(Translations of Mathematical Monographs 107) B M Makarov, M G Goluzina, A A Lodkin, A N Podkorytov - Selected Problems in Real Analysis-American mathematical Society (1992)

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TRANSLATIONS OF
MATHEMATICAL
MONOGRAPHS
VOLUME 107
TRANSLATIONS
MATHEMATICAL
MONOGRAPHS
.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 92-15594
515'.8—dc2O CIP
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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 recursive
sequences 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 Lebesgue
Integral 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 to
deepen their knowledge of mathematical analysis, and for lecturers conduct-
ing seminars in university mathematics departments. It is somewhat different
from the usual problem books in the greater difficulty of the problems, which
include a number of well-known theorems in analysis. Despite this, no special
preparation is required to solve the problems in Chapters 1—Vu and in §1 of
Chapter X, and many of them are accessible even to first-year students in the
second semester. All the facts needed to solve these problems are contained
in the standard university analysis texts, in particular, in the books of Zorich
[7], Kudryavtsev [16], Rudin [23], and Fikhtengol'ts [29]. The problems in
Chapters VIII and IX and §2 of Chapter X require a somewhat higher level
of preparation of the reader and presuppose that he is familiar with the fun-
damental concepts of measure theory. The corresponding facts can be found
in the last chapter of the cited text of Rudin and, in more complete form, in
the books [14] of Kolmogorov and Fomin and [4] of Vulikh.
The contents of the first seven chapters, which include about two thirds
of all the problems, do not go outside the framework of the classical topics
of analysis (functions, derivatives, integrals, asymptotics). Both here and
in 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 are
less traditional for a problem book in analysis. Besides the tastes of the au-
thors, the program of the mathematical analysis course adopted at Leningrad
University 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 not
included in the book. For example, we do not use many attractive problems
whose solution is based on the Lebesgue theorem on differentiation of an
integral with respect to a variable upper limit. Also;almost no reflections of
problems connected with complex analysis are found. We refer the readers
interested in this circle of questions to the widely known collection of Pãlya
and Szegô [21] and to the book of Titchmarsh [27].
FOREWORD
We tried to combine problems dealing with separate topics or methods
into cycles within whose confines it would be possible to exhaust, step by
step, various circles of questions with sufficient thoroughness. Partly for this
reason we were not able to avoid a certain lack of homogeneity in the degree
of difficulty of neighboring problems, which can grow markedly in the course
of a single cycle. Therefore, it is not rare that more difficult problems give
way to relatively simple problems, and the reader who has not solved some
problem must not feel disheartened and can hope in full for success in the
solution of subsequent problems.
Brief but often detailed solutions of most of the problems are given in the
second part of the book. However, we advise the reader not to be in a hurry
to use this part of the book and thus miss the chance of devising a better
solution 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 only
a few of the problems presented can pretend to be original. We saw it as
our goal first and foremost to try to systematize and introduce into everyday
practice problems contained (sometimes in implicit form) in almost inac-
cessible sources (especially for students) and inthe mathematical folklore.
The experienced reader will notice that along with the traditional material
are 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 exceptional
cases.
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 to
us. 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 first
denoting 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 (a
Roman numeral). For example, problem Vll.2.5 is problem 2.5 in Chapter
VII.
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 ball
is 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 a
nonincreasing (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 sequence
of 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 let
mean
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 function
fir is integrable on the set A;
2°°(A) is the set of functions f in 2°(A) such that the essential
supremum of fl is finite;
ess supAf is the essential supremum of a function f in
is Lebesgue measure on the space R";
2. is the measure
is 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 of
all 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) by
taking x y if x/y E Prove that the intersection of each equivalence
class 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 points
of any equivalence class coincides with S1.
1.11. Let S' = {z ECIIzI = 1}. Sets A, BC S1 aresaidtobecongruent
if there exists a number a E R such that B = E A}, that is, B "is
obtained 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 lying
on distinct rays (Figure 1). Prove that on the plane it is possible to arrange
an 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 formed
by 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 many
disjoint "hoops" (cylindrical rings) of fixed radius (the thickness of the hoops
is 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 sport
classes. In each class there were only finitely many devils, but there were so
many 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 of
finitely 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 that
card(B fl Am) � 2. Prove that:
a) the natural numbers belonging only to finitely many of the setsform
a 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 thickly
covering N (see problem 1.17), then there exist indices p and q such that
card(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 C
b) 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 cannot
have interior points.
b) A point a in a set E C R is said to be semi-isolated if there exists a
number 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 most
countable.
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. Does
there 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 his
worker Bakia (Cxa3Ka o none pa6oTHHKe cr0
I. INTRODUCTION
1.26. Let be a sequence of open subset of R that are unbounded
above. Prove that there is a number x0 > 0 such that each of the sets
contains 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 middle
third (1/3, 2/3) of [0, 1]. In other words, K1 is the union of the two
closed 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 the
indices es,..., es,, where takes the values 0 and 1. The indices of the
intervals of first and second ranks have already been indicated, and the rest
of the indexing is done by induction. Suppose that the intervals of nth rank
have already been supplied with indices, and / is one of them. The
two 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 possible
closed intervals of (n + 1)st rank.
1.27. Prove that
a) 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 Doesthe
closure 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 the
property? (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 as
follows. The set K1 is obtained by removing from the closed interval K =
[a, b] a nonempty open interval 5 = (p. q) whose endpoints are different
from 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 one
of which we denote by Ac0, and the right-hand one by Ac1. Thus, K2
is the union of four closed intervals A10, A11, which we call
the intervals of second rank. The subsequent construction is continued by
induction. Suppose that we have constructed the set made up of the
closed intervals of nth rank. It is convenient to "enumerate" the intervals of
nth rank with the help of the indices er..., , where can take the value
o or 1. The indices of the intervals of first and second ranks have already
been indicated, and the subsequent indexing is carried out as follows. In
constructing the (closed) intervals of (n + 1)st rank we remove from each
closed interval of nth rank a nonempty open interval whose
endpoints are different from those of Ac
c
The difference Ac £
consists of two intervals of (n + 1)st rank, the left-hand one of which we
denote 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 not
have 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) Acc
2) 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 is
an 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 that
a)
2n&�
1<k<n 1<k<n k
b)
2 1 2(m+M)2
4mM
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 numbers
a finite sequence of real numbers with > —1 for
k = 1 n. Prove that:
a) if S = a1 + + � 0, then
fi
1<k<n
with equality possible only in the cases when n = 1 or a1 = .. =
a a
1+a1
INEQUALITIES 9
then
a 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, andde-
note the nondecreasing (respectively, nonincreasing) rearrangements of those
sequences. Prove the inequalities:
E dzkbk � E akbk � E
1<k<n
2.7. Let and be finite sequences of real numbers. Prove
the inequalities:
� E � akbk.
In the solution of the following problems the Abe/transformation can turn
out to be useful:
E akbk = + (ak — ak+l)Bk,
where Bk = b1 + ... + bk (k = 1 n). It is especially convenient to use
this 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, with
nonincreasing. Consider an index m E {1, 2 n} such that
m maxl<k<fl bk � bk. Prove that:
E akbk � bk) E
1<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), then
the 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 precise
representations 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 of
the summation inequalities already considered. They can be proved by two
paths: 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 the
limit, the inequalities in problems 2.23 and 2.24 can be carried over to the
case 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]. If
the 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 following
inequalities 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)
(inf
1b
f(x) dx
<
f(x)g(x) dx
a<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], I
and 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)2ô2
1 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 � j
f2(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)
exp
1b
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
� exp
1b
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 any
functions f and g in K
2v'E
Ia
max(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 the
fractional part of a real number x.
3.2. Suppose that x E Prove that the sequence {ö(nx)} is dense
in (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 . . 3/2
a) {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 E
c) {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)dt
1<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 a
geometric progression: � 1 + 1/(8C) (thus, the lower estimate in
the 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 that
a) 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 isirrational;
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 that
forall 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
1
cn/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 algebraic
polynomials 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 algebraic
polynomials 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 limit
1.2. Let = — l)!!/n!!. Find
lim lim
the limit the sequence where
= [J sin
l<k<n
1.5. Let a ER and = >1<k<fl((a + Find the limit
1.6. Let
a) Find the limit
b) Determine the character of monotonicity of the sequence
c) 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 lim
1.9. Find the limits:
a) lim b) lim —
k . .r
c) 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 finite
limit 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 that
the 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 = + for
n>2;
e) 2x2 = + + +•• for x � 1, where
is 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 the
interval with endpoints and
§2. AVERAGING OF SEQUENCES 19
1.22. a) Suppose that the sequence of real numbers is such that
xfl.F1 + —, 0. Prove that the set of limits of convergent subsequences of
this 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 sequence
can be represented as a union u U U , where the i\ are closed
intervals in IL
1.23. Suppose that the sequence of real numbers is such that —
Prove that
1.24. a) Give an example of a bounded divergent sequence such
that
b) 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 assertion
in 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 is
called the density of the set A (or the density of the sequence {nk}). Prove
that:
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, then
2.5. 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 limit
x—x —
lim
yn
exists, then —, 1 (a theorem of Stoltz, a discrete analogue of UHôpital's
rule). Formulate and prove a similar assertion for indeterminants of the type
0/0.
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 subsequent
problem.
2.9. Prove that >i<k<fl = Inn + y + o(1) for some number y. The
number y(y = 0.5772. .)is called the Euler constant.
2.10. Prove that ln(n!) = = With
the help of the Wallis formula show that C = which implies the
Stirling 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) E
i<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). Prove
that the limit exists, and find it.
3.2. Let x0 = a and x1 = b. Find the limit in the following
cases:
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 the
asymptotic behavior of
b) For what values of x1 does converge? What is
3.4. 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 such
that —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 for
some 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), then
converges, 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 series
converges, 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
((1
0
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 following
properties (e,ö,x1,x2ER):
a) Ye 3ö>0
b) Ye>0 3ö
c) Ye>0 3ö>0
d) Ve>0 Vö>0
e) Ye>0 3ö>0
f) Ye>0 3ö>0
g) Ye>0 3ö>0
h) 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 C
c) 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 most
countable 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 of
EL, continuous only from the right at points of ER, and continuous at the
remaining points of [0,1].
1.5. Let be the set of points at which a function f on
R 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 most
countable?
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 closed
sets. 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 continuity
and a continuum of its points of discontinuity.
1.10. Prove that a function f defined on an interval (a, b) is continuous
if 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? An
arbitrary metric space?
1.13. Describe all the continuous functions on an interval [a, b] having
a local extremum at each point of (a, b).
1.14. Suppose that a function 1 is defined and one-to-one on (a, b) and
continuous at a point c E (a, b).
a) Is r' necessarily continuous at the point 1(c)? Does always
have 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) exists
for 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 without
isolated 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 have
a horizontal chord of length ö if there is a point x E [0, 1 — ö] such that
1(x) = f(x + ö). Prove that if f(0) = 1(1), then for any n E N there is a
horizontal chord of length 1/n. Show that for chords of a different length
this 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). Describe
all functions that are Cësaro-contiuous.
1.24. Let f(x) = 11k>O(1 + x < 1. Prove that
a) 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 of
functions 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) sernicontinuous
if 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 following
properties 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) � supflEN
e) 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 semicontinuous
if 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 takes
only finite values, then it attains the smallest value on [a, b] (and hence is
bounded below). Does it have to be bounded above?
2.4. Let g be an arbitrary function defined on an interval C R, and
let 1(x) = g(y) (x E - Prove that f is lower semicontinuous.
2.5. Let be a family of lower semicontinuous functions defined
on 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 function
h = 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 and
only 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 continuous
on
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, then
IEC'([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 limit
uniformly on [a, b] as h 0.
3.6. Let f E C((a, b)), and suppose that for any x E (a, b) the limit
f(x — h)) = g(x) exists and is finite.
a) Provethatif g�0 on (a,b),then f isincreasing,andif
then 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 I
is 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. Prove
that 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 all
possible 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 all
functions 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. Does
g'(O) exist for = 1?
3.13. Supposethat for 0<x�1
and 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 having
bounded 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 let
f(x — t)co(x)=j
s/iT
dt (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 Cantor
function (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,
'1
I
4 I
I I I
I .__i I I I
4 I I I
I III
O 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 value
1
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 will
be 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 point
of K).
3.21. Suppose that F C R is a closed set without isolated points. Prove
that 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 the
class Lip(a, c) (cr> 0) if there exist positive numbers L and ö such that
If(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 the
image P(F) of any closed set F C C is again a closed set. Is this true for
an 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 and
B such that each can be mapped onto the other in a one-to-one bicontinuous
manner (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_l
a2 + 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 the
square [0, 1] x [0, 1]. The mapping f is described schematically in Figure
3. Each of the intervals k = 1 9. is mapped onto the square with
the same index (Figure 3(a)). Figure 3(b) shows the circuit of the unit square
when 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 and
such 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 arbitrary
Peano curve defined on [0, 1]. Prove that if u, v E 1]), then
a � 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 between
them.
§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 is
representable as the sum of a linear function and a function not exceeding a
in 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 any
xER,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 < 1
5.12. 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))2
forall 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 any
XE[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 into
a group G' in the following cases (the notation introduced in the preceding
problem 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 decimal
expression 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 +P
iMu)
,,
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, and
ço(n) is the number of positive integers � n that are relatively prime to n.
1.6. Prove that for a> 1 the series
1
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 +m
LCM(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 and
m are relatively prime and 0 otherwise.
1.8. Verify that the sum of the series
11 111111
111111111111
(the group of terms + — is repeated times) is equal to 1. Find
a 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 of
the subsequent problems: if sequences and are such that the se-
quence converges (where = b1 +. • then the series
and — converge or diverge simultaneously. In the case when
the series E converges, another variant of this assertion is useful: if the
sequence converges (where = + then the series
and — converge or diverge simultaneously. These
assertions are easy to prove by applying the Abel transformation (see Chap-
ter I, §2) to the sums Ei<n<jv The Abel transformation is especially
convenient 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 the
series 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 the
following 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 the
series 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 n
and 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 series
always 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 series
converges, 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 ofnumbers 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 sequence
tending to zero, then E < +00.
3.9. Prove that the following two properties of the sequence are
equivalent:
a)
b) if the series converges, then so does the series
3.10. Suppose that the function is monotone and bounded on [0, +oo).
Prove that
c—'+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 is
such 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), not
even 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))P
3.16. 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 of
zero.
3.19. Suppose that Q is a countable subset of (0, 1) with infQ = 0 and
sup Q = 1. Prove that for any number x E (0, 1) it is possible to number
theset 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—4.9
can 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.
2
1
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/2
2 2N
I xcos xdx
fl>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 increasing
order, = 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 of
n.
§5. Function series
5.1. For what p. q >0 and x, y E R does the series
x .
E
converge?
5.2. Suppose that —' 0. Does the series converge uni-
formly on R? Estimate I
under the assumption that <
I forany nEN.
5.3. Does the series
xn
42 IV. SERIES
converge uniformly on [0, +oo)? Estimate
xn
sup>
2
x?o
5.4. Does the series min(xn!, l/xn!) converge for x > 0? Does it
converge uniformly on (0, +oo)? Estimate >min(xn!, 1/xn!).
5.5. Suppose that ER, >0, = and = +00, and
assume 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, then
A(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 method
for summation of series (computation of the limit a/) is
stronger 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 limit
lim(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 product
fl(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. Prove
that 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 any
finite 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 a
sequence —, +oo such that sin Bnx 0 on E.
6.4. Prove that +IbnI) <+00 if � const
On
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 1
flt/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—j
nsinflx
c) —; d)
6.9. Prove that:
In problems 6.10—6.21 the sequence decreases to zero. The Abel
transformation (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 E
Z).
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 mx
n�l
(m E Z) converge uniformly on ILL
6.13. 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 and
only if = o(1/n).
6.15. Prove that the partial sums of the series sin nx are uniformly
bounded if and only if = 0(1/n).
6.16. Let = 0(1/n) and suppose that the function is absolutely
integrable on [a, bJ. Prove that the series sin nx can be integrated
termwise on [a, bJ. Using the result of problem 6.8a), compute the sum of
the 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 and
only if <+00.
6.19. Suppose that < +oo f(x) = cosnx g(x)
= sinnx, and E C([0, irj). Prove that the series cos nx
and 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)SupposethatProvethat g�0 on
b) 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
for
6.25. 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, a
c) the partial sums SN(x) are uniformly bounded on ILL
6.29. 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 change
the 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 whose
partial 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 a
max =0(n);
1<k<n
b)
max > = 0(na);
k>n
c) the function
ikOf(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) dx
implies that of the following integrals:
a) j f3(x)dx; b) j
1.13. 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. [ Isinxldx
JO eX2Smn2X
13 r
+00
1.5. [ Isinx"Idx
J1
+00 dx
1.7. [
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" +1
dx
x
1.8. / xPlsinxlXdx
I'
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 expressed
in terms of the Euler constant y (see problem 11.2.9).
1.16. 1.17.
Jo x
118
lnx—dx. 1.19.Li-x
+00 x2dx
1.20. 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 mx
1.26. 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-Poisson
integral dx.
§2. COMPUTATION OF MULTIPLE INTEGRALS 49
1.34. Je_e dx k=O,1,2,3,4,5.
1.35. 1 ed
Jo
1.36.
j+OO (2 2/2)
1.37. a) I —dx; b) I
Jo 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;FOO
JInx —
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) E
I
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 mean
that the probability of choosing a point belonging to some set is proportional
to its measure (length, area, volume).
SO V. INTEGRALS
2.9. Two points are chosen randomly in the interval [a, bJ. Find the
mean value M of the distance between them. What is the probability P
that this distance is greater than M?
2.10. Three numbers are chosen randomly on the interval [0, aJ. What
is 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 do
not 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 them
as 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_1
n—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(xxdyi
J_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 then
fg(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,then
as x—'b—O;
d) if fg(t)dt <+00 and 1(t) =o(g(t)) as t—' b—0,then
o(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. Supposethat the function is continuous on R and has period
T> 0, and that = + ço(t)dt 0. Prove that f11°° c,/e
as 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)j
n/21 Ad) [ xl
J0 \sinx/ 3
1.10. Find the asymptotics as A —, +00 of the following integrals:
a) f
sin2x 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). the
function 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 that
JA
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. Prove
that:
§2. THE LAPLACE METHOD 55
a) if f°° 1(t) dt = +00 and f7' f(t) dt = dt), then
dt 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 has
period 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)dx
11b
(
sint)dt) dx.
1.16. Find the asymptotics as e —, +0 of the following integrals:
a) b)
1.17. Prove that as e —, +0:
3
a) jx
b) 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 the
integral 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 2
a) J x
dx
b) I max Isinkxl— Inn;
Jo 1<k<n x
Isinkxl dx
c) I max
Jo 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/'.TA
A
J
cos
2.4. Verify that the asymptotics as A —, +00 of the following integrals
do not depend on the parameter 0 > 0:
2A dx
a) j (1 —x ) dx (0 � 1); b)
J J
dx
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 phenomenon
of localization—the asymptotic behavior of the integral depends on the be-
havior of the integrand only in a neighborhood of a single point. This effect
also appears clearly in a more general situation in the study of important
integrals of the form
=
A
where is nonnegative and piecewise monotone. For large values of the
parameter 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 assumed
that is monotone on [a, bJ. Here it suffices to consider only the case
when is decreasing. Then the values of ç0A at points far from a are
negligibly small in comparison with its values at points near to a, which give
the main contribution to the integral c1(A). To find the principal part of
it remains to approximate in a neighborhood of a by a simpler
function 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 amounts
to the realization of the scheme presented.
Most often encountered is the case when the difference — is an
infinitesimally 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 the
Laplace asymptotic formula (for brevity of formulation it is assumed in
problem 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 for
piecewise monotone functions.
2.6. Suppose that —00 <a < b � +00, the function is positive and
decreasing on [a, b), and
x—.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 are
absolutely 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
J
g(x)co'4(x)dx
J
f(x)ço(x)dx.
a A'+OOa
2.9. Find the asymptotics as A +00 of the following integrals:
A cos(ax)
a)
J
sin x dx; b)
(1 + x2)A
dx (a E R);
çA
A sinx dx
c)
J
e (A — x) dx; d) (p <2);
e) 1' dx
f)
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 form
considered 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 neighborhood
of the point x4 at which 14 attains its largest value. Although the result in
problem 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 of
x4 by its Taylor expansion and the integral over (a, b) by the integral over
the neighborhood of x4, we find the principal part of c1(A). The choice of
the neighborhood is the basic difficulty here. On the one hand, it must not
be 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 error
due to the replacement of the integral over (a, b) by the integral over the
neighborhood, this neighborhood cannot be too smalL The main content of
the solution involves a successful choice of the neighborhood, allowing good
estimates 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 px
a) 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
I
Jo A—.+x A
2.13. Provethatas A—'+oo:
1' dx 1
l/2
1
a) j b) j x dx
In the following problems we study the asymptotic properties of the Euler
r-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:
•
_______
i
a) 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 1
c) 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 bn
d) 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 that
b) Can it be asserted that a, 1/n if inn?
3.6. Suppose that an 1 0, E an < +00, and >.k�nak n', where
y > 0. Prove that an
vi. ASYMPTOTICS
3.7. Suppose that > 0 and let
i(i)k
1<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 Euler
constant (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 closely
connected with that of the sequences {njç}; more precisely, there exists a
number 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, and
let = 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 that
f(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, and
f(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-1/2
Var(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 on
3.25. 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 bounded
below, but for any E (0, this is false.
What can be said about boundedness below for the partial sums of the
series , 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 the
following 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 the
following 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)2
3.30. 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 the
following series:
a) T(t) = b) S(t) =
where is the number of divisors and a(n) the sum of the divisors of a
number n.
§4. Asymptotics of implicit functions and recursive sequences
4.1. Suppose that z(t) = (t E R), and let t(z) be the
function 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 infinitely
differentiable 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 fixed
positive parameter, determines an infinitely differentiable function y defined
in 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 flnlnx
a) y(x) = lnx —lnlnx + + 0 1mx \ lnx
ln2x ln4x I ln'1x
b)
lnx lnx 1 ln2x 11n2x
c)
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 11
1)
4.6. Let 1),where for x�0 (a and
p are positive constants). Prove that
(.v0>0).
4.7. Suppose that x0 = and ; = — (ii E N). Prove
that 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) Find
the asymptotics of the sequence
4.12. Suppose that a0 = 1, and = + (>.0<k<n ak) (n E N).
Prove that
4.13. 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 all
points x1 , x2 E (a, b) and all numbers > 0 with + = 1 the
inequality
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 of
the graph of f with abscissae x1 and x2, x1 and x3, and x2 and x3. The
slopes 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 nonempty
open 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 the
supergraph 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 is
convex.
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. Prove
thatI 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 all
x 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' � 0
on (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, then
r' is either convex or concave.
1.13. Suppose that a function f is convex on [a, bJ, and a function
g is convex and increasing on [c, dJ. Prove that if the composition g o I
makes 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 the
limit 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 a
sequence 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 the
given class it is the triangles that have the greatest static moment and the
greatest moment of inertia with respect to the x-axis.
1.24. Suppose that the function f is nondecreasing on [0, 1/21, and
1(1 — x) = 1(x) on [0, 11. Prove the inequality
fI fl fl
I 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 following
conditions:
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, and
that 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 convex
polyhedron inscribed in a circle, with A0 and fixed. How should the
points A1,..., be selected in order to maximize the perimeter and the
area 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 of
f and from the x-axis according to the well-known law. Is the whole area
illuminated if 1(x) = 0?
Let f be a convex function, f: R —, L The Legendre transform of I is
defined 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 line
through the origin with slope x in order to make it a supporting line of the
supergraph 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 of
d) 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)) ,where
the infimum and the supremum are over those points t E at which
f'(t) exists, and for x E the supremum in the definition of f*(x) is
attained;
f) if s = f'(t) < f
f 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*)1
are mutually inverse;
h) if 1 � g, then
1.35. 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 I
JO JO
1.36. Find the Legendre transform and write Young's inequality (see
problem 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 on
satisfying 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 W
imply 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 Legendre
transform 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 of
2.2. 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 all
xE[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) are
real 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 inequality
holds:
� — 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 constant
cannot 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 that
if 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 set
into 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). The
Schwarzian derivative Sf of f is defined by
Sf(x) — f"(x) 3
— f'(x) 2
— 6
d ( 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 not
positivity) 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 is
piecewise 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 local
minimum 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<n
Ik/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] and
coincide 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
on
c) 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 any
function g E C2([0, 1]) equal to zero in neighborhoods of the points 0 and
1. Prove that f is a linear function.
3.11. Suppose that I E C([0, it]) and e > 0. Prove that there exists a
trigonometric polynomial T of the form T(t) = >0<fl<N cos nt such that
If(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 positive
constant 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 EN
and xE[0, 1].
b) If a continuous function f on [0, 1J is such that x) � 1(x) for
all 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 to
zero 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 convergent
sequence 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 most
n.
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
a
0 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 an
integer k such that Ix — y — kI <e.
A set X C R is said to be relatively dense if there exists an L > 0 such
that 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) and
any e > 0 there is an n E Z such that a (mod 1). Verify that the set
of 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, then
for 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 integer
solutions of the system (1) is relatively dense on the line.
4.3. What condition must the numbers E R satisfy in order
that 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 in
order 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 different
signs.
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 each
interval 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 if
and only if the factors .., are pairwise commensurable (that is, their
ratios are rational).
4.8. Let f E be a function having period 1 with respect to each
argument, 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 UAP
functions on R is a UAP function.
4.!2. Prove that if a UAP function has a derivative that is uniformly
continuous, then the derivative is also a UAP function.
4.!3. Prove that if a primitive F of a continuous UAP function f is
bounded, then it is also a UAP function.
4.14. Suppose that f E C(R), h E R, and fh(x) = f(x + h). Prove
that 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 again
a UAP function.
4.16. Prove that the mean value
1
çT
M(f) = lim —! f(x) dx
T—'+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. Prove
that:
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 0
for at most countably many values A. Can this set be an arbitrary countable
subset of R?
4.18. Let a nonperioclic sequence. Define the
sequences and as follows:
b J0 a—fl
ifa 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} intersects
every segment of the natural numbers of length L. Almost periodicity for
two-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 the
Cesàro 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 with
k an irrational number, and let Z = S n Z2 be the set of integer points in
S. Prove that the orthogonal projections of the points in Z on a side of the
strip (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 of
their 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 e
by determining its initial segments A,, recursively: A0 is an arbitrary word,
A0 = e, A, = 4 ,where is some word of 0's and
l's, A2 = E {0, 1}, and so on. Thus, A0 serves as
the initial part of A,, A, as the initial part of A2, and so on. The infinite
sequence determines some sequence e. Prove that e is periodic or
almost periodic.
4.24. A sequence offolds is defined to be a sequence s = of 0's
and l's whose initial segments of length 2k — 1 (k = n E N) have
the form . . •. (where = 1 — si). The name "sequence
of folds" is due to the fact that with the word it is possible to associate
a sequence of folds of a paper strip that are formed on the strip as a result
of bending it n + 1 times (first in halves, then once more in halves, and so
FIGURE5
a
84 VII. FUNCTIONS (CONTINUATION)
on). If upon moving along the strip a 1 is associated with each fold leading
to 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 by
specifying 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 a
number L >0 and numbers E N (m E N) such that:
a) if is the number of falling in the interval C R4, then
for 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 of
folds, then for any t E R the limit
1 --2nktzC =lim— xke
1<k�n
(Fourier coefficient) exists; moreover,
(m+2)= if t = (2! + 1)2 in 0.
and C, = 0 otherwise. Verify "Parseval's equality"
lim! IXkI
1(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 and
A1 = C1D1 = 00, while and are defined recursively for n> 1:
= = = is obtained from by
the substitution 0 1).
Prove the almost periodicity of this sequence. We note that the sequence
is 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 or
on the sphere 5m1 = {x E Rm
I lxii = 1}. Lebesgue measure on Rm is
denoted 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 corresponding
coordinates 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 integer
coordinates.
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 V
contains 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 fl
1.9. 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
I
x, 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 is
b) 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 set
E + 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 set
E1 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 operation
with and instead of we get four intervals of
second rank U C U C whose union is denoted by
E2. 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 set
E = is called the set of Cantor type constructed with the help of the
defining sequence {1,} on the interval (cf. the construction of the Cantor
set 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 what
is 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, and
find the coefficients a, b. and c.
1.17. Prove that every (nonempty) open Subset of Rh can be represented
as 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 sets
are not measurable.
1.19. The mapping x '—' {x + O} where {y} is the fractional part of a
number 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 a
nonmeasurable 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 C
having 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's
lemma (see, for example, [13], p. 33)).
Prove that the set E = {EkEA
2—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, that
is, 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 dense
subset of Rm. Prove that + E)) = 0.
Measurable functions
The symbol denotes the set of all Lebesgue-measurable functions
on 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.) on
E c Prove that if Ill < 1 on a set of positive measure, then there exist
mutually nonequivalent functions 11 , 12 E 5!°(E) such that
1111 � 1 and 1121 � 1a.e. on E, and I = + 12)/2.
2.2. Give an example of a monotone function on R that is not equivalent
to a continuous function, not on any (nonempty) open interval.
2.3. The Dirichlet function (equal to 1 at rational points and to 0 at
irrational points of the real line) is discontinuous at each point, but equivalent
to a continuous function. Does there exist a function in 5!0([0 1]) such
that any function equivalent to it on [0, 1] is discontinuous at each point of
this interval?
2.4. Let K be the Cantor set, f the Cantor function (see problem
111.3.17)), 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 measure
of 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 the
same distribution functions are said to be equimeasurable.
2.5. Prove that a distribution function is left-continuous. When is it
continuous 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. Assuming
that the distribution function F of a function f is continuous and strictly
increasing, 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 function
2 2 2 2f(x,y)=x +y (x +y �1).
2.13. Let f(x) = >1<k<flak/(x—ck), where a1 >0 >0, and
C1,..., çER. Provethatforany t>0
where A = ak.
§3. Integrable functions
In this section all the sets and functions under consideration are assumed
to be measurable. The set of functions integrable with respect to Lebesgue
measure 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,..., EN
coincides with [0, 1]. Let
S1 = and S2 = fl Ek).
Express the integral
dx
0
I
in terms of S1 and
3.2. 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 arbitrary
numerical 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 given
set 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 that
I 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 any
E (0, p).
3.14. Let E C Rm and f E Prove that
JE
dx
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
(JE
f"(x)g(x) dx)
p—.+O
efE
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 or
1, 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 the
decimal 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 and
1(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 I
t 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) is
integrable 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 is
assumed 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 quantity
is 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 that
the formula a) remains true also for n = 2 if S° is understood to be the
boundary 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 E
J
0 < 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), andlet
n—i
Prove 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 22
Ka{(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 of
the 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, the
positive 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 by
rotating 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
density
3.30. 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 a
and 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 let
ji be a probability measure on such that I(x, y)Idu(x) <+00 for
any 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=J
Q 4(Q)
Is this true if the determinant of A is equal to 1, but its elements are not
necessarily integers?
3.36. Let T be an arbitrary compact convex body in Prove that if
the (n — 1)-dimensional volume of the projection of T on any hyperplane is
not less than S. then diam T � nV/S. where V =
3.37. Suppose that T C is a compact convex centrally symmetric
body with respect to zero, and let 3 be an ellipsoid of maximal volume
contained in T:
3 C T, =
I
C T, an ellipsoid}.
Prove that T C
3.38. Let K C C be a closed bounded set of positive (planar) measure.
Define the function by
=JJ
Prove 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 B4
and its sum is equal to f, then following generalization of the Cauchy
integral 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 the
integral 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 = m2m
JE 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
jr
tm_lh(t) dt.
b)Forak>0 (k=1,2 m),let
E= (x1,..., xm)ERmI x1�0
1% 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}. Prove
that
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 the
integral fR2 dx.
4.6. Let a be the Cantor function (see problem 111.3.17). Compute the
following 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. Determine
the values of p E R for which the following integrals are finite:
a
11 ('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 the
function by
9,(z)= I (:€C).
Jo J0
Prove that (cf. problem 3.38):
a) E C(C),
'-'00
1;
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 called
an 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, then
N(A, e) <00 for any e >0. The function H(A, e)=log2N(A, e) is called
the e-entropy of A.
5.1. A set E C RI? is said to be e-distinguishab!e if the distance between
any two points in it is at least e. Let
M(A, e) = max{card(E)
I
E 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 that
N(A, e) � M(A, e) � N(A, e/2).
5.2. Let A be a bounded subset of
a) 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 the
following 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,where
b) 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}. Prove
that
H(A, e) (e —, +0).
5.11. Find the metric dimension and metric order of the set
I
Y—>.'1k3''lk—O, 1,2,
L €k and
17k
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<e
k�1
Theinfimum in the definition of e) is over all possible countable
coverings 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 and
b) = 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
I
JLq(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 C
b) 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 modified
p-dimensional Hausdorff measure by
= sup inf >(diam(Ck))" I A C U Ck, diam(Ck) <e
c>0 k�1
Prove that � � and hence can be used instead
of 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 the
following problems (cf. the definitions of the Cantor set and of a generalized
Cantor 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 an
open interval 5 symmetric with respect to its midpoint in such a way that
each of the closed intervals making up has length The left-hand
one 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 the
intervals of nth rank with length in, has already been constructed. From each
closed interval of nth rank remove an open interval symmetric
with respect to its midpoint in such a way that each of the closed intervals
making up the set \5c...c has length 11141. The left-hand one of these
closed intervals is denoted by and the right-hand one by
The 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)Find
5.26. 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 lakEEforanykEN
1%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. If
the 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 density
If 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 ihiEuclidean
norm of a vector x, that is, the quantity llxll2. Finally, as usual, the mean
value 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 to
use the formula (see problem VIII.3.22b))
j 1(x) dx
= j (j f(rw) dr, f E (*)
1 rflF2
a) —I dx=—n+2
b) _!_ I Ix
I
dx
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 volume
of the watermelon is filled by its rind: a) in 3-dimensional space? b) in
100-dimensional space?
§6. INTEGRALS OF HIGHER MULTIPLICITY 105
6.3. After falling into 1000-dimensional space, Alice was asked how to
make thin glass hoops (spherical zones) in such a way that their width is
equal to 1/10 of the diameter. "It couldn't be simpler," replied Alice. "You
need to blow up spheres, and then cut off the excess" (see Figure 7). What
percentage 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 assumed
that the probability of the point falling into a given set is proportional to the
measure of the set.)
6.5. Two points are chosen at random on the sphere S'1'. Let be
the 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' and
r E [0, 1], where h is an increasing continuous function on [0, 1] that
vanishes at zero. Prove that the mean values of f on the ball B'1 and on
the 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 . . have
a finite positive limit as n —' 00.
6.9. Prove that for any q E R and 0 <p < oo
f
6.10. 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 inproblems 6.14
and 6.15 and some other problems), then the expression x means that
<oo.nb
We use this notation in an analogous sense not only for sequences but also
for arbitrary families of positive numbers.
6.14. Suppose that q E R, m, n E N, 2 � m � n, in + q> 0, and
x=(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 \fl
and find a and b.
6.18. a) The ball B'(r) is tangent to all the edges of the n-dimensional
cube [—1, Compare for large n the volume of the cube and the volume
of 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 ball
B'1(p) tangent from the outside to all the balls inscribed in the cubes. For
what n is B'1(p) c [—1, 1J'1? For large n which is larger: the volume of
the cube [—1, iJ'1 or the volume of the ball B'1(p)?
6.19. Find the asymptotics of the radius of the ball whose
Gaussian 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-dimensional
octahedron: = {(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 octahedron
a) 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, then
for 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 the
set of all measurable functions defined on a set E and almost everywhere
finite.
We assume that the reader is familiar with the basic facts in the theory of
functions (the theorems of Lebesgue and Riesz about the connection between
convergence in measure and convergence almost everywhere, Lebesgue's the-
orem on taking the limit under the integral sign, the theorem on termwise
integration of positive function series, Bessel's inequality, the Riesz-Fischer
theorem, the completeness of the trigonometric system of functions). Some
of these theorems are constantly used in the solutions (for example, the the-
orem on termwise integration of positive series), and it seems to us that
without others (for example, the theorems of Lebesgue and Riesz) the very
formulation of a number of problems cannot be sufficiently understood, even
if these theorems are not formally used in the solutions.
§1. Convergence in measure and almost everywhere
1.1. Determine whether the function sequences below converge
in measure and almost everywhere on the intervals they have
integrable 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 be
the 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 —, I
in measure as n —, 00. Prove that f coincides almost everywhere with a
monotone function 10, and
—+
f0(x) at the points of continuity of
1.5. Let = I + where x E (0, 2ir), and
and are real numerical sequences such that > 0 and
—, +00. Prove that —, 0 in measure on (0, 2ir) as n —' It is
necessary 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 and
a.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 lit
1.9. 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 is
a 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) (a
regulator") 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 assertion
remain 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 any
e > 0 there is a set e c (0, 1) such that <e and f on (0, 1)\e
as 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 R
implies 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 1nk
k
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 measure
is 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 E
b) 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}
Thequantity )(f) is called the quantile (compare the definition of
with 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—see
problem VIII.2.9). Prove that:
a) for
b) 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 � ö
for some C, ö > 0 and any n E N. Prove that if the series
converges absolutely almost everywhere on E, then <ci.
2.8. Prove that if sin(nx + convergesin a set of positive
measure, 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 large
numbers.)
2.10. Prove that the following sequences converge in measure
on 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 � M
almost everywhere on E for any n E N, and let ; = fk• Prove
that:
a) Hd7k2112 <so;
b) —' 0 almost everywhere on E.
(The last statement is often formulated by asserting that the strong law of
large 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. on
E, that is, the strong law of numbers holds for the sequence
Prove that, moreover, the function h = belongs to
2.14. Let be the nth digit in the binary expansion of a number
x E (0, 11, and let
a —' 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 occur
equally 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 in
the 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-ary
expansion 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 numbers
written one after the other (in decimal notation)).
2.16. Generalizing the problems 2.14 and 2.15, prove that almost all the
numbers in the interval (0, 1) are normal.
2.17. Suppose that f E p2(0 1) and f(x)dx = 0. For arbitrary
n 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 functions
are 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) such
that rk(t) = €k for k = 1 n. What is the measure of the set of such
points?
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 J
1<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) j
c) (z,ckEC).
d)Provethatif ER,
J1
e1' 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 \1/2
2
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 law
of large numbers. Using the connection between the Rademacher functions
and the digits of the binary expansion (see problem 3.lc)), deduce from this
that the binary expansion of almost everywhere number contains "equally
many" 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 all
possible vectors in with coordinates equal to ±1. Prove that
a) y)4 � K1 . VE4(Cx. y)2)2.
b) � K2 where K1 and K2
are absolute constants.
3.12. Using the uniqueness theorem for Fourier transforms of finite Borel
measures, 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 the
Rademacher 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) L
ak%(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 series
1
sin 2kx converges in measure on
some set of positive measure, then akl <oo.
3.19. Let <00. Prove that:
a) if
esssup <oo
xEA k�1
for some (nonempty) interval then lakI <oo;
b) if
esssup sin 2kx <00
k�1
for some (nonempty) interval, then lakI <00.
In problems 3.22 and 3.21 we consider the Haar system {hm}n,>0. This
system is defined as follows. Each positive integer in can be
uniquely 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) the
Fourier 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, and
e112. 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 assertion
e) 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
2n
I ikx : 2
J
ICkI
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 the
series
+ .. +
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 very
weak 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):
1 222
a) 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 the
class Lip1_0.
In problems 4.3-4.15 the symbol J(n) denotes the Fourier coefficient of
a function f integrable in (—it, it) with respect to the orthogonal system
rInxl
1e
1(n) = j f dx (n E Z).
4.3. a) Let f, g E r), h = fg. Prove that if f(n) = =0
for n<0,then h(n)=0 for n<0.
b) Let f E it) and let c1 = e'. Prove that if 1(n) = 0 for
n<0,then also for n<0.
4.4. Let f E it). Prove that <00 if and only if
f(x) = g(y)h(x — y)dy almost everywhere on (—ir. it), where g and
h 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 the
class
f is a function of bounded variation on [—it, itJ , then
1(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 problem
111.3.13 a)), f(n) = 0(1/n).
§4. FOURIER SERIES AND THE FOURIER TRANSFORM 121
4.9. Let 0 < < 1 and = Prove that the function
1(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 Fourier
series 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 of
f. 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 a
2ir-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 depending
on a and f, but not on x;
b) if— = O(nU), where is the nth partial sum of the Fourier
series 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 become
acquainted with one of the numerous applications of this important concept
in the following two problems, borrowed from [41].
4.16. Let 1 = n0 < < be positive integers, let a0 am be
arbitrary 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, and
gk 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 IIfjfl2
j�k
4.17. Let 1 � < (n1 E N). Prove that for any complex
numbers 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 the
function 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 E
The 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 E
where 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 Cantor
function (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 uniqueness
theorems:
§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 on
with 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) = fl
d) —
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 is
invariant with respect to orthogonal transformations and is at the same time
a 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 inequality
f(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 inX). A mapping h: X —' X is called an involution
if ho h =
1.3. Let 1(x) = x+ 1 and g(x) = x— 1 (x E R). Choose two involutions
h 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 X
such 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], j
c) = (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 some
x0 E R, then 1 has a fixed point. Is this true for a continuous mapping
f: C-'C?
1.8. Prove that a finite group of affine transformations of the plane has
a fixed point (common for all transformations in the group).
126 X. ITERATES OFTRANSFORMATIONS OF AN INTERVAL
Mappings f: and g: are open intervals on
the line) are said to be topologically conjugate if there exists a continuous
bijection (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 h
will be called a conjugating mapping.
In problems 1.9—1.12 it is assumed that the mappings are continuous, are
defined on open intervals, and map them into themselves.
1.9. Prove that the relation of topological conjugacy is an equivalence
relation, 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 opposite
strict inequality holds for g (depending on the character of monotonicity of
h). Prove that if f and g are defined on R and one of the mappings f
and 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 topological
conjugates 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? Describe
the 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 mapping
1: 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 between
them, and g = is the corresponding Chebyshev polynomial determined
by 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 closed
intervals are replaced by open intervals.
1.17. Classify, up to topological conjugacy, the homeomorphisms of a
closed interval having a finite set of fixed points that includes the endpoints
of the interval.
Suppose that I is a mapping of an interval C R into itself, and let
f°(x)=x, f'(x)=f(x),and for n>1. Wesaythat
the 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 the
domain 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. If
for some neighborhood U of the fixed point and an arbitrary point
x0 E U, x0 x*, there is an index n0 such that U, then x* is
called 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 the
mapping 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 the
function We say that the Newton iteration process with initial point x0
converges to c if xk = —, c.
1.23. Suppose that E c E = 0, and > m > 0
for 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 the
Newton 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 of
1.26. 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 Newton
iteration process starting at x0 converges to a root of P, and all roots except
for 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. If
the trajectory is finite, then there is obviously an n E N such that the points
x0 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 if
x0 is an attracting fixed point for I", and repelling if x0 is a repelling fixed
point 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 infinitesequence of closed intervals such
that 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 has
a 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. Determine
whether these cycles are attracting.
1.33. Let 1(x) = 1 — ax2 (x E R). Find the infimum of the values of
the parameter a such that f has: a) an attracting 2-cycle; b) a repelling
2-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 infimum
of 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 are
negative and 1 , then 1 has a 2-cycle. In particular, a 2-cycle
exists if f'(x0) < —1.
1.36. Let g be a piecewise smooth mapping of [0, 1J into itself. Prove
that if Ig'(x)l � a > 1 at all points where the derivative exists, then g
has a for any n E N. Show by an example that this is false if
differentiability 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 itself
that belong to the class and are such that the Schwarzian derivative
Sf (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, where
f(x) 0). The class of such mappings is denoted by
1.37. Let f and suppose that the set of critical points of I is
finite. Prove that for each n E N the transformation I can have only finitely
many n-cycles.
1.38. Let g=ft (n� 1),andlet x<y<z besuccessive
fixed points of g. Prove that if [x, yJ does not contain critical points of
g,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 the
points 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) cannot
have more than one attracting cycle, and for a =2 it has an n-cycle for any
n E N, but none of them are attracting.
1.41. Let be a continuous mapping of the circle into itself having a
fixed 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 condition
f(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 any
three points, and an arbitrary orientation-preserving homeomorphism of the
circle 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 homeomorphism
then 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 this
homeomorphism. 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 function
f is said to be a density of p if p(A) = IA 1(x) for an arbitrary
measurable 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 almost
everywhere on X. By the well-known theorem, p has a
density if and only if it is absolutely continuous (with respect to Lebesgue
measure), that is, if and only if p(A) = 0 when = 0 (see [14J). A
measure p is said to be equivalent to Lebesgue measure if p(A) = 0 if and
only if = 0. It will be said that a mapping T: X —, X preserves
the measure p, or that the measure p is invariant with respect to T, if
p(A) = p(T'(A)) for any measurable set A C X. If ii is a measure on the
measurable subsets of a set Y C Rm, then ii is said to be the image of p
under a mapping T: X —, Y if i'(B) = p(T'(B)) for any measurable set
B 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 measure
2.3. 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 the
torus T2 = S' x S1 and TA with an automorphism of T2 as a group. This
lets us call the transformation 7'4 an automorphism of the torus.
We recall a fundamental result: on every compact group G there is a
nontrivial 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 E
A}. 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 the
following problem.
2.4. Let the mapping TA: X X be the same as in the preceding prob-
lem, but let det A E Z,
I
det 1. Such a mapping (now not one-to-one)
is called an endomorphism of the torus. Prove that it preserves Lebesgue
measure.
2.5. Verify that the mapping T(x) = f mod! (x E (0, 1J) preserves the
measure 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 power
TIZ (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 monotone
and smooth. Prove that if it has an attracting cycle (see §!), then T does
not 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 translations
if for each a E the condition p(A) = 0 is equivalent to the condition
p(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 to
2.!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 f
be 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) is
finite (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 if
Pf(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,..., of
values it takes on Let us denote it by The set of nonnegative
vectors 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 matrix
determined 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
(2x
1
Find the general form of an absolutely continuous measure invariant with
respect to T.
b) Let T: [0, 1J be the mapping defined by
(x+1
L.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) and
prove that there is no invariant finite absolutely continuous measure for T.
What is the limit of measures = dx as n —, 00 if is normalized
by 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 of
Q, that is, the mappings T(z) = (az+b)/(cz +d), where a, b, c, d ER
and
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 into
itself 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 invariant
with respect to all the transformations Lab and Ra ,b' respectively.
We remark that the transformations under discussion correspond to left
and 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 this
group.
2.18. Prove that if a transformation T: [a, b] [a, b] preserves some
finite absolutely continuous measure, then T is topologically conjugate to
some 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 Bogolyubov
and Krylov:
For every continuous mapping of a compact set into itself there exists a
finite invariant measure.
2.19. Prove that if T is a homeomorphism of the circle S' with an
irrational rotation number (see problem 1.42), then T is conjugate to the
rotation 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 satisfies
the normalization condition p(X) = 1 (such measures are called probability
measures).
Let T be a transformation of the set X. A set A C X is said to be
invariant if T'(A) = A. A transformation T preserving a measure p
is said to be ergodic if for each invariant set A C X either p(A) = 0 or
p(X\A) =0.
2.20. A measurable function f defined on a set X with measure p is
said 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 invariant
measurable 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 Lebesgue
measure. Prove that if T is ergodic, then for almost all x E X the trajectory
of x is dense.
2.23. Let X = [0, 1)1? be the n-dimensional torus (n E N), and let S
be 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 conjugate
if there exists a measurable mapping h: X —, Y such that hjz = ii and
Soh(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 measure
ji 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 for
almost all x E X (with respect to ji).
A fundamental theorem in the theory of transformations with an invariant
measure (or "ergoclic theory") is the Birkhoff-Khintchine ergodic theorem (see
[3], [15]):
Let T be a transformation of a set X preserving a probability measure
and let f be a measurable function with <+00. Then
for almost every x E X (in the sense of ii) the limit
lim! >2
f(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) and
equal to f(x)dji(x) (that is, to the mean value of f on X). The limit
on the left-hand side of (*) is called the mean value of I along the trajectory
of 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 limit
lim >0<k<fl €k exists (the mean number of times the trajectory hits the set
A) for almost all x E X, and find it.
Probability measures ii1 and on X are said to be mutually singular
if there exists a set A C X such that ii1(A) = 1 and ii2(A) = 0.
2.28. Let ii1 and ii2 be two probabilitymeasures on X, and let T be
a transformation that is ergodic with respect to both ii1 and ii2. Prove that
the 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 that
f(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, and
the equality (1) is called Weyl's criterion for uniform distnbutivity.
2.30. Let x = Xk E {0, 1}. be the binary expansion of a
number x E [0, 1). Prove that the limit lim Xk (the frequency of
l's) exists for almost all x E [0, 1), and find it. Sofve the analogous problem
for p-ary expansions.
2.31. Prove that if a transformation T: X —' X preserving a probability
measure 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: if
e1,..., 1} and
n)}
(here x = Exk/2k is the binary expansion of a number x). then
= fJ
l<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 the
measure
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 number
2k (k=O, 1,...),thatis, a0= 1, a1 =2, a2=4, a3=8, a4= 1
What 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 a
continued 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 many
terms if and only if x is rational. A finite continued fraction [a i,...,
can be written uniquely in the form of an irreducible fraction An
infinite continued fraction is a limit of finite continued fractions:
[a1, a2, . -.1 = lim[a1,... , =
and the number = is called a convergent. It is not hard
to verify that
—
The Gauss transformation
T(x) = mod 1 (x E (0, 1])
is connected with expansion of numbers in continued fractions. Indeed, let
x = [a1,a2, ...] and y = [a2,a3,...]. Then x = l/(a1 +y), which
implies 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 the
inequalities
� 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 to
1
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 the
mapping a = {a,,} —, E EN where = {,1(k) k)
1k) = a2k—(211) (k, J E N), and show that it is bijective (compare with the
solution 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 convergent
sequences 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 of
the form aER.
1.9. b) Consider the set = {A E .9(N)I maxA = n} (n E N), and
prove that = Setting = 0, define by induction with
the help of the equality (A E n E N) and
show that the function + 1 has all the required properties.
1.11. Using the Axiom of Choice, form a set E1 by choosing one point
from 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 off
zero. Use the same idea as in problem 1.12, taking points with rational
coordinates in all three angles corresponding to a bird track and sufficiently
close 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 to
its axis. On this axis fix points M and M' symmetric with respect to P
and separated by a distance equal to the diameter of the hoop. Consider
the balls about M and M' with radius ö (see Figure 10). We say that
these 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 balls
accompanying a hoop 0 intersect the balls accompanying a hoop c/, then
O 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 points
in the same set
b) Let = {nlk E By a), there is an index in1 such that the set
E1 = 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 a
2
FIGURE 11
and again using a), we see that there is an index m2 > m1 such that the
set 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 if
card(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 = 0
then forany nEN.
Let [p, q] C E and a E E, and assume for definiteness that a> q. We
prove that [p, q] C E. Let c = sup{x > x] C E}. It is clear that
Assumethat c<a. Considerapoint a0EE suchthat c<a0
and (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 all
points of the form (n E N), and E also contains arbitrarily small
numbers.
1.22. a) Let be the system of all possible disks with rational center
and radius. It is clear that if x E then there is a disk B' E such that
x E B' c For each disk in fix arbitrarily a disk Ba containing it (if
such a disk exists). Show that the system of disks chosen in this way is the
desired object.
b) For each n E N find an at most countable set E such that
E 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, ={x
To 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 there
exist 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] such
that 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 any
x E q12]. Continuing this process, we get a sequence of nested intervals
whose common point is the desired one.
1.26. Argue as in the solution of problem 1.25, replacing the set G at each
step 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 of
binary 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 the
set Q = has the cardinality of the set of binary sequences. Verify
that Q \A is at most countable.
1.33. Assume that (a, b) is a bounded interval, and is a sequence
of 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 the
nonempty sets we construct the intervals and Carrying out
this construction for we get the intervals and The intervals
and are disjoint from the set F1 U F2 and from each
other. Continuing this process by induction, we construct a family ....
of open intervals satisfying the conditions of problem 1.32, and also tcie
equality n any e1,..., It now remains
to refer to problem 1.32.
1.34. Find a straight line not containing points of tangency of the given
disks. Use the result of problem 1.33.
1.35. Let be the closure of Arguing by contradiction, construct
a sequence of nested closed intervals C (a, b) such that = 0 for
any 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 that
5,, 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 inequality
of 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. To
prove 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 � H
k
= fi
1<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 that
n weights of magnitude a1..., a lever rotating in the
vertical 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 if
the the weights increase in the direction away from 0. But if
the 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 and
2.8. 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 the
inequality 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 the
inequalities (j—k)(ak+l—ak) � � for 1 � k <J � n
are used instead of the inequalities ölk — fI � ak — � INk — ii.
2.12. a) Using the Cauchy inequality about the arithmetic mean and the
geometric 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 set
bk = 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 follow
from the result of the preceding problem. To prove b) for n = 11 use the
factthat 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 do
this, show that f first decreases and then increases, that is, f'(x) <0 for
x 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. Then
g(x) = + 1)P1 where
1>1.
Itisclearthat as t—' 1+0,and as t—'+oo. Itnow
suffices for us to verify that is increasing. Uncomplicated transformations
give 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 are
strictly 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. Now
consider the function v. Since
I 2r \fi_1\r
it remainsfor 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 to
represent in the form and investigate the sign of the function
ip 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 in
the form = iW(t) and show that � > 0.
b) It must be verified that the function = 12/n2 + et(1 — 1/n)'1 does
not exceed 1 on [0, n]. Since the inequality holds at the endpoints of
the interval, is suffices to establish that � 1 if = 0, that is, if
e'(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 to
prove 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 the
function = — 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 that
sgn = 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 it
suffices to observe that for p <0
and
For p E (—1, 0) the inequality d) is valid for x close to 1, but it does
not hold in a neighborhood of zero—for a proof it suffices to consider the
Taylor 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 than
two, then the function is easy to represent as half the sum of two different
functions 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 functions
f± having the same intervals of constancy and taking on them the values
c1(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 consisting
of 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 the
condition f5 E K: c1 = bf(s(b—a)) and c2 = af(s(b—a)). If a <1 <s <b
then
b
j
The 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 exist
integers p and q such that 1 � q � Q and Ix — < 1/(qQ) (since
the right-hand side of the inequality tends to zero as Q +00, while the
left-hand side is positive by condition a), this implies that there are solutions
of the inequality Ix < 1/q2 with arbitrarily large denominators q).
Fixing a number Q E N, we partition the interval [0, 1) into Q congruent
intervals [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 the
number of these points is greater than the number of intervals, at least one
of 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 get
that — —' 0 as k —, +00. Since — E Z, this implies that
— = 0, that is, p/q = for all sufficiently large indices k. and
this 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, then
4 = + where k,, E Z and —, 0.
§3. IRRATIONALITY 153
Consequently, 0 = — + —
Therefore, for sufficiently
large n we have = , that is, = A E But then the
equality 4fl = + ç, implies that 1 = irA. This is impossible, because the
number it is irrational (see problem 3.13).
If F 0, then we get from the equality
sin
4n+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 = + + it
follows that 0 = 2ir(4k1? — + + —
, and this is possible only
for = ±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, where
A E But then it follows from the equality 4' = 2irk1? — 2ir/3 + that
it E The case = 2ir/3 is handled analogously.
b) Represent the number 0 in the form 0 = irm + it where
m = [0/in and = 0 or 1.
3.5. Consider the numbers 0.101001000100001000001 ... (the nth 1 is
followedby n 0's) and 0.1234567891011121314151617181920... (allthe
natural 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 problem
11.1.21. 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 a
number m E N such that the numbers a = m +p and b = m + q satisfy the
inequality
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 interval
of 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. qJ
for 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, where
E 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 get
a sequence {[ak, bkj} of nested intervals. The desired point is taken from
their intersection.
In conclusion note that the numbersx satisfying the conditions of the
problem form a set having the cardinality of the continuum. For the proof
it 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 step
the number of intervals having the required properties. As a result we get a
generalized Cantor set with all points satisfying the condition of the problem.
3.9. Direct computations prove the assertion for functions of the form
f(t) = cos(2irmt) (m = 0, 1, 2,...). This implies that the assertion is valid
for all functions I of the form f(t) = >0<m<M am cos(2nmt). It remains
to use the Weierstrass theorem (see VII.3.1 1).
3.10. Use the same idea as in the proof of the assertion a) b) in
problem 3.1: for any number Q E N there exist integers q, p1,.., such
that 1 � q � Qm and I 1 m. Fora proof,
partition the cube [0, 1)m into Qm congruent disjoint cubes and consider
the 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 3.1
holds.
§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 enough
to that — < 1 and P(p/q) 0. Since E Z\ {0}, it
follows that � On the other hand, expanding the polynomial
P(t) in powers of 1— we get that
1<k<n
Thus, � — The assertion proved is called
Liouville's theorem.
3.16. b) Let = E ek'nk and prove that for any e > 0 there exist
solutions of the inequality — < with arbitrarily large q E N.
Since (see problem 3.14), this gives us by Liouville's theorem that
the 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 that
the sum of the series is not an algebraic number of degree N. It
remains 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!, we
get that
pn!—qn!
O<k<n
The left-hand side of this inequality is a positive integer, while the right-hand
side 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 for
A = 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 positive
integer, 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 find
a 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 side
tends 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. Suppose
that 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 Wallis
formula can be used to compute their limits.
1.3. Show that the sums under consideration differ little from a geometric
progression.
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 27 256
U 1++Il+ 3+ 4" \n n2 n31 (n+1) (n+1) (n+1)
(fk\k
V,,
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 monotonicity
and the inequality 1+ which is proved by induction. To compute
the 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:
fl
1<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) fl
1<k<n
it follows that 2— = Using the result of problem 1.11, we get
that +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, and
this 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 � 1
in example e); these inequalities are easy to prove by induction). To find the
limit 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 also
of n.) With the help of the equality c) and the result of problem 2.12a) show
that
that is, — const — 1)!). This implies immediately that
a—xN —xN — 1)!). Therefore, using the result
of 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 fixed
in 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+1
This is impossible, because the partial sums of the series
n+l
do 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 form
n = 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 implies
that = =1, that is, ar/n —i!.
1.22. Apply the result of the preceding problem to the sequences
and {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 a
sequence 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 clear
that if the lengths of the open intervals adjacent to the increase without
bound, then — —* 0. On the other hand, by choosing successively the
lengths of the to be sufficiently large it is possible to ensure that there
are numbers arbitrarily close to +1 and arbitrarily close to —1 among the
numbers + 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 contain
infinitely many numbers of the form [ck]. but this is incompatible with the
assumption 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 suffices
to consider the case when = 1 <+00. Let L >1 and let m = mL be
an 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
thiscase
2.4. 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 and
B = 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 the
solution of problem 1.24b): let {bk} form an alternating sequence of +1
and —1 on each closed interval and on the adjacent open intervals
let bk = 0. For a suitable choice of the lengths of the closed intervals
and 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 + ... + and
lim + + only the second exists.
b) The inequality a2 � b follows from the Cauchy inequality. To construct
a sequence with given limits a = lim((a1 + + and b =
+... + where 0 � a2 � b � a 1, consider the sequence
taking the value on a set A C N of density a (see problem 2.3) and the
value /3 on the set N\A. Prove that the required equalities can be ensured
by suitably choosing the parameters and /3.
§2. AVERAGING OF SEQUENCES 163
2.6. It can be assumed that / � 0. Here it suffices to consider only
the case / = 0: if 0 < / < +00, then the sequence — should be
considered instead of and if / + 00, then the sequences and
should be interchanged. (The conditions > and +00
hold 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. For
fixed 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 of
the 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 1
0<———<--—---——< 2[x] x x—1 X (x—1)
for x> 1 and observing that the improper integral
1\
I t———idx
J2 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. Since
1 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 inequality
can 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, it
is 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
'-' 1
x =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 then
S
S ak
Yn+l = Y1 + — = —
1<k<n k>n
— a a
1 — bk>l
This yields b). It remains to consider the case when 0 < Ibi < 1. We have
that
— 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 all
n E N, or � for n � n0 (since the inequality � implies
that � for n � n0).
Let 0 <p < 1, and define / = and L = Since the
sequence is bounded above (by the number max{x1 , L < +oo.
Assuming that / <L, fix two numbers 1' and L' with / <1' <L < L'; they
will 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 the
number 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 the
preceding 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 this
case � 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 last
inequality holds for n = 1, 2.
3.8. The proofs of these assertions are similar. We confine ourselves to
the cases c). Fix a fi E (0, cr/k). By assumption, there exists a number
c = > 0 such that
166 II. SEQUENCES
x
e
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 and
for sufficiently large M = Its validity for all subsequent indices n is
ensured by the choice of np.
3.9. Arguing just as in the solution of the preceding problem, we get that
it suffices to prove that fi(n — k) lnAfl k — (fin — —. +oo. For brevity
let a number y > 0; its choice will be made more precise
below. For sufficiently large indices j we have that j(1 — a11/a1) <y, that
is, 1 — y/j. Consequently,
aflk
H
J n--k<j�n
Thus,
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 for
1 � m <n, n > k. Then
C Ix�—I +...+
+ + + + 1).
Therefore,
C? 1
<
CeS
—
(the last inequality follows from the inequality_e' � 1 + t). Choosing the
factor C large enough that � CeS/yA for n = 1 k, we get
from this that the inequality holds for all indices n.
3.11. Use the result of the preceding problem. In example d) use the
result 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. It
is clear that A = where = {x E A1w1(x) � 1/ n}. If A is
assumed 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)), and
let {xk} C Xk —, x0, Xk x0 (k E N). It can be assumed without
loss of generality that Xk > x0 for any k E N. We prove that f does not
have a limit from the right at the point x0. Indeed, since Xk E there
are 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. To
prove 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 that
if 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 of
theform (EEL and SEER.
1.5. Assume that the set DR(f) is at most countable, and prove that the
set 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) is
uncountable, 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 such
that (x0, x0 + e) n A 0 for any e > 0 (see problem I.1.23b)). Fix a
sequence 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 and
passing 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, that
is, 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 E
Let 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 interval
centered 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 we
construct a sequence of indices and a sequence of closed nested
intervals such that Ii < 1/3 for x E '-'21—1 and < 1/3
for Itisclearthat 1/3 atapoint
and hence the sequence cannot have a limit.
1.8. a) Consider the function equal to 1 on an at most countable subset
of E with closure E (see problem I.1.22b)) and equal to 0 at the remaining
points.
b) Consider the series where is the function constructed as
described 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 that
f is locally constant on Use the result in problem I.1.22a). Another
solution can be obtained by modifying the arguments in the second solution
of problem 1.3.
The arguments remain in force for a separable metric space. The result
is not true in the case of a nonseparable space. To get a counterexample
consider the space XxS without isolated points, where X is a closed interval
with the discrete metric and S = {z E C I Izi = 1}, along with the function
f(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 < 1
f(x)= l+x'+[x']
1 + (2— x)' + [(2— x)'] for 1 <x < 2
2+(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 some
positive number. Use a).
1.17. Assume that the limit 1(x) does not exist. Then there
are 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 that
each 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)—1
1.0 foranynEN,
where is the sequence of prime numbers, and is a sequence whose
set of limits of convergent subsequences fills R (for example, enumerate the
set 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 and
the 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 exist
and such that and supa if(x + h) — f(x)i
e. Prove the last assertion by arguing by contradiction and constructing a
sequence of nestedclosed intervals and a numerical sequence
such that —, +00 and + h) — � e for h E
1.21. Prove that the function = f(x + 1/ n) — 1(x) takes a zero
value on [0, 1 — 1/n]. If 5 1/ n and S > 0, then consider the function
f(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 when
f 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) such
that 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+1
F(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 = is
a 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 way
from 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 and
use the result of problem 1.1.33. Prove that if the closure of Em contains an
interval 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 E
N).
3.2. Show that f(x) —. 0 as x —. +00, assuming that the
limit 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 a
sequence +00 such that =0, — a, and —, /3.
Another proof of the equality 1(x) = 1, extremely short but not
as instructive, can be obtianed by applying L'Hospital's rule to the pair of
functions 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 satisfy
the Bolzano-Cauchy criterion at infinity.
3.3. Assuming that f'(a) <0 < f(b), prove that the smallest value of
f 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 can
be 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 sequence
such that > 0, — 0, and + <0. Therefore,
g(x) = —
f is increasing. In the general case
consider the function f(x) = 1(x) + ex, where e is an arbitrary positive
number. If g 0, then the result already obtained should be applied to the
functions 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 respect
to h , show by using 3.6a) that 1" const. Using 3.6b), we see that this result
is preserved if f and g are continuous on (a, b). If only the continuity of
I 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]l
1(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 imply
a). A corresponding counterexample can be obtained by considering one of
the coordinate functions of the Peano curve constructed in problem 4.4 and
using the fact that a set of constancy of this function does not have isolated
points (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 an
endpoint of the smallest region of monotonicity of I whose length is not
less than h.
3.14. Consider the series with terms of the form — ak), where f
is 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, 1
3fl)
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 to
the boundedness of the sequence This is not true for arbitrary
polynomials in two variables (for example, for Q(x, y) = x). Verify that
Q(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 and
ip follows from the coincidence of any previously specified number of first
digits 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 obtained
only 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 manyof the digits are
even, then, replacing a2k by 2— for sufficiently large k, we get a point
in arbitrarily close to t. If a2k = 1 for k � k0, then consider the
point (k � k0) . • — obtained from
by replacing the digits = 1 and = 1 by 0's, and by the
number 2 — The arguments are similar for the consideration of the
set
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 is
an index k � in — 1 such that Let k is the smallest index
with this property. Then in view of the inequalities 3m — <
the expansions for t and ? must have the form t = .. .
and ? = — 1)22... ... (if at least
one 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 four
cases 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 number
of 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 homeomorphism
of 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 independence
of 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 graph
of 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. Consider
the function g(x) = +1(x) (x > 0) and use the arguments in the solution
of 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 of
problem 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 strictly
increasing 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(n
Pflk+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 as
X —, +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 0
and 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 see
from 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 be
the largest zero of I on (0, a). Then = 0 for any n E N, and
hence 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 the
number c. Note further that a can be chosen arbitrarily large, because
= 4hf(a) 0. Accordingly, 1(x) 0 for x 0. Assume that
f(1) = 1 and consider the set A = {x > 011(x) = x}. Verify that its
closure 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 of
the 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)) and
H(u) = (u ER).
§5. FUNCTIONAL EQUATIONS 177
b) Verify that 0 � I � 1 and that if 1(x0) = 0 (1(x0) = 1) at some
point 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 is
strictly increasing. Consider the function g(t) = in llnf(e')l and use the
result 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 the
function g(z) = f(z)/f(IzI) and by using the fact that g is one-to-one
on 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 that
g(z) coincides either with or with Prove that the function
h(t) = If(t)I is increasing on (0, 1). Represent f(t) in the form
and 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 induction
to see that f coincides with the Cantor function (see problem 3.17) on the
open 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 from
this 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 for
some m EN the equality = 1 holds for any E Prove that if
n 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 in
e) 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 they
cannot 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 have
that n2 = irk + e, (n — 1)2 = irk' + e', and(n + 1)2 = irk" + e", where
k, 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, and
1/ar = +00, then — = +00, where = (A0 = 0).
Let
0m2 A—A
Since the harmonic mean does not exceed the arithmetic mean, it follows
that
— = —
m 2<n�241
I 79
180 IV. SERIES
Consequently,
1 22m
1
—
=
Therefore,
1 1 1 2m+i
EAA A-A
n 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 follows
that N = U Nm. Let Nm = {n1, n2, . . . }. Since � <1/2n1, it
follows that n1 > 2Cm Using the result of problem L3.llb), we get that
1
n
2 1
+ �
m
Taking Cm = 2m—2 we get that
1 m(l—a)
= >2Sm � <+00.
1.7. Instead of double series it is convenient to consider sums of the form
In 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 used
in the solution of problem f).
1.8. Each fraction 1/2fl (n = 0, 1, ...) appears in the series times
with the sign + and times with the sign —. The desired rearrangement
is 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 by
the sequence 1 for n=2k, for
2.6. a) Let am = (a0 = 0). By changing x1 it can be
assumed 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 corresponding
counterexample is easy to construct by taldng = 1 for n E N.
b) Take a subsequence {nk} of indices such that 1/2, and
let = 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 Abel
to the series we get that this assertion is equivalent to
the relation = If the series converges, then this relation
follows 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 divergeS
if l/ln(l +n).
2.10. It is not hard to construct positive but nonmonotone sequences
and 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 suitable
choice of the sequence
2.11. a) Since = — (S0 = 0), the series diverges
if 74 1. But if —, 1, then divergence follows from the
divergence 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 the
original series converges simultaneously with the series
—
a 1, then this series converges, but if a = 1 then it diverges, since the
series — 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 the
divergence 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 that
x/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, we
get 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 into
the sum of the integrals over the intervals , S,], we get that the series
diverges. Therefore, the series also diverges—for a
proof 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 partial
series do not fall in the interval (cr, /3).
b) a). Fixing a number s E (0, A], use induction to construct a
sequence 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 index
N such that <1/(k+1) for k � N. Then in view of the definition of the
sequence we have that k � N. Therefore, the
sufficiently 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,
>Xk
J�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 that
Eekuk <+oo. From the sequence {ek} we construct a sequence of indices
Nk 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 Dirichlettest.
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 L1
N�n<M
N
M
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 such
convergent 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 over
the 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 problem
3.3b) gives us that
b 1
In <+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 follows
from 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 series
diverges. Since at least one of the series and a2m+l diverges, there
exists 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 sequence
of nested sets and a increasing sequence of indices such that
the density of is equal to and n(P � 1. The desired
sequence {n1, n2, ... } is obtained by numbering the elements of the set
A = 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, and
the 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 that
lami � >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 series
converges.
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 sum
ço(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 N
is sufficiently large.
3.11. In the sequence {lfn} rearrange the terms with 2k1 2k in
increasing 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 to
the 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 a
sufficiently large index L. The assertion g) a) is obvious if a, 1. The
example of the sequence 2, 3, 22, 32 33, shows that it is not true
for 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 that
c) 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 series
converges:
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. Assume
not. Then there exists a sequence —, 0 such that = 0.
Take a sequence C N such that 74 0. Then the series obtained
from the series a1 — a1 + a2 — a2 +.•. by repeating the nth pair of terms
times is convergent. However, the series consisting of the values of I
diverges, since 74 0.
It is clear that f is continuous at zero (see problem 3.17). Prove that
f is continuous in some neighborhood of zero. Assume the opposite: there
exists a sequence of points of discontinuity such that —, 0. For any
n 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 that
74 0, 1(ç) — > and — <+00. The series ob-
tained from the series a1 — t1 + a2 — t2 + ... by repeating times the
pair — 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 of
integers T +oo such that 74 0. The series obtained from the series
by repeating each group + + — — — of terms times
obviously converges. However, the series consisting of the corresponding
values 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 for
the relation i,, 0, that is, for the equality x = It is not hard to
verify that if (*) holds for some index, then it holds also for the preceding
index, and hence for all the preceding indices.
The sequence will be constructed by induction. It can be assumed
here that Q has already been numbered in some way. Therefore, we can
refer 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 and
sup Q = 1 that such elements exist. Assume that q1,..., have already
been constructed and that (*) holds for m = 0 n— 1. Let q be
the first element in the set Q\{q1,..., q E — 1,
then we let = q. The inequality (*) will be satisfied for m = n. If
q — 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, the
number satisfies the inequality (*) for m = M, and hence (*) holds
for 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
E
1<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 the
integral 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's
formula.
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 first
assertion 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), we
get 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 convergence
is not uniform, because the general term of the series does not tend
uniformly to zero: = 1. To estimate the sum of the series we
break it up into two sums: + where N E N is such that
N! � 1/x < (N+ 1)! (if x> 1 the first sum is equal to zero, and the
second 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 limits
superior (inferior) of the ratios A(t)/B(t) and >n>N a//B(t) are the same
for 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 by
selecting 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 relation
A(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 hence
A(t) I \
= exp > + o(1) —, 2
as
5.10. 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 problem
VI.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 get
that
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 terms
of 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
>: -1/2
n 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 use
the Abel theorem (see problem 5.6).
6.9. The problem can be reduced to the preceding problem with the help
of the equality
1 1( 1 1
4n2 — 1 — 2 2n 1
6.10—6.12. Using the Abel transformation, employ the result of problem
6.5.
6.13. Use the result of the preceding problem and the equality
in .
. 10
I 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 uniform
estimate of the remainder break this sum up into two sums, the first con-
taining the terms with "small" indices (not exceeding ir/2x). In estimating
it, use the inequality sin � ti. To estimate the second sum use the Abel
transformation 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 of
problem 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 integrals
i: 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 the
solution of the preceding problem. If dx < +00, then integrate the
series (see problem 6.16) sinnx termwise over [0, it] and use the
result 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 which
the sum = kx)/k has a nonpositive minimum. Using the
necessary condition for an extremum = 0), show that sin nx0 � 0,
and hence the sum also takes nonpositive values. Continuation of this
argument 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
Jlt
dx = +... 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 a
decreasing 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 difference
1(0) — 1(x) for x = 2ir/4m (m E N) or use the result of the preceding
problem.
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 consider
the 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
=it
0<n<N
Let S = supXER MEN SM(x). By assumption, S < +00. Therefore.
+ � +
j
� =
Thus, + � The sum + can be
estimated similarly, except that the product
fl
0<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 above
on 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 also
for 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 proved
follows 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 the
iiiequality a) use the result of problem 6.30. To prove the inequality b) use
the 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 prove
uniform 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 the
assertion 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—1
AlA 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 any
numbers A and B with 0 < A <B there is a number C E [A, B] such
that = e_4C Since the integral f°° f(x)dx
converges, 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 is
equivalent to the convergence of the integral f00
f is continuous on [0, +oo), then the assertion to be
proved 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 +00
I I -CX, f(x)dx—, e f(x)dx
JO JO
A +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 of
the third integral follows from the second mean value theorem). The first
integral 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 use
the result of problem IV.4.12. To justify termwise integration of the sum of
the 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 x
1.20. 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 VARIABLE203
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 parameter
use 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_x
I —dx—i —dx.
J0 x j1 x
with the help of the result in problem 1.24. Use integration by parts and the
integral 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 equality
ln(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 lnxdx
up to a quantity infinitesimally small as —' 0.
1.34. For k = 1, 3, and 5 reduce the given integral to the Euler-Poisson
integral. For k = 4 use the result of problem 1.14.
204 V. INTEGRALS
1.35. Integrating by parts in the indefinite integral, we get that
2 2 2 2
I e_X dx I 1 e_X I
2 2
—i---1-dx
J (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 —x2
I 2 2=21 e
Jo (x +1/2) Jo
1.36. Denote the desired integral by 1(a). Differentiating with respect
to a for a > 0 and making the change of variable x a/x, we get that
f'(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 problem
1.33):
2 +00 +00 2
I(s)
= sinx (j dy) dx
2 (+00 sinxdx\ 2 [+00 dy
= Jo Jo 1 + (e
Therefore,
[+00 sinx
d — 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), show
that
1 1 1 \ 1 1lim Ini
J2 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)) and
change 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 dx1
c)
dx1 d; = 1+00
[+00
f+OO
x1 x2
2.3. Use the Euler-Poisson integral (see problem 1.33).
2.4. a) Assuming that a2 + b2 > 0, let u = (ax + and
v = cx + dy, where c and d are chosen so that (x, y) —. (u, v) is an
orthogonal transformation. Then x2 + = U2 + v2, and hence
lax +
= IL2
= e_V2 /2
dv j
ue_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 sinOdO
i 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 dv
J 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 = JJJJ
dt1 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 computed
with 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 an
orthogonal transformation of variables the problem can be reduced to the
case 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 4ir
cosh(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 the
last 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) isthe
area of E.
2.10. Suppose for definiteness that x � y � z. These numbers are the
sides of some triangle if and only if z � x + y. Consequently, to find the
desired 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 Figure
12). Let S be the set of points in the square that lie above the parabola. The
equation 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 follows
that P1(a)= 1 as a—.+oo.
b) A biquadratic equation has both real and complex roots if and only if
v <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 double
integral sign in (**) is absolutely integrable on the disk {(u, v) I u2+v2 � 1}:
(1 1! lu—aldudv
JJu2+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} and
represent 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 the
set 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 finite
limit as a —. 1. Since f is clearly continuous on [0, +oo), it follows that
f 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 sin2
2 = 4a
(jnI2
+jn
0 1 — 2a cos ço + a 0 n/2
2 I.7tj2 sin2 dço=4a(1+a)i 22 2 2
Jo (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 for
a>!.
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 )dx
Jo \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 dx
i = (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 the
induction. Assume that rn < n and
rn
I —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
i
n—rn_ n—rn
I = (fin_mIni3 lna)—
n—rn (n—rn)2
and, moreover, — en_rn) = 0, because 0, if the degree
of the polynomial p is less than n. Therefore, it follows from (2) that
+ + xn)_rn —(/3 + + Xn)rn) dx
=
lnfl—a (3)
Since
0 as /3 —, +00 if k < n, (4)
we get the required result by passing to the limit in (3) as /3 —. +00. The
relation (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.
x
2.17. a) In the integral +(an—ao)xn) dx1 . .
make the change of variables y1 = i—x1, = x2/y1,..., Yn = xn/yi. Use
induction.
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••dx
I c+ akxk
For c = 0 this leads to the equality
p
1 1 dx1•..dx
11
lakI) J —i.\1<k<n
The last integral can be expressed in terms of the Euler f-function:
p/2
cos 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 11
lnxlnA+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 integral
f+°° / 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 integral
for 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 the
equality (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 to
that of problem 1.6d).
b) With the help of the change of variable y =r° and integration by parts
prove 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 integral
c) Let
ir/2
1(A)
= j cos2x/(1 + Ax)dx
and
,r/2
1(A)
= j sin2x/(1 +cos2 Ax) dx.
It is clear that
çit/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 from
below. 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 relation
11+T/A
J f(x)ço(Ax)dx = / (f(x)—f(t))ço(Ax)dx
I 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) dx
k
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 continuous
periodic function on R can be taken instead of sin Ax.
1.16. a) Make the change of variable x = where = is a small
positive parameter. We have that
j+°°
dx = exp ((i) —
dt.
Let us now select the parameter so that as e —, +0 the integrand tends
to 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 e
J0
it suffices to justify passage to the limit under the integral sign:
(+00 (ft \1_C
I 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 the
integrals 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 help
of 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 integrals
over 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<nif the nonincreasing sequence {ak} of positive numbers is such that the
sequence {kak} is nondecreasing.
Let M(x) = maxl<k<fl akl sin kxl. Since
I
sin � 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
x
2<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)A
dt.
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 problem
c) 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 numbers
I and L, 0 < I < 1 < L, and choose a positive number 0 such that
Ix" � 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 them
we 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 lim
A—'+oo A—'+OO
Since I and L are arbitrary numbers, 0 < I < 1 < L, this implies the
equality = 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 that
Ih(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
1b
dx � 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 of
variable 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 contribution
is 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 at
e
— 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 that
r 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 APX
c) 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 remaining
interval 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 change
of variables t = = —xlnx Then
1C
i x dx=—i
j0
where c =(In 3)/3 and ip is the function inverse to Since lnx In
as x — +0, it follows that In! In as I +0. Using the result of
problem 2.8, we get
113
Ax
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)x
dt
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, we
get that it is equivalent to the relation
— 1
Jo
u—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) is
equal to the number of points (u, v) E N2 not lying above the hyperbola
uv = n. Since the set {(u, v) E N2Iuv � n} is symmetric with respect to
the line u = v • to compute T(n) it suffices to calculate the number T(n) of
points in the set E = {(u, v) E N2Iuv � n, u � and take into account
that the points in the square [0, x [0, are also in the set symmetric
to 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 of
problem 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 index
n. 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 any
n E N, then = +00. Here it is not hard to ensure that the integral
ff00 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 +
1
çk+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 equality
fk 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 the
equality 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-Maclaurin
formula. 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—cr
n 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). We
that 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 that
the 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 exponential
along with the result of problem 2.9c).
3.21. Find the asymptotics of the sums = >1<k<n where p is
a fixed positive number. Since the numbers are rapidly increasing for
1 � k � n/2, the main contribution to the sum under consideration comes
from the terms with indices close to n/2 (as is clear from the following, it
suffices 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(2m
I m—k)'t
j = (1 + O(m"3)) + o(l)
1�k<m \ / /
2/3
= (1 + O(m"3)) I --pu2/m 1 (F
e
To obtain the final result it remains to verify that which
implies that
1 it
,z
3.22. Denote the sum of the series by coN(z). It is clear that
dt
e
____
n+t+>2l n+t )
di.
n�N°
Therefore,
°°
= J —i--- dt + >2
N 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 ii
e
— 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 the
integrals over the intervals [k — 1/2, k + 1/2] (k E N, k � N); to estimate
this sum use the result in the preceding problem.
3.24. Using the identity
n+1/2
cosnx = 2sinx/2 lfl_1/2
cos(xt) dt,
226 VI. ASYMPTOTICS
we get that
>2 = 2sin(x/2)
(j+cocos(xt))
= j V
dv + 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 preceding
problem, we get that for x E (0, ic)
>2
cosnxf'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]) boundedness
below of the sums n° cos nx is equivalent to nonnegativity of the
integral 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 1
1,r/2 cosv 1
i I
Jo v V Jo v v
Using the decrease of the function v on (0. ir/2), we get that
cosv 1
çn/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 the
integral = which is positive for all >0.
3.26. Let A = and C = sup,>a w'(t). Then
f+00 A +oo
J
dt — 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, show
that the summation can be replaced by integration (see problem 3.14). To
investigate 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 not
tend monotonically to zero, the inequality of problem 3.14 is not applicable
to these series. One should consider separately the partial sum in which the
general term increases, and the remainder of the series, in which the general
term 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, and
3.28b) 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 —3
10
Consequently,
0< —x1000
4.8. It is dear that 0. The asymptotic behavior of the sequence
can 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 function
f in a neighborhood of zero by the functions considered in problem 4.6, and
then compare their iterates.
Another solution is based on the following which can be
used also in more complicated situations (see problems 4.12—4.14). The
recursive sequence = i)' x0 > 0, clearly converges to zero if the
function f is such that 0 < f(t) < t for all t > 0. With the help of a
figure it is easy to see that the sequence converges rapidly to zero if
the difference t — f(t) is large, and converges slowly otherwise. Therefore,
to find the asymptotics of the sequence it is expedient to consider the
function = 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 the
derivative 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) for
k=1
To simplify the computations it is sometimes more convenient to consider
instead 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, we
x—'+O
have that
fXi 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 theidea 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 the
Taylor formula that
— (1
1\ 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) a
function f such that the sequence = coincides with a previously
given sequence strictly decreasing to zero. It can be assumed here
that = where is a strictly decreasing function on [0, +oo) with
= 0. Then the equality = = is equivalent
to the equality f(ço(n — 1)) = which is satisfied, for example, by the
function f(t) = +
4.11. Consider the sequence and
4.12. 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) is
chosen so that
2 "
1
(1—p)/(1+p)
_i) forallnEN.
From these relations obtain successively that
2 2 — 11 —p 2\
—an 2
1 —p
An
(1 _p\P/(P_l) 2p/(p—1)
2 )
and, finally, — = + o( 1), which implies
that
4.14. 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, and
hence (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 combination
falls 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. Prove
this first for n = 2k• Verify that if the inequality is true for the arithmetic
mean of n arbitrary numbers, then it remains true also when n is replaced
by 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}, where
C = max(f(c), 1(d), K), and, using the result of problem 1.1.19, show that
E = [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 midpoint
of 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 largest
integers 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 (see
1.3a)) for the points x, x +5, and x + no, and the left-hand one can be
proved similarly.
1.5. a) b) is obvious. To prove the implication b) c) consider
the 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 of
half-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 chord
between the graph points with abscissae X1 and X2. Let = I — g. It
is clear that L1(X) = L1(X) > 0, and for f to be convex it suffices that
f1(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, and
passtothe limit as e—'O.
1.10. The necessity is obvious. To prove sufficiency rewrite the inequality
in the form f(f(X+ t) + f(X — t) — 2f(X))dt � 0 and use the preceding
problem.
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 that
the 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 points
X1,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, since
zero can always be added as the last point. For a � b the function x
f(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 common
vertex (0, 0) by joining this point to the node points of the graph (see Figure
13a). 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 intersect
the x-axis, we get a partition of the subgraph into triangles (Figure 13b). We
introduce 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 sufficiency
use the fact that a piecewise linear convex function is a sum, with positive
coefficients, of functions of the form ç9(x) = max(0, a — x) and a linear
function.
1.26. a), b) Represent the given integrals as the sums of the corresponding
integrals 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 that
f12 � 0. Verify that the sum of two suchfunctions 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) are
equivalent.
Since f is convex (see problem 1.3b)), the one-sided derivatives and
exist 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 satisfies
the 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", where
A = 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 > 0
suitably.
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 + jis
ji > 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 way
of 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 get
from 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 p
and 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 a
proof 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 bethesuccessivepointsof
reflection of the ray from the graph, and let B. be the points of reflection
from the x-axis. Then to each ray there corresponds a (finite or possibly
infinite) sequence (trajectory) SA1B1A2B2• or SB1A2B2A3. (the point
A1 is absent if the first reflection is from the x-axis). We prove the existence
of a number M such that for all rays sup1 x1 � M (this means that only a
bounded 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 points
S, A?, and lie on a single line, and let 4° = Then for any
other trajectory containing A1 the inequality x, holds (it is very easy
to see this).
3) Suppose that n � 2. > E (0, ir/2) is the angle between
the 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
— tan
tan tan tan
from which
1 — —
= sin sinç9,1
sin
—
sin
Consequently,
sinç91— >
—-——
> sin > sin
Note that � = I � Consequently, the number
can be taken as M.
1.34. a) Observe that f* is the pointwise supremum of the family of
linear 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 and
only 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 the
equality ç9'(t) = 0 is a sufficient condition for a minimum. Therefore, if
f'(t) = x exists at the point t, then the supremum in the definition of
f*(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)), which
contradicts the strict convexity of 1. Similarly, differentiability of I im-
plies strict convexity of
f and for any x E the equality f*(x) = xt — f(t) is attained only
for 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 which
f(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 many
extrema.
2.7. Let x, ö E (0, 1). Then
(ö—1) 2.u_
f(öx) = 1(x) + (ö — 1)xj (x) +
2
x 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 by
suitably choosing x we make the first term smalL
2.8. Verify that
çT
—h(t)=—j f(x)g(t—x)dx
J—T
= j f(x)(g(x+ t) - g(x - t))dx
= I I f'(x)g'(y)dxdy
JO Jx--t
£ x+t T x+t
= (11+1 J J )P Oxt Tt2Tx
where 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, and
the 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 to
2M0M2. The inequality becomes an equality for the piecewise smooth func-
tion 1(x) = ço(u)du + 2) dt, where = —2sgn u for ui � 1 and
ço(u) = 0 for ui> 1. Replacing by a continuous function as shown in
Figure 15a), we get a Ca-smooth function and for it
as e —, 0.
The inequality proved admits the following mechanical interpretation: if
the 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
I
I
I
a)
b)
—1
—2
1
I
I
I I I—-C - -+E
k k k
FIGURE 15
cannot be too large at anytime. More precisely, it does not exceed
All are familiar with an analogue of this fact from a basic physics course: in
a free fall (with zero initial velocity) the instantaneous velocity of a body is
equal to where S is the length of the path passed by the body and
g is the acceleration of gravity.
b) Replace [x — h, x + h] by [x, x + h] in the precedingargument. As a
result the inequality M1 � is obtained.
d) Use an analogous device, choosing h in (0, (b — a)/2] and integrating
over [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 by
the 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 that
c = 0. Consider the function g determined by the conditions g'(x) =
— and g(0) = 1(0) (for definiteness assume that this number
is 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). Draw
the tangent to the graph of g at the point 0, and let be the point where
it intersects the x-axis, = f(0)/g'(O). Then we get from a comparison of
areas 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 by
the 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 hence
a) 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
that
I
f'(t)(l — 1)dlI = al. Applying the Cauchy-Schwarz-Bunyakovskii
inequality 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 are
proportional, 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 n
there is a constituent interval of the nonempty open set
f is not a polynomial.
§2. SMOOTH FUNCTIONS 241
The assertion of the problem is easily derived from (A): assuming that I
is not a polynomial on R, we construct a sequence of nested open intervals
= (ar, c c with nonempty intersection, and
this contradicts the assumption.
Assume that (A) does not hold, i.e., that for some n and the function
f is not a polynomial on but its restriction to any constituent interval of
the set n is a polynomial. Let a(x) (x E be a maximal interval
with the properties that x E a(x) C A and is a polynomiaL Verify
that if the restrictions of f to two adjacent intervals are polynomials, then
I 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) = 0
for 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 the
series.
2.15. It can be assumed that a = 0. Then
1(x)
= j (lx) di
= j >
= >jxj' (ix) di,
and canbetakenas
2.16. 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 substitution
u = 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
I
ln(C+j g(x)dx.
\ Jo / 0
242 VII. FUNCTIONS (CONTINUATION)
Then
ft IftC+j f(x)g(x)dx�Cexp( I g(x)dx
Jo \Jo
which 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<m
1<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>0
0
1=
s
I<k<m
Passing to the limit in this inequality as S —, 0, we get the inequality
— tk)ZJZk � 0.
1<j�m
1<k<m
§2. SMOOTH FUNCTIONS 243
2.24. The assertions c) and d) follow from b), b) follows from a) and the
fact that Sh(x) = 0 (x —ö/y), a) can be verified directly, and e) follows
from 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 the
possible combinations of signs of f(x), f'(x), and f"(x) on both sides
of 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 Figure
16. Choosing x1, x2, x3, and x4 as shown in the figure (with x2 and x3
sufficiently 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 to
observe 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 of
points x1 <x2 <x3 <x4 is equivalent to the condition Sf <0.
§3. Bernstein polynomials
3.1. b) To compute the sums and Sn i
use the binomial formula
and the equality
O<k<n O�k<n
Compute the sums 5n%2' and with the help of the recursion
formula 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 that
a)
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 inequality
is proved similarly). Since the terms in the sum are nondecreasing, it
suffices 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 thefunctions 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<n
Ik/n--xI<o
fn\ k n--k
+
O<k<n
Ik/n--xI�ö
— 252n
Thus, Bn(f, x) — f(x)J <2€ if n > M/(2e52) and x E
c) It suffices to consider the case when 1(x) = 1 for x � x0 and 1(x) = 0
for x > x0. Then x0) = — To compute this
sum, break it up into two parts S1 and S2 by singling out the contribution
of 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 more
precise below). Estimating the sum S1 with the help of the inequality in
3.2a), we see that S1 = 0(1/no2). With the help of Stirling's formula and
Taylor'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)j
1
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 for
n 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 R
and 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 the
result of problem 3.9a).
3.10. Show that it is possible to weaken the conditions on g, requiring
instead of membership in C2([0, 1]) that its first two derivatives vanish
at the endpoints of [0, 1]. Indeed, for any such function g and for any
number A E (0, 1/2) the function equal to g((x — A)/(1 — 2A)) for
x E [A, 1 — A] and equal to zero otherwise satisfies the conditions of the
problem. 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 valid
forfunctions g ofthe form p>2 and n�0.
Passing to the limit as p —, 2, we get that f0' —x)2)" dx = 0 for
any 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 function
f(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 the
equality
= + 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 ——x
O<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 such
that 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 the
form
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 (problem
VII.1.1) that x) � = 1(x).
b) Assume the opposite: exist numbers a, b. and x such that
0 � a <x <b � 1 and
x—a b—x
1(x)> r—f(b) +
§3. BERNSTEIN POLYNOMIALS 249
It will be assumed that 1(a) = 1(b) = 0 (this can be ensured by subtracting a
linear 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 equal
to 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), we
arrive at the equality
Now fix an arbitrary number e > 0, and let 5 = > 0 be such that
x)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 sum
in 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 that
11(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<bn
I
O<j<n O<k<n
Changing the order of summation and using the result of problem 3.lb), we
get that
n2—1
O<k<n
It remains to prove that the sums and are bounded. To estimate
write it in the form
an(j<bn
I
where = t E [0, n]. Using the result of problem
VL3.15b), 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 max
J —
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+IJI
I
________
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 — differs
little 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 of
problem 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 result
of 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 see
that
=
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)). SinceIp(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 if
the polynomial P is positive on the whole interval [0, 1]. In the general
case we represent P in the form P(x) = — x)1', where > 0
on [0, 1]. Since the coefficients of the expansions of the polynomial P
in the basis {xk(l — x)
b
are 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 equal
to 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 either
Pn+m or falls in the same interval. The same can be said of one of
the points Pn—m and Pnrm Thus, the distance between adjacent solutions
of the equation a (mod 1) does not exceed L = rimi, which proves
the 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 of
segments (see Figure 17) obtained by intersecting the square [0, 1) x [0, 1)
with the family of parallel lines having slope and passing through all
possible 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 reader
follow 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. It
follows 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 of
and 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 the
coordinates and j of the point E R2 satisfy the condition <a
and <a. It follows form the irrationality of that is irrational,
and hence the winding F' = E R} of the torus is dense in
thesquare 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 that
the 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 also
a solution of (1) (as is one of the numbers n — r1m) (see Figure 18). This
implies that in each interval of length L = max1<1<4 r1ImI on the line there
is 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 the
points = 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 ak
and e the numbers must be rationally independent (incommen-
surable), that is, the relation + ... + = 0 (n1 E Z) must hold
only for n1 = •.• = n E Z, and suppose that
the 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 a
relatively 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 view
of the arbitrariness of e; as earlier, this implies the equalities n1 = =
= 0. The proof of sufficiency is obtained by generalizing the arguments
in 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 is
equal 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 equations
k 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) and
differentiating 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. But
this 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 t
such 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 an
almost period taken in the interval (—.v1. —x1 + L). The answers to the
remaining 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). They
are 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 and
F(x2) > M — e/6. Define d = Ix — x21 and let L = L(e/3d) be a number
chosen 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)dx
x2
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 z2
such 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, using
the diagonal process, choose a subsequence k1 = such that the sequence
{f(x + k1)} converges for all x Q. To prove the uniform convergence of
the sequence {fk } on R use the uniform continuityof f.
Sufficiency. Assume the contrary: there exist an 0 and a sequence
of open intervals with lengths tending to infinity that do not contain
e0-almost periods of f. Choose a sequence {hk} and a subsequence
of 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 L
contains an e-almost period of f, and let C = sup,ER If(t)I. Using an
almost period taken from the interval (a, a + L). a E R, prove the
inequality
1(x) dx — <e+
Deduce from this that the estimate 19,(nT) — < e + (CL/T) (n E
N) 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 for
x E (a + b + Consequently,
1
T
2 1 2T1 e b—a
2TJ.T
c)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 for
each n E N. On the other hand, for each countable collection ,
... and
for the arbitrary absolutely convergent series ak the series is
uniformly convergent on R, and its sum I is a UAP function (see problems
4.7 and 4.11). In view of uniform convergence,
(a
for
The reader can become more familiar with the details of these questions, for
example, 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 to
the sides of the strip are equal to the width of the strip, which implies that
one and only one of the endpoints of each of these segments belongs to
Joining 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 the
successive 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 this
case = 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 contains
a number 1 such that E (1 — 5, 1) (see problem 4.1). It is easy to see
that the differences — i = 1 1, are then the same, and hence
the "words" and coincide.
The analogous problem in space leads to the construction of an interesting
partition 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 of
adjacent points in S n Z3 orthogonally on P. we get segments bounding
certain parallelograms. These parallelograms cover F, and their interiors
are pairwise disjoint (see Figure 19). The covering obtained is remarkable
in 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 the
sense that any finite part of it occurs infinitely many times. The proofs of
these 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 — that
is, 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 form
that 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 contains
and hence B.
It is interesting to note that besides the case mentioned above, e is periodic
only 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 almost
periodicity.
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 aa
satisfying 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 = (for
such a representation to be unique we assume that among the there are
always infinitely many zeros) we associate r = 0.s1s2..., where is
the sequence of folds given by the equalities = n = 0, 1
then we get a function on [0, 1) that is continuous at irrational points and
right-continuous at dyadic rational points, and takes only irrational values.
4.25. Take numbers of the form m2', in E N, for (these numbers
satisfy 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 solution
of problem 4.24). Since the length of the word is equal to 2', at
least (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 and
contains Therefore, any word in r of length contains and
hence w.
We mention some curious properties of sequences s of folds (see problem
4.24) and the Rudin-Shapiro sequence r.
If S is defined by the equalities 52k = k mod 2, k = 0, 1,2 then
coincides 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 the
polygonal 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 with
the sequence r as follows: the numbers r1, r2, ... correspond to the hori-
zontal components of the polygonal curve, taken in order, where = 0 if
the nth horizontal component goes from left to right, and = 1 if from
right to left (the sequence of "up-down" passages of vertical components is
also reminiscent of the Rudin-Shapiro sequence).
With the polygonal curve corresponding to the sequence s it is possible
to associate a new polygonal curve, pictured in Figure 20 by the dotted line. It
III I I
FIGURE20
262 VII. FUNCTIONS (CONTINUATION)
is obtained from the solid line as follows: we replace each component by two
new components serving as the legs of an isosceles triangle with the original
component as hypotenuse. Further, the vertex of this triangle deviates from
the solid polygonal curve alternately to the right and to the left. Note that
the new polygonal curve was obtained from the old one by rotation and
contraction by a factor From the second polygonal curve it is possible
to construct a third one in an analogous way, then a fourth, and so on. In the
limit such a sequence of polygonal curves fills a sector in the plane bounded
by 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 similar
way from sequences of folds with a different alternation of 0's and l's in the
sequence (for example, 1), and also of the corresponding limit
sets on the plane. More details on the questions touched upon in problems
4.24—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 problem
1.1, 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 of
the 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 the
sets E, = {x—lIx E EnQ,}. Since = = 1,
the sets E, cannot be pairwise disjoint. Consequently, there are distinct
vectors 1',!" E Z such that E, nE,, 0. Let a E E, nE,. Then
a = 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 the
sets E1. Deduce from this that E contains distinct points x1,..., XN+l
such that Xk — X1 E Zm.
Consider an index s such that does not belong to the convex hull of
the remaining points. Verify that the points ±(xk — with k s are all
distinct, and hence are what is required.
1.5. Prove that the measure of the set of points in (0, 1) with 0 not
among 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 same
idea as in the solution of problem 1.1.
b) Consider the sets E1? c (0, 1) consisting of the points with nth decimal
digits nonzero, and verify that E1?) = (9/10)m (cf. the solution of
problem 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 (*) that
o < >).(E n Therefore, for at least one index n = n0 the
inequality 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)/2
and 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 places
with 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 by
intervals of length 1/n!.
1.14. It follows from the definition of the set E of Cantor type on [0. 1J
with defining sequence that ).(E) + — = 1 , where
= 1. Therefore, ).(E) > 0 if and only if — < 1. Since
the last sum can be arbitrarily small, the measure of a set of Cantor type
constructed 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 let
0 < 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 intervals
with 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 the
open intervals making up the set and let be compact subsets of
these 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 to
the lemma (with 0 = 02). Continuing this process, use induction to get a
family of sets � 1.
Let A = K0 U U1 Verify that
(*)
j=0
Show that any (nonempty) interval C contains some interval and
hence
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 with
each segment (j = 1, 2 n, = M1) of the boundary
polygonal curve consider the open rectangle lying outside A such that
the 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 to
see that the sum of the central angles corresponding to these sectors is equal
to 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 approximating
it by convex polygons.
1.17. Obviously, it suffices to consider the case when <00. Let
the 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) 0
k=Oj=l
1.18. Denote by a Lebesgue measure on S'. If measurability of at least
one of the sets is assumed, then in view of their congruence and the
invariance of the measure a with respect to rotations all the sets will be
measurable, 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 when
a(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 a
set having precisely one point in each equivalence class, and let C0 be the
image of C under translation by 0 modulo 1. We number all the rational
numbers in [0, 1) in the sequence {Oj and let = C9 . Since translation
modulo 1 preserves the measurability and the measure of a set, the sets
are measurable or not measurable simultaneously, and they have the same
measures if they are measurable. Assuming that the are measurableand
arguing as in the solution of problem 1.18, we again arrive at the impossible
inequality 0< A(E1) <00.
1.20. It will beassumed without loss of generality that E C (0, 1). Let
the be the sets constructed in the solution of problem 1.19, and let =
§2. MEASURABLE FUNCTIONS 267
E n E that
at 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 x
the 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 the
midpoint 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 let
Q 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 used
in 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 the
union 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 equation
f(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=J
0 l<1<N
+ 3j + (x)) dv
6j > dx
l�1<J<I%<N
= S1 + 6S2 + 6S3.
where S3 = fl fl To compute 53 note that since
E,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 function
3.6. 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
+ J
d:
� JA B(O,r)\A
= JB(Or)
b) Show that
I
e1X dxl = fE cos(x — dx for some E [0, 2irJ and
use the same idea as in a).
c) Show that
dxdy — ff xdxdy
JJEx+iy — liE' x2+y2
where E' is a set obtained from E by a rotation. To estimate the integral
lIE xdxdy/(x2 +y2) consider a set = {(x, y) E R21x/(x2+y2) � t}, on
which the function x/(x2 + y2) is large and choose t such that the measure
of is equal to that of E'. Prove as in a) that
xdxdy < f[ xdxdy
liE' 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 case
when it is finite use the inequality — 1)/pi � ti +e" (t E R, 0 <p < 1)
to prove that
L
fP(x)
— 1g(x) dx
p-'+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). Assuming
that y E (ak' fik) , we get that
f dx dx 1b dx
JKIx_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 with
vertices at the origin of coordinates. To compute the integral over each of
these pyramids use Fubini's theorem.
3.22. To avoid the difficulties associated with verifying that the function
is measurable on subsets of dimension less than n it will be assumed that
the 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 easy
to establish the induction base. Suppose that the formulas a) and b) are
valid 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--i
S÷ ={(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,..., V
12)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 proved
similarly 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, after
which 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
= j
and 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 E
3.25. 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 of
freedom, 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 2
3 . 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
= J
42d} 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 of
the function f, prove the desired formula by induction.
3.31. a) Since the measure is rotation-invariant, it can be assumed
that 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 sphere
and hence is equal to afl 2(1 — Consequently,
.7 = 2j t"(l —
= I — du
0n—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 desired
result.
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
x
2 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 the
sign of —0) is indicated inside it). Therefore,
2ir 1 2f=2afl3¼10
and since = (2ir/(n — we get the required result.
3.33. The assertion a) is a special case of problem 3.34. Using a) and the
formula 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 the
equality, 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
•
dço
J0 + 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 — = 0
for 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 equality
consider the sets Q1 =1+ Q and E, = —1+ E fl Q, (1 E Zm) and show that
UE,=Q, E,nE,=ø
IEZ'"
Since I is periodic, it is clear that fE 1(x) dx = fEnQ, 1(x) dx. It remains
to 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 the
projection of T on the hyperplane = 0. Consider the cone K with
vertex at and base Q. Since
= �
it suffices to prove that V � To do this, show that the intersection of
K with any straight line F through Q parallel to the is a segment
of length not exceeding the length of the segment obtained by intersecting /
with T. Constructing the plane passing through I and the consider
a point M E T of it that is farthest away from this axis and is located on
the same side of it as 1. Prove that segments of equal length are obtained
by intersecting F with K and the triangle
3.37. Verify that the convex hull of two ellipsoids obtained from each
other by a translation contains an ellipsoid of larger volume. Deduce from
this that the ellipsoid is centrally symmetric with respect to zero. It can be
assumed 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 that
x0 = (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 set
T1 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 the
ellipse 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 follows
that 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, for
a 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, for
example, [1], p. 169).
THEOREM. Suppose that p > 1, E C R", < oo, and {fj is
a sequence of measurable functions on E. If f —' f0 in measure on E
and fE dx <00, then the functions f'0, (n = 1, 2, ...) are
integrable 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 C
and holomorphic on an open dense set. See problem 4.8 (an example of
Urysohn) for a refinement of this result.
22
§3. INTEGRABLE FUNCTIONS 279
3.39. a) Let p, q> —1,and Jq =
C2). Then
q
L2+y22+v2
2 2 p 2 2 q= (x +y) (u +v ) djz3(x,y,u,v)
2q dxdydu
2JJj
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 k
f(z1, z2)z1 z2 dp3(z1, z2).
n�O Q<k<n
k)d1c2 Is'
We compute the integrals on the right-hand side of (1):
Is3
f(z1, z2)z1 z2 dji3(z1, z2)
j
Ik ._n—k= z2z1 - z2 dp3(z1, z2)
j.I�O
2k 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 in
several 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) dx
E 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)dt
0�k<n a
Step 2. The function h is bounded on [a, b]. Consider a sequence
of nonnegative continuous functions on [a, b] that converges to h
almost everywhere on [a, b]. It can be assumed without loss of generality
that � suph. Show that hfl(maxlxkl) —, h(maxlxkl) a.e. on E. Using
Lebesgue's theorem on taking the limit under the integral sign, we see that
the equalities
L
IxkI) dx = rn2mj di (n = 1, 2,...)
lead to the required result.
Step 3. The