consider the functions = (x E (0, 1)) andH(u) = (u ER).

§5. FUNCTIONAL EQUATIONS 177

b) Verify that 0 � I � 1 and that if 1(x0) = 0 (1(x0) = 1) at somepoint x0 E (0, 1), then f 0 (f 1). Assuming that 0 < f < 1 in(0, 1), prove that = 0 and = 1 and that f isstrictly increasing. Consider the function g(t) = in llnf(e')l and use theresult of problem 5.10.

5.12. Assuming that the desired function h is odd, show that the function= ln(h(e')) (t E R) is strictly increasing and satisfies the equation= Find a solution of this equation.

5.13. Verify that 1(0) = 0 and 1(1) = 1. For z 0 consider thefunction g(z) = f(z)/f(IzI) and by using the fact that g is one-to-oneon each circle IzI = r prove successively that (for r E (0, 1]) g(r) = 1,

g(—r) = —1, g(ir) = ±i, = e±h7t14, etc. Deduce from this thatg(z) coincides either with or with Prove that the functionh(t) = If(t)I is increasing on (0, 1). Represent f(t) in the formand use the result of problem 5.11.

5.14. Prove successively that 1(0) = 0, 1(1) = 1 (and hence f is in-creasing), 1(1/2) = 1/2,and 1(x) = 1/2 for x E [1/3, 2/3]. Use inductionto see that f coincides with the Cantor function (see problem 3.17) on theopen intervals complementary to the Cantor set.

5.15. Prove successively that 02f/OxOy const, f(x, y) = Axy +g(x) + g(y), and g" const.

5.16. Verify that if f(a) = 1(b) for al Ibi and the function g exists,then 1(t) = f(ct) for some c with cl < 1 and for any t E ILL Deduce fromthis that f const.

5.17. a) Verify that for each E S' the mapping Z —' S' defined

by= is a character. Prove that each character has the form

where

5.18. a) Verify that for each t E R the mapping R —' S' defined

by = eutX (x E R) is a character. If is an arbitrary continuous

character, then Re > 0 for n � N. Let 9,(2M) = e'0°, 1001 <n

o = Verify that = e'02 for any n E N, and prove that=b)—d). Use a) and the analogy with problem 5.17b).e) Verify that for each n E Z the mapping S1 —, S' defined by

= r E S') is a character. To prove that each character has this

form, study the set Prove that it is finite if 1, and that forsome m EN the equality = 1 holds for any E Prove that ifn E N is the smallest such m, then = or =

f) Use the fact that the groups R are isomorphic.

g) Use the fact that the groups and x S1 are isomorphic.5.19. a) See problem 5.2.b) Consider

178 III. FUNCTIONS

c) Consider co(eX).d) Consider ine) See problem 5.18a).f) Use the fact that the groups and x S1 are isomorphic. Prove that

if S1 —, is a continuous homeomorphism, then 1. Use part d)and the result of problem 5.18g).

CHAPTER IV

Series

§1. Convergence

1.1. Find the number of indices n E [10k', 10N+1) whose decimal ex-pansions do not contain the digit 9, and get an upper estimate for the sum

1.2. a), b) Consider the numerators of adjacent terms and show that theycannot be simultaneously close to zero.

c) It suffices to prove that the series

1)21+Isinn2l+Isin(n+ 1)21)

diverges. To do this, show that the numerator is bounded below by a pos-itive number. Indeed, if not, then for arbitrarily large indices n we havethat n2 = irk + e, (n — 1)2 = irk' + e', and (n + 1)2 = irk" + e", wherek, k', k" E Z and e'I. e"I < Therefore, 2 = (n—1)2—2n2+(n+1)2 =ir(k' + k" — 2k) + e' + e" — 2€, which is impossible.

1.3. For a � 1 consider the sums am = cos(b Inn), where

Nm ={n ENI2lrm�blnn � 2,rm+ir/4}, andshowthat am 740.1.4. a) Consider separately the cases p> 1 and 0 <p < 1.b) To study the sums use the result of problem

VI.3.l3for 1/2<p�1.

_____

c) Use the relation arccosx —x) for x —, 1—0.1.5. a) Obviously,b) Use the result of problem I.2.5b).c) Use the equality = — 1.

d) Prove the following more general assertion: if > 0, and1/ar = +00, then — = +00, where = (A0 = 0).

Let

0m2 A—A

Since the harmonic mean does not exceed the arithmetic mean, it followsthat

— = —

m 2<n�241

I 79

180 IV. SERIES

Consequently,

1 22m1

—=

Therefore,

1 1 1 2m+i

EAA A-An n—i m�0 n n—i

(the last series diverges according to the Cauchy theorem; see problem 2.2a)).1.6. Let = minmEN — rn/ni. Since < 1/2n, it must be shown

that <+00. Let Nm = {n E NICm ifl2 <Cmfl2},where c0 = 1/4 and Cm T +00 (the choice of the sequence {Cm} will be made

more precise below). Since > c0n2 (see problem L3.lla)), it followsthat N = U Nm. Let Nm = {n1, n2, . . . }. Since � <1/2n1, itfollows that n1 > 2Cm Using the result of problem L3.llb), we get that

1

n

2 1

+ �m

Taking Cm = 2m—2 we get that1 m(l—a)

= >2Sm � <+00.

1.7. Instead of double series it is convenient to consider sums of the formIn examples e) and g) consider such a sum, extended

ihiprime values of rn, and use the result of problem 1.5.b).f) Use the inequality

GCD(rn,n) I

GCD(m,n)=I n + m ) 3)

h) Use the equality LCM(n, rn)GCD(n, rn) = nrn and the method usedin the solution of problem f).

1.8. Each fraction 1/2fl (n = 0, 1, ...) appears in the series timeswith the sign + and times with the sign —. The desired rearrangementis obtained by multiplying the series by —1.

§2. Properties of numerical series connected with monotonicity

2.1. If 74 1 , then the series diverges. Suppose that —, 1.Since arccosx — x) as x —, 1 —0, the given series converges simul-

§2. PROPERTIES CONNECTED WITH MONOTONICITY 181

taneously With the series — and the latter converges simulta-neously with the series = ln(al/aN).

2.2. a) Use the inequalities � �b), c) Use a).2.3. To prove the relation = 0(1/n) use the result of problem 2.2a).2.4. To prove the relation = 0(1/Inn) use the results of problems

2.2a) and 2.3.2.5. Using the result of problem 2.3, show that = The

assertion is false for nonmonotone sequences; a counterexample is given bythe sequence 1 for n=2k, for

2.6. a) Let am = (a0 = 0). By changing x1 it can beassumed that —, 0. > 0 and let lamI <e for m > M (M E N).Then setting a = maxmEN fr'mI' we get that

= am

1

n

n+1 n

am —, 0, it follows that iEamI � 2€. The required asser-tion follows from this, since e > 0 is

For nonmonotone sequences the assertion is false; a correspondingcounterexample is easy to construct by taldng = 1 for n E N.

b) Take a subsequence {nk} of indices such that 1/2, andlet = for n = nk and = 0 for n nk. It is clear that

= 1/k = +oo. Let nk � m < nk+I. Then

am � =1<n<m

To prove b) for a nonmonotone sequence take a subsequence

such that a,, = and repeat the argument above for it.

182 IV. SERIES

2.7. Apply the Abel transformation to the series >k>fl akxk =>.k>n ak(rk — rk+l), where rk = >.j�k and use the fact that rk 0.

2.8. a) b). Since

n>1 k�1

it suffices to prove that the series = — and =

converge or diverge simultaneously. Applying the Abelto the series we get that this assertion is equivalent to

the relation = If the series converges, then this relationfollows from the result in problem 2.6a). And if the series converges,

then En/2<m<n —, 0, and hence _.' 0.The assertion a) c) can be proved similarly.2.9. Use the Abel transformation. The series diverges, for

example, for = and 0 < e � 1/2. The series divergeSif l/ln(l +n).

2.10. It is not hard to construct positive but nonmonotone sequencesand such that = E = +00 and <+00.

The required monotone series are obtained by considering the series ++ + +... and fl1 + /3k/ak + /3k/ak (the

terms of the form and /3k/ak are repeated times) for a suitablechoice of the sequence

2.11. a) Since = — (S0 = 0), the series divergesif 74 1. But if —, 1, then divergence follows from thedivergence of the series 1/Sr) = limln(Sl/SN) =

b) Compare the given series with the series — Sr).2.12. c) Use the inequality � f:+d) Consider the counterexample =2.13. a) Use the inequality

n�L

Estimate the inside sum with the help of the Hãlder inequality.b) Apply the assertion a) to the remainder of the series

a > 0. Let = — a. It that theoriginal series converges simultaneously with the series

—

a 1, then this series converges, but if a = 1 then it diverges, since theseries — diverges (see problem 2.12a)). Suppose now that

§2. PROPERTIES CONNECTED WITH MONOTONICITY 183

j 0. It can be assumed that < 1 for all n E N. It follows from thedivergence of the series — 1) that the series

________________

L_d >.1<k<flak_1/ak

diverges too (cf. problem 2.1 la)). Using the inequality between the arith-metic mean and the geometric mean, we get that the series

4—jn

diverges.

If —, +00, then the series under consideration can converge (for exam-ple, = n'1) or it can diverge (for example, =

2.15. Note that it follows from convergence of each of the series thatx/f(x) —, 0 as x —, +oo (for a proof, get a lower estimate of the sum>.n<m<2n) The series

k�1

converges simultaneously with the series

k" 1 k" 1 1

kl n(n + 1)) — +[f(k)] — 1 +[f(k + 1)]

Using the Abel transformation and the relation x/f(x) —' 0 as x +00, weget that this series converges simultaneously with the series 1/(1 +[f(k)])that is, simultaneously with the series 1/1(k).

2.16. Let = a1+• =0. If <+00, then 1000 f(t)dt <+00. Consequently,

<f(t)dt

= j f(t)dt <+00.

But if >f(n) = +00, then 10400 f(t)dt = oo. Breaking up this integral intothe sum of the integrals over the intervals , S,], we get that the series

diverges. Therefore, the series also diverges—for aproof it suffices to observe that the series — converges(this follows from the boundedness of the sequence

2.17. a) b). If > for some index n0, then =

</3 = It is easy to see that the sums of all the partialseries do not fall in the interval (cr, /3).

b) a). Fixing a number s E (0, A], use induction to construct asequence ii = {n1, n2, ... } of indices such that

= min{n E Nm > + � s}.

Verify that A(v)=s.

184 IV. SERIES

2.18. Consider the series 2/ 3fl•

2.19. Take the as small as possible, that is, n1 = min{n E N I <1},and = min{n E N I > k � 2. Assume that

<+00. Then = o(1/k). Consequently, there exists an indexN such that <1/(k+1) for k � N. Then in view of the definition of thesequence we have that k � N. Therefore, thesufficiently remote remainders of the series and coincide,

which is incompatible with the condition of the problem.The assertion is false for nonmonotone sequences. The following example

was suggested by A. A. Shul'man.We consider the sequence {xk}, where Xk = l/(n2(2h1 — k)) if �

k It is clear that Xk —+ 0. Moreover, the series Xk diverges:

n�1 n�1

We now consider a sequence {k1} of indices such that Xk <1/f for j E N.

For arbitrary n E N let = {j E � <2h1}. Let m =M = and ji = — kM = — ks). Then

J

2 / 1\ 2 / m

1 1 1 1=m M M n2/1

therefore, n2 > m/ji, and hence xk � 4 ln(1 + n2). Thus,

>XkJ�L

2.20. a) If � 0, then it is possible to use the result of problem 2.12b).In the general case we take two sequences €k j 0 and 11k T +00 such thatEekuk <+oo. From the sequence {ek} we construct a sequence of indicesNk T +00 such that <€k for n > m > Nk. Let = /1k if

§3. VARIOUS ASSERTIONS ABOUT SERIES 185

Nk <i � Nk+l. Then for Nk <m � Nk÷l we have:

E � + +m<j<n

� /2k€k + /Lk+1€k÷1 +" —+ 0.

b) Consider the counterexample = Inn, =But if T +00, then the series diverges, since otherwise

the series ç would converge by the Dirichlet test.2.21. a) Using the inequality between the arithmetic mean and the geo-

metric mean, we get that

v' ! 1 + 1 1 +

n i M L1N�n<M

NM

1

t

M E N.b) Get a lower estimate of the partial sums of the series

with the help of the result of problem L2.8c).

§3. Various assertions about series

3.1. For example, = (—1)'1/n. But if >0, then

(an+ � = +00.

3.2. For example, + n). Moreover, there exists suchconvergent series ç that for any m = 2, 3, ... the series is

divergent. For example, a3k = 2/Ink, a3kl = a3k+l = —1/Ink (k =2,3,...).

3.3. a) Fixing a parameter 2 > 1 (its choice will be made more pre-cise below), break up into two sums: in the first the summa-tion is over those indices n such that � and in the second overthe remaining indices. It is easy to see that the first sum does not exceed

and the second does not exceed Minimizing the quantity

1) over all 2> 1, we get that � (1 +

186 IV. SERIES

b) Get an upper estimate of the sum of those terms for which � n3.3.4. Let = Then and hence

By assumption, <+00, which by the result of problem3.3b) gives us that

b 1In <+00,

and thus <+00.The converse assertion is false—consider a sequence such that =

1/2 for n = 2k and + n) < +00. But if j 0, then it followsfrom the convergence of the series + n) that + n) =0(1/n), which implies that =

3.5. It follows from the result of problem 2.6a) that +• • —, 0.It remains to use the proof of problem II.2.4b).

3.6. b) a). It can be assumed that � 0. Supposing that =+00, we construct a sequence with density 0 such that the seriesdiverges. Since at least one of the series and a2m+l diverges, thereexists a set N1 C N with density 1/2 such that

a sufficiently large P1 E N. Similarly, we construct a

set N2 C N1 with density 1/4 and an index P2 > P1 such that >nEN =+00 and ç � 1. By induction we construct a sequenceof nested sets and a increasing sequence of indices such that

the density of is equal to and n(P � 1. The desiredsequence {n1, n2, ... } is obtained by numbering the elements of the setA = n Ps]) in increasing order. The density 0(A) of A is

equal to zero, since 0(A) � 0(N1) = for any I E N.3.7. For in, n E N let = >k arnk, where the summation is only

over the indices k � 1 relatively prime to all numbers not exceeding n.Using the equality = if n + 1 is not a prime number, andthe equality = — + 1)rn) if n + 1 is a prime number,show by induction on n that S,1(rn) = 0 for any n. Derive from this thatlami � >k>mn laki for any in, n E N.

3.8. Assume that = +00. Let E N). Let

= (1/k) sgn if � Then —, 0, and it is obvious that= +00.

3.9. a) b) Use the Abel transformation to prove that the seriesconverges.

Assume that Let

IAkak+lI.l<k<n

§3. VARIOUS ASSERTIONS ABOUT SERIES 187

Then — = +00 (see problem 2.lla)). Consequently,

= +oo,

where =—

Using the Abel transformation, we get that

—= +00, although the series — converges.

3.10. It suffices to consider the case = 0. To estimate the sumço(en)a,, break it up into two sums: >1<fl<N and Using the

Abel transformation, show that the second suin is small for all e > 0 if Nis sufficiently large.

3.11. In the sequence {lfn} rearrange the terms with 2k1 2k inincreasing order.

3.12. In the sequence {1/n} rearrange the terms with (k — 1)! <n � k!in increasing order.

3.13. a) c). The inequality A, � = is equivalent tothe inequality � (1— It can be proved similarly that b) d).a) and c) b):

� A, — � C>2 � c >=

k>n k k�n k

b) and d) a):

>1<k<n k

q2

Thus, the first four assertions are equivalent. Let us now prove that a) g):

since a,/a,+L � CAfl/Afl+L � it follows that afl/a,+L � 1/Q for asufficiently large index L. The assertion g) a) is obvious if a, 1. Theexample of the sequence 2, 3, 22, 32 33, shows that it is not truefor nonmonotone sequences.

It remains to prove that a) e) and b) f). We prove the first asser-tion (the second has an analogous proof). It suffices to prove that a) e)

(the converse assertion is obtained by applying this assertion to the sequence= Since a g), it follows that

+•• + = +... + = O((a,L+L +... +

= = O(a).

3.14. The solution is analogous to that of the preceding problem.3.15. a) Let = — and prove that

cr,, � — (n —

(n — (take A0 = 0), it follows that

/ np \ ,, p(n—l) "—i 2= — 1) A,,— p — 1

. )

188 IV. SERIES

Finding the minimum of the function

ço(x)=

for x�O,weseethat

� (i—

+ !AP1.

Together with (2) this yields (1). Thus,

v' (i.L1

1<n<N

=1<n<N l<n<N

that is,

1<n<N 1<n�N

for all NEN.b) Using the Hälder inequality (problem VII.1.29), we get from the in-

equality a) that

� p (LAP) 1-1/P

which implies that � (..L1)P Hardy-Landau inequality.3.16. The divergence of the series

(a1 + the divergence of the harmonic series.To prove convergence of the series qa1 • . observe that

I + . + V''z

for p> 1 (see problem VII.L30). It follows from the Hardy-Landau in-equality (see problem 3.15b)) that

• �Letting p go to infinity, we get the Carleman inequality:

> cia1 • � e

3.17. The assertions a) b) and b) c) are obvious. We prove thatc) a). Assume the opposite: there exists a sequence —. 0 such that

—, +00. It can be assumed without loss of generality that 0 << l/2n2 and = > n. Consider the series obtained from

§3. VARIOUS ASSERTIONS ABOUT SERIES 189

ç by repeating each term times, where = This seriesconverges:

n>1 )However, the series consisting of the Values of f diverges:

( = =n>1 )

½ � = +00.

3.18. Prove first that f is odd in some neighborhood of zero. Assumenot. Then there exists a sequence —, 0 such that = 0.Take a sequence C N such that 74 0. Then the series obtainedfrom the series a1 — a1 + a2 — a2 +.•. by repeating the nth pair of terms

times is convergent. However, the series consisting of the values of Idiverges, since 74 0.

It is clear that f is continuous at zero (see problem 3.17). Prove thatf is continuous in some neighborhood of zero. Assume the opposite: thereexists a sequence of points of discontinuity such that —, 0. For anyn E N there exists a number > 0 such that sup1, a <5 — f(t)I >

for any 5 > 0. Take sequences C N and C R such that74 0, 1(ç) — > and — <+00. The series ob-

tained from the series a1 — t1 + a2 — t2 + ... by repeating times thepair — ç converges, and the series of corresponding values of f di-

verges, because +f and odd in some

neighborhood of zero.Prove now that f(x + h) + f(x — h) = 21(x) in some neighborhood

of zero. If this is not so, then there are sequences 0 and 0

such that + + — — = 0. Take a sequence ofintegers T +oo such that 74 0. The series obtained from the series

by repeating each group + + — — — of terms timesobviously converges. However, the series consisting of the correspondingvalues of f diverges, because

+ + — — = 0.

Thus, in some interval [—5, 5] the function f is odd and continuous,and satisfies the equality 1(a) + f(b) = 2f((a + b)/2) for al, IbI � 5.Consequently (see problem 111.5.2), f is linear on [—5. 5].

3.19. Obviously, if is the required sequence, then the numbers rm =

x— (r0 = x) satisfy the inequality 0 Therefore,

190 IV. SERIES

in Selecting the number (after the numbers q1 have already

been found) it is necessary to ensure the inequalities 0 < — 2--m--1q1 <2"' , that is, the inequalities

— 1 <2m+lrm. (*)

It is clear that the satisfaction of this inequality for all m E N suffices forthe relation i,, 0, that is, for the equality x = It is not hard toverify that if (*) holds for some index, then it holds also for the precedingindex, and hence for all the preceding indices.

The sequence will be constructed by induction. It can be assumedhere that Q has already been numbered in some way. Therefore, we canrefer to the first element in any nonempty subset of it.

As q1 take the first element of Q satisfying the inequality (*) for m = 0,that is, 2x —1 < < 2x. It follows from the conditions infQ = 0 andsup Q = 1 that such elements exist. Assume that q1,..., have alreadybeen constructed and that (*) holds for m = 0 n — 1. Let q bethe first element in the set Q\{q1,..., q E — 1,

then we let = q. The inequality (*) will be satisfied for m = n. Ifq — 1, then we consider two cases: a) q � 2n+1 b)

In case a) let M = Then � q <r2M+t In particular,M> n. Select numbers qM from Q\{q1 small enough

that rM = — + + q, we

get that rM > and 2M+trM_1 ,that is, thenumber satisfies the inequality (*) for m = M, and hence (*) holdsfor m=0 M.

In case b) consider the first index M> n such that — +• +2M) <(q + 1)2M1• Such an index exists, because as M —. +00 the left-hand side tends to — = — + <0 (see the inequality (*)with m = n — 1). Now select numbers from Q\{q1,...,close enough to 1 that rM =

Taking = q, we get that 2MHrM — 1 By the choice of M,

(1 + � — f... +

2M—1 + + 2MrM < 2M+lrM+I. Thus, (*)

holds for m=M, and hence also for m=0,..., M.

§4. Computation of sums of series

4.1. Use the equality — sin =4.2. Use the equality 3cos + cos = 4cos3

§4. COMPUTATION OF SUMS OF SERIES 191

4.3. Use the equality

((n+r—1\' (n+r\'r ) r )

4.4. Use the equality tdt = (2n)!!f(2n + 1)!!.4.5. Consider the partial sums Snm•4.6. Use the result of problem VII.1.22c).4.7. Change the order of summation.4.8. Show that

E1<n<2N

and replace the sums by the corresponding asymptotic expressions (see prob-lems 11.2.9 and II.2.13a)).

4.9. Show that

>I<k<N

where ak = and, transforming ak to the form ak =— >.2k_<j<2k 1/ j use the asymptotic behavior of the partial sums of the

series.4.10. Reduce the integration in each integral to the interval [2ir, +oo).

To justify changing the order of summation and integration, represent theintegral nx)/x dx in the form

r2nNsinnx / 1

J2n x nN

where N EN. Then use the results of problems 6.16 and 6.8a) and Stirling'sformula.

4.11. The solution is analogous to that of the preceding problem.4.12. Denote the right-hand side of the equality to be proved by 'N• It

suffices to prove that 'NI — 'N = 1/N2 and 'N —. 0. To prove the firstassertion use the known equality

2N ir (2N — 1)!!cos

(2N)!!

and integrate twice by parts. To prove the second assertion note that

'N = ir (2N—i)!!

(JC+ � +

Consequently, llm'N � Since e > 0 is arbitrary, 'N —. 0.

192 IV. SERIES

4.13. a) Using the equality (1 — = 1 + x + x2 + for x =prove that

� (1 (1 �1<n<m

b) The solution is analogous to that of problem a).4.14. Use the results of problems 4.12 and 4.13a).4.15. Representing the fraction t"/(l —t'1) as the sum of a geometric pro-

gression, prove the identity — = where = ak(kin means that n is divisible by k).

§5. Function series

5.2. If � 1, then, taking x = m + 0, m E Z, and 0 E [0, 1), weget that

I

� � e0 + e0 �n�I

5.3. Since

1 1

the partial sums of the series are equal to = 1—

+ xkyt.Obviously, —' 1 as n —, 00 if x � 1 ,and = If

o < x < 1, then S(x) = l—llk>I(l+X) . Thus, =1. The uniform convergence on [0, 1] follows from the inequalities

rn(rn- 1)

5.4. The convergence of the series for x> 0 is obvious. The convergenceis not uniform, because the general term of the series does not tenduniformly to zero: = 1. To estimate the sum of the series webreak it up into two sums: + where N E N is such thatN! � 1/x < (N+ 1)! (if x> 1 the first sum is equal to zero, and thesecond does not exceed 1/n! = e — 1). Then

n!�2xN!�2;l<n<N l�n<N

=1)! (N: 2)k <7flj �

Consequently, 1/n!x) � 7/2.

§5 FUNCTION SERIES 193

5.5. Prove that lim,,1 0 B(t) = +00 and that as 1 —, 1 — 0 the limitssuperior (inferior) of the ratios A(t)/B(t) and >n>N a//B(t) are the samefor any NE N.

5.6. By changing a0 it is possible to assume that = a0 +• • + 0.Then from the equality = (1 — t) we get that

EaI � (1 —t) E —

n>O O<n<N n>N

�(1—t)O<n<N

By selecting N to be large we make the second term small, and then byselecting t to be close to 1 we make the first term small.

Another solution can be obtained with the help of the equality

= (1 — t)ES/ =n�0

and the result in the preceding problem.5.7. Convergence of the series follows from the relation =— = 0(n). Since

= (1 = (1

n�0 n�0

the use of the result in problem 5.5 gives what is required.5.8. Use the equality

A(t)/B(t) = E(ao +... + an)tn/ +... +n�0

and the result in problem 5.5.5.9. Convergence of the product is equivalent to conver-

gence of the series + Since Iln(l + a/)I = this series(and with it also the product) converges for all t E (—1, 1). The relationA(t) —, A as t —' 1—0 is equivalent to the relation —,

0 as t — 1 — 0. By Taylor's formula, it reduces to the equality

— 1) + — t'7)) = 0,

which follows from the Abel theorem (see problem 5.6).The example = shows that the condition <+00

is essential. Indeed, in this case

In = — 1) + — 12n) + 0

194 IV. SERIES

and henceA(t) I \

= exp > + o(1) —, 2

as5.10. Show that instead of the given series it is possible to consider the

alternating series Sm(x), where

Sm(x) = >

=

To investigate the sum use the result of problemVI.3.14.

5.11. For x> 4 let N = [log4x]. Then

If(x)I � + < 1+ In N+ = 0(lnlnx).1<n<N n>N

On the otherhand, taking Xm = (mEN), we getthat

1 . /1 1 \= — sin it

+ ++ 0(1)

1<n<m

� + 0(1) �1<n<,n

The equality 1(x) = 0 follows from the easily verified relation

f E Lip1, then f(ir/2N) = 0(1/N). and hence

From this, >.1<n<N = 0(1).5.13. Use ihefact that the coefficients of the polynomial satisfy a

linear system of order m + 1, and prove that they can be expressed in termsof the values of at the fixed points tk E [cr, /3] (k = 0, 1 m).Derive from this that the coefficients of converge.

5.14. Prove that the series converges uniformly on any interval[cr, /3] C (a, b) (Dini's theorem).

§6. Trigonometric series

6.1. Construct a sequence of indices and a sequence of nested in-tervals on which + � 1 — 1/k.

§6. TRIGONOMETRIC SERIES

6.2. Represent C05 nx + Sin flX in the form cos(nx + and,assuming that 74 0, use the result in the preceding problem.

6.3. Compare with problem II.1.lOa).6.4. Integrate the partial sums of the series over the interval [cr, /3] and

show that for some y> 0

j cos nx + sin nxl dx �6.6. Use the boundedness of the function (1/(sinx) — 1/x) on the inter-

val (0, ir/2] and the result of problem VI.1.lOa) and c).6.7. Compare the sums

>: -1/2n exp(znir(x —

k2<n<(k+1)2

and

ak(x) = exp(iirn(x—lnk)).k2<n<(k+1)2

Show that 0k (x) 74 0 if the sequence {k1} is such that ln <x + 2m1 <ln(1+k1) forall jEN.

6.8. Find the sum of the series for E C, Izi < 1, and usethe Abel theorem (see problem 5.6).

6.9. The problem can be reduced to the preceding problem with the helpof the equality

1 1( 1 1

4n2 — 1 — 2 2n 1

6.10—6.12. Using the Abel transformation, employ the result of problem6.5.

6.13. Use the result of the preceding problem and the equalityin .

. 10I sinnxsinnmxdx =Jo form=n.

6.14. To prove the relation = 0(1/n) consider the sum>.n/2<k<n sin kx for x = ir/2n. If = o(1/n), then for a uniformestimate of the remainder break this sum up into two sums, the first con-taining the terms with "small" indices (not exceeding ir/2x). In estimatingit, use the inequality sin � ti. To estimate the second sum use the Abeltransformation and the result in problem 6.5.

6.15. The solution is analogous to that of the preceding problem.6.16. It can be assumed that [a, b] C [0, in (otherwise, break up [a, b]

into several intervals). If a > 0, then it is possible to use the result ofproblem 6.10. If a = 0, then, using the results of problems 6.10 and 6.5,estimate the integrals of the remainder of the series over [0, e] and [e, b]forsmall e>0.

196 IV. SERIES

6.17. With the help of the Abel transformation and the results in prob-lems 6.6 and 2.4, show that there is a C> 0 such that

J0 1<n<N

for all N E N. Using the result of problem 6.10, show that the integralsi: If(x)Idx are bounded uniformly with respect to e E (0, it).

6.18. The proof of the absolute integrability of g is analogous to thesolution of the preceding problem. If dx < +00, then integrate theseries (see problem 6.16) sin nx termwise over [0, it] and use theresult of problem 6.13.

6.19. Repeating the solution of problem 6.17, prove that for M � N

J ço(x) cosnx dx � 'ON'0

where 'ON —, 0. Since the series >k>N cos nx converges uniformly on

[e, irJ (see problem 6.10), this implies that J' cosnxl dx �'ON' and hence cosnxldx � Pr,.

6.20. a) Arguing by confradiction, consider a point x0 E (0, it) at whichthe sum = kx)/k has a nonpositive minimum. Using thenecessary condition for an extremum = 0), show that sin nx0 � 0,and hence the sum also takes nonpositive values. Continuation of thisargument leads to a contradiction of the fact that S1 (x) = sin x > 0 on(0,it).

b) Prove that the local minima of the function cos kx on [0, it]form an increasing sequence.

6.21. a) Use the Abel transformation and the result of problem 6.20a).b) Let

1 sin(n+1/2)x 1 1

2sin(x/2)Then

1(x) = +n>1

= —

n>0

Since

+...= (sin(nx/2))2

the Abel transformation gives us that

1(x) = —(sin(nx/2))2 � o.

§6. TRIGONOMETRIC SERIES 197

Note that

Jltdx = +... 1(x))dx =

Therefore (see problems 6.10 and 5.14),

jf(x)dx = = —Ar) =

6.22. Prove that the local maxima of the function on [0, it] form adecreasing sequence.

6.23. Use the fact that

Ck = J f (x

(x + e dx.

6.24. In studying the quantity If(x)—f(y)I represent the difference x—y

(assuming that x > y) in the form x — y = €k4,0, 1, 2, 3 and 0. Break up the series —

1(x) — 1(Y) = sin sin

into two sums: and and estimate each of them separately.To prove that I for 1/2, get a lower estimate of the difference1(0) — 1(x) for x = 2ir/4m (m E N) or use the result of the precedingproblem.

6.25. The solution is analogous to that of the preceding problem.6.26. Use the same device as in the solution of problem 6.24. The func-

tion I is not in the class Lip1, since 1(x) > (m/2)x for x =(m EN).

6.27. a) Get a lower estimate of 1(x) for x = 1/rn! (rn E N).b) For an arbitrary point x E R consider the intervals /3j (n � n0),

containing it, where = pn/n!, = + sn)/n!, = [xn!], E N,and Sn � n/2. To get a lower estimate of the quantity If(fln) — use

the inequality

If(en) — � sin +1�k<n

Icos +

—

I<k<n--2

Show that 5n can be chosen so that I cos + � Prove that in this

case ICOn) — 1(0!n)I� � —

198 IV. SERIES

6.28. b) c). Use the fact that for N> the difference —

SN(x) has period it/2N0 and hence maxa<x<b ISN(x) — SN(x)I =maxXER ISN(x) — SN(x)I, if N0 is sufficiently large.

c) a). Write the sum S2N in the form

S2N(x)=0<n<2N

where = and E [—it, it] for n = 0 2N, and considerthe nonnegative trigonometric polynomial

RN(x) = fl (1 + +0<n<N

(the F. Riesz product). It is easy to see that RN has the form

RN(x) = 1 + + + cos(kx + Wk),0<n<N

where = = = = ... = 0. Using the equalities

J m, k EZ. ml

(orthogonality of the trigonometric system), we get that

J = +0<n<N —ir

=it0<n<N

Let S = supXER MEN SM(x). By assumption, S < +00. Therefore.

+ � +

j� =

Thus, + � The sum + can beestimated similarly, except that the product

fl0<n<N

is considered instead of the polynomial RN(x).In conclusion note that the proof of the assertion c) a) used only

the boundedness of the sums SN(x) from above. Therefore, the series

§6. TRIGONOMETRIC SERIES 199

+ converges if the sums SN(x) are uniformly bounded aboveon some interval [a,bJ, a<b.

6.29. Verify the proof by induction.6.30. The assertion is obvious for N = 0, 1. Assume that the inequali-

ties are true for N < 2m for some m E N, and show that they are true alsofor 2m � N < 2m+1. Note first of all that RN(Pfl) = and RN(Qfl) =for n � m, and in this case the validity of the inequalities to be provedfollows from the result of problem 6.29c). Verify that for n � m +2

RN(Pfl) = RN(Qfl) = RN(Pm+I).

Therefore, it suffices to estimate RN(Pm+l) and RN(Qm+l). It is clear that

RN(Pm+l)(Z) = Pm(Z) + z2m RM(Qm)(Z), where M = N — 2m. If M <2m_1 � N/2, then we get from the induction hypothesis that IRM(Qm)(Z)I �

IRN(1°m+I)(Z)I �� iov'N. But if M � 2m1, then RM(Qm)(Z) = Pmi(Z) +

Z2RL(Qm_ 1)(z), where L = M — 2m--1 Therefore,

� + IRM(Qm)(Z)I

� + I1°m—i(Z)I + IRL(Qml)(Z)I

� + + iou:�

The polynomial RN(Qm+I) can be estimated similarly.6.31. Consider a sequence {ek}k>o whose terms for k < coincide

with the coefficients of the in problem 6.29. To prove theiiiequality a) use the result of problem 6.30. To prove the inequality b) usethe relations

22E(m + 1)=

ISm(O)12 dO � ISm(0)I j ISm(0)I dO.

6.32. Consider the series€k ikO

where the sequence {ek} is defined in the hint for problem 6.31. To proveuniform convergence of the series use the Abel transformation.

6.33. a), b). Use the Abel transformation.c) Let f(O) = SN(O)+RN(O), where SN(O) is the Nth partial sum of the

series defining f. Then

1(0) — f(0')I � ISN(0) — SN(0')I + IRN(0)I + IRN(O')I.

Taking N = [1/10— O'I], estimate the remainders RN with the help of theassertion b). To estimate the difference SN(O) — SN(O') use the inequality

ISN(O) — SN(0')I � 10— O'I maxXER and the assertion a).6.34. Use the results of problems 6.31 and 6.33c) for = 1/2.

CHAPTER V

Integrals

§1. Improper integrals of functions of a single variable

1.1. Consider the integrals over the intervals [kit, (k + 1)ir].1.2. To estimate the integrals over the intervals [kit, (k + 1)ir] use the

inequalities x E (0, ir/2).1.3—1.7. The solutions of these problems are analogous to the solution of

problem 1.2.1.8. Use the result of problem VI.2.9.1.9. Integrating by parts,

1A

cos(x —x)dx=2

J1 J1 3x—1AlA X 3=o(1)+6 I sin(x —x)dx,

ii

reduce the original integral to an absolutely convergent integral.1.10—1.11. The solutions of these problems are analogous to the solution

of problem 1.9.1.12. a) Consider the function with f(x) = n for x E [n, n + n3]

(n EN), and 1(x) = 0 otherwise.

b) Consider the function 1(x) = x2 sin(x"), where p> 3.1.13. In investigating convergence of the integral at infinity with the help

of the result of problem 1.16 and integrating by parts, show that 1(x)(ln(x2 — as x —* +00.

1.14. a) Find the limit of the integral

[+00 f(ax) — f(bx)dx = I as +0.

x Jca

b) If f(t)dt = 0, then see problem a). Otherwise, consider the function

f(t)dt.c) Find the limit of the integral

[ asA—'+oo, e—'O.x Jea t fAa

202 V. INTEGRALS

1.15. a) It follows from the second mean value theorem that for anynumbers A and B with 0 < A <B there is a number C E [A, B] suchthat = e_4C Since the integral f°° f(x)dxconverges, all the integrals of the form 1(x) dx are small if A is suffi-ciently large. Therefore, the integrals f dx are small, and this isequivalent to the convergence of the integral f00

f is continuous on [0, +oo), then the assertion to beproved is easy to get by integrating by parts: f e_cXf(x) dx =

e eF(x) dx (here F(x) = f f(t)dt).b) We represent the difference between the integrals in the form

+00 +00I I -CX, f(x)dx—, e f(x)dx

JO JOA +oo +00

I —tx I I —CX=, (1— e )f(x)dx +, f(x)dx —, e f(x)dx.JO JA JA

For large A the last two integrals are small for all e > 0 (the estimate ofthe third integral follows from the second mean value theorem). The firstintegral can be made small by letting e go to zero for a fixed A.

1.16. With the help of integration by parts, compute the sum of the in-tegrals over the intervals [0, a] and [1/a, +oo) for a E (0, 1).

1.17. Since I = ln(sinx) dx = dx, it follows that

fsin2x\ 1 fsint\21

= j In k—) dx= —J

In dt

=J12 (5lnt)

dt = I — 2.

1.18. After expanding (1 —x)' in a series, integrate it termwise and usethe result of problem IV.4.12. To justify termwise integration of the sum ofthe series, show that

( n. xNinx

lim i i x In x I dx = lim I dx = 0.N—+ooJ0 / N—+ooJ0 1 —x

To prove the last equality use the estimate

j = +0 (jelN)= o(1).

1.19. In the preceding problem make the change of variable x1.20. Integrating twice by parts, we get

[+00 x2dx= [ x2d(tanhx 1) = 4

xe_2Xdx

Jo coshx Jo Jo

= 2j ln(1 +e2 X) dx =j

§1. IMPROPER INTEGRALS OF FUNCTIONS OF A SINGLE VARIABLE 203

It remains to uSe the result of problem 1.18.1.21. Differentiate the integral with respect to the parameter a.1.22. Use the results of problems 1.21 and 1.15b).1.23. By integrating by parts, reduce the problem to the preceding one.1.24. Taking the factor under the integral sign and computing the

integral obtained by means of differentiation with respect to the parameteruse the result of problem 1.15b).1.25. Integrating by parts, we get that

— 1)

1 i(a+b)x. . lax . ibx

= j —(e z(a + b) — we — ibe ) dx

=f (asinax + b sinbx —(a+ b) sin(a +

= (lal + IbI — a + bI)J dx.

It remains to use the result of problem 1.22.1.26, 1.27. Replace e_x by and use the inequality in problem

I.2.16a).1.28. Reduce the original integral to the difference

(11 — e_X 1+00 e_xI —dx—i —dx.

J0 x j1 x

with the help of the result in problem 1.24. Use integration by parts and theintegral in problem 1.26.

1.29. Reduce the problem to the preceding one.1.30. Reduce the given integral to the difference

11 f+OOe_XI —dx—i —dx,J0 X j1 x

and use the result of problem 1.26.1.31. Using integration by parts and a change of variable, reduce the

given integral to the Frullani integral (see problem 1.14).1.32. Consider the integral over the interval [is, +oo) (is> 0) and prove

that it is equal to ln(1 + e_X)dx. Using the equalityln(1+e_X) = ln(1_e_2x)_ln(1_e_x), reduce the last integral to the integral

fln(1 — and show that it is equal to the integral lnxdxup to a quantity infinitesimally small as —' 0.

1.34. For k = 1, 3, and 5 reduce the given integral to the Euler-Poissonintegral. For k = 4 use the result of problem 1.14.

204 V. INTEGRALS

1.35. Integrating by parts in the indefinite integral, we get that2 2 2 2

I e_X dx I 1 e_X I2 2

—i---1-dxJ (x + 1/2) J X 2x +1 x(2x +1) J x

= xe_X2+2 1 dx.

x2+1/2 J

Consequently,

f +00 1+00 —x2I 2 2=21 e

Jo (x +1/2) Jo

1.36. Denote the desired integral by 1(a). Differentiating with respectto a for a > 0 and making the change of variable x a/x, we get thatf'(a) = —21(a), and hence 1(a) = Ce_2a. It remains to use the Euler-Poisson integral: C = 1(0) =

1.37. a) By the result in problem 1.15b), sinxdx =

1(e), where 1(e) = f°° x)e_CX dx. To compute the inte-

gral 1(e) for > 0, use the equality = 4 dy (see problem1.33):

2 +00 +00 2

I(s)= sinx (j dy) dx

2 (+00 sinxdx\ 2 [+00 dy

= Jo Jo 1 + (e

Therefore,

[+00 sinxd — F

— 2 [+00 dy —x—irn

b) The solution is analogous to that of problem a).1.38. Computing the integrals over the intervals [pk (where p1 =

2, p2 = 3, ... are the prime numbers, numbered in increasing order), showthat

1 1 1 \ 1 1lim IniJ2 x —x 22/ \ 3 /

and use the result of problem IV.4.14a).

Computation of multiple integrals

2.1. Represent the given integral in the form

2j (In dx.

In the inside integral make the change of variable y = ix (t E (0, 1)) andchange the order of integration.

§2. COMPUTATION OF MULTIPLE INTEGRALS 205

2.2. Using the symmetry of the domains of integration and of the inte-grands with respect to arbitrary permutations of the variables, show that:

a) max(x1,..., = n!f x1 dx1

b) min(x1,..., = n!f dx1c)

dx1 d; = 1+00[+00

f+OOx1 x2

2.3. Use the Euler-Poisson integral (see problem 1.33).

2.4. a) Assuming that a2 + b2 > 0, let u = (ax + andv = cx + dy, where c and d are chosen so that (x, y) —. (u, v) is anorthogonal transformation. Then x2 + = U2 + v2, and hence

lax +

= IL2= e_V2 /2

dv jue_u2 /2 du

= 2/s.b) Pass to a new coordinate system with one axis directed along the vector

a.2.5. Let lxii = a. It can be assumed without loss of generality that

x = (0, 0, a). Passing to spherical coordinates, we get that

(1 /' dy 2 sinOdOi rdri dçoi

lix — Jo Jo Jo — 2arcosO + a2

= r\/r2 — 2arcosO +

2ir 1'=—J

We leave it to the reader to finish the computations.2.6. It is clear that

1111 eX22_U2_1)2 dx dy du dvJ JJ

= 11 (11 dxdy./

To compute the double integrals use poiar coordinates.2.7. Choosing an orthogonal basis in R" in which the matrix A has

diagonal form, we get

I = JJJJdt1 dt2 dt3 dt4

J(Ax,x)�l

206 V. INTEGRALS

with the help of the corresponding orthogonal transformation, where 2k (k =1, 2, 3, 4) are the eigenvalues of A. Making the change of variables Uk =

we see that

I (Ax,x), e dx

J(Ax,x)z1= 1

1111 du du du du.ffff.(L<414�l

1 2 3 4

It remains to see that = det A, and the integral can be computedwith the help of polar coordinates (see problem 2.6).

2.8. Let t = (ti, t2, 13, 14). It is obvious that

K(i)

=e dx2 dx3 dx4) dx1

(*)

Let s = (12, 13, 14) and y = (x2, x3, x4). Then the inside integral in (*)

can be written in the form = f(s). With the help of anorthogonal transformation of variables the problem can be reduced to thecase when the vector s has the form (a, 0, 0), a = Ilsil, and thus

f(s)= JJJ

dx2 dx3 dx4 =J1

— dx2.

Computing the last integral, we see that it is equal to

4irx1 4ircosh(ax1) — —i- sinh(ax1).

After substitution of this result in (*), K(t) takes the form

K(t)=

(x1 cosh(ax1) —!sinh(ax1)) dx1.

This integral is finite if and only if a < that is (since a = Ilsil =

'/4 + + t E Int A. We leave it to the reader to compute thelast integral.

2.9. Show that

1 S(E)i I Ix—YIdxdY and P=

(b—a) Ja Ja (b—a)2

where E={(x,y)Ix,yE[a,b], Ix—yI>(b—a)/3},and S(E) isthearea of E.

2.10. Suppose for definiteness that x � y � z. These numbers are thesides of some triangle if and only if z � x + y. Consequently, to find thedesired probability it is necessary to compute the volume of the pyramid{(x,y,z)10�x�y�z�a, z�x+y}.

§2. COMPUTATION OF MULTiPLE INTEGRALS 207

—

FIGURE 12

2.11. a) The realness of the roots is determined by the sign of the clis-criminant, which is equal to u2 — 4v. For a > 4 consider the square[—a, a] x [—a, a] and the parabola v = u2/4 in the (u, v)-plane (see Figure12). Let S be the set of points in the square that lie above the parabola. Theequation z2 + uz+ v = 0 has real roots if and only if (u, v) S. Therefore,P1(a) is always > 1/2. Since the area of S is equal to (4/3)a312, it followsthat P1(a)= 1 as a—.+oo.

b) A biquadratic equation has both real and complex roots if and only ifv <0. Therefore, the corresponding probability is equal to 1/2.

2.12. Show that

(JJ1r u2+v<1 4

2.13. Show that:a)

b)

L = Ie" = Ie'° — lidO;

where

2.14. It is easy to see that c1(x, y) = 1(a) and tP(x, y) = g(a), where

a = f(a) = — a)2 + v2)"2dudv, and g(a) =

208 V. INTEGRALS

ffu2+v2<i ln((u — a)2 + v2)dudv. Prove that I E C'([O, +oo)) and f'(O) =0, clearly implies that the function C1is smooth. Show that

f'(a) =11 (((u — a)2 + V2)P/2)'a dudv. (**)

u2+v2_<1

For a > 1 this equality is easily established by considering the relation+ h) — f(a)). To justify passing to the limit as h —, 0 under the

integral sign it suffices to use the mean value theorem and the inequality(u—a)2+v2 �(a—1)2>0.

Suppose now that a E [0, 1). Note that the function under the doubleintegral sign in (**) is absolutely integrable on the disk {(u, v) I u2+v2 � 1}:

(1 1! lu—aldudvJJu2+v2<i l( )al dudv =

JJu2+v2<l ((u — a)2 +ft luldudv

= lPljj 2 2 1+p/2(u+a)2+v2cZl (u + V )

If dudv(u2 +

=2lrlPlf 7<+00

We consider the set E = {(u, v) I + V2 � 1, ui � 1, lvi � 1} andrepresent f in the form

1(a)= £' — a)2 + dudv — Jj((u — a)2 + v2)"2 dudv

= f1(a) -f2(a).

For a E [0, 1) the inequality (u — a)2 + V2 � (a — 1)2 > 0 holds on theset E, therefore, = ffE(((u — a)2 + dudv. Since f1(a) =fl;_aafll(1L2 it follows that E C'([O,

_J((1 _a)2 ((1

=—j j(((u — a)2 + dudv

=— a)2 +

The equality (**) follows from the representations obtained for f(a) and

§2. COMPUTATION OF MULTIPLE INTEGRALS 209

Thus, for all a � 0 with a 1

f(a)= JJ ((u — a)2 + dudv

u2+v2�1

=— JJ (((u — a)2 + dudv

u2+v2�'

= —J((u — a)2 + dv

-1

2 .2 —p,2=— j

— a) + sin ç0) cosq dç0.

This implies that f" is continuous on [0, 1) U (1, +oo) and has a finitelimit as a —. 1. Since f is clearly continuous on [0, +oo), it follows thatf E C'([O, +oo)). It is clear that f(0) = 0.

It is proved similarly that g is smooth and

g'(a)= —j cosç0ln(1 +a2)dç0.

Therefore,

g'(a) = 4aJ sin22 = 4a

(jnI2+jn

0 1 — 2a cos ço + a 0 n/2

2 I.7tj2 sin2 dço=4a(1+a)i 22 2 2Jo (1+a ) —4a cos ço

= +a2— 1 = 2irmin(a, a').Since

g(O)=jj ln(u2 + v2)dudv = 21rJ rlnr2 dr = —it,

u2+v2czl 0

it follows that g(a) = ir(a2 — 1) for a E [0, 1], and g(a) = 2irlna fora>!.

2.15. Use induction.2.16. Let = + x1 + + dx. It is clear that

dx

(a + x1+ +

On the other hand,

r1ft dx•••dx \J(a)=I Il 2 fl )dxJo \J(o,ir

= j

V. INTEGRALS

Therefore, = + t)dt = By induction, 1(a) =(k = 1, 2 n — 1). In particular,

= = An_i(ln(a + 1)— ma) =

that is,I dxi = (ma). (1)

(n—i)!

Thus, the required equality is proved for rn = n. Let us prove it for 1 �rn <n by induction from rn + 1 to rn. The equality (1) is the base for theinduction. Assume that rn < n and

rnI —rn—i n—rn—ifin dx= lna).

Integrating this equality over the integral [a, /3] (0 < a < /3), we get that

(J'a + x1 + + xnrrn_i da) dx

(—1) 1' f n—rn—i= I I a lnadal. (2)rn!(n—rn—i)!

We remark that

in—rn_ n—rn

I = (fin_mIni3 lna)—n—rn (n—rn)2

and, moreover, — en_rn) = 0, because 0, if the degreeof the polynomial p is less than n. Therefore, it follows from (2) that

+ + xn)_rn —(/3 + + Xn)rn) dx

=lnfl—a (3)

Since0 as /3 —, +00 if k < n, (4)

we get the required result by passing to the limit in (3) as /3 —. +00. Therelation (4) follows from the equalities = + Onn), where 0 <On < 1, and

k (n) k!(n — k — 1)! n—k—i(x mx) = n—k(—1) forn>k.

x2.17. a) In the integral +(an—ao)xn) dx1 . .

make the change of variables y1 = i—x1, = x2/y1,..., Yn = xn/yi. Useinduction.

b) Both integrals (see [31], p. 77, about Feynrnan integrals) can be com-puted by induction.

§2. COMPUTATION OF MULTIPLE INTEGRALS 211

2.18. a) Assuming that ab 0, make the change of variables x = U,ax + by = cv, where c = al + Ibi. This leads to the equality

1 If dxdy—ii f(ax+by)fiR2 (1 +x2)(1 +y2)

1100 [00 cdu=—, f(cv)dv,

2

lbl(1 + u2)(1 + —

Show that the inside integral is equal to ir/(1 + v2).b) Using a) for the induction step, prove by induction that

pif dx••dxI c+ akxk

For c = 0 this leads to the equalityp

1 1 dx1•..dx

11lakI) J —i.\1<k<n

The last integral can be expressed in terms of the Euler f-function:

p/2cos çodço

J—ool+t Jo

I —p—i .p—1 .2= j cos

= BQ'4!

—2 ) 2 ) — sin((p + 1)ir/2) —

2.19. Represent the integral as the sum of the integrals over the squares

x L] (1 k, j n).

To compute these integrals with an accuracy of o(n4), replace the func-tion I by its second-degree Taylor polynomial constructed for the points

1? 1?

CHAPTER VI

Asymptotics

Asymptot'ics of integrals

1.2. a) Use the relation 1 — a(1 — x) as x 1.b) Compare the integral X(t) with the integral

J(2(1 —x)(1

c) Compare the integral Z(t) with the integral obtained from it by re-placing cosx by 1 — x2/2.

1.3. a) Use the relation

1 1 1 11lnxlnA+O(l)

for x E [A, 2A].b), c) Use the result of problem 1.la).d), e) Integrate by parts and use the result of problem 1.ld).1.4. a) Make the change of variables y =b) Integrate by parts after the change of variables y = Ax2.c) Complete the square in the exponent.d) Represent the integral in the form

Ali?Aj (1+—)J2/A \ lnAj e

and prove that" lnt'°dt r+OOdI

J—r+o(1).

2/A mA e 2/A e

To do this for p > 0 use the relation 1(1 +z)" — = O(IzI + (z > —1).For p <0 verify that

lntYdt 1 1

J2/A

and estimate the integralf+°° / lntV' dt

I 11+—I —1 —,\ lnAj e'

214 VI. ASYMPTOTICS

using the fact that I(z + l)P— = 0(z) (z> —1/2).

1.6. a, b). Integrating by parts, prove the boundedness of the integralfor e � 0.

c) To estimate the integral over [ir/2, +oo) integrate twice by parts.d) Representing the integral in the form e(1 + e) f00(1 — dx,

prove that it is possible to pass to the limit under the integral sign. Use theequality (see problem V.1.22) f00(1 — cosx)/x2dx = dx =ir/2.

1.7. a) After the change of variable y = the solution is analogous tothat of problem 1.6d).

b) With the help of the change of variable y =r° and integration by partsprove that the integral over [1, +oo) is infinitesimally small.

1.8. Using the boundedness of the integrals f1(co(u) —CQ)dU show that

the integrals — dt are bounded for e> 0.1.9. a) Use the boundedness of the function 1.2 — x on (0, ir/2).b) Make the change of variable y = Ax and use the result of problem

V.1.37b). Integrate by parts to estimate the integralc) Let

ir/2

1(A)= j cos2x/(1 + Ax)dx

and,r/2

1(A)= j sin2x/(1 +cos2 Ax) dx.

It is clear thatçit/2 dx

2Jo 1+cos Ax— 1 dx dx it

— A Jo 1+ cos2 x A—'+oo Jo 1+ cos2 x =

Prove that 1(A) — 1(A) —' 0 as A —, +00. See also problem 1.14.d) Use the relation — x = 0(12) and the result of problem a).1.10. Represent the integral as the sum of the integrals over the intervals

[irn, ir(n + 1)] (n E N) and estimate each of them from above and frombelow. See also problem 1.13.

1.11. Use the relation

=—

� — T))j

1.12. Use the result of the preceding problem.1.13. The solution is analogous to that of problem 1.10.

§1. ASYMPTOTICS OF INTEGRALS 215

1.14. It can be assumed that c, = 0 (otherwise consider the function= — C,). Representing the given integral as a sum of integrals over

intervals of length T/A, use the relation11+T/A

J f(x)ço(Ax)dx = / (f(x)—f(t))ço(Ax)dxI JI

�where w1(ö) = 11(t1) — f(t")I is the modulus of continuity of f.

1.15. Let be the intervals of the form [2irk/A, 2ir(k + 1)/A], k =p q,contained in [a, b]; w1 is the modulus of continuity of f. Then

1(A) =0 + E j f(x, sin Ax) dxk

dx.

Consequently,

1(A) = o(1)j2fl

ç sint) dt

rj (j f(x, sint)dt) dx.

In conclusion note that under the conditions of the problem any continuousperiodic function on R can be taken instead of sin Ax.

1.16. a) Make the change of variable x = where = is a smallpositive parameter. We have that

j+°°dx = exp ((i) —

dt.

Let us now select the parameter so that as e —, +0 the integrand tendsto a positive integrable function. It is not hard to see that by taking =eln(1/e), we get that

1ft\1_c t\ —f

_a_)_40e

Therefore, to prove the relation(+00

1

i eJ0

it suffices to justify passage to the limit under the integral sign:(+00 (ft \1_CI expll—J ——idt —, i e dt=1.

Jo I c—.+O Jo

216 VI. ASYMPTOTICS

To study the integral on the left-hand side, we break it up into the sum of theintegrals over the intervals [0, 1/IlneI], [ln2e, +oo), and [1/IlneI, ln2e].Let us estimate the first two integrals from above. It follows from the in-equalities u u > 0 and 0 <e < 1 that

111\1C t\exp

—

� exp —

Therefore,

+00 1, 1

I exp I _! dt < I = — —, 0.JIn2c — C

Since the integrand is uniformly bounded above for t > 0 and 0 < e < 1 , the

integral over [0, Inel] is infinitesimally small. To compute the integral

over [1/I In In2 e] we represent the difference — in the form

— i) = + 0(212 = —t(1 +

Consequently,

£ ff \' \ in2 c dt in2 £ dt

J (—) ——I dt=J 1101 —, 1.

1/linci+ e

b) The solution of this example is analogous to that of example a).1.17. a) Use the relation

= = 1 +exlnx+ çx2ln2x+ 0(e3), 0< x � 1.

b) Making the change of variable t = we get that

1 +oo 1

J= 2e2J s—

= 2e2J

It remains to use the Taylor expansion of the exponential.c) Break up the interval [0, 1] into the intervals [0, 0] and [0. 1], where

o = = o(e/lne). The integral over [0, 0] does not exceed 0. On [0, 1]replace the exponential by its Taylor expansion, verifying first with the helpof integration by parts that

dx =oi-_L-_Jo \01n40

Show that for = c/In2 e the remainder term in the formula to be proved

has the estimate 0(e In I In el/In2 e).

§2. THE LAPLACE METHOD 217

1.18. Use the inequality

(e2eu foru>O,O<e<l'le

for u <0.

1.19. a) Make the change of variables I = nx and estimate the integralsover the intervals [(k — 1)ir, kit] from above and from below.

b)—c) We prove the following more general assertion:

• dx ak

J rnaxaklslnkxl—=— — 1<k<n

if the nonincreasing sequence {ak} of positive numbers is such that thesequence {kak} is nondecreasing.

Let M(x) = maxl<k<fl akl sin kxl. SinceIsin � it follows that

M(x) � a bounded contri-bution. To get an upper estimate of the integral use the fact that

M(x) � max(a1x, 2a2x,..., kakx, ak+l = kakx

if x � 1/k. Consequently,

j � 0(1)+

=0(1)+1<k<n

To get a lower estimate of the integral note that M(x) � sin kx �on the intervals [ir/2(k + 1), ir/2k] (k = 2 n), and thus

J � 0(1) +jn/2k 2dx

x2<k<n n/2(k+1) x

=0(1)+1<k<n

§2. The Laplace method

2.1. Since (n + = 1, it suffices to prove that

j o(1/n).

Fixing an arbitrary number e > 0, we get the estimate e) = o(1/n)for the integral over ir/2]. To estimate the integral over [0, note that

Ixf(x) — f(0)sinxl � sinx on this interval, where

= sup 1(0) — xf(x)/ sin xl —, 0.O<x�e

2.2. Make the change of variable y =

218 VL ASYMPTOTICS

2.4. Verify that the integral over each interval [a. 0], 0 < a < 0, de-creases exponentially as A —, +00.

2.5. a) It is clear that

(1P)Adt.

Therefore,

A" I[4

A—'+oo J0 PJ0

p \p/ \ p

On the other hand, for any positive number a

� (i — dt

Consequently,

— � dt = f(1 + 1/p).

It follows from the solution given that the integral over any interval [0, 0]with 0 < 0 < 1 has the same asymptotics.

b), c). The solution is analogous to the solution of problem a). In problemc) make a preliminary change of variable y =

2.6. It can be assumed without loss of generality that a = 0 and C = 1(otherwise make the change of variable y = C"(x — a)). Fix the numbersI and L, 0 < I < 1 < L, and choose a positive number 0 such thatIx" � 1 — � Lx" < 1 for x E (0. 0). Break up the integral c1(.-1)into the sum of the integrals over [0, 0] and [0, b). For the first of themwe the two-sided estimate

0 0 0

J (1 —Lx" y4dx � J � J (1 —Ix" )'4 dx.0 0 0

while the second is easily estimated from above:

dx � j dx =

Therefore,

limA"j(l � limA"(t)(A) �

� urn A" JO — Ix")4 dx.

§2. THE LAPLACE METHOD 219

Consequently (see the solution of problem 2.5a)),

lim limA—'+oo A—'+OO

Since I and L are arbitrary numbers, 0 < I < 1 < L, this implies theequality = f(1 +p).

2.7. The solution is analogous to that of the preceding problem.2.8. Let h = g — f and prove that

1b4= o (L1'ix x coA(x)dx).

Fix an arbitrary number e > 0 and let /3 be a number in (a, b) such thatIh(x)I � ef(x) for x E (a, /3). Then

I f �

e j dx +

Show now that = For this fix a number in(a, /3) and note that

1bdx � j jf(x)dx.

Since and f(x)dx >0, it follows that

A A(çb

Afr4)=o j (x)dx

Thus,i ff11' � 2e if A is sufficiently large.

2.9. Use the asymptotic formula of Laplace (problem 2.6).b) Using the result of problem 2.8, replace cos(ax) by 1.c) Make the change of variable x = Ay.d) The main contribution is given by the integral over [0, 1]. Using the

result of problem 2.8, replace sin x by x in it and then make the change ofvariable y =

f) Verify that the main contribution is given by the integral over [0, ir/4].In this integral replace cosx by cos(2x).

h) By means of the result of problem 2.7 show that the main contributionis given by the integral over [1/2, 1].

2.10. c) Make the change of variable x = y2.2.11. a) It is easy to see that xA —, 1 as A —' +oo. Making the change

220 VI. ASYMPTOTICS

of variable x= 1+t,wegetthat+oo A

( h+hl ate

— f_i +t2)

=1 +1 +1 =11+12+13.—C £ —1

Since

Ii = fexp — + O(At3)) dt

in the case when Ae3 —* 0 we have thatr 2e —u2

I e " du.v A

Therefore, if e = A215). Integral'2 is easy to estimate from above:

j dt = o

For an estimate of 13 verify that the integrand is nondecreasing on [—1. —e]

if e = A215. Therefore,

13 � (2 = Ae2/2 + O(Ae3)) = o

b) Using problem 2.8, replace the function under the integral by APXc) Make the change of variable x = Ay. Using Taylor's formula, study

the integral over [0, A213], and get an upper estimate of the integral over[A213, 1] by using the fact that the integrand is decreasing on [A213, 1]for large A.

2.12. Get an upper estimate of the integrals over the intervals [0,1/(AInA)] and [(2lnA)/A, 1]. To compute the integral over the remaininginterval use the relation I = (1 +o(1)) lnA for x E [1/(A ln.-1), (2lnA)/A].

2.13. a) The solution is similar to the solution of problem 2.1 Ib).b) It is clear that the contribution of the integral over the interval [1/3, 1/2]

is small. To study the integral over the interval [0, 1/3]. make the changeof variables t = = —xlnx Then

1C

i x dx=—ij0

where c =(In 3)/3 and ip is the function inverse to Since lnx Inas x — +0, it follows that In! In as I +0. Using the result ofproblem 2.8, we get

113Ax

cC e4' cc.4

I x dx'-—! —di=—!Jo j0 lni A j0 lnA — lnu

1 e" du I du 1

A Jo InA — In u A Jo mA AIn A

§3. ASYMPTOTICS OF SUMS 221

2.14. Writing + x) in the form

(x)XJ40° (1+t)xdt

use the asymptotic formula of Laplace (problem 2.6).2.15. Use Stirling's formula.2.16. The solution is analogous to that of problem 2.14.2.17. a) Let = x+0'Ji (0 = 0(x)). By the result of the preceding

problem the equality

1 1

lim I te dt=—x—'+ooJT(1+x)j0 2

is equivalent to the equality

1 I x—1lim i te dt=O.x—'+ooJT(l +x)

Making the change of variable t = x+ and using Stirling's formula, weget that it is equivalent to the relation

— 1

Jou—o( ),

which is valid only in the case when 0(x) — 0 as x — +00.b), c) The proofs of these assertions are analogous to the proof of assertion

§3. Asymptotics of sums

3.2. Obtain a lower estimate of the sum k°ak.3.3. Express the sums and ; in terms the numbers tm =

>.1<k<m k°xk and use the results of problems 11.2.7 b) and c).3.4. Express the sums in terms of the numbers tm = >.k>m k°xk and

use the results of problems 11.2.7 b) and c).3.5. a) Fix an arbitrary number M> 1. It is clear that

1—y

Since

it follows that for any M> 1

ny--i — 1

Therefore,

limna > lim =1.n—M....,i+o(M_1)(1_y)

222 Vi. ASYMPTOTICS

To prove the inequalities n7; � 1, show that for m E (0, 1)

(1 rn)(ly)b) Consider the sequence = (In 2)/2k for 2k < 2k+1

3.6. The solution is analogous to the solution of problem 3.5a).3.7. For � 1 use the relation (1 — = 1 + O(k/n), and for 1

the equality

k�1

= +

3.8. See [5]. It is easy to see that i(k) is equal to the number of points(u, v) E N2 lying on the hyperbola uv = k. Therefore, the sum T(n) isequal to the number of points (u, v) E N2 not lying above the hyperbolauv = n. Since the set {(u, v) E N2Iuv � n} is symmetric with respect tothe line u = v • to compute T(n) it suffices to calculate the number T(n) ofpoints in the set E = {(u, v) E N2Iuv � n, u � and take into accountthat the points in the square [0, x [0, are also in the set symmetricto E (make a sketch). This gives us that

T(n) = 2T(n)— [,f]2

= 2—

=2n

To compute this sum we can use the following refinement of the result ofproblem 11.2.9:

which is easy to verify with the help of the theorem of Stoltz (problem 11.2.6).3.9. a)Let a=supn/ç. Since N

k>N

� <

Taking N = I + [at], we get for t � 1/a that

1(t) �+

� 3a.

§3. ASYMPTOTICS OF SUMS 223

But if 1 E [0, 1/a), then f(i)< 2a. The inequalityb)can be proved analogously.

c) To get a lower estimate of the quantity 1(t) fix a large indexn. Then for all t in the interval l/jç] we have that

E �t > -4--1�k<n + 2t

Consequently, f(s) �3.10. Use the inequality f,° f(t)dt � f(t)dt. The equality

is not true for nonmonotone functions. For example, if f(n) = 1 for anyn E N, then = +00. Here it is not hard to ensure that the integralff00 1(t) dt converges.

3.11. Use the result of the preceding problem.3.12. It is clear that

- (2k÷2)P)

Using Taylor's formula, we get that f(p) = 0(p) + p 1/(2k +

1çk+1

dt 1 1

it follows that

f(p) pj +0(p) =

3.13. Use the inequalities f(k + 1) � f(t)dt � 1(k).3.14. See the hint for the preceding problem.3.15. It suffices to estimate the quantity Ak = f(k) — 1(t) dt (k =

M N). In part a) use the equalityfk pk+1/2

Ak = j (1(k) — f(t)) dt + / (1(k) — 1(t)) dt.Jk—1/2 Jk

In part b) use integration by parts to prove that

Ak = 2Jk1/2(t_k+

11k+1/2(t_ k—

We remark that by using integration by parts it is possible to get from theequality in b) estimates of the difference A that contain higher-order denva-tives.

224 VL ASYMPTOTICS

The equation b) is actually a special case of the classical Euler-Maclaurinformula. On this topic see, for example, [29].

3.16. Use the result of problem 3.13a) and Taylor's formula.3.17. By using the recursion formula

a(n—crn n

it is possible to reduce the problem to the case a < 0. Then In (fl;a) =

El<k<flln(l — ce/k). It remains to use the result of problem 3.13a). Wethat the equality c0 = l/(f(l — a)) follows from the properties of

the f-function (see problem 2.15).3.18. After taldng logarithms in parts a) and b), use the result of problem

3.14, and after taldng logarithms in part c), use the result in problem 3.15b).3.19. a)—c). Using the result of problem 3.14, replace the summation by

integration. The integrals thus obtained were considered in problems 2.llb)and 2.13a). In part c), first use Stirling's formula.

d) Find the asymptotics of the sum of the last terms and show thatthe sum of the remaining terms is sufficiently small.

3.20. It suffices to prove that >k>n flk/k! e'1/2. To do this, use the in-tegral representation of the remainder in Taylor's formula for the exponentialalong with the result of problem 2.9c).

3.21. Find the asymptotics of the sums = >1<k<n where p isa fixed positive number. Since the numbers are rapidly increasing for1 � k � n/2, the main contribution to the sum under consideration comesfrom the terms with indices close to n/2 (as is clear from the following, itsuffices to take into account about n213 central terms).

Assuming that n = 2rn, we have that

=(1+2

To find the principal part of the sum obtained, note that for k> 1

() rn 1 — f/rn — iii (1

1 — f/rn— rn+k 1<k 1+1/rn

— n 1+f/m

It follows from the relation 0< —21—ln((1—t)/(1+t)) = (0<1 < 1/2)that

— 1(1 + fork � rn213,

— 1 for k > rn213.

§3. ASYMPTOTICS OF SUMS 225

Thus (See problem 3.14),I(2mI m—k)'t

j = (1 + O(m"3)) + o(l)1�k<m \ / /

2/3

= (1 + O(m"3)) I --pu2/m 1 (Fe

To obtain the final result it remains to verify that whichimplies that

1 it,z

3.22. Denote the sum of the series by coN(z). It is clear that

dte

____

n+t+>2l n+t )di.

n�N°Therefore,

°°

= J —i--- dt + >2N n>N

—

+>2 n+t di

=+ >2

o(zt)di

I n�N

kV"

3.23. Using the result of problem 3.15b), we get that

12 --z,Ji iie

— 00

e di — ! — [t]—

(e di = I — I.n�N — IN—1/2 2 N—1/2

The integral I is easy to compute. To estimate I we write it in the form

dt

8lN 1/2

(t—[t]—-

The integral obtained can be estimated by breaking it up into the sum of theintegrals over the intervals [k — 1/2, k + 1/2] (k E N, k � N); to estimatethis sum use the result in the preceding problem.

3.24. Using the identityn+1/2

cosnx = 2sinx/2 lfl_1/2cos(xt) dt,

226 VI. ASYMPTOTICS

we get that

>2 = 2sin(x/2)(j+cocos(xt))

= j Vdv + 0(1).

It can be proved similarly that

>sinnx a_if snvd +0(1)

It follows from the result in problem VII.1.26 that the integrals A =f°°(cos v)/v° dv and B = f°°(sin dv are positive.

3.25. With the help of the device used in the solution of the precedingproblem, we get that for x E (0, ic)

>2cosnxf'cos(xt)dt+o(l)

1<n<N 0

=cosv

dv + 0(1).

From this, uniform (with respect to n E N and x E [0, ic]) boundednessbelow of the sums n° cos nx is equivalent to nonnegativity of theintegral 1(u) = for u � 0. It is clear that 1(u) =

I(3ir/2) = dv. It remains to show that the function =I 12(cosv)/va dv changes sign once on the interval (0, 1). Since 1(0) =—1 and 1(1 —0) = +00, it suffices to see that the function i is monotone.It is easy to see that

13n/2 cos v 11,r/2 cosv 1

i IJo v V Jo v v

Using the decrease of the function v on (0. ir/2), we get that

cosv 1çn/2

1

I I ln—dv>0.Jo v V — Jo V

Similarly, the study of the sums nx)/na reduces to the study of theintegral = which is positive for all >0.

3.26. Let A = and C = sup,>a w'(t). Then

f+00 A +oo

Jdt — A = —a + J — 1) dt + J

a a A

=—a+11 +12.

§4. ASYMPTOTICS OF IMPLICIT FUNCTIONS AND RECURSIVE SEQUENCES 227

Let us estimate the integrals and '2:

Ill � � j� eCj < = C;

0 � '2 �

� j dt

— C — C

3.27. a)—e). Using the monotone decrease of the general term, showthat the summation can be replaced by integration (see problem 3.14). Toinvestigate the integrals obtained use the results of problems 2.9g) and 3.26.

f) With the help of the inequality (n/3)'1 <n! < (n/2)'1 reduce the prob-lem to problem e).

3.28. Since for fixed t E (0, 1) the general term of the series does nottend monotonically to zero, the inequality of problem 3.14 is not applicableto these series. One should consider separately the partial sum in which thegeneral term increases, and the remainder of the series, in which the generalterm decreases.

In b) use the result of problem 1.4d).3.29. a)—f), h) The solutions are analogous to the solutions of problems

3.27 and 3.28. In b) and f) use the results of problems V.1.18 and V.1.19.g), i)—l) Expand the general term of the series in a power series and change

the order of summation.3.30. See the hints for problems 3.27—3.29.3.31. Use the results of problems IV.4.15 and 3.29e), f). Another solution

of problem a) is obtained by using the results of problems 3.8, IV.5.8, and3.28b) for p = 1.

§4. Asymptotics of implicit functions and recursive sequences

4.1. With the help of integration by parts, prove that z(t) = (1 + o(1))

x e12h12/t as t —' +00. Therefore,

—-i-.

Consequently, t(z) as z — +0.4.2. Show that = — icn — ir/2 is an infinitesimally small quantity.

To obtain the asymptotics of the sequence replace ; in the equality= tan; by + icn + ir/2 and pass to the equivalent quantities.

228 VI. ASYMPTOTICS

4.7. It is not hard to see that for any number a> 1

x x / 1

forxE

Iterating the functions and st', we get the two-sided estimate

xo xo . 1 1

<x <— ifx EI0,1——1+nax0 1+nx0 a

For n = 1000 and a = 1/0.999. we have that

—30.999 1 —310

Consequently,

0< —x1000

4.8. It is dear that 0. The asymptotic behavior of the sequencecan be found with the help of the device used in the solution of the

preceding problem. For this, get upper and lower estimates of the functionf in a neighborhood of zero by the functions considered in problem 4.6, andthen compare their iterates.

Another solution is based on the following which can beused also in more complicated situations (see problems 4.12—4.14). Therecursive sequence = i)' x0 > 0, clearly converges to zero if thefunction f is such that 0 < f(t) < t for all t > 0. With the help of afigure it is easy to see that the sequence converges rapidly to zero ifthe difference t — f(t) is large, and converges slowly otherwise. Therefore,to find the asymptotics of the sequence it is expedient to consider thefunction = t — f(t). Then the recursive formula takes the form

=

Assuming that is the value at n of some smooth function 0 defined on[1, +oo), we get that

0(n) — 0(n — 1) = — 1)).

If as an approximation the difference 0(n) — 0(n — 1) is replaced by thederivative O'(n — 1), then it turns out that the function 0 satisfies the "dif-ferential equation" 0' —ço(O). Solving it, we find that n,that is, = 0(n) where c1(y) =

To justify these heuristic considerations we must study the difference— and, using the given recursion relation, show that they

tend to 1. Then, adding the equalities CJ?(xk) — = 1 + 0(1) fork=1To simplify the computations it is sometimes more convenient to considerinstead of c1 a function c1 equivalent to it. For example, in the case

§4. ASYMPTOTICS OF IMPLICIT FUNCTIONS AND RECURSIVE SEQUENCES 229

1(x) — x where C and p are fixed positive constants, wex—'+O

have thatfXi dt fxi dt 1

Considering the difference — we get that

b(x ) — )= —

—1.

Cp

From this, 1/Cpx = n, that is,4.9. a) Use the idea described in the solution of problem 4.7.b) Refine the result in a). Since

it follows that x = np + O(ln n) = np(1 + O((lnn)/n)). Thus, =(np)lh'(1 + yr), where = O((lnn)/n). Substituting the expressions ob-tained for and into the recursion relation, we get by means of theTaylor formula that

— (11\ ip—i 0(ln2n

that is,ip—! Iln2n

—y

This implies that

= + 0(1).

4.10. It suffices to solve the following problem: construct on (0, +oo) afunction f such that the sequence = coincides with a previouslygiven sequence strictly decreasing to zero. It can be assumed herethat = where is a strictly decreasing function on [0, +oo) with

= 0. Then the equality = = is equivalentto the equality f(ço(n — 1)) = which is satisfied, for example, by thefunction f(t) = +

4.11. Consider the sequence and4.12. Prove first that T +00. Using for e E (0, 1) the inequality

0

prove that Deduce from this that lim(a1 + ... + � 1,and hence a1 + a2 + . + nan. Therefore,

2 2 2; 2

which implies that 2lnn.

230 VI. ASYMPTOTICS

4.13. Prove first that an 1 +00, an+1 and = o(An), where

= a0 + ... + an. For 0 � p < 1 the last two relations are obvious. For—1 <p <0 they follow from the two-sided estimate

0 � � (n E N).

The upper estimate follows from the obvious inequality

� 1 + (kak)" � 1 +1<k<n

The lower estimate can be proved by induction if the number m E (0, 1) ischosen so that

2 "1

(1—p)/(1+p)

_i) forallnEN.

From these relations obtain successively that

2 2 — 11 —p 2\—an 2

1 —pAn

(1 _p\P/(P_l) 2p/(p—1)2 )

and, finally, — = + o( 1), which implies

that4.14. Prove successively that 1 +00, an+l an, and =

where = +. + The last relation follows from the inequality

o � = J,kan_k k

which is valid for any fixed index k. For p —1 deduce from this that

— (p + (p + 1)(lnSn+i — lnSn).

Therefore, for p < —1 the limit lim 5n = a,, exists and is finite, that is,

an+i—ap=(l/ap)+o(l),from which Forp>—1 the

relation a°7' (p + l)lnSn —, +00 is valid. Using the mean value theorem,prove that this implies the equality

5nil — = (P + 1 + o( 1).

that 15, n(p + Therefore, Inn, andhence (p + 1)lnS, (p + 1)Inn

For p = —1 ,prove by induction that = (2n)!!/(2n — 1)!!.

CHAPTER VII

Functions (Continuation)

§1. Convexity

1.1. Use induction (verify first that a convex combinationfalls in (a, b)). — —

1.2. Let=

and 22 =x3—x1 x3—x1

and write Jensen's inequality for x2 = +1.3. a) Since some of the numbers E can be the same, it suffices

to prove Jensen's inequality for the arithmetic mean of n numbers. Provethis first for n = 2k• Verify that if the inequality is true for the arithmeticmean of n arbitrary numbers, then it remains true also when n is replacedby n — 1.

b) Use a).1.4. It can be assumed that p, q E (a, b). Let K = supp<x<q 1(x). First

verify that f is bounded above on any interval [c, d], (p. q) C [c, d] C(a, b). To do this, consider the set E = {x E [c, d]If(x) � C}, whereC = max(f(c), 1(d), K), and, using the result of problem 1.1.19, show thatE = [c, d]. If f is bounded above on (p. q)\Q, then show that f is

bounded also on (p. q), because each point of this interval is the midpointof some interval with endpoints in (p, q)\Q.

The continuity of f at a point x E (c, d) follows from the inequalities

!(f(x) - C) 1(x) - f(x -5) f(x + 5) - 1(x) !(C - 1(x)),

where S is an arbitrarily small positive number, and m and n are the largestintegers for which x—mS, x+nS E [c, d]. The middle inequality is obvious,the right-hand one can be proved with the help of Jensen's inequality (see1.3a)) for the points x, x +5, and x + no, and the left-hand one can beproved similarly.

1.5. a) b) is obvious. To prove the implication b) c) considerthe set K = — c), where c E R2 is a point of the graph. Show

that K is a convex cone and take a straight line containing one of its

232 VII. FUNCTIONS (CONTINUATION)

boundary rays. To prove that c) b) verify that is an intersection ofhalf-planes.

1.6. Let 0 � x <y 1/2. Represent y and 1 — y as convex combina-tions of the points x and 1 — x, and use the convexity.

1.7. Use Borel's covering lemma.1.8. Use the three chords lemma (problem 1.2).1.9. Consider first the case when L1(x) > 0 for all XE (a, b). Suppose

that x1, X2 E (a, b), x1 <X2, and y = g(X) is the equation of the chordbetween the graph points with abscissae X1 and X2. Let = I — g. Itis clear that L1(X) = L1(X) > 0, and for f to be convex it suffices thatf1(X) � 0 for all X E [X1 , X2]. Assuming the contrary, show that 0

at a point where f1 attains its largest value on [X1, X2]. In the general

case (L1(X) � 0) consider the functions f(X) + €X2, e > 0, andpasstothe limit as e—'O.

1.10. The necessity is obvious. To prove sufficiency rewrite the inequalityin the form f(f(X+ t) + f(X — t) — 2f(X))dt � 0 and use the precedingproblem.

1.11. a) Use the uniform continuity and the graph by chords.b) Using a), show that it suffices to consider only functions of the form

IX — al.1.12, 1.13. Use the three chords lemma (problem 1.2).1.14. Verify that Jensen's inequality for one function is carried by the

transformation X 1/X into Jensen's inequality (with different coefficients)for the other.

1.15. Use the existence and monotonicity of one-sided derivatives (prob-lem 1.8).

1.16. Reduce the problem to the case a = 0, f(0) = 0, and verify thatthe function f(X)/X is monotone.

1.18. Use the three chords lemma (problem 1.2) for the points 0, x —andX (0<h<X).

1.19. Itsufficestoverifythat forall x1 <X2. X2—X1 <b.This follows from the three chords lemma (problem 1.2). applied to the pointsX1,X2,x1+b and X2,X1+b,x.-I-b:

(v — + b) — + b) — f(X2)b — x1+b—X,

� + b) — f(X2)=

1.20. It will be assumed that n = 2m + 1 is an odd number, sincezero can always be added as the last point. For a � b the function xf(b + x) —f(a+X), x � 0, is increasing (see problem 1.19). Consequently,

§1. CONVEXITY 233

— � i ( — a2k+l)) — i ( (a2k — a2k+l)

/ ki<k<m

1=0 rn—i,

f(a2m) — f(a2m+l) � —

Adding these inequalities, we get what is required.1.21. Using the inequality — � — , prove that

<an+p+q—an+p

p+q — q

Interpret this inequality geometrically.1.22. See problem IV.2.3.1.23. Verify that only piecewise linear functions in the indicated classes

may be considered (see problem 1.1 la)). We explain the subsequent argu-ments for two inequalities.

a') Partition the subgraph of the function into triangles with the commonvertex (0, 0) by joining this point to the node points of the graph (see Figure13a). Suppose that the kth triangle is bounded by the graphs of the func-tions c°k and 0 f, and c9k+l(x) =outside the projection of the kth triangle on the x-axis. Then f =

— Prove the inequality a') for each function—

and then add these inequalities. Show that the convexity requirement is es-sentiaL

d') Continuing the rectilinear segments of the graph until they intersectthe x-axis, we get a partition of the subgraph into triangles (Figure 13b). Weintroduce the functions c°k (1 � k � N) just as in the preceding case. Then

= — Verify that

=

Mk is the height of the kth triangle.

FIGURE 13a FIGURE 13b

234 vii. FUNCflONS (CONTINUATION)

1.24. Replace the integration over [0, 1] by integration over [0, 1/2].Use the results of problems 1.6 and 1.2.22.

1.25. To verify necessity, consider the functions ç9(x) = C, ç9(x) =Cx (CE R), and ç9(x) = max(0, a —x) (a E (0, 1)). To prove sufficiencyuse the fact that a piecewise linear convex function is a sum, with positivecoefficients, of functions of the form ç9(x) = max(0, a — x) and a linearfunction.

1.26. a), b) Represent the given integrals as the sums of the correspondingintegrals over intervals of length 2ir/a. In b) use the results of problem 1.6.

1.27. a) Using the Cauchy-Schwarz-Bunyakovskii inequality, verify that

1,1 f2 � o.b) With the help of problem 1.llb) we can confine ourselves to the case of

a twice differentiable function. The logarithmic convexity of I means thatf12 � 0. Verify that the sum of two such functions f and g satisfies

the inequalities

(1" + g")(f+ g) � (f + g')2.

1.28. (See [18].) We confine ourselves to a proof that 3) and 4) areequivalent.

Since f is convex (see problem 1.3b)), the one-sided derivatives andexist at each point x, � and = everywhere

except on a countable set of points (see problem 1.8). We prove that f(1)exists. Let h > 0. Then

1 =f(1)�f(1+h)f(1/(1+h))= [f(1)+f(1)h+o(h)J[f(1)—f(1)h+o(h)]=

which implies that � and (considering the opposite inequality)= Let p = f'(l). Fix an x such that 1(x) exists. Consider

the function b(t) = f(t)f(x/t), t > 0. It is differentiable at 1 and satisfiesthe relation � 1(x) = for all 1. Therefore, = 0, that is

((1)f(x) — [(1)41(x) = 0,(*)

pf(x) = xf(x).Thus, the derivative f(x) , which exists on a dense subset E of (0, +oo),

coincides on it with the function pf(x)/x, which is continuous on (0, +oo).From this, since and 1' are monotone (see problem 1.8), it follows that

f E C'(O, +oo). It now follows easily from (*) that 1(x) = Ax", whereA = 1 since [(1) = 1, and the restriction on p is due to the convexity of

In connection with the condition 1'), consider the function 1(x) =1) (xE(0,+oo)).

§1. CONVEXITY 235

1.29. Write Jensen's inequality for the function f(x) = x" in the form

(!

where c = = c1 = and choose the numbers > 0suitably.

1.30. a) If 0 < r <s, then use the Hälder inequality with p = s/r (see

the preceding problem). If r < s < 0, then use the relation (,c(—t))' =If r <0 <s, thenuseb).

e) inequality for the function ln at the points r, s, Ar + jisji > 0, = 1) becomes the Hälder inequality if we set = xr,

= p = and q = 1/4u.

For a continuous positive function f we should define ic(p) to be(f( and consider

exp (j'lnf(x)dx)

instead of (x1,..., Assertions a)—e) remain in force.1.31. The inequalities a) and b) are continuous analogues of Jensen's

inequality and can be obtained from it by passing to the limit. Another wayof proving the inequality b) (which is more general than a)) is as follows.By 1.11, we can assume that ço" exists. Setting I = I(x)p(x)dx , we getfrom Taylor's formula that

ç9(f(x)) = ço(I) + ço'(I)(f(x) — I) + — J)2

� ço(I) + ço'(I)(f(x) — I).

The required result is obtained after multiplication of this inequality by pand integration over [a, b].

c) Use b) with ç9(t) = ln(1/t).1.32. Both quantities are as large as possible in the case when the points

A1,..., divide the arc joining A0 and into equal parts. For aproof use the concavity of the sine function on [0, it].

1.33. Consider an arbitrary ray emanating from the point S = (0, b),0<b <1(0). Let A1=(x1,y1), 1=1,2 bethesuccessivepointsofreflection of the ray from the graph, and let B. be the points of reflectionfrom the x-axis. Then to each ray there corresponds a (finite or possiblyinfinite) sequence (trajectory) SA1B1A2B2• or SB1A2B2A3. (the pointA1 is absent if the first reflection is from the x-axis). We prove the existenceof a number M such that for all rays sup1 x1 � M (this means that only abounded part of the subgraph is illuminated).

1) Note first that if for some n, then for all k > 0(the ray emanates from the domain).

236 VII. FUNCTIONS (CONTINUATION)

2) Let be the unique trajectory for which the three pointsS, A?, and lie on a single line, and let 4° = Then for any

other trajectory containing A1 the inequality x, holds (it is very easyto see this).

3) Suppose that n � 2. > E (0, ir/2) is the angle betweenthe components of the polygonal line that are incident to Bk and the x-axis,and = I arctanf(xk)I (see Figure 14). Then, since f is convex,

<(tan —xv).

Taking into account also that

2y, —xv) and—

= 20,.

we get the inequality

— tantan tan tan

from which

1 — —

= sin sinç9,1sin

—sin

Consequently,

sinç91— >—-——

> sin > sin

Note that � = I � Consequently, the numbercan be taken as M.

1.34. a) Observe that f* is the pointwise supremum of the family oflinear functions h, a(X) = xt — a over all pairs (1, a) such that a � f(t).

b) Use Jensen's inequality and problem 1.5.

FIGURE 14

§2. SMOOTH FUNCTIONS 237

c) Use b) and the fact that = xt — b � f(t) for all t E R if andonly if b � sup,ER(xt — f(t)) = f(t), along with the hint for a).

e) Obviously, for a convex function differentiable at the point t theequality ç9'(t) = 0 is a sufficient condition for a minimum. Therefore, iff'(t) = x exists at the point t, then the supremum in the definition off*(x) is attained at t. Let a <b, suppose that a = f'(a) and /3 = f(b)exist and are finite, and let x E (a, /3). Then the values of at the points

and /3 are finite, and hence f is finite on (a, /3) (see b)).g) The function f cannot have jumps and is strictly monotone, hence

continuous, and its range really is an interval. The failure of to be dif-ferentiable at the point x would mean that t = < ç =

and then the function f would be linear on [t, (see f)), whichcontradicts the strict convexity of 1. Similarly, differentiability of I im-plies strict convexity off and for any x E the equality f*(x) = xt — f(t) is attained onlyfor those t such that f'(t) = x. Analogously, for t E (a, b) the equal-ity 1(t) = f**(t) = xt — f(x) is attained only for those x such that(f*)I() = t. This means that f and (fe)' are mutual inverses.

1.35. Use problem 1.34.1.38. See problem 1.36a) andf). To prove uniqueness use problem 1.34d)

and h).

§2. Smooth functions

2.2. Integrate the given identity with respect to y over the interval [0, 1].2.3. Approximate by a polynomial uniformly on [a, b] with an ar-

bitrary degree of accuracy. (The possibility of such an approximation follows,for example, from 3.8b)).

2.4. Study the Taylor series of I at an endpoint of an interval on whichf(x) 0.

2.5. Use induction on n. Employ the formula

1<k<n

and, arguing by contradiction, study the derivatives of the function +p

k -

2.6. Prove that either f(x) —, 0 as x —* or 1' has infinitely manyextrema.

2.7. Let x, ö E (0, 1). Then

(ö—1) 2.u_f(öx) = 1(x) + (ö — 1)xj (x) +

2x j (x)

238 VII. FUNCTIONS (CONTINUATION)

where C (ox, x). Therefore,

1(x) — f(Ox) 1 —0 2 ,,, —xj(x)= +—1--xj (x)

=1—0 x

— 1(x) — f(Ox) (1—0— i—o +

By choosing 0 close to 1, we make the second term small, and then bysuitably choosing x we make the first term smalL

2.8. Verify thatçT

—h(t)=—j f(x)g(t—x)dxJ—T

= j f(x)(g(x+ t) - g(x - t))dx

= I I f'(x)g'(y)dxdyJO Jx--t

£ x+t T x+t

= (11+1 J J )P Oxt Tt2Txwhere t C (0, T) and P is the rectanglewithvertices (0, t), (t, 0). (T, T—t), and (T — t, T). Since P C [0, T]2, the first integral is nonnegative, andthe remaining two integrals are equal to zero because g' is odd.

2.9. Write 1(1/2) according to Taylor's formula [(1/2) = 1(x0) +...,setting x0 equal to 0 and then 1.

2.10. a) Starting from the identity

f(x) = j(f(x) - f(x + t))dt + f(x + h)-f(x - h)

and applying the mean value theorem to the integrand, we arrive at the in-equality

If(x)I � (f ItIM2dt + 2M0) = +

for all h > 0. The expression on the right-hand side has a minimum equal to2M0M2. The inequality becomes an equality for the piecewise smooth func-

tion 1(x) = ço(u)du + 2) dt, where = —2sgn u for ui � 1 andço(u) = 0 for ui> 1. Replacing by a continuous function as shown in

Figure 15a), we get a Ca-smooth function and for itas e —, 0.

The inequality proved admits the following mechanical interpretation: ifthe motion of a point takes place on the interval [—M0, M0], and the accel-eration never exceeds M2 in absolute value, then the velocity of the point

§2. SMOOTH FUNCTIONS 239

2

III

a)

b)

—1

—2

1

II

I I I—-C - -+Ek k k

FIGURE 15

cannot be too large at anytime. More precisely, it does not exceedAll are familiar with an analogue of this fact from a basic physics course: ina free fall (with zero initial velocity) the instantaneous velocity of a body isequal to where S is the length of the path passed by the body andg is the acceleration of gravity.

b) Replace [x — h, x + h] by [x, x + h] in the preceding argument. As aresult the inequality M1 � is obtained.

d) Use an analogous device, choosing h in (0, (b — a)/2] and integratingover [x, x+h] or over [x—h, h],dependingon the sign of x—(a+b)/2.

c) In the case it is convenient to use another argument. Adding the equal-ities

1(x) — f(0) = xf'(x) —

[(2) — f(x) = (2— x)f'(x) + —

termwise, we get that

2f(x) = 1(2) — f(0) + —(2—

which at once gives the desired estimate. That it is sharp is demonstrated bythe example f(x) = (x2/2) — 1.

240 VII. FUNCTIONS (CONTINUATION)

2.11. The convergence of the integral dx (if at least one in-tegral diverges, then there is nothing to prove) implies that the limit c =

f'(x) exists, and the convergence of If(x)I dx implies thatc = 0. Consider the function g determined by the conditions g'(x) =— and g(0) = 1(0) (for definiteness assume that this numberis positive). It is decreasing and convex, and it satisfies the inequality g(x) �f(x) (a consequence of the inequality g'(x) f(x) = — f° f"(t)dt). Drawthe tangent to the graph of g at the point 0, and let be the point whereit intersects the x-axis, = f(0)/g'(O). Then we get from a comparison ofareas that

� j max(g(x), 0)dx � j If(x)I dx,

which implies that

� j If(x)I dx Ig'(O)I.

First replacing by If"(x)Idx and then the geometric mean bythe arithmetic mean, we get a) and b).

To prove that the constants 2 and cannot be improved, we approx-imate the function h(x) = max(1 — kx, 0) (k > 0) by a convex function

E c2([0, +oo)) as shown in Figure 15b).Then 'i \

I f(x)dx= f(--e) =k,J1/k--e \" /

jf(x)dx 1(0) = 1,

and hencea) 1 = = 2. .k 2fg°If(x)Idxfg° If"(x)Idx:b) fo°°(If(x)I +If'(x)I)dx +k = for k =2.12. By Taylor's formula, f(1) = f"(t)(l—z) di .which implies

thatI

f'(t)(l — 1)dlI = al. Applying the Cauchy-Schwarz-Bunyakovskiiinequality to the integral, we get a lower estimate:

1/2

al < (j'((I(1))2d1) .

Equality is attained in the case when the functions f"(z) and 1 — t areproportional, which implies that f(z) = al(1 — 1)(i — 2)/2.

2.13. Let = = 0}. = Prove the auxiliary asser-tion (A): If f is not a polynomial on some open interval then for any nthere is a constituent interval of the nonempty open setf is not a polynomial.

§2. SMOOTH FUNCTIONS 241

The assertion of the problem is easily derived from (A): assuming that Iis not a polynomial on R, we construct a sequence of nested open intervals

= (ar, c c with nonempty intersection, andthis contradicts the assumption.

Assume that (A) does not hold, i.e., that for some n and the functionf is not a polynomial on but its restriction to any constituent interval ofthe set n is a polynomial. Let a(x) (x E be a maximal intervalwith the properties that x E a(x) C A and is a polynomiaL Verifythat if the restrictions of f to two adjacent intervals are polynomials, thenI is a polynomial on their union. Show that if a is an endpoint of a(x),then: 1) there exists a sequence Xk E Xk a, Xk —+ a; 2) J(m)(a) = 0for all m � fl; 3) vanishes on a(x). Then, in particular, = 0

for all x E fl which is impossible (the solution of D. Yu. Burago).2.14. By formal termwise differentiation obtain a recursion relation that

can be used to successively define and Note that by suitably choosing

fir, it is possible to ensure the possibility of termwise differentiation of theseries.

2.15. It can be assumed that a = 0. Then

1(x)= j (lx) di

= j >

= >jxj' (ix) di,

and canbetakenas2.16. Note that if f is a constant, then b) and c) imply that F(f) =0.

Two-fold use of the result of the preceding problem gives the representation

f(x) = 1(a) + —

+ >J(x1 — aJ)(xk — ak)hJk(x),j ,k

where g1(a) = It remains to use a), b), and c) once more.2.17. Rejecting the trivial case when f 0, we assume that the number

a = inf{tIf(t) > 0} is equal to zero (otherwise, we make the substitutionu = t — a). Then it follows from the assumption that for t > 0

f(t)g(t) <g(i).C+f0tf(x)g(x)dx —

Passing to primitives, we get the inequality

Iln(C+j g(x)dx.

\ Jo / 0

242 VII. FUNCTIONS (CONTINUATION)

Thenft IftC+j f(x)g(x)dx�Cexp( I g(x)dx

Jo \Jowhich immediately implies what is required.

To prove the corollary observe that, since Ih(t)I = I h'(x)dxl �h'(x)Idx, the condition of the first part of the problem is satisfied by

the functions 1(t) = Ih'(t)I and g(t) M and the number C = 0.2.18. a) Use the fact that 1(5 — tk) = f(j)f(tk).b) Use the fact that cos(a, t) is half the sum of two positive-definite

functions.2.20. Represent f and g in terms of their Fourier transforms.2.21. b) Use the identity

a cos(x, t)— 1

—IIxII =caJ n+a dt,11th

where ca > 0 (cf. problem 1.26b)). Then (see problem 2.18b))

—

I<j<m1<k<m

R

1<k<m

2.22. b) Use the fact that, as follows from the inequality (1), the matrix

11(0) 1(t)1(0)

is positive-definite for any t E R.c) Use the positive definiteness of the matrix

f 1(0) f(t1) 1(12)

1(t1) 1(0) f(t2—t1)f(t2—t1) 1(0)

forany t1,t2ER.2.23. Suppose that z1,..., Zm E C and z1 + + Zm = 0. Then for

s>00

1=s

I<k<m

Passing to the limit in this inequality as S —, 0, we get the inequality

— tk)ZJZk � 0.1<j�m1<k<m

§2. SMOOTH FUNCTIONS 243

2.24. The assertions c) and d) follow from b), b) follows from a) and thefact that Sh(x) = 0 (x —ö/y), a) can be verified directly, and e) followsfrom the equality g" = —gSf/2.

To prove f) use the fact that, by the conditions f"(x0) = 0 and Sf(x0) <0, f'(x)f"(x) <0 in some neighborhood of the point x0, and consider thepossible combinations of signs of f(x), f'(x), and f"(x) on both sidesof x0.

2.25. b) Let a1,..., be the roots of the polynomial f'(x). Then

1 31 1Sf(x) =2(x —

a = f(c) > 0 is the minimal value of 1' in a neigh-borhood of the point c. Then the graph of f looks as pictured in Figure16. Choosing x1, x2, x3, and x4 as shown in the figure (with x2 and x3sufficiently close to c), we have that

f(x3)—f(x2) — f(x4)—f(x3) . f(x2)—f(xl)x4—x1 x3—x2 x4—x3 x2—x1

=b(a+e1)—(b+e2)(b+e3),

where and €3 are small numbers, and b denotes the expression(1(x4)— f(x1 ))/(x4 —x1). Passing in this equality to the limit as x2, x3 —, c,we get that b(a — b) <0, which contradicts the assumption. It is useful toobserve that if a function f E is monotone, then the validity of the

FIGURE 16

Lb

x, x3

244 VIL FUNCTIONS (CONTINUATION)

inequality in the condition of the problem for an arbitrary quadruple ofpoints x1 <x2 <x3 <x4 is equivalent to the condition Sf <0.

§3. Bernstein polynomials

3.1. b) To compute the sums and Sn iuse the binomial formula

and the equality

O<k<n O�k<n

Compute the sums 5n%2' and with the help of the recursionformula a).

3.2. a) Comparing the sums an,o and 5n ,2(see problem 3.1), we get

that

O<k<n

1 x(1—x) I=—S (x)=n,2 —

b) Compare the sums Cnö and 5n,4 and use the inequality Sn,4(X) <

n2/4.3.3. a) Use the result in problem 3.1.3.4. b) Use the inequality

forA=Bn(f,x).

3.5. Show thata)

x)=

1) xk(1 _X)nkl (i — i

b)

B(f, x) =n(n— 1) >(n2)Xk(1 _X)nk2

O<k<n—2

3.6. a) Use the equality _X)nk = 1.

b) It suffices to consider the case when f = 0 (otherwise take the

§3. BERNSTEIN POLYNOMIALS 245

function J = f— Then f(k/n) � 0 if k/n E and hence

O<k<n

k n—k�(supf)2_d

[O,1J O<k<n

It remains to use the inequality a) in problem 3.2.c) Let I = 0. It can be assumed that I � —1 (this can be

ensured by multiplying f by a positive constant). Let = (a, b), =

x) � M xk(l — — > —

O<k<an an<k<an

—

=

Prove that <E2 for sufficiently large n E N (the inequalityis proved similarly). Since the terms in the sum are nondecreasing, itsuffices to prove that

Mn (['iI) — < —

x I nMni <I\1—xJ

The left-hand side of this inequality decreases as x increases. Therefore,we need only verify it for x = a. Since

....,

— p)Pl (see

problem 11.1.8), it suffices to show that aa(l — < — for<a, but this is obvious, because the left-hand side of the last inequality

decreases as a increases, and coincides with the right-hand side for a =3.7. Applying the inequality b) of problem 3.6 to the functions f— g

and g— f, show that x0) — x0) —, 0.3.8. a) Apply the inequality b) of problem 3.6 to the functions 1(x) —

f(xo) and — f(x).b) Fix an e > 0 and choose a number 5 = > 0 such that If(t)—f(x)I <

e if x E t E [0, 1], and It—xI <5. Let M = sup1011 fl. With the help

246 VII. FUNCTIONS (CONTINUATION)

of the inequality a) in problem 3.2 we get that

IBn(f, x) - f(x)I=

- 1(x)) -

O<k<nIk/n--xI<o

fn\ k n--k+

O<k<nIk/n--xI�ö

— 252n

Thus, Bn(f, x) — f(x)J <2€ if n > M/(2e52) and x Ec) It suffices to consider the case when 1(x) = 1 for x � x0 and 1(x) = 0

for x > x0. Then x0) = — To compute thissum, break it up into two parts S1 and S2 by singling out the contributionof the terms with "small" indices:

fn\ k n--k

O<k<n(x0—ö)

n(x0--ö)<k<nx0

(here 5 = is a small positive parameter whose choice will be made moreprecise below). Estimating the sum S1 with the help of the inequality in3.2a), we see that S1 = 0(1/no2). With the help of Stirling's formula andTaylor's formula we get that for the terms in

—

=exp

(+ 0(n03)).

We now assume that S = has been chosen in such a way that —, +oo

and —, 0 (for example, = n215). Then S1 = 0(1), and

= o(1)+ n(xo

= o(I) + (1 + o(1))tJ2( —

(ro exp(-n(x0—t)2 d,+0(!

\_2x0(1—x0)j1

otJn/2x0(1--x0) 2

=o(1)+7=j

§3. BERNSTEIN POLYNOMIALS 247

3.9. a) From the assumption it follows that x)dx = 0 forn E N. Since 4 f on [0, 1] (see problem 3.8b) for = [0,it follows that f2(x)dx = x)dx = 0, and hence

b) It is clear that + — x)2)" dx = 0 for all a, b E Rand n E N. Let J(x) = 1(x) — (ax + b), where a and b are chosen so that

f(x)dx = 0 and f(x)xdx = 0. Since = 0,the numbers = J(x)? dx satisfy the recursion relation

(n + 3)(n + = 2(n + 2)(n + — (n +

Using the equalities

Mo=jJ(x)dx=0, M1=jJ(x)xdx=0,

we get from this that = 0 for all n = 0, 1 It remains to use theresult of problem 3.9a).

3.10. Show that it is possible to weaken the conditions on g, requiringinstead of membership in C2([0, 1]) that its first two derivatives vanishat the endpoints of [0, 1]. Indeed, for any such function g and for anynumber A E (0, 1/2) the function equal to g((x — A)/(1 — 2A)) forx E [A, 1 — A] and equal to zero otherwise satisfies the conditions of theproblem. Therefore,

0=jf(x)g(x)dx=j dx

= f(A+x(1

This implies that f(x)g"(x) dx = 0. In particular, this equality is validforfunctions g ofthe form p>2 and n�0.Passing to the limit as p —, 2, we get that f0' —x)2)" dx = 0 forany n � 0, and this is equivalent (see problem 3.9b)) to the linearity of 1.

3.11. Apply the result of problem 3.8b) for A = [0, 1] to the functionf(x) = f(arccosx).

3.12. a) Formulate and prove the two-dimensional analogue of the in-equality a) in problem 3.2, and then modify the solution of problem 3.8b).

3.13. Prove by induction the identity

x) = >(n_r)

O<k<n--r

where Arf(y) = f(y + — 1(y) and = 21f(y)). Use theequality

= + Lo), where 0 E (0, 1).

248 VII. FUNCTIONS (CONTiNUATION)

3.14. a) Suppose that the number M> 0 is such that 11(x) — �Mix — yla for all x, y E [0, 1]. Then

x) -f(x)I = (i - f(x)) -O<k<n

k cxfn\ k n--k�M>2 ——xO<k<n

Let Pk = — It follows from the result of problem 3.lb) that

1k \2 x(1—x)= 1 and — x) =

O<k<n

Using the Holder inequality (see problem VII.1.29), we get that(2--a)/2 , a/2

Pk) ()

=

b) To get a lower estimate of Bn(fa, 1/2) consider the sums

= xC(i — — nxIu.

O<k<n

It follows from the HOlder inequality that sa+b(x) �if p, q> 1 and 1/p + 1/q = 1 Choose parameters a, b. and p suchthat a + b = 2, ap = a, bq = 4 (a = — cr), b = 4(2 — —

p = q = (4—a)/(2—4). In this case the last inequality takes theform

s2(x) �Since (see problem 3.lb)) s2(x) = = nx(1—x) and s4(x) = �n2x(1 — x), it follows that — x) � sa(x). Consequently, �x(1 — x = 1/2 we get that

Bn(fa,

3.15. a) We write x) in the form x) =where Pk = — x)hik. Since Pk = 1 and kpk = nx(see problem 3.lb)), we get with the help of inequality (problemVII.1.1) that x) � = 1(x).

b) Assume the opposite: exist numbers a, b. and x such that0 � a <x <b � 1 and

x—a b—x1(x)> r—f(b) +

§3. BERNSTEIN POLYNOMIALS 249

It will be assumed that 1(a) = 1(b) = 0 (this can be ensured by subtracting alinear function from f). Moreover, it can be assumed that 1(x) = max[a bJ

It follows from the result of problem 3.6c) that x) <1(x) for suffi-ciently large n, but this contradicts the assumption.

3.16. To estimate the quantity

= x) - 1(x)- x(1- x)t(x)

write the difference x) — 1(x) in the form

x) — fix) = xk(i — (i — 1(x)).O<k<n

Use the following variant of Taylor's formula:

f(t) = 1(x) + f'(x)(t — x) + — x)2 + ço(t, x)(t — x)2,

where the function is continuous on the square [0, 1] x [0, 1] and equalto zero for t = x. Setting t = k/n, we get that

1(t) = f(x)+f'(x) x)

Substituting this expression in (*) and using the result of problem 3.lb), wearrive at the equality

Now fix an arbitrary number e > 0, and let 5 = > 0 be such thatx)I <e if It—xI<o. Then

k/n—xj<ö k/n—xj>ö

M� +

< x(1—x) Mx(1—x)_e

where M= sup Thus, � 2ex(1 — x)/n, if n > M/e52.3.17. See [40] about problems 3.17 and 3.18. Let

—

1(f) = j f(x)(x(1 — x)g(x))" dx.

250 VII. FUNCTIONS (CONTINUATION)

It follows from problem 3.16 that —, I(ço) for any function E

c2([O, 1]). To prove convergence in the general case, write the sumin the form = where

fn\ fk\ I If

It suffices to see that the sums are bounded. Indeed, suppose

that � C, and let M = I(x(1 — Fix an

arbitrary number e > 0 and choose a function E C2([0, 1]) such that11(x)— <e (a Bernstein polynomial BN(f) with sufficiently

large rndex N can be taken as problem 3.8b) with = [0, 1]). Then

—I(f)I � 11(1— +

(M + C)e + —

Consequently, —, 1(f) for any function f E C([0, 1]) if>.0<j<n = 0(1). To estimate this sum use the smoothness of the func-tion g:

=g(L) +g'(L)

This gives us that

� (i-

+ k' 0<k<n

(i(kJ)2

Since g is equal to zero outside the interval (a, b), this implies that

� + +0<j<n

§3. BERNSTEIN POLYNOMIALS 251

where

ni I

'\nJ '\ flJ I

an<j<bnl I

"an�j<bnI

O<j<n O<k<n

Changing the order of summation and using the result of problem 3.lb), weget that

n2—1

O<k<n

It remains to prove that the sums and are bounded. To estimatewrite it in the form

an(j<bnI

where = t E [0, n]. Using the result of problemVL3.15b), we get that

= j + o (f dt).O<k<n

Since the function changes sign on [0, n] only twice,

J 19,.(t)Idt <4 maxJ —

Uncomplicated computations show that

max .(t) = and max W.(t)I = +

It follows from the equality = — t)/(t(n — t)) that for / E[an, bn]

COflSt(t) 0 1

max < — max = — n—f) ).By the equality

n+1 1

n+IJII

________

J ço1(t)dt = n t (1 — = n(n +0

252 VII. FUNCTiONS (CONTINUATiON)

this gives us that

= an<j<bn

+ o -J)j) - = 0(1).

The sum is estimated similarly.3.18. Use the results of problems 3.17 and 3.lOa).3.19. The necessity of the condition 1(0), 1(1) E Z is obvious. But if

it is satisfied, then the polynomial — differslittle from the polynomial x):

O<k<n O<k<n

1�k<n 1<k<n

It remains to use the result of problem 3.8b) with = [0, 1].

3.20. To compute 0) use the identity given in the solution ofproblem 3.13.

3.21. a) Show that L'&kf 0 for k > m and use the formula for x)in the preceding problem. The uniform convergence follows from the resultof problem IV.5.13.

b) Write x) in the form

x) = nm ktm)

= nm—F(0)O<k<n

where = — = (xe' + 1 — x)'. It is easy to seethat

=j�1

and all the coefficients are nonnegative. Therefore,

=l<i<n:

in (n) ni d F1� max(1. lxi ) = max(1, lxi )-pr(O).l<i<n,

It remains to observe that P1(z) = and hence = n".Regarding this problem see also paper [2J.

3.22. a) Obviously, we can assume that r> 1. Fixing an arbitrary num-ber e > 0, we choose an index N = large enough that lc,,11r?P <e.Let P(x) = c,,1x" and let p(x) = 1(x) — P(x). Then — I =

§4. ALMOST PERIODIC FUNCTIONS AND SEQUENCES 253

— P + — p. The difference — P is uniformly small on[—r, il if n is sufficiently large (see problem 3.21a)). Since

Ip(x)I � kmI Ixim �m>N m>N

it remains to estimate x). To do this, use the inequality in 3.1 ib):

x)I � Ixim)

m>N

ICmfr <€.m>N

A sharper result was obtained by Kantorovich in [9].b) Use the inequalities 3.15a) and 3.21b).3.23. a) Make the change of variable y = xf(1 — x).b) Use the fact that multiplication by x + (1 — x) does not change the

polynomial.c) The necessity of the condition is obvious. To prove sufficiency com-

pute the coefficients Let P(x) = >0<j<m Make the change of

variable y = xf(1 — x). Then it follows from the equality =

that = >.0<k<n Equat-ing the coefficients of equal powers of y, we get that

O<j�min(m ,k)

=

Therefore, > 0 for k = 0 n for all sufficiently large indices n ifthe polynomial P is positive on the whole interval [0, 1]. In the generalcase we represent P in the form P(x) = — x)1', where > 0on [0, 1]. Since the coefficients of the expansions of the polynomial Pin the basis {xk(l — x)

bare positive, this representation

implies that the coefficients of the expansion of P in the basis{xk(l _x)n_k}o<k<fl are positive for k = a, a + 1 n — b and are equalto zero for the k.

§4. Almost periodic functions and sequences

4.1. Let 5(x) = x — [x] and let = Since is irrational,

L'n for Note that mod! forany n.m EZ.For any N> If e there are two points among p separated by a

254 VII. FUNCTIONS (CONTINUATION)

FIGURE 17

distance less than a. Let k1, k2 (k1, k2 � N) be indices such that 0 <— Pk <a, and let m = k1 — k2. Then 0 <Pm <a. Let r E N be

such that 1—a <Prm <1. In this case if E(a—a,a+a), then eitherPn+m or falls in the same interval. The same can be said of one ofthe points Pn—m and Pnrm Thus, the distance between adjacent solutionsof the equation a (mod 1) does not exceed L = rimi, which provesthe second assertion.

4.2. a) Consider the set r = {(x(t) = y(t) = E R}.Verify that for irrational the sets = {x(t)Iy(t) = 0} and 4,, ={y(t)Ix(t) = 0} are dense in [0, 1) (use 4.1). The "curve" r consists ofsegments (see Figure 17) obtained by intersecting the square [0, 1) x [0, 1)with the family of parallel lines having slope and passing through allpossible points of the sets x {0} and {0} x 4,,. It is called an irrational

winding of the torus T2 = [0, 1)x[0, 1) R21Z2. We suggest that the readerfollow the motion of a point (x(t), y(t)) with, for example, x(t) = t mod 1,y(t) = mod 1 on a sufficiently large interval of the parameter t. Itfollows from the structure of r described above that this set is dense in[0, 1) x [0, 1), which is equivalent to solvability of the system (1).

Consider the sequences tk = (k + and Yk = Y(tk). As established

in problem 4.1, there exists a relatively dense set of k E Z such that Yk a2.At the same time, Xk = x(tk) = a1.

b) We pass to integer solutions of the system (1). The irrationality ofand leads to the fact that the points = (Xk, Yk) are pairwise distinct,where this time Xk = x(k) and Yk = y(k). Then the have an accumula-tion point. Consequently, for any a > 0 there exist k1, k2 E Z such that thecoordinates and j of the point E R2 satisfy the condition <aand <a. It follows form the irrationality of that is irrational,

and hence the winding F' = E R} of the torus is dense inthesquare T2. Let m=k1—k2. Thenthepoints Ppm' rEZ,forman

a part of F, and hence also in the square.Let = be a point such that —a11 <a and —a21 <a.For the four open squares Qi,..., Q4 in T2 with sides of length a and

§4. ALMOST PERIODIC FUNCTIONS AND SEQUENCES 255

abutting on the angles in T2 choose positive integers r1,..., r4 such thatthe point falls in Q1 (one of the numbers r1 can be taken equal to 1).Then for any n that is a solution of (1), one of the numbers n + r1m is alsoa solution of (1) (as is one of the numbers n — r1m) (see Figure 18). Thisimplies that in each interval of length L = max1<1<4 r1ImI on the line thereis at least one integer solution of our system.

4.3. Prove that the system is solvable for any Let =and = j = 1 k, n E Z. Then = 0, and thepoints = E Tk4 = [0, 1) x x [0, 1) have an

k--i times

accumulation point. Arguing as in the solution of the preceding problem,find an n0 � 1 such that <e.

4.4. Prove that for the real solvability of the system for any a1 akand e the numbers must be rationally independent (incommen-surable), that is, the relation + ... + = 0 (n1 E Z) must holdonly for n1 = •.• = n E Z, and suppose thatthe system has a solution t for any values of a i,..., ak and e. Then

<e, i= 1 k,forsome m1,..., EZ. Consequently,< e is

arbitrary, this implies that n1a1 E Z. Since also a1 ak are arbitrary,= ... = nk = 0. Using problem 4.2a), prove by induction that the rational

incommensurability of the numbers ensures the existence of arelatively dense set of solutions.

4.5. Show that a necessary and sufficient condition for a positive an-swer to the questions a) and b) is the rational independence of the numbers

mod 1. Assuming the existence of an integer solution m and fix-ing arbitrary numbers n i,..., nk E Z, we arrive as in the preceding problem

FIGURE 18

256 VIL FUNCTIONS (CONTINUATION)

at the inequality

If mod 1 = 0, then Z, and hence n1a1 Z in viewof the arbitrariness of e; as earlier, this implies the equalities n1 = =

= 0. The proof of sufficiency is obtained by generalizing the argumentsin the solution of problem 4.2b).

4.6. If a and b are commensurable (i.e., the number a/b is rational),then the function f is periodic, and the mean value of f on a period isequal to zero. If a/b is irrational, then there are arbitrarily close maximum(minimum) points of the two terms (see problem 4.2a)).

4.7. To verify the first assertion consider the system of equationsk 1 n, and use problem 4.3 (see also prob-

lem 4.15).The• assumption of periodicity for f implies the identity

— 0 is a period). Setting bk = ak(eL — 1) anddifferentiating this identity at zero, we get the equalities = 0, m =0, 1, 2 Since the are pairwise distinct, the Vandermonde determi-

nant is nonzero, and hence all the bk are equal to zero. Butthis is possible only if Z, that is, the are pairwise commensu-rable.

4.8. Use problem 4.4.4.9. Suppose that there is an L> 0 such that for any e (0, the

interval [1, 1 + U contains a for which

x€R. (1)

Since the closed interval is compact, there exist a sequence 1 0 and a tsuch that ç —' Passing to the limit in (1). we see that r is a period of

1.4.10. Uniform continuity and boundedness are proved in almost the same

way as in the periodic case. For example, to estimate If(x) — f(x,)I use analmost period taken in the interval (—.v1. —x1 + L). The answers to theremaining questions are as in the periodic case: they exist; it does not follow.

4.12. Consider the functions = (f(x+h)— 1(x)) (h > 0). Theyare uniformly almost periodic and tend uniformly on R to 1' as h —, 0.

4.13. For simplicity suppose that F is real, let M = F(x), in =infR F(x), and e > 0, and let and x, be such that F(x1) < m -i-e/6 andF(x2) > M — e/6. Define d = Ix — x21 and let L = L(e/3d) be a numberchosen according to the definition of almost periodicity for f. If is an

§4. ALMOST PERIODIC FUNCTIONS AND SEQUENCES 257

(e/3d)-almost period of f, then for z1 = x, + I = 1, 2,fZ2 p

F(z2) — F(z1) = F(x2)+ J

f(x)dx — F(x1)— J

f(x)dxx2

1x2

= F(x2) — F(x1)+ J

(f(u + — 1(u)) du

=M-m-e.This implies that each interval of length L + d contains points z1 and z2such that F(z1) < m + e and F(z2) > M — e. Then for any x and any

E (x, x + L + d) with F(z1) < m + e we have that

p: pz+T=F(z1+i)—F(z1)+J f(x)dx—J f(x)dx

= F(z1 + - F(z1)- JZI

_f(u))du

> m — (m + e) — (L + d)e = —e(1 + L + d).

It can be proved similarly that F(x + t) — F(x) <e(1 + L + d) for all x,and this yields what is required.

4.14. Necessity. Consider the sequences {f(x + x E and, usingthe diagonal process, choose a subsequence k1 = such that the sequence

{f(x + k1)} converges for all x Q. To prove the uniform convergence ofthe sequence {fk } on R use the uniform continuity of f.

Sufficiency. Assume the contrary: there exist an 0 and a sequenceof open intervals with lengths tending to infinity that do not contain

e0-almost periods of f. Choose a sequence {hk} and a subsequenceof intervals in such a way that > 3<k 1h1 — h31, hk+I —

for i = 1 k. Considering supXER Ifh(x) — fh(x)I, i < I, prove that

the sequence {fh} cannot contain a uniformly convergent subsequence.4.15. For the case of a sum use the criterion of almost periodicity from

the preceding problem. In the case of a product use the equality [g =((1+ g)2 —(1— g)2)/4.

4.16. Let L = L(e) be a number such that any interval of length Lcontains an e-almost period of f, and let C = sup,ER If(t)I. Using analmost period taken from the interval (a, a + L). a E R, prove theinequality

1(x) dx — <e+

Deduce from this that the estimate 19,(nT) — < e + (CL/T) (n EN) holds for ço(T) = 9jfTTf(x)dx. Then prove that —

<2€ + + for and T2 with T1/T2 E

258 VII. FUNCTIONS (CONTINUATION)

4.17. a)We explain only why the equality f) = 0 implies that I 0.Suppose that If(x)12 > e > 0 in some interval (a, b). Then in each

interval of length L = L(e/2) there is a such that f(x)12 > e/2 forx E (a + b + Consequently,

1T

2 1 2T1 e b—a

2TJ.Tc)Let dkEC, k=1 n.Then

0�(f—g,f—g)

_____

= 1)— — +

= + >2(dk — — —

= (1.

Setting dk = , we prove the inequality.The countability of the set of nonzero "Fourier coefficients" follows

from the fact that, since M(1f12) is finite, the set � I/n} is finite foreach n E N. On the other hand, for each countable collection ,

... andfor the arbitrary absolutely convergent series ak the series is

uniformly convergent on R, and its sum I is a UAP function (see problems4.7 and 4.11). In view of uniform convergence,

(afor

The reader can become more familiar with the details of these questions, forexample, in [17J.

4.18. Use the equality = +4.19. The uniform almost periodicity of f implies the periodicity of

{ek}kEz.4.20. False. It is convenient to describe a corresponding example in the

notation of problem 4.23. Let A0 = 0,

(PEN).

4.21. Let p=(m, Considerthe points (m±1, n), (m, n±1)adjacent to p. Note that the projections of the segments joining (m + 1, n)to (m, n + 1) and (m — 1, n) to (m, n — 1) on the line perpendicular tothe sides of the strip are equal to the width of the strip, which implies thatone and only one of the endpoints of each of these segments belongs toJoining each point in to the two neighbors by segments, we get an infinite

§4. ALMOST PERIODIC FUNCTIONS AND SEQUENCES 259

polygonal curve lying in S with components parallel to the coordinate axes,which easily implies the first assertion.

Let us prove the second assertion. Considering the distances from thesuccessive vertices p1 of the polygonal curve to the line y = kx, divided

by we get a sequence {xj}jEz, x1 E [0, 1). Then x1 =1/(k + 1) = a if the edge is vertical (it will be assumed that in thiscase = 1), and x1 = —k/(k + 1) = a — 1, if is horizontal(; = 0). Thus, if it is assumed that p0 = (0,0), then x1 = mod 1.Let €s+i be an arbitrary word, and let =5 = 1 — It is clear that each sufficiently long segment of Z containsa number 1 such that E (1 — 5, 1) (see problem 4.1). It is easy to seethat the differences — i = 1 1, are then the same, and hencethe "words" and coincide.

The analogous problem in space leads to the construction of an interestingpartition of the plane. Suppose now that S0 is the open layer in located

between two parallel planes P and Q with PflZ3 = {(0, 0, 0)} and QnZ3 ={(1, 1, 1)}, and let S = U P. Projecting the segments joining pairs ofadjacent points in S n Z3 orthogonally on P. we get segments boundingcertain parallelograms. These parallelograms cover F, and their interiorsare pairwise disjoint (see Figure 19). The covering obtained is remarkablein that it consists of parallelograms of all three sorts, is not periodic (that is,does not pass into itself under some translation), but is quasiperiodic in thesense that any finite part of it occurs infinitely many times. The proofs ofthese facts are left to the reader.

FIGURE 19

260 VII. FUNCTIONS (CONTINUATION)

4.22. First show that if A = = €1 •then e has the form

A°A'A'A°A'A°A°A' , where A° = A, A' = (1 —e,,..., 1 — thatis, the upper indices again form the sequence e. Let B be any word in e.If n is large enough that the word contains B. then, as is easy to see,

B is contained in each word of length 2n+2

4.23. It is useful to single out the case when there exists an n such that= . A,, for all p � n. In this case e has the form

and hence turns out to be periodic.In the opposite case, for a word B in e choose some word of the formthat contains B. Then find a k > 0 such that the words and

necessarily appear in Afl+k. Since e is obtained by concatenating A,+k and

, every word whose length L is double the length of containsand hence B.

It is interesting to note that besides the case mentioned above, e is periodiconly if = APA . . . for all sufficiently large p (verify this).

4.24. Obviously, s has the form As2kl4kAs6kAs8k. , where k =

(n E N), A = A = . . This at once yields almostperiodicity.

If a(n) denotes(_1)S (n E N) , then the sequence {a(n)} is characterized

by the following properties:

1) a(2n+ 1) = 2) = 1)

is again a sequence of folds. Assume that the sequence has period= 2a(2b + 1). Then a(n + 2a(2b + 1)) = a(n) for all n � n0. Setting n

equal to 2(2Hm, we arrive at the equalities + 2b + I)) =m � in0. Denoting by aa(rn), m � 1, we get a sequence aasatisfying 1) and 2). Consequently,

aa(2m + 2b + 1) = ( l)m+ba(l) = (7a+i(in), ifl � 1fl0.

But then aa+2(m) = = const, which is impossible.

As a consequence of aperiodicity we remark that if with a number x E

[0, 1) whose infinite binary expansion has the form x = (forsuch a representation to be unique we assume that among the there arealways infinitely many zeros) we associate r = 0.s1s2..., where isthe sequence of folds given by the equalities = n = 0, 1

then we get a function on [0, 1) that is continuous at irrational points andright-continuous at dyadic rational points, and takes only irrational values.

4.25. Take numbers of the form m2', in E N, for (these numberssatisfy the condition a)) and write s in the form ASk.4S2kAS3k.4S4k... with

k = where A and are words of length k — 1 (see the solutionof problem 4.24). Since the length of the word is equal to 2', atleast (1—1/k)n of then numbers /=1 n,willbe

§4. ALMOST PERIODIC FUNCTIONS AND SEQUENCES 261

zero (corresponding to those falling in some word of the form A or A).Consequently,

1 1

for sufficiently large 1, which implies the properties b) and c).4.27. Each finite word w in r is contained in some word Verify

that r has the form B1B2 where each word B1 has length andcontains Therefore, any word in r of length contains andhence w.

We mention some curious properties of sequences s of folds (see problem4.24) and the Rudin-Shapiro sequence r.

If S is defined by the equalities 52k = k mod 2, k = 0, 1,2 thencoincides with Si — n) mod 2, n E N (r1 = 0). An inter-

esting polygonal curve oiithe plane is also connected with these sequences(see Figure 20, the solid curve). It is composed of alternating "horizontal"and "vertical" components of equal length, and the turns (by 900) of thepolygonal curve to the right and to the left correspond, as in problem 4.24,to the 0's and l's of the sequence s. This polygonal curve is connected withthe sequence r as follows: the numbers r1, r2, ... correspond to the hori-zontal components of the polygonal curve, taken in order, where = 0 ifthe nth horizontal component goes from left to right, and = 1 if fromright to left (the sequence of "up-down" passages of vertical components isalso reminiscent of the Rudin-Shapiro sequence).

With the polygonal curve corresponding to the sequence s it is possibleto associate a new polygonal curve, pictured in Figure 20 by the dotted line. It

III I I

FIGURE 20

262 VII. FUNCTIONS (CONTINUATION)

is obtained from the solid line as follows: we replace each component by twonew components serving as the legs of an isosceles triangle with the originalcomponent as hypotenuse. Further, the vertex of this triangle deviates fromthe solid polygonal curve alternately to the right and to the left. Note thatthe new polygonal curve was obtained from the old one by rotation andcontraction by a factor From the second polygonal curve it is possibleto construct a third one in an analogous way, then a fourth, and so on. In thelimit such a sequence of polygonal curves fills a sector in the plane boundedby two sides of a 45° angle, that is, they determine a curve of Peano type.Try to carry out an independent investigation of curves arising in a similarway from sequences of folds with a different alternation of 0's and l's in thesequence (for example, 1), and also of the corresponding limitsets on the plane. More details on the questions touched upon in problems4.24—4.26 can be found in the papers [35J and [42J.

CHAPTER VIII

Lebesgue Measure and the Lebesgue Integral

§1. Lebesgue measure

1.1. Use the equality fll<k<m Ek = (0, ')\Ul<k<m where =(0, l)\Ek. Provethat 1.

1.2. Consider the sets Ek = (k = 1 m) and, using problem1.1, show that fll<k<m Ek 0. Take a point z0 in the set fll<k<m Ek.

1.3. a) Partition the space Rm into cubes with vertices at the points ofthe lattice Zm:Rm = Q,, where Q, = {! + E [0, 1)m}. Then

"translate" all the sets En Q1 into the cube(') [0, 1)m, that is, consider thesets E, = {x—lIx E EnQ,}. Since = = 1,the sets E, cannot be pairwise disjoint. Consequently, there are distinctvectors 1',!" E Z such that E, nE,, 0. Let a E E, nE,. Thena = x'— 1' = x" = 1", where x', x" E E. Thus, 0 x'— x" = F'— 1" E Zm.

b) Consider the set E = {x/21x E V} and use a).c) Consider the set E = {x/21x E V}, and, arguing by analogy with the

solution of a), show that there is a point belonging to at least N + 1 of thesets E1. Deduce from this that E contains distinct points x1,..., XN+lsuch that Xk — X1 E Zm.

Consider an index s such that does not belong to the convex hull ofthe remaining points. Verify that the points ±(xk — with k s are alldistinct, and hence are what is required.

1.5. Prove that the measure of the set of points in (0, 1) with 0 notamong the first n digits of their decimal expansions is equal to (9/10)1?.

1.6. a) Consider sets E1? such that — ).(E1?)) < 1 Use the sameidea as in the solution of problem 1.1.

b) Consider the sets E1? c (0, 1) consisting of the points with nth decimaldigits nonzero, and verify that E1?) = (9/10)m (cf. the solution ofproblem 1.5).

1.7. Consider the sets Hq = (q = 1, 2, ...)and verify that A C Uq>m Hq for any m E N.

1.8. Since Lebesgue measure is regular, there is an open set G j E such

(1)In what follows we shall call cubic cells the translator of cubes of the form [0, a)m

264 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

that(*)

Let G = where are disjoint cubic cells. It follows from (*) thato < >).(E n Therefore, for at least one index n = n0 theinequality n must hold. Setting A = we get the

required result.1.9. Let H = E — E, and let A be an interval such that 3).(A)/4 <

nA) (see problem 1.8). Prove that H ).(A)/2).

Let lxi Verify that (EnA)n(x+EflA) 0. Indeed, otherwise,

=

= 2).(EnA) > 3).(A)/2.

On the other hand, since lxi have that <3).(A)/2and hence � � contradicts

the inequality (*).Let y E (EflA)fl(x+EflA). Since y E x+EflA, there is a point x' E

such that y = x + x'. Thus, x = y — x', where v, x' E E fl A C E. that is.x E H. This proves the inclusion ).(A)/2) C H

b) This assertion follows immediately from a).1.10. Show with the help of problem 1.9c) that IntE 0, and use the

result of problem 1.1.19.1.11. a) If = 0, then = 0, and hence (see problem 1.1.34)

> + 1 infinitely many times. But if > + 1 for j E N,

then in the binary expansions of the points in the digits at the placeswith indices nk + 1 are equal to 0, and this implies that = 0 (cf. the

solution of problem 1.5).b) Verify that if — � 2+ log2 m, then in the binary expansions

of the points in E(m) the digits having indices + 1 are equal to 0.

c) Use the representation

H= U E2m)_ma,mn=l k�l

1.12. Verify that for any positive integer n the set E can be covered byintervals of length 1/n!.

1.14. It follows from the definition of the set E of Cantor type on [0. 1Jwith defining sequence that ).(E) + — = 1 , where

= 1. Therefore, ).(E) > 0 if and only if — < 1. Sincethe last sum can be arbitrarily small, the measure of a set of Cantor typeconstructed on [0, 1J can be arbitrarily close to 1.

1.15. Using the hint for problem 1.14, prove the following.

LEMMA. Let be an arbitrary finite (nonemptv) open interval, and let0 < 0 < 1. There exists a compact set K C A such that:

§1. LEBESGUE MEASURE 265

a) IntK = 0; b) = c) consists of disjoint open intervalswith lengths not exceeding half the length of

To get the Set A required in the problem, carry out the following con-struction. Let = (0, 1), > 0, and E1>0 < 1. Use the lemma (with

o = Oo) to construct a compact subset K0 of Let (n E N) be theopen intervals making up the set and let be compact subsets ofthese intervals constructed according to the lemma (with 0 = 0k). Next, let

(n E N) be the open intervals making up the set U Kr),and let be compact subsets of these intervals constructed according tothe lemma (with 0 = 02). Continuing this process, use induction to get afamily of sets � 1.

Let A = K0 U U1 Verify that

(*)j=0

Show that any (nonempty) interval C contains some interval andhence

and

Verify, moreover, that

j U

j=I+ln—I

and, replacing by and A by A' = U U1 prove that

the equality = holds together with the equality(*), which implies that > 0.

1.16. Assume first that A is a convex polygon with vertices M1,...,and let L be the length of the boundary polygonal curve. Together witheach segment (j = 1, 2 n, = M1) of the boundarypolygonal curve consider the open rectangle lying outside A such thatthe length of the side perpendicular to is equal to e. It is clear that

A U U,where is the sector of the disk B(MJ, e) included between the rectangles

and for j> 1, and between P1 and for j = 1. It is easy tosee that the sum of the central angles corresponding to these sectors is equalto 2ir. Therefore,

= + +l<j�n l<j<n

= 22(A) + + ire2

266 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

The case of an arbitrary compact convex set is exhausted by approximatingit by convex polygons.

1.17. Obviously, it suffices to consider the case when <00. Letthe set G be represented as a union of pairwise disjoint cubic cells: G =

In each cell we inscribe a ball It is clear that

= where = 1)). Therefore,

=

= (1 =

where q = 1 — < 1. Let G1 = The set G1 is open, and

by repeating the procedure described above, we get balls such that

a family of disjoint balls such that

UBr) 0k=Oj=l

1.18. Denote by a Lebesgue measure on S'. If measurability of at leastone of the sets is assumed, then in view of their congruence and theinvariance of the measure a with respect to rotations all the sets will bemeasurable, and the equality = is valid for any n, m E N.Using the countable additivity of the measure, we get that

a(S') = a (u = =n=l n�l

Thus, 0 < 1a(E1) <00, which is impossible both in the case whena(E1)>0 and iiithecasewhen a(E1)=0.

1.19. Partition the numbers in [0. 1) into equivalence classes by regard-ing two numbers as equivalent if their difference is rational. Let C be aset having precisely one point in each equivalence class, and let C0 be theimage of C under translation by 0 modulo 1. We number all the rationalnumbers in [0, 1) in the sequence {Oj and let = C9 . Since translationmodulo 1 preserves the measurability and the measure of a set, the setsare measurable or not measurable simultaneously, and they have the samemeasures if they are measurable. Assuming that the are measurable andarguing as in the solution of problem 1.18, we again arrive at the impossibleinequality 0< A(E1) <00.

1.20. It will beassumed without loss of generality that E C (0, 1). Letthe be the sets constructed in the solution of problem 1.19, and let =

§2. MEASURABLE FUNCTIONS 267

E n E thatat least one of them has positive measure, and hence

contains (distinct) points separated by a rational distance (see problem 1.9b)),which contradicts the construction of the sets

E if xthe x E an

the form (k + n = 0, 1, 2 k =0, 1 2" — 1, and x' is the point symmetric to x with respect to themidpoint of XEE ifandonlyif x'

E that =for any open interval a C (0, 1). The latter contradicts the result of

problem 1.8.1.22. Let Q=[0,1)tm, Q,=!+Q,and E,=—!+EflQ, (!EZm).

Verify that E, = Q and E, fl = 0 for /P = [—c, C)m. It suffices to verify that

+ E)) = 0. Let e be an arbitrary positive number, and letQ be a cube such that (see (problem 1.8)

(1)

Show that there exist points a1, a2 aN E A such that

P C IJ + Q) and + Q) (2)1<n<N

Then

P\U(a+E)C U (3)

aEA

and in view of (1) and (2)

+ Q)1<n<N (4)

�Since e is arbitrary, (3) and (4) give us the equality

'tm (P\Ua+E) =0.aEA

§2. Measurable functions

2.1. Prove that for some e > 0 the set E El lf(x)l 1 —

has positive measure, and consider the functions f+ and 12 =

f_2.2. Consider the function 1(x) = E where = x10, +oo)'

and is the set of rational numbers, enumerated in an arbitrary way.2.3. Consider the characteristic function of the set in problem 1.15.

268 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

2.4. b) Find the lengths of the images of the intervals of nth rank usedin the construction of the Cantor set (see problem 1.1.27).

c) Use the result of problem 1.20, which ensures the existence of a non-measurable subset of the set g(K).

2.13. We prove the first of the required equalities. Suppose that c1 <C2 <and let = +00. Since the function f is strictly decreasing from

+oo to —oo on each of the open intervals (ck, ck+l) (k = 1, 2 n — 1)

and from +oo to 0 on (ca, +oo), the set E, = {x E RIf(x) > t} is theunion of the intervals (ck, xi), where Xk is a point such that f(xk) = 1,Ck <Xk <Ck+l. Thus,

= — ck) = — Ck.

1<k<n

To find the sum note that the points are roots of the equationf(x) = t or of the equation

= IP(x), where P(x) = [J (x — ck).- k

The last equation can obviously be written in the form

l<k<n J

where the degree of the polynomial Q(x) is less than n — 1. By

theorem, Xk = + Alt, which leads to the required result.The second equality can be proved analogously.

§3. Integrable functions

3.1. It is clear that

f=J0 l<1<N

+ 3j + (x)) dv

6j > dxl�1<J<I%<N

= S1 + 6S2 + 6S3.

where S3 = fl fl To compute 53 note that sinceE,UEJuEk=[0, 11 for I

X[OIJ = XE, + XE, + — XE,XE, — — +

§3. INTEGRABLE FUNCTIONS 269

Consequently,

1

iiEk).

Summing these equalities over 1 �! <j <k � N, we find S3.3.2. It follows from the condition of the problem that >.1<j<N XE (x) �

k forany XE[0, 11.3.5. a) Prove the convergence of the series E dx, where

A is an arbitrary finite interval.b) Let = is the sequence of all rational

numbers, enumerated arbitrarily. Consider the function3.6. Consider the function

JO forx=0,1(x)

— x2sin(1/x2) for x 0.

3.10. a) Let E E}. It is clear that fE lix — dy =

fE dz. Let A = n B(0, r) and C = EX\A. Then =and

E C} � = E B(0, r)}.

Therefore, Ic � � and

j lix — dy= j dz

+ Jd:

� JA B(O,r)\A

= JB(Or)

b) Show thatI

e1X dxl = fE cos(x — dx for some E [0, 2irJ anduse the same idea as in a).

c) Show thatdxdy — ff xdxdy

JJEx+iy — liE' x2+y2

where E' is a set obtained from E by a rotation. To estimate the integrallIE xdxdy/(x2 +y2) consider a set = {(x, y) E R21x/(x2+y2) � t}, onwhich the function x/(x2 + y2) is large and choose t such that the measureof is equal to that of E'. Prove as in a) that

xdxdy < f[ xdxdyliE' x2+y2 liE, x2+y2

for such a choice of t, and compute the last integral.

270 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

3.15. It is clear that the integral fE g(x)lnf(x) dx exists. In the casewhen it is finite use the inequality — 1)/pi � ti +e" (t E R, 0 <p < 1)to prove that

LfP(x)

— 1g(x) dxp-'+O j g(x) Inf(x) dx.

If fE g(x)lnf(x)dx = —00, then consider the functions 15 = max(ö. 1),where 0 < ö < 1 , and pass in the inequality

(j g(x)f(x) dx) (j g(x)J7(x) dx)

first to the limit superior as p —. +0, and then to the limit as ö +0.3.16. b) Use the pairwise orthogonality of the functions — 1/2 (k =

1,2 ii).3.20. Represent the set (a, b)\K as a union fik) of disjoint open

intervals. Then

= >21 ix-' dy.

Therefore,

IKS() dx =dY) dx

(1)" dx

= >21 '-' dy.k KIXYI

Let us estimate the inside integral on the right-hand side of (1). Assumingthat y E (ak' fik) , we get that

f dx dx 1b dxJKIx_yIs+1 Ja J/3 IXYIS+l

if 1 1 \s + (/3

_y)S) � (i).

On the basis of this inequality it follows from (1) that

IKS)� � —a) <00.

3.21. It is clear that a) follows from b). To prove b) partition the cube[—a, aim into the 2m pyramids with the faces of the cube as bases and withvertices at the origin of coordinates. To compute the integral over each ofthese pyramids use Fubini's theorem.

3.22. To avoid the difficulties associated with verifying that the functionis measurable on subsets of dimension less than n it will be assumed thatthe function is continuous. (The general case can be exhausted with the

§3. INTEGRABLE FUNCTIONS 271

help of approximation.) By using polar and spherical coordinates it is easyto establish the induction base. Suppose that the formulas a) and b) arevalid for and ,respectively. Replacing 1(x) by f(x)x10 11(HxII),where X10,11 is the characteristic function of [0, 1], we see that

j f(x)dx= j {j2f(tW)dJLn2(CO)} dt. (1)

We proceed to the proof of the equality in a). Let

n--i n--iS÷ ={(x1 Ixn�O},

n—I n—iS = {(x1,..., xn) ES I; � O}.

We prove that

Isa—'

f(x1,..., xni, xn)duni(x)

=

{j f(u1 sinO Uni sinO, cosO)d/in 2(u)} dO.

Indeed,

f(x1,..., xni,+

= I• f(y1,..., V12)dY

Using (1), we get that

xn_I , xn) djini(x)

=

1' f(tco1,... , d/2n2(CO)} dt.

After the change of variable t = sinO we arrive at (2). It can be provedsimilarly that

J f(x1,..., xni,

= L,12{jf(u1 sinO Un_i sinO, cosO)d/in2(u)} dO.

Let us proceed to the proof of b). By Fubini's theorem,

j f(x)dx=j {J...j f(xi..., xn)dxi ...dxni} dxn.

272 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

We transform the inside integral by using the induction hypothesis, afterwhich we get that

j f(x) dx

= j {j f(ru1 dr} dxv.

We now pass to the polar coordinates = t cos 0. r = t sin0 in the half-plane (xv, r), r � 0. This gives us that

j 1(x) dx

=

{jfl

sin 0, tcos0)}

d0j}

dt.

Using the equality a) already proved, we see that

sin 0, tcos0) dO

= jand thus

j 1(x) dx j (j f(tw) (w)} dt.

3.23. a) Apply the formula b) in problem 3.22 to the characteristic func-tion of the set T, changing the order of integration on the right-hand side.

b) Use a).3.24. Use the formula given before problem 3.24, assuming that f is the

characteristic function of the set E3.25. Using the formula before problem 3.24. we get that

, n/2--l

= (27)_fl/2j dt

= r(t_)

(-)The function is called the x-square distribution with n degrees offreedom, and is frequently used in mathematical statistics.

3.26. By theorem,

Y4(1<a)= (2n)2 11° {jJj dy d_-dt} dx,

§3. INTEGRABLE FUNCTIONS 273

where (Ka)x = {(y, z, t) E R31\/y2 + z2 + � Therefore,

= j 4irr2e1'2dr) dx

arctana 1 00 23 . 2 --p/2

=p sin çoe dp dço.10 LO

We leave the rest of the computations to the reader.3.27. In the section of the set T by the hyperplane x1 = x we get the

set = {(x2, x3, E + + E where

EB}=[a_\1R2_X2,a+\1R2_X21.

Therefore,

=and hence

R RI;{4(T) =1 R3X = J J 4iry2 dy dx.

We leave the final computations to the reader.3.28. Let A' be the projection of A on the x1-axis, and let =

{.v I(x, y) E A}. It is clear that = IA' dx, where =

{(x2, x3, + + E AJ. Since = 4iry2 dy, it follows

that

= J42d} dx =

which is what was required.3.29. Use the equalities

=

1 1=

where A' is the projection of A on the x1-axis, = y) E A}, and

and show by using the spherical invariance of the set that

= j--(n--l)/2 I n—2 --y2/2

= (2ir)IA

e dy.

274 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

3.30. Let = . E < t"}, t > 0.Verify that = Using the equality and Fubini's theorem, prove

that = dt. Expressing the integral in terms ofthe function f, prove the desired formula by induction.

3.31. a) Since the measure is rotation-invariant, it can be assumedthat a = IIaHe1 , where e1 = (1, 0 0). Therefore

U'ZHJ

0n—1

=2 2_xi —...

= {L } dxi.

The inside integral on the right-hand side is the area of the sphereand hence is equal to afl 2(1 — Consequently,

.7 = 2j t"(l —

= I — du0n—1

— f((p + 1)/2). f(n/2) ,,

f((n+p+1)/2) a

Substituting the values of 0n—2 and in this equality, we get the desiredresult.

3.32. (See [36].) Since the measure is spherically invariant, we can as-sume that a = e1 and b = cos 0e1 + sin 0e2, where e1 = (1, 0, 0 0)

and e2 = (0, 1, 0 0). Thus, we should compute the integral .7 =where f(u, v)=(sgnu)•sgn(ucosO+vsinO). It

is clear that

.7=2 f f f(x1,dx1

2xi •

— [1 f(xl,x2)2 2

x2 2 2 3 n—I

§3. INTEGRABLE FUNCTIONS 275

The inside integral on the right-hand side is none other than the area of the

(n — 3)-dimensional sphere of radius — — Therefore,

=f(u, — — du dv

= j dçp.

Obviously, f(rcos rsin ç9) = f(cos sing,), and hence

= j sin dço j(1 —

= afl.3 j2

We leave it to the reader to show that the last integral is equal to 2ir — 40

(see Figure 21, where the sign of cos is indicated outside the disk and thesign of —0) is indicated inside it). Therefore,

2ir 1 2f=2afl3¼10and since = (2ir/(n — we get the required result.

3.33. The assertion a) is a special case of problem 3.34. Using a) and theformula given before problem 3.24, we get that

= j j Jr — pI(rp)' drdp.

0

+

FIGURE 21

276 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

Passing to polar coordinates in the integral on the right-hand side of theequality, we arrive at the equality

2 p/2 • n—i 100 2n u2/2

n2nr2(n/2)Jo dçoj u •e du

çp .= I I cos — — sin sin dço

2'T2(n/2) Jo 2 2

— + 1/2)d

— Jo +

It remains to observe that

[7r12•

dçoJ0 + n—'oo

(see problem VI.2.8) and to use Stirling's formula.3.34. Since A B(O, e) for some e > 0, it follows that � llxll/e,

and hence

— qA(y)l dp(x) dp(y) � j,j

dji(x) <kI<k<n /

Let X,A be the characteristic function of the set tA (t > 0). Then

j u(tA)[1 — u(tA)] dt= j {j j X,A(X)[l — X,4(Y)]d4u(x) d/i(v)}

= j j {j X,4(X)[1 djz(x)djz(y).

(1)If q4(x) > q4(y), then x,4(x)[l — x,4(y)1 = 0; if q4(x) < q4(y), then

— = 1 for q4(x) � t � and X,.1(x)[1 — = 0for t q4(y)J. Therefore,

j x,4(x)E1 — X,4(Y)]dt = max(q4(y) — q4(x) 0).

With the help of the last equality we get from (1) that

j u(tA)[1 — u(tA)]dt= j j max(q4(y)

— q4(x). 0) d4u(x) dji(y)

=1. L q4(x) — qA(y)I dji(x) dp(y).

3.35. Let E C Rm be a measurable set generating a tesselation (see prob-

§3. INTEGRABLE FUNCTIONS 277

lem 1.22). Prove that fE 1(x) dx = fQ 1(x) dx (from this we get the asser-tions a) and b) for E = a + Q and E = A(Q)). To verify this equalityconsider the sets Q1 =1+ Q and E, = —1+ E fl Q, (1 E Zm) and show that

UE,=Q, E,nE,=øIEZ'"

Since I is periodic, it is clear that fE 1(x) dx = fEnQ, 1(x) dx. It remainsto use the countable additivity of the integral.

In the case when A is not an integer matrix consider the example

A= f(x,y)=sin(2irx) (x,yER).

3.36. Let d = diam(T). It can be assumed that the points 0 =(0 0) and = (0 0, d) belong to the set T. Let Q be theprojection of T on the hyperplane = 0. Consider the cone K withvertex at and base Q. Since

= �it suffices to prove that V � To do this, show that the intersection ofK with any straight line F through Q parallel to the is a segmentof length not exceeding the length of the segment obtained by intersecting /with T. Constructing the plane passing through I and the considera point M E T of it that is farthest away from this axis and is located onthe same side of it as 1. Prove that segments of equal length are obtainedby intersecting F with K and the triangle

3.37. Verify that the convex hull of two ellipsoids obtained from eachother by a translation contains an ellipsoid of larger volume. Deduce fromthis that the ellipsoid is centrally symmetric with respect to zero. It can beassumed without loss of generality that 3 = B'1. In this case

� if C T and is an ellipsoid. (1)

Assume that x0 E T, 11x011 > It can obviously be assumed thatx0 = (c, 0 0). where c = 11x011. Consider the convex hull of the

ball B'1 and the points ±x0. (The two-dimensional section S of the setT1 by a plane through the x1-axis is pictured in Figure 22.) Obviously,

C T. Take a point a lying on the x1-axis, 0 < a <c, and consider theellipse E determined by the inequality + (y2/b2) 1, where b2 =

(c2 — a2)/(c2 — 1), and the y-axis is taken in the plane of the section perpen-dicular to the x1-axis. Verify that E C S. From this conclusion it followsthat T1 contains the ellipsoid obtained by rotating E about the x1-axis

278 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

in the space Let V(a) = and

I

2 (n—1)/2 2 2 (n--1)/2

(i_;) •a.

It is clear that V(1) = and V'(l) = 1) > 0. Therefore, fora close to 1 and a> 1 the inequality V(a) > holds although C T,which contradicts (1).

3.38. The continuity and boundedness of and its holomorphicity out-side K can be established with the help of the following theorem (see, forexample, [1], p. 169).

THEOREM. Suppose that p > 1, E C R", < oo, and {fj isa sequence of measurable functions on E. If f —' f0 in measure on Eand fE dx <00, then the functions f'0, (n = 1, 2, ...) areintegrable on E, and fE

If K is totally disconnected, then supplies an example (due to Pom-peiu) showing that Liouville's theorem ceases to be valid if instead of en-tire functions we consider functions that are continuous and bounded on Cand holomorphic on an open dense set. See problem 4.8 (an example ofUrysohn) for a refinement of this result.

22

§3. INTEGRABLE FUNCTIONS 279

3.39. a) Let p, q> —1,and Jq =C2). Then

qL2+y22+v2

2 2 p 2 2 q= (x +y) (u +v ) djz3(x,y,u,v)

2q dxdydu2JJj

2 2 p 2

______________

= (x +y)(1—x —y)

2 2q=4 1 (x2 + y2)P(1— x —y)

Jx2+y2�1

dux

— — — }

dxdy

= (2ir)2[1

—Jo

=2 f(p+1)f(q+1)

2ir2 [ ?(1 — 1)q dt = 2ir f(p+q+2)

c) Note that the series + + C2z2) is uniformly convergent

in the ball 1z112+1z212 1, because +c2;I a < 1,

where ai/1C112+1C212. Therefore,

dji3(z1, z2)z2)

(1— —

=E(n+ 1) [ f(z1, z2) (1)n�O

k n kf(z1, z2)z1 z2 dp3(z1, z2).

n�O Q<k<nk)d1c2 Is'

We compute the integrals on the right-hand side of (1):

Is3f(z1, z2)z1 z2 dji3(z1, z2)

jIk ._n—k= z2z1 - z2 dp3(z1, z2)

j.I�O2k 2(n--k)

= akflk 1z21 du3(z1, z2)

2ir2

280 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

Substituting this result in the right-hand side of (1), we get that

djz3(z1,z2)I f(z,z)Js'

n�0

akflkClf2 =n>00<k<n

§4. The Stieltjes integral

4.1. Suppose first that b <00. We carry out the solution in this case inseveral steps.

Step 1. The function h is continuous on [a, b]. Break up the interval[a, b] into parts by the points = a < t1 < •. < < = b. Let

F(t) = t]m) = (2t)m (t � 0). Then

J h(max IxkI) dx = J h(max IxkI) dxE 0�k<n

= >0<k<n

=0<k<n

where 7k' tk E [tk, tk+l] (k = 0, 1 n — 1). Thus,

L h(maxlxkl)dx

=m2m tk) ln2mJl?n_lh(t)dt0�k<n a

Step 2. The function h is bounded on [a, b]. Consider a sequenceof nonnegative continuous functions on [a, b] that converges to h

almost everywhere on [a, b]. It can be assumed without loss of generalitythat � suph. Show that hfl(maxlxkl) —, h(maxlxkl) a.e. on E. UsingLebesgue's theorem on taking the limit under the integral sign, we see thatthe equalities

LIxkI) dx = rn2mj di (n = 1, 2,...)

lead to the required result.Step 3. The function h is arbitrary (measurable, nonnegatiVe). The re-

quired result is obtained with the help of approximation by a sequence ofbounded measurable functions.

§4. THE STIELTJES INTEGRAL 281

The case b = +00 is exhausted by passing to the limit from finite intervalsto an infinite interval.

b) The arguments are according to the same scheme as in a). Here thefunction F(t) = (21)m is replaced by the function

F1(t) = Xm) E RtmIl � minxk, maxxk � t}=(t_1)m (t—1).

4.2. The arguments are according to the same scheme as in the solutionof problem 4.la). The function F(t) = (21)m is replaced by the function

F1(t) == (21)m

(t � 0).

4.3. The arguments are according to the same scheme as in the solutionof problem 4.la).

4.4. The arguments are according to the same scheme as in the solutionof problem 4.la).

4.5. It is clear that fR2 dx = + fR2\A dx. Arguing ac-

cording to the same scheme as in the solution of problem 4.la), show that

fR2\A e' dF(t), where F(t) = = UXEA B(x, t).Use the result of problem 1.16.

4.6. a) Show that the integral is the limit of the sums

£ cE{0,2} 1<k<n

b) Use the same method as in a);c) Use the fact that the graph of the function a is centrally symmetric

with respect to the point (1/2, 1/2).d) Prove that a(x) da(x) = a2(t)/2 (t > 0) or use the same device as

in the solution of a).e) Use the equality

a(x)+a(1—x)= 1 (x€R).

4.7. a) Let = [0, 3J] x [0, 3'] and E1 = QJ\QJ+I (j = 0, 1,

2,...). It is clearthat[I [1 da(x)da(y)

= da(x)da(y)H

Jo Jo (x2 + y2)P12 j�o JJEJ (x2 + y2)P12

Since � (x2 + y2)"2 �

3 4_-J

<JJda(x)da(y) <3u+i)p • (**)

4 — (x2+y )P12 — 4

Therefore, the series (*) converges simultaneously with the series

282 VIIL LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

(j = 0, 1,...). Arguing as in a), show that the integrals

[f da(x) da(y)

admit an estimate analogous to (**).4.8. The solution of this problem is based on the result of problem 4.7

and is carried out accordingly to the same scheme as the solution of problem3.38. In essence, the function was first considered in [28], p. 93, where itwas constructed as a limit of functions that are integral sums for the integral

— +

§5. The e-entropy and Hausdorif measures

5.1. Let E0 be an e-distinguishable subset of A with a maximal num-ber of points. Then it forms an e-net, and hence N(A, e) � card(E0) =M(A, e). On the other hand, distinct points in E0 can be approximated towithin e/2 only by distinct points, and thus M(A, e) � N(A, e/2).

5.2. a). Use the results of problem 5.1 in computing the minimal numberof points in an e-net for the cube.

b) Let > 0, and let the set E be a minimal e-net for A, that is,card(E) = N(A, e). Then A c UXEEB(X, e) and ).(A) � e))= N(A, which yields the desired inequality with C =where is the n-dimensional volume of the unit ball.

In the solution of problems 5.3—5.12 we omit for brevity the expression"e—'-i-O" when we use the symbols 0, o,and

5.3. a) Suppose that 0 and < 2—m that is m =[log2(1/€)J. It is obvious that the points 1, 2_rn are €-distinguish-able and form a 2€-net. Therefore (see problem 5.1), N(A, 2€) � m + 1 �N(A, €/2), which implies that H(A, e) log2 log2(1/€).

d)Supposethat >0 andm — 1 � is dear that the points 1, 2_a

are €-distinguishable. If we add the points of the form ke to them, wherek = 1 , 2 p and rn_a < (p + then we get an €-net of theset A = {n_aln E N}. Since p = [1/(€,na)], it follows that p = 0(m).Therefore, m N(A, €/2) and N(A, €) = 0(m). Thus, <N(A, €/2) + 1 and N(A, €) = which implies that

H(A, €) __!_log,!.

Problems 5.4, 5.5, and 5.6 can be solved similarly. We present a solutionof problem 5.4b).

5.4. b) Together with the given number 0 we consider a positivenumber 5 whose choice will be made more specific below. Let us break upthe graph A of the function sin ir/x on (0, 1] into parts and , where

§5. THE c-ENTROPY AND HAUSDORFF MEASURES 283

B5 is the part of A contained in the rectangle P5 = (0, 5) x [—1, 1], and= A\B5. The length of is

O(ö').

Therefore, the number N(C5, can be estimated by the quantity =The number N(B5, e) does not exceed N(P5, e), and hence

N(B5, = O(ö€2). Let ö be chosen such that the estimates for N(B5,and N(C5, e) have the same order; for this we take ö = We thereby

obtain the estimate N(A, = O(€_312). To see that the number of e-distinguishable points in A also has order €_3u12 we consider the intervals

= [1/(2k + 1), 1/2k] for k = 1,2 and the correspond-ing parts rk of the graph. Since = [0, 1] for any k, by using e-distinguishable points in [0, 1] it is possible to determine at least [1/e]distinguishable points in Since the distance between neighboring in-tervals is not less than e, points lying on different arcs will be e-distinguishable. The total number of e-distinguishable points constructed inA is [1/e] . Thus, H(A, log2(1/e).

5.7. Suppose that e > 0 and 3_(m+1) <3_rn i.e., m = [log3(1/e)].The endpoints of the intervals -c

of rank m (see the definition of theCantor set before problem 1.2.27) are obviously e-distinguishable and forman e-net. Therefore, N(C, e) � 2 2rn � N(C, e/2). This implies thatH(C, e) = (1og32).log2(1/e).

5.8. Suppose that 0 and 1/(m + 1)! <€ � 1/rn!. The points ofthe form €k/k! €k = 0 or 1 are €-distinguishable and form

a 2€-net. Therefore, N(A, 2€) � 2rn+1 � N(A, €/2). It follows from thedefinition of rn that ln(1/€) mlnrn, that is,

ln(1/€)m

lnIn(1/€)

Consequently,

H(A )ln(1/€)

5.9. The assertion a) follows from the result of problem 5.10 if it is takeninto account that in this case = x + rn! for rn! � x <(m + 1)! andthat (as is easily seen) = and = 1.

See problem 1.11 about assertion b).

5.10. Suppose that 0 and < i.e., rn =Obviously, the points >1<k<rn €k2, where €k = 0

or 1 , are €-distinguishable and form a 2€-net. Therefore, N(A, 2€) <2rn �N(A, €/2). Consequently,

H(A,2€)�rn�H(A,€/2). (1)

284 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

It can be assumed without loss of generality that w is linear on each interval[n — 1, n] (n E N). Then, as is easy to see, + h) — � h forx, h � 0 and hence

(y,?7�O). (2)

It follows form (1) that H(A, 2€) — 1 � 1/c) � H(A, e/2) + 1.Together with (2) this leads to the required result.

5.11. Suppose that e > 0, 3(m+1) < s-rn and

We consider the points (x, y) E A with coordinates x = Ck2 and

Y = El<k<n: + >,fl<k<p €k3, where Ck = 0 or I and = 0, 1, or2, and €k and 17k have the same parity for k � m. It is clear that thesepoints are e-distinguishable and form a 4e-net. The number of these pointsis equal to Therefore,

N(A. 4€) � 3?fl2J)fll � N(A.

that is,

H(A, � [log3 + [log2 � H(A,

Consequently, H(A, (2—log3 2) log2 1/e.5.12. a) Suppose that e > 0 and the set E is a minimal e/in-net for A,

and cardE = N(A, e/rn). Then the set E E (in terms) is anc-net for and � (N(A, e/rn))". Consequently, e) �mH(A, e/m) = o(ln 1/c), which is possible only if ).(A) = 0 (see problem5.2).

b) The solution is analogous to the preceding.c) No. The indicated set serves as a counterexample.5.13. a) If A is countable, A = {x1 x,. ... } then to compute

consider the covering of A by the balls B(xk.e > 0, and let be a covering of .1 such that rk <e

forall /,kEN and Then

�1>1 k>l

> =I�l

Since e > 0 is arbitrary, we get what is required.5.14. a) Let e > 0, and let {B(xk, be a covering of A such that

rk for all k E N and Ek>1 � + 1 Then

§5. THE c-ENTROPY AND HAUSDORFF MEASURES 285

E tfl(B(xk, rk))=k�1 k>n

1).

Therefore, t(A) � tfl(B(xk, rk)) � + 1), which yieldswhat was required, since e > 0 is arbitrary.

b) If = 0, then = 0, because A can be covered by a se-quence {B(xk, of balls such that the sum hence also thesum Ek>1 ).fl(B(xk, rk)), is arbitrarily small.

If = 0, then the set A can be covered by a sequence

= — h1, aL + hi), of cubes such that the sum =is arbitrarily small. Then it is clear that the balls

"circumscribed about the cubes" (where a1 = (ai,..., also form acovering for A, and the sum differs from the sumonly by the factor

5.15. Since (for r � 1)

B'(ö) — 11x02

where a = (0 0, 1) and is the projection of the set flB(a, r) on the plane = 0, it follows that there exist constants m > 0 andM (depending on the dimension) such that mrThi � flB(x, r)) �

for all x E By using this circumstance it is possible to get therequired results by a simple modification of the solution of problem 5.14.

5.16. a) This assertion is obvious, because if rk � 1 , then �b) The proof that = 0 if q > p and <+oo is analogous

to the solution of problem 5.14a). If > 0 and 1 < q < p. then/Lq(A) = 00, because otherwise we would come to a contradiction of the partof the assertion already proved.

5.17. a) Suppose that >0, —, 0, and H(A, —, r(A).Further, let q > r(A), and let be an ri-net for A such that card(E1) =N(A, Then A C B(x, ri). and H(A, <qlog2 1/r1 for suffi-

ciently large j. Therefore, � N(A, � . = 1. Conse-quently, JLq(A) < oo for q> r(A), which implies (see problem 5.16b)) thatdimH(A) � r(A).

b) The inequality supk dimH(Ak) � Ak) is obvious. On theother hand, if q > supk dlmJf(Ak), then for any e > 0 there is a cover-ing of such that < e (j, k E

N). Consequently, C B(4, rL), < e (j, k E N), and

<Ei�1 e2' = e. Thus, = 0, which by the choice

of q implies that A1) � supk dimH(Ak).

286 VIII. LEBE5GUE MEASURE AND THE LEBESGUE INTEGRAL

5.18. It is clear that

= IA C U Ck, diam(Ck)I. k=1

C rk), 2rkL%k�1 k=l

=

Passing to the limit as e —, +0, we get the inequality

�On the other hand, let C (k E N) be sets such that A C Ck°, =

<e, and > If Xk E Ck°, then Ck° C C

we get that � p,,(A) by passing to the limit as e —, +0.5.19. a) If [a, b] C — rk, Xk + rk), then b — a � Ek 2rk by

the countable subadditivity one-dimensional Lebesgue measure )..Conse-

quently, p1([a, b]) � (b — a)/2. On the other hand, by dividing [a, b]into equal parts of arbitrarily small length it is easy to see that u1([a, b]) �(b—a)/2.

b) Fix a number e >0, small enough that the difference 1—h between thelength / of the arc and the length h of the chord subtending it is less than€1 if!<e. Supposethat Lk=AflB(xk,rk),and let hk be the length of the chord subtending the arc Lk and Yk themidpoint of this chord. It is clear that hk � 2rk and Lk = An B(yk, hk/2).Therefore,

� /k(1k�1

where is the length of the arc Lk. Thus,

�k�1

andhence p1(A,e)�ir(1—e).On the other hand, dividing the circle into n equal parts of length 2ir/n <

and covering these arcs by discs with centers at the midpoints of the chordssubtending them and with radii equal to it/n, we see that u1(A, e) �ir/n=ir,andhence p1(A)�ir.

c) This is a special case of d).d) Let G be an open subset of and let e > 0. Find balls B(xk, rk)

(k EN) such that

k=1 k�1

§5. THE c-ENTROPY AND HAUSDORFF MEASURES 287

Then

rk))k>1

= e) + e).k>1

Since e is arbitrary,

(*)

Now take balls B(yk, Pk) (k EN) such that Pk B(yk, pk)flB(yJ, p1) =0 for k and Pk)' where 0 (see problem1.17). Further, fix balls B(zk, 5k) covering the set e and satisfying theconditions Ek>l <e and 5k <e (k E N) (see problem 5.14b)). ThenG C U1(B(zk, 6k) U B(yk, Pk)) and

k�1

Pk))+ ek>1

= (u B(yk, Ilk)) + =

Since e > 0 is arbitrary, this gives us the inequality j(G) �which together with (*) leads to the final result: = In

particular, = =5.20. Note that if p = log3 2, then (1 + � 1 + 1" for 0 � t � 1.

Therefore, if some interval of length / is divided into three parts with lengths, and /3 such that 12 , then

/p�4,+/3p. (*)

We use this inequality to compute 1/2). By definition,

/1p (ic IKCU(ak,bk), bk—ak<

It is clear that the condition bk — ak < 1 can be discarded, and the coveringscan be assumed to be finite and made up of closed intervals. Accordingly,

(ic = inf{ 1K C U[ak, bk]}.1<k<N k 1

Suppose that the intervals [a1, b1] [aN, bN] form a covering of K.Making them smaller if necessary, we can assume that no two of them have

288 '1111. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

common interior points. Let

min{bk—aklk=1,2 N}=bmam=&At least one of the intervals neighboring call it has distance fromnot exceeding 5. Let be the smallest interval containing and In

view of the inequality (*) we get that + (5')P, where and 5' arethe lengths of the respective intervals and is'. Thus, replacing andby we again get a covering of K consisting of N — 1 closed intervals

I I I I I Ip P[a1,b1J [aNl,bNlJ, andContinuing this process, we arrive af the covering of K consisting of thesingle interval [0, 11, without increasing the sum of the pth powers of thelengths of the intervals making up the covering of K at each step. Thisgives us that 1/2) � Since the converse inequality is obvious,

we get that 1/2) = It follows from similarity considerations that3_flJ; = = 2—(p+n) for any n EN. Let be

one of the intervals of rank n in the definition of the Cantor set. Obviously,

= 3_flj, = 2—(p+n)

Since the distances between the intervals of rank n are � it followsthat (see problem 5.18d))

U

=

Thus, = =

To find the function we observe that it is continuous on [0, 11 and isconstant on the open intervals complementary to K. while its increment on

is n ) = It is now easy to prove by induction thatthe values of at the endpoints of /.

£coincide with the value of the

Cantor function.5.21. a) See problem 5.20.b) See problem 5.3e) and 5.13a).5.22. Let = fl [0, xJ), 0 � .v � 1, and let , where= 0 or 1, be the intervals of rank n corresponding to the set E. Since all

the sets are congruent, it follows that ) =Consequently, the increment of on each interval i is proportional tothe increment of the function 0E Moreover, like aE, is constant on theopen intervals complementary to E. It is now easy to prove by inductionthat the values of the functions /LP(E)aE and coincide at the endpoints

§5. THE c-ENTROPY AND RAUSDORFF MEASURES 289

of the intervals Since the set of such points is dense in E, we getwhat is required.

5.23. Suppose that aE is the Cantor function corresponding to the setE, a = lim2'1!, and is a sequence of positive integers such that

—. a. Let e be an arbitrary positive number. Further, assume thatthe indices are such that < e and < a + e. Now consider acovering of E by closed intervals of rank (for fixed k). Then we getthat e) � < a + e. In particular, we see that if a = 0, then

= 0, while if a <00, then also <00.Nowlet 0<c<a<oo. Fixan index m suchthat !m<€ and

for n � m, and consider an arbitrary covering of E by open intervals— + r1) satisfying the condition r, <Im (j = 1, 2 ). Choose

indices s1 such that � Then < 1s2, > m, and

� � cj�1 j�1 j�1

j�1

cJE(xJ — r1)) � =i�1

The last of the inequalities above is valid because the measure generated bythe function aE is countably subadditive. Accordingly, we have proved that

� c/4 for any covering of E by open intervals (x1 — r1, + r1) ofsufficiently small length, which implies that � c/4> 0. In particular,if a = +00, then since c < a is arbitrary, = +00.

5.24. Verify first that 2 � � 2 (k = 1,2,...), and prove

the equalities lim(k/nk) = 1/2 and = 1.

5.25. Use the result of problem 5.23.5.26. Let e > 0, and let N be a positive integer such that <

Then the points akn , where ak E E (k = 1 N), form

an e-net for A with cardinality = Therefore, H(A, e) �log2 = Nlog2 m � p log2 1/e, where p = log2 rn/log2 n. Consequently(see problem 5.17), dim11(A) � r(A) p. On the other hand, if A c

— + then the sum can be estimated from belowas follows. Let a be the Cantor function corresponding to A, and let k1 be

290 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

indices Such that <2r1 (j E N). Then

(i)P= 1>2k

1?' j�1 fl

(a _a(xj_rj))

�1?'

�The last inequality is valid because the measure generated by the function ais countably subadditive. Thus, � >0 and dimH(A) � p.

5.27. To compute dimH(A) and dimH(A +A) use the result of the pre-ceding problem.

5.28. Consider the function

= 1111 da(u)da(v)(z E C)

j0 j0 z—(u+zv)where a is the Cantor function corresponding to the set K (see problem111.3.20). Use the same idea as in problem 4.8.

It is known (see, for example, [101) that a function with the indicatedproperties does not exist if dimH(K) < 1/2. For the case dimH(K) = 1/2

see [81, pp. 346—348.

§6. Asymptotics of integrals of higher multiplicity

6.1. c) Using Fubini's theorem, show that

IB"(r) 1x1dx =

jrt"(r2 — di.

Express the last integral with the help of the function f and use its properties.6.4. Using Fubini's theorem, we see that

<4 =

= J1The right-hand sides of these equalities are equivalent as n —. oo (see prob-lem VL2.4). Therefore, E 1x11 <e} —+ 1.

6.5. By definition, = a,21 lix' — xli

INTEGRALS OF HIGHER MULTIPLICITY 291

Since the measure is spherically invariant, we get that =x lie1 —xli 1(x),where e1 = (1, 0 0). Reduce the last in-

tegral to an integral over [—1, 1J and use the asymptotic theorem of Laplace.The result obtained = attests that radii drawn to two ran-domly chosen points on the sphere are almost orthogonal, "as a rule."

6.6. Assumethat Then

=j' v'ñdx1 . . .

0n—1 JV 1 n—i

Reduce this integral to the form

n—2 du

Making the change of variable u = — 12/n in the last integral, we get that

= (n— dt.

It remains to pass to the limit under the integral sign and to find the limit ofthe factor in front of it.

6.7. We represent the integral over the ball f(x) dx in the formf(xfiixii) dx where

f(x)dx,

(f(x)_f(-L.))dx,1—c�IIxII<1 lixil

dx

It is clear that M(1 — � M(1 — and <h(e). More-over,

j I dx = {L-1 f(w) d } dr

= J_.jSince = , it follows that

j IJi +

�2M(1

292 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

6.8. Represent the given integral in the form

q/n —x2/2e dx)

and use the fact that

1 x!/2 1 +00 qin lxi

_x2/2dx=1+— — •e dx+

n /6.9. Use the formula (*) from the beginning

of problem V.2.3.b) Use the result of problem VL2.16.c) Let .1 = Then

n/2 00

— l—n/2 100 dt.— f(nf2)2

Since1

(1+e)2 1 €2

and f(n/2) � it follows that

,_9 � [00

eit J1+c

We get similarly that

£

< —nc2/2— — •e•e e dt.

—— 0

Consequently,

Jxd5' (x)

<

nc /2 12/2— •e•e

Je dt

—1I�c — it 0

e= <2 —nc2/2

6.11. Using the formula (*) from the beginning of §6, we get that

n n/2 I q dx=

= (j dr.

§6. INTEGRALS OF HIGHER MULTIPLICITY 293

Moreover,(fl)n/2 [Co

dr2ir j0

f(n)(n)q/2—1

— 2) 0n—1

6.12. Use the formula (*) and the relation f(a + c) • f(a) (see

problem VL2.15).a +

6.13. Let

F(t)= J = (i — (t � 0).

Then

j dF(i)

= (i — vf 1

(1)

Let z(t) = e_U2 /2 du. It is clear that z'(t) =z(0) = 1, z(t) 0. Therefore, making the change of variable z =z(t) in

the integral on the right-hand side of (1), we get that

= nj — dz,

where t(z) is the function inverse to z(t). Since z(t)

e2h12, it follows that (see problem VL4.1) t(z) = . (1 +where z:+o° Consequently,

= (1- dz. (2)

Using the results of problems VI.2.8 and VI.2.12, we get that

+ — dz — dz

Together with (2) this leads to the required result.6.14. Arguing as in the solution of problem 6.11, replace the given inte-

gral by the integral with respect to Gaussian measure. Then use the result ofproblem 6.13.

6.15. Using the inequality lxii,, � n"ix1 and the result ofproblem 6.8, we get a lower estimate for q) in the case q > 0, andan upper estimate in the case q <0.

294 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

a) We estimate q) from above. Since lxii,, llxll2, it fol-

lows that .f(p, q) � Using the result of problem6.12, we get what is required.

b) In this caseq/p +00 q/p

q) � (j=

(Jc) Since lxii,, � it follows that

q) � j = j dy1(x).

We estimate q) from below for q <0. Let a = —q. Then

1

=llxli2. � a) q).

Therefore, q) � (p, a), and it remains to use the upper estimatefor q).

6.16. Use the results of problems 6.11 and 6.15.6.17. Using the equality

1—— f(n/2+1)

and Stirling's formula, we get that

==

(i +. + +o ()).6.18. a) Show that the radius of the ball is equal tob) Show that the radius of the ball is equal to 2 — 1; use Stirling's

formula.

6.19. Show that =• e' dt, and use the

result of problem VI.2.17a) to show that = n/2 + o(n"2).6.20. For t> 0 let = {(x, E t} and

F(t) = = Find the mean value of the functionon It is clear that

n! I' dF(u) n ftn—2 du

(n�2).

Let us estimate Since 11x111 � lixil, it follows that

—1 —1/2 f dxn i —

fIJ' lixil

j0 n—i

INTEGRALS OF HIGHER MULTIPLICITY 295

To get an upper estimate of we fix a number e > 0 to be made moreprecise later. Then

f dx f dxcxb<I —+1 —— II-'iIi 11x111

—

Thus,

b < 1 2 (2€) (n—1)/2 F(n/2+1)n/2 n!

By using Stirling's formula it is easy to see that the second term is

Taking e = we get that � 0(1/n).We proceed to the estimation of Let G(t) = Then

G(t) = and since 1x11 + + � n maxl<k<fl IxkI, it follows that

> J- 11 = dt =t j0 n—i

On the other hand, using the same device as in the determination of an upperestimate for the numbers we have that

— JO,(cn) 11x111 J(—i IJ'\O,(cn) en

�=

n—i n!

Stirling's formula gives us easily that the first term on the right-hand side ofthe last inequality is O((ee)'1). Taking e = 1/3, we get that � 3/n +o(1/n).

6.21. a) Let

J2dx=fThen

2 j0 11x111

= u2 (i ;i) du

= 1' u2 (1' duf—1 lul+t/

296 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

where F(t) = = — 1)!. Therefore,

1 1--u 1n--2 dtfn2(nl)j u2(j

=n2(n—u2du)

Since u2 � u2/(u + t) � u2/t, it follows that

n2(n— 1)1 dt � � n2(n — 1)j dt

b) Let IIxH2/11x111 dx = Since 11x111 it follows that

� j lixil dx = j =

The upper estimate for follows from the inequality x

1/11x111 dx and the estimate for in problem 6.20.c) It is clear that

K = 1 1x1 dx1 d;

J J

Using the inequality 1x11 + + � n maxl<k<fl IxkI, we get that

[——1,

The upper estimate of follows at once from the inequality

� 2flj[——1,

and the estimate for in problem 6.20.6.22. b) We remark first that

2h1_l.fl.f(fl/2) 1

—n!

(*)

Show that

It follows from these inclusions that

� v, �where = + eB'1), that is,

((i + — <v (i +n! \\ j — n! \

§6. INTEGRALS OF HIGHER MULTIPLICITY 297

Using (*), we get that

it/SI/nFrom this it follows, in particular, that ) —4

6.23. For 1 letA1(t) = [0, tJ, = for n � 1, and 1. Then =

— u)du for n � 1, and hence

= — t), = (n � 2).

Deduce from this that

=0,= (2k + 1)!

(*)

Let F(x) = . and G(x) = . Using (*),verify that 1 and G(x) cosx sinx, that is, F(x) = 1/ cosxand G(x) = tanx. Use the expansions

= cotanx=flit

nEZ

Replacing x by ir/2 — x in them, we get that

tanx=V' 2x1 (2k—1)ircosx

k�1 (2k — 1)2ir2/4 —L_i '2'- — 1)2ir2/4 — x2k�1

Consequently (for IxI <ir/2),

1 2x \cosx = it(2k— 1) (it(2k —1))k�1 m>O

= 4

it \lt — (2k_1)2m+l'm�O k>1

8x I 2x2

1) m�O

4 12x 2m+1

m�O(2k — 1)2m+2

298 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

Thus,

v 1

—(_1)kl

2m(

2m+1

From this, = +6.24. Since > the set contains at least (3/2)1?

points with integer coordinates (see problem 1.3b)). We remark that thenonzero coordinates of these points can be equal only to +1 or —1, because

C (—2, 2)1?. Let K1? be the number of points with coordinates 0 and ±1such that the number of nonzero coordinates does not exceed m = [n/10J.It is clear that = .+2m.(n). Verify that K1?for large n; this implies that the set contains the desired point.

6.25. a) Fix an arbitrary positive number e and consider the set

1 ,, 1

E1?(e)=1(x1,..., x1?)E[0,1JI >e

By Chebyshev's inequality and the pairwise orthogonality of the functions— 1/3 (k = 1, 2 n) on [0,

(>(xk2*))

(!ne ro,lrl<k<fl

(Thus, coincides" with on a set of measure as closeas desired to 1.) Consequently,

1! 1 1 lIxil 1

Jioirdx

< j dx

dx+[ edx\ E(i)

+ e).

Since ).1?(E1?(e)) —, 0, for sufficiently large n

L I—

§6. INTEGRALS OF HIGHER MULTIPLICITY 299

b) Consider the set

and prove that 0. Use the continuity of the function I atthe point 1/2.

c) Since . = and lnxdx = —1, consider theset

>€1<k<n

and use the same considerations as in a) and b).d) Remarking that

1(+00

—x2/2IxIe dx=V_,

consider the set

6.26. c) It obviously suffices to consider only the case when —, 00.

Let = {(x1,..., E >1<k<fl � and

= J dxv.

There is a number 5 > 0 such that f(x) dx < e1. For x = (x1,...,E let E(x) = {j E NI � ö}. It is clear that cardE(x)Let N = 3 = {B C {1,2 n}IcardB = N}, and CB ={(x1,..., E <ö for k B}. Then C Since

f... ff(x1). . .. .d; � it thus fol-CD

lows that

J .. .Jf(x1) . f(;)dx1 ..•dx (1)

CB

Estimating we get that

�Together with (1) this gives us that

. �

300 VIII. LEBESGUE MEASURE AND THE LEBESGUE INTEGRAL

Since —' 0 as n —, 00, we get for sufficiently large n that

6.27. It follows from the Hälder inequality that the quantity =(2_li I(x, is increasing as p increases. For p = 2 it coincides

with IIaII/3. Thus, the right-hand inequality is nontrivial only for p > 2,and the left-hand inequality is nontrivial for p <2. Let us prove the right-hand inequality. Obviously, it suffices to do this in the case when hail = 1.It will be assumed first that p = 2m is an even number. Using the inequality

� cosh t, we get that

J (x, a)2" dx � (2m)!J cosh(x, a)dx =

where ak (k = 1 n) are the coordinates of the vector a. Using the

inequality (sinh t)/t � e''2, we get that

a)2n dx � =

This implies that � 2mhlahl. In the case of an arbitrary p > 2 wefind a positive integer m such that 2m � p <2m +2. Then

(2m + 2)hlahl � (p + 2)hlahl.

Let us proceed to the proof of the left-hand inequality for p E (0, 2).Consider a number 0 E (0, 1) such that 2 = Op + 4(1 — 0) Using theHolder inequality with the exponent 1/0, we get that

= j l(x, a)12 dx= I(x, . l(x, dx

� l(x, aW') . J I(x. a)l4dx)

Consequently,

I (x, > Hall

) —

> 11a112 —

— 3B . hlahl4U6) —

that is, � 0)) l/p0hail.

CHAPTER IX

Sequences of Measurable Functions

§1. Convergence in measure and almost everywhere

1.2. b) Show that the series f( converges.

1.3. Consider the functions = (_1)fl.

.where the are the

functions in problem 1.2.1.5. Let E N. Then the functions + and Isinxl are

equidistributed on (0, 2ir). Therefore,

).{x E (0, 2ir) I

I Isinxl> —' 0.

If N, then consider the smallest interval containing (0, 2ir) and aninteger number of periods of the function and use analogousconsiderations.

1.6. To prove the first of the required equalities consider the sets

E (0, 2ir) I + < 1 — e} (e >0)and prove that for any > 0 there is a sequence of indices suchthat the intersection E N.

1.7. Let = f0°° + dx. It is clear that

1 °° 1(k+1)n

=— j + dt =

—elmlsin(t + dt

= f +k>O 0

= !1 (1 + 0 (i)) Isin(s +

k>0 0

=! + 0 I = 1 + 0(i)

' Jo 0

Since sds (see VI.2.9a)), it follows that

301

302 IX. SEQUENCES OF MEASURABLE FUNCTIONS

1.9. a) Use the result of problem 1.8.b) Consider the set e = E R I > e,j for sufficiently large

p.1.10. It can be assumed without loss of generality that the sequence

is decreasing. Consider a subsequence such that

E (0, 1)I � 11k2} � 1/k2 (k = 1, 2,...). Using the result

of problem 1.9a), show that the sequence where = k for

� <flk+1, is the desired sequence.1.11. Let be a sequence such that —, 00 and

— 1(x)) —, 0 a.e. on (0, 1). Prove that g = — 1)1

is the desired function.1.12. Consider the set e = {x E (0, 1) I g(x) > C}, where g is the

function in problem 1.11 and C is a sufficiently large positive number.1.15. Consider a sequence such that

and use the inclusion

{x E (0' 1)1 — > e}

C {x E (0' 1)1 Ifflk(x) — > e/2}

U{xE(0,

b) Use Egorov's theorem (problem 1.12).c) Use the result of problem 111.1.6.

§2. Convergence in the mean. The law of large numbers

2.1. Use the Holder inequality.2.2. a) Use the inequality

2.3. Use the Cauchy-Schwartz-Bunyakovskii inequality and the absolutecontinuity of the integral.

2.4. Prove that — f0(x))f0(x) dx—,

0.

2.6. a) To estimate the integral fE If(x)I dx represent as the productIIIIIIIII and choose the numbers t > 0 and s> 1 such that st = r and(1 — t)s' = 2, where / = s/(s — 1); use the HOlder inequality.

b) For > 0 let E E I If(x)I > It is clear that� and

lull1 J lf(x)l dx + {m(E)fa(f)E(a)

+

�

§2. CONVERGENCE IN ThE MEAN 303

Consequently, if CsJ& � 1/2,then � �2.7. We fix a number e > 0 whose choice is made more precise below,

and we let I = It is clear that there is a number t > 0 suchthat <e, E E I 1(x) > t}. Then fE �

< The number e > 0 is chosen so that < ö/2. Inthis case fE\E > ö/2. Consequently,

ttm(E) � J1(x) dx = J dx

E\Epz>1

�n>I

from which

�n>1

2.8. Prove that limfE Isin(nx + dx > 0 for any (measurable) setE C (0, 22E), and use the result of problem 2.7.

2.9. Use the result of problem 2.2a).2.10. Use the result of problem 2.9.2.11. Prove that the measure of the set

1 .2 flksin kx<-4

1% 1<k<n

tends to zero as n —. 00.2.12. b) Prove that � IakzI + 2M/k, where k =

2.13. Let (k=0, 1,2,...). Then

k3 =n�1 k2<n((k+1)2

Therefore,

�k�1

J�0

<2A0+ <00.J�1

Thus, akz(x) —, 0 a.e. on E. Suppose now that n = k2+p, where k =

and 0 �p � 2k, and let g = It is clear that g

and hence wk(x) = —, 0 a.e. on E. Moreover, �

304 IX. SEQUENCES OF MEASURABLE FUNCTIONS

Ian — + frJklI, and

k 1

< i/ .3/4

k2<j�(k+1)2 /

1 1 .3/21 1/2

/ J

J

� + 1)312(2k +k—.oo

almost everywhere on E.2.14. b) As established in problem VIII.3.16, — dx =

0(1/n2). Therefore, the series (;(x) — converges almost ev-erywhere, and hence — 1/2 —' a.e. on (0, 1).

2.15. The solution is analogous to that of the preceding problem.2.17. Use Chebyshev's inequality and the pairwise orthogonality of the

functions f(x1),f(x2) in the cube (0,2.18. The solution is analogous to that of the preceding problem.2.19. Consider the function 1(x) = x2 — lf2ir (x E R) and the measure

y with density Prove that the y x x y-measure (n factors) of theset

>e

tends to zero, and hence the integral fE(e) dx is small for

large n. The integral dx is close to 1/2ir) forsmall e>0.

2.20. a) Since IS — = it follows that

<00.k�l

b) It is clear that SA2(x)k

S(x) a.e. on E. Let n = k2 + p. where

k = and 0 � p � 2k. Then

<pk2<j�n

2 = 2Rk.

§3. THE RADEMACHER FUNCTIONS 305

Since the series '/J• converges almost everywhere, it followsthat Rk(x)

k—0 aTmost everywhere. Thus,

IS(x) — Skl(x)I

� IS(x) —Skz(x)I —, 0 a.e.

c) It follows from the_preceding inequality that — � g, where the

function g = + IS — SkZI2 is square-integrable. It

isclearthat

§3. The Rademacher functions. Khintchine's inequality

3.4. a)—c) Use the result of problem 3.2b).d) Use the fact that the infinite products ffn>i cos and

lln>i(1 — 2) converge only simultaneously.Show with the help of 3.3b) that for real Ck

1' (dx = + 6 � 3 (

2

0 1<k<n 1<k<n 1<j<k<n 1<k<n

b) Use the idea of the solution of problem 2.6a).3.6. Represent the double integral as an iterated integral and use the result

of problem 3.5.3.7. The solution is analogous to that of problem 3.5.3.8. The solution is analogous to that of problem 3.5.3.10. Use the results of problems 3.9a) and 3.5a).3.11. Use the equality (C = (cjk)1<Jk<fl)

dsdtx,yEA 0 0 1<j,k<n

and the result of problem 3.6.3.12. Compute F(s) by using the equality

fr(s)= J

eiSt dF(t)= j

where F is the desired distribution function, along with the results of a) andb) in problems 3.3 and 3.4.

3.13. a) Let = akrk. Use the equality 5m =

+ — am+irm+i) , the triangle inequality, and induction on n (n � m).b) Use the convexity of the functions s eXS and induction on n (n �

m).c), d) Represent E as a union of intervals of the form (J =

0, 1 2m — 1) and use a) and b).

306 IX. SEQUENCES OF MEASURABLE FUNCTIONS

3.14. a) It will be assumed without loss of generality that A = 1 (other-wise, t can be replaced by At'). We introduce the sets

XE(O,1)I

a positive number to be made more precise below. Then

dx

�= etS [I cosh(sak) � es2h12.

1<k<n

Choosing s so that the right-hand side of the inequality is minimal (s =

we get that � e'12. The measure of the set E can be estimatedanalogously.

b) Use a) and the inclusion

xE(O,1)I ckrk(x) >t

1<k<n

1<k<n

4xE(O,1)I flkrk(x)1<k<n

where = Re(ck) and fik = Im(ck).c) Represent E as a union of disjoint intervals of the form

m n, 0 <J <2m) chosen so that

max = for x E

I<k<1 l<k<m

Then it follows from the inequalities a) and b) of the preceding problem that

§3. THE RADEMACHER FUNCTIONS

� C' fE I >.1<k<n � AC' and

{(E) � 2. cosh dxE J

dx.0

JThe last integral was estimated in the proof of a).

d) Let F be the distribution function of g. Using the inequality 1 —

F(t) � established in c) and assuming that s < 1/2A2, we get that1 oo 00

J(x) dx = I eS1 dF(t) � 4s I --s)1 dt <00.

0 Jo Jo

But if s � 1/2A2, then we consider the function

g,(x) = supn�N

N is chosen so that = >k>N < 1/(4s). As already shown above,

10' <oo. Moreover, g(x) � K + g1(x), where K = >.1<j<Nand hence g2(x) � 2K2 + Consequently,

101 dx � dx <oo.

3.15. Obviously, all the numbers ak can be assumed to be reaL Let= a1r1(x) + Using the result of problem 3.14c), prove

that for any number e > 0 the measure of the set

{x E (0, 1)1 sup > e}pEN

does not exceed (>k>n Derive from this that

3.16. By shrinking the set E if necessary the problem can be reducedto the case when the sum f of the series is bounded on E. Suppose thatIll � C on E. Consider a subsequence {Sfl}k>l of partial sums of theseries that converge uniformly to I on E. Obviously, we can assume that

2C for any x E E and k E N. Fix a positive integer m to bemade more precise below, and let

2 2 2T = Uk = for > m.1�J�m

IX. SEQUENCES OF MEASURABLE FUNCTIONS

We prove that the sequence {uk}k>1 is bounded, which suffices for the solu-tion of the problem. It is clear thai

1/2�

(1 (airi(x))dx)_T.

E m<J<n&

Consequently,

E m<j,1�n&

(1)E

� —2 / (J dx).Vm<i<I�n&

£

Since the family is an orthonormal system (see problem 3.7a)),Bessel's inequality giveslis that

Om

=(jrj(x)r,(x)dx)2 —'0.

We take m large enough that 40m Then it follows from (1) that

(T+ � — 2UkOm �

Thus, Uk � for all > in for this choice of m.as required.

3.17. It is clear that a) b) c) (see problems 2.1 and 2.2). Theimplications c) a) d) were established in problems 3.16 and 3.15.Finally, d) c) by Lebesgue's theorem.

3.18. The solution is analogous to that of problem 3.16. Holding to thenotation introduced there and assuming that E C (0, 2ir). we get instead of(1) that

(icT + � cç'J sin2 — 2/E

V

(Jsin2xsin2'xdx)2.Nrn<i<1<n&

E

§3. THE RADEMACHER FUNCTIONS

SinceI . 2 j 1 1, sin 2 x dx = — — — i cos 2 x dx —,

JE 2 2JE 2

m can be taken large enough that fE sin2 dx> for j > m. Toestimate the sum sin sin 21x dx)2 use the equality

j sin21xdx = (j cos(2' — — jcos(2' +

and Bessel's inequality for the trigonometric system.3.19. a) Show that the boundedness of the sum of the series

on implies its boundedness on (0, 1). Suppose thatI

�C on(0, 1). Fix an arbitrary positive integer n and consider an

1),

/=1,2 n. Since= 0 for j> n, it follows that

� J = aJ =j�1

Thus, forany nEN.b) Use idea of the solution of problem IV.6.28.3.20. b) Use the idea of the solution of problem 3.13c),c) Use the idea of the solution of problem 3.14c).d) Use the idea of the solution of problem 3.15.3.21. a) Let be the nth partial sum of the Fourier series of and

let e be an arbitrary positive number. Prove that

E (0, 1)1 NISfl+P(x)Sfl(x)I >€} �

Derive from this that

E (0,>€} Si' j

To complete the proof it remains to repeat the arguments used in the solutionsof problems 3.15 and 3.20d).

b) Let E (0, 1)1 supl<m<n > t}. Using as a model thearguments in the proof of c) in the problem, prove that �

after which it remains only to pass to the limit as n —,

3.22. Prove that akrkllP � The second in-equality can be obtained wltlithe help of the in the solution of prob-lem 2.6a). Let F be the distribution function of the function I akrkl.

310 IX. SEQUENCES OF MEASURABLE FUNCTIONS

Using the estimate 1 — F(t) � obtained in problem 3.14a), whereA2 = >.1<k<n show that for any p > 0

J dx= J dF(t) � 2pJ le_12/2A2

0 1<k<n 0 0

= 2pA"J yP_le_Y2/2 dy.

Expressing the last integral in terms of the function F and using Stirling'sformula, we get the required estimate for B,,.

A proof of Khintchine's inequality without making explicit use of the prop-erties of the distribution function can be obtained by analogy with the solu-tion of problem VIII.6.27.

3.23. The implication e) b) is trivial, and the implication a) e)

follows from Khintchine's inequality.3.24. It will be assumed that ICkI2 = 1. Consider the function

P(x,t)= E ckrfl+k+l(t)e (xER,tE(0, 1)).IkI�n

a) Use the inequality IP(x, t)Idt � (where > 0 is an absoluteconstant), which follows from problem 3.22.

b) If we show that the measure of the set

E = {t E (0, 1)1 nIP(x, t)I>

is less than 1, then by taking a point t0 in (0, 1)\E (not dyadic rational)and letting €k = rfl+k+1(t0) we get the desired trigonometric polynomial.To estimate the measure of E we observe that maxXER IP(x, t)I is "almost

realized" at one of the points = 2icj/n2 , j = 0, 1 n2. Indeed, if� ic/n2, then

IP(x, t) — t)I �Ikj(n

ICkI .2. �Therefore, maxXER IP(x, t)I � + ')I. Consequently,

E max t)I >

Finally, it is clear that

EC U1<j<nh

§4. FOURIER SERIES AND THE FOURIER TRANSFORM 311

As established in problem 3.14b), the measure of each of the sets on theright-hand side of the last inclusion does not exceed = 4n9. Con-sequently,

� � n2 •4• n9 = < 1 for n � 2.

As is clear from the last inequality, supXER IP(x, t)I � for allbelonging to the set (0, 1)\E, the measure of which is close to 1. Thus,

arranging the signs €k randomly by means of coin tosses, we get polynomialssatisfying the required estimate in the overwhelming majority of the cases.

We remark that numerous results relating to the questions touched on thisproblem can be found in [11].

3.25. Let = 2 Inn. With the help of Khintchine's inequality showthat the series 1/

I I d

converges for any a > 2. In the proof of this use the estimate B,, �+ e)/e, valid for any e > 0 and sufficiently large p (see problem

3.22). It follows from the convergence of this series that the series2lnn

converges almost everywhere, which is possible only if

<1fl-'OO -

almost everywhere.3.26. Use the same idea as in the solution of the preceding problem.

§4. Fourier series and the Fourier transform

4.2. It is clear that for h > 0 and any x E R

19,(x+h)—9'(x)I

where L = ess supR If(x)I. Show that the integral on the right-hand side of

the inequality does not exceed (4/(1 —4.3. a) Use the equality

h(n) = lim JSk Sk are the kth partial sums of the Fourier series of the

respective functions f and g. Prove first that IIh — SkSkIIlk

0 (see

problem 2.2b)).b) Represent as the sum of the series fk/k! and use a).

312 IX SEQUENCES OF MEASURABLE FUNCTIONS

4.4. Usethe equality (nEZ).4.5. Compute the Fourier coefficientS p3(n) of the function = I *

(see problem 4.2) and Show that for n 0

J(n) f1\

4.6. Use the equality 1(n) = f(x + dx.4.7. Use integration by parts.4.8. Express 1(n) in terms of the integralS

xd

Jo \flJ \'Estimate the last integral by integrating by parts.

4.9. Suppose that � (n = 1, 2,...). Show that

1(x) — f(x')I � C > + 2C

n>Ix—x'r'

Prove that each of the sums on the right-hand side of the inequality is0(Ix —

For = 1 consider the function 1(x) = and verify that itsderivative is unbounded on (0, ic).

4.10. Let

= f(n)eWX and fk(x) =sin(2"Ixl)

InI�i 1<n<k n

Show that

Sm(0) = !j fk(x)Sin(x/2) dx, where mk=

Prove that+ 1/2)

dsin(x/2)

x

2 sin2 rnkxdx

— ln<k

sindx + 0(k).

Use the result of problem VI.1.10.4.11. Use the equality

— 1(x) = J(f(x — — dt. (*)

Prove that= 1 sin2(nt/2)

di —' 0iw sin2(i/2)

§4. FOURIER SERIES AND THE FOURIER TRANSFORM 313

for any 5 > 0, and deduce from this that for any x E R

— f(x)I � +

where co1 is the modulus of continuity of f.4.12. a) Using the formula (*) in the solution of the preceding problem,

prove that

— f(x)IC sin2(nt/2)

dt (x ER),

where c is a Lipschitz constant L for f.b) Use the result of a) and the extremal property of the partial sums of

the Fourier series.4.13. Using the Cauchy-Schwarz-Bunyakovskii inequality and the asser-

tion b) of the preceding problem, prove that If(n)I =

4.14. With the help of the inequality � applied to thefunction = f — show that for k 0 the Fourier coefficients f(k) off are equal to zero. —

4.15. Use the equality f(—n) = 1(n) (n E Z).

4.16. It is clear that � Ifk(x)Ie . Therefore, a)

follows from the result of problem To prove b) show that

- aq

=- >21k(x))

k�p

With the help of the result of problem 4.3b) derive from this that

= k�p

and hence

1f(nq) — aqi � J Ifk(x)I dx_Jtk�p

J Ifk(X)II11hJ(X)I dx.k�p

It remains to use the Cauchy-Schwarz-Bunyakovskii inequality.4.17. Use the inequality

cje"Yj dx � e

jg

(where e is a sufficiently small positive number, and F is the function con-

structed in the preceding problem with = (j = 1, 2 in).

314 IX. SEQUENCES OF MEASURABLE FUNCTIONS

The inequality in 4.17 was conjectured (in the particular case when C1 =c2 = = 1) by J. Littlewood. It was proved half a century later by S.Konyagin (On the Littlewood problem, Izv. Akad. Nauk SSSR Ser. Math.45(1981), no. 2) and simultaneously by 0. McGehee, L. Pigno, and B. Smith(see [41]).

4.18. a) Show that the function is a solution of the differentialequation y'(s) + csy(s) = 0.

4.19. We get by n-fold integration by parts that2 +00 2d1z .22

ha(s) = (_1)fleS /21 e' /) dx.

Using the equality

d ld=

prove that

ha(s) =j+OO

dx)

To compute the last integral use the result of problem 4.18.4.21. Since a(2/3 +z) = 1/2 + a(i) for 0 � z� 1/3. it foliows that

1/3 1

&(s) = I elSlda(t) + I e1St da(t)Jo J2/3

p1/3 çl/32

= j da(i) + / da(i)Jo Jo

1/3

= (1 elSlda(1).

Using induction, show that

a(s)

= ( fl (1

1<k<n 0

Derive from this that

&(s) = flk>1

It is clear from the last equality that

= fl 0k>2

for any n E N.4.22. a) Let 0 � x <y � 2. Since

a1(y)= J a(y — t)da(t) � / a(y — t)da(t)

0 Jo�jXax

— t)da(t) =

§4. FOURIER SERIES ANDTHE FOURIER TRANSFORM

the function a1 is nondecreasing. To prove that it is strictly monotone, showthat there exist closed intervals (k + and (1 +of rank n (see the definition of the Cantor set before problem 1.1.27) suchthat

(cf. problem L1.28b)). Then

fk+l+1\ fk+la1(y) — a1(x) � a1 3fl ) — a1

1 'k+l+l ' 'k+l ''� J — t) — a — t)) da(t)

k3 3 3

11+1 \ 11 \\= J0

— — a — da(r)

(1+1 1 I 1\\� J — ) — a da(t)

= > 0.

b) Show that = (8(s))2, and use the result of the preceding problem.4.23. Use the fact that the Fourier transform of a convolution of inte-

grable functions is equal to the product of their Fourier transforms.4.24. a) Show that the range of g is dense in [0, 1] and use monotonic-

ity.b) Study the intervals of constancy of g and find the measure of their

union.c) Use the same device as in the solution of problem 4.21.d) Study the Fourier transform of the measure corresponding to the func-

tion

h(x)= J g(x — t)dg(2t) (x ER).

4.25. The solution is analogous to the solution of the preceding problem.4.26. Prove that

fl(s)=

dF(u),

where = p{t ER'1 I (s, t) u} (u ER).4.27. Let p be a measure satisfying the conditions of the problem. Then

= where p1 and are Borel probability measures on subspacesL and M, respectively. Since p is rotation-invariant, it follows that dimL+dimM = n. Again since p is rotation-invariant, we see that fl(s) = fl(IIsIIa),where a is an arbitrary normalized vector. The invariance of ji with respectto orthogonal transformations implies that p1 and p2 are also invariant withrespect to such transformations (in L and M, respectively). Therefore,

= fl1(IIuIIu0) and fl2(v) = fl2(IIvII .v0), where U E L, v E M, and

316 IX. SEQUENCES OF MEASURABLE FUNCTIONS

u0, v0 are arbitrary normalized vectors in L and M, respectively. With thehelp of Fubini's theorem we get that

where uEL, v EM.

Since lu + vii = + llvll2, this implies the equality

+ llvll2ai = tL1(IIUIIUO) (1)

Let = fi(ta) for t > 0. Setting v = 0 and u = 0 in turn in (1), we seethat

= ui(lluHa) =9,(llvll) = =

Thus, it follows that (1) that

= for any t2 � 0.

Using the result of problem 111.5.9, we get that = for t � 0.Thus, = and c � 0 because < 1. If c = 0, then ji is

the unit measure concentrated at the origin of coordinates. The case c> 0

corresponds to the Gaussian measure with density

CHAPTER X

Iterates of Transformations of an Interval

§1. Topological dynamics

1.1. Prove by induction that n � 1(n) <min{f(k)Ik > n} for eachnEN.

1.2. a) Prove that f(O) = 0 and investigate the sign of 1(x) for x > 0.b) Prove that f is strictly increasing and that inff = a E (—oo, 0),

1(a) = 0. Fix an arbitrary strictly increasing function 10 E C([a, 0]) satis-fying the conditions f0(a) = 0 and = ea, and extend it to R by usingthe equation.

c) Prove that f must be strictly increasing on [0, +oo) and strictly de-creasing on (—oo, 0] and that 1(0) � 0, which is impossible, because F,and hence f, takes negative values.

1.4. Let 1: X —. X be a bijection. For an arbitrary point x E X letx0 = x, x1 = f(x), x1 = x2 = 1(x1), x2 = and soon. Denote by Orb(x) the set and call it the orbit of the elementx. Two cases are possible. a) There is a number n E N such that x x0

for I = 0 n — 1, and = x0. In this case x1 = for any i E Z,and the set Orb(x) consists of precisely n points. b) x0 for all n > 0.Then forall i, jEZ,andtheset Orb(x) isinfinite.

It is not hard to see that the orbits of two points x and y either are disjoint(if y Orb(x)) or coincide. Thus, the whole set X splits into disjointsubsets , each of which is the orbit of any of its points and is thus invariantwith respect to f. It suffices to define the desired involutions (denote themby h and k) for each set separately, taking an arbitrary point x0 in it.This can be done as follows: for x. E Xa = Orb(x0) let h(x1) = andk(x,) = in case a), and h(x,) = x11 and k(x,) = x1 in case b).

1.5. Show that the difference x — 1(x) cannot have constant sign, anduse the Bolzano-Cauchy theorem.

1.6. The sequence is monotone. The assertion is false for a de-creasing function (for example, 1(x) = 1 — x, x E [0, 1]).

1.7. The absence of a fixed point implies that the sequence is mono-tone. The rest goes as in the preceding problem. In the complex case afixed point can fail to exist: consider the mapping f given by the formula

1(z) = (Izi + (z E C).

318 X. ITERATES OF TRANSFORMATIONS OF AN INTERVAL

1.8. Consider the arithmetic mean of the images of an arbitrary pointwith respect to all the transformations of the group. 1.13. a) To constructa conjugating mapping h use the method in the solution of problem 111.5.12(see also Problem 1.16).

b) Suppose that h(sinx) = cosh(x), where h : R —, R is a continuousbijection. Prove that h([—1, 1]) = [—1, 1] and use the strict monotonicityof h.

c), e) Choose a linear function h.d) See the hint for b).1.14. a) Considering the results of problems 1.11 and 1.12, show that

since the point x = 0 is a minimum point and a fixed point of the pointx = —a/2 must play the same role for g in the case when f g, and thisimplies the necessity of the condition b = (a2—2a)/4. If this equality holds,then g(x) takes the form (x + a/2)2 — af2, and it is not hard to choose alinear function conjugating f and g.

b) Comparing the number of fixed points for f and g, we arrive at theconclusion that if f g, then a � —1/4. For all such nonzero a andfor b = 1 ± (let b = 0 for a = 0) there exists a linear functionconjugating f and g (h(x) = (4x — 2)/(b —2)). Note that here b can beanything except b = 2. For b = 2 the maximum point of the mapping g issimultaneously a fixed point, and this does not happen for f, whatever thevalue of a.

c) Assuming that g and g o h = h o we arrive at the conclusion(see problem 1.10) that h carries the fixed points of into the fixed pointsof g (that is, h(0) = 0 and h(1/3) = 2/3), and the intervals of monotonic-ity of into the intervals of monotonicity of g. However, g is in factmonotone on [0, 1/3], while is not monotone on [0, 2/3] = h([0, 1/3]).Contradiction.

d) First, it follows from the condition g([0, 1]) C [0, 1] that b E [0, 4],it follows from the condition C that a � 2, it follows from thecondition g(0) = g(1) that f(p) = f(q), that is, p = —q, and it followsfrom the condition g(0) = 0 that p or q is a fixed point of 1. Moreover,the constraints a � —1/4 and b 2 hold for the same reasons as in b).

Let x1, x2 (x1 � x2) be the roots of the equation 1 — ax2 = x. Weconsider the following cases:

a E [—1/4, 0], = [—x1, x1];

/3) a E [—1/4, 0), = [—x2, x2];

y)

a E (0, 2], = [—x2, x2];

A) bE[0,1];B) bE[1,2);f) bE(2,4].

Comparing the number of fixed points of f and g, we arrive at the con-

§1. TOPOLOGICALDYNAMICS 319

clusion that if f g, then case a is incompatible with B and F, and /3and y are incompatible with A. Arguing by analogy with c), we discoveralso that a and B are incompatible, and /3 and A are incompatible. Inthe case ô the condition c is not satisfied. Mappings I and grelating to cases a and A (respectively, /3 and B, y and F) are conjugateif a = b(b — 2)/4. The conjugating mapping is the same as in b).

1.15. a) This is the special case of b) for n = 2. A conjugating mappingis

c) A conjugating mapping is odd, and is given for x> 1 by the formulah(x)=cosha(x—1).

1.16. Obviously, the differences 1(x) — x and g(x) — x have constantsigns inside the intervals. Choose arbitrary points c0 E (a1, b1) and d0 E

(a2, b2), and let and be the trajectories of these points under theaction of f and g, that is, = = n E Z. Thesesequences are monotone, and they converge (since f and g are continuous)to fixed points, that is, to the endpoints of the intervals, as n —' ±00. Fordefiniteness assume that I and g are increasing.

We define a homeomorphism h : [a2, b2] —' [a1, b1] as follows. Leth(a2) = a1, h(b2) = b1, and = (n E Z). On [d0, d1] defineh so that it is a homeomorphism of this interval onto [c0, c1], and on theremaining intervals [dr, let h(x) = where f°(x) = x,11(x) = 1(x), ftZ(x) = f for n <0. Verify that the definition is unambiguous and that h is a homeomorphismconjugating I and g.

1.17. Suppose that I is a homeomorphism of [a, b], and a = x0 <x1 < ... = b are all its fixed points, and let e,(f) = sgn(f(x) — x),where x is an arbitrary point in (x11, x,), i = 1 n. With the helpof problems 1.11 and 1.12 show that if I g, then they have the samenumber of fixed points, and the collections of numbers e,(f) and e.(g) ei-ther coincide or (in the case of a decreasing conjugating function) satisfythe relation = —;(g), i = 1 n. To verify that this con-dition suffices for f and g to be topologically conjugate observe that Iimplements a homeomorphism of each interval x.], i = 1 n,onto itself. If [y,1, p.] is the ith closed interval bounded by fixed pointsof g, then (see the preceding problem). Verify that

if e.(f) = e1(g), then the conjugating mappings h. satisfy the conditionsh(y,1) = x11 and h(y1) = x1, which enables us to "glue together" fromthem a conjugating mapping defined on [a, b]. If = —e1(g). thenthis means that the restriction of g to the ith interval (starting from the pointa) is conjugate to the restriction of I to the ith interval (starting from b),and the conjugating homeomorphism changes the orientation, which againenables us to construct a single conjugating homeomorphism h on [a, b](moreover, h(a)=b, h(b)=a).

320 X. ITERATES OF TRANSFORMATIONS OF AN INTERVAL

1.18. By the monotone sequence theorem, the limit lim = c exists.By the continuity of f, the relation = leads to the equality1(c) = c. The behavior of the iterates is illustrated in Figure 23.

1.19. To determine the character of the behavior of the sequenceit is useful to picture the process of its construction graphically. This is donein Figure 24 for the functions under consideration. In the cases a)—d) useproblem 1.18. In e) the point x1 falls in [1, oo) whatever the position ofx0, and problem 1.18 can be used on this interval. In f) (see page 322) theinterval (0, 1] plays the same role.

In g) (see page 322) verify that the only fixed point is x = v's— 1. Theinequality x* <f2(x) = f(f(x)) <x holds for x >x <f2(x) holds for x <f. Therefore, = I and = texist and satisfy the relations f2(/) = F and f2(/1) = I', which implies that/ =jl=

a)

a)

FIGURE 24a)

b)

xI xo xo xi

FIGURE 23

2

—I

FIGURE 24b)

§1. TOPOLOGICAL DYNAMICS 321

—2 —1 0 1 2 3 4

FIGURE 24d)

3

0 1 2

FIGURE 24e)

4

322 X. ITERATES OFTRANSFORMATIONS OF AN INTERVAL

2 4

1.20. The solutions of these problems are analogous to the solution of theproblems in 1.19. In b)—d) the interval of values of x0 such that convergenceholds has the form [—p, p], where p = max(1t11, 1t21), and t1 and t2 arethe roots of the equation 1(t) = t. In d), e), and h) it is useful to considerthe sequences and (see the solution of problem 1.19g)). Inthe case of the mapping in h) the sequence is monotonically increas-ing for x0 E (1/ v3, while is decreasing for n less thann0 = > and for n > n0, is in one of the inter-vals (—oo, —1) or (1, +oo), depending on the parity of n0. Consequently,

1 or —1. For x0 E —1/v'3) the situation is symmetric.If 1x01 = 1/v'3, then = x0 and = —x0, and the limit does notexist. If 1x01 < 1/v5, then —, 0.

1.21. a) It follows from the relation

(1(x) — x*)f(x — x*)

that — x*I � 1x0 — x*I, 0 <c < 1, for x0 sufficiently close tob) The proof is analogous.c) Consider the mappings f(x) = 1 — x (x E R) and 1(x) = 1 + x2/4

(x E IR).1.22. Note first that if f is continuous, and the iterates converge to

a limit x*, then x* is a fixed point for f.a), b) Show with the help of problem 1.21 that for al > 1 the fixed

points are repelling, and hence does not have a limit unless stabilizes

FIGURE 240

2 3

FIGURE 24g)

3

§1. TOPOLOGICAL DYNAMICS 323

beginning from some n0. Considering the graphs, show that if lal < 1 , thenfor any x0 the unique fixed point is the limit of The cases a = ±1 aretrivial.

We study the mapping in d) (the same considerations can be used in thecase c)). Let a> 1. Investigating the sign of the minimum (which is equalto (1 + lnlna)/ln a) of the convex function = aX — x, we see that thefixed points x, � of f exist only for a E (1, else]. Further (seeproblem 1.18), x,, X for x0 —, 00 for x0 > and forxo =

If a < 1, then I has a unique fixed point If a then f(x*) <—1, and hence (see problem 1.21) cannot serve as the limit of ifx0 For a E 1) prove that the function = Jo I is increasing

and satisfies the inequality f2(x) > x for x < xx > (for this consider (f2)1 and (f2)11) and, using problem 1.18, checkthat and forall X0ER.

1.23. a) Verify that N(c) = 0 and use problem 1.21a).b) By Taylor's formula,

0 = = + - x) + - x)2,

where lies between x and c, which implies that

Setting x = Xk in this formula and writing Xk+1 in the form Xk —

co(xk)/co'(xk), we obtain the relation

1 2

which implies the estimate lxk+l — Cl � Llxk — Cl2 � ... � L2 1x0 —

1.24. Verify that = 1 — 1/n.1.26. Verify that on the interval between any two neighboring extremum

points of the polynomial P it has a root and an inflection point. Then useproblem 1.25.

1.27. Use the fact that = f(f . . .f(x0) == (f')'(xk), k = 0 n — 1, along with problem 1.21.

1.28. a) Use the following simple assertion:(A) If I, I are Closed intervals and J c f(I), then there is a closed

interval K C I such that f(K) = I.

324 X. ITERATES OF TRANSFORMATIONS OF AN INTERVAL

Indeed, let J = [m, M], and let x, y E I be points with x <y suchthat f(x) � m and 1(y) � M. Setting w = min{t E [x, y]If(t) � M},z = max{t E [x, w]If(t) � m}, and K = [z, w], we get the desired closedinterval.

We set K0 = and define the remaining closed intervals by induction: ifhas already been constructed and has the properties that C

and = 'ni' then, applying (A) to the closed intervalsC = and to the mapping we find a closed intervalC such that = 1

b) For any point x in the nonempty intersection we have that

1.29. Suppose that the points w <x <y < z form a 4-cycle. Verifythat, whatever the order in which these points are mapped one to the other,two of the intervals [w, x], [x, y], and [y. z] (call them I and J) satisfythe relations f(I) 3 J and 1(J) 3 1. Prove that there is a closed intervalK C I not containing the endpoints of I such that f2(K) 3 K, and useproblem 1.29.

1.30. It suffices to prove that c) follows from a) (the implications c)b) a) are obvious). Assume for definiteness that the points b, c,and d are in the order d � a <b <c. Denote [a, bJ by L and [b, c]by R, and for arbitrary m � 2 let = = = R, 'rn—i = L, and

'rn = R', where R' C R is a closed interval not containing the endpointsof R. It follows from a) that f(R) j L, 1(R) j R, and 1(L) 3 R',and hence always contains By the assertion in 1.28, there is aclosed interval Krn C R' such that fm(Krn) = 'm j If we apply tothe assertion formulated in Problem 1.5b), there is a point x0 E Km suchthat Xm = = x0. It remains to verify that = x0 for0<n<m.

If = x0 for 0 < n <m, then all the points Xk, k = 0, 1, 2 havethe number n as their period. Then since rn—i is representable in the formpn+q (p EN, 0 � q < n), Xrn_i = Xq ER', contradicting the inclusionXrni E L, which follows from the definition of x0. The case in = 1 (thatis, the existence of a fixed point) is trivial (see problem i.5a)).

The assertion proved is due toT. Li and J. Yorke [39].1.31. Consider the closed intervals = [cos(2ir/n), cos(ir/n)] and= [cos(ir/n), 1], which are mapped by f = in a one-to-one fash-

ion onto [—1, 1]. Take c to be the point cos2ir/n, b to be the uniquepoint in / such that 1(b) = c, a to be the unique point in such thatf(a)=b,and d tobel. Then f(c)=d and c<b<a<d,whichimplies(see problem 1.30) that f has cycles of any order.

1.32. b) Only the case when a > 0 is interesting. Removing the absolutevalue signs on the eight intervals into which the line is divided by the points

§1. TOPOLOGICAL DYNAMICS 325

—2

a0

a <a0

a>a0

FIGURE 25

0 ±!,

solve the equation

1—all —all—alxlll=x

and verify that its solutions are contained among the numbers of the formx = (1 +e1a+e2a2)/(1 +e3a3), where = ±1; moreover, for1 <a <a0 = (l+v'3)/2 only two of them, p = l/(l—a) and q =turn out to be roots of the equation (these are obviously fixed points of 1).For a � a0 all eight numbers of the indicated form are roots, and if werenumber them in increasing order, then x1 = p, x6 = q, and the triples{x2, x4, x8} and {x3, x5, x7} form cycles that coincide for a = a0. Thefact that they are repelling follows from the inequality lf(x)l> 1 for x = x,(1 = 1 8) (cf. the solution of problem 1.21). In the case —l <a � 1only the root q remains, and for a � —l there are no roots. The characterof the graphs for the three values of the parameter is shown in Figure 25.

1.33. a), b) Since each fixed point of f is a fixed point of any of its iter-ates fn,the fourth-order polynomial f2(x)—x is divisible by the quadratictrinomial 1(x) — x. As a result of the division we obtain the polynomiala2x2—ax+ 1—a, which has real roots for a � 3/4. If a � 3/4. then thefixed points of j2 reduce to the two fixed points of f. For a> 3/4 thereare two points x1 and x2 satisfying the relations f2(x) = x and 1(x) x.It is easy to see that 1(x1) = x2 and 1(x2) = x1, that is, {x1, x2} is a 2-cycle. The character of the graph of f2 is pictured in Figure 26 for a < 3/4,a=3/4, and a>3/4.

326 X. ITERATES OFTRANSFORMATIONS OF AN INTERVAL

Since (f2)'(x1) = f(x1)f(x2) = (—2ax1)(—2ax2) = 4a2x1x2 = 4(1 — a),the cycle turns out to be attracting in the interval 3/4 < a < 5/4, while it isrepelling for a> 5/4 (see problem 1.27).

c) Prove that f does not have 4-cycles for a � 5/4 The case a � 1is almost obvious, and we assume below that a> 1. Let P(x) = f2(x) =1 — a(1 — ax2)2 (x E R). To a 4-cycle of f there correspond two 2-cyclesof P. It is these 2-cycles we are looking for. To find a 2-cycle of P meansto find a solution of the system

J y = P(x),

1 x = P(y),(1)

satisfying the condition x y.Let L and be the graphs of the functions i' = P(x) and x = P(y),

respectively (Figure 27). It is clear that the curves L and are symmetricto each other with respect to the line y = x, and it is not hard to get theanswers of interest to us from accurately drawn sketches of these graphs.

Now let us give a rigorous argument. Show that to verify that the system (1)does not have nontrivial solutions it suffices to prove the absence of solutionswith abscissae in the intervals [1 — a, x0] and [x1, 1], where x0 and x1(x0 < x1) are points forming a 2-cycle of f. Consider first the interval[x1, 1]. Let Q be the function on 1] whose graph coincides with partof the curve In other words, Q is the function inverse to the restrictionof P to x0]. Our goal is to show that P(x) Q(x) for x E(x0, 1].

FIGURE 26

b)

§1. TOPOLOGICALDYNAMICS 327

\ y=P(x)

FIGURE 27

a)

y=P(x)

y=P(x)

328 X. ITERATES OF TRANSFORMATIONS OF AN INTERVAL

First note that

P'(x)<O, P"(x)<O, P"(x)<O

for x> and, moreover, IP'(x0)I � 1 (it is here that the conditiona 5/4 plays a role). Using the rule for differentiating an inverse functionand denoting Q(x) by t, we get that

Q'(x)= Q"(x)=

Q"(x) = +

(P'(t))4 (P'(t))5

Obviously,

P(x1) = Q(x1) = x1, P'(x1)— Q'(x1) � 0,

P"(x ) — Q"(x ) = P"(x1) +(x1)

1 1 (P(x1))3

Finally, let us see that P"(x) — Q"(x) > 0 for x E (x1, 1). Indeed, since

IP'(t)I� 1 and at2� 1 for xE(x1, 1),wehave that

,,, ,,, 3 24a3t (4a2(3at2 — 1))2P (x)—Q (x)=—24ax——————+3

IP'(t)14 IP'(')I

> — +— IP'(t)14 IP'(t)15

It follows from what has been proved that the difference P(x) — Q(x) isstrictly increasing on [x1, 11. Thus, the system (1) does not have solutionswith abscissae in (x1, 11. Then it does not have solutions with positiveabscissae at all. But if y) is assumed to be a solution of (1) such that1 — a <X <x0, then the pair 1(y)) is also a solution of (1), which,however, is impossible, because f(s)> [(1—a) > 0 (the condition a � 5/4is again used in verifying the last inequality).

If a> 5/4. then P'(x0) < —1 <Q'(x0) <0, and the existence of nontriv-ial solutions of (1) follows from considerations connected with the Bolzano-Cauchy theorem.

We remark that for a = 2 the system (1) has 12 nontrivial solutions (seeFigure 28), which form three 4-cycles (the points generating a particular cycleare marked by the same symbols).

d)(3) By the assertion in a) and problem 1.29, f does not have cycles fora � 3/4. Suppose that a >0 and let p and q be the fixed points of [.

(3)R. Sibilev helped in the solution of the this part of the problem.

§1. TOPOLOGICAL DYNAMICS 329

FIGURE 28

With the help of a sketch of the graph of y = f3(x) it is not hard to seethat the number of fixed points of can only increase as the parameter aincreases (follow the movement of the points corresponding to the extrema of

on Figure 29, and then try to make a rigorous argument). Verify that thefixed points of different from p and q generate a 3-cycle. For a � 3/4only p and q are fixed points of j3, but has eight fixed points forlarger values of a. Thus, there exists a minimal value a0 of the parameterfor which the number of fixed points of j3 is greater than two. Suppose thata=a0, f3(x1)=x1, x2=f(x1), x3=f(x2),and x1 q. The pointsx1, x2, and x3 form a 3-cycle, and each of them is a root of the equationf3(x) = x; moreover, x1, x2, and x3 must be multiple roots, because thegraphs of y = f2(x) and y = x are tangent at these points. Consideringthe number of roots and the degree of the polynomial P(x) = f3(x) — x, we

conclude that P(x) = A(x — p)(x — q)(x — x1)2(x — x2)2(x — x3)2 Dividing

P(x) by x—f(x) = a2x2+x—1 = a2 (x—p)(x—q) ,we arrive at the equality

66 55 5 44 3 43 3 4 22—a x + a x + (3a — a )x + (a — 2a )x + (3a — 3a — a )x

+a3 —2a2+a— 1 = —a6(x—x1)2(x—x2)2(x—x3)2

= —a6(x3+bx2+cx+d)2.

Equating coefficients of like powers of x, we get a system of six equations ina, b, c, and d. Elimination of b, c, and d from this system leads to the

330 X. ITERATES OF TRANSFORMATIONS OF AN INTERVAL

a=4 /

FIGURE 29

equation (4a — 7)2 = 0, which gives us that a must be equal to 7/4, anda check confirms that the system is consistent in this case. Thus, there areno 3-cycles if a < 7/4, has 5 fixed points if a = 7/4 (p. q, and threepoints generating a cycle), and f3 has eight fixed points if a> 7/4 (namely,p and q and six more points, each of which, as already mentioned, must bein a 3-cycle). Consequently, there are exactly two 3-cycles for a> 7/4.

1.34. Use problem l.14d).1.35. Let L be the curve symmetric to the graph of f over the interval

[x0, 11 with respect to the line y = x, and let

y(x) = (f(x), x), x � 1,be its parametrization. Show that by virtue of the conditions on the one-sidedderivatives there is a number 5 > 0 such that the points y(x) lie under thegraph of f if x0 <x <x0+ö. On the other hand, the point y(l) lies abovethe graph of f (or on it), and hence the graph of f intersects the curve L.The abscissa of the intersection point generates a 2-cycle.

1.36. Consider the mapping f = g2 and prove that f has a fixed pointsatisfying the condition of the preceding problem. Verify that the points ofa 2-cycle of f generate a of g. To construct an example, considerthe continuous function on [0, 11 whose graph is the polygonal curve withvertices at the points

1

1n' 16n

TOPOLOGICAL DYNAMICS 331

1.37. Let g = This mapping also belongs to the class S(A) (seeproblem Vfl.2.24c)). The assumption that g(x) = x for infinitely manypoints implies by the mean value theorem that g'(x) = 1 also on an infiniteset. As a consequence of this, Ig'I has infinitely many local minimum points.By the "minimum modulus principle" (problem VII.2.24f)), it follows thatg, hence also f, has infinitely many critical points.

1.38. There exist points u and v such that x < u <y <v <z andg (u) = g'(v) = 1. The assertion now follows from the "minimum modulusprinciple" (problem VII.2.24f)) for the interval [u, vJ.

1.39. Suppose that x E (a, b) is an attracting fixed point for g =(n EN) (if x = a or x = b. then there is nothing to prove). Then

a) Suppose first that < 1. Consider the maximal interval (r, s)containing x such that the trajectory {gm(y)} of any point y in it is at-tracted to x. If r> a, then for all m we have that gm(r) (r, s) (a conse-quence of the maximality of the interval), and if s < b, then gm(s) (r, s)for m E N. Thus, this interval has the form [a,s), (r,bJ, [a,bJ, or(r, s). The first three possibilities lead at once to the assertion to be proved.In the fourth case it follows from the obvious relations g((r, s)) C (r, s)'ir.d g(r), g(s) (r, s) that g(r), g(s) E {r, s}. In view of assertioni.38, the assumption that g(r) = r and g(s) = s leads to the inequalityg'(x) > 1 which is impossible. If g(r) = s and g(s) = r, then g2(r) = rand g2(s) = s, which is impossible for an analogous reason. Thus, we arriveat the conclusion that g(r) = g(s) ; but then, by Rolle's theorem, the interval(r, s) contains a critical point p of g that is attracted to x. Since g =p must have the form k E N, which at once implies the assertion tobe proved.

b) Suppose now that g by g2 if necessary, wecan assume that g'(x) = 1. Let (r, s) be the maximal interval contain-ing x and not containing other fixed points of g. It is nonempty, becausethe set of fixed points of g is finite (problem 1.37). Then either g(y) > yfor all y E (r, x) or g(y) < y for y E (x, s), since otherwise each ofthe intervals (r, x) and (x, s) contains infinitely many points t at whichg'(t)> 1, which leads to a contradiction with the minimum modulus prin-ciple (VII.2.24f)) and the finiteness of the set of critical points of g.

Assume for definiteness that g(y) > y for y E (r, x). If r = a andg is increasing on [a, x], then the trajectory of a is attracted to x (seeproblem 1.18). If r = a and g is not monotone on [a, xJ, then somecritical point p of g falls in (a, x), and g is increasing on [p, xJ. Thenthe trajectory {gm(p)} is attracted to x. If r > a, then g(r) = r, and bythe mean value theorem the interval (r, x) contains a point z such thatg'(z) = 1. Using the facts that g(z) > z, g(x) = x, and g'(x) = 1 andagain using the minimum modulus principle, we arrive at the conclusion that

332 X. ITERATES OFTRANSFORMATIONS OF AN INTERVAL

some critical point p lies between z and x and is attracted to x. Theargument concludes in the same way as in a). The assertion proved is due toD. Singer [43J(4)•

1.40. Let p <0 beafixedpointof f,andlet q = —p. Then f([p, q]) C[p. qJ for a � 2, and p is a repelling fixed point, ffl(q) = p for n � 1,and an attracting cycle must attract the trajectory of the point 0 in view ofassertions VII.2.25a) and 1.39. For a = 2 the mapping f is the Chebyshevpolynomial T,, and we can use problems 1.31 and 1.39.

1.41. (5) a) Choose the direction of traverse of the circle so that upon sucha circuit the fixed point t and the points of the cycle are encountered in theorder z—'a—'b—'c—'t,where ço(b)=c,and ço(c)=a. Ifx and y are two points of the circle, denote by [x, y] the arc which in thechosen direction begins at x and ends at y. We say that the condition A0holds if a]) 3 [a, bJ, and that the condition A1 holds if ço([t, aJ) D

[b. aJ (obviously, one of these conditions always holds). This will be writtensymbolically as follows: A0 = (9,([1, a]) 3 [a, bJ) and A1 = a]) 3[b, aJ). In exactly the same way, one of the two conditions in each of thefollowing pairs is always valid:

B0 b]) j [b, c]), B1=(9'([a, bJ) 3 [c, bJ);

C0 ci) [c, aJ), cJ) [a,D0 tJ) j [t, aJ), D1 t]) [a, t]).

We always have at least one of the sixteen propositions of the form

CC&DC, -

Observe that the intervals with endpoints 1, a, b. or c always containeither two intervals I and I with the properties that ço(I) I and DJul. or three intervals 1, 1, and K with the properties that D I,

K,and j JUK. Modifying the arguments used in the solutionof problem 1.30, verify that there is always a point x whose iterates visitsuccessively the intervals I, J I , 1, 1 or K, K K , 1, 1, K

n—I n—2

and which is such that = x but cok(x) .v for k <n.b) Parametrizing the circle in the usual way, we identify it with the interval

[0, 2ir). Consider the mapping 1: [0, 2ir) —' [0, 2ir) given by the formula

and carry it over to the circle. Corresponding to the point ir/3 is a fixedpoint of the circle, and corresponding to the points 0, 2ir/3, and 4ir/3 isa 3-cycle, but it is easy to see that there are no 2-cycles.

(4)The book [33] contains extensive material about iterates of transformations of a closedinterval.

(5)The solution of this problem was communicated to us by Yu. Khokhlov.

§2. TRANSFORMATIONS WITH AN INVARIANT MEASURE 333

1.42. b) Verify that the inequality m x —x0 < m + 1 (m E Z) implies

c) Let com(z0) = = Then fm(X0) = x0 + k for some integer k.Using the fact that f(x0) forn=lrn+r (lEN, 0�r<rn),provethat

d) In this case fm(x) — x cannot be an integer, whatever the values ofrn � 1 and x E IlL Then for a fixed rn and some k E Z

k<ftm(x)—x<k+1.

Setting x equal to 0, fm(0) in this equality and adding theinequalities obtained, we get that

1k <1(k + 1) (1 E N),

in particular,k<ftm(0)<k+1.

These inequalities give us that

2

Im m rn

Interchanging the roles of I and rn and adding the inequalities obtained, weget that

rn I mlwhich at once implies the existence of a limit for x0 = 0, and hence forall XEIR.

It remains to verify that for rational some power of has a fixed point.Suppose first that = 0. If it is assumed that there are no fixed points,

then {ffl(0)} is a monotone sequence (say increasing). The assumption thatfm(o) � 1 for some m � 1 implies that fkm(0) � k, k E N, whichcontradicts the stipulation that = 0. But then the sequence {fm(0)} isbounded, and its limit is a fixed point, contrary to assumption.

The case = k/rn can be reduced to the preceding case by passing to thefunction g(x) = ftm(x) — k, which gives the mapping çotm.

e) Show that and 12 differ by an integer.

§2. Transformations with an invariant measure

2.1. It suffices to compute the measures of the inverse images of openintervals (or arcs).

2.2. It is convenient to represent the baker's transformation as the com-position of the mappings T1 and represented in Figure 30 (which justifiesthe name of the transformation). It is obvious that each of T1 and T2 isarea-preserving.

334 X. ITERATES OF TRANSFORMATIONS OF AN INTERVAL

2.3. a) If T(x1, x2) = then z1 = a11x1 + — +E Z (i = 1, 2). Solving this system, we get (thanks to the condition

IdetAl = 1) that x1 — Z and x2 — Z, but this is possible only forand

b) Consider the linear transformation TA : given by the ma-trix A. It is invertible and preserves Lebesgue measure. In particular,

= for B C X. The intersections C;, = T'(B) n (X + v)(v E Z2) are nonempty only for finitely many vectors v, and translations ofthem lie in X, are pairwise disjoint (see a)), and make up the set T'(B).See also the solution of the following problem.

2.4. Consider the measure ii = on X = R2/Z2. Like Lebesguemeasure, it is translation-invariant (all sums are assumed to be modulo 1):

v(B+y) = ii(B + TA(x)) TA(x)))(1)

(BcX, yEX)Therefore (see the assertion before the statement of the problem), ii = ct.Setting B = X in (1), we get that c = 1. Thus, =

2.5. The measure is a probability measure when c = 1/In 2. It sufficesto verify that jz(T'(A)) = jz(A) for all sets A of the form (0, aJ, 0< a1. Since

the measure of the inverse image is equal to

1/k d= — lnk + ln(1 +k+a) — ln(k+a))

= —lim I ln(1 + a) —in 1 +1n2 \ \ ii+1ln(1+a) 1

ça dx= =—i —=u(A).

1n2 1n2j0 1+x2.6. a) Use the topological conjugacy of and the sawtooth mapping(problem 1.15b)), which preserves Lebesgue measure (this is obvious). We

find the image of Lebesgue measure with respect to a conjugating mapping:= where h(x) = sin (x E [—1, 1]) (it will clearly be the desired

FIGURE 30

§2. TRANSFORMATIONS WITH AN INVARIANT MEASURE 335

invariant measure). For intervals,

ji([a, x]) h' (x)]) =

and this implies that djz = (lxi < 1). The measures )and ji can be multiplied by an arbitrary constant, of course.

b) Use problem 1.15c).2.7. To prove sufficiency consider the measure zi(A) = + ... +

Where ji is a measure invariant with respect to Tn.2.8. Suppose first that T is increasing. Denote by F the set of fixed

points of T, and by G the complement of F. If A = (cr, /3) is somemaximal interval in G, then /3 E F, and hence T(A) = A. Therefore, itsuffices to prove the assertion for each such interval.

Let c be an arbitrary point in If We let xn = n E Z (T isinvertible), then the closed intervals An having and as endpoints

form a covering of A. For any positive function h0 E We define themeasure = h0 dx on and extend it by zero to The measures

= T"ti0 are nonzero on An and are also absolutely continuous. Theequality ti = >flEz tin defines the desired (infinite) measure.

If T is decreasing, then consider T2 and use the assertion in 2.5.2.9. Suppose that T has an attracting n-cycle. Then the mapping S =

TIZ has an attracting fixed point x*. Let f> 0 be the density of somemeasure ji. We show that there is a set A such that S'(A) J A and

> 0. These relations contradict either the invariance with re-spect to S or the finiteness of the measure

If SF(x*) > 0, then S is monotonically increasing in some neighborhoodof x*, and then there exist points c and d such that E (c, d), andA = (c, d) satisfies the condition above. If Sl(x*) <0, then we replace Sby S2. If Sl(x*) = 0, then IS(x) _S(x*)l � lx _x*I/2, for lx _x*l(e sufficiently small), which implies the existence of the desired set A.

It follows from problem 2.7 that there is no measure with the desiredproperties with respect to T.

2.10. a) Suppose first that a > 0. The mapping considered sepa-rately on each half-line (—oo, 0) and (0, +oo), is topologically conjugate tothe translation x x + ln a on R (h(t) = —et is a conjugating homeomor-phism in the first case, and h(t) = et in the second). Verify that the measureon R\{0} determined by the density 1(x) = 1/Ixi is the image of Lebesguemeasure with respect to h, and that this measure is invariant with respect toall the

b) Prove that in this case the desired measure ji does not exist. Since jimust be, in particular, translation-invariant, its values on intervals must beproportional to the length, that is, ji = kt (k > 0). But such a measure

336 X. ITERATES OFTRANSFORMATIONS OF AN INTERVAL

is obviously not invariant with respect to homotheties (the cases k = 0 ork =00 lead to the trivial measures).

2.11. Using the fact that these transformations are topologically conju-gate to suitable transformations in problem 2.10, verify that in the case a)the invariant measure has the form (xl dx, while in the case b) thereis no nontrivial invariant measure.

2.12. a) Let A C be such that ji(A) = 0. Thenand ji(A + y) (y E are also equal to zero.

b) The quasi-invariance of ji means the existence of a function I on

x R'1 such that for all y E R'1 and x C

(1)

and f(x, y) > 0 for almost all x, y. It suffices to verify just for boundedsets that the equalities ji(A) = 0 and = 0 are equivalent. We integrate(1) over a set Y such that > 0. Let XE denote the characteristicfunction of a set E. On the one hand,

1= ju(Y+A)dY=jdYjX4(x—Y)dJL(x)

= j dyj XY(y)XA(x —y)dp(x)=j dif Xy(x — t)XA(t)d/1(x)

= j dtjXy(xt)dIL(x).

On the other hand,

j p(y + A)dy=j dyf f(x, y)dji(x)

=j dp(x) j f(x, y) dy.

Equating the right-hand sides of these equalities, we conclude that if =0, then I = 0, and then we get that ji(A) = 0, because f(x, y) > 0 and

By the theorem, the measure p is determined by somedensity g. We show that the set e = {xlg(x) = 0} has zero Lebesguemeasure. Assuming the opposite, consider the family of translates e + Xkof e by vectors xk forming a countable dense subset of The set C =

Uk(e+xk) has zero Lebesgue measure (see problem VIII.1.23), and thenalso u(C) = 0. But /L(e+xk) = jz(e) = 0, that is, = 0. Contradiction.

2.13. a) If f � 0, then A i'(A) = fT_(A)f(x)dx (A c X) is anabsolutely continuous measure on X, and the existence of follows fromthe theorem. If f is arbitrary, then it can be decomposedinto a difference of nonnegative functions.

b) Let f � 0, and let U and V be neighborhoods of x and y = T(x)with Then

jf(x)dx = j f(y)dy= j f(T(x))IdetT'(x)Idx.

§2. TRANSFORMATIONS WITH AN INVARIANT MEASURE 337

Using "differentiation with respect to the domain," that is, dividing bothsides by and passing to the limit as U shrinks to the point x, we getwhat is required.

c) This is proved similarly.d)Verify(1)forthesets A=[O,aJ,2.14. a)

1/' ifd1

= fd(x)dx= '(A) JT'(A,)

—n

— l<j<n

Which implies that

= = n T (A1)).

Therefore, the transformations

(m) 1 kand =— irum

O<k<m

map the compact set K = {u E Rj >.1<i<n u. = C} (C > 0) into itself.Consequently (for fixed u E u 0), the sequence {V(m)} has a limit

point v = lim Since it is continuous,

(m) . 1 k= lirn irv = lim — it ul<k<m

= lim + — U)) = limv(m) = v = (v1 , v2 va).

It follows from Perron's theorem that if > 0 for all i and j, then sucha limit point is unique and independent of u E K, and also v. > 0 for all1. Frobenius extended this result to certain nonnegative matrices (see, forexample, Gantmakhe?s book, "The theory of matrices").

2.15. a) It is easy to find (for even n) by algebraic methods a fixedpoint of the discrete Perron-Frobenius operator (see problem 2.14), that is,an eigenvector v of the matrix it that corresponds to the eigenvalue 1.Furthermore, it turns out that Vk = 0 for k � n/2, and Vk = V3n/2kfor k > n/2. This suggests the following answer: if Pf = f (P is thePerron-Frobenius operator; see problem 2.13), then 1(x) = 0 for x < 1/2,and 1(x) = f(3/2 — x) for x> 1/2. To verify it we write the equality

Pf( = j for y <

338 X. ITERATES OF TRANSFORMATIONS OF AN INTERVAL

(see problem 2.13c)), which implies that

J 1(y) dy= J

P1(y) dy= J

1(z)dz, n = 1, 20 0 0

which is possible only for

f(y)dy = 0(n EN).

Consequently, f(y) = 0 almost everywhere on [0, 1/21. It follows fromthis immediately that f(y) = Pf(y) = 1(3/2 — y) for y> 1/2.

b) For n = 3 the discrete Perron-Frobenius operator is written by thematrix

fO 0 1/2it= 0 1/2

1 0

(see problem 2.14a)). The vector v = (1/5, 2/5, satisfies the equalityirv = v. Let 1(x) = 1/5 for x < 1/3 and 1(x) for x> 1/3, andshow that the measure = Idx is invariant with respect to T.

c) The transformation T does not have an absolutely continuous invariantmeasure, because it has an attracting fixed point (see problem 2.9). Thefixed points v E for the operator it have the following form: v1 = c,Vk = 0 for k> 1. The measures on the interval that correspond to themare determined by the densities = n for x � 1/n and = 0 forx> 1/n. If is the measure determined by such a density, then it is clearthat —. 0 if A = (e, 1), e > 0, and 11) = 1. This enablesus to compute the limit measure, which is concentrated at the point 0 (it isinvariant with respect to T, but does not have a density).

2.16. Let f be a function on fl satisfying the condition P1 = 1 (prob-lem 2.13b), e)). Denoting 1(0, 1) by C and writing the equation Pf = Ifor the mappings z yz (y > 0), we get that [(0, y) = C/y2. The appli-cation of the same device to the mapping z z + x (x E R) leads to theequality f(x, y) = 1(0, y) = C/y2. It remains to see that an arbitrary au-tomorphism T is a composition of automorphisms of the form considered,and hence Pf=[ forall T.

2.17. By the results in 2.13b) and e), the density 1L of the measuremust satisfy the relation

IL(ax,ay+b)=—IIL(x,y). (1)

Setting x= 1, y=0, and fL(a,b)=A/a2 for any (a, b) E P. Obviously, such a function satisfies (1). Thedensity 1R of the measure 11R is found similarly from the equation

IR(ax, bx + y) = y).

§2. TRANSFORMATIONS WITH AN INVARIANT MEASURE 339

2.18. Let f be the density of an invariant measure for T. Consider thefunction

b xh(x)=a+(b_a)(j f(t)dt) L f(t)dt

and verify that the mapping S = ho To has the required property.2.19. It Suffices for us to confine ourselves to the case when E [0, 1)

(see 1.42e)). Let ji be a finite measure on S' that preserves T (it exists bythe Bogolyubov-Krylov theorem). It can be assumed that ji(S') = 1. Iden-tifying S' with the interval [0, 1), consider the function h(x) = ji([0, xJ).The assertion 1.42c) ensures that T does not have periodic points. Conse-quently, the trajectory of each point is infinite, and then the measure of eachpoint is equal to zero, the function h is continuous, h(0) = 0, and h(x) —. 1

as x—+ 1. Let /3=u([0, T(0)J). Then

h(T(x)) = ji([0, T(x)J)

= (ji([0, T(0)J) + ji([T(0), T(x)J)) mod 1

= (/3 + ji([0, xJ)) mod 1 = (h(x) + /3) mod 1.

From this it is clear that h o T = o h, where is the rotation of thecircle through the angle /3 E [0, 1).

To prove that /3 = we consider some function f: R R such that1(x) = T(x) for x E [0, 1) and f(x + 1) = 1(x) + 1 for all x E R, andverify that limf(0)/n = /3 (see 1.42b), e)).

Let ii be a measure on R whose restriction to each interval of the form[k, k + 1) (k E Z) is obtained from the measure ji by translation byk. Then it follows from the invariance of ji that =i'([O, 1(0)]) = /3 for n E Z. Therefore, /3 = But since

Iv([0,y])—yI< 1 forall y'

/3= lim !v([0, f(0)J) = lim !f(0) =

2.20. If T is not ergodic, then there is an invariant set A such thatji(A) > 0 and ji(A) > 0, where = X\A. Then a function f taking thevalue a on A and b a) on A is invariant and not constant. Conversely,if f is an essentially nonconstant function, then there is a number c suchthat the sets A = {x E XIf(x) � c} and have positive measure. If f isalso invariant, then A and A are invariant, and T is not ergodic.

2.21. a) Use the preceding problem and note that to prove ergodicity itsuffices to check the functions I in 1]). As is known, every suchfunction can be expanded in a Fourier series:

2ninxf(x) = f(n)),

nEZ

and the sequence {cfl}flEz of coefficients determines f uniquely (up to valueson a set of measure zero). Corresponding to the function g(x) = f(T(x)) is

340 X. ITERATES OFTRANSFORMATIONS OFAN INTERVAL

the Fourier series >flEZ If 1(x) = g(x) almost everywhere,

then = for all n E Z. For irrational this implies that = 0

for n 0, that is, the Fourier series for f reduces to a constant. It followsthat 1(x) = c0 almost everywhere, which means ergodicity. If = k/I E

then the coefficients cml (m E Z) can be nonzero, that is, there existnonconstant invariant functions.

b) Arguing as in a), we take the function f(x) and get

from the equality 1(x) = f(T(x)) the relations = =

= for all n E Z. The condition I E 22(X), orleads to the fact that = 0 for n 0, which implies ergodicity of T as ina).

c) This is analogous to b), since the transformation T can be interpretedas the mapping x nx mod 1 of the interval [0. 1).

d), e) The functions in 22(X) can be in a double Fourierseries: f(x, y) ,,2EZcfl,2ee =Then

f(TA(x, y))n1 ,n2EZ

=

Denote the vectors (n1, n2) and (m1, m2) by n and m, respectively. Thenm = A*n, where A* is the matrix obtained by transposing A. If I isinvariant, then c, = = = ..., and the convergence of the series

>nEz2 leads to the fact that if the trajectory {(A*)kn} of the vector nis infinite, then = . = 0.

If the eigenvalues of A are not roots of unity, then (A*)kn n for allk E N and n E Z2\{0}. In this case 1(x) = almost everywhere as above,that is, TA is ergodic. In the opposite case there exist a number k E Nand a vector v E R2 such that (A*)kv = v. Since A* is an integer matrix,a vector v satisfying this relation can be assumed to be an integer vector.As a nonzero invariant function we can take the sum of the exponentialscorresponding to the vectors v, A*v Consequently, TA isnot ergodic.

2.22. Suppose that the trajectory {Tm(x)} of the point x is not dense inX. This means that there exists an open ball B (with rational coordinatesof the center and with rational radius) such that x U,flEN T_m(B) = GB.Since GB is open, it follows that > 0. It follows from ergodicitythat = 0 (consider the invariant set c GB). Sincethe balls with rational parameters form a countable family {Bk}, the unione = has zero measure. And every point in X\e has a densetrajectory.

§2. TRANSFORMATIONS WITH AN INVARIANT MEASURE 341

2.23. If S is ergodic, then in view of the preceding problem the trajectoryof almost every point is dense, and this, by the result in VII.4.5a), impliesthat i,..., are linearly independent over Q modulo 1. The sufficiencyof this condition can be proved with the help of Fourier series (modify thesolution of problem 2.21a)).

2.24. Show that the transformations T and S admit the same "symbolicrepresentation." Let D be the set of dyadic rational numbers in I = [0, 1],and let X = I\D. Since = 0, it suffices to present a measurableinvertible transformation h : X —. X that preserves Lebesgue measure andis such that Soh=hoT.

Let = [0, l/2]nX and = [1/2, 1]nX. Define successively thesets and 4=S'(4'), k=2, 3 Provethatthesesets have the following properties:

k k k —n1) n n n = 2 for any k1 < k2 < •. < and

i1 E {0, l} (can be verified by induction);

2) the set consists of a single point for any sequence 1k E

{0, 1} , and each point in X is uniquely representable as such an intersection.The last property allows us to identify the points in X with binary se-

quences,andthen 1, i2, ...)=(i2, 13,...).Let x = 0.x1x2... be the binary expansion of a point in X. Setting

1k = Xk, we get a binary sequence determining (with the help of 2)) a pointy E X. Let h(x) = y. Obviously, T(x) = 0.x2x3..., which yields theequality Soh=hoT.

It remains to see that h is measurable and preserves Lebesgue measure(the invertibility of h is obvious). This follows from the fact that for anyinterval of the form = {x = = es,..., = e,j where

E {0, 1}, the equality£

X) = n n ... n and prop-erty 1) are satisfied. The measurability of h follows from the fact thatthe inverse image h'(A fl X) of any dyadic rational interval A =(k + of rank n coincides with A' = [1f2n, (1+ 1)12n].

2.25. a) Use the fact that by a linear substitution T can be transformedinto the transformation S in the preceding problem, and hence is metricallyconjugate to the mapping x (2x) mod 1 (x E [0, 1)), which is ergodic(see problem 2.21b)).

b) The problem can be reduced to the preceding with the help of thesolution of problem 2.6a), in which it is established that the mappings x

1 — 2x2 and x 1 — 21x1 are metrically conjugate.c) With the help of problems 1.15b) and 2.6a) replace the mapping Tn

by the saw-tooth mapping Then, modifying the assertion in problem2.24, replace by the mapping x (nx) mod 1 (x E [0, 1)), which wasactually proved to be ergodic in 2.21c).

2.26. For every sufficiently large k EN the set Ak = {x E XIf(x) � 1/k}

342 X. ITERATES OFTRANSFORMATIONS OF AN INTERVAL

has positive measure. Consider the set B of points x in Ak such thatT'1(x) n E N (nonrecurrent points). Verify that Tm(B) n B =

= 0 for all m, n E N. The finiteness of the measure ofX leads to the conclusion that jz(B) = 0, and this implies that almost everypoint x E Ak returns to Ak infinitely many times, that is, � 1/kfor infinitely many indices, and = +00.

To finish the proof, exhaust X by the sets Ak.2.27. Apply the ergodic theorem to the characteristic function of A.2.28. Let B c X be a set such that u1(B) ji2(B), and let XB be its

characteristic function. By the ergodic theorem, for i = I or 2

XB(Tk(x))_/uj(B)O�k<n

for all x in a set A. with u1(A1) = I. Since A1 nA2 = 0, this implies theassertion to be proved.

2.29. a) First suppose that f(0) = 1(1). Then f can be thought of asdefined on the circle, where it is uniformly continuous. For any e > 0 thereexists a ö >0 such that if I(x — x')I mod ii <ö. then

I(Sk(x) — Sk(xt)) mod ii <ö and If(Sk(x)) — f(Sk(xI))I <e(k E N).

Let at(x) = (Sk(x)), and suppose that x' is a point at which, in

addition, (1) holds. Then — <e for all n, which implies that

j f(x)dx+e.

Since e is arbitrary, this implies (1) for x.If f(0) f(1), then we can select functions g, hE C([0, 1]) such that

g(0) = g(1), h(0) = h(1), g � f � h, and — g(x)) dx <e. It thenremains to use the inequalities

� <e (x E [0, 1]).

To prove b) replace f in this argument by the function XA.See problem 1.3.9 for a proof of a) without use of the ergodic theorem.2.30. Let T(x) = (2x) mod 1. For those x E [0, 1) that are not dyadic

rational numbers (as always, we can neglect a set of measure zero whenconsidering questions involving measure) the equality Xk = 1 means that

T is ergodic (problem 2.21b)), the ergodic theoremgives us that the desired limit is 1/2 for almost all x (problem 2.27). It canbe proved in exactly the same way that the corresponding frequency is equalto 1/p for the p-ary number system. The following fact can be deduced atonce from this: the set of all x such that for any p all the digits in the p-aryexpansion of x occur identically often (such numbers are said to be normal)has measure 1. See also problems IX.2.14 and IX.3.10.

§2. TRANSFORMATIONS WITH AN INVARIANT MEASURE 343

2.31. If A is invariant, then n A = A for all n E N, and therelation (*) leads to the equality u(A) = (u(A))2, that is, u(A) is equal toOorl.

2.32. Let x = EXk/2 and y = EYk/2 (Xk,Yk E {O, 1}) be thebinary expansions of numbers x and y in [0, 1). If T(x) = y, then

Yk = Xk+l (k E N). We prove first the relation (*) from the previousproblem for sets of the form = {x E [0, l)Ixk = ek, k = 1 m}

E {0, 1}) (we call them If A = AC Cand B = ,, , then

(for n > m)

A n = {xlx1 = = = =

Computing the measures of these sets, we get

1 1 1= = n T (B)) =

and hence (*) holds. In almost the same way (*) can be verified for setsA and B representable as finite unions of cylinders (in particular, for allintervals with dyadic rational endpoints). Every measurable set A can beapproximated to any accuracy e by a set A' representable in such a form(this means that + < e). In this case we Write A A'.If B B', then also and An A' nFrom this, n � 2€. Since e is arbitrary, (*) isproved.

2.33. a) The proof is the same as in the case of Lebesgue measure (seethe preceding problem).

b) Verify that the measures coincide on intervals of the form [kf2'1,(k+

c), d) Use the ergodicity of (see problem 2.31) and assertion 2.28.

2.34. Suppose that the first digit of the number 2kis p. This means

that for some m E N

m+log10p <m+1og10(p+1). (1)

Consider the transformation S(x) = (x + mod 1 on [0, 1), with =log10 2. The relation (1) means that Sk(0) E [log10p. 1og10(p + 1)). Sinceis irrational, the desired limit is equal to 1og10(1 + lfp) (see 2.21a), 2.29b)).

2.35. a) Consider the sets Aa.. a, = {x E (0, 1J1a1(x) = a1, , =

which are intervals of monotonicity of Let ip = Wa a, be the

function on [0, 1) that is the inverse of on A = Aa...a• Using the

explicit expression

= 1

1

=t E 1)

344 X. ITERATES OF TRANSFORMATIONS OF AN INTERVAL

for ip (we Omit the elementary computations), we can compute the lengthof A:

= +

If 1],then

flA) = - = - 1

+ +

Since the quotient on the right-hand side is between and wearrive at the inequality

� nA) �for an arbitrary interval A C [0, 1), hence also for an arbitrary measurableset A. Ifwe replace). by a Gauss measure jz with density between 1/(21n 2)and 1/In 2, then, replacing the constant 2 by C = 4/In 2, we can write theinequality

� nix) � (1)

Since the value of the measure of any interval can be recovered from itsvalues on all possible intervals of the form L'&a..a (this follows from the

fact that < 2fl+l) we can replace in (1) by an arbitrary set.

Suppose now that A is an invariant set, and take the = (0, 1]\A in placeof A. Then A)u(A) � jz(A nA) = 0, which implies that jz(A) = 0 or

=0.bi) Use assertion 2.27.b2) Since

ak=— > a1(T"(x)),O<k<n O<k<n

the ergodic theorem should be applied to the function 1(x) = a1 (x).However, this function is not integrable, because 1(x) = k for x E(1/(k + 1), 1/k). Therefore, we replace f by the smaller function 1N =min(f, N), for which

1 1 f'fN(x)hm — (x))=

dx.O�k<n

The expression obtained increases unboundedly as N —, 00. and hence thelimit of interest to us is infinite.

b3) Apply the ergodic theorem to the function 1(x) = lna1(x).b4) Verify first the equality = which implies that

p1(TIZi(x))— 1—'

§2. TRANSFORMATIONS WITH AN INVARIANT MEASURE 345

and hence1 1 fPflk(T(x))

——lnq(x)=— mlnO<k<n

If we set 1(x) = mx, then the last expression differs from the sum

1f(Tk(x)) by the quantity

a - Infl (Tkx

O�k<n V

which tends to zero, because for all x E (0, 1)

in < -1 = - = _L <±..— qk —

— 2k1

(the last inequality can be proved by induction, for which it is useful towrite out the recursion relations connecting and withand Khintchin&s book "Continued fractions" contains a wealthof information about continued fractions).

Finally,

1 1 ftlnx— f(i

O�k<n

Answers

Chapter I

1.6. Yes. 1.7. Yes. 1.8. Yes. 1.14. N � ha2. 1.20. a) {0, 1,1/2, 1/3, ...}; b) c) R; d) [—1, 1]. 1.25. Yes. 1.28 b) [0, 2],[—1, 1]. 1.29. No. 1.30. For example, the set of vertices of regularn-polygons inscribed in circles of radii 1 — 1/n with center at a particularpoint. On the line—the set of midpoints of the intervals complementary tothe Cantor set.

2.10. If and only if is an arithmetic progression. 2.15. c)b). 2.18. f) No. 2.19. c) Reverse the inequality for p < 0; d) forp E (—1, 0) the inequality is satisfied for all x E (0, 1).

3.6. a) [—1, 1]; b) R; c)—e) [—1, 1]. 3.7. a) {0, 1/q (q — 1)/q}if x = p/q is an irreducible fraction; b)—d) [0, 1]. 3.8. Yes.

Chapter II

1.1. 1/2. 1.2.

1.3. a), b) 2. 1.4. 1.5. e(2H/(e_ 1). 1.6. a)

for 1.7. 4e'. 1.8. —plnp—(1—p)ln(1—p) if pE(0, 1); 0if p=0 or p= 1. 1.9. a) (1/2)1n2;b) 2E/8;c) (1/2)1n5;d) 2Eln2;e)2E. 1.10. a) 0; b) 22E; c) +oo; d) 0. 1.12. b) C2 =

3.2. a) e'(x0 + (e — 1)x1); b) — 1)x0 + x1). 3.3. a)— 1 + 21n2) for x1 1 — 21n2; b) —, 1 for x1 = 1 — 21n2.

Chapter III

1.1. a), e) The empty set; b) the set of all functions; c), d) the set ofconstant functions; f) the set of functions bounded on each finite interval;g) the set of functions uniformly continuous on R; h) the set of functionsbounded on R; i) the set of nondecreasing functions uniformly continuouson IlL 1.2. No. 1.5. Yes. 1.6. b) No. 1.7. Yes. 1.12. The assertionis true for a separable space and is not true for a nonseparable space. 1.14.

a) No; b) yes. 1.23. The functions of the form ax + b. 1.25. Yes.

347

348 ANSWERS

2.3. No. 2.5. For a finite family of functions it is true, for a countablefamily it is false.

3.9. a) b), c) does not imply a). 3.12. Yes, g'(O) = 2/a. 3.18. 2.3.19.

4.1. No. 4.2. = (1 — 12)1(1 + 12) + 1(21)1(1 + 12). 4.3. Yes. 4.4.

c) For example, = 2/9, = 1/3.5.1. a) The constants; b) 1(x) = g(lnx). where g is an arbitrary contin-

uous function on R having period In 2.5.3. f(x)=Alnx. 5.4. f(x)=Axlnlxl for XER, f(0)= 0.

5.5. 1(x) 0 or 1(x) = x. 5.6. 1(x) 0. 1(x) = x, and, moreover,1(x) = —x if n is odd. 5.10. 1(x) = Alnx + h(lnx), where h is anarbitrary continuous function on (—oo, +oo) with period T = ln(a/b) andA = — 1(b)). 5.11. a) = 0 and = (lnx)H(lnIlnxI)for 0 < x < 1, where H is an arbitrary continuous function on R having

w(InIInxI)

period 1n2; b) 1(x) 0, 1(x) 1, or 1(x) = xe for 0 <x < 1,where w is a continuous function on R having period 1n2 and satisfying thecondition w(t+h)— w(t) > —h for t, hEIR, h >0. 5.12. For example,

for f(0)=0. 5.13. f(_-)=: or

for Izi = 1, 1(0) = 0, and f(z) = for0 < 1:1 < 1, where w and H are continuous real functions with period1n2, and co(t+ h) — w(t)> —h for t, h ER and h > 0. 5.15. f(x, y) =a(x2+4xy+y2)+b(x+y)+c, where a, b, and c are arbitrary real numbers.5.16. If f const, then g exists if and only if I is strictly monotone onR or f is even and strictly monotone on [0. +oo).

In the answers to problems 5.17 and 5.18 the letter stands for a char-acter.

((k,I)EZ2), where C2 E S'; c) ço(n) = C? (n E where is an nith root ofunity. 5.18. a) = eUX (x ER), where I E R;b) ç9(x) = where(x, t) is the inner product of the vectors x and I in c) n) =(x E R. n E Z) where t E R and E S1; d) = Re :+s Im:) (:E C)where t,sER;e) 9'(C)=C' nEZ; f)(x >0), where I ER; g) ço(:) = (:E ce). where t ER andn E Z. 5.19. An arbitrary continuous homomorphism is defined by therespective equalities: a) = ax; b) = c) = alnx; d)ç9(x) = xa; e) = f) = where a and bare arbitrary real numbers, and n E Z.

Chapter IV

1.1. Converges. 1.2. a)—c) Diverges. 1.4. a) p > 0; b) p> 1/2; c)p > 2. 1.5. a) Converges; b—d) diverges. 1.7. a), b), g) Diverges: c),d), f), h) converges.

ANSWERS 349

2.5. For nonmonotone sequences the assertion is false. 2.6. a) Yes; b)no. 2.9. Not always. 2.10. Not always. 2.12. d) Not always. 2.14.For a = +00 either convergence or divergence can occur. 2.18. Yes.2.19. Monotonicity is essential. 2.20. b) The series >2 can converge,but if T +00, then it diverges. 2.21. b) True.

3.1. Yes. It is impossible to choose a positive sequence. 3.2. Yes. 3.4.The converse assertion is true if j 0. Monotonicity is essential.

4.1. — sinx), 4.2. cosx — cos3x, 4.3. hr. 4.4.

4/(4 — z + z12) di. 4.5. kim. 4.6. e/2. 4.7. 1 — y. 4.8.(y—(1n2)/2)1n2. 4.9. y+1n2. 4.10. 4.11.

—ln2E).

5.1. 2y <x2, q > 2p. 5.2, 5.3. Yes. 5.4. The series converges nonuni-formly. 5.9. The condition < oo is essentiaL

6.3. For example, E = = 0 or 1}, = 22En!. 6.6. Thelimits are equal to 2/2E and 1/2; 2(lnn). 6.8. a)(2E—x)/2 for 0<x<2,E; b) for 0<x<22E;c)—x/2 for lxi d) sin cot + cos for 0< x <

Chapter V

1.1. The integral converges. 1.2. The integral converges for p> 2. 1.3.The integral converges in the following three cases: q < —1, r and p

1.4. The integral converges in the following three cases: q <—1 r and parbitrary; q�—1, p>0; r>0, 1+q<p/r. 1.5.

Theintegralconvergesifq>1, p�0 or q>1+p, p<0. 1.6.Theintegral converges for q < —1 and p > 0. 1.7. The integral converges forp < 1 and q> 1. 1.8. The integral converges if p <—1, q � 0 or q > 0,2(p + 1) <q. 1.9.—i.!!. The integrals converge. 1.12. a), b) No. 1.13.

For any e > 0. 1.14. a) f(0)ln(b/a); b) (f(0)—+ 1T f(x)dx)ln(b/a); c)(1(0) — 1)ln(b/a). 1.16. 0. !.i7. —(,E/2)1n2. 1.18. 1.19.

1.20. 2E2/6. 1.2!. 2E/2—arctana. 1.22. ir/2. 1.23.

1.24. 0. 1.25. ,E(lal + ibi — Ia + bI). 1.26. —y. 1.27. V2 + 2E2/6

1.28. y. 1.29. (!/q— !/p)y. 1.30. (!/q— !/p)y. 1.3!. 1n4. 1.32.

—(!/2)1n22. 1.33. 1.34. 3/4;1n2; !)2E. 1.35. 1.36. 1.37. a), b)

1.38.

2.!. 21n2. 2.2. a) n/(n+!) ; b) !/(n+!); c) 2.3.

2.4. a) + b2), b) + 1)12). 2.5. 27E —

2,E11x112/3 for lxii � 1; 42E/(311x11) for lxii � 1. 2.6. 2E2(cosh! — 1).

2.7. 2.8. t E Int(A); K(t) = — — — for= (ti, t2, t3, t4) E Int(A). 2.9. M = (b — a)/3 P = 4/9. 2.10. 1/2.

350 ANSWERS

2.11. a) P1(a) > 1/2 for any a, P1(a) 1; b) 1/2. 2.12.

2.13. a) 4R/,E; b) 5,E/6. 2.14. 'P(x, y) = 2E(x2 +y2 — 1) for x2 +y2 � 1and 'P(x, y) = 2EIn(x2 + y2) for x2 + � 1. 2.17. (n!a1a2•2.19.

jChapter VI

1.2. a) a'IIn(l —t)I ; b) In(1 —t)I ; c) . 1.3. a) A/mA;b) A; c) 2A/2E; d) e) 1.4. a) (mnA)/p; b)c) d) AIifA. 1.5. a) (22En2)1; b) c)

1.10.

1

forpczl,

b) (eImneI)'.2.1. 1(0). 2.3. dt. 2.5. a), b) f'(1 + c)

f'((1 2.9. a) b) c)

d) f'(2 — e) f) A2; g) where

h) A'. 2.10. a)

b) c) + 2.11. a) b)

c)

3.11. a) b) 3.27. a), c) f(1 + l/p)(1 b)

((1—

!)S(t) = In ti) (IP) (p(1 —

d) IIn(1 —t)I/Ina; e), f) IIn(1 —t)I/InIIn(1 3.28.

af'(l+p)

b

ANSWERS 351

3.29.

1n2

________

1a)—; b) ;1 —t 12(1 _t)2 2(1 —1)

d)In(1 — 1)1

e)In(1 — 1)1

. f) g),E2

ma 1—i ' 6(1_t)2'lmn(1—t)I . 1n2 . 1

1) —————. —(1_t)2 ' 1—i' 2(1—i)'

k),E2•

I)12(1 —t)2 (1_t)2

3.30.

a) 4(1-i)' b) -tv'; c)

d' !Imn(1 —1)11'2 1—i

, e) --i)

3.31. a)IIn(1 — 1)11(1 —1); b) (,E2/6)(1 —

4.3.

2 33a)y=x—x +•••;

b)y=x—1 9 2 109 3

x—

2 3

x 8s i+a7x +....

4.8. a) 1/n; b) c) d) e), f) 2/n; g) 4/n2;h) 4.10. For example, 1(x) = x — x2e_hhIX for 0 <x � 1,f(x) = f(1) for x> 1. 4.11. If the sequence does not become

stationary, then or (—1) depending on the choiceof x0. In particular, the first case is realized if 0 <x0 < 1. 4.13. a,

where

K =" I

352 ANSWERS

4.14. ((p + for p> —1, ; for p = —1, andC,,n for p <—1, where is a constant depending on p.

Chapter VII

1.33. No. 1.36. a) x2/2; b) where q = p/(p — 1); c) +00outside the interval [mini', max 1']; the piecewise linear function havingas break points the values of f' on the interval [mini', maxf'] ; the valuesof (f)' are the break points of the function f; d) +00 for x � 0,for x<0,where q=p/(p—1); e) f) f(—x); g) x(lnx)—xfor x>0, 0 for x=0, and +00 for x<0.

2.12. 3a2.3.3. a) x) 1, , x) x, x) x2 + x(1 — x)/n; b)

(1 +x(a'm —4.3. For arbitrary 4.4. a), b) ). are linearly independent

over 4.5. a), b) ). are linearly independent over modulo1. 4.19. Only if the sequence {ek} is periodic. 4.20. No.

Chapter VIII

1.4. a) 1/10; b) 4/10; c) 1/100; d) 1/2. 1.5. 1. 1.6. a) Yes;b) no. 1.11. a) if

— � 2. 1.12. Yes. 1.13. a) =b) c = 1/k!; c) if > + 1 infinitely many times; in this case

Im = >k>m2. 1.16. a = ).2(A), b = L, c = 2E, where L is the lengthof the curve bounding the set A.

2.2. For example, where is the sequenceof all rational numbers. 2.3. Yes. 2.5. If and only if

E Elf(x) = = 0. 2.6. a), b), d) F(x) = + 2arcsinxfor lxi � 1, F(x) = 0 for x < —1, F(x) = 22E for x � 1; c) F(x) = 0

for x < —1, F(x) = + 2 arcsinx) for —1 � x � 0, F(x) =+ 4arcsinx) for 0 � x � 1 , and F(x) = 22E for .v> 1. 2.7. a) Yes;

b) no. 2.8. a) F(t—c), F(tfc), b) F(t). 2.9. c) No in general.2.10. f*(t) = F' — z), 0 � z � 2.11. a) sin((2E — z)f2)(0 < z � 2,E); b) cos(z/4) (0 � z � 22E); c) cotz (0 < r < 2E). 2.12.f*H\/17

3.1. 3.6. Yes. 3.7.a) p> 1; b)1<p<2;c)p<1;d)p>2;e)—1<p<2. 3.8.a),b)Yes;0,0;c)no; 0, 0; d) no; 3.9. a) p <2; b) p < 1: c) p < 3. 3.15. No,as is clear from the example E = 1), g 1, f(x) = 3.16. a)1/2, 1/4;b) 1/4n. 3.17.a) 1/10; b) 1/100. 3.18. 1/11. 3.19.a)3; b) F(t)=0 for 1<1, for n<t<n+1; f*(r)=nfor (4)fl � <(2)Thi 3.25.

n/2--l

lin/2)Joe.

ANSWERS 353

3.26. — a/(1 + a2)). 3.27. ir2R2(4a2 + R2). 3.30.+ l/p)/f'(1 + n/p).

331 a)F((p+1)/2)F2(n/2)

— 1)/2)f'((n + p + 1)/2) laD.

3.35. b) No.

3.39. a)2ir2 (m +

n J

1 1 1 14.2.b) I 1+1+—+...+eala2•..am \ 2! (rn—i)!

4.5. 2ir, where L is the length of the curve bounding A. 4.6.a) 1/2; b) 3/8; c) 1/2; d) 1/2; e) 1/2. 4.7.a),b) p<21og32.

5.3. a), b) c) d) e)5.4.a) 1;b) 3/2. 5.5.a),b) 5/3; c) 4/3. 5.6.a),b),c) 2. 5.7. 1og32.5.8. 5.11. 2—1og32. 5.12 c) No. 5.19. a) p = 1,

= (b — a)/2; b) p = 1, = it; c) p = 2, jz2(A) = d)

p = n, = cc'. 5.20. p = log3 2, = 5.21. a)dimH(A) = M-dim(A) = log3 2; b) dim11(A) = 0, M-dim(A) = 1; c)

dim11(A) = log3 2, M-dim(A) = 1. 5.24. a) Im = b) 1/2; c)00. 5.26. (lnm)/ln n. 5.27. dimH(A) = (In 5)/In 10, dim11(A + A) =(lng)/ln 10.

6.2. a) About 14%, b) more than 99%. 6.3. Less than 0.25%. 6.4.The probability that 1x11 <e tends to 1 as n —. oo for any e > 0. 6.5.

6.10. b) 1/2. 6.18. The ratio of the volumes of the cube and theball is equivalent to: a) b) In case

b) the ball is contained in the cube only for n � 9. 6.19. =6.22. b)

?

Chapter IX

1.1. Convergence in measure holds in all cases except e). There is anintegrable majorant only in cases b) and d). 1.10, 1.11. The assertionis true also for R. 1.12. The theorem ceases to be true When (0, 1) is

replaced by R. 1.14. If the condition is violated, then the assertion isfalse. 1.15. a), b) Yes; c) no.

2.11. it2. 2.19.

3.1. d) 1 — 2x. 3.4. a) (sin 1)/i; b) e"2 eUt da(x); c)

cosh(zck). 3.12. a) F is the Cantor function; b) F(t) =(t + 1)/2for ti � 1, F(t) = (1 + sgnt)/2 for � 1.

354 ANSWERS

4.1. a), b) c) . 4.9. No. 4.18. a), b) = c)

4.27. The measure concentrated at zero or the measure with density

(c >0).

Chapter X

1.2. a) No; b) yes; c) no. 1.3. h(x) = 1 — x, k(x) = —x. 1.13. a)

Yes; b) no; c) yes; d) no; e) yes. 1.14. a) b = (a2 — 2a)/4, a E R; b)for a�—1/4, a=0:bER\{2};d) fora = b(b — 2)/4 with b E [0, 2) U (2, 4); moreover, = [—x1, x1J for a E[—1/4, 0J and b E [0, 11, = [—x2, x2J for a E [—1/4, 0J and b E [1, 2),and = [x1, —x1J for a E (0, 2J and b E (2, 4), where x1, x2 (x1 � x2)are the roots of the equation 1(x) = x. 1.19. a) 0; b) 2; c) —2; d) 1;e) 1; f) 1; g) 1. 1.20. b) = 0 for x0 E (—1, 2), = 2

for x0 E {—1, 2}, = +00 for the remaining values of x0; c) the limitisequalto

(1 for the remaining x0; d) the limit isequal to (—1 for 1x01 <(1 (—1

for IxoI = (1 + and +00 for the remaining x0; e) the limitis equal to (1 — for 1x01 < (1 + (1 +for 1x01 = (1 + and +00 for the remaining x0; f) the limit isequal to c for x0 E [c — 1, cJ and +00 for the remaining x0; g) lim isequal to 1/3 when x0 1 and 1 when x0 = 1; h) the limit is equal to 0for 1x01 < ±1 for 1x01> and does not exist for 1x01 =1.22. a), b) —. 1/(1+a) for al < 1 and any x0,and for the remaining athe sequence either diverges or stabilizes; C) see the answer to problem 1.20c)for a � 3/4, and the sequence either diverges or stabilizes for a > 3/4; d)for a E [e_e, 1J and for all x0 E R the sequence converges to theunique fixed point of for a E (1, we have that = forx0<L for and for x0>L where

< are the fixed points of 1. and for the remaining a the sequenceeither diverges or stabilizes. 1.32. a), b) For a = 1 there is a continuumof 2-cycles {z, 1— t} (t E [0, 1/2)), for a> 1 there is one repelling 2-cycle

fl—a l+a1. l+a2

and from one to three repelling 4-cycles; c) for a > (1 + = 1.618...there are two repelling 3-cycles

fl—a—a2 l+a—a2 l+a+a2l+a3 ' l+a3 ' l+a3

andf l—a+a2 l+a—a2 1—a—a2

1—a3 ' 1—a3 ' 1—a3

ANSWERS 355

and for a = (1 + they merge into one. 1.33. a) a = 3/4; b), c)

a = 5/4; d) a = 7/4. 1.34. a) b =3; b), c) b = 1 + = 3.449...; d)1.41. b)No.

2.6. a) d/1 = b) d/1 = for lxi < 1,

djt = B >0). 2.10. a) Yes, djt = dx/ixi(x 0); b) no. 2.11. a) Yes, djt = dx/(xilnxl) (x 0, 1); b) no. 2.14.

a) = n 2.15. djt = Idx, where a) f(x) = 0 forx < 1/2, 1(x) = f(3/2—x) for x> 1/2; b) 1(x) = 1/5 for x < 1/3,1(x) = 2/5 for x> 1/3; c) the limit measure is concentrated at the point0. 2.16. djt = dxdy/y2. 2.17. dILL = dxdy/x2, = dxdy/x.2.21. a) Yes, if and only if is irrational; b) yes; c) yes; d), e) yes, if andonly if there are no roots of unity among the eigenvalues of the matrix A.2.23. must be linearly independent over modulo 1. 2.28.jz(A). 2.30. 1/2, 1/p. 2.34. 1og10(1 + l/p).

Appendix I

We establish here an interesting fact showing that the arithmetic propertiesof a real number that were discussed in Chapter I can unexpectedly play adecisive role in investigations far removed at first glance from this topic.

A trigonometric polynomial in two variables x, y E R is defined to be asum

i(nx+my)T(x,y)= cn,men,mEZ

in which only finitely many coefficients m E C are nonzero. We considerthe projection acting in the set of all such polynomials as follows:

y) = cnmei(nx+mY)

(fl is a subset of Z2). We say that the projection is bounded if thereexists a number > 0 such that

(*)

for every polynomial T.We consider the simple case when is a strip, that is,

It turns out that this projection is bounded if and only if the coefficient ).is rational (see the paper of E. S. Belinskii in the collection "Theory ofmappings and approximation of functions," "Naukova Dumka," Kiev, 1983,

pp. 18—20).To prove this assertion, consider first of all the analogous problem in the

one-dimensional case: for any trigonometric polynomial Q(t) = cke(only finitely many of the coefficients Ck E C are nonzero), let PQ(t) =

kI<Ncke. It will be assumed that N � 2, N E N. Since PQ is a

partial sum of the Fourier series of Q, it follows that

— 1 l/2)SdQ( / — Sj

2sin(s/2)S•

358 APPENDIX I

Therefore, the projection P is bounded: IPQ(t)I � CmaxIQ(t)I, and

1 sin(N+ 1/2)sds

— it 2sin(s/2)

It follows from the result of problem IV.6.6 that C � const N. Theexample of the polynomial

iNt 1. 1 1 ikiQ0(t)=e

shows that the estimate for C cannot be essentially improved: on the onehand (see P1.6.15) 1Q0(t)I � A (the number A is independent of t and N),and, on the other hand,

PQ0(t) =l<k<N

hence

maxIPQ0(t)I = IPQ0(O)I =

Thus, C x In N in the one-dimensional case.For a better understanding of the subsequent arguments we point out the

principle difference between a strip with rational sloje and a strip with irra-tional slope. It is this difference that gives rise to the effect under discussion.In the case of a rational slope the following alternative holds for any lineL C R2: either L C fl or the set L n fl contains a bounded number NLof integer points (the upper bound for NL depends on the strip fl). But if

Q, then the situation is different: it is possible to draw a line L suchthat the set L fl fl contains an arbitrarily long series of integer points withcoordinates forming an arithmetic progression.

Now consider the projection on a strip with rational slope. Let ). = p/q,where p E Z, q E N, and p and q are relatively prime, and choose integersp0 and q0 such that pp0 + qq0 = 1. The integer matrix determines

a bijective linear mapping of Z2 onto itself (since the determinant is equalto 1, the inverse matrix is an integer matrix). Let us make the substitution

n'=np—mq, fx'=xpo_yqo,

m e' x+my) = T'(x' y')n' ,m'EZ

APPENDIX 1 359

(here rn' = Since the inequality rn — � is

equivalent to the inequality In'I � the projection is written as followsin the new coordinates:

y) =n

Consequently, the action of the projection reduces to the action of theone-dimensional projection with respect to the variable x', and thus

� const•ln(2+

Let us now consider the case when Q. Choose an irreducible fractionp/q such that —p/qI

<(2 Let L be the line through the points (0, 0)and (q, p). Since p and q are relatively prime, all the integer points on Lhave the form (kq. kp), k E Z. It is clear that flnL = {(kq, kp)Ik E Z,IkI � N} for some N> 0. We show that N � To do this we representthe number in the form = p/q + 0/q2, 101 � 1. For IkI � we

have that Ikp — = IkI — (p/q + 0/q2)qI = k01/q � � that is,(kq, kp) E flfl L for such k.

Consider the action of the projection on polynomials T with coef-ficients Cnrn equal to zero for (n, m) L, that is, on the polynomials

T(x, = It is clear that

= E ik(qx+Py)

kEZ, IkI�N

It follows from the treatment of the one-dimensional case that is bounded,that is, the inequality (*) holds, only if

� const•lnN �Since the numerator q can take arbitrarily large values, it follows that =+oo for a strip with irrational slope.

Appendix II

Suppose that the partial sums of a trigonometric series a0/2 +>2(ak coskx + bk sin kx) with real coefficients are nonnegative on a set E C

[0, 2ir) with ).(E) > 0. Then = o(n2) almost everywhere on E (in

particular, = o(n2)). Thus, if the partial sums of a trigonometricseries are bounded below then they are not too large above. This result,obtained by W. Darsow (see J. London Math. Soc. 35 (1960), 237—238) wasessentially strengthened by S. V. Konyagin (see Math. Notes 44(1988), 910—920), who proved that if the partial sums are nonnegative on a set of positivemeasure, then they are bounded almost everywhere on this set. Consequently,a trigonometric series cannot diverge to +00 on a set of positive measure.Moreover, for an arbitrary real trigonometric series with partial sums notnecessarily positive on a set E with > 0 the following alternativeholds: either the sequence is bounded for almost all x E E or

= —oo and = +oo for almost all x E E. But

if = 0 and E is closed, then, as shown by Olevskii (see Colloq. Math.60/61 (1990), pp. 709—712), there exists a Fourier series with sum +oo onE and finite off E.

We prove only the simpler assertion formulated at the beginning of thisappendix. For the proof we use the well-know fact that almost all points of ameasurable set E are points of density: fl(x — e, x + e)) = 1

for almost all x E E.Let x be some point of density of E. Then 0 is a point of density of

the set F = {t — xlt E E} fl {x — E E}. Fix numbers u and v with

ir/2 < u < v < their choice is made more precise below. Since 0 is apoint of density of F, for sufficiently large n the set F fl [u/n, v/nj is notempty, that is, there is an index N such that for all n � N there exists a

with Usingthefactthat for n�N,weget that ± � 0. Therefore,

0 � = a0/2+ (ak coskx+bk sin

l<k<n

= + — cos(k +O�k<n

362 APPENDIX II

Thus, for n � N

>O<k<n

>ncosu kO�k<n

Let w = —v/cosu (w> 1), and = S0(x)+•• Then �and hence � (1 + � (1 +

1\W

an...2

Thus,

�wn aN...I0(WI)

It remains to see that cos u) <3, and hence the parameters uand v canbechosensothat w=—u/cosv <3.

Appendix III

Consider the chord obtained by intersecting a convex closed curve F and aline L. What is the rate of decrease of the chord length as the line approachessome support line without changing direction? This is clearly determined bythe order of contact of the support line and the curve f. The length of thechord can fail to tend to zero in general if the curve f contains a segmentparallel to L. It is intuitively clear that this is an exceptional case. Thesame can be surmised about contact of order greater than two. In otherwords, the order of contact of "most" support lines with f is at most two,and hence the lengths of the corresponding chords have the estimate(e is the distance between L and the support line parallel to it). A possibleinterpretation of this vague assertion, one which gives a quantitative estimate,is the inequality

e) is the length of a chord at a distance e from the supportline; the angle gives the direction of the support line (the vector =(cos sin is perpendicular to it and directed toward the curve f); theconstant depends only on f. Thus, in the mean over all rotations of thesupport line the rate of decrease of the chord length for the curve f is notless than for a circle.

We prove that the constant on the right-hand side of the inequality isdetermined only by the two quantities r and R if the curve f is containedin a disk of radius R and contains a disk of radius r. In proving such aninequality it can be assumed that f is a smooth strictly convex curve. It

suffices to prove that for small e (for example, e <r)

� CrR/€•

364 APPENDIX III

This yields the required inequality, because

= (JCI(o)do) dço

�I do

�Denote by the point on f at which the vector is the inner normal

to f For each denote by + (respectively, — the slopeof the normal to r at the endpoint (respectively, at the initial point) of thechord of length e) orthogonal to the vector (the angles and

depend on e). It is not hard to see that 0 �for e E (0, r). Since e) = + it follows that

+ cot2 o(ço)) dço.

To estimate the resulting integral we inscribe in f a polygonal curve witheach component subtending an arc of "height" e. This polygonal curve isnot closed in general, and hence its first and last components intersect. Weconsider three successive components. Suppose that they subtend arcs oflengths m m0, and Assume that the vector is perpendicular tothe middle chord (that is, is the "vertex" of the middle arc). The point

divides the arcs of length m_ and in0, while the pointdivides the arcs of length m0 and We prove that

2 2

J(cot R(fh_ (*)

çQ0—ö(çQ0)

It suffices to estimate the integral over the interval Further,it can be assumed without loss of generality that = 0. Let = Wenote first the estimates

� BrR(mo+m+)fe, �which are uniform with respect to E [0. is]. We verify only the first one.For this we introduce a rectangular system of coordinates u, v with centerat the point and such that the positive direction of the OU-axis coincideswith the vector e,. In the new coordinates the curve r can be representedas the union of the graph of a concave function M and the graph of a convexfunction N, both defined on some interval [0. � 2r. It is not hardto see that IN(u)I. IM(zi)I for u E [0, rJ. If � ir/2,then, using the concavity of the function M. we get that

0 = M'(e) <M(e)/e �� +

APPENDIX III 365

But if > ir/2, then, as is not hard to see from a sketch,

I = IM'(e)l � (IN(e)I +

� (1

To estimate the integral of cot2 note that the function = +is increasing, and hence � that is, � — This

implies that = — p)). Thus

� p,), (in0 +

Consequently,

�To estimate the integral of we observe that =

Indeed, since <it, it suffices to consider the case <ir/2,and hence 0 < = IN'(e)I � IN(e)I/e. Since IN(e)I sing' � 2€, wehave that 0 � 2/sin = Thus,

ICotö(9')I � Lr,Rmin(l/cO, (m0+m_)/e).

Consequently,

� 2Lr,R(mO+m_)/€.

The inequality (*) is proved. Summing all the inequalities corresponding tothe different components of the polygonal curve, we get is the length ofthe curve fl:

� 2j + cot2 dço

8 CR�

Bibliography

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Subject Index

A mapping, 132Abel method, 42Abel transformation, 9, 36, 43Abel's theorem, 42, 48Absolutely continuous, 132Almost periodic sequence, 82

in the Besicovitch sense, 84Attracting, 128, 129Automorphism of the torus, 132

Bernstein polynomial, 77Bessel's inequality, 82Binary sequence, 3Birkhoff-Khintchine ergodic theorem, 137Bogolyubov-Krylov theorem, 136

Cantor function, 30, 90, 99, 102, 122Cantor set, 6, 38, 90, 94, 99, 102

generalized, 7, 31, 33Cantor staircase, 30Cantor type, 103Carleman inequality, 188Cesàro method, 42Cesãro-Fejér sums, 121Character, 34Chebyshev polynomial, 127, 130, 133, 137Chebyshev polynomials, 18Chebyshev's inequality, 93i-square distribution, 272Concave, 67Conditionally positive-definite, 75Congruent, 4Conjugate function, 121Conjugating mappings, 126Convergent, 139Convex, 10, 13, 69Convex (concave), 78Convex (concave) sequence, 44Convex strictly convex, 67Cubic cells, 263Cycle, 129Cylinders, 343

Defining sequenceS 88, 103

369

Density, 132Density of a sequence, 19, 38Density of the set, 19Diagonal sequence theorem, 111Diameter, 102Dini's theorem, 194Dirichlet function, 26, 90Discrete, 5Distribution function, 90, 116, 122

Egorov's theorem:, 110Endomorphism of the torus, 133e-almost period, 81e-distinguishable sets, 99e-entropy, 99e-net, 99Equimeasurable functions, 90Equivalent, 132Ergodic transformation, 136Euler I-function, 58Euler constant, 20, 40, 60Euler-Poisson integral, 48

F. Riesz product, 198Feynman integrals, 210Fixed point, 125Fixed points, 129Fourier transform, 76, 116, 122Fractional, 14Fractional part of a number, 139Frullani integral, 47Function of bounded variation, 29Function of the first Baire class, 27

Gauss transformation, 139Gaussian measure, 96, 104Gibbs phenomenon, 45Gronwall's inequality, 75Guldin's theorem, 96

Haar functions, 118Haar measure, 133Haar measures. 135Hardy-Landau inequality, 188

370 SUBJECT INDEX

Hausdorfldimension, 102Hausdorffmeasure, 101, 102Hironaka curve, 101Holder inequality, 71Homomorphism, 34

Image, 132Invariant measure, 123, 132, 136

invariant, 132Invariant set, 136Involution, 125

Jensen's inequality, 67

Kepler's equation, 64Khintchine's inequality, 119

Laplace asymptotic formula, 57Laplace method, 56Law of large numbers, 112, 113Law of the repeated logarithm, 120Legendre transform, 72Liouville numbers, 15Liouville's theorem, 155Littlewood conjecture, 314Logarithmically convex, 70, 71Lower semicontinuous, 28

Mean value off along the trajectory, 137Measure, 122Median of a function, 91Method of successive approximations, 127Metric dimension, 100, 101Metric order, 100Metrically conjugate mappings, 136Minimum modulus prindple, 76Mixing, 138Morse sequence, 83Mutually singular, 137

Newton iteration process, 129Nonincreasingrearrangement of a function,

91, 94, 112Normal, 342Normal numbers, 113

Outer measure, 100

Peano curve, 32, 171, 172Period, 129

Periodic point, 129Perron-Frobenius operator, 134Pompeiu, 278Positive-definite, 75Preserves the measure, 132Probability measures, 136

Quantile, 112Quasi-invariant measure, 133

Rademacher functions, 114theorem, 132

Relatively dense, 80, 81Repelling, 128, 129Respective measures, 136Rotation number, 131, 136Rudin-Shapiro polynomials, 45, 85Rudin-Shapiro sequence, 85

Schwarzian (derivative), 131Schwarzian derivative, 76Semi-isolated, 5Sequence of Fibonacci numbers, 18Sequence of folds, 83Set of Cantor type, 88, 102, 103Sharkovskii theorem, 130Standard Gaussian measure, 95Stirling formula, 20, 48Stirling's formula, 58Supergraph, 68System of sets, 3

Theorem of Stoltz, 20Three chords lemma, 67Topologically conjugate mappings, 126Trajectory, 129Translation by 0 modulo 1, 89

Uniformly almost periodic, 81Uniformly distributed sequence, 138Upper semicontinuous, 28Urysohn's example. 278, 282

Variation of a function, 29

Winding, 254

Young's inequality, 72, 73