Problem Books in Mathematics
Edited by P.R. Halmos
Springer New York
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Problem Books in Mathematics Series Editor: P.R. Halmos
Polynomials by Edward J. Barbeau
Problems in Geometry by Marcel Berger, Pierre Pansu, Jean-Pic Berry, and Xavier Saint-Raymond
Problem Book for First Year Calculus by George W. Bluman
Exercises in Probability by T. Cacoullos
An Introduction to Hilbert Space and Quantum Logic by David W. Cohen
Unsolved Problems in Geometry by Hallard T. Croft, Kenneth J. Falconer, and Richard K. Guy Problems in Analysis by Bernard R. Gelbaum
Problems in Real and Complex Analysis by Bernard R. Gelbaum
Theorems and Counterexamples in Mathematics by Bernard R. Gelbaum and John M.H. Olmsted Exercises in Integration by Claude George Algebraic Logic by S.G. Gindikin
Unsolved Problems in Number Theory (2nd ed.) by Richard K. Guy
An Outline of Set Theory by James M. Henle (continued after index)
Gabor J. Szekely Editor
Contests in Higher Mathematics Miklos Schweitzer Competitions 1962-1991
With 39 Illustrations
Springer
Gabor J. Szekely Department of Mathematics Eotvos Lorand and Technical University Mtizeum krt. 6-8 1088 Budapest, Hungary Series Editor: Paul R. Halmos Department of Mathematics Santa Clara University Santa Clara, CA 95053 USA Mathematics Subject Classification (1991): 15-06, 05-06, 26-06, 51-06
Library of Congress Cataloging-in-Publication Data Contests in higher mathematics:Miklos Schweitzer competitions, 1962-1991 / Gabor J. Szekely (ed.). p. cm. - (Problem books in mathematics) Includes bibliographical references and index. ISBN 0-387-94588-1
1. Mathematics - Competitions - Hungary. 2. Mathematics - Problems, I. Schweitzer, Miklbs, 1923-1945. II. Szekely, Gabor J., 1947- . III. Series. QA99.C62 1995 exercises, etc.
510' .76 - dc20
95-25361
Printed on acid-free paper. © 1996 Springer-Verlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone.
Production managed by Hal Henglein; manufacturing supervised by Jeffrey Taub. Camera-ready copy prepared from the editor's LaTeX files. Printed and bound by R.R. Donnelley & Sons, Harrisonburg, VA. Printed in the United States of America.
987654321 ISBN 0-387-94588-1 Springer-Verlag New York Berlin Heidelberg
Preface "I had the opportunity to speak with Leo Szilard about the contests of the Mathematical and Physical Society, and about the fact that the winners of these contests turned out later to be almost identical with the set of mathematicians and physicists who became outstanding ... "
(J. Neumann, in a letter to L. Fejer, Berlin, Dec. 7, 1929)
The solutions to deep scientific problems rarely come to us easily. Thus, it is important to motivate students to begin efforts on these kinds of problems. Scientific competition has proved to be an effective stimulant toward intellectual efforts. Successful examples include the "Concours" for admission to the "Grandes Ecoles" in France, and the "Mathematical Tripos" in
Cambridge, England. At the turn of the century, mathematical contests helped Hungary become one of the strongholds of the mathematical world. With the revolution in 1848 and the Compromise in 1867, Hungary broke free from many centuries of rule by the Turks and then the Hapsburgs, and became a nation on equal footing with her neighbor, Austria. By the end
of the 19th century, Hungary entered a period of cultural and economic progress. In 1891, Baron Lorind Eotvos, an outstanding Hungarian physicist, founded the Mathematical and Physical Society. In turn, the Society founded two journals: the Mathematical and Physical Journal in 1892 and the Mathematical Journal for Secondary Schools in 1893. This latter journal offered a rich variety of elementary problems for high school students. One of the first editors of the journal, Lasz16 Ritz, later became the teacher of John Neumann and Eugene Wigner (a Nobel prize winner in physics). In 1894, the Society introduced a mathematical competition for high school students. Among the winners there were Lipot Fejer, Alfred Haar, Todor
Kirmin, Marcel Riesz, Gibor Szego, Tibor Rado, Ede Teller, and many others who became world-famous scientists. The success of high school competitions led the Mathematical Society to found a college-level contest. The first contest of this kind was organized in 1949 and named after Mikl6s Schweitzer, a young mathematician who died
in the Second World War. Schweitzer placed second in the High School Contest in 1941, but the statutes of the fascist regime of that time prevented his admission to college. Schweitzer Contest problems are proposed and selected by the most prominent Hungarian mathematicians. Thus, V
PREFACE
vi
Schweitzer problems reflect the interest of these mathematicians and some aspects of the mainstream of Hungarian mathematics. The universities of Budapest, Debrecen, and Szeged have alternately been designated by the Society Presidium to conduct the Schweitzer Contests. The jury is chosen by the mathematics departments of the universities in question from among the mathematicians working in the host city. The jury sends out requests to leading Hungarian mathematicians to submit problems suitable for the contest. The list of problems selected by the jury is posted on the bulletin boards of mathematics departments and of local branches of the Mathematical Society (copies are available to anyone interested). Students may use any materials available in libraries or in their homes to solve the
contest problems. In ten days the solutions are due, with the student's name, faculty, course, year, and university or high school recorded on the solution set. The Schweitzer competition is one of the most unique in the world. Winners of the contests have gone on to become world-class scientists. Thus, the Schweitzer Contests are of interest to both math historians and mathematicians of all ages. They serve as reflections of Hungarian mathematical trends and as starting points for many interesting research problems in mathematics. The Schweitzer problems between 1949 and 1961 were previously published under the title Contests in Higher Mathematics, 1949-1961 (Akademiai Kiad6, Budapest, 1968; Chapter 4 of this book summarizes the mathematical work of M. Sch veitzer). Our book is a continuation of that volume.
We hope that this collection of Schweitzer problems will serve as a guide
for many young mathematicians and math majors. The large variety of research-level problems may spark the interest of seasoned mathematicians and historians of mathematics. I wish to close by acknowledging the outstanding work of Dr. Marianna Bolla as Managing Editor. InI addition, without the constant assistance of Dr. Dezso Miklds as Technical Editor, we could not have this book.
Bowling Green, OH August 26, 1995
Gdbor J. Szekely
Contents Preface Chapter 1. Problems of the Contests Chapter 2. Results of the Contests Chapter 3. Solutions to the Problems 3.1 3.2 3.3
3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11
Algebra (J6zsef Pelikan) Combinatorics (Ervin Gyori) Theory of Functions (Janos Bognar and Vilmos Totik) Geometry (Balazs Csik6s) Measure Theory (Janos Bognar) Number Theory (Imre Z. Ruzsa) Operators (Janos Bognar) Probability Theory (Gabriella Szep) Sequences and Series (Jeno Torocsik) Topology (Gabor Moussong) Set Theory (Peter Komjath)
Index of Names
v 1
49 55 55 121 150
244 330 362 388 404 465 517 552
565
Vii
1. Problems of the Contests The letter in parentheses after the text of a problem refers to the section
in Chapter 3 containing its solution. The topics include these areas of mathematics: A: Algebra C: Combinatorics F: Theory of Functions G: Geometry M: Measure Theory N: Number Theory
0: Operators P: Probability Theory S: Sequences and Series T: Topology R: Set Theory Thus, for example, P.3 refers to problem in "Probability Theory" section. When available, the names of proposers are in brackets at the very end of each problem.
1
1. PROBLEMS OF THE CONTESTS
2
1962 1. Let f and g be polynomials with rational coefficients, and let F and G
denote the sets of values of f and g at rational numbers. Prove that F = G holds if and only if f (x) = g(ax + b) for some suitable rational numbers a # 0 and b. (N.1) [E. Fried] 2. Determine the roots of unity in the field of p-adic numbers. (A.1) [L. Fuchs]
3. Let A and B be two Abelian groups, and define the sum of two homomorphisms q and x from A to B by a(17 + x) = a77 + ax
for all a E A.
With this addition, the set of homomorphisms from A to B forms an Abelian group H. Suppose now that A is a p-group (p a prime number). Prove that in this case H becomes a topological group under the topology defined by taking the subgroups pkH (k = 1, 2, ...) as a neighborhood base of 0. Prove that H is complete in this topology and that every connected component of H consists of a single element. When is H compact in this topology? (A.2) [L. Fuchs] 4. Show that
(x2 + y2) = (-1)[] 1<x
(mod p)
2
for every prime p =_ 3 (mod 4). ([.] is integer part.) (N.2) [J. Suranyi]
5. Let f be a finite real function of one variable. Let D f and D f be its upper and lower derivatives, respectively, that is,
f (x+h)- f ( x h+k h,k-0
Df(x)=limsup
Df(x)=liminf
h,k>O h+k>O
h+k>O
h,k-.0
f (x+h) - f (x-k) h+k
h,k>O
Show that V f and D f are Borel-measurable functions. (M.1) [A. Csaszar]
6. Let E be a bounded subset of the real line, and let S2 be a system of (nondegenerate) closed intervals such that for each x E E there exists an I E S2 with left endpoint x. Show that for every e > 0 there exist a finite number of pairwise nonoverlapping intervals belonging to S2 that cover
E with the exception of a subset of outer measure less than e. (M.2) [J. Czipszer]
7. Prove that the function
f0)
VI_(_X2
dx - 1)(1 -192x2)
(where the positive value of the square root is taken) is monotonically decreasing in the interval 0 < 19 < 1. (F.1) [P. Turin]
1. PROBLEMS OF THE CONTESTS
3
8. Denote by M(r, f) the maximum modulus on the circle I z I = r of the transcendent entire function f (z), and by M,,, (r, f) that of the nth partial sum of the power series of f (z). Prove the existence of an entire function fo(z) and a corresponding sequence of positive numbers rl < r2 < -> +oo such that lim sup M" (r., fo) _ +00. n-co M(rn, fo)
(F.2) [P. Turan] 9. Find the minimum possible sum of lengths of edges of a prism all of whose edges are tangent to a unit sphere. (G.1) [Miiller-Pfeiffer] 10. From a given triangle of unit area, we choose two points independently with uniform distribution. The straight line connecting these points divides the triangle, with probability one, into a triangle and a quadrilateral. Calculate the expected values of the areas of these two regions. (P.1) [A. Renyi] 1963 1. Show that the perimeter of an arbitrary planar section of a tetrahedron is less than the perimeter of one of the faces of the tetrahedron. (G.2) [Gy. Hajos] 2. Show that the center of gravity of a convex region in the plane halves
at least three chords of the region. (G.3) [Gy. Hajos] 3. Let R = Rl ®R2 be the direct sum of the rings Rl and R2, and let N2 be the annihilator ideal of R2 (in R2). Prove that R1 will be an ideal in every ring R containing R as an ideal if and only if the only homomorphism from R1 to N2 is the zero homomorphism. (A.3) [Gy. Pollak] 4. Call a polynomial positive reducible if it can be written as a product of two nonconstant polynomials with positive real coefficients. Let f (x)
be a polynomial with f (0) # 0 such that f (x') is positive reducible for some natural number n. Prove that f (x) itself is positive reducible. (A.4) [L. Redei]
5. Let H be a set of real numbers that does not consist of 0 alone and is closed under addition. Further, let f (x) be a real-valued function defined on H and satisfying the following conditions:
f(x) < f (y) if x < y and f (x + y) = f (x) + f (y)
(x, y E H).
Prove that f (x) = cx on H, where c is a nonnegative number. (F.3) [M. Hosszu, R. Borges]
6. Show that if f (x) is a real-valued, continuous function on the half-line
0<x
= f2(x)dx < oo,
then the function
g(x) = f (x) - 2e-x J X et f (t) dt 0
1. PROBLEMS OF THE CONTESTS
4
satisfies
j 9'(x)dx =
f2(x)dx. 0
(F.4) [B. Sz6kefalvi-Nagy] 7. Prove that for every convex function f (x) defined on the interval -1 < x < 1 and having absolute value at most 1, there is a linear function h(x) such that
If(x)-h(x)ldx<4-08.
J (F.5) [L. Fejes-T6th] 8. Let the Fourier series
2 + E(ak cos kx + bk sin kx) k>1
of a function f (x) be absolutely convergent, and let 2
2
2
2
ak + bk > ak+1 + bk+l
(k = 1,2,...).
Show that 1
12, (f(x + h) - f(x - h) )2 dx
(h > 0)
h
is uniformly bounded in h. (S.1) [K. Tandori] 9. Let f (t) be a continuous function on the interval 0 < t < 1, and define the two sets of points At = { (t, 0) : t E [0, 111,
Bt = { (f (t), 1): t E [0, 1] }.
Show that the union of all segments AtBt is Lebesgue-measurable, and find the minimum of its measure with respect to all functions f. (M.3) [A. Cs szar] 10. Select n points on a circle independently with uniform distribution. Let P,,, be the probability that the center of the circle is in the interior of the convex hull of these n points. Calculate the probabilities P3 and P4. (P.2) [A. Renyi] 1964
1. Among all possible representations of the positive integer n as n = Ek 1 ai with positive integers k, a1 < a2 < < ak, when will the product nk1 ai be maximum? (C.1) 2. Let p be a prime and !et
lk(2 x,y) = akx + bky
(k = 1,...,p
),
1. PROBLEMS OF THE CONTESTS
5
be homogeneous linear polynomials with integral coefficients. Suppose that for every pair rl) of integers, not both divisible by p, the values lk (1=, 77), 1 < k < p2, represent every residue class mod p exactly p times.
Prove that the set of pairs {(ak, bk) 1 < k < p2} is identical mod p with the set {(m, n) : 0 < m, n < p - 1}. (N.3) 3. Prove that the intersection of all maximal left ideals of a ring is a (twosided) ideal. (A.5) :
4. Let A1i A2, ... , An be the vertices of a closed convex n-gon K numbered
consecutively. Show that at least n - 3 vertices Ai have the property that the reflection of Ai with respect to the midpoint of Ai_1Ai+1 is contained in K. (Indices are meant mod n.) (G.4) 5. Is it true that on any surface homeomorphic to an open disc there exist two congruent curves homeomorphic to a circle? (G.5) 6. Let y1(x) be an arbitrary, continuous, positive function on [0, A], where A is an arbitrary positive number. Let
yn+l (x) = 2 J
y
yn(t) dt
(n = 1, 2,
... ).
0
Prove that the functions yn(x) converge to the function y = x2 uniformly on [0, A]. (S.2) 7. Find all linear homogeneous differential equations with continuous coefficients (on the whole real line) such that for any solution f (t) and any real number c, f (t + c) is also a solution. (F.6) 8. Let F be a closed set in the n-dimensional Euclidean space. Construct a function that is 0 on F, positive outside F, and whose partial derivatives
all exist. (F.7) 9. Let E be the set of all real functions on I = [0, 1]. Prove that one cannot define a topology on E in which fn -* f holds if and only if f,,, converges to f almost everywhere. (S.3) 10. Let E1, E2, ... , E2n be independent random variables such that P(Ei =
1)=P(e,=-1)=1/2foralli,and define Sk=X:k 1Ei, 1
0, or Sk = 0 and Sk_1 > 0. Compute the variance of N2n. (P.3) 1965
1. Let p be a prime, n a natural number, and S a set of cardinality pn. Let P be a family of partitions of S into nonempty parts of sizes divisible by p such that the intersection of any two parts that occur in any of the partitions has at most one element. How large can Pl be? (N.4) 2. Let R be a finite commutative ring. Prove that R has a multiplicative
identity element (1) if and only if the annihilator of R is 0 (that is, aR = 0, a c R imply a = 0). (A.6) 3. Let a, b0i b1, ... , bn_1 be complex numbers, A a complex square matrix of order p, and E the unit matrix of order p. Assuming that the eigenvalues
1. PROBLEMS OF THE CONTESTS
6
of A are given, determine the eigenvalues of the matrix b0E
b1A
b2A2
..
b,,,-,An-1
abn_iAn-1
b0E
b1A
...
bn_2An-2
abn iAn-1
b0E
...
ab2A2
ab3A3
B=
abn_2An-2
\ab1A
bn_3An-3
boE
f
(0.1) 4. The plane is divided into domains by n straight lines in general position, where n > 3. Determine the maximum and minimum possible number
of angular domains among them. (We say that n lines are in general position if no two are parallel and no three are concurrent.) (G.6) 5. Let A = A1A2A3A4 be a tetrahedron, and suppose that for each j : k, [Aj, Ask] is a segment of length p extending from Aj in the direction of Ak. Let pj be the intersection line of the planes [AjkAj1Ajn,.] and [AkAIAm]. Show that there are infinitely many straight lines that intersect the straight lines P1, P2, p3, p4 simultaneously. (G.7)
6. Consider the radii of normal curvature of a surface at one of its points PO in two conjugate directions (with respect to the Dupin indicatrix). Show that their sum does not depend on the choice of the conjugate directions. (We exclude the choice of asymptotic directions in the case of a hyperbolic point.) (G.8) 7. Prove that any uncountable subset of the Euclidean n-space contains an uncountable subset with the property that the distances between different pairs of points are different (that is, for any points P1 : P2 and Q1 # Q2 of this subset, P1-P2= Q1Q2 implies either P1 = Q1 and P2 = Q2, or Pl = Q2 and P2 = Q1). Show that a similar statement is not valid if the Euclidean n-space is replaced with a (separable) Hilbert space. (T.1)
8. Let the continuous functions fn (x), n = 1,2,3,..., be defined on the interval [a, b] such that every point of [a, b] is a root of fn (x) = fm(x) for some n # m. Prove that there exists a subinterval of [a, b] on which two of the functions are equal. (S.4) 9. Let f be a continuous, nonconstant, real function, and assume the existence of an F such that f (x + y) = F[f (x), f (y)] for all real x and y. Prove that f is strictly monotone. (F.8) 10. A gambler plays the following coin-tossing game. He can bet an arbitrary positive amount of money. Then a fair coin is tossed, and the gambler wins or loses the amount he bet depending on the outcome. Our gambler, who starts playing with x forints, where 0 < x < 2C, uses the following strategy: if at a given time his capital is y < C, he risks all of it; and if he has y > C, he only bets 2C - y. If he has exactly 2C forints, he
1. PROBLEMS OF THE CONTESTS
7
stops playing. Let f (x) be the probability that he reaches 2C (before going bankrupt). Determine the value of f (x). (P.4) 1966
1. Show that a segment of length h can go through or be tangent to at most 2[h/v/-2-] + 2 nonoverlapping unit spheres. ([.] is integer part.) (G.9) [L. Fejes-Toth, A. Heppes] 2. Characterize those configurations of n coplanar straight lines for which the sum of angles between all pairs of lines is maximum. (G.10) [L. FejesToth, A. Heppes] 3. Let f (n) denote the maximum possible number of right triangles determined by n coplanar points. Show that n lim f lim n3 0. noo n2 = oo and n-oo
f n=
(G.11) [P. Erdos] 4. Let I be an ideal of the ring of all polynomials with integer coefficients such that (a) the elements of I do not have a common divisor of degree greater than 0, and (b) I contains a polynomial with constant term 1. Prove that I contains the polynomial 1 + x + x2 + + x''-1 for some natural number r. (A.7) [Gy. Szekeres]
5. A "letter T" erected at point A of the x-axis in the xy-plane is the union of a segment AB in the upper half-plane perpendicular to the xaxis and a segment CD containing B in its interior and parallel to the x-axis. Show that it is impossible to erect a letter T at every point of the x-axis so that the union of those erected at rational points is disjoint from the union of those erected at irrational points. (M.4) [A. Csaszar] 6. A sentence of the following type is often heard in Hungarian weather reports: "Last night's minimum temperatures took all values between -3 degrees and +5 degrees." Show that it would suffice to say, "Both -3 degrees and +5 degrees occurred among last night's minimum temperatures." (Assume that temperature as a two-variable function of place and time is continuous.) (T.2) [A. Csaszar] 7. Does there exist a function f (x, y) of two real variables that takes natural numbers as its values and for which f (x, y) = f (y, z) implies x = y = z ? (N.1) [A. Hajnal] 8. Prove that in a Euclidean ring R the quotient and remainder are always uniquely determined if and only if R is a polynomial ring over some field and the value of the norm is a strictly monotone function of the degree of the polynomial. (To be precise, there are two more trivial cases: R can also be a field or the null ring.) (A.8) [E. Fried]
9. If
a.,,,.I < oo, then what can be said about the following expres-
sion? lim
1
+oo
2n + 1
m=-.
lam-n + am-n+l + ... + an,,+n l
1. PROBLEMS OF THE CONTESTS
8
(S.5) [P. Turan] 10. For a real number x in the interval (0, 1) with decimal representation
0.a1(x)a2(x)...a,,,(x).... denote by n(x) the smallest nonnegative integer such that afl(X)+lan(x)+2al(X)+3al(X)+4 = 1966.
Determine fo n(x)dx. (abcd denotes the decimal number with digits a, b, c, d.) (P.5) [A. Renyi]
1967
1. Let f(x) = ao+alx+a2x2+a10x10+allx11+a12x12+a13x13
(a13
0)
and g(x) = by + b1x + b2x2 + b3x3 + bllxll + b12x12 + b13x13
(b3
1
0)
be polynomials over the same field. Prove that the degree of their greatest common divisor is at most 6. (A.9) [L. Redei]
2. Let K be a subset of a group G that is not a union of left cosets of a proper subgroup. Prove that if G is a torsion group or if K is a finite set, then the subset
n k-1K kEK
consists of the identity alone. (A.10) [L. Redei] 3. Prove that if an infinite, noncommutative group G contains a proper normal subgroup with a commutative factor group, then G also contains an infinite proper normal subgroup. (A.11) [B. Csakany] 4. Let al, a 2 ,. .. , aN be positive real numbers whose sum equals 1. For a natural number i, let n2 denote the number of ak for which 21-2 > ak > 2-2 holds. Prove that 00
<4+
loge N.
(A.12) [L. Leindler] 5. Let f be a continuous function on the unit interval [0, 1]. Show that
li
1... 0
f fo
dx1 .
.
.
dx ,
f (2) \
and 1
limo
n-o
J0
1
f (" xl ... x.,)dxl ... dxn, = f (e )
.
1. PROBLEMS OF THE CONTESTS
9
(P.6) 6. Let A be a family of proper closed subspaces of the Hilbert space H = l2 totally ordered with respect to inclusion (that is, if L1, L2 E A, then either L1 C L2 or L2 C L1). Prove that there exists a vector x E H not contained in any of the subspaces L belonging to A. (T.3) [B. Sz6kefalvi-Nagy]
7. Let U be an n x n orthogonal matrix. Prove that for any n x n matrix A, the matrices Am
1
m
1
m
U ' AU'
j=0
converge entrywise as m -+ oo. (0.2) [I. Kovacs]
8. Suppose that a bounded subset S of the plane is a union of congruent, homothetic, closed triangles. Show that the boundary of S can be covered by a finite number of rectifiable arcs. (G.12) [L. Geher] 9. Let F be a surface of nonzero curvature that can be represented around one of its points P by a power series and is symmetric around the normal planes parallel to the principal directions at P. Show that the derivative with respect to the arc length of the curvature of an arbitrary normal section at P vanishes at P. Is it possible to replace the above symmetry condition by a weaker one? (G.13) [A. Moor] 10. Let a(SS,, k) denote the sum of the kth powers of the lengths of the sides of the convex n-gon SS, inscribed in a unit circle. Show that for any natural number greater than 2 there exists a real number ko between 1 and 2 such that a(Sn, ko) attains its maximum for the regular n-gon. (G.14) [L. Fejes-Toth] 1968 1. Consider the endomorphism ring of an Abelian torsion-free (resp. torsion) group G. Prove that this ring is Neumann-regular if and only if G is a discrete direct sum of groups isomorphic to the additive group of the rationals (resp., a discrete direct sum of cyclic groups of prime order). (A ring R is called Neumann-regular if for every a E R there
exists a,3 E R such that a/3a = a.) (A.13) [E. Fried] 2. Let al, a2 i ... , an be nonnegative real numbers. Prove that
H i=1
Z
a-1
i=1
a7
Z
i=1
(S.6) [J. Suranyi] 3. Let K be a compact topological group, and let F be a set of continuous functions defined on K that has cardinality greater than continuum. Prove that there exist xo E K and f j4 g E F such that .f (xo) = g(xo) = max f (x) = maxg(x). xEK
(T.4) [I. Juhasz]
xEK
1. PROBLEMS OF THE CONTESTS
10
4. Let f be a complex-valued, completely multiplicative, arithmetical func-
tion. Assume that there exists an infinite increasing sequence Nk of natural numbers such that f(n) = Ak # 0
Nk < n < Nk + 4
provided
Nk.
Prove that f is identically 1. (N.5) [I. Katai] 5. Let k be a positive integer, z a complex number, and e < 1/2 a positive number. Prove that the following inequality holds for infinitely many positive integers n:
Cn - kil
ze
>
(2-e)n.
o«< k+1
(F.9) [P. 111r6n]
6. Let 21 = (A; ...) be an arbitrary, countable algebraic structure (that is, 21 can have an arbitrary number of finitary operations and relations). Prove that 21 has as many as continuum automorphisms if and only if for any finite subset A' of A there is an automorphism 7rA' of 2t different
from the identity automorphism and such that (x)'rA' = x
for every x E A'. (A.14) [M. Makkai]
7. For every natural number r, the set of r-tuples of natural numbers is partitioned into finitely many classes. Show that if f (r) is a function such that f (r) > 1 and limr . f (r) = +oo, then there exists an infinite set of natural numbers that, for all r, contains r-tuples from at most f (r)
classes. Show that if f (r) 74 +oc, then there is a family of partitions such that no such infinite set exists. (C.2) [P. Erd6s, A. Hajnal] 8. Let n and k be given natural numbers, and let A be a set such that IAI < n(n + 1)
k+1 For i = 1, 2, ... , n + 1, let Ai be sets of size n such that
IAinA;I
(i A
n+1
A= UAi. i=1
Determine the cardinality of A. (C.3) [K. Corradi] 9. Let f (x) be a real function such that lim
fx =1
x-.+00 ex
1. PROBLEMS OF THE CONTESTS
11
and If"(x)l < cjf' (x) for all sufficiently large x. Prove that lim f' ( x = 1. X-+00 ex
(F.10) [P. Erdos] 10. Let h be a triangle of perimeter 1, and let H be a triangle of perimeter A homothetic to h. Let hl, h2, ... be translates of h such that, for all i, hi is different from hi+2 and touches H and hi+1 (that is, intersects without overlapping). For which values of A can these triangles be chosen so that the sequence h1, h2i ... is periodic? If A > 1 is such a value, then determine the number of different triangles in a periodic chain hl, h2i .. .
and also the number of times such a chain goes around the triangle H. (G.15) [L. Fejes-T6th] 11. Let A1, ... , A.,, be arbitrary events in a probability field. Denote by Ck the event that at least k of A1, . . . , A,,, occur. Prove that
[JP(Ck) <_ k=1
fP(Ak)
k=1
(P.7) [A. Renyi]
1969 1. Let G be an infinite group generated by nilpotent normal subgroups. Prove that every maximal Abelian normal subgroup of G is infinite. (We call an Abelian normal subgroup maximal if it is not contained in another Abelian normal subgroup.) (A.15) [J. Erdos]
2. Let p > 7 be a prime number, C a primitive pth root of unity, c rational number. Prove that in the additive group generated by the numbers 1, C, (2, r3 + C-3 there are only finitely many elements whose norm is equal to c. (The norm is in the pth cyclotomic field.) (A.16) [K. Gyory]
3. Let f (x) > 0 be a nonzero, bounded, real function on an Abelian group G, 91, ... , gk are given elements of G and A1,. .. , Ak are real numbers. Prove that if k
Ai.f (gix) > 0 i=1
holds for all x E G, then k
EAi>0. i=1
(S.7) [A. Mate] 4. Show that the following inequality holds for all k > 1, real numbers a1i a2, ... , ak, and positive numbers x1, x2, ... , xk. k
k
Exi In
i=1
k
1-a; xi
aixiln xi
< i=1 k
?xi 1
1. PROBLEMS OF THE CONTESTS
12
(S.8) [L. Losonczi] 5. Find all continuous real functions f, g and h defined on the set of positive real numbers and satisfying the relation f(x + y) + g(xy) = h(x) + h(y)
for all x > 0 and y > 0. (F.11) [Z. Daroczy] 6. Let x0 be a fixed real number, and let f be a regular complex function in the half-plane Re z > x0 for which there exists a nonnegative function F E Li (-oo, oo) satisfying If (a + i0)I < F(/3) whenever a > xo, -oo < ,Q < +oo. Prove that
I -ioo
f(z)dz = 0.
(F.12) [L. Czach] 7. Prove that if a sequence of Mikusinski operators of the form pe-a9 (A and µ nonnegative real numbers, s the differentiation operator) is convergent
in the sense of Mikusinski, then its limit is also of this form. (0.3) [E. Gesztelyi]
8. Let f and g be continuous positive functions defined on the interval [0, oo), and let E C [0, oo) be a set of positive measure. Prove that the range of the function defined on E x E by the relation F(x, y) = foo f (t)dt + J
y
g(t)dt
0
has a nonvoid interior. (M.5) [L. Losonczil 9. In n-dimensional Euclidean space, the union of any set of closed balls (of positive radii) is measurable in the sense of Lebesgue. (M.6) [A. Csaszar]
10. In n-dimensional Euclidean space, the square of the two-dimensional Lebesgue measure of a bounded, closed, (two-dimensional) planar set is equal to the sum of the squares of the measures of the orthogonal projections of the given set on the n-coordinate hyperplanes. (M.7) [L. Tamassy]
11. Let Al, A2.... be a sequence of infinite sets such that IAi n A; 2 for i # j. Show that the sequence of indices can be divided into two disjoint sequences it < i2 < ... and j1 < j2 < ... in such a way that, for some sets E and F, I Ain n El = 1 and lA;,, n F1 = 1 for n = 1, 2,.... (C.4) [P. ErdSs]
12. Let A and B be nonsingular matrices of order p, and let and g be independent random vectors of dimension p. Show that if , g and A + r1B have the same distribution, if their first and second moments exist, and if their covariance matrix is the identity matrix, then these random vectors are normally distributed. (P.8) [B. Gyires]
1. PROBLEMS OF THE CONTESTS
13
1970 1. We have 2n + 1 elements in the commutative ring R: 011 011...,ani01i...iOn-
Let us define the elements n
ai.i
Qk=ka+ i=1
Prove that the ideal (oo,01i...,vk,...) can be finitely generated.
(A.17) [L. Redei]
2. Let G and H be countable Abelian p-groups (p an arbitrary prime). Suppose that for every positive integer n, PnG
Pn+1G .
Prove that H is a homomorphic image of G. (A.18) [M. Makkai] 3. The traffic rules in a regular triangle allow one to move only along segments parallel to one of the altitudes of the triangle. We define the distance between two points of the triangle to be the length of the shortest
such path between them. Put (nz 1) points into the triangle in such a way that the minimum distance between pairs of points is maximal. (G.16) [L. Fejes-loth] 4. If c is a positive integer and p is an odd prime, what is the smallest residue (in absolute value) of P-1
. (2:)cn
(mod p)?
n=0
(N.6) [J. Suranyi] 5. Prove that two points in a compact metric space can be joined with a rectifiable arc if and only if there exists a positive number K such that, for any s > 0, these points can be connected with an e-chain not longer than K. (T.5) [M. Bognar] 6. Let a neighborhood basis of a point x of the real line consist of all Lebesgue-measurable sets containing x whose density at x equals 1. Show that this requirement defines a topology that is regular but not normal. (T.6) [A. Csaszar] 7. Let us use the word N-measure for nonnegative, finitely additive set functions defined on all subsets of the positive integers, equal to 0 on finite sets, and equal to 1 on the whole set. We say that the system 2t of sets determines the N-measure µ if any N-measure coinciding with p on all elements of 2t is necessarily identical with µ. Prove the existence of an N-measure µ that cannot be determined by a system of cardinality less than continuum. (M.8) [I. Juhasz]
1. PROBLEMS OF THE CONTESTS
14
-
8. Let 7rn (x) be a polynomial of degree not exceeding n with real coefficients
such that
-x2
for
- 1 < x < 1.
Then 2(n - 1).
(F.13) [P. Turan] 9. Construct a continuous function f (x), periodic with period 27r, such that the Fourier series of f (x) is divergent at x = 0, but the Fourier series of f2 (x) is uniformly convergent on [0, 27r]. (S.9) [P. Turan] 10. Prove that for every 19, 0 < 19 < 1, there exist a sequence an of positive
integers and a series E' 1 an such that (1) An+1 - An > ()n)9, (ii)
, lim 1-n >2 anr" exists, n=1 00
(iii) >2 an is divergent. n=1
(S.10) [P. Turan] 11. Let 1, C2, ... be independent random variables such that m>0 and Var(Cn) = Q2 < 00 (n = 1, 2, ... ). Let {an} be a sequence of positive numbers such that an --+ 0 and F-° 1 an = oo. Prove that Jim
P
n- oo
n
>2 ak4 = oo = 1. k=1
(P.9) [P. Revesz] 12. Let 191 i ... , i9n be independent, uniformly distributed, random variables in the unit interval [0, 1]. Define
h(x) = n #{k: 19k < x}. Prove that the probability that there is an xo E (0, 1) such that h(xo) _ x0, is equal to 1 - n. (P.10) [G. Tusnady] 1971 1. Let G be an infinite compact topological group with a Hausdorff topol-
ogy. Prove that G contains an element g # 1 such that the set of all powers of g is either everywhere dense in G or nowhere dense in G. (A.19) [J. Erdos] 2. Prove that there exists an ordered set in which every uncountable subset contains an uncountable, well-ordered subset and that cannot be represented as a union of a countable family of well-ordered subsets. (N.2) [A. Hajnal]
1. PROBLEMS OF THE CONTESTS
15
3. Let 0 < ak < 1 for k = 1, 2, .... Give a necessary and sufficient condition for the existence, for every 0 < x < 1, of a permutation iry of the positive integers such that 00
ary (k) 2k
k=1
(S.11) [P. Erdos] 4. Suppose that V is a locally compact topological space that admits no countable covering with compact sets. Let C denote the set of all compact subsets of the space V and U the set of open subsets that are not contained in any compact set. Let f be a function from U to C such that f (U) C_ U for all U EU. Prove that either (i) there exists a nonempty compact set C such that f (U) is not a proper subset of C whenever C C U EU, (ii) or for some compact set C, the set
f-1(C) = U{U E U: f(U) C C} is an element of U, that is, f
(C) is not contained in any compact
set.
(T.7) [A. Mate] 5. Let Al < A2 < ... be a positive sequence and let K be a constant such that n-1
k=1
Prove that there exists a constant K' such that n-1
EAk
(S.12) [L. Leindler] 6. Let a(x) and r(x) be positive continuous functions defined on the interval [0, oo), and let
liminf(x - r(x)) > 0. X-00
Assume that y(x) is a continuous function on the whole real line, that it is differentiable on [0, oo), and that it satisfies y'(x) = a(x)y(x - r(x))
on [0, oo). Prove that the limit lim y(x)exp -
X-00
f
exists and is finite. (F.14) [I. Gydri]
a(u)du fx
1. PROBLEMS OF THE CONTESTS
16
7. Let n > 2 be an integer, let S be a set of n elements, and let Ai, 1 < i < m, be distinct subsets of S of size at least 2 such that
AinA,#O,AinAk
O,A,nAk#0 imply AinA;nAk#0.
Show that m < 2n-1 - 1. (C.5) [P. ErdBs] 8. Show that the edges of a strongly connected bipolar graph can be oriented in such a way that for any edge e there is a simple directed path from pole p to pole q containing e. (A strongly connected bipolar graph is a finite connected graph with two special vertices p and q having the property that there are no points x, y, x # y, such that all paths from x to p as well as all paths from x to q contain y.) (C.6) [A. Adam] 9. Given a positive, monotone function F(x) on (0, oo) such that F(x)/x is monotone nondecreasing and F(x)/xl+d is monotone nonincreasing for some positive d, let An > 0 and an > 0, n > 1. Prove that if n
00
E AnF an n=1
E AAkn
)
< 00,
k=1 An
or 00o
n=1
AF
(Eak) <00, k=1
then EO° 1 an is convergent. (S.13) [L. Leindler] 10. Let {¢n(x)} be a sequence of functions belonging to L2(0,1) and having norm less than 1 such that for any subsequence {(pnk(x)} the measure of the set N 1 {x E (0, 1) : L `ink (x) I y} I
E
tends to 0 as y and N tend to infinity. Prove that On tends to 0 weakly in the function space L2(0,1). (M.9) [F. Moricz] 11. Let C be a simple arc with monotone curvature such that C is congruent to its evolute. Show that under appropriate differentiability conditions,
C is a part of a cycloid or a logarithmic spiral with polar equation r = ae'. (G.17) [J. Szenthe] 1972 1. Let .1 be a nonempty family of sets with the following properties:
(a) If X E F, then there are some Y E F and Z E F such that Y n Z= 0 and Y U Z = X. (b) If X E F, and Y U Z= X, Y n Z= 0, then either Y E T or Z E Y. Show that there is a decreasing sequence Xo D X1 D X2 ... of sets Xn E .F such that 00 IIXn=O
n=0
1. PROBLEMS OF THE CONTESTS
17
(C.7) [F. Calvin]
2. Let < be a reflexive, antisymmetric relation on a finite set A. Show that this relation can be extended to an appropriate finite superset B of A such that < on B remains reflexive, antisymmetric, and any two elements of B have a least upper bound as well as a greatest lower bound. (The relation < is extended to B if for x, y E A, x < y holds in A if and
only if it holds in B.) (tt.3) [E. Fried] 3. Let ).i (i = 1, 2, ...) be a sequence of distinct positive numbers tending to infinity. Consider the set of all numbers representable in the form 00
niAi, i=1
where ni > 0 are integers and all but finitely many ni are 0. Let
L(x) _ E 1
and
M(x) _ E 1.
a; <x
µ <x
(In the latter sum, each p occurs as many times as its number of representations in the above form.) Prove that if L(x + 1) lim x-.oo L(x) then
lira x-.oo
M(x + 1) M(x)
--
1.
(F.15) [G. Halasz] 4. Let G be a solvable torsion group in which every Abelian subgroup is finitely generated. Prove that G is finite. (A.20) [J. Pelikan] 5. We say that the real-valued function f (x) defined on the interval (0, 1) is approximately continuous on (0, 1) if for any x0 E (0, 1) and e > 0 the point x0 is a point of interior density 1 of the set
H = {x : If (x) - f (xo) I < e}. Let F C (0, 1) be a countable closed set, and g(x) a real-valued function defined on F. Prove the existence of an approximately continuous function f (x) defined on (0, 1) such that f (x) = g(x)
for all x E F.
(M.10) [M. Laczkovich, Gy. Petruska] 6. Let P(z) be a polynomial of degree n with complex coefficients,
P(0) = 1,
and
IP(z)I < M for IzI < 1.
1. PROBLEMS OF THE CONTESTS
18
Prove that every root of P(z) in the closed unit disc has multiplicity at most cv,n-, where c = c(M) > 0 is a constant depending only on M. (F.16) [G. Halasz] 7. Let f (x, y, z) be a nonnegative harmonic function in the unit ball of R3 for which the inequality f (xo, 0, 0) < E2 holds for some 0 < xo < 1 and
0 < E < (1 - x0)2. Prove that f(x,y,z) < E in the ball with center at the origin and radius (1 - 3E1/4). (F.17) [P. Turin] 8. Given four points A1, A2, A3, A4 in the plane in such a way that A4 is the centroid of the LA1A2A3i find a point A5 in the plane that maximizes
the ratio
minl
(T(ABC) denotes the area of the triangle AABC.) (G.18) [J. Surinyi] 9. Let K be a compact convex body in the n-dimensional Euclidean space. Let P1, P2, ... , Pn+l be the vertices of a simplex having maximal volume
among all simplices inscribed in K. Define the points Pn+2, P+3,... successively so that Pk (k > n + 1) is a point of K for which the volume of the convex hull of P1, . . . , Pk is maximal. Denote this volume by Vk. Decide, for different values of n, about the truth of the statement "the sequence Vn+l, Vn+2,... is concave." (G.19) [L. Fejes-Toth, E. Makai] 10. Let T1 and T2 be second-countable topologies on the set E. We would like to find a real function a defined on E x E such that
0< a(x,y) < +oo, a(x, z) < a(x, y) + a(y, z)
a(x,x) = 0, (x, y, z E E),
and, for any p E E, the sets Vi(p,E) _ {x : a(x,p) < E}
(6>0)
form a neighborhood base of p with respect to T1, and the sets V2(p, E) _ {x : a(p, x) < E}
(s>0)
form a neighborhood base of p with respect to T2. Prove that such a function a exists if and only if, for any p E E and Ti-open set G 3 p (i = 1, 2), there exist a Ti-open set G' and a T3_i-closed set F with p E G' C F C G. (T.8) [A. Csaszir] 11. We throw N balls into n urns, one by one, independently and uniformly. Let Xi = Xi(N, n) be the total number of balls in the ith urn. Consider the random variable
y(N, n) = min 1Xi - N I. 1
Verify the following three statements:
n
1. PROBLEMS OF THE CONTESTS
19
(a) If n -> oo and N/n3 -+ oo, then
P
(y(N,
- ' 1 - e-'
n) < nyVn
21"
for all x > 0.
(b) If n -* oo and N/n3 < K (K constant), then for any e > 0 there is an A > 0 such that
P(y(N, n) < A) > 1 -
(c) If n->ooand N/n3-4Othen P(y(N, n) < 1) -> 1. (P.11) [P. Revesz] 1973
1. We say that the rank of a group G is at most r if every subgroup of G can be generated by at most r elements. Prove that there exists an integer s such that for every finite group G of rank 2 the commutator series of G has length less than s. (A.21) [J. Erdos] 2. Let R be an Artinian ring with unity. Suppose that every idempotent element of R commutes with every element of R whose square is 0. Suppose R is the sum of the ideals A and B. Prove that AB = BA. (A.22) [A. Kertesz]
3. Find a constant c > 1 with the property that, for arbitrary positive integers n and k such that n > ck, the number of distinct prime factors of (k) is at least k. (N.7) [P. Erdos] 4. Let f (n) be the largest integer k such that nk divides n!, and let F(n) _ max2<,n
1.
(N.8) [P. Erdos] 5. Verify that for every x > 0, F'(x + 1)
F(x + 1)
> log x.
(F.18) [P. Medgyessy] 6. If f is a nonnegative, continuous, concave function on the closed interval [0, 1] such that f (0) = 1, then
f
0
[f 1 1xf(x)dx <
2 3
0
20
1. PROBLEMS OF THE CONTESTS (F.19) [Z. Daroczy]
7. Let us connect consecutive vertices of a regular heptagon inscribed in a unit circle by connected subsets (of the plane of the circle) of diameter less than 1. Show that every continuum (in the plane of the circle) of diameter greater than 4, containing the center of the circle, intersects one of these connected sets. (G.20) [M. Bognar]
8. What is the radius of the largest disc that can be covered by a finite number of closed discs of radius 1 in such a way that each disc intersects
at most three others? (G.21) [L. Fejes-Toth] 9. Determine the value of
sup [log E - E log ],
1
where is a random variable and E denotes expectation. [Z. Daroczy]
(P.12)
10. Find the limit distribution of the sequence Tin of random variables with distribution
P (un = arccos(cos2
(2j
1)7r
2n
)J = 1n
(j = 1, , ... , n).
(arccos(.) denotes the main value.) (P.13) [B. Gyires] 1974 1. Let .F be a family of subsets of a ground set X such that UFE.FF = X, and
(a) ifA,BE.F,then AUBC_Cforsome CE.F; (b) if An E .F (n = 0,1, ... ) , B E Y, and Ao C Al C ... , then, for some k>_O, An nB=AknB for all n>_ k. Show that there exist pairwise disjoint sets X.y (y E 1'), with X = U{X.y : -y E r}, such that every X.y is contained in some member of .F, and every element of .F is contained in the union of finitely many X.y's. (R.4) [A. Hajnal] 2. Let G be a 2-connected nonbipartite graph on 2n vertices. Show that the vertex set of G can be split into two classes of n elements each such that the edges joining the two classes form a connected, spanning subgraph. (C.8) [L. Lovasz]
3. Prove that a necessary and sufficient condition for the existence of a set S C {1, . . . , n} with the property that the integers 0, 1, ... , n - 1 all have an odd number of representations in the form x - y, x, y E S, is that (2n - 1) has a multiple of the form 2 4k - 1. (N.9) [L. Lovasz, J. Pelikan] 4. Let R be an infinite ring such that every subring of R different from {0} has a finite index in R. (By the index of a subring, we mean the index of its additive group in the additive group of R.) Prove that the additive group of R is cyclic. (A.23) [L. Lovasz, J. Pelikan]
1. PROBLEMS OF THE CONTESTS
21
5. Let {fn}°°_o be a uniformly bounded sequence of real-valued measurable functions defined on [0, 1] satisfying
f
1
f"2`
0
Further, let {cn} be a sequence of real numbers with
cn = +oo. n=a
Prove that some re-arrangement of the series E°° 0 cn fn is divergent on a set of positive measure. (M.11) [J. Komlos] 6. Let f (x) = En 1 an/(x + n2), (x > 0), where En 1 < oc for IanIn-a
some a > 2. Let us assume that for some 3 > 1/a, we have f (x) _ O(e-xo) as x --+ oc. Prove that an is identically 0. (S.14) [G. Halasz] 7. Given a positive integer m and 0 < 6 < it, construct a trigonometric polynomial f (x) = a0 + 1(an cos nx + bn sin nx) of degree m such that f(0) = 1, fb<1x1<,, If(x)I dx < c/m, and max_,<x<,r I f'(x)I < c/6, for some universal constant c. (S. 15) [G. Halasz]
8. Prove that there exists a topological space T containing the real line as a subset, such that the Lebesgue-measurable functions, and only those, extend continuously over T. Show that the real line cannot be an everywhere-dense subset of such a space T. (T.9) [A. Csasz6x]
9. Let A be a closed and bounded set in the plane, and let C denote the set of points at a unit distance from A. Let p E C, and assume that the intersection of A with the unit circle K centered at p can be covered by an arc shorter than a semicircle of K. Prove that the intersection of C with a suitable neighborhood of p is a simple arc of which p is not an endpoint. (T.10) [M. Bognar] 10. Let It and v be two probability measures on the Borel sets of the plane. Prove that there are random variables S1, 2, r11,172 such that (a) the distribution of (t1i 2) is p and the distribution of (771,712) is v, (b) 1 < 771, 2 < 772 almost everywhere, if and only if p(G) > v(G) for all sets of the form G = U,=1 (-oo, xi) x (-oo, yz). (P.14) [P. Major] 1975 1. Show that there exists a tournament (T, ->) of cardinality R1 containing no transitive subtournament of size R1. (A structure (T, --*) is a tournament if -+ is a binary, irreflexive, asymmetric, and trichotomic relation. The tournament (T, -4) is transitive if -+ is transitive, that is, if it orders T.) (R.5) [A. Hajnal] 2. Let An denote the set of all mappings f : {1, 2, ... , n} -a {1, 2, ... , n}
such that f -1(i) {1,2,...,i}. Prove
{k
:
f (k) = i} 54 0 implies f -1(j) I'An I =
kn E 2k+1 00
k=0
0, j E
1. PROBLEMS OF THE CONTESTS
22
(C.9) [L. Lovasz] 3. Let S be a semigroup without proper two-sided ideals, and suppose that for every a, b E S at least one of the products ab and ba is equal to one of the elements a, b. Prove that either ab = a for all a, b E S or ab = b for all a, b E S. (A.24) [L. Megyesi] 4. Prove that the set of rational-valued, multiplicative arithmetical functions and the set of complex rational-valued, multiplicative arithmetical functions form isomorphic groups with the convolution operation f o g defined by
.f(d)g(d)
(f og)(n) _ din
(We call a complex number complex rational, if its real and imaginary parts are both rational.) (N.10) [B. Csakany] 5. Let I fn} be a sequence of Lebesgue-integrable functions on [0, 1] such that for any Lebesgue-measurable subset E of [0, 1] the sequence fE fn is convergent. Assume also that limn fn = f exists almost everywhere. Prove that f is integrable and fE f = limn fE fn. Is the assertion also true if E runs only over intervals but we also assume fn > 0? What happens if [0, 11 is replaced by [0, oo) ? (M.12) [J. Sziics]
6. Let f be a differentiable real function, and let M be a positive real number. Prove that if
If(x + t) - 2f(x) + f(x - t)l <Mt2 for all x and t, then
If'(x+t)- f'(x)I <MItI. (F.20) [J. Szabados]
7. Let a < a' < b < b' be real numbers, and let the real function f be continuous on the interval [a, b'] and differentiable in its interior. Prove that there exist c E (a, b), c' E (a', b') such that
f (b) - f (a) = f ' (c) (b - a),
f(b') - f(a') = f'(cl) (b' - a'), and c < c'. (F.21) [B. Sz6kefalvi-Nagy]
8. Prove that if
m Ean
< Nam
(m=1,2.... )
n=1
holds for a sequence {an} of nonnegative real numbers with some positive integer N, then ai+n > pai for i, p = 1, 2, ... , where iN
ai =n=(i-1)N+1 E an (S.16) [L. Leindler]
(i = 1, 2, ... ).
1. PROBLEMS OF THE CONTESTS
23
9. Let 10, c, a, g be positive constants, and let x(t) be the solution of the differential equation
([lo+ct,]2x')'+g[l0+cta]sinx=0,
t>0, -2 <x< 2
satisfying the initial conditions x(to) = x0i x'(to) = 0. (This is the equation of the mathematical pendulum whose length changes according to the law 1 = to+ct' .) Prove that x(t) is defined on the interval [to, 00); furthermore, if a > 2 then for every x0 # 0 there exists a to such that liminf I x(t)I > 0. t-100
(F.22) [L. Hatvani] 10. Prove that an idempotent linear operator of a Hilbert space is self-adjoint
if and only if it has norm 0 or 1. (0.4) [J. Szucs] 11. Let X1, X2,..., X,,, be (not necessary independent) discrete random vari-
ables. Prove that there exist at least n2/2 pairs (i, j) such that
H(Xi+Xj) >
1 min {H(Xk)}, 3 1
where H(X) denotes the Shannon entropy of X. (P.15) [Gy. Katona] 12. Assume that a face of a convex polyhedron P has a common edge with every other face. Show that there exists a simple closed polygon that consists of edges of P and passes through all vertices. (G.22) [L. Lovasz]
1976
1. Assume that R, a recursive, binary relation on N (the set of natural numbers), orders N into type w. Show that if f (n) is the nth element of this order, then f is not necessarily recursive. ('2.6) [L. Posa] 2. Let G be an infinite graph such that for any countably infinite vertex set A there is a vertex p joined to infinitely many elements of A. Show that G has a countably infinite vertex set A such that G contains uncountably infinitely many vertices p joined to infinitely many elements of A. (C.10) [P. Erdds, A. Hajnal]
3. Let H denote the set of those natural numbers for which r(n) divides n, where r(n) is the number of divisors of n. Show that (a) n! E H for all sufficiently large n, (b) H has density 0. (N.11) [P. Erdos]
4. Let Z be the ring of rational integers. Construct an integral domain I satisfying the following conditions:
(a) Z # I; (b) no element of I \ Z is algebraic over Z (that is, not a root of a polynomial with coefficients in Z); (c) I only has trivial endomorphisms. (A.25) [E. Died]
1. PROBLEMS OF THE CONTESTS
24
5. Let S = E 1 bjz (v = 0, ±1, ±2.... ), where the bj are arbitrary and the za are nonzero complex numbers. Prove that I So I < n max
o
I S,,
(S.17) [G. Halasz]
6. Let 0 < c < 1, and let 77 denote the order type of the set of rational numbers. Assume that with every rational number r we associate a Lebesgue-measurable subset Hr of measure c of the interval [0, 1]. Prove the existence of a Lebesgue-measurable set H C [0, 1] of measure c such
that for every x c H the set
{r:xEHr} contains a subset of type q. (M.13) [M. Laczkovich] 7. Let fl, f2, ... , f,,, be regular functions on a domain of the complex plane, linearly independent over the complex field. Prove that the functions Iii k, 1 < i, k < n, are also linearly independent. (F.23) [L. Lempert] 8. Prove that the set of all linear combinations (with real coefficients) of xn2}O°_0 the system of polynomials {xn + is dense in C[0,1]. (F.24) [J. Szabados]
9. Let D be a convex subset of the n-dimensional space, and suppose that D' is obtained from D by applying a positive central dilatation and then a translation. Suppose also that the sum of the volumes of D and D' is 1, and D fl D' # 0. Determine the supremum of the volume of the convex hull of D U D' taken for all such pairs of sets D, D'. (G.23) [L. Fejes-Toth, E. Makai] 10. Suppose that T is a metrizable topology on a set X of cardinality less
than or equal to continuum. Prove that there exists a separable and metrizable topology on X that is coarser than T. (T.11) [I. Juhasz] 11. Let S1, S2.... be independent, identically distributed random variables with distribution -1) = P(e1 = 1) =
2
(n=1,2,...), So=0, and
Write T,,, =
1
max Sk .
V fn- O
Prove that lim of (log n) Tn = 0 with probability one. (P.16) [P. Re-oo vesz]
1. PROBLEMS OF THE CONTESTS
25
1977 1. Consider the intersection of an ellipsoid with a plane or passing through its center O. On the line through the point 0 perpendicular to a, mark the two points at a distance from 0 equal to the area of the intersection.
Determine the loci of the marked points as a runs through all such planes. (G.24) [L. Tamassy] 2. Construct on the real projective plane a continuous curve, consisting of simple points, which is not a straight line and is intersected in a single point by every tangent and every secant of a given conic. (G.25) [F. Karteszi]
3. Prove that if a, x, y are p-adic integers different from 0 and pIx, paIxy, then 1 (1 + x)Y - 1 log(1 + x) (mod a). y
x
x
(N.12) [L. Redei]
4. Let p > 5 be a prime number. Prove that every algebraic integer of the pth cyclotomic field can be represented as a sum of (finitely many) distinct units of the ring of algebraic integers of the field. (A.26) [K. Gydry]
5. Suppose that the automorphism group of the finite undirected graph X = (P, E) is isomorphic to the quaternion group (of order 8). Prove that the adjacency matrix of X has an eigenvalue of multiplicity at least 4.
(P = {1, 2, ... , n} is the set of vertices of the graph X. The set of edges
E is a subset of the set of all unordered pairs of elements of P. The group of automorphisms of X consists of those permutations of P that
map edges to edges. The adjacency matrix M = [m=j] is the n x n matrix defined by mij = 1 if {i, j} E E and mz3 = 0 otherwise.) (A.27) [L. Babai]
6. Let f be a real function defined on the positive half-axis for which f (xy) = xf (y) + yf (x) and f (x + 1) < f (x) hold for every positive x and y. Show that if f (1/2) = 1/2, then
f(x)+f(1-x)>-xlog2x-(1-x)log2(1-x) for every x E (0, 1). (F.25) [Z. Daroczy, Gy. Maksa]
7. Let G be a locally compact solvable group, let c1, ... , cn be complex numbers, and assume that the complex-valued functions f and g on G satisfy
n
1: ck f (xyk) = f(X)g(y)
for all x, y E G.
k=1
Prove that if f is a bounded function and inf Re f (x)X(x) > 0
xEG
for some continuous (complex) character X of G, then g is continuous. (F.26) [L. Szekelyhidi]
26
1. PROBLEMS OF THE CONTESTS
8. Let p > 1 be a real number and R+ = (0, oo). For which continuous functions g : R+ ---> R+ are the following functions all convex?
L
1
g(
)X+1
Z 1g()
n=1,2,... (S.18) [L. Losonczi]
9. Suppose that the components of the vector u = (u0,. .. , u,) are real functions defined on the closed interval [a, b] with the property that every nontrivial linear combination of them has at most n zeros in [a, b].
Prove that if or is an increasing function on [a, b] and the rank of the operator b
A(f) = f u(x)f (x)da(x), a
f E C[a, b],
is r < n, then or has exactly r points of increase. (F.27) [E. Gesztelyi]
10. Let the sequence of random variables {Xm, m > 0}, Xo = 0, be an infinite random walk on the set of nonnegative integers with transition probabilities
pi=P(Xm+1=i+1IXm=i)>0, i>0, Qz=P(Xm+1=i-11 Xm.=i)>0, i>0. Prove that for arbitrary k > 0 there is an ak > 1 such that
Pn(k)=P( max 0<j
L-. L n=1 Pn Ma'k < oo . lim
1
(P.17) [J. Tomko] 1978 1. Let N be a family of finite subsets of an infinite set X such that every finite subset of X can be represented as the union of two disjoint sets
from N. Prove that for every positive integer k there is a subset of X that can be represented in at least k different ways as the union of two disjoint sets from N. (C.11) [P. Erdos] 2. For a distributive lattice L, consider the following two statements: (A) Every ideal of L is the kernel of at least two different homomorphisms. (B) L contains no maximal ideal.
1. PROBLEMS OF THE CONTESTS
27
Which one of these statements implies the other? (Every homomorphism cp of L induces an equivalence relation on L: a - b if and only if acp = bcp. We do not consider two homomorphisms different if they imply the same equivalence relation.) (A.28) [J. Varlet, E. Fried]
3. Let 1 < al < a2 < . . . < a,,, < x be positive integers such that En 1/ai < 1. Let y denote the number of positive integers smaller than x not divisible by any of the ai. Prove that
y>
cx
logx
with a suitable positive constant c (independent of x and the numbers ai). (N.13) [I. Z. Ruzsa] 4. Let Q and R be the set of rational numbers and the set of real numbers, respectively, and let f : Q --+ R be a function with the following property. For every h E Q, xo E R, f(x + h) - f(x) -> 0
as x E Q tends to x0. Does it follow that f is bounded on some interval? (F.28) [M. Laczkovich] 5. Suppose that R(z) = E°° _00 anzn converges in a neighborhood of the unit circle {z : IzI = 1} in the complex plane, and R(z) = P(z)/Q(z) is a rational function in this neighborhood, where P and Q are polynomials of degree at most k. Prove that there is a constant c independent of k such that 00
lanl < ck2 max IR(z)J. n=-oo
I
=
I
-
(S.19) [H. S. Shapiro, G. Somorjai] 6. Suppose that the function g : (0, 1) -> R can be uniformly approximated by polynomials with nonnegative coefficients. Prove that g must be
analytic. Is the statement also true for the interval (-1,0) instead of (0, 1)? (F.29) [J. Kalina, L. Lempert] 7. Let T be a surjective mapping of the hyperbolic plane onto itself which
maps collinear points into collinear points. Prove that T must be an isometry. (G.26) [M. Bognar]
8. Let Xl,... , Xn be n points in the unit square (n > 1). Let ri be the distance of Xi from the nearest point (other than Xi). Prove the inequality
ri+
+r2 <4.
(G.27) [L. Fejes-Tdth, E. Szemeredi] 9. Suppose that all subspaces of cardinality at most R1 of a topological space are second-countable. Prove that the whole space is secondcountable. (T.12) [A. Hajnal, I. Juhasz]
1. PROBLEMS OF THE CONTESTS
28
10. Let Yn be a binomial random variable with parameters n and p. Assume
that a certain set H of positive integers has a density and that this density is equal to d. Prove the following statements: (a) limn . P(Yn E H) = d if H is an arithmetic progression. (b) The previous limit relation is not valid for arbitrary H.
(c) If H is such that P(Yn E H) is convergent, then the limit must be equal to d. (P.18) [L. P6sa]
1979
1. Let the operation f of k variables defined on the set {1, 2, ... , n} be called friendly toward the binary relation p defined on the same set if f (al, a2, ... , ak) p f (bl, b2, ... , bk)
implies ai p bi for at least one i, 1 < i < k. Show that if the operation f is friendly toward the relations "equal to" and "less than," then it is friendly toward all binary relations. (C.12) [B. Csakany] 2. Let V be a variety of monoids such that not all monoids of V are groups.
Prove that if A E V and B is a submonoid of A, there exist monoids S E V and C and epimorphisms cp : S --> A , cpl : S --+ C such that ((e)cpi 1)cp = B (e is the identity element of C). (A.29) [L. Marki] 3. Let g(n, k) denote the number of strongly connected, simple directed graphs with n vertices and k edges. (Simple means no loops or multiple edges.) Show that
na-n
(-1)kg(n, k) = (n - 1)!. k=n
(C.13) [A. A. Schrijver] 4. For what values of n does the group SO(n) of all orthogonal transformations of determinant 1 of the n-dimensional Euclidean space possess a closed regular subgroup? (G < SO(n) is called regular if for any elements x, y of the unit sphere there exists a unique cp E G such that cp(x) = y.) (A.30) [Z. Szab6] 5. Give an example of ten different noncoplanar points P1,. .. , P5, Q1, .... Q5 in 3-space such that connecting each Pi to each Qj by a rigid rod results in a rigid system. (G.28) [L. Lovasz] 6. Let us define a pseudo-R.iemannian metric on the set of points of the Euclidean space E3 not lying on the z-axis by the metric tensor 1
0
0
1
0
0
0 0
,
- x2+y2
where (x, y, z) is a Cartesian coordinate system in E3. Show that the orthogonal projections of the geodesic curves of this Riemannian space
1. PROBLEMS OF THE CONTESTS
29
onto the (x, y)-plane are straight lines or conic sections with focus at the origin. (G.29) [P. Nagy] 7. Let T be a triangulation of an n-dimensional sphere, and to each vertex of T let us assign a nonzero vector of a linear space V. Show that if T has an n-dimensional simplex such that the vectors assigned to the vertices of this simplex are linearly independent, then another such simplex must also exist. (C.14) [L. Lovasz]
8. Let Kn(n = 1, 2, ...) be periodical continuous functions of period 27r, and write 2n
k,,, (f ; x) =
J0
f (t)Kn (x - t)dt.
Prove that the following statements are equivalent: (i) f0 I kn (f; x) - f (x) I dx -> 0 (n ---> oo) for all f E Ll [0, 27r]. (ii) kn (f ; 0) - + f (0) for all continuous, 27r-periodic functions f . (S.20) [V. Totik] 00
9. Let us assume that the series of holomorphic functions E fk (z) is abk=1
solutely convergent for all z c C. Let H C_ C be the set of those points where the above sum function is not regular. Prove that H is nowhere dense but not necessarily countable. (S.21) [L. Kerchy]
10. Prove that if ai (i = 1, 2,3,4) are positive constants, a2 - a4 > 2, and alai - a2 > 2, then the solution (x(t), y(t)) of the system of differential equations a1 - a2x + a3xy,
y = a4x - y - a3xy
(x, y E R)
with the initial conditions x(O) = 0, y(O) > al is such that the function x(t) has exactly one strict local maximum on the interval [0, 00). (F.30) [L. Pinter, L. Hatvani] 11. Let 1=1 be a double sequence of random variables such that ESijSkl = O((log(2Ii-kI+2) log(2I j -1I+2))-2)
(i,7, k, l = 1, 2.... ).
Prove that with probability one,
m n 1
EGI mn k=11=1 (P.19) [F. Moricz]
-+ 0
as max(m, n) --> oo.
1. PROBLEMS OF THE CONTESTS
30
1980
1. For a real number x, let lixil denote the distance between x and the closest integer. Let 0 < x,,, < 1 (n = 1, 2, ... ), and let e > 0. Show that there exist infinitely many pairs (n, m) of indices such that n # m and IIx,,, - x,,,,IH < min
e,
1
21n - mI
(C.15) [V. T. Sos] 2. Let H be the class of all graphs with at most 2k0 vertices not containing a complete subgraph of size R1. Show that there is no graph H E 1-l such that every graph in 7-l is a subgraph of H. (N.7) [F. Galvin] 3. In a lattice, connect the elements aAb and aVb by an edge whenever a and b are incomparable. Prove that in the obtained graph every connected component is a sublattice. (A.31) [M. Ajtai]
4. Let T E SL(n, Z), let G be a nonsingular n x n matrix with integer elements, and put S = G-1TG. Prove that there is a natural number k such that Sk E SL (n, Z). (N.14) [Gy. Szekeres] 5. Let G be a transitive subgroup of the symmetric group S25 different from
S25 and A25. Prove that the order of G is not divisible by 23. (A.32) [J. Pelikan]
6. Let us call a continuous function f : [a, b] - R2 reducible if it has a
double arc (that is, if there are a < a < (J < ry < 6 < b such that there exists a strictly monotone and continuous h : [a, /3] -+ [^/,b] for which f (t) = f (h(t)) is satisfied for every a < t < /3); otherwise f is irreducible. Construct irreducible f : [a, b] --p l[(2 and g : [c, d] - ][82 such that f ([a, b]) = g ([c, d]) and
(a) both f and g are rectifiable but their lengths are different; (b) f is rectifiable but g is not. (F.31) [A. Csaszar]
7. Let n > 2 be a natural number and p(x) a real polynomial of degree at most n for which max Ip(x)I < 1,
-1<x<1
p(-1) = p(1) = 0.
Prove that then Ip (x)I <
n cos 2n
1x cos 2n
1
C- cos 2n
<x< cosn 1 /
(F.32) [J. Szabados] 8. Let f (x) be a nonnegative, integrable function on (0, 27r) whose Fourier series is f (x) = ao + F,k 1 ak cos(nkx), where none of the positive integers nk divides another. Prove that jak I < ao. (S.22) [G. Halasz] 9. Let us divide by straight lines a quadrangle of unit area into n subpolygons and draw a circle into each subpolygon. Show that the sum of the perimeters of the circles is at most -7r / (the lines are not allowed to cut the interior of a subpolygon). (G.30) [G. and L. Fejes-Toth]
1. PROBLEMS OF THE CONTESTS
31
10. Suppose that the T3-space X has no isolated points and that in X any family of pairwise disjoint, nonempty, open sets is countable. Prove that X can be covered by at most continuum many nowhere-dense sets. (T.13) [I. Juhasz] 1981 1. We are given an infinite sequence of 1's and 2's with the following properties: (1) The first element of the sequence is 1. (2) There are no two consecutive 2's or three consecutive 1's. (3) If we replace consecutive 1's by a single 2, leave the single 1's alone, and delete the original 2's, then we recover the original sequence.
How many 2's are there among the first n elements of the sequence? (S.23) [P. P. Palfy] 2. Consider the lattice L of the contractions of a simple graph G (as sets of vertex pairs) with respect to inclusion. Let n > 1 be an arbitrary integer. Show that the identity n
n
X A V yi =
V
i-0
j=0
xA
V
Y.
o
holds if and only if G has no cycle of size at least n+2. (C.16) [A. Huhn] 3. Construct an uncountable Hausdorff space in which the complement of the closure of any nonempty, open set is countable. (T.14) [A. Hajnal, I. Juhasz] 4. Let G be a finite group and 1C a conjugacy class of G that generates G. Prove that the following two statements are equivalent: (1) There exists a positive integer m such that every element of G can be written as a product of m (not necessarily distinct) elements of 1C. (2) G is equal to its own commutator subgroup. (A.33) [J. Denes] 5. Let K be a convex cone in the n-dimensional real vector space R' , and
consider the sets A = K U (-K) and B = (IRn \ A) U {0} (0 is the origin). Show that one can find two subspaces in Rn such that together they span Rn, and one of them lies in A and the other lies in B. (G.31) [J. Sz{ics]
6. Let f be a strictly increasing, continuous function mapping I = [0, 1] onto itself. Prove that the following inequality holds for all pairs x, y E I:
1 - cos(xy) < fox f (t) sin(t f (t))dt +
Jo
y
f -1(t) sin(t f -1(t))dt
(F.33) [Zs. Pales]
7. Let U be a real normed space such that, for any finite-dimensional, real normed space X, U contains a subspace isometrically isomorphic to
1. PROBLEMS OF THE CONTESTS
32
X. Prove that every (not necessarily closed) subspace V of U of finite codimension has the same property. (We call V of finite codimension if there exists a finite-dimensional subspace N of U such that V + N = U.) (F.34) [A. Bosznay] 8. Let W be a dense, open subset of the real line R. Show that the following two statements are equivalent: (1) Every function f : III - R continuous at all points of R \ W and nondecreasing on every open interval contained in W is nondecreasing on the whole R. (2) IR \ W is countable. (T.15) [E. Gesztelyi]
9. Let n > 2 be an integer, and let X be a connected Hausdorff space such that every point of X has a neighborhood homeomorphic to the Euclidean space R. Suppose that any discrete (not necessarily closed) subspace D of X can be covered by a family of pairwise disjoint, open sets of X so that each of these open sets contains precisely one element of D. Prove that X is a union of at most N, compact subspaces. (T.16) [Z. Balogh]
10. Let P be a probability distribution defined on the Borel sets of the real line. Suppose that P is symmetric with respect to the origin, absolutely continous with respect to the Lebesgue measure, and its density function p is zero outside the interval [-1, 1] and inside this interval it is between the positive numbers c and d (c < d). Prove that there is no distribution whose convolution square equals P. (P.20) IT. F. M6ri, G. J. Szekely] 1982
1. A map F : P(X) -* P(X), where P(X) denotes the set of all subsets of X, is called a closure operation on X if for arbitrary A, B C X, the following conditions hold: (i)
A C F(A);
(ii) A C B = F(A) C F(B); (iii) F(F(A)) = F(A). The cardinal number min{ IAA : A C X, F(A) = X} is called the density of F and is denoted by d(F). A set H C X is called discrete with respect
to F if u V F(H - {u}) holds for all u E H. Prove that if the density of the closure operation F is a singular cardinal number, then for any nonnegative integer n, there exists a set of size n that is discrete with respect to F. Show that the statement is not true when the existence of an infinite discrete subset is required, even if F is the closure operation of a topological space satisfying the T1 separation axiom. (T.17) [A. Hajnal]
2. Consider the lattice of all algebraically closed subfields of the complex field C whose transcendency degree (over Q) is finite. Prove that this lattice is not modular. (A.34) [L. Babai] 3. Let G(V, E) be a connected graph, and let dG(x, y) denote the length of the shortest path joining x and y in G. Let rG(x) = max{dG(x, y) :
y E V} for x E V, and let r(G) = min{rG(x) : x E V}. Show that if
1. PROBLEMS OF THE CONTESTS
33
r(G) _> 2, then G contains a path of length 2r(G) - 2 as an induced subgraph. (C.17) [V. T. Sos] 4. Let
.f (n) _ E p' PI°
Prove that lim sup f ( n
n-oo
)
log log n = 1. n log n
(N.15) [P. Erdos] 5. Find a perfect set H c [0, 1] of positive measure and a continuous function f defined on [0, 1] such that for any twice differentiable function g defined on [0, 1], the set {x E H : f(x) = g(x)} is finite. (M.14) [M. Laczkovich]
6. For every positive a, natural number n, and at most an points xi, construct a trigonometric polynomial P(x) of degree at most n for which 2ir
P(xi) < 1,
P(x)dx = 0,
and
max P(x) > cn,
fo
where the constant c depends only on a. (F.35) [G. Halasz]
7. Let V be a bounded, closed, convex set in Rn, and denote by r the radius of its circumscribed sphere (that is, the radius of the smallest sphere that contains V). Show that r is the only real number with the following property: for any finite number of points in V, there exists a point in V such that the arithmetic mean of its distances from the other points is equal to r. (G.32) [Gy. Szekeres]
8. Show that for any natural number n and any real number d > 3n/(3n - 1), one can find a covering of the unit square with n homothetic triangles with area of the union less than d. (G.33) 9. Suppose that K is a compact Hausdorff space and K = U°_OAn, where An is metrizable and An C An for n < m. Prove that K is metrizable. (T.18) [Z. Balogh] 10. Let p0, p,.... be a probability distribution on the set of nonnegative integers. Select a number according to this distribution and repeat the selection independently until either a zero or an already selected number is obtained. Write the selected numbers in a row in order of selection without the last one. Below this line, write the numbers again in increasing order. Let Ai denote the event that the number i has been selected and that it is in the same place in both lines. Prove that the events Ai (i = 1, 2, ...) are mutually independent, and P(A2) = pi. (P.21) [T. F. Mori]
1. PROBLEMS OF THE CONTESTS
34
1983 1. Given n points in a line so that any distance occurs at most twice, show
that the number of distances occurring exactly once is at least [n/2]. (C.18) [V. T. Sos, L. Szekely]
2. Let I be an ideal of the ring R and f a nonidentity permutation of the set {1, 2, ... , k} for some k. Suppose that for every 0 a E R, aI # 0 and Ia # 0 hold; furthermore, for any elements x1i x2, ... , xk E I,
x1x2...xk = xlfx2f ...xkf holds. Prove that R is commutative. (A.35) [R. Wiegandt] 3. Let f : R -+ R be a twice differentiable, 27r-periodic even function. Prove
that if 1
fr(X) + f(x)
f (x + 37r/2)
holds for every x, then f is it/2-periodic. (F.36) [Z. Szabo, J. Terjeki] 4. For which cardinalities n do antimetric spaces of cardinality ,c exist? (X, o) is called an antimetric space if X is a nonempty set, o : X2 ->
[0, oo) is a symmetric map, o(x, y) = 0 holds if x = y, and for any three-element subset {al, a2, a3} of X
o(alf,a2f) + o(a2f,a3f) < o(aif,a3f) holds for some permutation f of {1, 2, 3}. (1Z.8) [V. Totik]
5. Let g : R -> R be a continuous function such that x + g(x) is strictly monotone (increasing or decreasing), and let u [0, oo) --> R be a :
bounded and continuous function such that t
u(t)+
j_ g(u(s))ds
is constant on [1, oo). Prove that the limit limt.,,. u(t) exists. (F.37) [T. Krisztin]
6. Let T be a bounded linear operator on a Hilbert space H, and assume that IIT111 < 1 for some natural number n. Prove the existence of an invertible linear operator A on H such that IIATA-111 < 1. (0.5) [E. Druszt]
7. Prove that if the function f : R2 -+ [0, 1] is continuous and its average on every circle of radius 1 equals the function value at the center of the circle, then f is constant. (F.38) [V. Totik] 8. Prove that any identity that holds for every finite n-distributive lattice also holds for the lattice of all convex subsets of the (n -1)-dimensional Euclidean space. (For convex subsets, the lattice operations are the settheoretic intersection and the convex hull of the set-theoretic union. We call a lattice n-distributive if n
n
x A (V yi) = V (x A( V yi)) i=O
j=0
O
ii4j
1. PROBLEMS OF THE CONTESTS
35
holds for all elements of the lattice.) (A.36) [A. Huhn] 9. Prove that if E C JR is a bounded set of positive Lebesgue measure, then for every u < 1/2, a point x = x(u) can be found so that
i(x-h,x+h)nEI>uh and
i(x-h,x+h)n(IR\E)l >uh for all sufficiently small positive values of h. (M.15) [K. I. Koljada]
10. Let R be a bounded domain of area t in the plane, and let C be its center of gravity. Denoting by TAB the circle drawn with the diameter AB, let K be a circle that contains each of the circles TAB (A, B E R). Is it true in general that K contains the circle of area 2t centered at C? (G.34) [J. Sziics] 11. Let Mn C Rn+1 be a complete, connected hypersurface embedded into the Euclidean space. Show that Mn as a Riemannian manifold decomposes to a nontrivial global metric direct product if and only if it is a real cylinder, that is, M" can be decomposed to a direct product of the form Mn = Mk x ]fin-k (k < n) as well, where Mk is a hypersurface in some (k + 1)-dimensional subspace Ek+1 C Rn+l, Rn-k is the orthogonal complement of Ek+1. (G.35) [Z. Szabo] 12. Let X 1 ,.. . , Xn be independent, identically distributed, nonnegative random variables with a common continuous distribution function F. Suppose in addition that the inverse of F, the quantile function Q, is also
continuous and Q(O) = 0. Let 0 = Xo:n < Xl:n < . . . < Xn:n be the ordered sample from the above random variables. Prove that if EX1 is finite, then the random variable 1 [ny]+1
O= sup 0
n
y
(1 - u)dQ(u)
(n + 1 - i)(Xi:n - Xi-l:n) fo
i=1
tends to zero with probability one as n -+ oo. (P.22) [S. Cs6rg6, L. Horvath]
1984 1. Let r. be an arbitrary cardinality. Show that there exists a tournament TK = (VK, EK) such that for any coloring f : EK -* ,c of the edge set E", there are three different vertices x0, X1, x2 E VK, such that xox1, xlx2, x2xo E EK and
I{f(xoxl),f(xlx2),f(x2xo)}I < 2.
(A tournament is a directed graph such that for any vertices x, y VK, x # y exactly one of the relations xy E EK, yx c EK, holds.) (C.19) [A. Hajnal]
1. PROBLEMS OF THE CONTESTS
36
2. Show that there exist a compact set K C R and a set A C R of type FQ such that the set
{xER:K+xCA} is not Borel-measurable (here K + x = {y + x : y E K}). (M.16) [M. Laczkovich]
3. Let a and b be positive integers such that when dividing them by any prime p, the remainder of a is always less than or equal to the remainder
of b. Prove that a = b. (N.16) [P. Erd6s, P. P. Palfy] 4. Let x1, x2, Yi, y2, Z1, z2 be transcendental numbers. Suppose that any 3 of them are algebraically independent, and among the 15 four-tuples only {x1, x2, yl, y2}, {x1, x2, z1, z2}, and {y1, y2, z1, z2} are algebraically
dependent. Prove that there exists a transcendental number t that depends algebraically on each of the pairs {x1, x2}, {y1, y2}, and {z1, z2}. (A.37) [L. Lovasz]
5. Let ao, a1, ... be nonnegative real numbers such that co
E an = 00.
n=0
For arbitrary c > 0, let k
nj(c) =min k : c j < E ai
j = 1, 2, ...
.
i=0
Prove that if Ea ° o a1 < oo, then there exists a c > 0 for which ani (,) < 00, and if I a ° o ai = oo, then there exists a c > 0 for which I:j' 1 ani (c) = 00. (S.24) [P. Erdos, I. Joo, L. Szekely] 6. For which Lebesgue-measurable subsets E of the real line does a positive constant c exist for which sup
-00
L
sup
Leinxf(x)dx
n=O,±1,...
for all integrable functions f on E? (M.17) [G. Halasz] 7. Let V be a finite-dimensional subspace of C[0,1] such that every nonzero
f E V attains positive value at some point. Prove that there exists a polynomial P that is strictly positive on [0, 1] and orthogonal to V, that is, for every f E V, 1
J
f (x)P(x)dx = 0.
(F.39) [A. Pinkus, V. Totik] 8. Among all point lattices on the plane intersecting every closed convex region of unit width, which one's fundamental parallelogram has the largest area? (G.36) [L. Fejes-Toth]
1. PROBLEMS OF THE CONTESTS
37
9. Let X0, X1, ... be independent, identically distributed, nondegenerate random variables, and let 0 < a < 1 be a real number. Assume that the series
cc
1: akXk k=0
is convergent with probability one. Prove that the distribution function of the sum is continuous. (P.23) [T. F. Mori] 10. Let X1i X2, ... be independent random variables with the same distribution: (i = 1, 2, ... ).
P(Xi = 1) = P(Xi = -1) = 2 Define
Sn=Xl+X2+...+Xn (n=1,2,...), (x,n)=I{k: 0
and
a(n)=I{x: (x,n)=1}1
(n=0,1,...).
Prove that
P(liminf a(n) = 0) = 1 and that there is a number 0 < c < oc such that P(lim sup a(n)/log n = c) = 1. (P.24) [P. Revesz] 1985
1. Some proper partitions P1,. .. , Pn of a finite set S (that is, partitions containing at least two parts) are called independent if no matter how we choose one class from each partition, the intersection of the chosen classes is nonempty. Show that if the inequality
A < IP11 ... IPn1
(*)
holds for some independent partitions, then P1,.. . , Pn is maximal in the sense that there is no partition P such that P , P 1 ,. . , Pn are independent. On the other hand, show that inequality (*) is not necessary for this maximality. (C.20) [E. Gesztelyi] 2. Let S be a given finite set of hyperplanes in R', and let 0 be a point. Show that there exists a compact set K C Rn containing 0 such that the orthogonal projection of any point of K onto any hyperplane in S is also in K. (G.37) [Gy. Pap] 3. Let k and K be concentric circles on the plane, and let k be contained inside K. Assume that k is covered by a finite system of convex angular
domains with vertices on K. Prove that the sum of the angles of the
38
1. PROBLEMS OF THE CONTESTS
domains is not less than the angle under which k can be seen from a point of K. (G.38) [Zs. Pales] 4. Call a subset S of the set {1, ... , n} exceptional if any pair of distinct elements of S are coprime. Consider an exceptional set with a maximal sum of elements (among all exceptional sets for a fixed n). Prove that if n is sufficiently large, then each element of S has at most two distinct prime divisors. (N.17) [P. Erdds] 5. Let F(x, y) and G(x, y) be relatively prime homogeneous polynomials of degree at least one having integer coefficients. Prove that there exists a number c depending only on the degrees and the maximum of the absolute values of the coefficients of F and G such that F(x, y) # G(x, y) for any integers x and y that are relatively prime and satisfy max{ Ix I, IyI } > c. (A.38) [K. Gyory]
6. Determine all finite groups G that have an automorphism f such that H f (H) for all proper subgroups H of G. (A.39) [B. Kovacs] 7. Let pl and P2 be positive real numbers. Prove that there exist functions fi : R -* R such that the smallest positive period of fi is pi (i = 1, 2), and fl - f2 is also periodic. (A.40) [J. Riman] 8. Let 2/(/+1) < p < 1, and let the real sequence {an } have the following property: for every sequence {en} of 0's and ±1's for which >
1 e,,p'n =
0, we also have E°° 1 0. Prove that there is a number c such that an = cpn for all n. (S.25) [Z. Daroczy, I. Katai]
9. Let D = {zEC: IzI <1}andD={wEC: IwI=1}. Prove that if for a function f : D x B -- C the equality
az+b aw+b _ b aw+b f bz+a' bw+a - f z w+ f bw+a (d,
holds for allzED,wEBanda,bEC, Ia12=1+ b12' then there is a function L : ] 0, oo[--* C satisfying
L(pq) = L(p) + L(q) for all p, q > 0
such that f can be represented as 2
f(z,w)
L(1-IZI l \Iw-z12/
for all
z E D,w E B.
(F.40) [Gy. Maksa] 10. Show that any two intervals A, B C_ R of positive lengths can be countably disected into each other, that is, they can be written as countable unions A = Al U A2 U ... and B = B1 U B2 U ... of pairwise disjoint sets, where Ai and Bi are congruent for every i E N. (M.18) [Gy. Szabo] 11. Let l; (E, ir, B) (7r : E -* B) be a real vector bundle of finite rank, and let be the tangent bundle of E, where Vl; = Ker d-7r is the vertical subbundle of TE. Let us denote the projection operators corresponding to the
1. PROBLEMS OF THE CONTESTS
39
splitting (*) by v and h. Construct a linear connection V on V such that VX V Y - Vy V X = v[X, Y] - v[hX, hY]. (X and Y are vector fields on E, [., .] is the Lie bracket, and all data are of class C°°). (G.39) [J. Szilasi] 12. Let (Il, A, P) be a probability space, and let .Fn) be an adapted sequence in (SZ, A, P) (that is, for the v-algebras we have .Fl C_ .F2 C_ C_ A, and for all n, Xn, is an .Fn-measurable and integrable random variable). Assume that
(n = 2, 3 ... ).
E(Xn+l lFn) = 1 Xn + 1 Xn-i 2 2
Prove that supra EIXnI < oo implies that Xn, converges with probability one as n -+ oo. (P.25) [I. Fazekas]
1986 1. If (A, <) is a partially ordered set, its dimension, dim (A, <), is the least
cardinal i such that there exist n total orderings { ko, then there exist disjoint A0i Al C A with dim(Ao, <), dim(Al, <) > k0. (N.9) [D. Kelly, A. Hajnal, B. Weiss] 2. Show that if k < n/2 and .F is a family of k x k submatrices of an n x n matrix such that any two intersect then
n-1 2
IYI < (k - 1)
.
(C.21) [Gy. Katona]
3. (a) Prove that for every natural number k, there are positive integers < ak such that ai - aj divides ai for all 1 < i, j < k, al < a2 <
i#j
(b) Show that there is an absolute constant C > 0 such that al > kCk for every sequence a1, ... , ak of numbers that satisfy the above divisibility
condition. (N.18) [A. Balogh, I. Z. Ruzsa] 4. Determine all real numbers x for which the following statement is true: the field C of complex numbers contains a proper subfield F such that adjoining x to F we get C. (A.41) [M. Laczkovich] 5. Prove the existence of a constant c with the following property: for every composite integer n, there exists a group whose order is divisible by n
and is less than nC, and that contains no element of order n. (A.42) [P. P. Palfy]
6. Let U denote the set If E C[0, 1] If (x) I < 1 for all x E [0, 1]}. Prove that there is no topology on C[0,1] that, together with the linear struc:
ture of C[0,1], makes C[0,1] into a topological vector space in which the set U is compact. (T.19) [V. Totik]
1. PROBLEMS OF THE CONTESTS
40
7. Prove that the series EP cP f (px), where the summation is over all primes, unconditionally converges in L2 [0, 1] for every 1-periodic function f whose restriction to [0, 1] is in L2[0,1] if and only if EP cpj < oo. (Unconditional convergence means convergence for all rearrangements.) (F.41) [G. Halasz] 8. Let ao = 0, a1, ... , ak and bo = 0, b1, ... , bk be arbitrary real numbers.
(i) Show that for all sufficiently large n there exist polynomials pn, of degree at most n for which
i = 0,1,...,k,
p ) (-1) = ai, p(i) (1) = bi,
(1)
and
max Ipn(X)I
(2)
n2
where the constant c depends only on the numbers ai, bi. (ii) Prove that, in general, (2) cannot be replaced by the relation
lim n2 max IN(x)I = 0.
n--*oo
(3)
1xI<1
(F.42) [J. Szabados] 9. Consider a latticelike packing of translates of a convex region K. Let t be the area of the fundamental parallelogram of the lattice defining the packing, and let tm;n(K) denote the minimal value of t taken for all latticelike packings. Is there a natural number N such that for any n > N and for any K different from a parallelogram, ntm;n(K) is smaller than the area of any convex domain in which n translates of K can be placed without overlapping? (By a latticelike packing of K we mean a set of nonoverlapping translates of K obtained from K by translations with all vectors of a lattice.) (G.40) [G. and L. Fejes-Toth] 10. Let X1, X2, ... be independent, identically distributed random variables such that Xi > 0 for all i. Let EXi = m, Var(Xi) = Q2 < oo. Show
that, forall0
lim
n Var
([X1+a2or 2 n J
J
m2(1_")
(P.26) [Gy. Michaletzki]
1987 1. Let us color the integers 1, 2, ... , N with three colors so that each color is given to more than N/4 integers. Show that the equation x = y+z has a solution in which x, y, z are of distinct colors. (C.22) [Gy. Szekeres] 2. A binary relation -< is called a quasi-order if it is reflexive and transitive. The infimum of the quasi-order (Q, ) is the greatest subset J C_ Q such
that (i) for every B E Q there is an A E J with A (ii) A B, A, B E J imply B -< A.
B, and
1. PROBLEMS OF THE CONTESTS
41
Let X be a finite, nonempty alphabet, let X* be the set of all finite words from X, and let P be the set of infinite subsets of X*. For A, B E P, let A -< B if every element of A is a (connected) subword of some element of B. Show that (P, -<) has an infimum, and characterize its elements. (N.10) [Gy. Pollak]
3. Let A be a finite simple groupoid such that every proper subgroupoid of A has cardinality one, the number of one-element subgroupoids is at least three, and the group of automorphisms of A has no fixed points. Prove that in the variety generated by A, every finitely generated free algebra is isomorphic to some direct power of A. (A.43) [A. Szendrei] 4. Let the finite projective geometry P (that is, a finite, complemented, modular lattice) be a sublattice of the finite modular lattice L. Prove that P can be embedded in a projective geometry Q, which is a coverpreserving sublattice of L (that is, whenever an element of Q covers in Q another element of Q, then it also covers that element in L). (A.44) [E. Fried]
5. Let f and g be continuous real functions, and let g 0 be of compact support. Prove that there is a sequence of linear combinations of translates of g that converges to f uniformly on compact subsets of R. (F.43) [Sz. Gy. Revesz, V. Totik] 6. Is it true that if A and B are unitarily equivalent, self-adjoint operators in the complex Hilbert space f, and A < B, then A+ < B+? (Here A+ stands for the positive part of A.) (0.6) [L. Kerchy] 7. Let x : [0, oo) --+ R be a differentiable function satisfying the identity t
x'(t) = -2x(t) sin 2 t + (2 - cos tl + cos t)
I
x(s) sin2 s ds
on [1, oo). Prove that x is bounded on [0, oo) and that limt . x(t) = 0. Does the conclusion remain true for functions satisfying the identity
(t) _ -2x(t)t + (2 - cos t+ cos t)
f
x(s)s ds ?
-1
(F.44) [L. Hatvani]
8. Let c > 0, c # 1 be a real number, and for x E (0,1) let us define the function
00
f(x) =
fI(1+cx2k).
k=0
Prove that the limit lim
f ( x3 )
x-.1-0 f (x)
does not exist. (F.45) [V. Totik] 9. Show that there exists a constant ck such that for any finite subset V of the k-dimensional unit sphere there is a connected graph G such that the set of vertices of G coincides with V, the edges of G are straight line
1. PROBLEMS OF THE CONTESTS
42
segments, and the sum of the kth powers of the lengths of the edges is less than Ck. (G.41) [V. Totik] 10. Let F be a probability distribution function symmetric with respect to the origin such that F(x) = 1 - x-1K(x) for x > 5, where if x E [5, oo) \ U°°_5(n!, 4n!), if x E (n!, 2n!], n > 5,
1
K(x) _
,
3- 2,
n > 5.
if x E (2n!, 4n!),
Construct a subsequence Ink} of natural numbers such that if X1, X2, ... are independent, identically distributed random variables with distribution function F, then for all real numbers x
x
li
00
P
n j=1, Xj < 7rx
}
= 1 + 1 arctan x. 7r
(P.27) [S. Csorgo] 1988 1. Define a partial order on all functions f : R -+ R by the relation f -< g if f (x) < g(x) for all x E R. Show that this partially ordered set contains a totally ordered subset of size greater than but that the latter subset cannot be well-ordered. (R.11) [P. Komjath] 2. Suppose that a graph G is the union of three trees. Is it true that G can be covered by two planar graphs? (C.23) [L. Pyber]
3. Let G be a finite Abelian group and x, y E G. Suppose that the factor group of G with respect to the subgroup generated by x and the factor group of G with respect to the subgroup generated by y are isomorphic. Prove that G has an automorphism that maps x to y. (A.45) [E. Lukacs] 4. Let lb be a family of real functions defined on a set X such that k o h E 4)
whenever fi E -b (i E I) and h : X -+ RI is defined by the formula h(x)i = fi(x), and (1) k: h(X) -> R is continuous with respect to the topology inherited from the product topology of RI. Show that f = sup{g3 : j E J, gj E -D} = inf{h,,,.: m E M, h,,,, E 4)) implies f E P. Does this statement remain true if (1) is replaced with the following condition? (2) k: h(X) -> R is continuous on the closure of h(X) in the product topology.
(T.20) [A. Csiszar] 5. Let us draw a circular disc of radius r around every integer point in the plane different from the origin. Let Er be the union of these discs, and denote by dr the length of the longest segment starting from the origin and not intersecting Er. Show that lim (dr - 1) = 0. r-.o r
1. PROBLEMS OF THE CONTESTS
43
(G.42) [M. Laczkovich] 6. Let H C R be a bounded, measurable set of positive Lebesgue measure. Prove that A((H Il) t \ H) > 0 o liminf
where H + t = {x + t : x E H} and A is the Lebesgue measure. (M.19) [M. Laczkovich]
7. Let S be the set of real numbers q such that there is exactly one 0-1 sequence {an,} satisfying CK)
1: a,4
= 1.
n=1
Prove that the cardinality of S is 2s0. (S.26) [P. Erdos, I. Joo]
8. Let f and g be holomorphic functions on the open unit disc D, and suppose that J12 + 1912E Lipl. Prove that then f, g E Lip2. A function h : D -> C is in the Lipa class if there is a constant K such
that Jh(z) - h(w)l < KJz - wia for every z, w E D. (F.46) [L. Lempert] 9. We say that the point (al, a2, a3) is above (below) the point (b1i b2, b3) if a1 = b1, a2 = b2 and a3 > b3 (a3 < b3). Let el, e2, ... , elk (k > 2) be pairwise skew lines not parallel with the z-axis, and assume that among their orthogonal projections to the (x, y)-plane no two are parallel and no three are concurrent. Is it possible that going along any of the lines the points that are below or above a point of some other line ei alternately follow one another? (G.43) [J. Pach]
10. Let a E C, lal < 1. Find all values of b E C for which there exist probability measures with characteristic function 0 satisfying 0(2) = a and 0(1) = b. (P.28) [T. F. Mori] 1989 1. Let p be an arbitrary prime number. In the ring G of Gaussian integers, consider the subrings
An ={pa+pm'bi:a,bEZ},
n=1,2,....
Let R C G be a subring of G that contains An+1 as an ideal for some n. Prove that this implies that one of the following statements must hold: R = An+1i R = An; or 1 E R. (A.46) [R. Wiegandt] 2. Let n > 2 be an integer, and let f 2n, denote the semigroup of all mappings g : {0,1}n -> {0,1}n. Consider the mappings f E 52,,, which have the following property: there exist mappings gi {0,1}2 , {o, 1} (i = 12 n) such that for all (a a a) E {0 1}n :
f(a,,a2i...,an) _
(91(al,a,),92(al,a2),...,9fl(afl-1,an))
1. PROBLEMS OF THE CONTESTS
44
Let An denote the subsemigroup of 1l, generated by these f's. Prove that An contains a subsemigroup Fn such that the complete transformation semigroup of degree n is a homomorphic image of rn. (A.47) [P. Domosi]
3. Let n1 < n2 < ... be an infinite sequence of natural numbers such that /2k n1k tends to infinity monotone increasingly. Prove that E00 k=11/nk is irrational. Show that this statement is best possible in a sense by giving, for every c > 0, an example of a sequence ni < n2 < ... such that nk/2k > c for all k but EO° k=1 1/nk is rational. (N.19) [P. Erdos] 4. Cancelled. 5. Characterize the sets A C R for which
A+B={a+b:aEA, bEB} is nowhere-dense whenever B C R is a nowhere dense set. (T.21) [M. Laczkovich]
6. Find all functions f : ][83 -> R that satisfy the parallelogram rule f (x + y) + f (x - y) = 2f (x) + 2f (y),
x, y
and that are constant on the unit sphere of RI. (F.47) [Gy. Szabo] 7. Let K be a compact subset of the infinite-dimensional, real, normed linear space (X, II
-
II).
Prove that K can be obtained as the set of
all left limit points at 1 of a continuous function g : [0,1 [-+ X, that is, x belongs to K if and only if there exists a sequence to E [0, 1 [ (n = 1, 2, ...) satisfying limn. to = 1 and limn- 0 11g(tn) - xli = 0. (0.7) [B. Garay] 8. For any fixed positive integer n, find all infinitely differentiable functions f : ][8n -> R satisfying the following system of partial differential equations: n
bask f = 0,
k=1,2,....
i=1
(F.48) [L. Szekelyhidi]
9. Suppose that HTM is a direct complement to a vertical bundle VTM over the total space of the tangent bundle TM of the manifold M. Let v and h denote the projections corresponding to the decomposition TTM = VTM ® HTM. Construct a bundle involution P : TTM -TTM such that Poh = voP and prove that, for any pseudo-Riemannian metric given on the bundle VTM, there exists a unique metric connection V such that
VxPY-VyPX =Poh[X,Y] if X and Y are sections of the bundle HTM, and
VxY - VYX = [X,Y]
1. PROBLEMS OF THE CONTESTS
45
if X and Y are sections of the bundle VTM. (G.44) [J. Szilasi] 10. Let Y(k), k = 1, 2.... be an m-dimensional stationary Gauss-Markov process with zero expectation, that is, suppose that Y(k + 1) = A Y(k) + e(k + 1),
k = 1,2,...
Let Hi denote the hypothesis A = Ai, and let Pi (0) be the a priori probability of Hi, i = 0, 1, 2. The a posteriori probability P1(k) = P(H1IY(1), ... , Y(k)) of hypothesis H1 is calculated using the assumptions P1(0) > 0, P2(0) > 0, P, (0) + P2(0) = 1. Characterize all matrices A0 such that P{limk . P1 (k) = 1} = 1 if Ho holds. (P.29) [I. Fazekas] 1990 1. Let A be a finite set of points in the Euclidean space of dimension d > 2.
For j = 1,2,...,d, let Bj denote the orthogonal projection of A onto the (d - 1)-dimensional subspace given by the equation xj = 0. Prove that d
IB;l >
IAld-1
j=1
(G.45) [I. Z. Ruzsa] 2. Prove that for every positive number K, there are infinitely many positive integers m and N such that there are at least KN/ log N primes among the integers m + 1, m + 4, ... , m + N2. (N.20) [I. Z. Ruzsa] 3. Let n = pk (p a prime number, k > 1), and let G be a transitive subgroup of the symmetric group Sn. Prove that the order of the normalizer of G in Sn is at most (A.48) [L. Pyber] 4. Let P be a polynomial with all real roots that satisfies the condition P(0) > 0. Prove that if m is a positive odd integer, then lGlk+1.
f k=0
(k)
k!
(0) xk > 0
for all real numbers x, where f = P-"`. (F.49) [J. Szabados] 5. We say that the real numbers x and y can be connected by a b-chain of length k (where 6 : R -+ (0, oo) is a given function) if there exist real numbers xo, X1, ... , xk such that x0 = x, xk = y, and (xi_i+xz) lxi - xi-ll < b
,
i = 1,...,k.
Prove that for every function 6 : R -+ (0, oo) there is an interval in which any two elements can be connected by a 6-chain of length 4. Also, prove that we cannot always find an interval in which any two elements could be connected by a 6-chain of length 2. (F.50) [M. Laczkovich]
1. PROBLEMS OF THE CONTESTS
46
6. Find meromorphic functions 0 and z#i in the unit disc such that, for any
function f regular in the unit disc, at least one of the functions f - 0 and f - 0 has a root. (F.51) [G. Halasz] 7. Denote by B[O,1] and C[O,1] the Banach space of all bounded functions and all continuous functions, respectively, on the interval [0, 1] with the supremum norm. Is there a bounded linear operator
T : B[0,1] -* C[0,1]
such that T f = f for all f E C[0,1]? (0.8) [G. Halasz] 8. Let Al°), ... , A(°) be a sequence of n >_ 3 points in the Euclidean plane Rz. Define the sequence A111, ... , An (i = 1, 2, ...) by induction as follows: let be the midpoint of the segment where A( Z+1) = A('-1). Show that, with the exception of a set of zero Lebesgue 1 measure, for every initial sequence (Al°), ... , A(°)) E (R2)", there exists a natural number N such that the points AlN)7 ... , (N are consecutive vertices of a convex n-gon. (G.46) [B. Csikos] 9. Prove that if all subspaces of a Hausdorff space X are Q-compact, then X is countable. (T.22) [I. Juhasz] 10. Let X and Y be independent identically distributed, real-valued random variables with finite expectation. Prove that
A)
EJX +YJ > EIX -Y1. (P.30) [T. F. Mori] 1991
1. To divide a heritage, n brothers turn to an impartial judge (that is, if not bribed, the judge decides correctly, so each brother receives (1/n)th of the heritage). However, in order to make the decision more favorable for himself, each brother wants to influence the judge by offering an amount of money. The heritage of an individual brother will then be described by a continuous function of n variables strictly monotone in the following sense: it is a monotone increasing function of the amount offered by him and a monotone decreasing function of the amount offered by any of the remaining brothers. Prove that if the eldest brother does not offer the judge too much, then the others can choose their bribes so that the decision will be correct. (F.52) [V. Totik] 2. Suppose that n points are given on the unit circle so that the product of the distances of any point of the circle from these points is not greater
than two. Prove that the points are the vertices of a regular n-gon. (G.47) [L. I. Szabo] 3. Prove that if a finite group G is an extension of an Abelian group of exponent 3 with an Abelian group of exponent 2, then G can be embedded in some finite direct power of the symmetric group S3. (A.49) [G. Czedli, B. Csakany]
1. PROBLEMS OF THE CONTESTS
47
4. Let n > 2 be an integer, and consider the groupoid G = (Z U {oo}, o), where
xoy= Ix+1 ifx=yEZn, 00
otherwise.
(Z denotes the ring of the integers modulo n.) Prove that G is the only subdirectly irreducible algebra in the variety generated by G. (A.50) [A. Szendrei]
5. Construct an infinite set H C_ C[0,1] such that the linear hull of any infinite subset of H is dense in C[O, 1]. (F.53) [V. Totik]
6. Let a > 0 be irrational. (a) Prove that there exist real numbers a1, a2, a3, a4 such that the function f : R -> ]E87
f (x) = ex [al + a2 sin x + a3 cos x + a4 COs(ax)1
is positive for all sufficiently large x, and lim inf f (x) = 0. X_+00
(b) Is the above statement true if a2 = 0? (F.54) [T. Krisztin] 7. Given an > an+1 > 0 and a natural number p, such that lim sup n
an < µ, aµn
prove that for all e > 0 there exist natural numbers N and no such that, for all n > no the following inequality holds: Nn
rn
1:ak <EEak. k=1
k=1
(S.27) [L. Leindler] S. Prove that if {ak} is a sequence of real numbers such that 00
00
l ak l /k = oo
and E I
2"-1
n=1
k=1
1/2
k(ak - ak+1 )z/
<00,
(k=2n-1
then 00
E ak sin(kx) dx=oo. 0
k=1
(F.55) [F. Moricz] 9. Let h : [0, oo) -> [0, 00) be a measurable, locally integrable function, and write
H(t) :=
J0
t h(s)ds
(t > 0).
1. PROBLEMS OF THE CONTESTS
48
Prove that if there is a constant B with H(t) < Bt' for all t, then 00
L
e- H(t)
eH(" )dudt = oo. 0t
(F.56) [L. Hatvani, V. Totik] 10. Consider the equation f'(x) = f (x + 1). Prove that (a) each solution f : [0, oc) --* (0, oo) has an exponential order of growth, that is, there exist numbers a > 0, b > 0 satisfying I f (x) I < a ebx,
x>0;
(b) there are solutions f : [0, oo) -p (-oo, oo) of nonexponential order of growth.
(F.57) [T. Krisztin] 11. Does there exist a bounded linear operator T on a Hilbert space H such
that
00
00
n Tn(H) = {0} but
n Tn(H)-
n_1
n-1
{0},
where - denotes closure? (0.9) [L. Kerchy] 12. Let X1, X2, ... be independent, identically distributed random variables such that, for some constant 0 < a < 1, P {X 1 = 2k/« } = 2-k,
k=1,2,...
Determine, by giving their characteristic functions or any other way, a sequence of infinitely divisible, nondegenerate distribution* functions Gn such that sup
P{
-00<S<00
(P.31) [S. Csorgo]
nl/«
<x -Gn(x)
-* 0
as
n--4oc.
2. Results of the Contests SUMMARY
Year of contest
Number of problems
Number of competitors
Number of correct solutions
1962
10
23
1963
10
56
50 402
1964
10
34
188
1965
10
52
321
1966
10
43
1967
10
34
333 258
1968
11
12
1970
12
37 33 27
169
1969 1971
11
26
159
1972
11
18
95
1973
10
35
184
1974
10
14
47
1975
12
63
441
1976
11
17
104
1977
10
30
138
1978
10
25
74
1979
11
34
84
1980
10
19
64
1981
10
42
137
1982
10
29
56
1983
12
28
138
1984
10
17
41
1985
12
34
183
1986
10
20
1987 1988
10
21
10
28
1989
10
1990
10
40 17
1991
12
17
62 84 128 224 46 74
49
191 161
2. RESULTS OF THE CONTESTS
50
List of Prize Winners and Honorably Mentioned Competitors (Notations: 1. First prize, 2. H. Honorably mentioned.)
Second prize, 3.
Third prize,
1962 1. Gabor Halasz
2. 3. H. Bela Bollobas, Arpad Elbert, Istvan Juhasz, Gyorgy Petruska, Domokos Szasz
1963 1. Arpad Elbert 2. Gyula Katona, Domokos Szasz
3. H. Gabor Halasz, Janos Komlos, Miklos Simonovits, Jozsef Szfics 1964 1. Bela Bollobas, Peter Vamos
2. Gabor Halasz, Istvan Juhasz, Gerzson Kery, Miklos Simonovits, Domokos Szasz
3. 1965
1. Bela Bollobas, Jozsef Fritz, Miklos Simonovits 2. Gerzson Kery, Attila Mate, Jozsef Pelikan
3. H. Laszlo Gerencser, Elod Knuth, LAszlo Lovasz, Lajos Posa, Gyorgy Vesztergombi
1966 1. Be1a Bollobas, Endre Makai, Miklos Simonovits 2. Laszlo Lovasz, Lajos Posa 3. Laszlo Gerencser, Miklos Laczkovich, Jozsef Pelikan, Ferenc Szigeti 1967 1. Laszlo Lovasz, Attila Mate 2. Laszlo Gerencser, Eors Mate
3. H. Laszlo Babai, Robert Freud, Miklos Laczkovich, Miklos Simonovits, Ferenc Szigeti 1968 1. Peter Gacs, Laszlo Lovasz, Endre Makai 2. Miklos Laczkovich, Lajos Posa 3. Leszlo Babai 1969 1. Peter Gacs, Laszlo Lovasz 2. Miklos Laczkovich
3. Laszlo Babai, Endre Makai, Jozsef Pelikan, Lajos Posa, Imre Z. Ruzsa
2. RESULTS OF THE CONTESTS
51
1970 1. LAszl6 Lovasz 2. LAszl6 Babai, Mikl6s Laczkovich
3. Peter Gacs, Endre Makai, J6zsef Pelikan, Lajos P6sa, Imre Z. Ruzsa 1971 1. Zsigmond Nagy, LAszl6 Babai 2. Lajos P6sa 3. J6zsef Pelikan, Imre Z. Ruzsa, Jen6 Deak 1972 1. Imre Z. Ruzsa 2. Zsigmond Nagy 3. LAszl6 Babai, Peter Frankl, Janos Pintz 1973 1. LAszl6 Babai, Peter Komjath 2. Janos Pintz 3. Peter Frankl, Zsigmond Nagy, Imre Z. Ruzsa H. Ervin Bajm6czi, Zoltan Balogh, J6zsef Beck, Ervin Gyori, Emil Kiss, Tamas M6ri, Zsolt Tuza, Endre Boros, LAszl6 Lempert, Janos Revizcky
1974 1. LAszl6 Lempert, Imre Z. Ruzsa 2. Peter Frankl 3. Ervin Gyori 1975 1. Imre Z. Ruzsa 2. Peter Komjath, LAszl6 Lempert, Vilmos Totik 3. Zoltan Fiiredi, Gabor Somorjai H. Ervin Bajm6czi, Ervin Gyori, Tamas M6ri, Peter Pal Palfy, Maria Szendrei
1976 1. Imre Z. Ruzsa 2. Peter Komjath
3. H. Vilmos Totik, Ervin Bajm6czi, Zoltan Furedi, Mihaly Gereb, Ervin Gyori, Emil Kiss, Janos Kollar 1977 1. Ferenc Gondocs, Vilmos Totik 2. Zoltan Fiiredi 3. Gabor Czedli, Peter Pal P61fy H. Tamas Bara, Vilmos Komornik, Andras Sebo, Nandor Simanyi, Sandor Veres 1978 1. Vilmos Totik 2. Zoltan Furedi, Janos Kollar, Nandor Simanyi 3. Emil Kiss H. Tibor Krisztin, Zoltan Magyar
2. RESULTS OF THE CONTESTS
52
1979 1. Janos Kollar, Peter Pal Palfy
2. 3. Mihaly Gereb, Gabor Ivanyos, Akos Seress H. Emil Kiss, Tibor Krisztin, Zsolt Pales, Nandor Simanyi 1980 1. Janos Kollar 2. Akos Seress 3. Nandor Simanyi H. Gabor Ivanyos, Zoltan Magyar 1981 1. Gabor Ivanyos, Zoltan Magyar 2. Balazs Csikos, Akos Seress 3. Peter Hajnal, Dezso Miklos, Mario Szegedy H. Gabor Moussong, Zalan Bodo, Andras Zempleni 1982 1. Zoltan Magyar, Gabor Tardos 2. Akos Seress 3. Balazs Csikos, Andras Szenes H. Peter Hajnal, Dezso Miklos, Mario Szegedy, Andras Zempleni 1983 1. Zoltan Magyar 2. Balazs Csikos, Gabor Tardos 3. Mario Szegedy, Jozsef Varga H. Peter Hajnal, Zoltan Buczolich, Andras Szenes, Akos Seress 1984 1. Gabor Tardos 2. Mario Szegedy 3. Zoltan Buczolich H. Balazs Csikos, Andras Szenes 1985 1. Gabor Tardos 2. Andras Szenes 3. Geza Bohus, Gabor Elek, Gyula Karolyi H. Ferenc Beleznay, Laszlo Erdos, Miklos Mocsy, Tibor Odor, Endre Szabo, Laszlo Szabo, Zoltan Szabo 1986 1. Gabor Tardos 2. Gabor Elek, Endre Szabo, Zoltan Szabo 3. Laszlo Erdos, Jeno Torocsik H. Gyula Karolyi, Akos Magyar 1987 1. Akos Magyar 2. Istvan Sigray 3. Gyula Karolyi H. Laszlo Erdos, Gabor Hetyei, Sandor Kovacs, Jen6 Torocsik
2. RESULTS OF THE CONTESTS
53
1988 1. Geza K6s, Zoltan Szab6 2. Laszl6 Erd6s 3. Istvan Sigray, Jend T6rdcsik
H. Sandor Kovacs, Akos Magyar, Andras Benczur, Tamas Keleti, Mikl6s M6csy
1989 l.- Lasz16 Erdds
2. Sandor Kovacs
3. H. Andras Benczur, Gydrgy Birkas, Andras Biro, Gabor Drasny, Geza Kos, Akos Magyar, L6szl6 Majoros, Miklos Mocsy, Tibor Szab6, Zoltan Szab6 1990 1. Andras Benczur 2. Gabor Drasny 3. Andras Bir6 H. Tam" Hausel, Geza Makay 1991 1. Andras Bir6
2. 3. Tamas Fleiner, Geza K6s H. Matyas Domokos, Gabor Hajdu, Gergely Harcos, Tamas Keleti, Vu Ha Van
3. Solutions to the Problems 3.1 ALGEBRA Problem A.1. Determine the roots of unity in the field of p-adic numbers.
Solution. The p-adic number a_,,,,p-' + + a_lp-1 + ao + alp + + (0 < ai < p) is a p-adic integer if all the coefficients with a,,.p' + negative index are equal to 0, and it is a unit in the ring of p-adic integers
if furthermore ao # 0. It is clear that every p-adic number a can be written in the form a = 3p', where 3 is a p-adic unit and r an integer. Since the product of p-adic units is again a p-adic unit, we deduce that a p-adic number can be a root of unity only in the case when it is a p-adic unit. Every root of unity can be written in the form e = a + ap', where r is a positive integer, a is a p-adic unit, and 0 < a < p. Now if e" = 1, then a' - 1 (mod p) holds true. So the exponent of a is a divisor of n. Consider the case p # 2. Let exp(a) = k; we are going to show k = n. Let 3 = ek, and suppose Q # 1. Then 3 is of the form ,3 = 1 + ryp', where -y is a p-adic unit. If k 54 n were the case, then some power of Q would be 1. To show the impossibility of this, it is enough to show that a power with prime exponent of a number of the form Q = 1 + ryp' (-y a p-adic unit) cannot be 1. Let q be a prime number; using the binomial theorem, we get Oq = 1 + q yp' + Sp23 = 1 + apps
if q 54 p;
while in the case q = p, using p > 2 we get Op = 1 + 7p9+1 + bp23+1 = 1 + cpp'+1 where -y is a p-adic integer, whereas cp, as we can easily see, is a p-adic unit.
This guarantees that 311 is different from 1. We have proved that in the field of p-adic numbers every root of unity is necessarily a (p - 1)th root of unity. Now we show that such roots of unity indeed exist. Let go be a primitive root of unity mod p. We are going to prove that it is possible to determine numbers 0 S ai < p in such a way that the p-adic number 6=go+alp+...+a,,,p"+...
55
3. SOLUTIONS TO THE PROBLEMS
56
should be a (p - 1)th primitive root of unity. It is clear that any power of e with exponent less than p - 1 is different from 1. So it is enough to show that for a suitable choice of the numbers aj, the natural numbers gi = go+alp+ +ajpj satisfy gP-1 EE 1 (mod pi+1) This will be proved by induction. For j = 0 the statement is true, as go is a primitive root. Suppose that the statement holds for some j, that is, gP-1 = 1 + cep3+1
Let aj+1 be the solution of the congruence gox - cj (mod p). Then
1P-1 +1 +1 g3+1 = (gi + aj+17jl )P- = 1 + cjp' 1
1 + (ca -
- gj aj+17 +1
(mod pj+2).
Therefore a and its powers are different (p - 1)th roots of unity; their number is p - 1. There can be no other roots of unity since the polynomial xp-1 - 1 can have at most p - 1 roots in a commutative field. Let us now deal with the case p = 2. Similarly to the case of odd prime
numbers, we can prove that a power with an odd exponent of a dyadic number of the form 1 + can be 1 only in case the number itself is 1. So if a dyadic number is a primitive n-th root of unity then n can have no odd prime divisors. There are two second roots of unity, 1 and the number as well. There can be no more second roots -1 = 1 + 2 + 22 + .. 2'' +
of unity, because the polynomial x2 - 1 can have at most two roots in a commutative field.
We show that there are no other roots of unity in the field of dyadic numbers. Suppose there were another root of unity; then this would be necessarily a primitive root of unity belonging to some power of 2, so there would surely exist a primitive further root of unity, that is, a dyadic number g with 772 = -1. Clearly, g must have the form 77 = 1 + 2' + where r > 1. This gives
-1 - (1 + 2r)2 - 1 + 2''+1 +
22r
(mod 2r+2)
Therefore, -2 =- 2' +1 +2 2''
(mod 4),
that is, 2' + 22' -1
(mod 2),
which is clearly impossible. This concludes the proof.
Problem A.2. Let A and B be two Abelian groups, and define the sum of two homomorphisms q and x from A to B by a(r1 + x) = ar7 + ax
for all a E A.
With this addition, the set of homomorphisms from A to B forms an Abelian group H. Suppose now that A is a p-group (p a prime number).
3.1 ALGEBRA
57
Prove that in this case H becomes a topological group under the topology defined by taking the subgroups pkH (k = 1, 2, ...) as a neighborhood base of 0. Prove that H is complete in this topology and that every connected component of H consists of a single element. When is H compact in this topology?
Solution. H is clearly a commutative group whose 0 element is the homomorphism mapping every element of A to 0 E B, and for i E H, -77 is the mapping defined by a(-77) = -a77. To prove that H is a topological group, we have to check the following: (i) The intersection of the neighborhoods of 0 is 0 alone. Suppose namely
that i E pkH(k = 1, 2.... ). As A is a p-group, the order of any a E A is of the form p" for some n depending on a. Because 77 E p'H, we have 77 = p'X for some X E H, and so a77 = a(p"X) = (p"a)X = OX = 0, proving 77 = 0.
(ii) To any neighborhood U of 0 there is a neighborhood V of 0 such that V + (-V) C U, a suitable choice being V = U. (iii) If a is contained in some neighborhood U of 0, then there exists a neighborhood V of 0 such that a + V C U. Again the choice V = U will do.
(iv) The intersection of any two neighborhoods of 0 contains a neighborhood of 0, because for any two neighborhoods, one of them contains the other. We now come to the proof of completeness. We have to prove that if 771, 772, ... , ?In.... is a Cauchy sequence, then it converges to some element of H. By repeating members of the sequence, if necessary, we can assume that 71i+1 - 77i c p'H, that is, 77i+1 = 77i +p119i. Furthermore, define 770 as 0.
Now define mappings Xi (i = 0, 1, 2, ...) in the following way: If the order of a E A is pk then let aXi = a(19i + pt9i+1 + ...+pk-179i+k-1)
It is easy to check that Xi is indeed a homomorphism from A to B and Xi = 19i + pXi+1
We will show that the limit of the sequence 770, 771, ... , 77,,, ... is Xo. For this it is enough to show Xo - 71i E piH, actually Xo - 71i = p'Xi, which we prove by induction. For i = 0, we have Xo - 770 = Xo - 0 = pOXo. Using our previous observations and the induction hypothesis, we have
Xo - 77i+1 = Xo - 77i - pit9i = pZXi - pil9i = pl(Xi - 7i) =
proving the completeness of H.
pi+1Xi+1,
3. SOLUTIONS TO THE PROBLEMS
58
We now prove that the connected component of 0 consists of 0 alone; to prove this, it is enough to show that the intersection of all open-closed sets containing 0 is 0. As p k H is an open subgroup, it is closed as well, and as the intersection of all of them (for k = 1, 2, ...) is already 0 alone, we get the desired result. For compactness we are going to prove the following result: H is compact if and only if the index of pkH in H is finite for every k. Necessity of this condition is easy to establish: the cosets X + pkH(X E H) cover H. If H is compact, then already a finite number of them have to cover H, which is precisely what the condition says. Suppose now that all the subgroups pkH have finite index. We are going to show that if a family of closed subsets of H has the property that any finite number of them have a nonempty intersection, then the whole family has a nonempty intersection. First, we show the following: let UA(A E A) be subsets of the set H such that the intersection of any finite number of them is nonempty.
Suppose that H is the disjoint union of two subsets S and T. Then at least one of the families s n UA(A E A) and T n UA(A E A) inherits the same intersection property. (In particular, either S or T meets all the UA.) Suppose that neither family inherited the property. Then we would have values )\1 i .'2, ... , .\,. and )r+1, )tr+2, , \r+9 such that the intersection of the s n Ua; and also of the T n Ua,,+j would be empty (i = 1, 2, ... , r, j = 1, 2, ... , s). In other words, the sets
tnUa,) n S and (uA+) n T are empty, or equivalently
fl Ua; C T and
(UA+.) C S.
Consequently,
/ c T n s= O,
UA: I (rn+ x=1
a contradiction. We then get the same type of statement for any decomposition of H into a finite number of pairwise disjoint subsets. Let us now consider closed subsets U,\ (A E A) of our topological group H
satisfying the finite intersection condition. Repeatedly applying the above procedure, we see that there exists a sequence 711, 772.... , 77k.... of elements of H for which
,q1+pH
772+p2H2...
1)k+pkH
...
(1)
and such that for any k the intersection of the UA(A E A) with 77k + pkH also satisfies the finite intersection condition; in particular, any UA and r!k +pkH have a common element Xk,A
3.1 ALGEBRA
59
In view of (1) and the completeness of H, the sequence il, iJ2, ... , rik, ... has a limit q for which 7ik + pkH = ri + pkH. For any A we have Xk,a E ri +P k H; thus the limit of the sequence Xl,a, X2,a, . , Xk,a, . is r). Since U. is closed and Xk,A E Ua, we get that r) is contained in every Ua, which completes the proof of the compactness of H.
Problem A.3. Let R = Rl ® R2 be the direct sum of the rings R1 and R2, and let N2 be the annihilator ideal of R2 (in R2). Prove that R1 will be an ideal in every ring R containing R as an ideal if and only if the only homomorphism from R1 to N2 is the zero homomorphism.
Solution. First suppose there is a ring R containing R as an ideal and such that Rl is not an ideal of R. Since R is an ideal in R, we have Rlr C R
and rR1 C_ R for every r E R. Suppose for example, that rR1 C Rl for some f E R. (The case Rlr C Rl can be treated similarly.) Using the element r we will define a nonzero homomorphism from R1 to N2. The direct sum property implies that for every rl E R1 the element rr1 E R can be uniquely decomposed in the form rr1 = g(rl) +h(r1), where g(rl) E Rl and h(ri) E R2. h will be the desired homomorphism. It is clearly defined for all elements of R1. Furthermore, (i) h(ri) E N2 for every r1 E R1. Namely, for every r2 E R2 we have h(rl)r2 = g(rl)r2 + h(rl)r2 = (g(r1) + h(ri))r2 = (rrl)r2 = r(rlr2) = 0
and
r2h(rl) = r2(frl) = (r2r)rl E RR, C R1, but on the other hand, r2h(rl) E R2, proving r2h(rl) = 0, which gives (i).
(ii) h(ri +r'1) = h(ri) +h(ri) for every rl,ri E R1. Since r(r1 + ri) = g(r1 + ri) + h(ri + ri) = rr1 + rr'
= g(rl) + h(ri) + g(ri) + h(ri) = g(r1) + g(ri) + h(r1) + h(ri), where g(rl)+g(ri) E R1 and h(r1)+h(ri) E R2, we get the validity of (ii).
(iii) h(rir') = h(rl)h(ri) for every rl,ri c R1. We will show that both sides of the equality are equal to zero. For the right-hand side, this is clear because h(ri) is an annihilator of R2 and h(ri) E R2. For the left-hand side, we have to consider the product r(rlri): r(rlri) = (rrl)ri = (g(rl) + h(r1))ri = g(rl)ri E R1,
so h(rlri) = 0. This establishes the first part of the statement. For the reverse statement, we use the same idea. We assume that there is a nonzero homomorphism h from Rl to N2, and we construct the element r in such a way that it should produce h.
3. SOLUTIONS TO THE PROBLEMS
60
We denote by Ro the zero ring over the additive group of the integers.
The element of R0 corresponding to n E Z will be denoted by n. The additive group of the ring R is defined by R+ = Ro ®Ri ® R2 , and for the elements of R
r=n+rl+ r2,
r'=n'+r'1+ r2,
we define the multiplication by rr' = r1r1 + r2r2 + nh(ri) + n'h(ri).
It is routine to check that R is a ring and R is a subring of it. In fact, R is even an ideal of R in view of RR C R. To verify that R1 is not an ideal of R, take an r1 E R1 with h(ri) # 0.
Then 1r1 = h(rl) V R1 as h(ri) E R2, and this shows that R1 is not an ideal of R.
Problem A.4. Call a polynomial positive reducible if it can be written as a product of two nonconstant polynomials with positive real coefficients. Let f (x) be a polynomial with f (0) # 0 such that f (x') is positive reducible for some natural number n. Prove that f (x) itself is positive reducible.
Solution. Consider the product representation of f (xn):
.f (x") = flgj (x),
(1)
j=1
where the gj (x) are polynomials with positive coefficients. Suppose that some of the polynomials gj(x) contain a term cjkxk with nonvanishing cjk such that n f k. Then the product on the right-hand side of (1) contains the term cjk fli#j gi(0)xk. The coefficient of this term does not vanish, since r['=1 gi (0) = f (0) # 0. Since the polynomials gj (x) all have positive coefficients, the right-hand side of (1) will contain the term xk with a nonvanishing coefficient, which is a contradiction in view of n { k. This means that all the polynomials gj (x) have the form I; gj(x) _ >bjlxmm
(bjl >, 0,l = 0,l,...,lj),
i=O
that is, gj (x) is actually a polynomial of xn: gj (x) = gj (x"). So upon substituting x" = y, we get a
f (y) = f9j (y), j=1
proving the statement of the problem.
3.1 ALGEBRA
61
Problem A.5. Prove that the intersection of all maximal left ideals of a ring is a (two-sided) ideal.
Solution. Let us denote by B the intersection of all maximal left ideals of the ring R (if there are no maximal left ideals, the statement is void). B is clearly a left ideal, so we only have to prove that b E B and r E R imply br E B, or equivalently br V B implies b V B. Since an element is not in B if and only if there is a maximal left ideal not containing it, we actually have to prove the following statement: If for some elements b E B, r E R there is a maximal left ideal M with br V M, then there exists a maximal left ideal N with b V N. Let
N= {xeRI xrEM}. We are going to prove that N has the desired properties. 1. N is a left ideal: if s E R and x E N, then xr E M implies sxr E M (M being a left ideal), which in turn implies sx E N. Also x E N, y E N clearly imply x - y E N.
2. bVNasbrVM. 3. N is maximal. To prove this, we have to show that for any element a of R with a N the only left ideal of R containing both a and N is R itself. We have ar V M since a V N. As M was maximal, every element of R is contained in the left ideal generated by ar and M; in particular, for every c E R, cr can be written in the form
cr = yar + nar + in, where y E R, m E M, and n is an integer. This implies
(c-ya-na)r=inEM, so
d = c - ya - na E N, but this shows that c = ya+na+d is contained in the left ideal generated by a and N, which was to be proved.
Remark. If R has an identity element, then B is of course the Jacobson radical.
Problem A.6. Let R be a finite commutative ring. Prove that R has a multiplicative identity element (1) if and only if the annihilator of R is 0 (that is, aR = 0, a E R imply a = 0).
Solution 1. If R has an identity element e, then aR = 0 (a E R) implies 0 = ae = a, so the annihilator of R is indeed 0.
3. SOLUTIONS TO THE PROBLEMS
62
Conversely, suppose R is a finite commutative ring whose annihilator is
This implies that to any element a of R different from 0 there is an element b of R such that ab # 0. If R = (0), then R surely has an identity element. So suppose R :A (0) and ao is an arbitrary element of R different from 0. Then the remark made above shows that for any natural number n we can find an element an E R such that the relations 0.
aoai # 0, aoala2 # 0,..., aoai...an# 0,...
hold true. Since R is finite, we have numbers m and n (0 < m < n) such that anal ... an = (aoal ... am)(am+i ... an) . (1)
This means that the set E of elements e # 0 of the ring R for which there exists a 0# d E R such that de = d
(2)
is nonempty. Choose an element e from E so that the number of elements d E R corresponding to e that satisfy (2) should be maximal. (Such 0 an e exists because R is finite.) We will show that e is an identity element of R.
Suppose, on the contrary, that for some a E R we have ae - a # 0. Then in view of (1) there are elements r, s E R such that 0 # (ae - a)r = (ae - a)rs
(3)
holds. (We allow r to be an empty product.) Equation (3) implies
ar(e - es + s) = ar #0
(4)
consequently, 0 # e - es + s E E E. If de = d, then
d(e-es+s)=de-des+ds=d-ds+ds=d;
(5)
are - ar = (ae - a)r # 0.
(6)
furthermore,
Equations (4), (5) and (6) together show that the element e - es + s E E satisfies condition (2) for more d E R than e. This contradiction shows the existence of an identity element in R.
Solution 2. Suppose the commutative ring R with n(,> 2) elements has annihilator 0. Then R cannot be nilpotent. If R were nilpotent, there
would exist an integer k >, 2 such that Rk = 0, Rk-1 # 0 would be satisfied, but this would imply that Rk-1 is some nonzero annihilator of R, a contradiction.
3.1 ALGEBRA
63
Next, we show that there is an element a of R such that no power of a is equal to 0. Suppose that to every element ai (i = 1, 2, ... , n) of R there exists a natural number li with ai' = 0. Let l be the maximum of the li (i = 1, 2, ... , n). If a product in R does not vanish, then every ai can occur in it at most (1 - 1) times. So every product with n(l - 1) + 1 factors Rn(,-1)+1 = 0, contrary to the nonnilpotency vanishes; in other words, of R.
So suppose a E R is such that all the elements aj (j = 1, 2.... ) are nonzero. Since R is finite, these powers cannot be all different, say ak = ak+l (1 > 0). This implies ak = ak+1 = ak+2i = ak+3i = ... , Choosing m big enough to satisfy ml > k, we have (aml)2 = ami . aml = ak+ml . aml-k = ak . aml-k = aml v
so aml is a nonzero idempotent of R. Let us write aml = e for simplicity. We can write R as a direct sum of the ideals A and B: R = A E B, (7) where A is the set of all elements of the form re (r E R) and B is the set
of all elements of the form r - re (r E R). Clearly, A and B are ideals of R. Any element r of R can be written in the form r = re + (r - re), thus R = A + B. Furthermore, ea = a for all a in A, whereas eb = 0 for all b in B. This implies A n B = 0, so (7) holds indeed. If B = 0 then R = A, so e is the desired identity element of R. If B # 0, then in view of 0 0 e E A, R is a direct sum of two rings, each of which contains fewer than n elements. Consider this case and suppose by induction that all commutative rings with fewer than n elements and with annihilator 0 have an identity element. (The one-element ring clearly has an identity element.) If z E A is such that
zA = 0, then in view of AB = 0 we have zR = z(A + B) = zA + zB = 0, implying z = 0. Similarly, from w E B, wB = 0 we get w = 0. This means that the annihilators of both A and B are 0, so by induction A has an identity element e1 and B has an identity element e2. But then e1 + e2 is an identity element of the ring R = A ® B. Remark. The statement of the problem is true for commutative Artinian rings, too (see R. Baer, Inverses and zerodivisors, Bull. Amer. Math. Soc., 48 (1942), 630-638).
Problem A.7. Let I be an ideal of the ring of all polynomials with integer coefficients such that (a) the elements of I do not have a common divisor of degree greater
than 0, and (b) I contains a polynomial with constant term 1. Prove that I contains the polynomial 1 + x + x2 + natural number r.
+ x''-1 for some
Solution. According to assumption (b), for a suitable f E Z[x] we have
1+xf EI.
3. SOLUTIONS TO THE PROBLEMS
64 Because of
(1+xf)xr+xr+l(_f) = xr
for any r > 0, we have xr E
(l+xf,xr+1).
Repeatedly applying this observation, we get (1+xf,xr+1) D (1+x f xr)
...
(1+xf,x)
1,
consequently,
l,x,...,xr E (1
+xf,xr+1),
which in turn means that we can find gr, hr E Z[x] for which 1 + x + ... + xr = (1 + xf)gr + xr+lhr
(1)
holds true. We claim that gr and hr can be chosen in such a way that
deg hr < deg f
(2)
holds. To prove this, observe that the polynomial gr figuring in (1) can be written as gr = xr+lq + p
where p, q c Z[x] and deg p < r. Rom (1) we get
1+x+...+xr = (1+xf)p+xr+1[(1+x f)q+hr], and choosing p (resp., (1 + x f )q + hr) as the new gr (resp., hr) we clearly get an hr satisfying (2) because of degp < r. Suppose s > r and
(1+xf)gs+xs+lh9, where deg hs < deg f. Subtracting from (3) xs-r times (1), we get
1 + x + ... + xs-r-1 = (1 + x.f) (g9 - grxs-r) + xs+l (hs - hr). This means that if there are indices s > r such that
hs - hrEI holds, then 1+x+...+x9-r-1
is also true.
EI
(3)
3.1 ALGEBRA
65
We prove that the ideal contains a nonzero constant polynomial. Take a p E I such that p # 0 and degp is minimal. (Such a p exists in view of
1+xf 54 0.) For any gellet
q=pu+v, where u, v E Q [x] and deg v < degp. Multiplying by a suitable nonzero integer a, we get aq = put + vi, where ul, vi E Z[x] and of course deg vi < degp still. This gives
v1 = aq - pul E I,
so by the choice of p and in view of deg vl < degp, we have v1 = 0. This means that to an arbitrary q E I we can find a u E Q[x] for which q = pu. Let p = opl, where pl is a primitive polynomial and cp E Z. Then q = pl(pu), which in view of Gauss's lemma implies Wu E Z[x]. This in turn implies that for any q E I we have pi q. In view of condition (a), this is only possible if pl = 1; consequently cp E I, thus (cp) C I. Finally, we show that there are values s > r such that I
h9 - hr E (p).
Indeed, any coefficient of a polynomial can take on at most cp values mod W; furthermore, deg hr <, deg f (r = 1, 2, ... ), so there exist values r < s( ,deg f+1 + 1) for which all coefficients of hs - hr are divisible by W. This proves the statement.
Remark. The following, more general, statement can be proved: Let R be a unique factorization domain whose proper factor rings are all finite. Let I be an ideal of the polynomial ring R[x] satisfying the following two conditions: (a) The elements of I do not have a common divisor of degree greater than 0. (b) I contains a polynomial with constant term 1.
Then to every natural number N there exists a natural number r > N for which N I r + 1 and E I.
Problem A.8. Prove that in a Euclidean ring R the quotient and remainder are always uniquely determined if and only if R is a polynomial ring over some field and the value of the norm is a strictly monotone function of the degree of the polynomial. (To be precise, there are two more trivial cases: R can also be a field or the null ring.)
3. SOLUTIONS TO THE PROBLEMS
66
Solution. We shall work with the following definition of a Euclidean ring: let R be a ring and N the set of nonnegative integers. R is called a Euclidean ring if there exists a map cp : R -+ N having the following properties: i)
ii) For any a, b E R (b 54 0) there exist q, r E R such that
a=bq+r and
W(r)
iii) The unicity of the quotient and remainder means that if a = bq + r = bql + r1 and V(r) < cp(b), W(r1) < cp(b), then r = r1 and q = q1. With these definitions, we will prove that if R is a (commutative) Euclidean ring whose quotient and remainder are unique, then R is the null ring (which consists of the single element 0), a (commutative) field, or a polynomial ring over a (commutative) field. Note that although commutativity of R is usually included in the definition of a Euclidean ring, we did not assume it, and in the following 16-step proof the first 15 steps never use the commutativity of R. In the remarks, following the proof we shall devote some space to the noncommutative case. Before embarking on the proof, let us make a simplification. If the values
actually taken on by cp are 0 = no < nl <
< nk < .... then instead
of the value W(a) = n,k take the value cp'(a) = k. cp' is equivalent to cp in the sense that W(a) < W(b) if and only if cp'(a) < cp'(b) and co' also satisfies conditions i), ii), and iii). So from now on we can assume nk = k. And now to the proof. 1. R has no zerodivisors. ab+0 If ab = 0, a # 0, b # 0 held, then the decomposition 0 = would contradict iii). 2. If c 0 0, then cp(ac) > cp(a). Suppose W(ac) < cp(a), then ac = ac + 0 = a 0 + ac would contradict iii).
Introduce the following notation:
T i={a: aER, cp(a),
C Tk C ...
and
R = U Ti. i-o
3. If R = To, then R is the null ring. This is clear. In the following, we assume R # To.
4. R has an identity 1 and 1 E T1.
Since R # To, there is an a # 0 with a E T1. If a = aq + r, we have W(r) < W(a) = 1, so r = 0, a = aq. Multiplying by some b and using step 1, we get ab = aqb, b = qb, so q is a left-sided identity. Proceeding in the same way with an arbitrary c: cb = cqb, c = cq, so q is a two-sided identity; let us denote it by 1. If W(a) < W(1), then a = 1 a + 0 = 1 - 0 + a, so by unicity a = 0; consequently cp(1) = 1.
3.1 ALGEBRA
67
5. W(a - b) < max{cp(a), cp(b)}. Suppose max{co(a), cp(b)} < c,(a - b).
Then a = (a - b) 1 + b =
(a - b) 0 + a, together with 1 # 0, contradicts iii). As a consequence, we have
6. Ti is an additive group. 7. w(-a) = cp(a). w(-a) = co(0 - a) < max{cp(0), p(a)} = cp(a). In a similar way, we have W(a) <, w(-a), so W(a) = w(-a). 8. If W(b) < cp(a), then co(a - b) = cp(a). Step 5 gives cp(a-b) <, cp(a). On the other hand, W(a) = cp(b-(b-a)) max{cp(b), cp(b - a)} = max{cp(b), W(a - b)}, so because W(b) < W(a) we have w(a) <, cp(a - b), and thus W(a - b) = cp(a).
9. An element a of R is a (two-sided) unit if and only if W(a) = 1 (equivalently, a E Ti and a # 0). Let W(a) = 1, and let b be an arbitrary element of R. We have b = aq + r with W(r) < W(a) = 1, so r = 0, b = aq. If, in particular, b = 1, then 1 = aal, which also implies a1 = alaal, 1 = ala, and thus a1 = a-1. Conversely, if ab = 1, then step 1 gives a # 0, b # 0 and step 2 gives W(a) 5 cp(ab) = cp(1) = 1, thus W(a) = 1. Steps 6 and 9 together imply 10. Tl is a (skew) field. If R = T1, then we are again finished (and we see that actually every skew field - not only the commutative ones - satisfies i), ii), and iii) with W(a) = 1, whenever a# 0). Suppose now that R T1. We will prove that R is the polynomial ring over T1. 11. If co(b) = 1, then cp(ab) = co(a). Step 2 implies namely co(ab) > W(a) = cp((ab) b-1) >, co(ab), so cp(ab) = cp(a).
The next statement is the converse of this. 12. If for some a 54 0 cp(ab) = cp(a), then W(b) = 1.
Let a = (ab)q+r, where co(r) < cp(ab). Then r = a(l-bq). If co(b) # 1, then 1 - bq # 0, so in view of step 2, W(r) > cp(a), so W(a) < cp(ab), a contradiction. 13. If W(x) = 2, then cp(xk) = k + 1. We prove the statement by induction on k. For k = 1 the statement is clearly true. Suppose that cp(x1c-1) = k. Then, using steps 2 and 12, we have co(xk) = cp(xk-1.x) > cp(xk-1), so co(rk) 3 k+1. On the other hand, let W(a) = k+ 1 and a = xk-1 b+r where cp(r) < cp(xk-1) = k.
Then co(xk-1b) = W(a - r) = W(a) = k + 1 > k = cp(xk-1). Using step 12, we get b ¢ Ti. So if b = xc + s with W(s) < W(x) = 2, then co(xc) = co(b - s) = W(b) because W(b) >, 2 > cp(s). Thus co(xc) >, 2, and consequently c # 0. By substitution, we get a = xkc + xk-1 s + r. Here W(a) = k+1; in case W(s) = 1, we have cp(xk-1s) = co(xk-1) = k; in case W(s) = 0, that is, s = 0, we have cp(xk-1 s) = 0, finally, we know W(r) < k. All in all W(a) > max{cp(xk-1s), cp(s)}, so using step
8, (p(xk) , co(xkc) = cp(a - xk-ls - r) = W(a) = k + 1. 14. If W(a) = k + 1, then a can be written in the form a = ao + xa1 +
3. SOLUTIONS TO THE PROBLEMS
68
...+X k ak, where ai E T l (i = 0,1, ... , k) and ak # O.
If k = 0, the statement is clear. Suppose it is true for k - 1. For k > 0, let a = xkb + r, where W(r) < cp(xk) = k + 1, so cp(xkb) = cp(a - r) = W(a) = k + 1 = cp(xk), thus W(b) = 1. This means b E T1, b # 0. Because W(r) < k, we have r = xk-lak_1 + +xal +a0 with ai E Ti. 15. x is transcendental over T1. Suppose xkak + + xal + ao = 0 (ai E T1). Then in view of cp(xkak) = cp(xk-lak_1 + + xal + ao) < k, ak = 0 follows. By repetition of the argument, we get ak-1 = 0, ... , al = 0, ao = 0. 16. If R is commutative, then R = Tl [x], and the Euclidean value of the norm is a strictly monotone function of the degree of the polynomial. This follows from steps 10, 14, and 15. Finally, we remark that in a polynomial ring over a commutative field quotient and remainder are indeed unique - the only possible quotient and remainder are the ones that we get using the usual division algorithm. Remarks.
1. The noncommutative case is tricky. The point is that although as a set R is the same as Tl [x], R is not actually isomorphic to Tl [x] equipped with the usual multiplication. If we define the product of two polynomials in the usual way, that is, the product of two polynomials n
m
E xiai and > xibi i=0
i=0
is defined as the polynomial
where ci is defined by the equation ci = aobi + albi-1 + ... + aibo
= b_1 = b_2 = . . . = 0), the statement of the (with a_1 = a_2 = problem does not remain true. But by suitably defining the product of two polynomials, the statement remains true. For arbitrary a E R there exist a° and aT which satisfy ax = xaa + aT
where
p(aT) < W(x) = 2,
so aT E T1. Clearly,
(a + b)a = a° + b° and (a + b)T = aT + bT . Furthermore, (ab)' = a°b° and (ab)T = aTbT.
3.1 ALGEBRA
69
So a - a° is an endomorphism and a : T1 -+ Ti is an automorphism, because for a E T1 (a
0) there exists an inverse of a, and since 1x = x1
implies 10' = 1 (so Ti # 0), we get 1 =
(aa_1)Q
= a°(a-1)°, which
shows a° E T1 and T1 being a field, each endomorphism of it is actually
an automorphism. If we now take a suitable automorphism a -* a° of T1 and a suitable derivation a -- aT (with respect to v; by this we understand a map of T1 into itself, satisfying (a + b)T = aT + bT and (ab)T = aT b° + abr) then, defining in the polynomial ring the product of two polynomials by ax = xa° + aT, we can see that T1 [x],,,. has the required properties. 2. In the formulation of the problem the words "strictly monotone" function of the degree are not superfluous. If, for instance, we have polynomials f and g with deg f = deg g, but cp(f) > cp(g), say, then R is still Euclidean, but with a suitable constant a we have deg (a f + g) = deg f
and a f +g= f q + r, where deg r < deg f , but also of +g= f a + g where co(g) < cp(f ), so in view of r g the remainder is unique. (We have in mind a co with V(r) < co(f ), of course.)
Problem A.9. Let f (x) = ao + a1x + a2x2 + alox1o + al1xu + a12x12 + a13x13
(a13 $ 0)
and g(x) = bo + b1x + b2x2 + b3x3 + bl1xll + b12x12 + b13x13
(b3 54 0)
be polynomials over the same field. Prove that the degree of their greatest common divisor is at most 6.
Solution. Let fi(x) = ao + a1x +.a2x2,
f2(x) = alo + a11x + a12x2 +
91(x) = b0 + blx + b2x2 + b3x3,
a13x3
92W = blo + bllx + b12x2 +
b13x3
(the problem says blo = 0, but we will not use this). This implies f(x) = fl (x) + x10f2(x),
9(x) = 91(x) + x1092(x),
so
f(x)92(x) - 9(x)f2(x) = fl(x)92(x) - f2(X)91(X)-
The right-hand side is divisible by the greatest common divisor of f (x) and g(x), since the left-hand side is divisible by it. But the polynomial on the right-hand side has degree 6, because f t (x)g2 (x) has degree at most 5 but f2 (x)gl (x) has degree precisely equal to 6 since a13 b3 $ 0. So the greatest common divisor of f (x) and g(x) divides a polynomial of degree 6; therefore its degree can be at most 6.
3. SOLUTIONS TO THE PROBLEMS
70
Remarks.
1. The upper bound of 6 for the degree of the greatest common divisor cannot be improved as shown by the polynomials x10 +x13 and X3+xl2, whose greatest common divisor is x3 + x6. 2. It is possible to prove in a similar manner the following generalization:
if f (x) and g(x) are polynomials with coefficients in the same field, grad f = n, grad g <, n, and g(x) has a term of degree k, but the terms of degree k + 1, ... , k + r - 1 are missing from both polynomials (k + r 5 n), then the greatest common divisor of f and g has degree at
most n - r. Problem A.10. Let K be a subset of a group G that is not a union of left cosets of a proper subgroup. Prove that if G is a torsion group or if K is a finite set, then the subset
k-1K I
I
kEK
consists of the identity alone.
Solution. Suppose there is an element a E G, different from the identity element such that a E f1 k-1K. Then for every k E K we have ka E K, kEK that is, Ka C K. First we observe that if we have equality here, then K is the union of some left cosets of the subgroup H generated by a. Indeed, Ka = K
implies Kai = K for all integers i, so KH = K, which in turn means K = UkEKkH. Now we only have to prove that in the cases mentioned in the problem we indeed have Ka = K. This is clear if K is finite since Ka has the same
cardinality as K and is a subset of it. On the other hand, if a has finite order n, then K D Ka implies
KDKaIKa2D...DKa"=K, again proving Ka = K. Remark. If we assume that G is an Abelian group, then the assumption that G is a torsion group can be replaced by the weaker assumption that every element of K has finite order.
Problem A.11. Prove that if an infinite, noncommutative group G contains a proper normal subgroup with a commutative factor group, then G also contains an infinite proper normal subgroup.
Solution. First we prove a lemma:
3.1 ALGEBRA
71
Lemma. If the infinite group G has a proper normal subgroup N satisfying N Z Z(G) (Z(G) denotes the centre of the group), then G has an infinite proper normal subgroup.
Proof. To prove the lemma, we can assume that N is finite. For every g E G, the mapping coy
: n F--* g 1ng
(n EN)
is clearly an automorphism of N. Furthermore, the mapping g --> cog is a homomorphism of G to some subgroup of the full group of automorphisms of N, and 4) is clearly finite in view of the finiteness of N. Because N Z Z(G), 4) contains nonidentity automorphisms so the kernel F of the homomorphism g H cog is a proper normal subgroup of G that is infinite in view of G/F _- 4D.
Now let H be a proper normal subgroup of the infinite, noncommutative
group G such that G/H is commutative. If H Z Z(G), then we can apply the lemma. If, on the other hand, H C Z(G) and h is any element of G not contained in Z(G), then using the commutativity of G/H we see that (H, h) is a normal subgroup of G. Clearly, (H, h) Z Z(G); on the other hand, (H, h) # G since G is noncommutative. Applying the lemma with N = (H, h) completes the proof. Remark. The condition of noncommutativity cannot be dropped from the statement of the problem, as shown by the group Zp-. Conversely, the alternating group of countably infinite degree shows that the condition of the existence of a proper normal subgroup with commutative factor group cannot be dropped either.
Problem A.12. Let a1, a2i ... , aN be positive real numbers whose sum equals 1. For a natural number i, let ni denote the number of ak for which 21-i > ak > 2-i holds. Prove that 00
ni2_i < 4
+
loge N.
i=1
Solution. We know that N
00
ni
22
i=1
00
aj=1 and
<
j-1
ni=N, i=1
3. SOLUTIONS TO THE PROBLEMS
72
so, by applying Cauchy's inequality, we get the following estimate: 00
ni
[logN]
00
ni +
2i = L
L
2i
ra2il-
i=[1ogNj+1
(logNl
1/2 (IogN]
\ \ i=1 1)
1/2
(
+
i=1 2Z)
1/2
00
( \
ni)
2
i=[1ogN]+1 z
o0
logN+V
Eli
v1 2
2
1
1/2
2Z)
i=[1ogN]+1
1/2
z
logN+v.
i=1
(1ogN +11 2l
00
)
This proves a somewhat stronger form of the statement with f in place of 4.
Remarks.
1. We can get the stronger estimate nt
00
log N + O
log log N
2i
i=1
log N
in a similar way if we start with the decomposition 00
nt 2
i=1 2
2
- i=1
nt C!2!ill
+
i=K+1
2
2
where K = [log N + log log NJ.
2. If we use Holder's inequality instead of Cauchy's inequality, we get a similar estimate: °°
.ni
(
)
1/P
-
(log N)
1/q
+ C(p),
i=1
where p > 1, 1/p+ 1/q = 1, and C(p) is a positive constant depending only on p.
Problem A.13. Consider the endomorphism ring of an Abelian torsionfree (resp. torsion) group G. Prove that this ring is Neumann-regular if and only if G is a discrete direct sum of groups isomorphic to the additive group of the rationals (resp., a discrete direct sum of cyclic groups of prime
3.1 ALGEBRA
73
order). (A ring R is called Neumann-regular if for every a E R there exists
a 0 E R such that a/3a = a.)
Solution.
In the following, "group" will always mean "commutative group", where the operation is written as addition. Instead of Neumannregularity, we shall just write regularity. First, we prove the necessity of the condition. If the group G is torsionfree, call it Go, if it is a torsion group, then it can be written as a (discrete)
direct sum: G = G1 ® ® Gi ®... , where in the group Gi all elements have order a power of pi (pi denotes the ith prime number). Let p be a prime number, and let 4tp be the map of G defined by 4tp: a pa (a E G). cp is clearly an endomorphism. Because of the regularity, there is an endomorphism Tp with (DpPAp = 4'p. Then for any a E G, pa =
p2,P p(a).
(1)
It follows from (1) and the fact that Tp is an endomorphism that if pea = 0, then pa = 0. Therefore every element (# 0) of Gi (i > 0) has order pi. In case G = Go, we can cancel (1) by p, which shows that every element a of Go is divisible by every prime, thus Go is a divisible group. Furthermore - as Go is torsion-free - the quotient is uniquely determined. Let po = 0, and denote by KO the field of rational numbers, whereas for i = 1, 2, ... , denote by Ki the field with pi elements. From our previous observations we see that Gi is a vector space over the field Ki (i = 0, 1, 2.... ). Invoking
Zorn's lemma, we see that Gi has a basis that is just equivalent to the direct sum decomposition formulated in the problem. Sufficiency will follow if we prove the following stronger statement: The endomorphism ring of the direct sum G = Go ® G1® ® Gi ®... is regular if Gi is a vector space over Ki (i = 0, 1, 2.... ). Here "®" can mean either discrete or complete direct sum. First, we prove that the endomorphism ring of each Gi is regular. Let a be an endomorphism of Gi. Then a is a linear transformation of
Gi as a vector space. Since in a vector space every subspace is a direct summand, there exist subspaces Ui, V such that
Gi=Keroz ED UU=U®Ima. It is clear that a maps the elements of Ui onto Im a in a 1-to-l way. Let us define 0: Gi -+ Gi by the conditions /3(a) = 0 for a E V and /3(a) = b for a E Im a, where b is the unique element of Ui with a(b) = a. /3 is clearly an endomorphism of Gi with a/3a = a. This proves the regularity of the endomorphism rings of each of the Gi (i = 0, 1, 2.... ). Now let a be an endomorphism of G. Since for every prime p, if a E G
has order p, then a(a) again has order p or it is 0, a maps Gi to itself if i 3 1. Now let a E Go, and suppose a(a) E Gi (i 3 1). We know pia(a) = 0. As Go is a vector space over KO, there is an x E Go with pix = a. So pl a(x) = 0, thus a(x) E Gi, so pia(x) = a(a) = 0. Applying
3. SOLUTIONS TO THE PROBLEMS
74
this argument not only to a itself, but to a and subsequent projections to Gi (i > 1), we see that each component of a(a) belonging to some Gi (i >, 1) is zero, so a maps Go to itself.
These observations imply that every a uniquely determines a vector (ao, al, a2, ...) where ai is an endomorphism of Gi, and conversely every such vector uniquely determines an endomorphism a of G from which we get back the original vector. We clearly have (a,3)i = a10i. Now let a be an endomorphism of G, (ao, al, a2i ...) the corresponding vector. We have already proved the existence of endomorphism Ni of Gi such that ai/iai = ai (i = 0, 1, 2.... ) . The endomorphism / i belonging to (,do, 01, /32, . . . ) clearly satisfies afa = a, so the endomorphism ring of G is regular.
Problem A.14. Let 2t = (A; ...) be an arbitrary, countable algebraic structure (that is, 2t can have an arbitrary number of finitary operations and relations). Prove that 2t has as many as continuum automorphisms if and only if for any finite subset A' of A there is an automorphism 7rA' of 2t different from the identity automorphism and such that (x) 7rA' = x
for every x E A'.
Solution. Suppose first that the condition mentioned in the problem is not satisfied. Let A' be a finite subset of A with the property that every automorphism of 2t fixing A' pointwise is necessarily the identity. Then for any two automorphisms ir1 and ire that satisfy (x)7r, = (x)7r2 for all x E A', irlir2 1 is the identity, so 7r1 = ire. The number of automorphisms of 2t is therefore less than or equal to the number of mappings of A' to A, and so the number of automorphisms is countable. Suppose now that the condition mentioned in the problem is satisfied. We can assume that A = {1, 2.... }. We define recursively an increasing sequence of finite subsets of A: A0 C Al C . . . C_ An C ... and a sequence of automorphisms of %: 7r1, ... , irn, .... Let A0 = 0. Now let n>, 0, and suppose we have already defined An, which is finite, and 7ri for all i < n. Let 7rn be an automorphism of 2t that fixes An pointwise and is not the identity, and let
An+1 = {1, ... , n} U U (An)iri1 ... 7rIn U {min{k I
(k)7rn
k}}
.
(E1,....En eiE{0,1}
Then An+1 is again finite, and by this procedure we have defined An and ir,,, for all n. The sequence so defined has the following properties: a) 1rn is the identity on An but is different from the identity on An+1 ,/a) If ei E {0,1} (i = 1, ... , n), then
n 1 ... (An)-7r1
i nn E
C An+1
3.1 ALGEBRA
75
'y) U°°_1An = A and all the An are finite.
Let (e1.... ,en) ...) be an arbitrary infinite sequence with en E {0, 1} (n = 1, 2, ... ). Let us define the product fn° 1 7rn = 7r in the following way. If k E A, let (k)7r = (k)7r" ...7rnkk,
(1)
where nk denotes the smallest number for which we have k E Ank . Because
of y), it is indeed a map of A into itself. For any n > nk, we have E' E (k)7r = (k)7r1 ...7rn" = ((k)7r1
Enk+1
E1
.
.
Enk
E Enk . 7rnk )ink+1 ... 7rnn = (k)7r1 ...7rnk
(2)
because in view of 0) we have (k)7r11 7rnk+11
7r nkk E Ank+l and in view of a)
, 7rn are all equal to the identity map on Ank+1. Taking into
consideration that 2i contains only finitary operations and relations, we see that in view of (2) it preserves operations and relations. For every 1 E A there exists an n with 1 E An+1. Thus, there exists a k with (k)7ri1 ...7r.Enn = 1. Because of a) we have for any m > n (k)7ri1 ... 7r /2 = 1 so in view of (2) we have (k)7r = 1. Thus it is an automorphism. We are going to show that for (E 1' ... , En, ... ) 7r = 11' 7r' = 11(E'1, ... , E'n ... n=17rEn n n=1 7rn is satisfied. Let n be the smallest number for which En e . We may assume En = 0, E'n = 1. Then 7ri1 .. 7rnn i' = 7r11 ... 7r " i' . Let 1 E An+1 be an element with (l)7rn 1. Let k be the number for which (k)7ri' ... 7rnn it = 1. In view of a) and (2) we have (k)7r = (k)7r1' ... 7rTEyn = (k)7r1' ... 7rnn i' = 1,
so (k)7r' = (l)7rn $ (k)7r = 1. This means that the number of different automorphisms of 21 is at least as much as the number of infinite sequences
of 0's and 1's. Taking into consideration the fact that A is a countable set, we see that the cardinality of the set of automorphisms of 21 is indeed continuum.
Remark. The following example shows that the statement of the problem is no longer true if we allow relations with infinitely many variables. Let 2l = (A, R), where A is the set of natural numbers and the relation R(al,... , an, ...) with countably many variables is true if and only if we have an = n with finitely many exceptions. Then for any automorphism it of 21, (n)7r = n holds with finitely many exceptions. Thus 21 has only countably many automorphisms although the condition for finite subsets formulated in the problem is clearly satisfied by this 21.
Problem A.15. Let G be an infinite group generated by nilpotent normal subgroups. Prove that every maximal Abelian normal subgroup of G is infinite. (We call an Abelian normal subgroup maximal if it is not contained in another Abelian normal subgroup.) Solution. Let the group G be generated by nilpotent normal subgroups, and suppose it has a maximal Abelian subgroup A that is finite.
3. SOLUTIONS TO THE PROBLEMS
76
The centralizer C of A in G is a normal subgroup of G that has finite index in G. Indeed, if we map every element g of G to the mapping x " g-1xg of A, then this is a homomorphism from G to the automorphism group of A and the kernel of this homomorphism is C. Therefore G/C is finite.
Suppose B is an Abelian normal subgroup in G. Then we have IBI n, where n = IAI IG : Cl. Indeed, we have C n B = An B, because otherwise the Abelian normal subgroup A(C n B) of G would strictly contain A. Therefore I B : A n BI < IG : CI because B/A n B = B/C n B -_ CB/C G/C, which gives Let H be a nilpotent normal subgroup of G. We claim that the nilpotency class of H is at most 2n - 2. Consider namely the lower central series of H:
H=Hl>H2>...>Hk>Hk+1=1.
It is well known that for the members of the lower central series, [Hi, Hj] <
Hi+j holds, so as soon as r > k/2 we have [Hr, Hr] < H2r = Hk+1 = 1, that is, the characteristic subgroup Hr of H is Abelian. By what we said previously, this implies J Hr I < n. This clearly implies that there can be
at most n Hr such that r falls in the range k + 1 >, r > k/2, that is, k+1 < 2n-1 holds. The group G is nilpotent. Indeed, for any elements 91, 92.... ,92n-1 E G, [... [[91,921,931,-
is true, because 91, 92,
,
92n-1] = 1
, 92n-1 belong to some subgroup of G whose nilpo-
tency class is at most 2n - 2. It is true that C = A. If this were not true, then the nilpotent group G/A would contain the nonidentity normal subgroup C/A. By a well-known theorem, this implies that the center of G/A has a nonidentity intersection with C/A; in other words, there is an element g E C \ A for which Ag is contained in the center of G/A thus the subgroup (A, g) would be an Abelian normal subgroup properly containing A. All this would imply that G is a finite group. Indeed, the subgroup C being identical with A is finite and we know that it has finite index in G, is finite.
Problem A.16. Let p > 7 be a prime number, ( a primitive pth root of unity, c a rational number. Prove that in the additive group generated by the numbers 1, (, (2, S3 + (-3 there are only finitely many elements whose norm is equal to c. (The norm is in the pth cyclotomic field.)
Solution. Let L(k)(x) = x1 + (kx2 + b2kx3 + (3k + (-3k )X4
(k = 1,...,p - 1).
It is enough to show that the inequality I L(1) (x) ...
L(P-1)(x)
I < ICI
(1)
3.1 ALGEBRA
77
has only a finite number of rational integer solutions x1i x2, x3, x4. For any such solutions 2P-11cl
>
f
P-1 I L(k>(x) I> 2P-1 JJ I Im L(k) (x)
(r
-2k)x3 - (-k )X2 + ((2k - (-2k
P-1
2P-1
k=1
P-1
= [f l
Sk
k=1
k=1
P-1
fl(r k- (-k) = k=1
P-1 fl(x2+((k+(-k)x3) l
k=1
=P IF(x2,x3)I.
But for any k with p f k we know that rk + C-k is an algebraic number Thue's theorem we see that the homogeneous polynomial F(x2i x3) with rational integer coefficients takes on rational integer values only at finitely many places, so there are only a finite number of possible values for x2 and x3. Therefore, it is enough to show that for any fixed x2 and x3, the number of possible values of x1 and x4 is also finite. For x2 = x3 = 0, the proof of this statement is analogous to the previous proof. If one of x2 and x3 is different from 0, then the absolute value of each of the L(k) (x) is
of degree (p - 1)/2 > 3, so by applying
bounded from below in view of IL (k) (x) I > IIm L(k) (x) I by a positive bound
independent of xl and x4i and as the product of them is bounded from above in view of (1), their absolute value is also bounded from above by a bound independent of xl and x4. But then the system of equations x1 + (r3 + (-3)x4 = L(1) (x) - ((x2 + 2x3), x1 + ((6 + C-6)X4 = L(2) (x) - ((2x2 +(4 X3)
guarantees that Ix1I and Ix4I are bounded from above, so the number of possible pairs xl, x4 is indeed finite.
Problem A.17. We have 2n + 1 elements in the commutative ring R:
a,a1i...,an,P1,...7On Let us define the elements n k
Qk = ka +
aiei i=1
Prove that the ideal (ao, or1, ... , Qk, ...) can be finitely generated.
Solution. First, we are going to show that we can assume that R is a ring with identity. There is a standard way to embed an arbitrary ring R as
3. SOLUTIONS TO THE PROBLEMS
78
an ideal into a ring with identity R+: as a set R+ consists of the ordered pairs (a, n) with a E R and n an integer, and the operations are defined in the following way:
(a, n) + (b, m) = (a + b, n + m),
(a, n) (b, m) = (ab + ma + nb, nm).
R can be identified with the ideal of R+ formed by the elements of the form (a, 0), and a subset of R generates the same ideal in R and R+. To verify
this statement, take a subset H of R and denote the ideals generated by H in R and R+ by I and J, respectively. Clearly, I C_ R n J c J. On the other hand, I consists of elements of the form (a, 0), and multiplying these elements by elements of the form (b, m), we again get elements of I so I is an ideal of R+, too. This proves J C I. This shows that the ideal of R mentioned in the text of the problem is an ideal of R+, too, and if it is finitely generated as an ideal of R+, then the same elements clearly generate it as an ideal of R, too. Therefore, from now on we can safely assume that R is a ring with identity. We are going to prove the following generalization of the problem: Theorem. Suppose we have polynomials fi (x) (i = 0, ... , n) with coefficients in the ring (with identity) R. Using these polynomials and fixed elements eo, ... , en of R, we form the elements n (k = 0,1,2,...).
ak = >fi(k)e i=o
Then the ideal (co, a1 i ... , ak, ...) is equal to the ideal (uo, ... , a9) where
s = E(deg fi + 1). We get the original problem as the special case where the polynomials
are fo(x) = x, fi (x) = ai (i = 1,... , n), and go = 1. Proof. Let ri be the degree of fi(x). If n
9-1
i=o
j=o
F(x) = fl(x - ei)r:+1 = x9 - E,Qjx and D is the operator of taking the derivative with respect to x and then multiplying by x, then Dm(xkF(x)) vanishes at the place ei if m <, ri. (We can see this if we take into consideration the fact that a root with multiplicity t of a polynomial P(x) is also a root with multiplicity at least (t - 1) of the polynomial D(P(x)).) As Dm(xk) = kmxk, we have Dm(xkF(x))
s-1
= (s + k)mxs+k - EOj (7 + k)mxj+k, j=o
and consequently s-1
(s +
(7 + j=o
k),,,,ej+k
3.1 ALGEBRA
79
Let
fi(x) = E`Ym,ixm
.
M=0
Then n
n
_ > f ( s + k)0 i=0 n
+k
ri
=
_Y.,, (s + k)mQ?+k i=0 m=0
r; 8-1 E7'm,ioj (j + k)
8-1
n
ri
j:0j E E 1'm,i (j+ k)m [Ji+k
i=0m=0j=0 8-1
n
j=O
i=om=O
9-1
I:OjI:fi(j + k)e'+k = I:OjO'j+k j=0
i=0
j=0
This means that Qs+k is contained in the ideal generated by Qt with smaller indices, and this proves the theorem.
Problem A.18. Let G and H be countable Abelian p-groups (p an arbitrary prime). Suppose that for every positive integer n, pnG 54 pn+1G.
Prove that H is a homomorphic image of G.
Solution. We say that an element g E G (g : 0) is of infinite height if for every natural number n there is an x E G that satisfies pnx = g. The elements of G that are of infinite height together with 0 form a subgroup A of G. Let G* = G/A be the factor group of G with respect to A. Clearly, G* is a finite or countably infinite Abelian p-group. G* contains no elements
of infinite height. To see this, pick a g* E G* such that pnx* = g* has a solution x* E G* for every natural number n. Take an element g E G which is mapped to g* by the natural homomorphism G -> G* and an x E G which is mapped to x* (for some fixed n). We see that pnx - g is mapped to 0, that is, pn - x E A, so pnx - g = pny for some y c G, which gives g = pn(x - y), and this in turn implies g E A, that is, g* = 0. Now a well-known theorem of Priifer gives that G* is a direct sum of cyclic groups (in the case where G* is finite, this follows more easily from the fundamental theorem of finite Abelian groups). Now we show that for every natural number n, pnG* # pn+1G*
We know that pnG # pn+1G, so for some h E G the equation pnh = pn+lx
has no solutions in G. We want to show that pnh* = pn+ly* has no solutions y* E G* (h* denotes the image of h in G* with the natural homomorphism). Otherwise, for a suitable y E G, pnh - pn+ly would be
80
3. SOLUTIONS TO THE PROBLEMS
mapped to 0 E G*, so p1h - pl+ly = pn+lz would hold for some z E G, giving the contradiction p'h = pfl+l (y + z). (This, incidentally, shows that G* cannot be finite.) Suppose the direct sum decomposition of G* is 00
G* = ®Ci, i=1
where Ci is a cyclic group of order pki . Then the set {k1, k2, ... } cannot be bounded. This follows from the fact that if, for some n, all Ci were of order at most p', then pEG* = pn+IG* = 0 would hold, contrary to our previous observation. Suppose ci is an element generating Ci. Then every element g* E G* can be written uniquely in the form g* = EO° 1 aici, where ai is an integer taken modulo pki and, with finitely many exceptions, all the ai
are equal to 0. Now take a finite or countably infinite Abelian p-group H, the elements of which are 0, h1, h2, .... Choose a series cil, ci2,... in such a way that the order of ci, is greater than the order of hi. Then we have a homomorphism of Ci, onto the cyclic subgroup generated by h3 . We map all the other Cl to 0. Since G* is a direct sum of these cyclic subgroups, there is a homomorphism of G* to H extending the above mentioned homomorphisms
of the cyclic subgroups. As every hj is the image of some element of Ci,, this extended homomorphism is clearly onto. The composition of the natural homomorphism G -+ G* and our homomorphism is the required homomorphism of G onto H.
Problem A.19. Let G be an infinite compact topological group with a Hausdorff topology. Prove that G contains an element g 1 such that the set of all powers of g is either everywhere dense in G or nowhere dense in G. Solution. Suppose G is an infinite compact group with the property that the closure of every cyclic subgroup (# 1) has an inner point, that is, every
closed subgroup (# 1) of G is open. We are going to prove that G has a dense cyclic subgroup. First, let G be commutative. The closure of a cyclic subgroup A :/- 1 is open so it has a finite index n in G; therefore, the continuous isomorphism
cp : x , xh (x E G) maps G to this subgroup. Gn # 1 is a compact, thus closed, thus open subgroup of G, therefore Gn n A is dense in Gn. So the cyclic subgroup cp-1(Gn n A) is dense in G. Now we show that the index of every open subgroup of G is a power
of the same prime number. If p is a prime divisor of n, then because GP, (i = 0, 1, 2, ...) form a strictly decreasing GP : G the open subgroups chain, so their intersection is a closed subgroup of infinite index, and so this intersection is the identity. The topology of G that we get by choosing GPi as a base for neighborhoods of 1 is coarser than the the subgroups
3.1 ALGEBRA
81
original compact topology of G, so it is equal to it. Consequently, every open subgroup of G has index a power of p, since it contains a subgroup some Gpi .
In the following, we do not suppose that G is commutative. The center Z of G is an open subgroup. Indeed, the closures of the cyclic subgroups (# 1) of G are open subgroups and cover G, so finitely many of them cover G already, the intersection of these is a subset of Z, and therefore Z is an open subgroup. The group G/Z is a p-group. Indeed, if every open subgroup of Z has index a power of p, and H < G is a subgroup that contains Z as a subgroup of index q (q is prime, therefore H is commutative), then all elements (# 1) of the factor group H9/H12 (0 1) formed using open subgroups of Z have order q; thus p = q and consequently the finite group G/Z is a p-group. Finally, we prove that G is actually commutative. Let V be a subgroup
with Z < V < G such that the center of G/Z is VIZ. Then for every g E G, taking the endomorphism of V defined by x -> [x, g] (x E V), the image of V will be finite (as the kernel of the endomorphism contains a subgroup Z that itself has finite index) so this image is the identity, thus V = Z and thus G = Z, because G/Z is a finite p-group. 0 Remarks.
1. Paragraphs 3, 5 and 6 of the elementary solution described above can be omitted if we use a theorem of Baer which states that if the factor group with respect to the center of a (discrete) group G is finite, then the commutator subgroup of G is also finite. 2. Jozsef Pelikan noticed the following. If we use the fact that every infinite compact group has a proper closed subgroup, then the statement of the problem can be reformulated in the following way:
Statement. If G is an infinite, compact, topological group, every closed subgroup (# 1) of which is open, then G is topologically isomorphic to the topological group of the p-adic integers (p prime).
Problem A.20. Let G be a solvable torsion group in which every Abelian subgroup is finitely generated. Prove that G is finite.
Solution. A finitely generated Abelian torsion group is finite. Therefore, the problem is equivalent to the following one: If in a solvable torsion group
every Abelian subgroup is finite, then G itself is finite. We are going to prove the following stronger statement: If in a solvable torsion group every Abelian normal subgroup is finite, then G itself is finite. First we prove a lemma: If a group G has a commutative normal sub-
group with a commutative factor group, and furthermore every Abelian normal subgroup of G is finite, then G itself is finite. Let H be such a normal subgroup. Using Zorn's lemma, we see that the partially ordered set of all Abelian normal subgroups of G containing
82
3. SOLUTIONS TO THE PROBLEMS
H has a maximal element F. If an element g is permutable with every element of F, then F and g generate a commutative normal subgroup (it is normal because G/H is commutative), so by the maximality of F we have g E F, that is, F is its own centralizer. Consequently, G/F is isomorphic to a subgroup of Aut (F) that is itself finite in view of the finiteness of F. Thus G is also finite. Now we prove the statement using induction on the length of the derived series of the group G. If this length is 1, there is nothing to prove. If
the length is 2, the statement follows from the previous lemma. In the general case, consider P, the last term of the derived series, different from the identity. P is a commutative normal subgroup (hence finite) and the length of the derived series of G/P is one less than the length of G. We
only have to check that every commutative normal subgroup of G/P is finite. But if N/P is a commutative normal subgroup of G/P, then N satisfies the conditions of the lemma, so N is finite (hence N/P is also finite), completing the proof. Remark. In the problem, the condition on the solvability (or some weakened form of it) cannot be dropped: Novikov and Adian constructed examples of infinite groups where the order of the elements is finite (in fact bounded) but every Abelian subgroup is finite (actually, cyclic).
Problem A.21. We say that the rank of a group G is at most r if every subgroup of G can be generated by at most r elements. Prove that there exists an integer s such that for every finite group G of rank 2 the commutator series of G has length less than s.
Solution. We start by showing that if G is a finite group of rank 2, N a normal subgroup of it, and N* the intersection of the normal subgroups of N that have prime index, then the following statement is true:
Statement. If in every rank-2 automorphism group of N/N* the length of the commutator series is at most n(> 0), then h(G), the length of the commutator series of G, is at most n + 2. Proof. The condition implies that the group obtained by restricting the inner automorphism of G/N* to N/N* has a commutator series whose length is at most n. In other words, taking the member G(n) of the commutator series G >, G' 3 G" > ... and forming the commutator group [N, G(n)], this will be a subgroup of N*. Now in case N < G(n), applying this observation to N' in place of N we see that the group N'/N'* is cyclic, for if a and b generate N, then the commutators [a, b] and [N', N] generate
N', so N'/N" is cyclic. Thus N'/N"* is commutative, because - again by the first observation - N"/N"* is a subgroup of the center of N'/N"*, which gives N"* = N", and thus N"' = N". Therefore, h(G) n + 2. We now show that for every subgroup A of the group L = GL2 (p) of 2 x 2 invertible matrices over the field of p elements (p prime), we have h(A) < 5 (this bound is not sharp).
3.1 ALGEBRA
83
In the group L of order (p2 - 1)p(p - 1), the matrices of determinant 1 form a normal subgroup S = SL2(p) and the factor group L/S is commutative. If c is a generating element of the multiplicative group of the field with p elements, then the diagonal matrix whose diagonal entries are c and c-1 is an element of order p - 1 in S. By multiplying with a generator element of the multiplicative group of the field with p2 elements, we get an automorphism of order p2 - 1 of the same field; this implies that S has an element of order p + 1. As a consequence, we see that every Sylow subgroup of odd order of S is cyclic, since any odd prime power dividing (p - 1)p(p + 1) divides precisely one of the three factors. Therefore, if A is a subgroup of rank 2 of L, then for every normal subgroup N of An S, any automorphism group of N/N* has a commutator series of length at most 2 (because N/N* is a direct product of a cyclic group of odd order and at most two groups of order 2); therefore, by the statement first proved we have h(A fl s) < 4, and consequently h(A) < 5. Now for every normal subgroup N of a finite group G of rank 2, it is true that every Sylow subgroup of N/N* is either of prime order or the direct product of two groups of prime order. Therefore every automorphism group B of rank 2 of N/N* is isomorphic to a subgroup of the direct product of groups A of the abovementioned type, which implies h(B) < 5, and then, once more using the first proved statement, we get h(G) < 7 (which is again not the exact bound). Remarks. 1. Instead of the straightforward proof described above, we could have obtained the statement of the problem by applying the theorem of Black-
burn on finite p-groups of rank 2 and the theorem of Zassenhaus on solvable matrix groups. 2. The statement of the problem does not remain true if we consider finite groups of rank 3 instead of finite groups of rank 2. A counterexample is provided by the p-Sylow subgroups of the automorphism group of
the direct product of two cyclic groups of order p" (p > 2 prime, n = 1, 2,
).
Problem A.22. Let R be an Artinian ring with unity. Suppose that every idempotent element of R commutes with every element of R whose square is 0. Suppose R is the sum of the ideals A and B. Prove that AB = BA.
Solution. We will use two lemmas. Lemma 1. Every idempotent e of R lies in the center of R.
Proof. Let r be an arbitrary element of R. Then (er - ere)' = 0, and consequently er - ere is permutable with e. Therefore, er - ere = e(er - ere) = (er - ere)e = 0,
er = ere.
3. SOLUTIONS TO THE PROBLEMS
84
We get re = ere in a similar fashion. Thus er = re, proving the lemma.
Lemma 2. If A is a nonnilpotent right ideal of a (right) Artinian ring
R, then A can be written as a direct sum A = eR ® N, where e is an idempotent element of A and N is a suitable nilpotent right ideal of R. Proof. The proof of this lemma can be found, for instance, in A. Kertesz,
Vorlesungen fiber artinsche Ringe, Akademiai Kiad6, Budapest, 1968; VEB, Deutsche Verlag, Berlin, 1975, Theorem 6.23, p.155. Turning to the proof of the statement, we see that it is enough to prove that ab E BA for any a E A and b E B. Since R is the sum of the ideals A and B, we have elements ao E A and bo E B such that 1 = ao + bo. This gives
ab = 1 ab = (ao + bo)kab = aoab + Ebiai
(bi E B, ai E A).
If A is nilpotent, then for k sufficiently large we have aoab = 0, that is, ab E BA. If A is not nilpotent, then using Lemma 2, we have A = eR ® N; in particular ao = er + n (r E R, n E N). Using Lemma 1, we get ao = exk + nk for suitable elements xk of R.
Since n is nilpotent, we have nk = 0 for sufficiently large k, that is, ao = exk, and consequently
ab = exkab + Ebiai = b'e + Ebiai E BA
(b' E B)
.
Problem A.23. Let R be an infinite ring such that every subring of R different from {0} has a finite index in R. (By the index of a subring, we mean the index of its additive group in the additive group of R.) Prove that the additive group of R is cyclic. Solution. Choose an element a E R such that the order of a in the additive group R+ is either infinite or a prime number p. Depending on these two cases, let M denote the ring of the integers or the finite field of p elements. Then the subring Ra, generated by a consists of the elements of the form f (a), where f is a polynomial with coefficients from M whose constant term is 0. Suppose f (a) 0 for every such f 0. Then the elements of the form f (a2) form a subring that has infinite index in Ra,, so also in R. Hence,
there exists a polynomial f # 0 over M with constant term 0 such that f (a) = 0. Take such an f whose degree is the least possible. We have f (x) = x(c + h(x)),
where h # 0 (because for c E M, c # 0 we know ca 0) and h is also a polynomial over M with zero constant term. Let b = h(a). Since the degree of f was minimal, we have b # 0. Then
b2 = h(a) h(a) = h(a)(c + h(a)) - c h(a) = -c h(a) = -c b.
3.1 ALGEBRA
85
This implies that the subring Rb generated by b consists of the elements of the form c b (c E M) alone; therefore, the additive group R6 is cyclic. By the assumption of the problem R6 has finite index in R+, so R+ is finitely generated and, as R+ is infinite, R5+ is also infinite. So a could not have finite order in R+, and consequently R+ is torsion-free. By the fundamental theorem of finitely generated Abelian groups, R+ can be written as a direct sum of infinite cyclic groups. But R+ contains an infinite cyclic group of finite index, namely R6 , so the rank of R+ is 1. Thus R+ is an infinite cyclic group.
Problem A.24. Let S be a semigroup without proper two-sided ideals, and suppose that for every a, b E S at least one of the products ab and ba is equal to one of the elements a, b. Prove that either ab = a for all a, b E S
orab=b for alla,bES. Solution 1. S is clearly an idempotent semigroup. Applying a theorem of McLean (see A. H. Clifford, and C. B. Preston, The Algebraic Theory of Semigroups, vol. I., AMS, Providence, R.I., 1961, P. 129), we get that there is a congruence relation O on S such that S/O is a semilattice and each class of the congruence relation O is a rectangular band. It can be seen easily that S/O itself has no proper two-sided ideals, so it consists of one element; in other words, S is a rectangular band. Recalling the definition, this means that there are sets X and Y such that S is isomorphic to the semigroup (X x Y; ), where multiplication is defined by the rule (X 1, yi)(x2, y2) = (x1, y2)
(xi E X, yz E Y).
But then the condition formulated in the problem can hold only in the case where either X or Y consists of a single element, and this proves the statement.
Solution 2. From the second and first conditions of the problem, we see in turn that (1) S is idempotent, (2) for any elements a, b E S, there exist elements x, y E S such that b = xay. Let us define relations L, R on S in the following way:
aLb 44 ab = a, aRb
ab = b.
L is clearly transitive, and in view of (1) it is also reflexive. Suppose for some a, b E S we have aLb. Then using (1) and (2), we see a = ab = axay = axay2 = (axay)y = ay, b = xay = x(ay)(ay) = (xay)(ay) = ba,
3. SOLUTIONS TO THE PROBLEMS
86
therefore bLa. Thus, we have proved that L (and similarly R) is an equivalence relation. The definition of L and R and the second condition of the problem give in turn that (3) aLb and aRb are both satisfied only in case a = b, (4) for any elements a, b E S, either aLb or aRb is satisfied. But for equivalence relations L, R, (3) and (4) can be true simultaneously only in the case when one of L and R is the full relation S x S.
Problem A.25. Let Z be the ring of rational integers. Construct an integral domain I satisfying the following conditions:
(a) ZCI; (b) no element of I \ Z is algebraic over Z (that is, not a root of a polynomial with coefficients in Z); (c) I only has trivial endomorphisms.
Solution 1. Choose a transcendental real number a, and let A be the set of numbers of the form f (a)/g(a), where f, g E Z[x] and the polynomial g is primitive, that is, the greatest common divisor of its coefficients is 1. A is an integral domain with identity that satisfies a) A \ Z contains no algebraic number, b) to every a E A (a # 0) there is a 3 E A (/3 0) such that a/3 E Z. Let M be the set of all integral domains I with Z < I < R that satisfy
a) and b). M satisfies the conditions of Zorn's lemma, so it contains a maximal element J. We are going to show that
c) for every a e J (a > 0) there exist elements /3k, /32, ... , I3k E J and 'y1,'y2...... yk E J such that k
a
0ti 2 =
1:n ryj
2
.
Indeed, let a c J, a > 0 be arbitrary. If a is an integer or VG- E J, then the statement holds trivially. Now let a E J \ Z, a # /32 (/3 E J), and look at the integral domain Jam; this satisfies b). If /3 + ry f # 0, (/3, ry E J), then there is a 6 E J (6 0) such that 6(/32 - y2a) E Z holds (we can assume /32 - -y2C, # 0) and so (/3 + -y/) (6/3 - 6\) E Z. Since J J[/, the maximality of J implies that J[/] cannot satisfy a). So there are ,3, ry E J with y # 0 such that /3 + ryvra- is algebraic, that
is, there is a polynomial f (x) = EN0 atixz E Z[x] with f (/3 + ry / ) = 0. We have
N
f (0 +
E ai (0 + 7v/-a-)z = C + D,/-a , i=o
where C and D, being polynomials with integer coefficients of /3, ry and a,
belong to J. If D = 0, then C = 0, so f (/3 - ry /) = C - D/ = 0, that is, /3 - 'yi is also algebraic. This implies that -yam, consequently ry2a, is algebraic, so by a) ry2a = n is an integer, and thus the statement of c)
3.1 ALGEBRA
87
holds for a. If, on the other hand, D 54 0, then C + Dvfa- = 0 implies aD2 = C2, so c) is satisfied again. Now we show that J satisfies the requirements of the problem. Let q' be a nontrivial endomorphism of J. Clearly, 0(1) = 1, and so 0(n) = n
for every n c Z. If a E J (a # 0), then b) implies q(a) 0. If a > 0, then it is easy to deduce from c) that 0(a) > 0, which means that 0 is order-preserving. But then 0 can only be the identity that we easily verify if we extend in an operation-preserving way to the quotient field of J and observe that in this way 0 becomes an order-preserving map, fixing all rational numbers.
Solution 2. We prove that the integral domain
J-zIx1 'x'x+3'x+10 1
1
1
L
satisfies the conditions of the problem. The elements of J \ Z are clearly
transcendental, so we only have to show that J has no nontrivial endomorphism. J is evidently the set of rational functions of the form f (x)/(xk(x + 3)"(x + 10)1), where f (x) E Z[x] and k, n, m >, 0. This shows that the invertible elements of J are precisely the elements of the form exk(x+3)"(x+10)1, where e = +1 and k, n, m are arbitrary integers. Let 0 be an endomorphism of J. If qi is not identically zero, then ¢(1) = 1, so the image of an invertible element is invertible. This implies ¢(x) = e1xk1(x + 3)"1 (x + 10)'"1,
(1)
O(x + 3) = e2xk2(x + 3)"2 (x + 10)m2 = O(x) + 3,
O(x + 10) = e3xk3(x + 3)"3(x + 10)"`3 = O(x) + 10,
and
(2) (3)
where Ej = +1 and ki, ni, mi are suitable integers (i = 1, 2, 3). If k1 < 0, then (1) and (2) imply that k2 = k1 (the right-hand side of (2) has a pole of order -k1 in 0, thus the left-hand side too), which gives 3x-k1 . e2xk2 (x +3) -2 (x + 10)m2 = el (x + 3)"1 (x + (4) 10)'"'1 +
Substituting x = 0, we get E23"210"`2 = E13"110'"
,
which gives E1 = E2, n1 = n2, m1 = m2. But this is impossible in view
of (4); therefore k1 > 0. We get similarly ki > 0, ni >, 0, and mi >, 0 (i = 1, 2, 3).
If nl > 0, then substituting in (3) x = -3 we get n3 = 0 and (-3)k3 713 = 0, which is clearly impossible. Therefore, n1 = 0 and similarly m1 = 0. We can't have k1 = 0, because then ¢(x) = E1 would give E3
O(x + 3) = 2 or 4, contradicting (2). Therefore k1 > 0, and substituting x = 0 in (2), we get k2 = O and E2 3"2 . 1012 = 3. This gives E2 = 1, n2 = 1, and m2 = 0; therefore ¢(x + 3) = x + 3 = O(x) + 3, that is, O(x) = x. Therefore, 0(a) = a holds for all a c J, which was to be proved.
3. SOLUTIONS TO THE PROBLEMS
88
Problem A.26. Let p > 5 be a prime number. Prove that every algebraic integer of the pth cyclotomic field can be represented as a sum of (finitely many) distinct units of the ring of algebraic integers of the field.
Solution. Put ( = e27z/r, and let us denote the pth cyclomatic field by K r = Q((). It is well known that 1 + C,1 + (2, ... , 1 + (P-1 and ( are units in K. So E1 = (c + C-1)2 and E2 = -((2 + b-2) are also units. Furthermore, E1 + E2 = 2.
(1)
We show that E1 and E2 are independent, that is, there are no rational integers a and b, at least one of them is nonzero, and such that E1E2 = 1.
(2)
Suppose the contrary. With the notation 4a = a', -2b = b', we deduce from (2) (Sr +
(-1)a' = -(r2 + (-2)b'
(3)
Let d denote the smallest positive integer such that 2d - 1 (mod p). Clearly, 3 < d 5 p - 1. It is well known that KP is a normal extension of Q and the automorphisms of K P are determined by ( - * c (i = 1, 2, ... , p-1). Repeatedly applying the automorphism (-4 r2 to (3), we get (2 r-2)a' ((22 + r _22 )b' (r(r22
+rS5-22)a'
+
=
(S23 +
rSS-23)b' (4)
((2 d-1
d-1
d
)ai
+ S-2
= (S 2
+(-2
d )b'
= (S
+
(-1)bi
(3) and (4) impliy
( + -1)a'd-b'd = 1 . If a'd ; b'd, then ( + C-1 is a root of unity. But ( + (-1 is a real number, so the only way it can be a root of unity is for it to be equal to 1 or -1. But this would imply that ( is a sixth or third root of unity, which is not possible. So a'd = b'd and consequently a' = +b' # 0. But then (3) implies [(S +
(-1)(S2 + -211 a' = 1
(5)
rrS+(-11a = 1
(6)
or
Cb2+ 2/
.
But the left-hand side of both (5) and (6) is a power with exponent a' 54 0 of a real number that can be equal to 1 only if the numbers themselves are equal to 1 or -1. But this would imply that c is a root of a polynomial of degree at most 6 from Z[x], which is different from the seventh cyclomatic polynomial. Since this is impossible, e1 and E2 are indeed independent.
3.1 ALGEBRA
89
As is well known, S1 = 1, (2 = (, ... , (P_1 = (p-2 form an integer basis of KK, which means that every algebraic integer of KK can be represented as a linear combination of these units with integer coefficients. Therefore, every algebraic integer of KK can be written as a sum of units of the form ±(O-I2 (1 < k < p - 1; i, j E Z), and it follows from the previous observations that these units are pairwise different.
Now let a be an arbitrary algebraic integer in Kp, and suppose that
a=
m
p-1-1
q
E E !:aijk(ke1E2,
(7)
i=1j=nk=1
where aijk, 1, m, n, q E Z. We can assume that some aijk $ 0 and that (7) is a representation for which the value of
M =laijkl
(8)
i,j,k
is minimal. Using induction on M, we prove that a has a representation of the form r s p-1
a=
ai jk(kElE22
i=1 j=n k=1
where r, s E Z and aijk = -1, 0, or 1. This is trivially true for M = 1. Suppose M > 2 and that the statement is true for all a such that laijkl < M. i,j,k
Let a be an algebraic integer in KK that can be represented in the form (7) with property (8) (if such an element exists at all). In view of (1), clearly 2(kEjE2 = rkeJ+1d + (keje'+1 . (9)
Applying (9) repeatedly to (7), any aijk with laijkl > 2 can finally be reduced to -1, 0, or 1. Furthermore, by the minimality of M, Ei j k laijkl remains unchanged during the repeated applications of (9). So, after a finite number of steps, we arrive at t
a=
r
p-1
r U
L: a,j k(ke E2 +
j=n k=1
L:
V p-1 E E bljk(k 63102 ,
(10)
i=1+1 j=n k=1
where bijk, t, u, v E Z and a=jk = -1, 0, or 1. In addition,
laijkl+Elbijkl=M j,k
i,j,k
and
laijkl j,k
so we can apply the induction hypothesis to the second term of (10), and this concludes the proof.
3. SOLUTIONS TO THE PROBLEMS
90
Problem A.27. Suppose that the automorphism group of the finite undirected graph X = (PE) is isomorphic to the quaternion group (of order 8). P r o v e that the adjacency matrix of X has an eigenvalue of multiplicity at least 4. ( P = {1, 2, ... , n} is the set of vertices of the graph X. The set of edges E is a subset of the set of all unordered pairs of elements of P. The group of automorphisms of X consists of those permutations of
P that map edges to edges. The adjacency matrix M = [mid] is the n x n matrix defined by mil = 1 if {i, j} E E and mi.7 = 0 otherwise.)
Solution. Let 7r be a permutation of the set P, and let A be the corresponding permutation matrix, that is, the matrix that has 1 in the ith row and jth column if i7r = j, and 0 otherwise. We see that An is an orthogonal matrix: AT = A- 1. We also see easily that the element in the ith row and jth column of An 1MAIT is just mi,rjr. This means that (1) it E Aut(X) <==>. A,; 1MA7 = M. We are going to use the following well-known (and trivial) fact:
(2) If AB = BA (A and B are n x n matrices), then the subspace belonging to some eigenvalue of B ("eigensubspace") is an invariant subspace of A.
Let the eigensubspaces of M (in ll ) be V1, ... , V3 and dim Vi = ni. The subspaces V are pairwise orthogonal (with respect to the usual scalar multiplication), and their direct sum is IRT. (This follows from the fact that M is a real symmetric matrix.) So, applying (2) we see that the group of the orthogonal matrices that are permutable with M is a subgroup of the group O (V1) x ... x O (V9) (here x denotes direct product, O (V) the group of orthogonal transformations of the Euclidean space V; later 0(k) will denote the group of k x k real orthogonal matrices). In view of (1) Aut(X) is isomorphic to a subgroup of the group 0(ni) x x 0(n3). Since 0(1) < 0(2) < 0(3), it will be enough to prove the following: (3) The quaternion group is not isomorphic to any subgroup of 0(3) x x 0(3). To prove this, suppose that the quaternion group H = {fl, fi, ±j, ±k} has an embedding
f : H ti 0 (3)x...x0(3), and let us denote by p, the rth projection of the group 0(3) x x 0(3) to 0(3). Now f,. = f o p,.: H --* 0(3) is a homomorphism. We claim the following:
(4) The kernel of any homomorphism h : H -* 0(3) contains -1 E H. This implies (3) as it gives -1 E ker f, for all r, thus -1 E ker f, proving that f cannot be an injection. To prove (4), let us recall the full list of finite subgroups of 0(3): (5) A finite subgroup of 0(3) is isomorphic to one of the following: A5 x Z2, S4 x Z2, A4 x Z2, Z,,, x Z2, D. X Z2, A5, S4, A4, D., Z..,
3.1 ALGEBRA
91
where At denotes the alternating group of degree t, St the symmetric group
of degree t, Zt the cyclic group of order t, and Dt the dihedral group of degree t (thus order 2t) (see H. S. M. Coxeter, Introduction to Geometry, Wiley, New York, 1969, Table III).
(6) Cosequence: 0(3) has no subgroup isomorphic to H. So h : H 0(3) cannot be an embedding. Therefore ker It contains an x E H different
from the identity. But then either x or x2 is equal to -1, so necessarily -1 E kerh. This concludes the proof.
Problem A.28. For a distributive lattice L, consider the following two statements: (A) Every ideal of L is the kernel of at least two different homomorphisms. (B) L contains no maximal ideal. Which one of these statements implies the other? (Every homomorphism cp of L induces an equivalence relation on L: a - b if and only if acp = bcp. We do not consider two homomorphisms different if they imply the same equivalence relation.)
Solution. (A) IMPLIES (B). Assume, by contradiction, that L contains a maximal ideal M. Suppose we have a homomorphism cp whose kernel is equal to M. We will show that the image of L \ M by cp is a single point. Since we also know that under the equivalence relation induced by cp all elements of M are equivalent and no element of L \ M is equivalent to an element of M,
we get that there is at most one homomorphism whose kernel is M. Let a and b be two elem ents of L \ M. The following description of the ideal (M, a) generated by M and a is well known:
(M,a)={xELI ImEM : x
The operations on L should be the set-theoretic union and intersection. In this way, L is a sublattice of the lattice of subsets of R and therefore L is distributive. We claim that it has no maximal ideal, but on the other hand, the ideal {0} is the kernel of a single homomorphism. Let J be a proper ideal of L. Not all half-lines of the form (-oo, x] can occur in J because otherwise all finite subsets of R would be elements of J forcing J = L. So there is an x E JR such that (-oo, x] V J which of course implies (-oo, y] V J for all y > x. So the ideal J* = J(-oc, z] U V I z<, x, V C R finite} U { V I V C 1[8 finite}
3. SOLUTIONS TO THE PROBLEMS
92
is a proper ideal properly containing J. Consequently, J is not a maximal ideal. To prove the other claim, let cp be a homomorphism whose kernel is {0}.
We show that cp is a monomorphism, which means that the equivalence relation induced by cp is such that each element is equivalent only to itself, and this contradicts (A). Assume, on the contrary, that a, b E L, a b but acp = bcp. The symmetric difference of a and b is nonempty, so there is a p E R with, say, p E a and p V b. This implies {p}cp = ({p} A a)cp = {p}cp A acp = {p}cp A bcp = ({p} A b)cp = OW = 0,
in contradiction with ker cp = {0}. Thus cp is indeed a monomorphism as claimed.
Problem A.29. Let V be a variety of monoids such that not all monoids
of V are groups. Prove that if A E V and B is a submonoid of A, there exist monoids S E V and C and epimorphisms cp : S -4A , cp1 : S -+ C such that ((e)cpj')cp = B (e is the identity element of C). Solution. Take D E V, which is not a group. Then there exists an element x E D that is not invertible. The powers of x with positive integer exponents and the identity element form a submonoid of D, and the dichotomy underlying the previous description of this submonoid is a congruence relation of it. Taking the natural homomorphism, we see that the variety V contains the monoid E = { f, g} (f # g) with operation defined by f f = f ,
.f .g=g.f =9, Consider the set
S={(a,g) I ac A}U{(b, f) I bE B}. On the direct product A x E of the monoids A and E, we have (ai, 9)(a2, 9) = (aia2, 9), (ai, 9) (bi, f) = (ai bi, 9)
(b1, f)(b2, .f) = (b1b2, f ), (ai E A, bi E B),
so S is actually a submonoid of A x E proving S E V. We see at the same time that the two sets defining S give rise to a congruence relation on S. Let cpl be the natural homomorphism with respect to this congruence relation and denote the image of S by C. Clearly, the set of the elements of S that are mapped to the identity element of C equals {(b, f) I b E B}. If we denote by cp the projection of an ordered pair to its first component, we see that cp : S --> A is an epimorphism, and we have {(b, f) I b c B} cp = B, which concludes the proof.
3.1 ALGEBRA
93
Problem A.30. For what values of n does the group SO(n) of all orthogonal transformations of determinant 1 of the n-dimensional Euclidean space possess a closed regular subgroup? (G < SO(n) is called regular if for any elements x, y of the unit sphere there exists a unique cp E G such that cp(x) = y.) Solution. The answer is that such a subgroup exists precisely for n = 2 and n = 4, namely for n = 2 the group SO(2) itself and for n = 4 the symplectic group denoted by Sp(1), which we get by taking the multiplications with quaternions whose absolute value is 1. We show that there are no other solutions of the problem. For odd n, SO(n) certainly does not contain a regular subgroup since in this case every transformation cp E SO(n) has a fixed vector. Suppose therefore that n is even, n = 2m. In this case, for every cp E G (G is subgroup in question) the space R2'' decomposes as the direct sum of m pairwise orthonal two-dimensional subspaces, invariant with respect to cp: I[82m, = h1 ® ... ® h,,,,. Clearly, on every subspace hi, cp induces a rotation cpi with an angle ai. Now we prove two lemmas:
Lemma 1. For every cp E G there is a Jordan decomposition R2"' _ that a one-dimensional subgroup 4 of G such that cp E (this is well known). It is also well known that there exists an element g E 4D whose powers are dense in (D. Let I[82m = h1 ® .. ® h,,,, be a Jordan decomposition with respect to g, and let the restriction of g to hi be gi.
Then gi is a rotation with angle Qi on hi. At least one of the /3is is an irrational multiple of 27r, and consequently every other f3j is an irrational multiple of 27r; otherwise for some k, gk # id would fix an element of h3. This gives that the restriction 4Di of (D to hi is the group SO(2) for all i. Let us consider the group homomorphism A: Di -> 4)j, which maps every (where g* E 4D) to g,* E 1,. Thus A is continuous and element gi E 1-to-1 in view of the regularity and compactness of G, so it produces an automorphism of the topological group SO(2). But there are just two such automorphisms: the identity and (identifying SO(2) with the unit circle of C) the conjugation. These map a rotation with angle a to a rotation with angle a, so - using A(Vi) = cps - we get ai = aj, which was to be proved. This lemma immediately implies
Lemma 2. If some cp E G maps a unit vector to a vector orthogonal to it, then c,2 = -id and cp maps any other vector y to a vector orthogonal to it; furthermore, the vectors y, V(y) span a plane that is invariant with respect to V. Proof. Let I[82m = h1® .. G h,,,, be the decomposition whose existence is
assured by Lemma 1. Suppose that x is orthogonal to cp(x) and x = x1 + +x,,,,, cp(x) = cp(xl)+ +V(x,,,,) the Jordan decomposition with respect
3. SOLUTIONS TO THE PROBLEMS
94
to 0 and consider the inner product (x, cp(x)) = Ei'j(xi, cp(xi)) = 0. Lemma 1 guarantees that either (xi, cp(xi)) > 0 for all terms or (xi, cp(xi)) for all terms, therefore necessarily (xi, cp(xi)) = 0 for all terms. Thus, Lemma 1 implies that cp rotates with an angle 7r/2 in every subspace hi, and from this the statements of the present lemma follow immediately. Now the question formulated in the problem is answered by the following Proposition. If the group SO(2m) (2m > 4) has a closed, regular sub-
group G, then 2m = 4 (furthermore, it is true that G is the universal covering group of SO(3), that is, Sp(1)). Proof. Let the unit vectors e0i el be arbitrary, but orthogonal, and let
01 E G be the group element with cpl(eo) = el. Let e2 be a unit vector that is orthogonal to the plane spanned by e0, el, and let e3 = cp1(e2). By Lemma 2, the vectors e0i el, e2, e3 are pairwise orthogonal (the subspace orthogonal to an invariant subspace is itself invariant) and denoting by cpi E G the group element with cpi(eo) = ei, then the following formulas are already true: c01 = -id,
cP2 = -id, cP3 = -id, O1W3 = -P301 = -P2, W2co3 = - 302 = 01 -cpjcpi in the last row follow immediately from the relation (cpicpj ) 2 = -id. In the case 2m = 4, the proof is finished. P1 P2 = The formulas cpicpj
It remains to show that the case 2m > 4 cannot occur. Suppose 2m > 4, and let K be the subspace spanned by e0i e1, e2, e3. If we choose a unit vector e4 that is orthogonal to K and define e5 = cpl(e4), e6 = cP2(e4), e7 = W3(e4), then the vectors e4, e5, e6, e7 are pairwise orthogonal and span a subspace K*, orthogonal to K, consequently e0i e1, ... , e7 is an orthonormal system of vectors. Denoting by cpi E G the group element with cpi (eo) = ei (i = 0, 1, ... , 7), the transformations cpi have the following multiplication table: APO
coo
id
cot
cot
W1
co2
W3
W4
W5
W6
co7
cot
co2
co5
cob
co7
co3
W4
co3 -W2 W2 -co3 -id c1 co2 -cot -id (P3
co5
co4 W4 -W5 -cob -°7
-id
co2
co3
W5 coy
-id
co4
co7
cob
cob
co4
co7
co7
cob -W5
co7
co6 -W7 -co4 c°7
co5
c6 -co5 -c4 co'
cot
co3
-cob -c1 -id W3 -co2 co5 -cot -°3 -id cot co4
-co3
co2
-cot
-id
From this multiplication table, we deduce the following identities: (co4co1co4)co6 = (-co4co4co1)co6 = co1co6 = co1(co2co4) = (co1co2)co4 = co3co4 = W7 w4(co5co6) = co4(co1w4co2co4) = W4(-co1co2co4co4)
_ co4(co1w2) _ co4w3 = -co3co4 = -co7
3.1 ALGEBRA
95
This contradicts the associativity of the multiplication, so the case 2m > 4 cannot occur.
Problem A.31. In a lattice, connect the elements a A b and a V b by an edge whenever a and b are incomparable. Prove that in the obtained graph every connected component is a sublattice.
Solution. Let us denote the fact that two elements a and b belong to the same connected component of the graph by a - b. The statement to be proved is that whenever b1, ... , b, is a path in the graph, we have b1 - b1 V b,,,, which in turn reduces to b1 V b,,,_1 - b1 V bn. We will denote the fact that two elements x and y are incomparable by x 11 y. Lemma. Let x 11 y, a = x V y, b = x A y, z arbitrary. Then one of the following two cases holds:
(a)aVz-bVz; (b) one of a, b, z is - a V z and one of a, b, z is - b V z. This will indeed prove the statement of the problem. In the case b,,,_1 =
x V y and bn = x A y, we can apply the lemma with a = bn - 1, b = bn, z = b1. In case (a), the proof is finished immediately, and in case (b) we get that one of b1, bn_1i bn is - b1 V bn, which suffices to conclude the proof. In the case bn_1 = x A y and bn = x V y, we can proceed similarly.
We will prove the lemma in six steps. First, we deal with that part of the statement that refers to a V z. 1. If a < z, then a V z = a, z so the first part of (b) holds. 2. Suppose a z V y. This gives x z V y since x>, z V y would imply x > y, whereas in case x < z V y we would have a = x V y < z V y. Therefore a V z = x V y V z > x A(y V z). In view of x A a= x, we can 11
further deduce x A (y V z) = x A (a A (y V z)) < x V (a A (y V z)) as soon as we
establish x 11 a A(z V y). To prove the latter statement, suppose first x < a A(z V y). Then in view of y < a A(z V y), we have a = x V y < a A(z V y), contrary to the initial assumption. On the other hand, if x > a A(z V y), then y < a A(z V y) implies y < x, which is not the case. But x V (a A(y V z)) = a since x, a A(y V z) < a and a A(z V y) >, y, thus x V (a A (y V z)) > x V y= a. We arrived at the conclusion a- a V z. The case a z V x can be treated similarly. 3. So we can assume a < z V x, a < z V y, and a 11 z. Thus x < z V y, which implies z V a = z V y. Similarly z V a = z V x consequently z V x = z V y = z V a. We can further assume x I I z and y z. (Suppose namely x < z. This gives a V z = y V z - a. The case y < z is similar, and in case x > z or y > z we get a > z.) Suppose now z 11 (z V b) A a. This implies
zVa-zAa=((zVb)Az)Aa=((zVb)Aa)Az <((zVb)Aa)Vz=zVb.
96
3. SOLUTIONS TO THE PROBLEMS
As to the last equality, < is clear and we get the other direction from
b<(zVb)Aa. What if z and (z V b) A a are comparable? z < (z V b) A a < a is excluded, so this can happen only if z > (z V b) A a > b, and consequently z = z V b. 4. Thus, we can assume z = z V b, z V x = z V y = z V a. If we have
z A x= b, we get z V a= z V x> z A x= b, giving z V a- b, and similarly in the case z A y = b. 5. In the remaining case let t = ((x A z) V (ynz)) . Then z V a = z V x >
z A x= x A t since z> t and z A x< t. Certainly, x 11 t since x< t< z would imply x< z and x > t implies x A y> y A z > b, which is impossible.
So xAt < xVt = xV(ynz). We have x 11 ynz since x < ynz < y is impossible and x y A z gives x A y y A z > b, a contradiction. So, finally, x V (ynz) > x A(y A z) = b since b < z. This gives a V z b, proving the part of the lemma that refers to a V z. 6. Now let us deal with the statement concerning b V z. In case b z, we have b V z = b, z. Otherwise, b V z > b A z, and applying the part of the
statement that has been already proved to the dual lattice, we get that
bAz-tooneofa,b,zorbAz-aAz. In case aZ z, we haveaAz=a,z, whereas in case a 11 z using a V z> a A z we get a V z- b V z, and this finally concludes the proof of the lemma.
Problem A.32. Let G be a transitive subgroup of the symmetric group S25 different from S25 and A25. Prove that the order of G is not divisible by 23.
Solution. We assume that 23 JGI, and we try to reach a contradiction from that assumption. Denote the stabilizer of a point x by Stab (x). By the transitivity and our assumption, all the Stab (x) are isomorphic permutation groups containing a 23-cycle. Suppose that Stab (x) is not transitive. Then taking the fixed point y of Stab (x), we see that the fixed point of Stab (y) is x. That would give a pairing of the 25 points, which is impossible. So G is 2-transitive, and as Stab (x, y) contains a 23-cycle G is actually 3-transitive. We have seen that the 23-cycles of G act transitively so we can take G to be the group generated by them. We know then that G < A25 and G is 3-transitive. By a well-known theorem (see H. Wielandt, Finite Permutation Groups, Academic Press, New York, 1964, 13.10), the order of G cannot be di-
visible by 11. We know that Stab (x, y) has prime degree and is transitive. According to Burnside's theorem (Wielandt, 7.3) it is either 2transitive or isomorphic to a subgroup of the group of all affine mappings: z " az + b (a, b E GF(23)). In the first case, G is 4-transitive, so its order is divisible by 25.24.23.22, therefore by 11, which is excluded. So Stab (x, y) is a subgroup of a group of order 2 11 23. Again 11 is excluded. An element of order 2 would
be of the form z H -z + b, which is a product of 11 transpositions and is thus not contained in A25. Therefore, the order of Stab (x, y) is 23, G
3.1 ALGEBRA
97
is sharply 3-transitive, but then by the theorem of Zassenhaus (Wielandt, 20.5), 25 - 1 = 24 has to be a prime power, which is not the case.
Problem A.33. Let G be a finite group and K a conjugacy class of G that generates G. Prove that the following two statements are equivalent: (1) There exists a positive integer m such that every element of G can be written as a product of m (not necessarily distinct) elements of K.
(2) G is equal to its own commutator subgroup.
Solution. Suppose first that (1) is satisfied, and consider the natural homomorphism cp : G -+ GIG'. Since for any h, k E K there exists a g E G
such that k = g-'hg we have cp(h-1k) = cp(h-1g-'hg) = 1. Thus K is contained in the coset hG' = V(h)
(h E K). h' (h1 .. hm, h'1 .. any two elements of G. By our previous remarks, Now
h' EK)be
p(g) = P(hi) ... cp(hm) = cp(h)m = V(hi) ... V (h') = V(9 ) In other words G/G' has only one element, that is, G = G'. Conversely, suppose (2) is satisfied and let us denote IGI by n. Since G = G', every element g of G can be written as a product of commutators: 9 = a1 -lb1 -la1b1 . . . aT-lbT-laTbT = a1"-lb1"-la1b1
. . .
arT
laTbT(1)
Since K generates G and the inverse of an element of K can be replaced
by its (n - 1)th power, it is true that every ai, bi (i = 1,... , r) can be written as a product of elements of K. Substituting these representations in (1), we can deduce the existence of a natural number k9 such that g can be written as a product of k9 n elements of K. But then for any k > k9, we can also write g as a product of k n elements of K by simply adding a suitable number of factors of the type h'" = 1 (h E K). Let l = max{k9 I g E G}. Then every element of G can be written as a product of m = n l elements of K. Remark. In the proof of (2) =:>. (1) the only property of K that we used was the fact that K is a system of generators of G.
Problem A.34. Consider the lattice of all algebraically closed subfields of the complex field C whose transcendency degree (over Q) is finite. Prove
that this lattice is not modular.
Solution. For any set of numbers H C C, let us denote by A(H) the smallest algebraically closed subfield of C containing H. Let us consider a system of numbers a, ,d, y (t; < w1) that are algebraically independent (over Q). Since the algebraic closure of a countable set of numbers is
countable, such a system does exist. We are going to show that for a
3. SOLUTIONS TO THE PROBLEMS
98
A( a,(33 D A(
a () A() O ) Figure A.1.
suitable < wl the following diagram is actually the Hasse diagram of a sublattice of the lattice of algebraically closed subfields of C, and this will prove the statement of the problem. The containments indicated by the lines are evident. Furthermore, A (A({a}) U A({-y£, a + 07C})) = A({a, /3, yg})
since the right-hand side clearly contains the left-hand side, while the left-
hand side contains each of the numbers a, /3, ye. We have A({a}) # A({a, /3}) in view of the original assumption, so all that remains to be proved is to show that for some
we have
A({a, )3}) fl A({N, a + /3'yy}) = A(O).
For this, it is enough to prove A({a, l3}) n (A({^%, a + /3'yC}) \ A(0)) = 0. Suppose
(1)
6. Then the transcendency degree of
A (A({7£1, a + )37£1 }) U
a + )37C2 })) = A({7£1, N2, a, Q})
is equal to 4; therefore, the transcendency degree of A({7f1, a + Q7£1 }) n A({7C2, a + )372 })
is equal to 0, which means (A({7£1, a
+)3-y01 }) \ A(0)) n (A({yC2, a + N7{2 D\ A(0))
Now the set A({a, /3}) is countable and the sets A(0) are pairwise disjoint, so there exists a concludes the proof.
= 0.
a + /3y£}) \
for which (1) holds, and this
3.1 ALGEBRA
99
Problem A.35. Let I be an ideal of the ring R and f a nonidentity permutation of the set {1, 2, ... , k} for some k. Suppose that for every
0 # a E R, al' 54 0 and Ia # 0 hold; furthermore, for any elements x1,x2,...,xk E I, x1x2...xk = xlfx2f...xkf
holds. Prove that R is commutative.
Solution. Let m E {1, 2, ... , k} the smallest number that is not fixed by f. that Clearly, m < mf. We then have x1x2.. Xk
0 for any x1,...,xk E I. Fixing
is,
the elements x2i ... , xk we see that x2 xm_1(xm xk - xm f Xkf) annihilates I so has to be equal to 0. Repeating the argument, we get xm ... xk = xm f ... xk f for any xm, ... , xk E I. Let n = m f . Then for every a E R and xm . . . xk E I, we get
axmxm+l ... Xk = axmfx(m+1)f ... Xkf = xmxm+l ... xn-laxnxn+l ... xk. (We get the second equality by using axmf E I.) Rearranging,
(axm ... xi_1 - xm ... xn_la)xn ... xk = 0 Xk E I. Again applying our previous argument,
for any a E R and Xm we get
axm ... xn-1 = xm ... xn-la
EI.
for
Now let a, b E R and Xm
xn_1 E I be arbitrary elements. Then we
have
(ab)xm ... xn-1 = Xm ... xn_l(ab) = (Xm ... xn_la)b = (axm ... xn-1)b = (axm)xm+l ... xn-lb = b(axm)xm+l ... xn-1 = (ba)xm ... xi_1.
Applying once more our previous argument to the resulting identity
(ab - ba)xm .. xn-1 = 0
(a, b E R, xm . . xn-1 E I),
we get ab - ba = 0 or ab = ba, which was to be proved.
Remark. The proof did not use the fact that I was a two-sided ideal of the ring R, only the fact that I was a left ideal in the multiplicative semigroup of the ring R.
3. SOLUTIONS TO THE PROBLEMS
100
Problem A.36. Prove that any identity that holds for every finite ndistributive lattice also holds for the lattice of all convex subsets of the (n 1)-dimensional Euclidean space. (For convex subsets, the lattice operations are the set-theoretic intersection and the convex hull of the set-theoretic union. We call a lattice n-distributive if n
n
x A (V yi) = V (x n( V yi)) j=0
i=0
O
holds for all elements of the lattice.)
Solution. Let us introduce some notation. Let En-1 denote the Euclidean space of n - 1 dimensions; for X C_ En-1, let C(X) denote the convex hull of X; and let L be the lattice of convex subsets of En-1. For an arbitrary H C_ En-1, denote by L(H) the partially ordered set ({X n H: X E L}, C). L(H) is a complete lattice since it is closed with respect to forming arbi-
trary intersections and it has a greatest element. If we denote the lattice operations in L(H) by VH and AH, respectively, then we see that for
X,YEL(H)wehaveXAHY=XnYandXVHY=C(XUY)nH.
For a finite H, L(H) is also finite since IL(H)l S 2IHI. First, we show that for a finite H C En-1, L(H) is n-distributive. Let X, Yo, ... , Yn E L(H) be arbitrary elements of L(H). By Caratheodory's theorem for an arbitrary subset U of En-1, C(U)
vcu C(V). wJ< n
Therefore
C(YoU...UYn)=
U j=0
C
i=0
Yi
i#j
also holds and so
X
\i=0
i=o
/
j=0
i=0
i
ii4j
=Hn u (XnHnC(.0)) i#j
n j=0
XAHVHnYi
i=0 i#j
That means that L(H) is n-distributive.
Let p be a lattice-theoretic term in k variables, let H C En-1, and denote by p (resp., PH) the term function induced by the term p on L
3.1 ALGEBRA
101
(resp., L(H)). For arbitrary X 1 ,-- . , Xk E L, pH(Xi n H,.. -, Xk n H) is a monotone function of H, that is, Hl C H2 implies
PH1(X1 n H,. -, Xk n Hl) c pH2 (X1 n H2, ... , Xk n H2). This follows easily from the monotonicity of u, n, and C, using induction on the length of the term p. We claim that for arbitrary X1,. .. , Xk E L and h E p(X1i ... , Xk) there is a finite H C En-1 such that h E H and h E pH(X1 n H,. - -, Xk n H). This statement will be proved by induction on the length of p. If p is just a variable, then we can choose H = {h}. Suppose that the statement is proved already for terms whose length is less than the length of p,,,,
and let p = p' A p". Then there are finite sets Hl and H2 such that h E p'H1(X1 nH1,...,Xk nH1), h E p'H2(X1 nH2,...,Xk nH2), and h E H1, h E H2. Let H = H1 U H2. Then we have
hEp'H1(X1nH1i...,XknHl)npH2(X1nH2i...,XknH2)
cpH(XinH,...,XknH)np"(X1nH,...,XknH) PH(Xl nH,...,XknH). Now let p = p' V p". Then h E p(X1, ... , Xk) C C (p(X1, ... , Xk) U P "(X1, ... , Xk))
By Caratheodory's theorem, there exist finitely many points h'1, ... , h' E
p'(Xl, . . . , Xk) and hl , . . . , by E P" (X1, ... , Xk) such that
hEC({h i,...,h',h',...,h';}) and u+v
j < v). Let H = {h} U Hl U
G;,...,Xk n G;) (1 < i < n,1
U
Now
h E C({hi,...,hu,hi,...,h';})nH c C(p'H(X1 nH,...,Xk nH) up" (X1 nH,...,Xk nH)) n H = pH (X1 n H,.. -, Xk n H). Now let p(x1, ... , xk) = 4(x1, ... , xk) be an identity of lattices that is satisfied by every n-distributive lattice, and let X1, ... , Xk E L be arbitrary. We have to prove p(X1i .... Xk) = q(X1,... , Xk). To prove the inclusion p(X1, ... , Xk) C q(X1, ... , Xk), take an arbitrary h E p(X1, ... , Xk). Then for a suitable finite H C Ere-1 we have h E pH (X1 nH, ... , Xk n H). Since L(H) is finite and n-distributive we have
h E gH(Xl nH,...,Xk nH)
gEn_1(Xl nE7L-1,...,Xk nEn-1) = q(Xl, ... 7 Xk) .
This establishes the inclusion p(X1, .... Xk) C q(Xi, ... , Xk), and the reverse inclusion can be proved in the same way.
102
3. SOLUTIONS TO THE PROBLEMS
Problem A.37. Let x1i x2, yl, Y2, z1, z2 be transcendental numbers. Suppose that any 3 of them are algebraically independent, and among the 15 four-tuples only {x1, x2i yl, y2}, {xl, x2i z1, z2}, and {y1, y2i z1, z2} are
algebraically dependent. Prove that there exists a transcendental number t that depends algebraically on each of the pairs {x1 i x2 }, Y1, Y2 }, and {z1, z2}.
Solution. Take (nonzero) polynomials p, q, r with rational coefficients such
that p(x1, x2, y1, y2) = 0,
q(x1, x2, zl, z2) = 0 , r(yl, y2, z1, z2) = 0 .
Since the polynomials in one variable, pi (X) = p(X, x2, yl, y2) and
ql (X) = q(X, x2, z1, z2),
have a common root, their resultant is zero:
R(pl, ql) = 0. But it is well known that R(pl, ql) is a polynomial with integer coefficients of the coefficients of p1 and ql. Therefore, it is a polynomial with rational coefficients of the numbers x2, y1, Y2, z1, z2:
R(pl, ql) = f (x2, yl, y2, zl, z2)
By the conditions, we know that x2 does not depend algebraically on the numbers yl,Y2,z1 whereas z2 does depend on them; consequently, x2 does not depend algebraically on y1, Y2, z1, z2. Therefore, A X2, y1, Y2, z1, z2) = 0
implies that f (X, yl, Y2, z1, z2) is the (identically) zero polynomial of X. Now take an arbitrary rational number u. Then we have f (u, yl, y2, zl, z2) = 0,
and consequently the polynomials p2(X) = p(X, u, yl, y2)
and
q2(X) = q(X, u, zl, z2)
have a common root t (since the resultant of the polynomials p2(X) and q2 (X) is equal to f (u, yl, y2i z1, z2) ). Define now polynomials p*, q* in three variables as follows:
P*(X,1'i,l'2) =p(X,u,x'1,1'2), q*(X,Y1,Y2) = q(X,u,Y1,Y2)
3.1 ALGEBRA
103
It is clear that we can choose u in such a way that none of the polynomials is the zero polynomial (if, for every rational number u, at least one of them were the zero polynomial, then for every complex number u the same would hold, in particular for u = x2 which is certainly not the case). Then we have
p*(t,yi,y2)=0, so t, y1i y2 are algebraically dependent. But y1i y2 are algebraically independent, so t is a transcendental number and depends algebraically on the pair { y1 i Y2 }. Similarly, t depends algebraically on the pair {z1, z2 }.
To prove that t depends algebraically on the pair {x1, x2}, suppose the contrary. Then {x1, x2, t} would be a transcendency base of the algebraically closed field generated by {x1, x2, y1, y2}, and similarly it would
be a transcendency base of the algebraically closed field generated by {xl, x2, z1, z2}. But these two fields cannot be isomorphic, since for example [X1, x2, yl, z1 } are algebraically independent.
Problem A.38. Let F(x, y) and G(x, y) be relatively prime homogeneous polynomials of degree at least one having integer coefficients. Prove that there exists a number c depending only on the degrees and the max-
imum of the absolute values of the coefficients of F and G such that F(x, y) G(x, y) for any integers x and y that are relatively prime and satisfy max{JxJ, Jyj > c. Solution. In the course of the proof, we will denote by c, c1, c2, c3 numbers that depend only on the degrees and the maximum of the absolute values of the coefficients of F and G. Let x and y be arbitrary relatively prime integers that are solutions of the equation F(X, Y) = G(X, Y) (1) .
We will show that max{xJ, yJ} < c for a suitable constant c, and this is just the statement of the problem. We can suppose xy 0 since otherwise in view of (x, y) = 1 we have max{xJ, JyI} < 1. Let us denote the degrees
of F and G by m and n, respectively, and let us assume m > n, say. Let f (X) = F(X, 1) and g(X) = G(X, 1). By assumption, F and G are relatively prime homogeneous polynomials so at least one of f (X) and g(X) is a nonconstant polynomial. Actually, we can assume that none of them is a constant. If f (X) were a constant, then denoting the coefficient of X'n
in G by a, we would get a # 0. Now (1) together with (x, y) = 1 implies y I a, and since F(X, a) - G(X, a) is not the identically zero polynomial, we get max{1x1, 1y1} < c1 .
f and g are relatively prime because F and G are relatively prime. Thus, denoting by R the resultant of f and g, we have R # 0. Furthermore JRI <, c2. By a well-known theorem (see, for example, L. Redei, Algebra, Akademiai Kiado, Budapest, 1954, Thm. 196, p. 376), there exist polynomials with integer coefficients A(X) and B(X) such that deg A < deg g < n,
3. SOLUTIONS TO THE PROBLEMS
104
deg B < deg f <, m, and
A(X) f (X) + B(X)g(X) = R. This implies that there exist homogeneous polynomials with integer coefficients A1(X, Y), B1(X, Y) and A2 (X, Y), B2 (X, Y) such that
and
A1(X,Y)F(X,Y) +B1(X,Y)G(X,Y) = R. Xm+n-1
(2)
A2 (X, Y)F(X, Y) + B2 (X, Y)G(X, Y) = R. ym+n-1
(3)
But then, using (1), (2), (3), and the fact that (x,y) = 1, we see that F(x, y) and G(x, y) are different from 0 and both are divisors of R. So, in case F(x, y) = G(x, y) = d we have d I R, and therefore Idi <, c2. This implies that x/y is a root of the polynomial with integer coefficients h(t) =
dm-n fn (t) - 9'" (t),
which is not identically zero since f and g are relatively prime. But then by Rolle's theorem we have max{jxj, IyI} < c3i and then with the choice c = max{c1i c3} we get the statement we wanted to prove.
Remark. It is not hard to get explicit expressions for c1, c2, c3, and c using the degrees of F and G and the maximum of the absolute values of their coefficients.
Problem A.39. Determine all finite groups G that have an automorphism f such that H V= f (H) for all proper subgroups H of G.
Solution. If cp has a fixed point a E G, a # 1, then cp((a)) = (a) so G = (a), and as every subgroup of (a) is also fixed, G = (a) cannot have any proper subgroups, consequently it is a cyclic group of prime order. Suppose now that cp is fixed-point-free. By a well-known result (see, for example, D. Gorenstein, Finite Groups, Harper & Row, New York, 1968, Theorem 10.1.2), for every prime p there exists a p-Sylow subgroup P of
G with W(P) = P, so G has to be a p-group. Furthermore, as the center is a characteristic subgroup and p-groups have nontrivial center, G has to be Abelian. Also, in an Abelian p-group, the elements of order p form a characteristic subgroup so every element of G (# 1) must have order p, that is, G is an elementary Abelian group: n
G=®ZP.
(1)
i=1
We now prove that groups of the form (1) indeed possess an automorphism cp with the required property. G can be considered as a vector space
3.1 ALGEBRA
105
of dimension n over the finite field GF(p), and automorphisms can be identified with the invertible linear transformations of this space. Furthermore, subgroups are the same as subspaces. Now for a subgroup H the condition H V(H) means that H is not an invariant subspace of cp. It is well known that a linear transformation with an irreducible characteristic polynomial can have no invariant subspace. It is also well known that for every n there exists a polynomial of degree n over GF(p) that is irreducible over GF(p). Finally, it is an elementary result that to every polynomial there is a linear transformation with the given polynomial as characteristic polynomial. Combining these observations, we see that a cp with the required properties exists indeed.
So the groups with the required property are the elementary Abelian groups (including the one-element group).
Problem A.40. Let pl and pz be positive real numbers. Prove that there exist functions fi : IR --4R such that the smallest positive period of fi is pi (i = 1, 2), and fl - fz is also periodic.
Solution. If pl/pz is rational, there exist positive integers m, n such that np1 = mpz. Let us take for i = 1, 2 the functions
fi (x)
-
1,
ifx=kpi,
0,
otherwise.
kEZ,
The smallest positive period of fi is pi. But fl, fzi and consequently fl - fz are also periodic with period np1 = mpz = p. Suppose now that pl/p2 is irrational. Define for i, j = 1, 2
aj, ifx = alpl + azpz, al, az E Z, i # j, A (x)
1,
otherwise.
fi is well defined, because x = alp, + azpz = 011p1 + a2p2 Z, ai # ai) would imply that pi /P2 is rational.
(ai, ai E
Now fi is periodic with period pi, since f i (x + pi)
= Jl aj = fi(x), ifx = alpl + azpz, 1 = fi (x),
otherwise.
It is clear from the definition that fi (x) = 0 4==> x = aipi, so the smallest
positive period of fi is pi. On the other hand, fl - fz is periodic with period pl + P2, since
(az+l)-(a1+1)=(fl-fz)(x)+ (fl-fz)(x+pl+pz)= {1-1=(fl-fz)(x),
if x=alpl+azpz, otherwise.
106
3. SOLUTIONS TO THE PROBLEMS
Problem A.41. Determine all real numbers x for which the following statement is true: the field C of complex numbers contains a proper subfield
F such that adjoining x to F we get C. Solution. We will prove that a field F with the desired properties exists if and only if either x is transcendental or x is algebraic but some conjugate of x is nonreal. First, if x belongs to one of the classes described above, then there is an automorphism cp of C for which x V cp(R). Then F = cp(R) will be the suitable proper subfield: C = F(x). Conversely, we show that if x is algebraic and all conjugates of x are real, then no subfield has the required property. Suppose on the contrary that C = L(x) for some proper subfield L of C. Then x is also algebraic over L, so C is a finite extension of L. We use the following well-known theorem. Theorem. If K is an algebraically closed field, char K = 0, and K is a finite extension of some proper subfield L, then I K : L 1= 2. In view of this theorem, we have I C : L 1= 2. Next we prove that
i V L. Suppose we have x2 + cx + d = 0 with c, d E L and i E L. This implies x = c' + d' for some c', d' E L, and consequently C = L(d'). But this in turn implies = a + Wrd-, with suitable a, b E L. Thus vfd- = a2 + bed + 2ab/, so a2 + bed = 0 and 2ab = 1, and therefore 4a4 + 4a2b2d = 0 and 4a2b2 = 1. This gives 4a4 + d = 0, d = -4a4, E L. This leads to L = L(-Vfd-), and this so f = ±i2a2, showing contradiction proves i V L. So C = L(i), which implies that L is real closed, thus L can be ordered.
Let us consider A = {y E L I y algebraic}. Then A can be ordered, too. But a well-known theorem states that every ordering of any subfield of the field of all algebraic numbers is Archimedean, so A is a field with an Archimedean ordering, thus there exists an embedding cp : A -4R. But then there exists an extension cp' of cp such that cc' is an embedding of all
algebraic numbers to C. Now L(i) = C implies that A(i) is the set of all algebraic numbers. Clearly, cp'(i) = ±i. It is also clear that cp' permutes the conjugates of x among themselves, so cp'(x) E R. Now we surely have W'(x) E cp'(A(i)), so cp'(x) E cp'(A(i)) n JR = cp'(A). Thus cp'(x) E cp'(A), that is, x E A, which is a contradiction, proving our statement.
Problem A.42. Prove the existence of a constant c with the following property: for every composite integer n, there exists a group whose order is divisible by n and is less than n°, and that contains no element of order n.
Solution. If n = p"(k >, 2), then G = ZP is suitable (ZZ denotes the cyclic group of order p). Suppose n = pilp22 prr (r >, 2). We take
G = PSL2(q) xP1P2Z,
3.1 ALGEBRA
107
where q is a prime satisfying q =_ 1 (mod pi) and q - -1 (mod P2). Then we have IGI = (q - 1)q(q + 1) 2
q-1 q+1
n pipe
pi
P2
l qn 2
which is divisible by n and IGI < (q3n/pip2). If there were an element of order n in G, then PSL2(q) would contain an element of order p1p2. But a well-known theorem of Dickson (see, for example B. Huppert, Endliche Gruppen I, Springer, Berlin, 1967, Hauptsatz 11.8.27., p. 213) states that the order of any cyclic subgroup of PSL2(q) divides one of q, (q - 1)/2 and (q + 1)/2. As these numbers are pairwise relatively prime and pi I (q - 1)/2 and p21(q + 1)/2, we see that PSL2(q) contains no element of order p1P2; consequently, G contains no element of order n. In our construction, q was an arbitrary prime number in a prescribed residue class mod p1p2. A well-known theorem of Linnik (see, for example,
K. Prachar, Primzahlverteilung, Springer, Berlin, 1957, Satz X.4.1., P. 364) states that the smallest such prime number satisfies q < (p1p2)C with some constant C. Then we have
G <
n (P1P2 )sc
P1Pz
(P1P2)3c-1n \ nsC
=
.
Problem A.43. Let A be a finite simple groupoid such that every proper subgroupoid of A has cardinality one, the number of one-element subgroupoids is at least three, and the group of automorphisms of A has no fixed points. Prove that in the variety generated by A, every finitely generated free algebra is isomorphic to some direct power of A.
Solution. We will show that Every subalgebra of a finite direct power of A is isomorphic to some direct power of A. For an arbitrary natural number k
def
(*)
{1,...,k} and for any I C k,
denote by prl the projection
Ak --> AI : (a1,..., ak)
(ai)ie,
It is clear that if B is a subgroupoid of Ac then prjB is a subgroupoid of AI
Denote by U the set of those elements u of A for which Jul is a subgroupoid of A. Furthermore, we shall use the following notation: if B C Ak,
1 <,i
= {(x1i...,xi) E Az I
(x1,...xi,ai+1i...,ak) E B}
108
3. SOLUTIONS TO THE PROBLEMS
It is clear that if B is a subgroupoid of A' and ai+1 i ... , ak E U, then B(x11... xi, ai+1i ... , ak) is a subgroupoid of Ai. We shall denote the operation in the groupoid A by "o". Our first observation is that o is surjective. It cannot be constant because then A would have a single one-element subgroupoid. On the other hand, the set of the elements that we get as results of the operation o is a subgroupoid, so it can only be A itself. Now we prove two lemmas.
Lemma 1. For every natural number n > 1, if B is a subgroupoid of An with Un C B, then B = An. Proof. Suppose for some n there exists a subgroupoid B of An for which Un C B C An holds, and choose the minimal value of n, for which this can happen. Clearly, n >, 2. Now for an arbitrary u c U, B(xl,... , xn_1, u) is a subgroupoid of An-1, containing Un-1. So, by the minimality of n, we have B(xl,... , xn_1, u) = An-1. So U is a subset of the set
S= {a E A I An-lx{a} C B1,
which implies SI 3 3. On the other hand, in view of B C An, we have S C A. Finally, using the surjectivity of o, we can easily check that S is a subgroupoid. The contradiction proves Lemma 1. Lemma 2. A2 has only the following subgroupoids: (a) subgroupoids with one element;
(b) Jul x A, respectively, A x Jul (u E U); (c) automorphisms of A; (d) A2. Proof. Suppose C is a subgroupoid of A2 not in the list above. Clearly,
priC = A (i = 1, 2); furthermore, for every element u E U, C(x, u) (and similarly C(u, x)) is one of the sets A, {v} (v E U). By Lemma 1, U2 S4 C, so by the previous observations only the following two cases are possible:
C=({a}xA)U(Ax{b})UC'for (1)
some C' C (A \ U)2 and a, be U or
C = {(u, uQ) I u E U} U C' for some
C' C (A \ U)2 and some permutation a of U.
(2)
In case (1), let T = {a E A I (a, a) E C}. Clearly, T is a subgroupoid of A. As T fl U = {a, b}, T is a proper subgroupoid, so DTI = 1 and a = b. By our assumptions, there is an automorphism 7r of A for which a 54 a7r. Consider the following set:
0 = {x E A I I y E A such that (x, y), (x7r-1, y7r-1 E C}
.
3.1 ALGEBRA
109
We can easily see that C is a subgroupoid of A and C fl u = {a, a7r}. Thus 2 < Cl S< JAI, which is impossible. In case (2), let
D= {(x, y) E A2 ] z E A such that (x, z), (y, z) E C} , and denote by i the diagonal of A2, that is, A = {(a, a) la E A}. It is easy to verify that D is a subgroupoid of A2 and D fl U2 C A. Since,
pr1C = pr2C = A and C itself is not a permutation, we have A C D. So the transitive closure of D is a nontrivial congruence of A, which is impossible.
Now we turn to the proof of (*). We prove by induction on n that if B is a subgroupoid of An, then B is isomorphic to some direct power of A. The case n = 1 being trivial, suppose n > 2. If B = An, there is nothing to prove, and if for some index i E n we have IpriBJ = 1, then B -and by induction we are ready. So we can suppose B An and priB = A for every i E n. Choose a minimal index set I C_ n such that pr,B # AI, and let k = III, I = {i1 < < ik}, and C = pr,B. Clearly, k > 1 and prk_{3}C = prI-{i3}B = Ak-1 for every j E k.
(3)
Consider the subgroupoid of A2 of the form
C(u1i... , uk-2, x, y) (u1,.. , uk-2 E U) .
.
Because of (3), C(u1,... , uk_21 x, y) has the property that both projections
of it is A, so in view of Lemma 2 - it is either A2 or an automorphism and our assumption on C imply that Uk 54 C, of A. Since Lemma 1 there exists (ui, ... '42) E Uk-2 for which C(u'l, ... , uk_2, x, y) is an automorphism. Suppose that there exists also some (ui, ... , uk_2) E for which C(ui, ... , uk_2, x, y) = A2. Then in the sequence
C(ui, ... , ui', ui+ll
Uk-2
(i = 0, ... , k - 2)
there are two consecutive terms such that the first is an automorphism of A and the second is A2. So we can assume u2 = u2, ... uk_2 = uk_2. Look at the subgroupoid C' = C(x, u2, ... , uk_2, y, ui) of A2. Clearly
C'(u'1, y) = 1, C'(ui, y) = A. But by Lemma 2 we see that this cannot happen. So we know that
C(u1,... , uk_2, x, y) is an automorphism of A for every u1,... , uk_2 E U. Now let
D ={(x, y) E A2 13 z1i ... , zk-1 E A, such that (z1, ... , zk-1, x), (z1, ..., zk-1, y) E C1.
(4)
110
3. SOLUTIONS TO THE PROBLEMS
It is easy to check that D is a subgroupoid of A2 and A C_ D (to prove this we use the following consequence of (3): prkC = A). So by Lemma 2 either D = A or D = A2. Suppose that D = A2, and fix arbitrary elements u, v E U, u v. Since (u, v) E D, there exist elements al, ... , ak_1 E A such that (a1, ... , ak-1, u), (a1, ... , ak_1, v) E C .
So for the subgroupoid C* of Ak-1 defined by
C = Q X1, ... , xk-1, U) n (Xi, ... , xk-1, v), we have (a1, ... , ak_1) E C*. On the other hand, because of (4) we have
C* n Uk-1 = 0,
(5)
which implies that not all of a1, . . . , ak_1 are elements of U. For example, let a1 E A \ U. Since a1 is contained in the subgroupoid pr1C* of A, we have pr1C* = A. So for every u1 E U, C* (u1 i xi, ... , xk_2) # 0 holds. Let 1 < k-2 be the greatest index such that there exist elements u1, ... , ul E U with the property that the set
E = C*(u1,...,ul,x1,...,xk_l-1) is nonempty. In the case 1 < k - 2, the maximality of 1, while in the case l = k - 2 (5) shows that (pr1E) n u = 0. On the other hand pr1E # 0 since E # 0. Thus pr1E is a proper subgroupoid of A which is different from the one-element subgroupoids {u} (u E U). This contradiction proves D = A. The equality D = A shows that the projection C -* prk_{k}C (= Ak-1) is injective, so it is an isomorphism. Consequently, the same is true for the projection B -+ prn-{ik}B. Using the induction hypothesis, we immediately get our claim for B. This concludes the proof of (*). Since in the variety generated by a finite algebra each finitely generated free algebra is isomorphic to a subalgebra of some finite direct power of the original algebra, (*) proves the statement of the problem.
Problem A.44. Let the finite projective geometry P (that is, a finite, complemented, modular lattice) be a sublattice of the finite modular lattice L. Prove that P can be embedded in a projective geometry Q, which is a cover-preserving sublattice of L (that is, whenever an element of Q covers in Q another element of Q, then it also covers that element in L).
Solution. Let us denote by 0 the smallest element of P. Let L' = {a E L 0 < a}. Then L' is a (clearly modular) cover-preserving sublattice of L. Evidently, it is enough to solve the problem for L' in place of L. Let Q be the sublattice of L' generated by those atoms of L' which are less than 1p (1p denotes the greatest element of P). Let 6 be the dimension function
3.1 ALGEBRA
111
on Q. Take a q E Q. Then there is a finite set al, . L' such that 1Q = q V al V V.
. . , an E Q of atoms of V an and such that the value of n is minimal.
Then
b(a1V...Van) =n (Q being modular) and 6(1Q) = 6(q) + n. Using the well-known relation b(x V y) + b(x A y) = b(x) + b(y)
(1)
we get b(q A(al V ...Van)) = 0, that is, a1 V V an is a complement of q, and therefore Q is a projective geometry. Now let Q E Q be an atom of Q. Then there exists an atom a E L' of L' with a < ,3(< lp). So a E Q, and since a was an atom, /3 = a, and thus /3 is an atom in L', too. Now, if q1, q2 E Q and q1 Q Q2 then there is an atom a E Q such that q2 = ql V p (using the fact that Q is a projective geometry), but then the modularity of L' implies that qi -
that r thus a
and rAp2 =0. Then there is an atom a EL' with arand
f (pi) and a A f (p2) r A p2 = 0, proving f (p1) $ f (P2) . Next we prove f (p1 V P2) = f (pl) V f (P2). The definition of f gives immediately f (p1 V p2) < f (pl) V f (p2). To prove the inequality in the other direction, let p'2 < P2 be an atom of P such that P1 V p'2 = P1 V P2 and p1 A p2 = 0. Clearly, f (p2) < f (p2) and f (pl V p2) = f (p1 V P2), which means that for proving the inequality in the other direction we can assume p1 A p2 = 0. Let a < P1 V p2 be an atom of Q. We have to prove a
f (p1) V f (p2). In case a < P1 or a < P2, this is clear. Suppose a Apt = a Ape = 0. Let al = p1 A(aV p2) and a2 = P2 A(aV p1). Using the modularity of L', we see that a1 and a2 are atoms and a V p2 = al V P2 and a V p1 = a2 V p1 are true. Again using the relation (1) (where 6 denotes the dimension function of L') for the elements x = P1 V a and Y = P2 V a,
we get that the dimension in L' of the element (pl V a) A(p2 V a) is two.
Since a1 # a2 and al, a2 < (p1 V a) A(P2 V a), we see that al V a2 = (p1 V a) A(p2 V a) . But this implies a < al V a2, thus a < f (PO V f (p2) . It remains to be proved that f (pi) A f (p2) = f (pi A p2). It is evident
from the definition of f that f (pl A p2) < f (p1) A f (P2). Since Q is a projective geometry, it is enough to prove that if /3 < f (pi) A f (P2) is an atom, then 3 < f (pi A P2). But if for an atom /3 we have 3 < f (pl) and a < f (p2), then 0 < p1i /3 < P2, and consequently 3 < Pi A p2, therefore a < f (pi A P2), concluding the proof.
Problem A.45. Let G be a finite Abelian group and x, y E G. Suppose that the factor group of G with respect to the subgroup generated by x and the factor group of G with respect to the subgroup generated by y are isomorphic. Prove that G has an automorphism that maps x to y.
112
3. SOLUTIONS TO THE PROBLEMS
Solution 1. It is enough to deal with the case where G is a p-group. Denote the subgroup generated by x (resp., y) by X (resp., Y). Suppose a1i ... , an is a base of G. Then every element of G can be written uniquely + knan, where 0 < ki < o(ai). For the given element in the form k1a1 + x, choose a base so that the number of nonzero coefficients ki should be minimal. By multiplying with suitable integers relatively prime to p, we can achieve that all these nonzero coefficients are powers of p; let us arrange them in decreasing order:
x = petal
+... +Pekak,
(1)
where el > e2 > > ek > 0 (k < n,). Let us denote the order of ai by pdi if i > k and by pei+fi, where fi > 0 if i <, k.
> ek. Suppose, on the contrary that We show that e1 > e2 > ei = ei+1i and let us say fi > fi+1 holds. Then, by replacing ai by ai + ai+1, we would get a base giving fewer nonzero terms in representation
(1). Next we show f1 > f2 >
> fk. Otherwise, assuming fi < fi+1,
replacing ai+1 by p ei -e;+l ai + ai+1 we would get a base giving less nonzero
terms in the representation (1). Now we determine the invariants of G/X (that is, the orders of the cyclic groups whose direct sum is isomorphic to G/X). Take i el-e: al + ... + p e,-i-e. ai-1 + ai (2) ai = P
for i <, k and a' = ai for i > k. Then a', a2, ... , a' is a set of generators. Furthermore, pei+fi}1a2 = pfi+lx c X, so the order of the image of a2 in the factor group is at most pe:+f:+1 (we define fk+1 to be 0). Since o(x) = pfl,
we get that the images of the elements ai form a base of G/X and their respective orders are pet+f2 pe2+f3I ... , pek-1+fk, pek, pdk+l ... , pd,
Therefore, if we know the isomorphism type of G and G/X, we can uniquely determine the exponents f i i el, ... , fk, ek, namely, we omit the common invariants and we use the fact that
el + fl > el + f2 > e2 + f2 > e2 + f3 > ... > ek-1 + fk > ek + fk > 0 . Consequently, if G/X and G/Y are isomorphic, then in a suitable base with o(bi) = o(ai) we have y = pet b1 + ... + pek bk with the same exponents el, ... , ek as in (1).
This means that the automorphism of G mapping ai to bi maps x toy. Solution 2. We assume again that G is a p-group and proceed by induction. We distinguish two cases according to whether X is contained in pG or not.
3.1 ALGEBRA
113
In the first case, we have p(G/X) = pG/X and (G/X)/p(G/X) ^G/pG. In the second case, p(G/X) = (pG+X)/X and (G/X)/p(G/X) ^ G/(pG + X) has an order smaller than G/pG; furthermore, pG/pX = pG/(X n pG) -- (pG + X)/X = p(G/X). Since G/X ^- G/Y, either the first case or the second case applies to both of them.
In the first case, pG/X = p(G/X) ^- p(G/Y) = pG/Y, so by the induction hypothesis there is an automorphism of pG mapping x to y. It is easy to see that every automorphism of pG can be extended to an automorphism of G so in this case the proof is finished. In the second case, pG/pX ^-, pG/pY so by the previous statements we can assume px = py. It is easy to see that in the elementary Abelian group G/pG (which is actually a vector space over the field with p elements) there is a maximal subgroup M containing the image of neither x nor y. We have
IG:MI=p,MnX=pX=pY=MnY,andG=M+X=M+Y.
Now it is clear that G has an automorphism that acts as the identity map on M and maps x to y.
Remark. For p-groups a third possible way to solve the problem is to prove that the automorphism exists if h(p'x) = h(pcy) for each k, where h(g) = h means that there is an element z of the group such that phz = g and there is no z such that ph+1z = g (h is the "height" of g). An integer d is called a divisor of the element g of an Abelian group if there is an element
z of the group such that dz = g. It is easy to see that for finite Abelian groups the above statement means that the automorphism exists if the divisors of x and the divisors of y are the same. One of the contestants, Balazs Montagh, later proved the following generalization: If G is a finitely generated Abelian group and xi, x2, ... , xn, yl, y2, ... , yn E G, then there
exists an automorphism f of G such that f (xi) = y2 for each i if the divisors of alx1 +
+ anxn and the divisors of aryl +
+ anyn are the
same.
Problem A.46. Let p be an arbitrary prime number. In the ring G of Gaussian integers, consider the subrings
An={pa+pnbi:a,bEZ}, n=1,2,.... Let R C G be a subring of G that contains An+1 as an ideal for some n. Prove that this implies that one of the following statements must hold: R = An+1i R = An; or 1 E R.
Solution. Since p E An+1 and An+1 is an ideal in R, we have for any x = a + bi E R, pa + pbi = p(a + bi) E An+1. This implies pb = pn+lb', that is, b = pnb' for a suitable integer V. Therefore, all elements of R are of the form x = a +pnbi. Now we distinguish two cases. First, let us assume that for all elements x = a+pnbi of R we have p I a.
Then a = pa' for a suitable integer a', that is, x = pa' + pnbi E A. Since
3. SOLUTIONS TO THE PROBLEMS
114
x is arbitrary, this shows R S An. Clearly, An+1 is an ideal in An, and the factor ring An/An+1 has p elements. Thus An+1 < R <, An implies either
R=An+1 orR=An. Now let us assume that there exists an element x = a + pnbi in R such that p fi a. Then for suitable integers u and v we have
pu+av=1. This implies
xv = av + p"bvi = 1 - pu + pn'bvi. In view of p E An+1, this gives
1 + pnbvi = xv + pu E R + An+1 = R.
With the notation z = xv + pu, we have
pnbvi=z-1, and therefore
p2nb2v2
= z2 - 2z + 1.
Since n >, 1 and z E R, we finally get
1 = -p(p2n-1b2v2)
- z2 + 2z E An+1 + R + R = R,
that is, 1 E R.
Problem A.47. Let n > 2 be an integer, and let Stn denote the semigroup of all mappings g : {0,1}n --+ {0,1}n. Consider the mappings f E Stn, which have the following property: there exist mappings gi : {0,1}2 _ {0,1} (i = 1, 2, ... , n) such that f o r all (al, a2i ... , an) E
0,1 n,
f(al,a2,...,an) = (91(an,al),92(al,a2),...,9n(an-1,an))
Let An denote the subsemigroup of Stn generated by these f 's. Prove that
An contains a subsemigroup Fn such that the complete transformation semigroup of degree n is a homomorphic image of Fn-
Solution. We prove a generalization of the problem, namely the one when {0, 1} is replaced by any group (G, +) of order at least two. (The set {0, 1} with addition mod 2 is such a group.) This generalization causes no additional complication. So let Stn be the semigroup of all mappings Gn -- G' with the operation
f,9 E Stn, a1,...,an E G: (f9)(ai,...,an) =9(f(al,...,an))
3.1 ALGEBRA
115
Let H denote the set of those mappings f E Stn for which there exist mappings gl, . . . , gn : G2 , G such that for every al, ... , an E G, f (al, ... , an) = (gl (an, al), 92(al, a2), ... , gn(an-1, an))
Let On denote the subsemigroup of f2n generated by H. Finally, let Tn denote the semigroup of transformations p: {1, ... , n} -> 11.... , n} with the operation P, q E Tn,
1
(pq)(i) = q(p(i))
Let us consider the mapping b : Tn -+ Stn defined by P E In, a,, ... , an E G :
0(p)(al, ... , an) = (ap(l), ... , ap(n)) .
We shall prove that has the following properties: (1) ip is a homomorphism; is injective; (2)
(3) b(Tn) S An. Then 0(Tn) = Pn will be a subsemigroup of On, and in view of the above-mentioned three properties the inverse of 0 maps ?/'(Tn) homomorphically onto In, which was the statement of the problem. Let us now prove properties (1), (2), and (3). (1) Let p, q E Tn, al, ... , an E G. Then V (Pq)(a,, ... , an) = (a(pq)(1), ... , a(pq)(n)) = (aq(p(1)), ... , aq(p(n))) =(q)(a(p(1), ... , ap(n))
= VG(q)[1(P)(al,...,an)] =
[b(P)t(q)J(al,...,an),
which shows that 0 is a homomorphism. (2) Let p and q be different elements of Tn. We show that '(p) 54,0(q); in other words, ip is injective. p # q means that there is an i (1 < i < n) for which p(i) # q(i). Since G has at least two elements, it has elements al, . . . , an such that ap(ti) 0 aq(j). But then t'(P)(a,,...,an) _ (a(p(1),...,ap(n))
(a(q(1), ... , aq(n)) = '(q)(a1,... , an) .
(3) The proof that, for p E In, 0(p) E An holds will be broken down into the following steps (a) toy (e). (a) First, we prove the statement for the case when p is a transposition interchanging two (cyclically) neighboring element, that is, one of the transpositions (1, 2), (2, 3), ... (n - 1, n), (n, 1). For instance, let p = (1, 2). For the other cases, the proof is similar. We represent the mapping w(P)(a,,a2,a3,...,an) = (a2,a1,a3,...,an)
3. SOLUTIONS TO THE PROBLEMS
116
as a product of elements of H. In what follows, `--' always denotes a mapping in H.
(al,a2,a3,a4,a5,...,an) -- (-an, al, a3 + a2i a4 - a3, a5 - a4, ... , an - an-1) -
(-an, al,a3+a2,a4+a2ia5-a4,...,an-an-1)- ...-4 - (-an,al,a3+a2,a4+a2,a5+a2.... ,an+a2) ---i (a2,al,a3 + a2, a4 + a2, a5 + a2i...,an + a2) -* (a2, a2 + al,a3 + a2 + a1ia4 - a3, a5 - a4,
an - an-1) -
- (a2,a1,a3,a4 - a3, a5 - a4,...,an - an-1) -r
(a2,al,a3,a4,a5 -a4i...,an -an-1) --* ... -4 (a2,a,,a3,a4,a5,...,an).
This procedure is correct for n > 5, and for n < 5 we can use a similar but simpler procedure.
(b) If p is an arbitrary permutation, then it can be represented as a product of transpositions interchanging neighboring elements. Since V) is a homomorphism and the images of these transpositions are products of elements in H, the same is true for V)(p).
c) Suppose that p E Tn is such that p(i) # i for some i, but p(j) = j for every j # i. Without loss of generality, we can assume that p(i) = 1. Let us consider the following representation:
(a,,a2,a3,...,ai-1,ai,ai+1, ...,an) -4 (a1, a2 + a1, a3 - a2, ... , ai-1 - ai-2, -ai-1, ai+1,... , an) - (al, a2 + a1i a3 + a1, ... , ai_1 - ai-2, -ai-1, ai+1, ... , an) --*
(al, a2 + al, a3 + a1i ... , ai_1 + al, -ai-1, ai+l,... an) --a
- (al, a2i a3 - a2, ... , ai-1 - ai-2, a1, ai+1, ... , an) --p -
-*
al, a2, a3i ... , aZ_1 - ai-2, a1, ai+l, ... , an)
(a1,a2,a3,...,ai-1,al,ai+1,...,an)
The case i = 2,3 can again be handled in a similar but simpler way.
d) Now let p be a transformation that is identical on its image set: p(p(i)) = p(i) (i = 1, . . . , n). Let Pk,i denote the following transformation of type (c): Pk,i(2) = k,
Pk,i(j) = j, if j # i .
Then this p can be represented in the following form: P = PP(1),1 ...Pp(n),n .
To prove this, observe that on the right-hand side the first transformation from the left that changes i is pp(i) i, and it maps i to p(i).
3.1 ALGEBRA
117
If some later Pp(k),k (k > i) changes this element, then necessarily k = p(i). But then Pp(k),k maps k = p(i) to p(k) = p(p(i)) = p(i); thus, nevertheless, it actually fixes p(i). Now in view of (c), each of the Pp(k),k can be represented as a product of elements of H, so p has such a representation as well. e) To conclude the proof, let p be an arbitrary transformation from Tn. We shall show that there is a permutation q such that r = qp is the identical map on its image set. This will establish the statement, since in this case p = q-lr (a permutation has an inverse), V'(p) = Vi(q-1)1li(r), and (b) and (d) show that 1li(q-1) and fi(r) belong to An, so the same is true for O(p). To construct the suitable q to the given p, we use induction. Let po = p and qo be the identical permutation. Suppose that we have already constructed pi and qj in such a way that qj is a permutation
and if k < i belongs to the image of p, then pi(k) = k. Now if i + 1 does not belong to the image of p, let pi+i = pi, qi+1 = qi However, if i + 1 does belong to the image of p then it also belongs
to the image of pi, so pi (j) = i + 1 for some j and in this case let pi+1 = (i + 1J )pi, qi+1 = (i + 1, j)qi, where (i + 1j) denotes the transposition of i + 1 and j. Then qi+1 is again a permutation, pi+1(i + 1) = i + 1, and if for some element k < i + 1 of the image
of p it was true that pi(k) = k, then pi+1(k) = k also holds. The reason for this is that k < i + 1 trivially implies k # i + 1 and k = j cannot hold either since pi(k) = k < i + 1 = pi(j). Finally, we choose q = qn. Then q is a permutation, and pn = qnp = qp is the identical map on its image set.
Problem A.48. Let n = pk (p a prime number, k > 1), and let G be a transitive subgroup of the symmetric group S. Prove that the order of the normalizer of G in Sn is at most IGIk+1
Solution. First, we show that if T is a minimal transitive subgroup of G, then T can be generated by k elements. Since T is transitive, we have pk I I T I. Let P be a Sylow p-subgroup of T; we are going to show that P
is transitive, thereby proving that T is necessarily a p-group.
Denoting the stabilizer of a point a in the group G by Gq, we have
IG:G,, I=pk,so pkI IG:G,,
nPI.
Since (IG:PI,p)=1, we get pkI IP:GgnP1,that is,pkl P:P" proving that P is indeed transitive. Now let M be a maximal subgroup of T. Then M is not transitive, so MT,, cannot be transitive either. Therefore MT,,, = M, that is, Tq < M. This shows that Ta is contained in the intersection of all maximal subgroups of T, the Frattini subgroup of T. Therefore I T : 4) (T) I < I T : Ta I = pk,
3. SOLUTIONS TO THE PROBLEMS
118
so by a well-known theorem of Burnside, T can indeed be generated by k elements: T = ( 9 1 ,---,9 0We next observe that the elements of N(G), the normalizer of G in S", act by way of conjugation on the elements of G, permuting the elements of G among themselves. The number of possible images of the k-tuple (gl, ... , gk) is at most IGIk. Conjugation by the elements n, m E N(G) has the same effect on (91, ... , gk) if and only if n-1m E C(T), the centralizer of Tin S. Thus IN(G)I IGIk IC(T)l. But the group C(T) is easily shown to be semiregular. Suppose that s e C(T) fixes the point a and for any other
point /3 choose a t E T with at = /3. Then /3 = at = (a)sts-1 = (/3)s-1, that is, /3s = /3, so s is the identity, proving semiregularity. This shows IC(T))I < pk, consequently
IN(G)l < IGIk pk , IGlk+l
Problem A.49. Prove that if a finite group G is an extension of an Abelian group of exponent 3 with an Abelian group of exponent 2, then G can be embedded in some finite direct power of the symmetric group S3. Solution. Applying the Schur-Zassenhaus theorem, we see that the extension is actually a split extension, that is, a semidirect product. Using the fundamental theorem of finite Abelian groups, we get that G actually has C
the form G = Z3 Z2, where is a homomorphism : Z2 -> Aut(Z3) _ GL,n(3). (GL, (3) is the group of automorphisms of Z3 , considered as a vector space over the field of three elements.) Applying Maschke's theorem
to the representation , we see that there is a basis b1, b2, ... , b,,, of the vector space Z3 , the elements of which are eigenvectors of all elements of (Z2). Leta1i a2, ... , an, be a basis in Z2, then bj is an eigenvector of (ai) with eigenvalue ci.7 E {1, 2}, say. Denote the elements (1, 2, 3) and (1, 2) of S3, by a and T, respectively. For an arbitrary b = EE", Aibi E Z3 , let b = (cya1, ... , aam,1, ... 1) E S3 +n
,
and let a. = (,rci1-1 .rCi2-1
m}i
.rcim-1 1
T
1
p 1) E S3 +n
Then B = {b I b E Z3 } is a normal subgroup of S3 isomorphic to Z3 . ,g'3 +,, The mapping ai r--* ai can be extended to a homomorphism Z2
and then A = {a
I
a E Z2} is a subgroup of S3 +n, isomorphic to Z.
Let us consider the homomorphism g : A ---> Aut(B) mapping each x E A
to the conjugation by x. Then for the subgroup BA of S3 +" we have BA ^ B >4 A. On the other hand, for any a E Z2 and b E Z3 we have V(a) = b7(a)
3.1 ALGEBRA
119
(it is enough to check this for basis elements bi and aj). This shows that indeed G ^ BA. Remark. Instead of referring to the Schur-Zassenhaus theorem, we can simply consider a Sylow 2-subgroup.
Problem A.50. Let n > 2 be an integer, and consider the groupoid G = (Zn, U {oo}, o), where
xoy=
(x+1 ifx=yEZn, 1.00
otherwise.
(Z,, denotes the ring of the integers modulo n.) Prove that G is the only subdirectly irreducible algebra in the variety generated by G. Solution. (9 will denote the groupoid G considered as a member of the variety of the problem.) The unary operation taking the constant value oo is a polynomial expression in g, for example, 00 = (x o x) o X.
(0)
(This unary operation will also be denoted by oo.) Furthermore, it is easy to check that for the polynomial expressions
f(x)=xox
and
xVy=f"-1(xoy)
(1)
of G the following hold:
V is a semilattice operation and o0 is the greatest element with respect to it,
f
is an automorphism with respect to
(2)
V,
(3)
and the following identities hold: fn(X) = x,
fk(x) V
fl (X)
=oo if k54l (0
(4)
It is also true that the original operation o can be expressed using V and f since the identity
xoy=f(xVy)
(5)
holds true in g. Now take any algebra C = (C; o) in the variety generated by 9, and consider the algebra C' = (C; V, f, oo) whose basic operations are the polynomial expressions of C defined according to (0) and (1). As all the properties discussed above are described by identities, they hold simultaneously for C (resp. C'). Identity (5) (together with the definitions included in (0) and
3. SOLUTIONS TO THE PROBLEMS
120
(1)) ensures that a map cp : C -> Zn U {oo} produces a homomorphism cp : C -+ G if and only if cP : C' - G' is a homomorphism. So instead of 9 and C, we can work with the algebras g' and C', which can be handled more
comfortably. Also denote by , the natural partial order of the semilattice (C; V). Let a be an arbitrary element of the set C, different from oo, and consider the following subsets of C:
Ci={cECI c< ft(a)}
(i(=- Z.),
Coo=C\UCi. iEZ
We will show that {C0,.. . , C.-1, C.} is a partition of C. Clearly, the union of these sets is C, and none of them is empty, since f '(a) E Ci (i E Z.n) and oo E C,,,, . If c e Ci f1 Cj for some 0 , i , j < n, that is, c , f i (a) and c , f i (a), then using (2) and (3) we get f3
(c) V c , fi-z (fa (a)) V fj (a) = fj (a) V fj (a) = fi (a) < oo,
which, in view of (4), implies j - i = 0. So Co,... , Cn_1 are pairwise disjoint, and C,,. is clearly disjoint from each of them. So the following map cp : C' --+ 9' is well defined:
Wa = i
if
CECi.
Clearly, Wa is surjective, it is permutable with the operation f, and oocpa = oo. For arbitrary c, d E C and I E Zn, we have
(cVd)cpa = l
t= cV d, f1(a) t c,d, ft(a)
ccpa = l,dcpa =1.
This implies that Wa is permutable with V too, so Wa : C'
is a
surjective homomorphism.
As acpa = 0, we have for every element b e C the following: acpa = So if a, b are different elements of the algebra C' (say a # oo), then in case b a we have acpa # bcpa, whereas in case b < a (this
bcpa q b , a.
oo) we have acpb # bcpb . Consequently, the homomorphisms implies b cpa(a E C\{oo}) give a representation of C' as a subdirect power of g'. This
shows that C' can be subdirectly irreducible only in case it is isomorphic
to g'. It is easy to check that the algebra 9' is simple, so necessarily subdirectly irreducible. Namely, if for some congruence relation a of g' we have
aor b,alb(say a#oo),then a=aVaor aVb=ooso for any 0
3.2 COMBINATORICS
121
3.2 COMBINATORICS Problem C.1. Among all possible representations of the positive integer n as n = Ek 1 ai with positive integers k, a1 < a2 < < ak, when will the product
Ilk 1 ai be maximum?
Solution. First, we are going to investigate the properties of the extremal integer sets. Suppose that An = {a1, a2, ... , ak} is extremal.
Claim 1. There are no integers i and j such that a1 < i < j < ak and
iVAn,iVAn-
If it is not the case, then there are some elements a,., a8 E An such that ar + 1 ¢ An, a8 -1 V An, and ar + 3 < a8. Then (ar + 1)(a8 - 1) = a,.a8 + (a8 - a,.) - 1 > a,.a8 + 2 > ara81
so replacing the elements ar, a8 of An by ar + 1 and a8 - 1, the sum of the elements of An does not change, but their product increases, a contradiction to the extremality of An.
Claim 2.21. If a1 = 1, then replacing the elements a1, ak of An by ak + 1 gives us a "better" system, a contradiction to the choice of An.
Claim3. a1=2ora1=3ifn>5. If a1 > 4, then replacing the element a1 of An by 2 and a1 - 2, the sum of the elements of An does not change, but their product increases since 2(al - 2) = 2a1 - 4 = a1 + (a1 - 4) > a1,
a contradiction to the extremality of A. If al = 4, then k > 1 by n > 5, and replacing the elements a1, a2 of An by 2, a1 - 1, and a2 - 1, the sum of the elements of An does not change, but their product increases, since 2(al - 1)(a2 - 1) = 2ala2 - 2(al + a2) + 2 = ala2 + (al - 2)(a2 - 2) - 2
> ala2 + 6 - 2 = a1a2 + 4 > a1a2,
a contradiction to the extremality of An.
Claim 4. If a1 = 3 and i is an integer such that a1 < i < ak, i ¢ An
then i=ak-1. If not, then i + 2 E An by claim 1, and replacing i + 2 by 2 and i, the sum of the elements of An does not change, but their product increases, since
2i > i + a, + 1 = i + 4 > i + 2, a contradiction to the extremality of A.
122
3. SOLUTIONS TO THE PROBLEMS
Now, we are ready to determine the extremal sets A. Let A(i, j, l)
denote the set {i,i+1,...,1 - 1,1+1,...,i+j - 1,i+j}, and let s(i, j, 1) denote the sum of the elements of A(i, j, 1). For k = 2, 3, ... , let s(2, k +
2, k+2) = 2+3+ +k+(k+1) = (k22) -1 = Lk. By claims 1-4, An has at least two elements if n > 5, and if k > 2 then the possible extremal sets are as follows: A(2, k+2, k+2), A(2, k+2, k+1), ... , A(2, k+2, 3), A(2, k+ 2, 2), A(3, k + 3, k + 1). Obviously, s(2, k + 2, k + 2) = Lk, s(2, k + 2, k + 1) = Lk + 1, ... , s(2, k + 2,3) = Lk + (k - 1), s(2, k + 2,2) = Lk + k and s(3, k + 3, k + 1) = Lk + (k + 1) = Lk+1 - 1, so the sums of the elements of the above sets are the integers in the interval [Lk, Lk+1), and each integer
appears exactly once. Thus, for any n > 5, we have exactly one set of type above and this is the extremal set An for this n. If 1 < n < 4, then obviously An = {n} is the only extremal set. Remark. Consider the following generalization of the problem: Let f (x) be an arbitrary function that is strictly concave from below in the interval
(0, +oo) (f (x) = log x in the problem above). Let n = Ek 1 ai, where < ak are positive integers, k is not fixed. When will the a1 < a2 < sum 1 f (ak) be maximum for a given n? It can be proved that also in this case every extremal set can be obtained from a sequence of consecutive
natural numbers deleting at most one element.
Problem C.2. For every natural number r, the set of r-tuples of natural numbers is partitioned into finitely many classes. Show that if f (r) is
a function such that f (r) > 1 and lim,._0 f (r) = +oo, then there exists an infinite set of natural numbers that, for all r, contains r-tuples from at most f (r) classes. Show that if f (r) 74 +oo, then there is a family of partitions such that no such infinite set exists.
Solution. Let N denote the set of natural numbers. We will use the following well-known theorem of Ramsey. If the r-tuples of natural numbers are divided into finitely many classes, then there is an infinite subset N' of N such that every r-tuple in N' is contained in the very same class. To prove the first statement, we define a sequence No D . . . 2 N,. D ...
of subsets and a sequence X1.... , x,., ... of natural numbers by induction
on r. Let No = N. Suppose that r > 0 and that N, and xi (i < r) are defined, N, is infinite. Let x,. be an arbitrary element of N,. and let N. = N,. - {X1, x,.}. For any set A' C_ {xl,... , x, }, divide the (r - s) tuples of N. into finitely many classes where s = IA'J. Put two (r - s)-tuples
into the same class if and only if adding the elements of A' to them the resulting r-tuples are in the same class of the original partition. Applying Ramsey's theorem 2'' times, we get that there is a subset N,.+1 of the set N. such that JAI = IBI = r, (1)
A,BC {x1i...,x,.}UNr+1,
An{x1,...,x,.}=Bn{x1,...,x,.}
imply that A and B are in the very same class.
(2)
3.2 COMBINATORICS
123
D N,. D ... ; xl,... , x,., .... Let So, we defined the sequences No D X = {X11 ... , x,,... }. Now, (1) and (2) imply that X has the following properties. (*) Suppose that CAI = IBI = r,
A, B C X,
A n {xl,... , x,.} = B n
{xl,... x,.} for any r. Then A and B are in the very same class. ,
Since f (r) > 1 and f (r) ---4 +oo as r ---* +oo, thus there exists a monotone subsequence xrk such that I {xr,, : rk < r} 1 < loge f (r).
Let X' _ {x,.l, x,.2, ... }. This set X' is obviously infinite and meets the requirements of the problem by (*). Now, we are to prove a statement that is stronger than the second part of the problem. We show that a set of cardinality continuum has the desired partition, as well. Let S = [0, 1]. We divide the set of r-element subsets
of S into r classes for any r. Let X = {x0,. .. , x,._1 } be an r element subset of S with x0 < < xr_1. We put X into the ith class if exactly i of the intervals (xj, xj+1) (j = 0,1, ... , r - 2) are longer than 1/r. If S' is an arbitrary infinite subset of S, then using the fact that S' has an accumulation point, it is easy to see that for any i there is a number ro such that for r > ro the set S' contains a set out of the ith class of the r-tuples, and so our proof is complete.
Problem C.3. Let n and k be given natural numbers, and let A be a set such that JAI
k+1
)
For i = 1, 2, ... , n + 1, let Ai be sets of size n such that IAi n Aj J < k
(i 34 j),
n+1
A= U Ai. i=1
Determine the cardinality of A.
Solution 1. Let 0x denote the number of sets Ai containing the point x. Obviously, n+1
!bx=1: JAinAjl =n+ZIAinAjl, i¢j
i=1
xEAj
and so
¢x
Add these up for all j. On the left-hand side, we get n+1
1=
Ox = j=1 xEA3
xEAxEAj
Wx
xEA
(1)
3. SOLUTIONS TO THE PROBLEMS
124
Estimating it by the inequality between arithmetic and quadratic means, we get
0.
z
=
- (A) ij1
02 > JAI
xE
TA
n 2 (n+ 1) 2
=
1
JAI
xEA
2
JAI
1
Adding up the right-hand sides of (1) for all j, we get n(n + 1)(k + 1). Thus, n2(n + 1)2 < n(n + 1)(k + 1),
JAI
that is, > n(n + 1)
A I
k+1
Since the opposite inequality was supposed, it implies the equality above. Naturally, k + 11n(n + 1) is needed for the existence of such a family of sets.
Solution 2. W e will use the following theorem: Let Fl, ... , F N be arbitrary polynomials of some events A1 , . , An, and let cl, ... , CN be arbitrary real numbers. Then, the inequality N
E ckp(Fk) > 0 k=1
holds in any probability space and for any events Al, ... , An provided that it holds in the trivial probability space. Thus, it is sufficient to verify that the inequality n+1
p(AI U ... U An+l)
(k + 1)2
2
Ep(Ai) - (k + 1)2
E p(Ai n Aj) 1
i=1
holds when Ai = 0 or Q. If Ai = 0 for all i, then we get 0 > 0, which holds. If exactly 1 of the Ai's are 1 (l > 1), then we have to prove that 2
(1)
(k + 1)2
2
2k+1 1
- (k
+
'
1)2l
that is, 2
12
(k + 1)2
(k+1)21+1 =
1J
> 0,
(i = 1, ... , n+ 1) be the sets given in the problem, and consider the probability space such that S2 = A = Ui 11Ai which holds as well. Now, let Ai
3.2 COMBINATORICS
125
and the probability of each point is 1/JAJ. So, even in this space, 2k + 1 n+1
p(A) = 1 > - (k + 1)2
(k + 1)2 1
CAI
2k+1 (n+1)n (k + 1)2
JAI
AAA n Ail
2
JAi 1
2
n+1
k
(k + 1)2
2
JAI
JAI
Reducing this, we obtain that JAI > n(n + 1)/(k + 1). We supposed that Al C< n(n + 1)/(k + 1), so JAI = n(n + 1)/(k + 1).
Problem C.4. Let A1, A2, ... be a sequence of infinite sets such that Ai n A; I < 2 for i # j. Show that the sequence of indices can be divided
into two disjoint sequences it < i2 < ... and ji < j2 < ... in such a way that, for some sets E and F, JAi n El = 1 and JAj n F1 = 1 for n = 1,2,.... Solution. Suppose that for k > 3, there are some finite disjoint sets Ek and Fk such that either JAi n EkI = 1 or IAi n FkI = 1,
(1)
if i > k then JAi n EkJ <1orJAinFkI <1.
(2)
For k = 3, it is easy to construct such sets E3 and F3. If we show that Ek and Fk+1 ? Fk satisfying (1) and (2) for Ak+l, then the sets E = U' 1Ek F = Uk 1Fk have the desired there are some finite disjoint sets Ek+1
properties. First, suppose that Ak+1 meets both Ek and Fk. According to (2), it is possible only if, say, IAk+1 n Eke = 1, and then Ek+1 = Ek and Fk+1 = Fk satisfy the conditions (1) and (2). Now, suppose that Ak+1 does not meet at least one of the sets Ek and
Fk, say, Ek. Consider all the sets Ai that meet Ek U Fk in at least three elements. Since the given three elements can be contained in finitely many
sets Ai only, the number of these sets is finite. Let us denote them by Ail, ... , Aim . Let ek+1 denote an arbitrary element of the infinite set m
k
Ak+1 - (U Air) U(U Ai) U Fk r=1
i=1
and let Ek+1 = Ek U {ek+1}, Fk+1 = Fk. These sets satisfy (1) obviously.
Furthermore, if i > k + 1 and ek+1 V Ai, then (2) holds as well, and if ek+1 E Ai then Ai differs from the sets Ail, ..., Aim by the choice of ek+l, and so JAi n (Ek u Fk)l < 2, which yields (2).
3. SOLUTIONS TO THE PROBLEMS
126
Problem C.5. Let n > 2 be an integer, let S be a set of n elements, and let Ai, 1 < i < m, be distinct subsets of S of size at least 2 such that
AinA;#0,AinAk540,A;nAk#0 imply AinA,nAk#0. Show that m < 2n-1 - 1. Solution 1. We will prove the statement by induction on n. It obviously holds for n = 2. Assume that n > 2, and let Al # S be a maximal element of the set
K= {A1:1
that is, if Al C_ Ai, then either i = 1 or Ai = S. Choose an arbitrary element x of the set S - A1, and let
K1= {A1EK:x0Ail, K2={Ai EK:xEAi,A1nAi=0}, K3={A1EK:xEAi,A1 cAi}, K4={AiEK:xEAi,A1nAi54 0,A12¢Ail. Obviously,
K=KIUK2UK3UK4.
(1)
We will estimate the cardinality of the set K. The inductional hypothesis implies that (2) jK1j < 2n-2 - 1.
Let I = IA1J. Then IK21 < 2n-L-1 - 1
(3)
since every element of K is a set of at least two elements. By the maximality
of A1, the only element that may be contained in K2 is S, that is, (4)
IK3I<1.
Finally, the elements of the set K4 can meet the sets S - Al and Al in at most 2n-1-1 and 21 - 2 distinct sets, respectively (they cannot meet Al in Al or 0). Pairing these intersections in all possible ways, we get IK41 <
2n-1-1(21 - 2) = 2n-1 -
2n-!
Actually, this estimate can be sharpened. If X E K4, then
Y=(X-A,)U(AA-X)0K4 since the sets A1, X, Y do not satisfy the intersection conditions for the sets
Ai (the sets Al n X, Al n Y, X n Y are not empty, but the set Al n X n Y is empty). Since the mapping
X-yY=(X-A1)U(AA-X)
3.2 COMBINATORICS
127
is one-to-one for a fixed set Al - and if x E X, Al n x # 0, and Al C X, then the same holds for the set Y so the cardinality of K4 is at most half of the estimate above, that is, IK4I <
2n-2
- 2n-1-1
(5)
Summarizing (1) to (5), we get the desired inequality, IKI < 2n-1 - 1.
Solution 2. We will prove the statement by induction on n. It obviously holds for n = 2. Assume that n > 2. We distinguish two cases.
Case 1. There are no i and j such that Ai U A; = S and Ai n A; is of one element.
In this case, it is not difficult to show the statement. Considering an arbitrary element of the set S, the number of sets Ai not containing x is at most 2n-2 - 1 by the inductional hypothesis. The number of sets X C_ S containing x is 21-1. At most half of them, that is, at most 2n-2, appear as a set Ai, since if X = Ai, then according to the assumption above, there
is no j such that A; _ (S - X) U {x}. Thus, the number of sets Ai is at most 2n-1 - 1. Case 2. There is an element x E S such that Al UA2 = S and Al nA2 = {x}.
Let r and s denote the cardinality of the sets Al and A2, respectively. Clearly, r + s = n + 1. The number of the sets Ai such that Ai C Al is at most 2''-1 - 1 by the inductional hypothesis. Similarly, the number of the sets Ai such that Ai C A2 is at most 2s-1 - 1. If Ai is not a subset of Al or A2, then Al n Ai # 0, A2 n Ai # 0, and, of course, Al n A2 # 0. Thus Al n A2 n Ai # 0 by the conditions of the problem, that is, x E Ai. Now, Ai = {x} U (Ai - A1) U (Ai - A2), and since the nonempty sets Ai - Al and Ai - A2 can be chosen in 2s-1 -1 and 2''-1 - 1 ways, respectively, the number of these sets Ai is at most (2s-1 - 1)(2r-1 - 1). Adding up the partial results obtained, we get that the number of all sets Ai is at most 2n-1 - 1. Remark. It can be shown that if k = 2n-1-1, then, necessarily, the sets Ai are exactly the sets of at least two elements containing a given element x E S. It can be shown by induction, say, like we proceeded in the first solution.
Problem C.6. Show that the edges of a strongly connected bipolar graph can be oriented in such a way that for any edge e there is a simple directed path from pole p to pole q containing e. (A strongly connected bipolar graph is a finite connected graph with two special vertices p and q having the property that there are no points x, y, x # y, such that all paths from x to p as well as all paths from x to q contain y.) Solution. First, we prove that there is a real-valued function f defined on the vertices such that
3. SOLUTIONS TO THE PROBLEMS
128
(1) f (p) = 1, f (q) _ 0,
(ii) if f (x) = f (y), then x = y, (iii) if x i4 p, then there is an edge xy such that f (y) > f (x), (iv) if x # q, then there is an edge xy such that f (y) < f (x). The graph G is connected so there is a path p = X0, . . . , x,,, = q from p to q. In this subset, f (xk) = 1 - k/m is a desired function. Now, suppose that f is a desired function on a proper subset V1 C V(G).
We show that f can be extended for some bigger set. Let z E V2 = V(G) - V1. The graph G is connected so there is a path from z to p. Let x1 be the last point of this path belonging to V2, and let x0 be the next point of this path. The graph G is strongly connected so there is an xl - p or x1 - q path not containing x0. Let X1, X2.... be such a path, and let n > 1 be the smallest index i such that xi E V1 (there is such a point since p, q are in V1). The points xn, x0 are distinct elements of V1 so the open interval f (xo), f (xn) is nonempty by (ii) and it contains infinitely many elements even if we delete the values of f in V1. Thus, we can assign a strictly monotone subsequence of n - 1 members to the points x1, ... , xn-1 so that the sequence f (xo), f (x1), ... , f (xn) is strictly monotone. Now, (i)-(iv) hold in the extended domain as well: (i) and (ii) hold obviously, as well, as (iii) and (iv) for x E V1. Finally, if x = xi (1 < i < n - 1), then xi_1 and xi+1 is an appropriate choice for y, respectively. Since G is finite, it implies that there is a function f satisfying (i)-(iv). Now, let us orient the edges of G as follows. An edge joining x and
y should be oriented from x to y if and only if f (x) > f (y). We show that this orientation has the desired properties. Let e be an edge from some x1 to some yl. For n > 1, let xn+1 be a neighbor of xn such that f(xn+l) > f(xn) if xn 54 p. (Such a point exists by (iii).) Similarly, let Yn+1 be a neighbor of yn such that f (yn+1) < f (yn) if yn # q. (Such a point exists by (iv).) Since G is finite, after a while we get that x,. = p, y, = q for some r, s > 1. Then x,., x,._1, ... , x2, x1, y1, y2, ... ,Ys_1,Ys is a desired path through e.
Problem C.7. Let F be a nonempty family of sets with the following properties:
(a) If X E F, then there are some Y E F and Z E F such that Y n Z = 0
andYUZ=X.
(b) If X E .F, and Y U Z= X, Y n Z= 0, then either Y E.F or Z E .F. Show that there is a decreasing sequence Xo
Xn E F such that
X1 D X2 D ... of sets
00
(1Xn0. n=0
Solution. We will show that the statement of the problem holds even if the condition (a) is replaced by the following weaker condition: (a') If X E F, then there are some Y E F and Z E F such that Y n z = 0,
YUZCX.
3.2 COMBINATORICS
129
We will prove it by contradiction. Suppose that the statement does
not hold, that is, if X0 J X1 nn_oxn 0.
X2 2 ... for some sets Xn E F, then
1. For any set A E F, there is a set A' E F such that A' C_ A and A' = U°-1t where the sets Ai are pairwise disjoint elements of F. By condition (a'), there exist sets A1, A'1 E )c' such that Al U A'1 C A, A1nA'1 = 0. Similarly, there exist sets A2, A'2 E .F such that A2UA'2 C
A1, A2 n A'2 = 0, and so on. We are done if U9_1Ai E.F. If it is not
the case then by condition b), the sets AO = A - U°_IAi E F and A' = A = U°_0Ai have the desired properties.
2. For any set A E F there are some sets B, C E F such that B n C = 0,
BUCCAandifBCPCBUCthenPEF.
Consider the sets Ai existing by 1. We try to define a sequence of sets Ni as follows. Let N1 = A', and if Na_1 is defined, then let Ni be an arbitrary set satisfying the following four conditions: Ni EJ7, Ni C Ni_1, cc
NicUA 7=i
Ni n A3 E F for infinitely many
i.
According to the third condition, n°O1Ni = 0 if the sequence is infinite, so there is an n such that Nn is defined but Nn+1 cannot be defined. Let B = Nn n An and C = Nn n Ak for some index k > n such that Nn n Ak E F.
Suppose that there is a set P such that B C P C B U C but P 19.F. Then
Nn - P E F, and it is easy to see that Nn+1 = Nn - P would be an appropriate choice. (B E F by the choice P = B.) Thus, for A E F, there are some sets B1, C1 E F such that B1 UC1 C A, BI nC1 = 0. Similarly, for C1 E F, there are some sets B2, C2 E F such that B2 U C 2 C C1i B2 n C 2 = 0, and so forth. Let P i = U J t i B B f o r i = 1, 2, ... .
Then Pi Pi+1, n°O1Pi = 0, and Pi E F because Bi C_ Pi C Bi U Ci, a contradiction.
Remark. The problem was motivated by the following set theoretical game. There are two players: Black and White. Let X1 be a given set. White partitions it into two sets, Yl and Z1. Then Black chooses one of the sets Y1, Z1, let us denote this set by X2, and partitions it into some sets Y2 and Z2. Then White chooses one of these sets Y2 and Z2 (let us denote it by X3) and partitions it into some sets Y3 and Z3, and so on. Thus, the players construct a countably infinite sequence X1, X2, X3,. .. of sets. White wins if n°OXi # 0 and Black wins if n9°1Xi = 0. If the statement of the problem were false, that is, if there were a family F of sets satisfying (a) and (b) such that n°O=1Xi # 0 for any decreasing sequence of sets Xi, then White would have a winning strategy for any set X1 E .F: White would partition X2k+1 so that Y2k+1, Z2k+1 E F and, from among
130
3. SOLUTIONS TO THE PROBLEMS
Y2k and Z2k, would choose a set contained in F. Finding this kind of set families is motivated by the fact that we do not know who has a winning strategy in sets X1 of different cardinalities.
Problem C.S. Let G be a 2-connected nonbipartite graph on 2n vertices. Show that the vertex set of G can be split into two classes of n elements each such that the edges joining the two classes form a connected, spanning subgraph.
Solution. Let F be a spanning tree of the graph G. The vertices of F can be colored with red and blue so that if two vertices are of the same color then they are not adjacent. Actually, F has exactly two colorings like this, which can be obtained from each other by interchanging the colors red and blue. Notice that the statement of the problem is equivalent to the following one: There exists a spanning tree F in G such that the numbers of the red and blue vertices are the same.
Let F1 and F2 be two spanning trees in G, and let x be a vertex that is a leaf (a vertex of degree one) in both spanning trees. We say that the spanning trees F1 and F2 are neighboring at x if F1 - x = F2 - x. We will prove the following lemma, which implies the statement above, as we will see.
Lemma. Let F and F' be two spanning trees of G, and let T be a common subtree of them. Then there is a sequence F0 = F, F1, . . . , Fk = F' of spanning trees such that T C_ Fi and Fi and Fs+1 are neighboring at some vertex not in T.
Proof. We prove the lemma by "backward" induction on the number of vertices in T. If T has 2n - 1 vertices, then F and F' are neighboring. Suppose now that T has 2n - 1 vertices (l > 2). Let xy and uv be an edge of F and F' leaving T, respectively, such that x, u E T. If xy = uv then T' = T+xy is a common subtree and there is a desired sequence of spanning trees by the inductional hypothesis. If y # v, then let F" be a spanning tree containing the tree T+xy+uv as a subgraph. Then T+xy C Ff1F", and so by the inductional hypothesis, there is a sequence F = F0, F1, . . . , F9' = F" of spanning trees such that F i and Fi+1 are neighboring (i = 0, 1, ... , rn-1) and T C T+xy g Fi. Similarly, T+uv C_ F'f1F" and so by the inductional hypothesis, there is a sequence F" = F,,,.+1, ... , Fk = F' of spanning trees such that F i and Fi+1 are neighboring (i = m, m + 1, ... , k - 1) and T C T + uv C Fi, and F = FO, Fl, ... , Fk = F' is the desired sequence of spanning trees. Finally, suppose that xy # uv but y = v. Since G - y is connected, it
has a vertex z ¢ T joined to a vertex w E T. Now, let F" and F"' each be a spanning tree containing T + xy + wz and T + uv + wz, respectively. Then, as above, there is a desired sequence of spanning trees from F to F", from F" to F"', and from F"' to F', and the union of these sequences is a desired sequence from F to F'.
3.2 COMBINATORICS
131
Since G is nonbipartite, it contains a cycle C of odd length. Let x be a vertex of C, and let y and z be its neighbors in C. Let To be a spanning tree of G - x containing C - x. Let F = To + xy and F' = To + xz. Color F so that x is red.
According to the lemma, there is a sequence F = FO, F1, ... , Fk = F' of spanning trees such that Fi and Ft+i are neighboring at some vertex xi # x. Color Fi with colors red and blue so that the colorings of Fi and Fi+1 differ from each other at most in the color of xi. It defines some colorings of the spanning trees F1, . . . , Fk = F' starting with the coloring of FO = F such that x is red in every coloring. Clearly, every vertex got different colors in the colorings of F and F', except x. Let ai denote the number of red vertices in the coloring of Fi. Then lai+1 - ail < 1, ao + ak = 2n + 1.
Thus,
either
ao < n < ak
ak < n < ao,
or
so there is an index 0 < i < k such that ai = n, and this is what we wanted to prove.
Problem C.9. Let An denote the set of all mappings f : {1, 2, ... , n} -> ... , n} such that f -1(i) := {k : f (k) = i} # 0 implies f -1(j) # 0, j E
11, 2,
{1, 2,...,i}. Prove kn
0o
lAnl=Elk+1 k=0
Solution 1. Let an denote the cardinality of the set An. Then the number of mappings f E An such that f (i) = 1 holds for exactly l integers i is equal to (1) an_i. Since 1 is positive for any function f, we have n-1
n
an = t=1
al (7) an-1 = E t=0
forn>1. Using the notation bn = an/n!, we have bo = 1 and n-1
bn = 1=o
1
(n
-l)! bl.
Clearly, bn < 2n, so if Izl < 1/2, then the series E°° 0 bnzn is absolutely
3. SOLUTIONS TO THE PROBLEMS
132
convergent. For the sum f (z) of this series, we have 0o n-1
00
1
f(z)=Ebnzn=l+> n=0
(n - l)!
n=1 1=0
00
blzn
00
+E
1
(n - l)!
n=1+1
1=0
zn
b,
00
= 1 + 1:(ez - 1)blzt = 1 + (ez - 1V (z). 1=0
It implies that 00
A z) = 12 1 - 1ez/2 00
k=O
2k+1
00
k=0 n=0
1 kn n z n! 2k+1
0o
kn
1
n=0
00
ekz
(;n;-.l t
k=0
zn
2k+1
Thus, 1
bn
= n!
kn
00
E
k=0
2k+1
which we wanted to prove.
Solution 2. Let sn,i denote the number of surjective mappings of the set {1, 2, ... , n} onto the set {1, 2, ... , i}. Then the cardinality of the set An is
Counting the mappings of the set {1, 2, ... , n} onto the set {1, 2, ... , k}, we get the equality k
kn=
(k)
Sn,i.
Now, we have to prove that
M k
n
i=1
/k\ 2
Sn,i -
2k+1 Sn,i, k=1 i=1
that is, using Sn,n+l = Sn,n+2 = . . . = 0 and supposing that the rearrangement does not change the sum, we have to prove that n i=1
S n,i
-=
k) 2k+1
n i=1
k=i
S n,i.
3.2 COMBINATORICS
133
In the parentheses, we have the sum of the terms of a negative binomial distribution ((k) /2k+1 is the probability of the event that flipping a coin, we get the (i + 1)st head in the (k + 1)st flip), and so the coefficient of every term s,,,,i on the right is 1. Since the series in the parentheses are absolutely convergent, the rearrangement did not change the sum, and the proof is complete.
Problem C.10. Let G be an infinite graph such that for any countably infinite vertex set A there is a vertex p joined to infinitely many elements of A. Show that G has a countably infinite vertex set A such that G contains uncountably infinitely many vertices p joined to infinitely many elements of A.
Solution. We prove the statement by contradiction. Assume that G is a graph with vertex set V such that (1) for any countably infinite vertex set A C V, the set HA of vertices p E V - A joined to infinitely many elements of A is a nonempty, countable set. The set HA is obviously infinite since if not, then for the set B = AUHA,
there is no vertex p c V - B joined to infinitely many elements of B. It implies, that deleting finitely many vertices from G, the resulting graph has property (1). Notice that the vertex set V is clearly uncountable since (1) holds for A=V as well. Now, we show that we can delete finitely many vertices from G so that
the resulting graph G1 has no finite vertex set X covering V with the exception of a countable set. (A vertex set X covers the union of the sets of neighbors of the elements of X.) Suppose it is not the case. Then there is a finite set X0 covering all elements of V except a countable set. Deleting X0, the resulting graph still must have a finite set X1 with the same property, and so on. So if we do not get a desired graph after finitely many steps then we can define the pairwise disjoint finite sets Xn covering V with the exception of some countable sets. Let A = U1 0Xn. Then every element of V (except the elements of a countable set) is joined to infinitely many elements of A, that is, V - HA is countable. Since HA is countable by (1), it implies that V is countable, a contradiction. As we have seen, C1 has property (1) as well, so we may assume that G1 = G. Let A C V be an arbitrary countably infinite set. We show that there is a countably infinite set F(A) C V - A such that HA C F(A) and for any finite subset {a1, . . . , an} of A, the set F(A) has infinitely many elements
not joined to any vertex ai (i = 1, ... , n). Let A = jai, a2.... } and for each n, put infinitely many elements of V - A not joined to any element of Jai,-, an} into F(A). It can be done, as we have seen. Then take the union of HA and the set of the elements obtained.
134
3. SOLUTIONS TO THE PROBLEMS
Now, let A0 C V be an arbitrary countably infinite set, and let
Ai = F(A0 u Al u ... u Ai-1)
(i = 1, 2, ... ),
00
A,,, = F U Ai
.
For any element p E & and for any index i, p is joined to finitely many
elements of A. Really, p E A, C V - U°=°0Ai C V - Ai+1, and so p F(Ao uAl u... UAi_1) D HA,. Let u°_oA1 = {al, a2.... }, A,,, = {bl, b2, ... }. Let us define an infinite set C = {c1, c2, ... } such that if C C U°_0Ai and n < m, then cn,, is joined to neither an nor bn. It yields the desired contradiction since p V U°oAi
for p E HC (if p = an, then it is joined to at most n elements c,n), and so, p E HC C HuOOOAi C F (U°OoAi) = A. However, it is a contradiction again, since if p = bn, then it is joined to at most the elements c1i c2, ... , cn of C. Thus, let,cl be an arbitrary element of A0, and suppose that C1, c2, ... , cn are given so that {al, a2i ... , an, C1, C2, ... , Cn} C Ao U Al U U AN for some N. Then, AN+1 has an infinite subset T such that the elements of T are not joined to any of the elements al, a2,. .., an. As we have seen above, each of the elements bl, b2i ... , bn is joined to finitely many elements of T, so by choosing cn+1 out of the rest of T, we meet the requirements.
Remark. Assuming the continuum hypothesis, it is possible to construct a graph G satisfying the conditions of the problem in which every countably infinite set A has a countably infinite subset B such that HB is countable.
Problem C.11. Let f be a family of finite subsets of an infinite set X such that every finite subset of X can be represented as the union of two disjoint sets from R. Prove that for every positive integer k there is a subset of X that can be represented in at least k different ways as the union of two disjoint sets from f. Solution. The solution is based on the following theorem of Ramsey.
Theorem. Let Y be an arbitrary infinite set and let n be an arbitrary natural number. Subdivide the family of all subsets of n elements of Y into two classes. Then, there exists an infinite set Y' C Y such that every subset of n elements of the set Y' belongs to the very same class. For a set A, let [A] n denote the set of subsets of n elements of the set A. For a given natural number k, we show that there is a natural number
m > k and an infinite set Z C Y such that [Z]' C X It implies the statement of the problem, since then any subset of 2m elements of the set Z can be obtained as the union of two disjoint members of f in at least 2n'') > k different ways. We find a desired set Z as follows. Subdivide the set [X] 2 into two classes depending on whether or not the elements are in H. By Ramsey's
3.2 COMBINATORICS
135
theorem above, there exists an infinite set X2 C X such that either
[X2]2 c H
or
[X2]2 n h = 0.
Then classify the subsets of three elements of X2 based on whether or not they belong to fi. Again by Ramsey's theorem above, there exists an infinite set X3 C X2 such that either
[X3]3 C f
or
[X3]3 of = 0.
Proceeding further, we obtain a sequence X D X2 subsets of X with the following property:
either
[X,,,.]' c H
or
X2k of infinite
[X,,,,]' n R = 0
for 2 < m < 2k. For such an m, we naturally have either
[X2k]' C fl
or
[X2k], n x = 0
as well. We are done if we can prove that the second possibility cannot be the case. Actually, choose a subset A of 2k elements of the set X2k. By the conditions of the problem (which are used only at this point), A = B U C for some sets B, C E One of these sets, say B, has at least k elements. Then for m = JBI, we have B E [X2k]m n f, that is, [X2k]m o f # 0, and so [X2k]m C which we wanted to prove. Remarks.
1. We did not use the fact that the finite subsets of X can be obtained as the union of two disjoint sets in fi; this part of the condition can be omitted. Furthermore, it is sufficient to assume that for a fixed integer
c, the finite subsets of X can be obtained as the union of at most c members of fi. 2. Considering natural numbers instead of sets and sum instead of union, we get the following analogous problem: let H be an arbitrary set of natural numbers. Suppose that every natural number is the sum of two elements of H. Does it imply that for any natural number k, there is a number that can be obtained as the sum of two elements of H in at least k different ways? This problem of S. Sidon is open. However, the multiplicative version of the problem (where we say "product" instead of "sum" in all cases) has been answered affirmatively. It follows from the statement of the original problem, as follows: let X be the set of all prime numbers, and assign the set of prime divisors to any natural number in H. Thus, we obtain a family 71 meeting the conditions of the original problem. Thus, there exists a set A C X that can be obtained as the union of two disjoint members of 'H in at least k different ways. Then, the product of the prime divisors in A can be obtained as the product of two elements of H in at least k different ways.
3. SOLUTIONS TO THE PROBLEMS
136
Problem C.12. Let the operation f of k variables defined on the set ... , n} be called friendly toward the binary relation p defined on the
{1, 2,
same set if
f(a1,a2,...,ak) p f (bi, b2, . . ., bk) implies ai p bi for at least one i, 1 < i < k. Show that if the operation f is friendly toward the relations "equal to" and "less than," then it is friendly toward all binary relations. Solution. Since f is friendly with the relations "equal to" and "less than," we have
f(1,...,1) < f(2,...,2) <... < f(n,...,n) and thus f (i, ... , i) = i for all i. Hence, the following property holds:
(*) If f(al,...,ak) = a, then there is an i such that ai = a (since f (a1i . , ak) = a = f (a, ... , a) and f is friendly with the relation .
.
"equal to")-
It is sufficient to prove that if f (al, ... , ak) = a and f (bl, ... , bk) = b then there is a subscript i such that ai = a and bi = b. Let I = {i I ai = a} and J = j j j bj = b}. Property (*) shows that I, J 0; however, what we need is that I fl J # 0. This is obviously true for n = 1. Now, let n > 2, and assume that I fl J = 0. We distinguish two cases: Case 1. a b. Let ci = b for i E I and ci = a otherwise. Furthermore, let di = a if i E J and di = b otherwise. Then f (cl, ... , Ck) #
f (al, ... , ak) = a (since ci = b # a = ai if i E I and ci = a # ai
if
i V I), and similarly f (d1i . . . , dk) # f (b1, . . . , bk) = b. Property (*) implies that f (C1, ... , ck) and f (d1, . . . , dk) could only be a or b and thus f (c1 .... ck) = b and f (d1 .... dk) = a. Without loss of generality, we may assume that a < b, that is, f (d1, ... , dk) < f (c1, ... , ck). Since f is friendly with the relation "less than" we must have an i such that di < ci.
On the other hand, however, if i E I or i E J then ci = di; otherwise, ci = a < b = di, a contradiction.
Case 2. a = b. Let ci = c(# a) if i E I and ci = a otherwise; similarly di = c if i E J and di = a otherwise; and finally ei = a if i E J and ei = c otherwise. Similarly to the previous case, we have f(c1,...,ck) f(a1,...,ak) = a, f(d1,...,dk) f(b1,...,bk) = b = a and f(e1,...,ek) # f(d1,...,dk). However, all the values f(c1,...,ck), f (d1i ... , dk) and f (e1, ... , ek) must be either a or c, and thus A C1, ... , ck) = f (d1, ... , dk) = c, and so f (e1, ... , ek) = a. Again, it is enough to see the case when a < b, that is, f (el i ... , ek) < f (cl,... , Ck), which implies the existence of an index i such that ei < ci. However, we
haveci=ei=cifiEI,Ci=ei=aifieJ,andc2=a
3.2 COMBINATORICS
137
Problem C.13. Let g(n, k) denote the number of strongly connected, simple directed graphs with n vertices and k edges. (Simple means no loops or multiple edges.) Show that n2-n
(-1)kg(n, k) = (n - 1)!. k=n
Solution 1. We prove the statement by induction on n. For n = 2, the statement is obviously true. Let us write the left-hand side in the form
E(-1)IE(G)I ,
(1)
G
where G runs over the graphs described above. For every G in the summation (1), consider the edges of G having one end vertex at n. We distinguish two cases:
Case 1. G has exactly two edges incident to n and the other endpoint of both of them is the same vertex p. In this case, G\{n} is strictly connected as well, and so (-1)IE(G\{n})1 = (_l)IE(G)I.
Thus, for a fixed p, the sum of these terms is (n - 2)! according to the induction assumption and summing them up for p = 1, ... , (n - 1), we get exactly (n - 1)!.
Case 2. There are more than two edges of G incident to n, or there are two such edges of it having different other endvertices. There must be two vertices p # q, 1 < p, q < n -1, such that pn E E(G)
and nq E E(G). In this case, the graph having the same edge set as G, except that it has the edge pq if G does not have it, or it does not have this edge if G does have it, is strongly connected if and only if G is strongly connected; thus we can pair these graphs such that every such graph G not containing the edge pq has its pair G+pq. For every graph, we have a pair, and the sum in (1) is 0 for such a pair; thus, the sum is 0 for the graphs in this case. Summing up (-1)IE(G)I for the graphs in Cases 1 and 2, we get that (1)
is (n - 1)!. Solution 2. If g is a set of graphs satisfying the properties of the problem, let us define µ(9) = 1: (-1)IE(G)I GEG
For an arbitrary graph G, assign the permutation {al = 1, a2, ... , an} of the vertex set such that if ak is already chosen we pick ak+l as the smallest
3. SOLUTIONS TO THE PROBLEMS
138
number having an edge from the set {al, a2i ... , ak} to it. (We always have such an edge and vertex since the graph G is strongly connected.) For any given permutation 7r, let G, denote the set of graphs having ir as the assigned permutation. Since we have
1(-1)IE(G)I = G
µ49r), ITESn_1
and since the number of permutations of the set {2, ... , n} is (n - 1)!, it suffices to prove that µ(gn) = 1 for all 7r. Let us color the edges of these graphs with blue and red as follows: let the edges going along the permutation 7r be colored blue and the other ones colored red (that is, the blue edges are of form aiaj for i < j). Clearly, (*) for every vertex v # 1, there is a blue edge pointing to v, but there is no blue edge akal if a1 < aj for some k < j < 1. If Ga is the set of graphs corresponding to a fixed set a of blue edges, that is, having exactly those edges blue that belong to a (provided we have the fixed permutation 7r), then we have
ii(Gir) = E µ(9a), 9.99. Let now us consider these sums µ(!9a) now. If in any graph in Ga there is
a red edge aiaj such that i - j > 1, then pick the edge having the smallest j and among them that having the biggest i. Let g j be the subset of Ga in which the graphs have aiaj as the red edge picked, and let G E 9,;j. The way we chose j ensures that the edges aiaj, a3 a.j_1, ... , a2a1 are edges in G
and furthermore - because of the way we constructed 7r - we can reach ai_1 from the vertex al = 1 through blue edges only. The graph, having the same edge set as G with the exception of the edge alai-1, is strongly connected if and only if G is strongly connected. Thus, we can pair the graphs of g4j so that the pairs differ only in the edge aiai_1. Such a pair contributes 0 to the sum; thus, µ(9',j) = 0 and Eµ(G J) = 0. i,,j
There is exactly one set of red edges not containing any red edge to pick up, namely {aiai-1 1 i = 2, ... , n}, and so µ(Ga) =
(-1)IaI(-1)n-1= (-1)IaI-(n-1).
It is easy to see that any set of edges satisfying property (*) and having all edges going along the permutation can be chosen as the set a of blue edges. Thus, we have (_1)Ia1-(n-1)
r
c. Sgn P2
P3
12=113=1
...
L (P,) PP
1n=1
7
Pi (-1)l'-1
= fT
9=213=1
(-1)13-1
3)
7
=
3.2 COMBINATORICS
139
where Pk denotes the maximal in-degree of vertex k. Since we have n,
E(-1t) t j=1
1
(2) = (-1)
'
(-1)t tj =0
1j)
C-
('))
( -1)(0 - 1)
1,
we get p(G,) = 1, and topthe proof is complete.
Problem C.14. Let T be a triangulation of an n-dimensional sphere, and to each vertex of T let us assign a nonzero vector of a linear space V. Show that if T has an n-dimensional simplex such that the vectors assigned to the vertices of this simplex are linearly independent, then another such simplex must also exist.
Solution 1. Let A = (ao, . . . , an) be a simplex in T such that the vectors . . , v(an) assigned to the vertices of A are linearly independent. We will call an (n - 1)-dimensional simplex in T red if for any index 0 < i < n - 1, only one vector in the set
v (ao), .
Xi = (v(ao),...,v(ai)) - (v(ao),...,v(ai-i))
is assigned to the vertices of it. (Here, (Vi, ... ) vk) denotes the subspace generated by the vectors Vi, ... , vk.) Obviously, (ao, ... , an) is red. Lemma. If the vectors assigned to the vertices of an n-dimensional simplex in T are not linearly independent, then the number of its red faces is 0 or 2.
Proof. If the set v(S) of vectors assigned to the vertices of S does not contain any vector in one of the sets Xo, . . . , Xn_1i then obviously S has no red face. If v(S) contains an element of each of the sets X0, . . . , Xn_1 and U Xn_1 then the vectors assigned to the vertices of a vector not in Xo U S are linearly independent, a contradiction. So, we may assume that v(S) contains an element of each set Xj (j = 0, 1, . . . , n - 1) and that it contains exactly two elements of, say, Xi and exactly one element of the other XD's.
Then S has two red faces by which can be obtained from S, deleting one of the vertices with vectors in Xi assigned to it. The proof of the lemma is complete.
Now, consider the graph G whose vertices are the n-dimensional simplices of T and where two vertices are joined by an edge if they share a common red face. In this graph, the degree of the simplex A is 1. Thus, G contains one more vertex of odd degree. By the above lemma, the vectors assigned to the vertices of the corresponding simplex are linearly independent which we wanted to prove.
Solution 2. We prove the following, slightly more general, statement: If K is an arbitrary n-dimensional complex with boundary 0 (that is, K is a cycle mod 2), and if we assign nonzero vectors of the linear space V to
140
3. SOLUTIONS TO THE PROBLEMS
the 0-dimensional simplices (that is, the vertices) of K, then the number of n-dimensional simplices with linearly independent assigned vectors cannot be 1. We prove the statement by induction on n. It obviously holds for n = 1.
Suppose that n > 2. Suppose that dimv(A) = n + 1 for A E K. We have to show that K contains one more such simplex. Let B be an (n - 1)dimensional face of A. If C A is a simplex such that B is a face of it, then v(C) V= v(B) implies that dim v(C) > dim v(B) = n, that is, C has the desired properties. So, we may assume that if C A and B is a face of A, then v(C) C v(B). Let Ko = IS E KIv(S) C_ v(B)} and let K1 = 6(Ko), where 6 denotes the boundary. Obviously, 6(K1) = 6(6(Ko)) = 0, and since A 0 Ko and 6(K) = 0, B is thus the face of an odd number of simplices in K0, that is, B E K1. By the induction hypothesis, there exists a simplex B1 B in K1 such that dimv(B1) = dimv(B), and so v(B1) = v(B). Since B1 E 6(Ko)
but B1 ¢ 6(K), there is a thus simplex Al in K such that B1 is a face of Al but v(Al) V. v(B), and so dimv(Al) > dimv(BI) = n. So, the vectors assigned to the vertices of Al are linearly independent. To make the proof complete, notice that it is impossible that A = Al because if it is the case then A has two faces B and B1 such that v(B) = v(Bi) and so dimv(A) = dimv(B) = n, a contradiction.
Problem C.15. F o r a real number x, let IIxII denote the distance between x and the closest integer. Let 0 < xn < 1 (n = 1, 2, ... ), and let 6 > 0. Show that there exist infinitely many pairs (n, m) of indices such that n 54 m and 1
Ilxn -
xmll <minIE,2In-ml
Solution. Let us fix E > 0. We show that if n is large enough, 0 < xi < 1 (i = 1, ... , n), then there exists a pair (i, j) of indices such that IIxi xjII < min(E,1/over2l i -ii). It suffices since it implies that by subdividing the infinite sequence x1i X2.... into blocks of n terms, we can find a desired
pair of indices in each block and the difference Ii - j I does not change if the indices in the block are replaced by the indices in the whole sequence. Let f : {1, . . . , n} -+ {1, . . . , n} be a permutation such that
0 <xf(1) <... <xf(n) < 1. We may assume that Ilxf(i) -xf(i+l)II <min
1
E,
21f(i) - f(i+1)I
for i = 1, ... , n since if it is not the case then we are done. Let A denote the set of the indices 1 < i < n such that Ilxf(i) - xf(i+1)Il > E. Since the
3.2 COMBINATORICS
141
cyclically taken intervals [xf(1), X f(2)), [x f(2), X f(3)), ... , [xf(n), xf(1)) cover the cyclic interval [0, 1)
1
EE+ - i=1 llxf(i) - xf (i+1) ll ? iEA iA
21f(i) 1- f (i + 1)1'
and JAI < 11E. That is,
1-IAIE>_ 2y If(z)-f(2+1)I Applying the inequality between the arithmetic and harmonic means, we get the inequality _
EIf(i)-f(z+1)I > iV A
2(1
2
-JAII )
In the sum I:n 1 I f (i) - f (i + 1)J, every integer j E {1, ... , n} appears exactly twice with positive or negative signs, so that the total sum of the signs is 0. Obviously, the sum is maximum when the large numbers have coefficient +2 and the small numbers have coefficient -2. Thus, n
If(i)-f(i+1)I < i=1
2
2
when n is even, and
i=1
If(i)-f(i+1)I
when n is odd. Combining this fact with the preceding inequality, we obtain 1 - IAIE > (1 - IAI/n)2, which is a contradiction if JAI > 1 and n is sufficienly large. If A = 0, then we get 1 = 1, which means that n is even and the numbers If (i) - f (i + 1) I are equal (mean inequality). However, if f (i) > n/2, then f (i) appears twice with positive sign, that is, f (i) > f (i - 1), f (i) > f (i + 1), implying that f (i - 1) = f (i + 1), which is possible only if n < 2. But it is easy to see that equality cannot hold in this case either, and we got the desired contradiction in all cases.
Remark. The existence of a pair (i, j) of indices such that Ilxi - xj II < min(e,1/cli - jI) can be proved if c < f (the proof is much more complicated). The statement is false if c > f since it would imply that for an irrational number a, there are infinitely many rational numbers p/q such that J a - p/qI < 1/cq2.
3. SOLUTIONS TO THE PROBLEMS
142
Problem C.16. Consider the lattice L of the contractions of a simple graph G (as sets of vertex pairs) with respect to inclusion. Let n > 1 be an arbitrary integer. Show that the identity n
xn V yi
n
=V
i-0
xA V
yi
0
j=0
i#j
holds if and only if G has no cycle of size at least n + 2.
Solution. It can be seen that a, b E Vk ozi holds if and only if there is a natural number 1 and a path a = u0, u1, ... , ui = b such that for
every 0 < j < 1, (ui, uj+1) E zi for some 0 < i < k. Furthermore, (a, b) E zl A Z2 holds if and only if there is a natural number I and a path a = uo, ul, ... , ui = b such that (ui, ui+1) E z1 and (ui, ui+1) E z2 for every 0 < i < 1. The inequality > always holds in equality (1), so what we have to prove is that the inequality < holds if and only if G has no cycle of at least n + 2 vertices. Suppose that v0, ... , vk is a cycle with k > n + 1. If x contracts (v0, vk), yi contracts (vi, vi+1) (0 < i < n), and y,i contracts the vertices vn, vn+1, .... vk, then the left-hand side of (1) is x and the right-hand side of it is not since (vo, vk) is not contracted in the right-hand side.
Conversely, suppose that G has no cycle of at least n+2 vertices, and let (a, b) E x AVa oyi. We show that (a, b) is an element of the right-hand side of (1) as well. The assumption (a, b) E xAV oyi implies that there is a path a = v0, v1, ... , vk = b in G such that (vt, vt+1) E x and (vt, vt+l) E A oyi for 0 < t < k. Let us fix t. There is a path vt = uo, ul, ... , ul = vt+1 such that for every 0 < s < 1, we have (u87 ue+1) E y n for some 0 << n. Since the vertices uo, u1, . . . , ui constitute a cycle, 1 + 1 < n + 2, that is, 1 < n and some of the y's do not appear among these y,,,,. Thus, (Vt,Vt+1) E
V
Yi
i=0,...,j -1,j+1,...n
for some j, and hence (a, b) is contained in the right-hand side of (1).
Problem C.17. Let G(V, E) be a connected graph, and let dG(x, y) denote the length of the shortest path joining x and y in G. Let rG(x) _ max{dG(x, y) : y E V} for x E V, and let r(G) = min{rG(x) : x E V}. Show that if r(G) _> 2, then G contains a path of length 2r(G) - 2 as an induced subgraph.
Solution 1. Let n = r(G), and let x be a vertex of G such that rG(x) = n and the cardinality of the set {y : d(x, y) = n} is minimum. Let xn be a vertex such that d(x, xn) = n, and let x = xo, x1i y = X2.... , xn be a path
3.2 COMBINATORICS
143
joining x and x,,,. We show that if G does not contain a path of length 2n-2 as an induced subgraph then rc(y) = n and that d(x, z) < n(z E V) implies that d(y, z) < n, which contradicts the choice of x, and d(y, xn) = n - 2.
rc(y) = n follows from the other statement, so it is sufficient to prove that one. Since d(y, z) < d(x, z) + 2, it is sufficient to consider the cases d(x, z) _
n-2andd(x,z)=n-1. Letx=z0,zl,...,zk=z(k=n-2orn-1)be a path of length d(x, z) between x and z. By the minimality of the paths,
if xi = yj (i > 0, j > 0) then i = j, but then the length of the path y = x2i .... xi = z , . .. , zk is at most k and we are done. Similarly, if xizj is an edge of the graph, then i - 1 < j. If this is an edge of G and i > 2, then the length of the path y = x2i . . , xi, zj, ... , zk is at most k (< n) and if i = 1 then the length of the path y, x1, zj, ... , zk is k - j +2. We are done
if either j > 2 or k = n - 2, so we may assume that j = 1 and k = n - 1 and then x , . .. X is an induced path of length 2n - 2. If G does not contain any edge xizj, then x , , . .. , X17 x0i z1, ... , zk is an induced path of length 2n - 2 or 2n - 1, which we wanted to prove.
Solution 2. (sketch) Every graph G with r(G) = n contains a connected, induced subgraph G' such that r(G') = n and r(G") < n for every connected, induced, proper subgraph G" of G', and it is sufficient to take such a graph G'. 1. If G' is a path, then its length is n - 1. 2. If G' - {x} is not connected, then it has two components and one of them is a path. 3. Let C be the set of vertices x such that there is no vertex y such that G - {y} is not connected and the component of G - {y} containing x is a path. Then C induces a 2-connected subgraph. 4. For every vertex x E C, we get a path of length k if we delete the vertices
of the component of G - {x} containing C - {x}. 5. For every vertex x E C, there is exactly one vertex y E C such that d(x,y) = n - k. 6. For such a pair of vertices {x, y}, the graph induced by C - {z, y} is not connected, that is, joining x and y in two components, we get a cycle of
length at least 2(n - k). 7. For some 0 < k < n - 2, the graph G' is a cycle of length 2(n - k) with a pending path of length k at each vertex of it. Remarks.
1. The second solution shows that the statement is sharp; it is false for path of length 2n - 1. 2. There is an infinite graph G such that r(G) = 3, and every induced path in G is of length at most 3.
Problem C.18. Given n points in a line so that any distance occurs at most twice, show that the number of distances occurring exactly once is at least [n/2].
3. SOLUTIONS TO THE PROBLEMS
144
Solution. Suppose that the points are on the real line at the numbers < Pn. For any index 1 < i < n, let us take the set Ai of P1 < P2 < lengths of segments from pi to the right, that is, Ai = {pj - pi i < j}. Obviously, IAiI = n - i. We prove (by contradiction) that IAi n Aj 1 < 1 for i < j. Suppose that :
u,vEAinAj,u#v,and, say, u=Pk1-pi=Pk2-pj and vPmt -pi= Pmt - Pj. But then, the distance pj - pi = Pk2 - Al = Pmt - Pml occurs three times, a contradiction. Now we estimate the number of distances occurring from below. As we have seen,
IAi-(A,U...UAi-1)I=IAi-((Ain A,)U...U(AinAi-))I = IAiI -
n - i - (i - 1) = n - 2i + 1. Hence,
IA1u...uAn,I =IAiu(A2-Al)u(A3-AAluA2))U...U(An-(Alu...UAn-1))I
that is, n2/4
IA1 U . . . U An
(n2 - 1)/4
if n if n
is even, is odd.
Now let d1 and d2 denote the number of distances occurring once and twice, respectively. Obviously,
d1 + 2d2 =
n
(2)
and
d1 + d2
>-
n2/4 if n is even, (n2 - 1)/4 if n is odd.
Subtracting the equality from the double of the inequality, we obtain the desired inequality, d1 > [n/2].
Problem C.19. Let rc be an arbitrary cardinality. Show that there exists a tournament T,, = E,.) such that for any coloring f : E, -+ n of the edge set E,,, there are three different vertices x0, x1, x2 E [/k, such that x0x1, x1x2, x2xo E Ek and
I{f (xoxl), f (xlx2), f (x2xo)}I < 2.
3.2 COMBINATORICS
145
(A tournament is a directed graph such that for any vertices x, y E V,,, x y exactly one of the relations xy E E,,, yx E E,, holds.)
Solution. By a famous theorem of Erdos and Rado, there exists a trianglefree graph G = (V, F) with chromatic number 2". Order the vertex set V of this graph G in an arbitrary way, and let < denote this ordering. Now we define a tournament on V as follows. For x, y c V, x < y, let xy E E if
xy E F and let yxEEifxy¢F. We show that the resulting tournament (V, E) meets the requirements of the problem. Let f : E -> n be an arbitrary coloring of E. For every x E V,
consider the set Ax = If (yx) : y < x, yx E E} c n. Let us fix a vertex x E V. The cardinality of the set {y E V : y < x, yx E F or x < y, xy c F} is greater than 2", so there is a vertex x' such that Ax = A,,, and x'x E E if x' < x, xx' E E if x < x'. Let x1 and x2 denote the smaller and the bigger element from among x and x', respectively. Then f (xlx2) E Ay1 = A,, that is, there exists a vertex xo E V such that xo < xi,
xoxl E E,
and
f(xoxl) = f(xlx2)
However, by the definition of E, xox1, xlx2 E F,
and so x2xo V F
and
x2xo E E,
that is, the vertices xo, x1i x2 have the desired properties. Remark. It is easy to see that we cannot replace 2 by 1 in the statement.
For any tournament T = (V, E) and for any ordering < of V, define a coloring f of E as follows: if x < y, then let f (xy) = 0 if xy E E and let f (yx) = 1 if yx E E. Obviously, there are no three vertices x, y, z E V such that xy, yz, zx E E and f is constant on these three edges.
Problem C.20. Some proper partitions P1,..
.
, P,,
of a finite set S
(that is, partitions containing at least two parts) are called independent if no matter how we choose one class from each partition, the intersection of the chosen classes is nonempty. Show that if the inequality
A
< IP11 ...IPn1
(*)
holds f o r some independent partitions, then P 1 ,. .. , P,, is maximal in the sense that there i s no partition P such that P , P 1 ,.. . , PP are independent.
3. SOLUTIONS TO THE PROBLEMS
146
On the other hand, show that inequality (*) is not necessary for this maximality.
Solution. We prove the statement by contradiction. Suppose that (1) holds and that there is a proper partition P0 such that P0, P1, . . . , P, are independent. Let us choose one of the classes of each partition and take the intersection of them. For the sake of brevity, let us call these intersections class intersections. Obviously, the total number of class intersections is IP0 P1 ... IPnI and any two class intersections are disjoint since among the classes constituting any two class intersections there are two classes belonging to the same partition, and so they are disjoint. On the other hand, all the class intersections are nonempty since Po, Pl, . . . , P,' are independent. Thus, we have IPOIIPII ... IPT < ISI.
(2)
Combining (1) and (2) and using the fact that ISI # 0, we obtain that IPnI < 2, which is a contradiction to the assumption that P0 is a proper partition. So, we proved that (1) implies the maximality. On the other hand, the next example shows that (1) is not necessary for the maximality. Let S = {al, . . . , a8}. Let P1={S11,S12},
P2={S21,S22}
be the partitions where Sll = {al, a2, a3, a4}, S12 = {a5, a6, a7, as},
S21 = {al,a5}, S22 = {a2, a3, a4, a6, a7, a8}.
Then Sll n S21 = {al},
(3)
and the other class intersections are not empty either, that is, P1, P2 are independent. Equation (1) does not hold since IP1IIP2I = 4 = ISI/2, and we show that {P1, P2} is maximal. Suppose that P0, P1, P2 are independent
for some proper partition P0 = {S01, S02, ... }. Then the intersection of Sol with the set (3) is nonempty and so al E S01. We similarly obtain al E S02. But then Sol fl S02 0, a contradiction to the assumption that P0 is a partition.
Problem C.21. Show that if k < n/2 and Y is a family of k x k submatrices of an n x n matrix such that any two intersect then
3.2 COMBINATORICS
147
Solution. For any k x k submatrix M E F, let RM and CM denote the k-tuple of its rows and columns, respectively. Obviously, RM and CM determine M in a unique way. The condition of the problem says that RMl n RM2 # 0
and
CM, n CM2 , 0
for any two matrices Ml, M2 E F. Let us take the families
R={RM:ME.F}
and
C={CM:ME.F}.
Then R and C are families of subsets of k elements of a set of n elements such that any two members of R and C have nonempty intersection, respectively. Thus, IR1, 10
(k - 1)
by the famous Erdos-Ko-Rado theorem, and so
(I)2. (n_1)2
Obviously, the bound is sharp since this is exactly the number of k x k submatrices containing a given element.
Problem C.22. Let us color the integers 1, 2, ... , N with three colors so that each color is given to more than N/4 integers. Show that the equation x = y + z has a solution in which x, y, z are of distinct colors.
Solution. We prove the statement by contradiction. Suppose that we can color the integers 1, 2, ... , N with red, green, and blue so that each color is
given to more than N/4 integers and there are no x, y, z of distinct colors, where x = y + z. We may assume that 1 is red. Then by our hypothesis, there are no green and blue integers such that their difference is 1. We will call a nonempty set S C {1, . . . , N} an interval if it consists of consecutive
integers, that is, if there are some integers 1 < a < b < N such that S = Is : a < s < b}. If we delete the red integers, then the remaining integers can be partitioned into intervals. Furthermore, each interval is monochromatic by the observation above, since if an interval contains some integers x < y of distinct colors, then there are two consecutive integers of distinct colors among x, x + 1.... l y. Suppose that there exist intervals of at least two integers in both colors
green and blue. Let A and B denote the longest green and blue interval, respectively.
If a E A and b E B, then ja - bl(= a - b orb - a) is not
red. Hence, the set C = {ja - bI : a C A, b E B} does not contain any red integer. If A = [al, a2], B = [b1, b2], then
r,=
( [b1 - a2, b2 - al] 1. [al - b2, a2 - b1]
if a2 < b1, if b2 < al,
148
3. SOLUTIONS TO THE PROBLEMS
so it is an interval of JAI + 1131 - 1 > JAI, JBI integers. The set C does not contain any red integers, so it is monochromatic, as we have seen above. But it is a contradiction to the maximal choice of the intervals A and B. On the other hand, at least one of the colors green and blue contains two consecutive integers, since otherwise the number of red integers would be greater than or equal to the number of green or blue integers. In that case, the number of blue or green integers would be at most N/2, a contradiction to the assumption that each color is given to more than N/4 integers. Suppose now that there are no two consecutive green integers but there
are two consecutive blue integers, that is, JBI > 2 for the longest blue
interval B. If 1 < s < N is green, then s > 2 and s - 1 ands + 1 are red. Suppose that the distance between any two green integers is at least 3. Then the intervals [s - 1, s + 1] are pairwise disjoint for the green integers
s and each of them contains two red integers. If N is not green, then it implies that the number of red integers is at least double the number of the green integers, which is impossible if each color is given to more than N/4 integers. If N is green, then it implies that nr > 2ny - 1, where n, and n9 are the number of red and green integers, respectively. If 2 is
not green, then 1 is not counted in the intervals above, so n,. > 2ny, a contradiction again. If 2 is green, then take a blue interval B = [b1, b2] such that IBI > 2, b2 < N. Now, b2 - 1 is blue, b2 + 1 is red, 2 is green, and they constitute a solution to the equation x = y + z such that x, y, z are of distinct colors, a contradiction. Thus, we may assume that there is a green integer s such that s + 2 is green as well. Either bl > s + 2 or b2 < s, since IBS > 2 and Bf1{s, s + 2} _ 0. We may assume that bl > s + 2. Consider the set
C={b-s:bEB}U{b-s-2:bEB}. Since bl < b2 so C is an interval such that ICI = JBI + 2. The interval C does not contain any red integers, so it is monochromatic. But C is not green because ICI > 2 and not blue because of the maximal choice of B, a contradiction.
Problem C.23. Suppose that a graph G is the union of three trees. Is it true that G can be covered by two planar graphs?
Solution. The answer is no. We construct a graph that is the union of three acyclic graphs but that cannot be covered by two planar graphs. Any acyclic graph can be extended to a tree, so it implies the statement.
Let A and B be a set of n and (3) elements, respectively, where the value of n will be determined later. Let A U B denote the vertex set of the graph. Let us choose three elements of A in all possible ways and join each of these triples to an element of B so that distinct triples are joined to distinct vertices in B. Let G denote the resulting bipartite graph. It can be obtained as the union of three acyclic graphs in the following way: for any vertex b E B, the three edges incident to b are put into three distinct
3.2 COMBINATORICS
149
subgraphs. Any vertex b E B is of degree one in each of the three resulting subgraphs, so these subgraphs do not contain any cycle. Suppose that G is the union of two planar graphs Gl and G2. We may assume that Gl and G2 have no common edge. For every vertex b E B, let us proceed as follows. Delete one of the edges of G incident to b so that the remaining two edges belong to the same graph Gi. (It can be done since two of the edges incident to B always belong to the very same subgraph.) Now, replace the vertex b and the remaining two edges incident to b by one edge joining the end vertices of these two edges in A. If Gl and G2 are planar, then the resulting graphs are as well. The resulting graphs are defined on A. The number of replacements is (3), and an edge is obtained in at most n - 2 different ways since a couple can be extended to a triple in at most this many ways. So, the resulting graph H has at least / l3))
n(n - 1)
n-2
6
edges. It is known that a planar graph of n vertices has at most 3n-6 edges
(if n > 3). Thus, H has at most 6n - 12 vertices. From these estimates, we obtain
n(n-1) <6n-12, 6
that is,
n2-37n+72 <0. But this inequality does not hold if n > 35. Thus, G cannot be covered by two planar graphs if n > 35.
3. SOLUTIONS TO THE PROBLEMS
150
3.3 THEORY OF FUNCTIONS Problem F.1. Prove that the function f (0)
dx
T
-
(x2 - 1)(1 - 192x2)
1
(where the positive value of the square root is taken) is monotonically decreasing in the interval 0 < 19 < 1.
Solution 1. Substitute
x2-1
+-
1-192x2
), the value t increases in (0, +oo).
While x increases in the interval 11, 1 19
Differentiating the relation 192t2x2
we obtain
dx dt
- t2 + x2 - 1 = 0, _ t(1 -192x2) x(1 + 02t2)
and, consequently, $
dx
f ('d)
(x2-1)(11)2x2) J1
dx t(1 - 192x2)
J0
dt
J0
dt
1+00
=
+°° dx +00
x(1 + 192t2) =
dt
t(1 - 02x2) dt (1 + t2)(1 + 192t2)
JO
Now, for increasing 19 the integrand decreases. Since the limits of integration are independent of 19, the integral is also monotonically decreasing.
Solution 2. Map the first quadrant of the z-plane (excluding the points z = 1 and z = 1/19 by semicircles open from below) to the w-plane with the help of the function w
do
=
Jo
(1 - (2)(1 - 1922)
(taking the value of the square root that is positive on the positive halfaxis). If, starting from 0, z = x + iy runs through the segment 0 < z < 1, then starting from 0, w = u + iv obviously runs through the segment
0<w< 0
i
dx (1 - x2)(1 - 192x2)
A.
3.3 THEORY OF FUNCTIONS
151
If 1 < z < -, then it is clear that w runs through the segment 79
dx
W
u=A, O
(x2 - 1)(1 -192x2)
o
B.
If z runs through the segment 1/19 < z < oc, then w runs from the point A + Bi along the horizontal line v = B to the point (A - C) + Bi, where C=f+00
19
dx (x2 - 1)(192x2 - 1)
On the other hand, if, starting from 0, z runs through the positive part of the imaginary axis, then, starting from 0, w runs through the segment
u=0,
0
where
dy
D = 1+00 o
(1 + y2) (1 + 192y2)
Since the mapping is domain preserving, we have the relations
B=D,
A=C,
the latter of which implies the statement.
Problem F.2. Denote by M(r, f) the maximum modulus on the circle zI = r of the transcendent entire function f(z), and by Mn(r, f) that of the nth partial sum of the power series of f (z). Prove the existence of an entire function fo(z) and a corresponding sequence of positive numbers rl < r2 < --> +oo such that
Mn(rn, f0) _ +oc. n-.oo M(rn, fo)
limsup
Solution. By a theorem of Fejer, there exists a power series anzn,
f (Z) n=o
which defines a regular function in the disc I zI < 1 such that and such that the sequence of the partial sums anzn
Sr(z) nL=moo
is unbounded at the point z = 1.
I f (z) I
3. SOLUTIONS TO THE PROBLEMS
152
We shall use this function f (z) for constructing the function fo(z) that meets the requirements of the problem. We define sequences of numbers nk, mk, and Ck as follows. Let no = mo = co = 0. Then suppose that nk-1, Mk-1, and ck_1 have already been defined.
We begin by defining mk. Since lim sup s(1) = oo, there is an Mk with n-+oO
k-1 Ismk(1)I > kEmaxIsn,(z)I; 1=0
Izl=k
in addition, we may assume that Mk > nk-1 We now define Ck. Since Sink (z) is continuous, for Ck sufficiently close to 1 we also have k-1 Sink (Ck) I > k E max I Snt (z) I l=0
jzj=k
We additionally assume that Ck is real, further and
Ck > Ck-1
ck > 1 - 2k.
Finally, we define nk in the following way. Since sn(z) tends to f (z) in the disc IzI < 1, for sufficiently large nk we have max Isnk(z)I < 1.
lzl=Ck
On the other hand, the absolute convergence of the power series of f (z) implies the convergence of E°° o Ianlck; thus, for sufficiently large nk,
E Ianlck < n>nk
In addition to these two inequalities, we require that nk satisfy nk > MkNow consider the power series n2
nl
zln
anzn+n=ni+1 > an \2/ n=1 ((
00
= 1:
nk
1:
k=1n=nk_1+1
+... 00
an
k/n -
L 18n,. k=1
\k/ _ Snk-1 ( Z ) } .
We show that the function fo(z) defined by this power series has the desired property.
Prescribing an arbitrarily large positive R, for k > 2R and IzI < R we have Ian (z/k)ni < Ianl(1/2n). Since E 0 Ianl(1/2n) is convergent, the
3.3 THEORY OF FUNCTIONS
153
power series of fo(z) is absolutely convergent in the disc IzI < R. Thus fo(z) is a transcendental entire function. We give an upper estimate for fo(z) on the circle IzI = kck. In view of our previous remarks, k-1
Ifo(z)I <
is-, (k) - s-'-1 (k) - S---- Gz) n
nk
00
E E an (k)
1=k+1 n=ni_1+1
-1
<2E
00
maxk Is..(z)I+ max Isnk(z)I+ > IzI=Ck
1=01zl=kc
i_.\i
,
Ianl
1=k+1 n=ni-1+1
k-1
<2Zmax i_ 1=0
nk
-___ I.
i_.\i
IzI=k
C-` i_
.
(
kck l
i_n
n>nk
k-1
<2EmaxIsn,(z)I+1+1. Setting E 1o maxlzl=k I sn, (z) I = Tk, we obtain
M(kck, f0) < 2Tk + 2.
Next we give a lower estimate for the corresponding partial sum on the circle IzI = kck; since we are concerned with the maximum, it is sufficient
to do this at the point z = kck. The modulus of the mkth partial sum = k-1
smk (k)
-Snk-1
(k) +
{sn,
l (k) -sn,-1 (k)}
k-1
> ISmk(ck)I -2EmaxIsn,(z)I ? (k-2)Tk; IzI=k 1=0
that is, Mmk (kck, fo) > (k - 2)Tk. Now, since lim sup Isn, (1) I = +oo, the numbers 1-+00
k-1
Tk = 1: Ima Isn,(z)I tend to +oo. Thus, for sufficiently large k we have Tk > 1 and, consequently, Mmk (kck, fO)
(k - 2)Tk
k-2
M(kck, fo)
-2(Tk+1)>
4
.
3. SOLUTIONS TO THE PROBLEMS
154
Therefore, taking r,,,,k = kck, and for n different from the 'Mk choosing r,,, with the consideration of monotonicity but otherwise arbitrarily, we obtain lira sup
Mn (rn, fo)
n-+,, M(rn, fo)
>_ lim sup Mmk (rmk , A) >_ lim sup k--+-k-+.
k-2
M(rmk, A)
4
= +oo .
Problem F.3. Let H be a set of real numbers that does not consist of 0 alone and is closed under addition. Further, let f (x) be a real-valued function defined on H and satisfying the following conditions:
f(x)f(y) if x < y
and f (x + y) = f (x) + f (y) (x, y E H).
Prove that f (x) = ex on H, where c is a nonnegative number.
Solution. Let xo be an element of H other than 0, and let c = f (xo)/xo. It can be seen by induction that f (nx) = n f (x) (n = 1, 2, ...) for every x in H. Therefore f (nxo) = cnxo.
(1)
Let y be an arbitrary element of H. Then there is a positive integer no such that (y+noxo)xo > 0. If n is a sufficiently large positive integer, then there exist two positive integers Mn and An such that
mnxo < n(y +noxo) < µnxo and
1 mn - µn1 = 1.
(2)
In view of the monotonicity of the function f (x), we have f (mnxo) < f (n(y + noxo)) < f (µnxo),
so by (1), crnnxo < n f (y + noxo)
It follows that
c1Lnx0.
Mn
c n xo < f(y + noxo) < c
n
xo.
(3)
According to (2),
Mn
nxo - y + noxo,
nxo --> y + noxo.
Thus, from (3) we obtain that f (y + noxo) = c(y + noxo). Consequently, f (y) = f (y+noxo)-f (noxo) = c(y+noxo) -cnoxo = cy. The monotonicity of the function f (x) yields c > 0.
3.3 THEORY OF FUNCTIONS
155
Problem F.4. Show that if f (x) is a real-valued, continuous function on the half-line 0 < x < oc, and 00
f2(x)dx < oo,
fo
then the function g(x) = f (x)
-
2e-x
J 0 et f (t) dt 0
satisfies
00
"0 f2 (x)dx.
g2(x)dx =
J0
Solution. We assume that the function f (x) is square integrable in the Lebesgue sense on the half-line (0, oo). The relation
f (x) - g(x) =
2e-'
J
x et f
(t) dt
(1)
0
implies that
(f (x) - g(x))' = f (x) + g(x) almost everywhere.
(2)
Using the Schwarz inequality, we obtain w
e-"
etf(t)dt <e-"
r
w/2
etf(t)dt +
(LW/ezt dt
f w fz (t) dt j/2
(L12 <e
/2
El et f (t) dt
f z (t) dt
J
e2t dt
+e
e-w
OF f(x)dx+ 100/2 f2(t)dt, o
whence it follows that lim
W---00
e
r etf(t)dt=0. Jo w
(3)
Relation (1) assures that f (x) -g(x), and therefore (1/2) (f (x) - g(x))2, is absolutely continuous on each bounded subinterval of (0, oo). By (1) and (2),
(fz(x)
- g2(x)) dx =
"
((f(x)_9(x))2\' J
2
(f (x) -2g(x))21J= 0
Making use of (3), the statement follows.
dx
2e - z"
et f (t) dt (Jow
2
3. SOLUTIONS TO THE PROBLEMS
156
Problem F.5. Prove that for every convex function f (x) defined on the interval -1 < x < 1 and having absolute value at most 1, there is a linear function h(x) such that
if(x)-h(x)ldx<4-V8-.
Solution. We prove more than stated. We establish the existence of a constant k such that
Without loss of generality, we may assume that f (x) is continuous even
at the endpoints of the interval [-1,1] and that f (-1) > f (l). Since f (x) is continuous and convex on [-1, 1], there is a largest interval [cl, c21 (-1 < c1 < c2 < 1) on which f (x) is minimal. Introduce the notation min f (x) = p, 1]
x
max11 f (x) = q.
Let 01(y) be the inverse of the restriction to [-1, cl] of the function f (x), and let
f-1(y) if p <- y <- f(1),
02(y)
1
if
f(1)
where f -1(y) denotes the inverse of the restriction to [c2, 11 of the function
f(x). Obviously, the function q5(y) _ 02(y) - 01(y) is continuous and strictly increasing on the interval [p, q]; further 0(p) = 02(P) - 4i (p) _ c2 - cl and O(q) = 02 (q) - 01(q) = 2. We distinguish between two cases. a. If c2 - c1 < 1, then by the above properties of the function 0(y) there is one and only one number k (E [p, q]) with 0(k) = 1. It can be shown
that k satisfies (1). Put 41(k) = d, 02(k) = e, D = (d, k), E = (e, k), G = (-1, k), H = (1, k), A = (-1,1), and B = (1, 1). The lines AD and BE intersect at a point F. By the convexity of f (x) and the relation If (x) I < 1, it is obvious that the graphs of the restrictions of f (x) to the segments [-1, d], [d, e], and [e, 1] lie in the triangles AGD, DFE, and EHB, respectively. Therefore, r1
J
1
d
e
1
If(x)-kldx= J If(x)-kIdx+ f If(x)-kldx+ f If(x)-kldx 1
d
e
< t(AGDA) +t(DFEA) +t(EHBo), where t stands for area. If F lies above the line y = -1, then - DE being the mid-parallel of the triangle ABF - the altitudes perpendicular
3.3 THEORY OF FUNCTIONS
157
to GH of the triangles in question are equal, so the sum of the areas of the triangles is 2
2
1m(GD+DE+EH) =m < 1, (2) 2
2
where m = BH. If F lies below the line y = -1, then (denoting by J the point of intersection of the line y = -1 and the segment AF, and further denoting by K the point of intersection of the line y = -1 and the segment BF) it is obvious that the graph of the restriction to [d, e] of f (x) remains in the trapezoid T determined by the vertices D, J, K, E and, consequently,
J
1
If (x) - kj dx < t(AGDn,) + t(T) + t(EHBL),
1
while m = BH > 1. Then
t(AGD1) +t(EHBA) = 2m GD + 2m EH = 2m. Since JK = (2m - 2)m-1, we have
t(T) =
2 (2mn 2
\
+ 1) (2 - m).
It follows that P1
=4- (-2 +m If(x)-kIdx< 4m-2-m2 m m 1/2
(M
m)
(3)
Relations (2) and (3) imply (1).
b. If c2 - c1 > 1, then let k = p. Setting A = (-1,1), B = (1,1), D = (cl,fp), E = (c2i p), G = (-1, p), and H = (1, p), it follows as before that
J
1
If (x) - pI dx < t(AGDn,) + t(EHBA) <
2
2
which completes the proof. Remarks.
1. Several participants have noted that the estimate cannot be improved in general. For instance, in the case of the function
(x){ 1+vf8-(x-1) ifif 0<x<1--,f2-/2, 1-/2<x<1, 1
the estimate is the best possible.
3. SOLUTIONS TO THE PROBLEMS
158
2. Paul Turin called the attention of the organizing committee to the fact that S. Bernstein (Doklady Akad. Nauk SSSR, 1927, 405-407) had proved the following theorem: Theorem. Let f (x) be an n+ 1 times differentiable function on the interval [-1, 11 satisfying the condition f (n+l) (x) > 0 on [-1, 1]. The expression
J If (X) - Rn(x)I dx, where Rn(x) denotes a polynomial of degree n, is minimal when Rn(x) is the Lagrange interpolation polynomial of degree n that coincides with f (x)
at the points cos (hir/(n + 2)) (h = 1,2,...n+ 1). Problem F.6. Find all linear homogeneous differential equations with continuous coefficients (on the whole real line) such that for any solution f (t) and any real number c, f (t + c) is also a solution.
Solution. As usual in the literature, we restrict attention to differential equations with the coefficient of the term of highest order identical to 1. Let the differential equation
y(n)(x) + fl(x)y(n-1)(x) + ... + fn(x)y(x) = 0
(1)
have the desired properties, and let ¢(x) be a solution. Let c be an arbitrary real number. Obviously, d'0(x + c) dxi
(\
dtt
) t=x+c
(i = 1, 2, ... , n);
(2)
since together with t(x) O (x + c)also satisfies (1), it follows that (d12
(t)1
dt
J
-
+ fl(x)
_'0(t)
dt
t_x+c
J
/ t=x+c
.+.... + fn(x) (Y'(t))t=x+c =
that is, 0(n) (t) + f1(t - )O(n-1) (t) +... + fn(t - C)t (t) = 0, and this is true for any real constant c and all real values of t. This means that all solutions of (1), so for example, n linearly independent solutions of (1), satisfy the differential equation y(n)(x)
+
f, (X
-
c)y(n-1)(x) +
... + fn(x - c)y(x) _ 0;
(3)
hence - as is well known - it follows that the coefficient functions of the differential equations (1) and (3) coincide: f:(x) = fs(x - c)
(i = 1,2,...,n).
(4)
3.3 THEORY OF FUNCTIONS
159
Choosing x = 0, we obtain fi(0) = fi(-c) for every c (i = 1, 2, ... , n). Thus fi(x) __ fi(0), and so the differential equation (1) has constant coefficients. Conversely, if (1) has constant coefficients, then from (2) it follows easily
that together with any solution 1(x) O(x + c) is also a solution. Remarks.
1. Assuming the conditions for a single c only, from (4) we obtain that the coefficient functions are periodic with period c.
2. If, for any solution O(x), 1(x + ci) and t(x + c2) are also solutions, then together with O(x) obviously 0((x + Cl) + c2) = q(x + c1 + c2) is also a solution. Consequently, assuming the conditions for a (finite or infinite) set {cQ}, the condition will also be fulfilled by the elements of the smallest additive semigroup that contains the numbers ca. In view of the previous remark, this implies that it is sufficient to assume the conditions only for values of c that generate an additive semigroup containing a sequence that tends to zero.
Problem F.7. Let F be a closed set in the n-dimensional Euclidean space. Construct a function that is 0 on F, positive outside F, and whose partial derivatives all exist. Solution. Define a function lpr(y) in the following way:
(y) =
eye 0
if if
jyj < r2, IyI ? r2.
Obviously, q5r(y) is positive if jyj < r2, and infinitely differentiable in the
intervals (-oo,r2) and (r2,+o0). Further, it is clear that (dkQJr(y)/dyk) has the form Rk(y)!ar(y), where Rk(y) is a rational fractional function; so lim dkq5r(y) x-.r2+0 dyk
= lim dkqir(y) = 0. x-.r2-0
dyk
Therefore (dk¢r(y)/dyk) is continuous in (-oo, r2) and (r2, +oo) and has right-hand and left-hand limits at the point y = r2, whence it follows by induction that Or (y) has continuous derivatives of any order at y = r2 and, consequently, at all points y. Denote by En the n-dimensional Euclidean space. For a E En, x E En, a= (a1, a2, ... , an), x = (x1, x2, ... , x,) Put ya(x) = (xi - a1)2 + (x2 - a2)2 + ... + (xn - an
)2.
Obviously, for every x, all partial derivatives of ya(x) exist. Finally, let G C EEC,, be an open ball of center a and radius r chosen arbitrarily, and let fG(X) _ Or(ya(x))
160
3. SOLUTIONS TO THE PROBLEMS
Since 0,(y) is infinitely differentiable for any y and all partial derivatives
of ya(x) exist for any x, all partial derivatives of fG(x) exist for any x. Moreover, it is evident that if x E G then 0 < ya(x) < r2, so fG(x) > 0, while if x 0 G then fG(x) = 0. Finally, each partial derivative of fG (x) is bounded since it is continuous everywhere and zero outside G.
The complement F of F relative to E, is an open set, so it can be represented in the form F = Uk 1Gk, where G1, G2, ... are open balls. Since each partial derivative of the functions fGl (x), fG2 (x), ... is
bounded, there exist positive constants ek (k = 1, 2, ...) such that all partial derivatives of order not higher than k of the function fGk (x) (including fGk(x) itself) have absolute value less than (1/ck)(1/2k) for every x. Then, obviously, the function series Co
1: Ck.fGk (x) k=1
is absolutely convergent; we set 00
F(x) _ E ckfGk (x) k=1
Forming a partial derivative of order i of the terms of the series, the definition of ck ensures that, beginning with the ith term, the absolute value of each term can be majorized by 1/2k, so the sum of the partial derivatives is absolutely convergent. It follows that all partial derivatives of F(x) exist. Finally, it is clear that F(x) = 0 for x E F and F(x) > 0 for x ¢ F, thus the function F(x) has the required properties. Remark. Let F1 and F2 be disjoint closed sets in the n-dimensional Euclidean space. By a similar method, one can construct an infinitely differentiable function that is equal to 0 on F1, equal to 1 on F2, and positive and less than 1 outside F1 and F2. Problem F.B. Let f be a continuous, nonconstant, real function, and assume the existence of an F such that f (x + y) = F [f (x), f (y)] for all real x and y. Prove that f is strictly monotone.
Solution. Suppose that f is not strictly monotone. Then by the continuity
of f there exist real numbers sl < s2 with f(si) = f (S2)- If E > 0 is arbitrary, then by the continuity of f there are values t1 < t2 in the closed interval [sl, 52] such that t2 - t1 < E and f (t1) = f (t2). Then, however,
f[t+(t2-t1)]=f[(t-t1)+t2]=F[f(t-t1),f(t2)] = F[f (t - t1), f (t1)] = f [(t - t1) + t1] = f (t) for all real values t. Thus 'r = t2 - tl is a period of f. Since r < E where e > 0 is arbitrary, it follows that the continuous function f has arbitrarily small periods. Hence f is constant, contrary to the assumption.
161
3.3 THEORY OF FUNCTIONS
Problem F.9. Let k be a positive integer, z a complex number, and e < 1/2 a positive number. Prove that the following inequality holds for infinitely many positive integers n:
(n-Wl l e J
zQ
> (2 - e)n .
o<e<
Solution. Put Cn - kt) e
an(z) = an = E
f
z ,
n = 1,2,....
0
We have to prove that limsup /anl > 1/2. If 1W + wk+1zI < 1, then 00
( k+1z ) = 1 1- w+w
m
M=1a=o
00
(T)zwm+tk
+
aw.
(1)
n=1
Therefore, the series 1 + E°O 1 anwn is the power series of the function 1/(1 - (w + wk+1z)) at the point w = 0. According to the CauchyHadamard criterion, it is sufficient to establish that the radius of convergence of the power series is not greater than 2. For this purpose, it is sufficient to show that some zero of the polynomial
1 - w - wk+1z (k > 1) has absolute value not greater than 2. If I1/zI < 2k+1, then this follows from the fact that the product of the zeros of the polynomial equals 1/z. If I1/zI > 2k+1, then Iwk+1zl < 1 < 11 - wI on the circle IwI = 2, so by Rouche's theorem the polynomial considered and the polynomial 1 - w have the same number of zeros inside this circle.
In the case z = -1/4, k = 1, using relation (1), it is easy to see that
an(-1/4) = (n + 1)/2n. This shows that the assertion concerning the lim sup cannot be improved. Remarks.
1. One participant proved the following, stronger, statement: lim sup Ian11/n > k/(k + 1), and this is sharp for every k; (We limsup Ian(z)I1/n assumes its minimum for z = (-kk/(k + omit the proof, which is rather long.) 2. Several participants proved that the an satisfy the following recursive 1)k+1).
definition: ao =
= ak = 1,
an+1 = an + an-kz
if n > k.
3. SOLUTIONS TO THE PROBLEMS
162
Problem F.10. Let f (x) be a real function such that lim
f(x = 1
x-+00 ex
and I f"(x)I < cl f'(x) I for all sufficiently large x. Prove that lim f 1(x) = 1. x-.+oo ex
Solution 1. Suppose that I f"(x)I < cl f'(x)I for x > xo. We first show that x > xo and t < 1/c imply
If'(x+t) <
1
1
If'(x)I
(1)
We may assume that I f' (x + t) I > I f' (x) 1, since otherwise (1) is trivially fulfilled. Put
to = min{t': t' > 0, If(x+t')I = If (x + t)IJ. Since the function I f'(l;) I is continuous and does not intersect the horizontal line of ordinate I f' (x + t) I for l; E [x, x + to) while it remains under this line at the point x, therefore I f'(6)I < I f'(x+t)I on the whole interval [x, x+to]. Prom the Lagrange mean value theorem,
If'(x+t)I - If'(x)I < If'(x+to)I - If'(x)I =tolf"(6I < tocl f'(6I < toclf'(x + t)I, which gives (1).
From (1) we see that if f'(x) = 0 for some x > xo, then I f'(x') I = 0 for all x' > x; but this is impossible since then e_ ' f (x) would tend to 0. Consequently, f'(x) is either positive for all x > xo or negative for all x > xo. However, because e-x f (x) -> 1, the function f (x) cannot be monotone decreasing. Thus f'(x) > 0
for all x
(2)
x0.
From (1) and (2), we obtain that in case x > xo + 1/c and t < 1/c,
(1-ct)f'(x+u) < f'(x) <
ctf'(x-u)
for 0 < u < t. Hence, by integration, (1 - ct) If (x + t) - f (x)] < tf' (x) < 1
1
If (x) - f (x - t)],
3.3 THEORY OF FUNCTIONS
163
that is,
1-ct rf(x+t)et- f(x)1 t
IL
ex+t
< f' (x) ex
J
ex
< t(1
1
Ct)
-
r f (x) L ex
f (x - t) a-tJ ex-t
(3)
We first consider the left-hand side of (3). For fixed t,
f `x+t) et -
lim
ex+t
Hence
lim inf X-00
f'(x) x e
x f (x)
= et - 1.
> (1 - ct) et - 1, t
and therefore
limnff (y)>tlimo1(1-ct)ett 1I=1. From the right-hand side of (3), it similarly follows that lim sup f f (x) ex
X-00
1 - e-t 1 = 11 t-.+0 L 1 - ct t J
< lim
1
1
which proves the statement.
Solution 2. We shall make use of the following theorem, which is well known and easy to prove.
Theorem. Let g be a two times differentiable function such that lima.+C,o g(x) exists and is finite, whereas Ig"(x)I < C if x > xo for suitable real numbers C and x0. Then limx_+0o g'(x) = 0.
Now let g(x) = (f (x)/ex). Then
f(x) 9 (x) =
f'(x)ex
= g"(x)
f"(x) - 2e (x) + f (x)
By assumption, lima-+co g(x) = 1. From the other assumption, it follows that for sufficiently large x the derivative f'(x) is of constant sign. Since limx..+c,. f (x) = +oo, it must be positive. Thus, if xl is sufficiently large and x > x1, then x
If'(x) - f'(xi)I 5 fxl f"(t)l dt < cJxlx f'(t) dt = f(x) - f(xi). Therefore I f"(x) I < c' f (x) and I f'(x) < c' f (x) with a suitable constant c'
and, consequently, Ig"(x)l < (3c' + 1)g(x) if x > x1. It follows that, for some C and x0, Ig"(x) I < C if x > x0.
164
3. SOLUTIONS TO THE PROBLEMS
By the theorem stated at the beginning,
lim g'(x) = x-++oo lim fi (x) ex- f
(x)
x-+oo
= 0,
whence limx.+,,o(f'(x)ex) = 1. Remark. In some sense, it is necessary to assume that I f"/ f I and Ig"I are bounded. To show this, let g(x) = 1
Then
g (x) =
+
sin x2
- sin x2 x2
x
+ 2 cos x 2
and
liminf g'(x) = -2 < 2 = limsupg'(x). x-+oo
x-+oo
Problem F.11. Find all continuous real functions f, g and h defined on the set of positive real numbers and satisfying the relation f(x + y) + g(xy) = h(x) + h(y)
for allx>0andy>0. Solution. Choose y = 1 in the equation, and then g(x) = h(x) - f(x + 1) + h(1)
(1)
for x > 0. Substituting (1) into the original equation, we obtain
h(x) + h(y) - h(xy) = f(x + y) - f(xy + 1) + h(1).
(2)
Put H(x, y) = h(x) + h(y) - h(xy). Then H(xy, Z) + H(x, Y) = H(x, yz) + H(y, z)
3)
for any triple of positive numbers x, y, z. By relation (2),
H(x, y) = f (x + y) - f (xy + 1) + h(1); putting this into (3), we find that
f(xy+z)- f(xy+1)+f(yz+1)= f(x+yz)+f(y+z)- f(x+y) (x, y, z > 0).
(4)
3.3 THEORY OF FUNCTIONS
165
Since f is continuous on the set of positive numbers, passing to the limit z -+ 0 (z > 0), from (4) it follows that f(xy) - f(xy + 1) + f(1) = f(x) + f(y) - f(x + y)
(5)
Introduce the notations f * (t) = f (t) - f (t + 1) + f (1)
(t > 0)
an d F(x, y) = f(x) + f(y) - f(x + y).
T h en
F(x + y, z) + F(x, y) = F(x, y +- z) + F(y, z)
(x, y, z > 0)
(6)
and, by (5),
F(x,y) = f*(xy)
(7)
Putting (7) into (6), we obtain f*(xz + yz) + f*(xy) = f*(xy +xz) + f*(yz)
(x,y,z > 0).
(8)
Hence, choosing z = 1/y and writing x
u = -, y
v = xy,
(9)
it follows that f*(u + 1) + f*(v) = f*(u + v) + f*(1)
(10)
for all positive values of u and v, since for positive u and v the system of equations (9) can be satisfied by suitable positive values x and y. From (10), interchanging u and v,
f*(u+1)+f*(v)= f*(v+1)+-f*(u), whence, taking v = 1, f*(u + 1) = f*(u) + f*(2) - f*(1).
Substituting this into (10), f*(u+v) + f*(1) = f*(u) + f*(v) + f*(2) - f*(1),
whence, by the continuity of f *,
f*(t)=at+/j,
3. SOLUTIONS TO THE PROBLEMS
166
where a and /3 are constants. Then in view of (5),
axy+/3= f(x)+f(y)-f(x+y), and therefore, with the notation j (x) = f (x) = (a/2)x2 - /3, we find that f (x + y) = f (x) + f (y). Thus f (x) = 'yx and
f(x) _ -2x2+yx+/3.
(11)
Putting (11) into (2), we see that the function
h(x) = h(x) + 2 x2 - yx - b (6= 2
- ry - 0 + h(1))
satisfies the equation
(x,y > 0),
h(x) + h(y) = h(xy) which yields h(x) = k In x. Consequently,
h(x) _ -2x2+yx+klnx-6.
(12)
Finally, from (1), using (11) and (12), it follows that
g(x)=kIn x+ax -2b-/3.
(13)
We have shown that the solutions of the equation can only be the functions of the forms (11), (12), and (13). On the other hand, it is easy to see that the functions (11), (12), and (13) are solutions of the equation for any choice of constants a, 0, y, b, and K.
Remark. Most participants first show that f, g, and h are two times continuously differentiable functions and then reduce the problem to a dif-
ferential equation. Several of them note that when using this method it is sufficient to assume the integrability of f, g, and h on every bounded, closed subinterval of the set of positive numbers.
Problem F.12. Let xo be a fixed real number, and let f be a regular complex function in the half-plane Re z > x0 for which there exists a nonnegative function F E Ll (-oo, oo) satisfying If (a + i/3) < F(/3) whenever
a > X0, -00 < /3 < +oo. Prove that
I
a-ioo
f(z)dz = 0.
Solution. Let x0 < al < a2. Let {i3,,} and {y,,} be sequences of real numbers tending to +oo such that F(O,,) -> 0 and F(-y,,) -> 0 . By the Cauchy integral theorem, al+i/jn
az+i(jn
f (z) dz+J i'Yn
ai+io3
-i'Yn
f (z) dz+
Ei)3n
dz+a
i-i'Yn
f (z)
f (z) dz = 0 la2 -iY n
3.3 THEORY OF FUNCTIONS
167
(the path of integration is always the connecting segment). Since
faz+ip
Jp01z
f (z) dz a1+iRn
f (a + ion) da <(a2-ai)F(/3 )-40,
a1
and similarly fcrl -i'Yn
J
f (z) dz < (a2 - ai)F(-'y) -+ 0,
az-iry, therefore
J
az+ioo
al+ioo
f (z) dz =
a1-ioo
J 12-ioo
f (z) dz,
(1)
which means that the integral in question is independent of a. (Here the improper integrals exist since the integrand admits an integrable majorant.) Denote by A the common value of the integrals appearing in (1). Apply our result to the function f (z)/z (which is analytic, for instance, in the half-plane Re z > 1). Then
B
= pa+boo f(z) dz = i f°° f(a + ij3) d
J
z
Ja-ioo
a > max{1, xo}
a'
a+
°°
is independent of a. Let a -* oo. Since f (a + i13)
a+i/3
< F(3)
(a > 1),
integration and transition to the limit can be interchanged by the theorem of Lebesgue, and we obtain B = 0. Then
A = A - aB
=
j
a
a+ioo
-zoo
f
(z) (1 - z) dz = -
foo
J
00
f (a + i,3) a
13
i'3
d,3.
Here, again, F(Q) is a common majorant of the integrands, and for each fixed value of /3 the integrand tends to 0 as a -a oc. Consequently, by the theorem of Lebesgue, the integral tends to 0, whence A = 0. Remark. All participants first show that the integral under consideration
is independent of a. Then some of them choose the function f (z)/z and the way described above, while others work with the function e-tz f (z), where t > 0 is a real number. They establish similarly (using the fact that this function also satisfies the conditions of the problem and tends to 0 as Re z -+ oo) that
a+i°o
e-"f (z) dz = 0;
a-ioo hence, letting t -+ 0, the desired equation follows.
3. SOLUTIONS TO THE PROBLEMS
168
Problem F.13. Let 7rn(x) be a polynomial of degree not exceeding n with real coefficients such that
Iirn(x)I<
1-x2
<x<1.
for
Then
Iirn(x)I < 2(n - 1).
Solution. The background of the problem is provided by the Markov inequality: If the polynomial P(x) of degree n has absolute value less than 1 through the interval (-1, 1), then its derivative has absolute value less than n2 on the same interval. Of similar type is Bernstein's theorem: If a trigonometric polynomial of order n has absolute value not greater than 1, then its derivative has absolute value not greater than n. We have strengthened the hypothesis of Markov's theorem and wish to prove an estimate much sharper than that of the Markov theorem. To this end, we shall need Bernstein's theorem cited above as well as the following theorem: Theorem. If a polynomial Q(x) of degree k satisfies
IQ(x)I<
11
x2
-1<x<1,
for
then
I Q(x) I < k + 1
for the same x.
For both theorems, see for example, I. P. Natanson, Constructive Function Theory 1-3, 1964-65 (translated from the Russian), sections V.1 and VI. 6.
Now we prove the statement of the problem. We may assume that 7rn (x) is nonconstant. From the relations irn (±1) = 0, it follows that 7rn(x) = (1 - x2) f (x), where f (x) is a polynomial of degree not exceeding
n-2 and
If(x)I <
11
x2
for
-1<x<1,
whence If (x) I < n - 1. Let x = cos 19. It is well known that in this case f (x) = f (cos 19) = F(19) is a trigonometric polynomial of order not exceeding n - 2. Write G(19) = F(19) sin 19. Since G(19) arises from f (x) 1 - x2 by the substitution x = cos 19, therefore I G(19) I < 1. The trigonometric polynomial F(19) has order not greater than n - 2, so G(19) has order not
greater than n - 1, that is, IG'(o)I < n -1.
(1)
We calculate the derivative of 7rn(cos19) with respect to 19 in two ways. On the one hand, we have 01rn(cos19) = 7rn(cos19)(-sin 19).
(2)
3.3 THEORY OF FUNCTIONS
169
On the other hand, d
ir,,,(cos'O) = d
sin V) = G'(19) sin O + G(t9) cos 19
= G'(t9) sin O + F(V) sin O cos V.
(3)
Comparing (2) and (3), -Ir'n (cos t9) = G'(0) + F(V) cos V,
(4)
that is, (n- 1)+(n- 1)lxl = (1 + Ixl)(n - 1) < 2(n- 1).
(5)
The proof is complete. Remark. It should be noted that for sin t9 = 0, we cannot divide by sin t9, but the continuity of 7rn(x) ensures that the assertion of the problem is true
also in this case. Equality can only hold if irn(x) = (1 - x2)Qn_2(x) or the negative of this, where Qn_2(x) stands for the Chebyshev polynomial of the second kind of degree n - 2. This case can also be characterized by the relation G(t9) = ± sin(n - 1)0. Then in the Bernstein inequality, for the values 19 corresponding to jxj = 1, we have equality, and in the triangle inequality applied in (5) the terms have equal sign, so equality really holds for these 7rn.
Problem F.14. Let a(x) and r(x) be positive continuous functions defined on the interval [0, oo), and let
liminf(x - r(x)) > 0. X-00
Assume that y(x) is a continuous function on the whole real line, that it is differentiable on [0, oo), and that it satisfies y'(x) = a(x)y(x - r(x))
on [0, oo). Prove that the limit li
y(x) exp -
f a(u)du
exists and is finite.
Solution. Integrating (2), we obtain y(x) = y(u) +
x
Ju
a(t)y(t - r(t)) dt (x > u > 0).
(3)
170
3. SOLUTIONS TO THE PROBLEMS
If y(x) is any continuous function on the interval (-oo, 0], then it is easy to see that y(x) can be uniquely extended to the whole real line so that (3) is valid. In fact, suppose that x0 is the supremum of those values up to which unique extension is possible. Then in the neighborhood of x0 of some radius 6, the function r(t) is greater than some positive E. Denote by y the smaller of the numbers a and 6. Then, by (3), the values of y(x) taken for x < xo -q/2 uniquely determine the values of y(x) in the interval (xo - 77/2, xo + 7)/2). This contradicts the choice of xo.
A similar reasoning shows that if y(x) is positive on (-oc, 0] then it is positive on the whole real line. In this case, with the help of (2) we obtain that y(x) is monotonically increasing on [0, oo). Put
z(x) = y(x)e- fo a(t) dt
(x > 0).
(4)
Differentiating and using (2) we obtain z'(x) = a(x)e- fo a(t) dt [y(x - r(x)) - y(x)]
(5)
To prove the statement of the problem, first suppose that y(x) is positive on (-oo, 0]. Then, as we have seen, y(x) is positive everywhere and increasing for x > 0. Thus, z(x) is positive for all x > 0; further, z(x) is monotone decreasing for sufficiently large x. Equation (1) implies that x - r(x) > 0 if x is large; so from (5) by the monotonicity of y(x) it follows that z'(x) is negative. Since z(x) is positive and decreasing, z(x) exists.
To prove the general case, represent the function y(x) on (-oo, 0] in the form y(x) = y1(x) - y2(x),
(6)
where yl(x) and Y2 (x) are positive, continuous functions. As we have seen, yl(x) and y2(x) can be extended to the whole real line so that they satisfy the differential equation (2) for x > 0. By the uniqueness of the solution of (2) mentioned above, it is also clear that (6) remains valid on the whole real line. Defining the functions zi(x) and z2(x) in analogy with (4), it follows as above that the limits lima-,,,, zl(x) and lim.,. z2(x) exist. This implies the existence of the limit
linnz(x) _ X-co lira (zl(x) - z2(x)) The solution of the problem is complete.
Problem F.15. Let Ai (i = 1, 2, ...) be a sequence of distinct positive numbers tending to infinity. Consider the set of all numbers representable in the form 00
µ= i-1
ni)i,
3.3 THEORY OF FUNCTIONS
171
where ni > 0 are integers and all but finitely many ni are 0. Let
L(x) _
M(x) _
and
1
a; <x
1. A <X
(In the latter sum, each p occurs as many times as its number of representations in the above form.) Prove that if lim
then
L(x + 1) L(x)
M(x + 1) - 1
lim Xoo M(x)
Solution. (When in the solution we speak of a p, we mean not only its value but also its representation by a fixed sequence {ni }, keeping in mind that a number can possibly be represented in several ways.)
If pi = >,
Ai and µ2 = E
then let µl Iµ2 mean that
0) (i = 1, 2, ... ). Then
E
µ=
Ai
n>1 integer a; na; Iµ
and
µ<x
Eµ=L: E Ai. µ<x n>1,a, nAiIµ
In the inner sum, p' = p - n)i is also a p-number; let us sum with respect to this. Then
Eµ'_µ'<x E
µ<x
n>1,a; nAi <x-µ'
Ai.
With the notation n> 1,aj nai
(replacing p' by p) we have
Ep=E£(x-µ). µ<x
(1)
µ<x
Apply this to x + 1 and subtract the two relations from each other:
E x<µ<x+l
1-t=E(G(x+1-1-)-L(x-p))+ E G(x+1-p). µ<x
x<µ<x+l
3. SOLUTIONS TO THE PROBLEMS
172
The left-hand side is at least x (M(x + 1) - M(x)), and the second sum on the right-hand side is at most L(1) (M(x + 1) - M(x)) (since L(y) is increasing). Consequently,
(x - L(1)) (M(x + 1) - M(x)) < E (L(x + 1 - µ) - L(x - µ)) µ<x
Here L(y) is expressed through the sequence Ai alone, and from the hypothesis relating to the latter we shall deduce that L(y)
whence for sufficiently large K, L(y+1)-L(y) < eL(y) if y > K. Therefore,
decomposing the sum based on whether µ < x - K or µ > x - K, and using (1) again,
E (L(x + 1 - µ) - L(x - µ)) < e E L(x -
µ<x-K
µ<x-K
<eEL(x-µ)=eEp<exM(x) µ<x
µ<x
and
E (L(x + 1- µ) - L(x - µ)) < L(K + 1)M(x). x-K<µ<x
We have thus obtained
(x - L(1)) (M(x + 1) - M(x)) < exM(x) + L(K + 1)M(x), M(x + 1) - M(x) < ex + L(K + 1) M(x) x - L(1) M(x + 1) - M(x)
-
< e,
lim sup X-00
and since e > 0 is arbitrary, lim sup X-00
M(x + 1) - M(x) M(x)
= 0.
It remains to prove (2). We may write L(x)
E 1> = ai<x Ai 1
A{<2
Ai
2+
Ai
> Z L(2) + 2 (L(x) - L(2 x)) = 2 xL(x). Now
ai.
L(x + 1) - L(x) n>_1,,\; x
(3)
3.3 THEORY OF FUNCTIONS
173
For a fixed e > 0, decompose the sum according to the cases Ai < e(x + 1) and Ai > e(x + 1):
E .\i E 1< a;<E(x+l) m
x- < n <
E Ai
I
a;<E(x+1)
-1
(since the interval corresponding to n has length not greater than
1/ min.\i= constant). Further, using (3), 1
min Ai
ai
< e(x + 1)
-
a; <e(x+1)
<
min Ai 2,-x
min)li
L (e(x + 1))
L(x) <
4e
mini L(x).
In the other part, n < (x + 1)/e(x + 1) = 1/e, so
xn1 1
/
\ (xn
1
n The inner sum consists of finitely many terms (for fixed e > 0) and, by the assumption, each of them satisfies the relation
L(x +
L(x)
=o(L(x))
as x -4 oo. Therefore, again by (3), the entire sum is o ((x + 1)L(x)) _ o (L (x)). As a result, G (x
+
whence
1)
- G (x ) <
m
ne
2 G( x ) + o (G( x ))
G(x+1)-1(x) G(x)
,
0
The proof is complete.
Remark. The solution above is based on relation (1). Application of this relation is motivated by the following argument. If A = log pi, where pi is the ith prime, then the numbers i are the log n where n > 1 are integers, each appearing once. Although the condition on
L(x) does not hold in this case, namely L(x + 1)/L(x) -* e, it is natural to start from the formula applied in prime number theory (for example, in the proof of Chebysev's theorem), the formula corresponding to the prime factorization of n!.
3. SOLUTIONS TO THE PROBLEMS
174
Problem F.16. Let P(z) be a polynomial of degree n with complex coefficients,
P(O) = 1,
IP(z)I < M for Izi < 1.
and
Prove that every root of P(z) in the closed unit disc has multiplicity at most cvfn-, where c = c(M) > 0 is a constant depending only on M. Solution 1. It is sufficient to examine the multiplicity of number 1. In fact, if we prove something for 1 then we may apply the result to the polynomial
p(z) = P(az) with Ial < 1, and in this way we obtain the same estimate for all roots lying in the unit disc. The idea of the solution is the following. We consider the integral
F(P) =
log
10
IP(eio)I do 0
and show that it exists and is nonnegative. Then we estimate it from above, once in the neighborhood of 1 with the aid of the multiplicity of 1 and the degree of P, and once at other points using the condition IP(z)I < M. It is sufficient to prove the existence of the integral for polynomials of
the form z - z0. If
n
P(z) = cJJ(z - zi), i=1 then
n
logIP(z)I = loglcl +loglz - zil. i=1
The existence of
I logleio - zoldo 0
1. Next let I zo I = 1. Without loss of generality, we may assume that z0 = 1 (a substitution 0 = 77 + 00 takes them into each other). Then log Ie`0 - 11 = 2 sin 2
is evident if I zo
and
j27r log 12 sin
\
I do
2/
really exists and is equal to 0. Next, compute the integral j27r
f (a) =
log
do
(a # 0)
3.3 THEORY OF FUNCTIONS
175
for a # 1. Obviously, its value depends on the absolute value of a only (again, a substitution as above may be applied), so it is the same for the numbers ael, aE2, ... aEn, where the Ej are the nth unit roots. Therefore, n
27r
77f(a) _ E f(aej) = f .9=1
n
log
i-1
0
= f27r
11-aEj eio
do
\\\
eino log 1-
do.
an
0
Now, if lal > 1, then for n - oo the integral on the right-hand side tends
to zero since 1 - eino/an - 1 uniformly; thus f (a) = 0. On the other hand, if Ial < 1 then 27r
n(f(a)+27rloglal)= flog a-end since 1 - IaIn < lan - einol < 1 + Ialn; so in this case f (a) = -27rlog I a I is only possible. In each of the three cases, we have f (a) _> 0. In our case, the relation P(0) = 1 implies
P(z)=f 1- -ziz n
,
j-1
whence
n
F(P) = E f (zj) < 0.
(1)
j_1
Now let P(z) = (z - 1)kQ(z) = ao + a1z + + anzn, where Q(1) # 0. We estimate F(P) with the help of k, n, and M. Let
f = f +f 27r
F(P) =
e
o
Then
F2 <
E
f
27r-E
=F1+F2.
E
27r
log M do = 27r log M.
(2)
0
We split F1 again into two parts:
F1=
flogl(z-1)kldo+ flogIQ(e2O)Ido=F3+F4. E
E
(3)
E
Clearly,
F3 = k f E log (2 sin = 2ke(log e - 1).
ICI
2
l
do < 2k
f
log 0 do
0
(4)
3. SOLUTIONS TO THE PROBLEMS
176
For estimating F4, we need an estimate of Q, which we obtain from the coefficients of the expansion of Q about 1. Let Q(1 + z) = R(z). We calculate the coefficients of R from those of P using the formula R(z) _ P(z + 1)/zk:
P(z + 1) = > aj (z + 1)i = E E aj j=0
j=0 m=0
n
m=k
n
j=m
aj
('m
I zm
(i),
since for m < k the coefficient of zm is 0 by our assumption. Thus
R(z) _
n-k
n
bm = E aj (m j+ k)'
bmzm,
(5)
j =m+k
M=0
Further, by the Cauchy inequalities, jajj = jP(j)(0)j/j! < M. Putting this into (5), we find
n+1 Ibml!5 M - (m+k) _ M(m+k+1) j =m+k
(6)
If jzj = 6 and 8(n - k)/(k + 2) < 1, then in view of (6),
amn+1 (m+k+1 )
IR(z)I <
M
n+1 k+1 n+1
n-k +b 2n-kn-k-1
k+2 k+3 + n-k 3_ n+1
1+ak+2 °°
1
(k+1)6k+2 ( ) - (k+1) 1-6=k k+2
j=0
Since Iei0 -1I = 2 Isin(0/2)1 < 101, therefore if e < ((k +2)/2(n - k)), then for 101 < e we have
i'i'_
IQ(e )I = IR(e
1)I < M
n+1
1
k+1
1 _ n-k
< 2M
n+1 k+l
If k > 2, then using the relation t! > tte-t we obtain n + 1
k+1)
(n + 1)n(n - 1)...(n - k + 1)
(k+1)! _ (n2 - 1)n(n - 2) ... (n - k + 1) nk+1 < (k+1)! < (k+1)!
en
(k+1)
k+1
3.3 THEORY OF FUNCTIONS
177
Thus
log lQ(ei0)I
F4 < 2e Llog(2M) + (k + 1) log
en 1
k+1
Collecting everything, by (1), (2), (3), and (4) 2en
en >- 0. (ir+e)logM+elog k+1 +eklogk+1
(7)
Now if n = k2/2c, let e = c/k (this fulfills the condition e < ((k + 2)/2(n k))). Then (7) becomes
+ k) log M +
c
2
log c(k
1)
+ clog 2(k
k 1)
> 0.
We only make things worse if we also replace k + 1 by k. Moreover, c/k =
k/2n < 1/2 gives -7r + c/k < 4, c/k log(ek/c) < (c/k) (ek/c) = e < 3, so finally
4log M + 3 > c log 2,
c<8logM+6.
(8)
Since k = 2cn, relation (8) means that we have proved the assertion of the problem with c(M) = Vf16 11ogM+12. Solution 2. It is sufficient to study the multiplicity of 1 (see the previous solution). We establish the following lemma:
Lemma. Assume that the polynomial wn(z) of degree n satisfies iwn(z)l < M if Izi < 1. Let z = 1 be a root of multiplicity f for wn(z). Then there exists an absolute constant cl (cl = (4e/7r) + e) such that for z = ei'o, 101 < f7r/2n we have wn(z)
clnl
<M(8 (z-1)1
I
Proof. Setting wn(z) = w2n(z), the assertion takes the form W2n(z)
< M2 (cln)2l
(z - 1)21
in which form we shall prove it for all polynomials of degree 2n. Let the roots of the equation z2n + 1 = 0 be 2k-!
zk = ex 2n "
(k = 1, 2, ... , 2n).
3. SOLUTIONS TO THE PROBLEMS
178
Proposition. There exist complex numbers ak (k = e, e+1, ... , 2n-e+1) such that
(z - 1)2t =
2n-2+1 i2
2n-C+1 zk
fj
_ i + 1 2 W2n(Zk)
k=e
(z - zj).
j=Q
Z1,
j34k,2n+1-k
0 (t < v < 2n-e+l). (Such v obviously exists Suppose that since wen has at most 2n - 2e roots different from 1.) For e < i < 2n - e + 1, i v, put 1
ai =
2n-1+1 (zi - 1)21
j=1
j#i,2n-i+l
(zi - zj)
Then for z = zi (i # v) the left-hand side is equal to the right-hand side. Indeed, the terms of the sum appearing on the right-hand side are zero except for the term with k = i; if k # 2n - i + 1, then this is true because of the factor 2n-1+1 (zi - zj) = 0,
11 j=1
j#k,2n+1-k
while if k = 2n - i + 1, then because of the factor zi - 1 Zk
+1=0.
zk
For k = i, the two sides are equal because of the definition of ai. Now choose the remaining a so that the coefficients of z 2n-21+1 and 1/z are 0. This can be done because the two coefficients are the negative of each other since 2n-1+1
11
(-zj) = 1
j=e
j#k,2n+1-k
for all k; together with each -zj appearing in the product, its reciprocal -z2n+1_j also appears.
Thus, the condition 2n-e+1 E akw2n(zk) = 0 k=1
is to be fulfilled and, as wen
0, this is possible if we choose a properly.
But then both the left-hand and the right-hand members are polynomials of degree 2n - 2e which coincide at 2n - 2e + 1 points (the points z = zi, e < i < 2n - 2e + 1, i $ v), so they are equal identically. (Of course, hence
3.3 THEORY OF FUNCTIONS
179
it follows that they are equal for z = z, and therefore, similar to the case
i# v, 1
a _
2n-1+1
H
(zv - 1)21
(zv - zj)
j=1
l
j#v,2n-v+l also holds.)
Since lz? - 11 = Izi - 1/zil = Izi - z2n+1-il and 2n
ll(zi - zj) = I(z2n + 1)z=z.l
= 2n,
j=1
joi
we have 2n
(zi - zj)
fl j=1
)
= 2n.
joi,2n+1-i
Relying on this, and setting ¢ = 7r/n, we obtain the following estimate of Jail (for t < i < n, but also for i > n because of the symmetry):
1
jail =
2n-1+1
H
(zi - 1)21
(zi - zj)
j=1
j54i,2n+1-i 1-1
2n (zi - zj)
(zi - zj)
fi
(zi - 1)
j=2n-1+2
j=1
1
(zi - 1)21
2n
1-2
t-1
[J[(i -A0) 11 [(i+j)0] 2
< j=1
j=0
1
21-1
[l2
<
1i
- 2)
21-2 - 1) 2
.2 21-1
Li - 1)21-1 2
2n
2] 021-2 .
. 1621-1 . n
<2 21
(We have made use of the relation r/2 < sin r < T, r < it/2.)
3. SOLUTIONS TO THE PROBLEMS
180
Now we have to give an upper estimate of the expression 2n-E+1
2n-P+1
fi
fi (z - zj )
(z - zj) <4
9=t
j=2
j54i,2n-i+l
for each z = ei'p, ICI < ($r/2n). Considering pairs and setting z = eiv', zj = ei1o, we find that
0 I(z - zj)(z - z2n-j+1)I = 4 sin
?p
sin
2
2
= 2(cos 0 - cos?)) < 2(1 - cost);
that is, in the interval Iz/iI <_ ($7r/2n) the expression attains its maximum
at the point z = 1. On the other hand,
f j='1(1-zj)
2n-t+1 4
fi (1-zj)
=4
j=1
4(12n+1) 2)21 2
2
zj)
l
< [llj=1(j
n)21-2
8 <-
-
(coear
{(_2)!()
e-1 2 <-
($/2) (ir/n), we have
Since z = ei'P, where ICI
2 Imz
2 Imzi
<1
for all i > B, and therefore 222
W2n(z) (z - 1)21
2
con 22-2 e7r)
2
<M (
cln 22 f ) ,
which proves the lemma. To prove the assertion of the problem, let 9(t) = Iwn(e279)12, wn(z) = c 11 (z Iz.,I<1
11 (z - zµ), 1z"1>1
w*(z)=c fi (1-z21) 11 (z-zµ). 1z,,1>1
Then by Example 48 in Gy. Pdlya, and G. Szeg6, Problems and Theorems in Analysis, Springer, Berlin, 1976, vol. 2, p. 82, Iwn(z)I = Iwn(z)I for Izi = 1 (this is evident), and wn (0) Wn
(o) =
f
Izl<1
Iz" I < 1,
3.3 THEORY OF FUNCTIONS
181
whence
j4(0)l ? 1W.(0)1 = 1, and by Example 53 in the same vol. 2, p. 84, 1
f2,
-
d19 = log 1W-(0)12 > log 1 = 0.
l
27r
On the other hand, g(z9) < M2 for all 79, and for 1791 < (t7r/2n)
< M2 (c ng)2e
11)22
1W2n(exi0)1 < M2
9(i9) =
(c n l z -
so in this interval log g(V) < 2log M + 2f log
cn19.
We use this relation only for 1191 < (f/ncl) (< (t/n) (7r/2)). At the remaining points, we apply log g(V) < 2log M to obtain
ft
27r
0<
log g(19) d19 < 27r
2 log M + 2 2f
0
f cin
1
J0
n,, log (cin z9) dz9
f2 logrdTr=47rlogM+4-(-1),
elm
0
that is, 4
$2
cin
< 47r log M,
f2 < 7rc1 log M n,
f<
7rc1
log Mv/n = c(M) V/n.
The proof is complete. Remark. If n is large, then we may choose c1 = 4e/7r + E.
Problem F.17. Let f (x, y, z) be a nonnegative harmonic function in the unit ball of R3 for which the inequality f (xo, 0, 0) < e2 holds for some
0 < xo < 1 and 0 < e < (1 - xo)2. Prove that f (x, y, z) < a in the ball with center at the origin and radius (1 -
3e1/4).
Solution. We regard f only in the interior of the unit ball. We write x0 instead of (xo, 0, 0) and x instead of (x, y, z), and let lxl denote the length of
the vector (x, y, z) in 1R3. Let 0 < A < 1, 0 < B < 1, max(A, B) < R < 1, and suppose that 1x1 < A and lxol < B. By Poisson's formula (see, for example, the reference in Remark 1 later) the values of f inside the sphere
3. SOLUTIONS TO THE PROBLEMS
182
S(0, R) with center at the origin and of radius R are given by the values
on S(0, R):
AX) =
R2 - 1x12
_
R2 - 1XI2 47rR
IS(O,R) Ix - C13
47rR
d5
(ixo - el 13 J
f(
)
Ixo f(o,R) ix -j where SS denotes the surface element on S(0, R). Since Ixo - (I < R + Ixol, Ix - (I >- R - IxI, we obtain from the nonnegativity of f that
) R+IxoI ) (R-
fW
R2 - ixi2
3
f(x) < I R+Ixol R-IxI
47rR
13
dS
s(0 R)
R2-Ix12f(x0) R2-Ix0I2 R+IxI (R+IxoI)2 . (R-Ixl)2f(xo)
jxj
R-Ixol
R-B (R-A)2'
This holds for all R < 1, hence with the notations 1- A = a and 1- B = /3 we get by letting R tend to 1 - 0, BB
f (x) < f (xo) (1±
)2
(1
+A)2 = f (x0)
(2 - a)2a - a) < f(xo) ago.
Thus, for a2/3 > 8E we have f (x) < (1/E) f (xo). In the problem e < (1 - x0)2, hence Ixoi < 1 - e1/2 and IxI < 1 - 3E1/4, so we can choose a = 3E1/4 and /3 = E1/2, because then a2/3 = 9E > 8E. Then from the preceding estimate
f(x) <
E2=E.
Remarks. 1. The above solution is virtually the same as the usual proof for Harnack's inequality (see, for example, Theorem 1.18 in the book W. K. Hayman, and P. B. Kennedy, Subharmonic Functions, Math. Soc. Monographs, 9, London, Academic Press, 1976) according to which if f is a nonnegative harmonic function in the unit ball of Rm, then for ICI < p < 1 (1 +
p)--1
f (0) < f (0 <
(1
1
f (0)
2. The statement of the theorem is a certain strengthening of the maximum principle for harmonic functions. In fact, the maximum principle asserts
that if f is nonnegative and harmonic in the unit ball, then f (xo) = 0 implies that f vanishes identically. The problem yields that if f is nonnegative and harmonic in the unit ball and f (xo) is "small," then in a disk of radius "almost 1" around the origin f is also "small." In this form, the statement is valid on an arbitrary, bounded, connected domain (this version is also often called Harnack's theorem), which can be seen by a standard argument using chains of overlapping disks.
3.3 THEORY OF FUNCTIONS
183
Problem F.18. Verify that for every x > 0,
r'(x+1) >logx. r(x + 1)
Solution. It is known (see, for example, 1.7. (3) in Erdelyi et al., Bateman Manuscript Project, McGraw-Hill, New York, 1953) that if x > 0, then
I'(x)>0and r, (X) F(X)
1
1
°O
x
(1 v
v_1
x+v
where C is Euler's constant. From this, we get r' (x) C
°°
1
V)
r(x) / r'(x)
F(X)
is strictly increasing. On the other hand,
f
rI (t) dt = log r(x + 1) - log r(x) = log x,
where we used r(x + 1) = xr(x) for all x. Hence, by the mean value theorem, Jlogx=Jx+l
1-
forsomex
x
and so r'(x + 1)
r(x+1) > r()
0
=logx.
Problem F.19. If f is a nonnegative, continuous, concave function on the closed interval [0, 1] such that f (0) = 1, then
r
J
1
x f (x)dx
3
0
r1
112
f (x)dx] 0
Solution. Let A = fo f (x)dx and B = fo x f (x)dx. Integrating by parts, we obtain
B=A
(JX
-J
o
1
f (t)dt/ l dx.
(1)
3. SOLUTIONS TO THE PROBLEMS
184
Since f is concave, its curve lies above the chord joining the points (0, f (0))
and (x, f (x)), that is, for 0 < t < x,
f(t) > f(x)
- It + 1,
(2)
where we have also taken into account that f (0) = 1. If we integrate (2), we obtain
ff
( t ) dt
>
f (x)
2 + x = 2 xf (x) + 2 x
-1
0
(3 )
Using (1) and (2), we arrive at
A - B= f
(jf(t)dt) >
o
that is,
B<2 (A-4)
(4)
.
From the inequality 0 < (2A-1)2 = 4A2 -4A+1, it follows that A- 1/4 < A2. Substituting this into (4), we finally arrive at
B<3(A-41
2A2,
(5)
and this is what we had to prove. Remarks.
1. The proof does not use the positivity of f. 2. The equality B = 3 A2
(6)
occurs if and only if
f(x) = 1 - X.
(7)
Clearly, (7) implies (6). Conversely, if (6) holds, then by (5)
A=2
(8)
Furthermore, because of the continuity of f, we must have equality in (2) for all 0 < t < x. But then f (t) - 1 = f (x) - 1 = c = constant, t x and so f is of the form f (x) = cx + 1. Using (8), we can conclude i
2
A=
f f(x)dx
2
+ 1,
0
from which c = -1 follows. Thus, f (x) = 1 - x, as we have claimed. 3. The statement is a special case of the following theorem that can be found in the book I. M. Jaglom, and V. I. Boltjainskii, Convex Figures (in Russian), Gostehizdat, Moscow, 1951: If the boundary of a convex domain contains a segment of length 1, then the distance of the weight point of the domain from that segment is at most 2/3 times the area.
3.3 THEORY OF FUNCTIONS
185
Problem F.20. Let f be a differentiable real function, and let M be a positive real number. Prove that if If (x + t) - 2f (x) + f (x - t)I <Mt2 for all x and t, then
If'(x+t)- f'(x)I <MItI. Solution. We shall prove more; namely, we shall not assume the differentiability of f in advance. It will be enough to assume its continuity. The inequality
If (x + t) - 2f (x) + f (x - t)I means that for all x and t we have
f(x+t) - 2f (x) + f (x - t)
(1)
and
f(x + t) - 2f (x) + f(x - t) > _M. t2.
(2)
Let hi(x) = f(x) - (M/2)x2. For this function, we obtain from (1)
hl (x + t) - 2h1(x) + hl (x - t) = f (x + t) - 2f (x) + f (x - t) - Mt2 < 0. This means that h1 is a continuous function with nonnegative second-order symmetric differences. We know that then h1 is concave. The same argu-
ment based on (2) yields the convexity of hl (x) = f (x) + (M/2)x2. It is well known that h1 and h2 then have one-sided derivatives for which h(-)(x)
h2-)(x)
> h(+)(x),
(3)
< h2+)(x).
(4)
It follows that f (-) and f (+) also exist at every point and satisfy f(-)(x) = h(-)(x) + Mx,
(5)
f(+)(x) = h(+) (x) + Mx,
(6)
f(-)(x)
=
h2-)(x)
- Mx,
(7)
f(+)(x) = h2(+) (x) - Mx.
(8)
By (3), (5), and (6) we have f (-) _> f (+), while (4), (7), and (8) yield f (-) < f (+). Thus, f (-) = f (+), which means that f is differentiable, and so both h1 and h2 are also differentiable.
3. SOLUTIONS TO THE PROBLEMS
186
Since the derivative of a differentiable concave (convex) function is decreasing (increasing), we can conclude for x < y that
hi(y) - hi(x) = f'(y) - f'(x) - M M. (y - x) < 0
(9)
ha(y) - h2(x) = fe(y) - fi(x) + M '(y - x) > 0.
(10)
and
Equations (9) and (10) can be summarized as
If'(y)-f'(x)I <MIy-xI, and this is equivalent to the statement. Remark. The statement can be generalized as follows. Let r
Dtf(x)=>2(-1)kI k I f (x+(2 - k)t) k=O
be the rth symmetric difference of f. If
Iot(x)I < Mtr for all x and t, then Di-i f(i)(x)I < Mtr-', I
1 < j < r.
This can be derived from the formula t
...
Def(x)= It
0
k l du1...du, (X_t+1+...+Uk J
provided the ( k - 1)th derivative of f is absolutely continuous.
Problem F.21. Let a < a' < b < b' be real numbers, and let the real function f be continuous on the interval [a, b'] and differentiable in its interior. Prove that there exist c E (a, b), c' E (a', b') such that f (b) - f (a) = f ' (c) (b - a),
f(b')-f(a)=f'(cl)(b'-a), andc
Statement. Let p, q, q' be real numbers such that p < q < q'. Suppose that the real function f is continuous on the interval [p, q'] and differentiable in its interior. Then for every r E (p, q) for which
f (q) - f (p) = f'(r) (q - p)
3.3 THEORY OF FUNCTIONS
187
holds, there exists an r' E (p, q') with
f(4) - f(p) = f'(r')(4 - p) and r' > r. Proof. To prove this, we may assume f (p) = f (q') = 0, since otherwise we can work with the function p)f(g') - f(p)
f(x) - f(p) - (x -
q -p
If f (q) = 0, then the statement is obvious, so we can also assume that f (q) > 0. Let us choose the point r E (p, q) so that f(q) = f(q) - f(p) = f'(r) (q - p).
If now f (r) < 0, then in the interval [r, q) there is a point p' such that f (p') = 0. Therefore in the interval (p', q') there is an r' with f (r') = 0, and this r' satisfies the requirements. We have to examine the case when f (r) > 0. Since
fl(r) = f (q) - f (p) >
q-p
0,
there is a p' E (r, q') with the property that the ratio
f (p') - f (r)
p' - r is bigger than 0, that is, f (p') > f (r). But then there is a point q" in the interval (p', q') for which f (q") = f (r), and so for a suitable point r' of the interval (r, q") we have f (r') = 0, and with this the statement is verified. Now we turn to the solution of the problem. Let us choose the point d E (a', b) so that f (b) - f (a') = f ' (d) (b - a'). On applying the above statement, we get a c' E (a', b') with the property
f(b')-f(a') =f'(c')(b'-a) and c' > d, and it similarly follows that there is a c E (a, b) for which
f (b) - f (a) = f'(c)(b - a) and d > c. This proves the assertion of the problem. Remark. It can be seen from this proof that the assertion is valid in the
casesa
3. SOLUTIONS TO THE PROBLEMS
188
Solution 2. Let D = f (b) - f (a)
b-a
r = inf{c E (a, b) : f'(c) = D},
D' = f (b') - f (a') b' - a' r' = sup{c' E (a', b') : f'(c') = D'}.
Our aim is to prove that r < r'. Suppose, on the contrary, that r > r'. Then
a
f'(x) > D. Likewise, in (r, b')(C (r', b')) the derivative f'(x) lies on one side of D'. We distinguish two cases.
Case 1. f'(x) > D' for all x E (r, b'). By the mean value theorem, we have
f (r) - f (a) > D(r - a),
f (b') - f (r) > D'(b' - r).
Since by definition
f (b) - f (a) = D(b - a),
f (b') - f (a') = D'(b' - a'),
we get by subtraction
f (b) - f (r) < D(b - r),
f (r) - f (a') < D'(r - a').
(1)
On the other hand, again using the mean value theorem
f (b) - f (r) > D'(b - r),
f (r) - f (a') > D(r - a').
(2)
Since b - r > 0 and r - a' > 0, we can conclude D > D' from the left sides of (1) and (2), while from their right sides it follows that D' > D. This contradiction proves our claim in Case 1. Case 2. f'(x) < D' for all x E (r, b'). Similarly as before, we get
D'(b-a') < f(b) - f(a')
f'(x) > D > T) nor in (r, b) (because there f'(x) < D' < T). Neither can t be equal to r (because of the choice of T), and we have arrived again at a contradiction. This proves the claim in Case 2.
3.3 THEORY OF FUNCTIONS
189
Problem F.22. Let lo, c, a, g be positive constants, and let x(t) be the solution of the differential equation ([lo + ct']2x')' + g[lo + ct'] sinx = 0,
t > 0,
-2
< x < 2,
satisfying the initial conditions x(to) = xo, x'(to) = 0. (This is the equation of the mathematical pendulum whose length changes according to the law l = to+ct'.) Prove that x(t) is defined on the interval [to, oo); furthermore, if a > 2 then for every x0 # 0 there exists a to such that liminf Ix(t)I > 0. t-.oo
Solution. With the notation y = [1o + ct' ]2x', equation (1) transforms to x =
y
[1o + ct']2
(t > 0, 1xI < 2 , y
y' = -9[lo + ct'] sin x
E
1E8)
Applying the Picard-Lindelof theorem to the latter system, we get that the solution satisfying the given initial conditions exists in a right neighborhood
of to. From the theorem of "continuation up to the boundary" (see the book L. Sz. Pontriagin, Ordinary Differential Equations, Addison-Wesley, London, 1962), we can conclude that if x(t) exists on the interval [to, T)
but it cannot be continued beyond T, then we must have jx(t)l -* 7r/2 as t -> T - 0, since it follows from equation (1) that x'(t) is bounded on bounded intervals. Let us introduce the notation l(t) = to + cta and consider the function
V(t,x,x') = 40 (x')2+2(1 - cosx), which is nothing else than the mechanical energy of the pendulum divided by l(t)/2. From equation (1), we get by differentiation that v(t) _ V(t, x(t), x'(t)) is a nonincreasing function. Therefore,
2 > v(to) = 2(1 - cosxo) > v(t) > 2(1 - cosx(t)), which makes Ix(t)I --> 7r/2, as t --> T-0 is impossible. Thus, x(t) is defined on the whole interval [to, oo). To prove the second statement, let us start from the equation
x(t) = xo - g
f
t
t"
[1 to+csa
]2
1110 + cT'] sin x(T)drds,
which follows from (1) by integrating twice and taking into account the initial conditions. If a > 2, then t 4,0C,
[lo +
to
[l0 + cs']dsdt < oo,
3. SOLUTIONS TO THE PROBLEMS
190
and so for large to I x(t) I ? Ixo H -g
[lo + cT']dTds > 12
Jto [lo + csa]2 Jto
II
for all t on the interval [to, oo), and this is what we needed to prove.
Remark. It can be proved that if 0 < a < 2, then for any solution of (1) that is defined on the whole interval [to, oo), we have
lim x(t) = 0
t-.oo
(see L. Hatvani, On absence of asymptotic stability with respect to a part of the variables, J. Anal. Math. Mech., 40 (1976), 223-225).
Problem F.23. Let fl, f2i ... , fn be regular functions on a domain of the complex plane, linearly independent over the complex field. Prove that the functions fi f k, 1 < i, k < n, are also linearly independent.
Solution. First, we prove the following lemma:
Lemma. Let fi, gi, 1 < i < n be regular functions in the domain D, and figi = 0, then the suppose that not every gi vanishes identically. If 1 fi's are linearly dependent.
Proof. Indeed, for every h# 0 and z E D 0
fi(z + h)9i(z + h) - fi(z)9i(z)
h
i=1
_9i(z+h)fi(z+h) - fi(z)
_
h
i=1
h + n fi(z) (gi(z+h)-gi(z)\ h h' i=1
Letting h tend to zero first on the real and then on the imaginary axis, we obtain n n
fi(z)9i(z) = 0
fi(z)9i(z) + and
i=1
i=1
n
n fi(z)9(z) - E fi(z)9i(z) = 0
i=1
i=1
for all z E D, from which E 1 figi = 0 follows. Repeating this argument we get E 1 f(')gi = 0 for every m. Without loss of generality, we can assume that 0 E D. Let oc
A (z) = 1: anti zi
and
gi (z) = 1: b(') z'
3.3 THEORY OF FUNCTIONS
191
These series converge in some disk {z : jzj < 8}. We may also assume that bo'i 54 0 for at least one i (in the opposite case, we may factor out an appro-
priate power of z from every gi). For every m, the function takes the value
Ei
1
ff("")gi
n
m! E a,;,>boxi =0
Ei
i=1
at z = 0, and so boxi fi = 0 because every coefficient in its power series 1 vanishes. This proves our lemma. Now we prove that if, besides the assumption of the theorem, the functions gj, 1 < j < m are also regular and linearly independent on D, then the same is true of the system figs, 1 < i < n, 1 < j < m. This is obviously stronger than what the problem asked for. In fact, suppose that En 1 1 c jfigj = 0. Then 0, and so it follows from our lemma that Ei=1 f(>1 j= Ej=1 cjgj = 0 for all i. From this, we get ci.9 = 0 for every i and j because of the assumed linear independence of the functions gj, and this is exactly
E
what we wanted to prove.
Problem F.24. Prove that the set of all linear combinations (with real coefficients) of the system of polynomials {x"+xn2}°' 0 is dense in C[0, 1].
Solution. Let P be the set of the linear combinations. Since the set of polynomials is dense in C[0, 1] (Weierstrass's theorem), it is enough to prove that every power xn, n = 0, 1, 2,..., can be uniformly approximated by polynomials from P. For n = 0,1, we have xn E P. If n > 1, then let f .n
m.-1
m a (xn2' + xn2j}1 i=0
= xn +
1
M
1:(-1)i+1xn2' .= xn + 1 . A(x). m
i=1
Obviously, fn E P. The sum defining A(x) consists of terms that alternate in sign and decrease in absolute value, and the first term lies in between 0 and 1. Hence, 0 < A(x) < 1. As a consequence, Ixn
- ff(x)I
for all x E [0, 1], which proves our claim.
<
ET
Remark. The problem can be generalized as follows. Let g(n) be a strictly increasing function defined on and taking values from the set of nonnegative integers. Then the linear combinations of the system {xn + x9(n)} form a dense set in C[0,1]. The proof goes along the lines presented above.
3. SOLUTIONS TO THE PROBLEMS
192
Problem F.25. Let f be a real function defined on the positive halfaxis for which f (xy) = x f (y) + y f (x) and f (x + 1) < f (x) hold for every positive x and y. Show that if f (1/2) = 1/2, then
f(x)+f(1-x)>-xlogex-(1-x)loge(1-x) for every x E (0, 1).
Solution. Let co(n) = f (n)/n. Then co(nm) = co(n) + W(m) and (n + 1)cp(n + 1) < ncp(n) hold for all n, m E N. Hence, there exists a c E R, for which W(n) = c loge n for every n E N (see p. 19 in J. Aczel, and Z. Dardczy, On Measures of Information and Their Characterizations, Academic Press, New York, 1975). From here, because f (1/2) = 1/2 and
0=f(1) = 2f(2)+2f ( 2) = 2(f(2)+2), we get
f (n) = -n loge n (n E N). Since for every positive x and y
f (x + y) - AX) -AY) =Y
[f (y + 1) - f ( ) ]
< 0,
we have
f(x+y) C f(x) +f(y) Let n E N and 0 < x < 1. Then using what we have obtained so far, we can write
0 =f(1) =f[(x+1 -x)n]
()i -
=f k-0
x)n-k < E f [(n)k(1 k=0
n) . xk(1 - X).-k +
I k 1 loge I k k =0
=I+II.
\
///
x)"-kl
\/
k= 0
(k) f (xk(1 - x)n-k) \
From the functional equation on f, we obtain f (xk) =
kxk-1 f(x)
and
f ((l
- x)k) = k(1 -
x)k-1 f (1
- x),
3.3 THEORY OF FUNCTIONS
193
hence the sum representing II can be seen to be equal to n f (x) +n f (1- x), which yields 1 (x) + f (1 - x) > n
()Xk(1
-
(nk
loge
J
Sn(x)
(1)
Now let pkW (x)
=
()xk(l_x)n_k,
k=0,1,...,n.
Since these (for a fixed x) take their maximum for either k = [nx] or k = [nx] + 1, we easily obtain from Stirling's formula for the factorials appearing in the binomial coefficients that limn. pkn' (x) = 0 uniformly in k. Using this and limt.o+o t loge t = 0, we get lim 1 n-oo n E N (x) loge pkn) (x) = 0.
(2)
k=O
On the other hand, n
n 1: p(kn) (x) 1092 pkn' (x) k=o
)Xk(1
- x)n-k
(n)
+ k 1092 x + (n - k) 1092 (1 (k =Sn(x)+xlog2x+(1-x)log2(x-1),
=n
kn
Llog2
x)]
from where, with the aid of (2), we can conclude that
lim Sn(x) = -xlog2 x - (1 - x) 1092 (1 - x).
n--.oo
Taking (1) into account, we get the statement.
Problem F.26. Let G be a locally compact solvable group, let c1,..., cn be complex numbers, and assume that the complex-valued functions f and g on G satisfy n
E ck f (xyk) = f(X)g(y) for all x, y E G. Prove that if f is a bounded function and inf Re f (x)X(x) > 0
XEG
for some continuous (complex) character X of G, then g is continuous.
3. SOLUTIONS TO THE PROBLEMS
194
We will give two solutions. Neither of them will use the local compactness, and the second solution will not use the solvability of G either. Hence, the claim is true on any topological group.
Solution 1. It is known (see for, example, F. R. Greenleaf, Invariant Means on Topological Groups, Van Nostrand, Princeton, 1969) that because of the solvability of G there exists a right-invariant mean m on the space of complex-valued, bounded functions on G. That is, m has the following properties: m is (complex) linear; m(1) = 1; m(f) = m(f); if f > 0, then m(f) > 0; furthermore, m(fx) = m(f) for every bounded function f defined on G, where f, (t) = f (tx), x, t E G. Let f, g, and X be as in the problem. Since Re f X > infxEG Re f(x)X(x) > 0, we have m(Re f X) > m(inf Re f (x)X(x)) = inf Re f (x)X(x) > 0, xEG
xEG
and from this m(f X) = m(Re f X) + im(Im f X) # 0.
Now let x, y E G be arbitrary, and let us multiply the equality n
1: Ck.f (xyk) =
f(X)g(y)
k=1
by X(x):
n
E ckf (xyk)X(x) = .f (x)X(x)9(y) k=1
On the left-hand side, we make use of the identity X(x) = X(xyk)X(y k) = X(xyk)Xk(y),
by which we obtain n
1: ckf(xyk)X(xyk)Xk(y) = f(x)X(x)9(y).
(1)
k=1
This holds for all x, y E G. Apply now to both sides as functions of x the mean m, and make use of mx(f(xyk)X(xyk)) = m(fX) 36 0
(here mx denotes that the argument has to be considered as a function of x). It follows that n
EckXk(y) = 9(y), k=1
from which the continuity of g is obvious.
(2)
3.3 THEORY OF FUNCTIONS
195
Remark. One can prove with the same method the generalization that one obtains by allowing different functions fk on the left of the assumed equality.
Solution 2. In the second solution, we will not use the existence of the mean m; otherwise, the argument that follows is similar to the one above. We start from the identity n
E Ck
f(Xyk)X(Xyk)Xk(y) = f(x)X(x)g(y)
k=1
that we obtained in the first solution in (1). Applying this with x = y-11 ... , x = y-m and adding the resulting equalities together, we obtain with
m
Sm(y) = 1: f (xy-i)X(xy i) j=1
the identity
E ck
m
k-1
n
(Sm(Y)
k=1
+ j=0 E -j=m-k+1 E
f (xya)X(xyi)
k(y) = Sm(y)g(y) (3)
By the assumption, Re Sm (y) > me -> oo
as m -> oo,
where c denotes a positive lower bound on the real parts of the products f (x)X(x). Hence, the numbers ISm(y)I tend to infinity as m -* oo. Now if we divide (3) by Sm(y) and let l tend to infinity, then we again arrive at (2) in view of the boundedness of f. This completes the proof.
Problem F.27. Suppose that the components of the vector . . . , u,,) are real functions defined on the closed interval [a, b] with the property that every nontrivial linear combination of them has at most n zeros in [a, b]. Prove that if a is an increasing function on [a, b] and the rank of the operator
u=(u0,
b
A(f) =
Ja
u(x) f (x)do(x),
f E C[a, b],
is r < n, then a has exactly r points of increase. Solution. By the assumption that there is a 0 i4 c E R'°+1 orthogonal onto the range of A, that is, (c, A(f )) = 0 for every f E C[a, b]. This means that
Ja
b(c, u(t))
f (t)da(t) = 0
196
3. SOLUTIONS TO THE PROBLEMS
holds for every f continuous on [a, b]. In particular,
j
b
(c, u(t))2da(t) = 0,
from which it follows that every point of increase of a is a zero of (c, u(t)).
Since c # 0, the assumption in the problem implies that a has at most n points of increase. Let these be xl < < x,,,,, m < n, and let the corresponding jumps be dal, . . . , Au,,,,. Then m
A(f)
u(xi)f (xi) Auj. i=1
Because of the assumption made on u, the vectors u(xl),... , u(xm) are easily seen to be linearly independent - furthermore Lvi > 0 - for every i = 1, ... , m. Hence the range of A is the subspace spanned by the vectors u(xl),... , u(x,,, ). Then its rank is m, and this was to be proved.
Problem F.28. Let Q and R be the set of rational numbers and the set of real numbers, respectively, and let f : Q -+ R be a function with the following property. For every h E Q, xo E R, f(x + h) - f(x) -+ 0
as x E Q tends to x0. Does it follow that f is bounded on some interval ? Solution. The answer is no. Consider the following function:
f( )
= log log 2q,
where p/q is a rational number written in the form where the denominator is positive and the fraction cannot be simplified. This is not bounded on any interval. On the other hand, if x = p/q and h = k/m, then x + h = (pm + qk)/qm, where the last fraction may be simplified. In any case, f (x + h) < log log 2qm, and so
f (x + h) -f(x) < log
log 2q + log m log 2q
Now if x --* x0i then q - oo, and so
lim sup f(x+h) - f(x) < 0. On applying this with h and x0 replaced by -h and x0 + h, respectively, we obtain the opposite inequality, lim inf f (x + h) - f (x) > 0. x--..To
Thus, this f is a counterexample.
3.3 THEORY OF FUNCTIONS
197
Problem F.29. Suppose that the function g : (0, 1) -* R can be uniformly approximated by polynomials with nonnegative coefficients. Prove that g must be analytic. Is the statement also true for the interval (-1, 0) instead of (0, 1)? Solution. Let p1i p2i ... be polynomials with nonnegative coefficients, and p,, (x) = g(x) pointwise on (0,1). We shall prove that assume that g is analytic. This is stronger than what the problem asked for, since we will not use the assumption that the limit is uniform. We claim that the polynomials p,,, are uniformly bounded on every disc {z : IzI < p} for every p < 1. In fact, let p.,, (z) = > akzk. Then for I z I < p, Ipn(z)I C
E akIzlk
< > akPk = pn(P) < K = K(P),
since the sequence {pn(p)} is convergent, and hence it is bounded.
We also know that the sequence {p,,} is convergent on a segment of the unit disk, hence we can invoke Vitaly's theorem to conclude that {p"} is convergent inside the unit disc and that the convergence is uniform on every compact subset of the open unit disk. This implies that G(z) = limn-oo pn(z) = g(z) is analytic in the unit disk, and so its restriction to (0, 1) also has this property. This proves the first part of the problem. The answer for the second part is negative. In fact, we shall prove much more, namely that the following result holds: Result. A continuous function g : [-1, 0] -* R can be uniformly approximated by polynomials with nonnegative coefficients if and only if g(0) > 0. The necessity of the condition is obvious, hence we only have to deal with
its sufficiency. Let G be the set of all continuous functions on [-1, 0] that are the uniform limit of some sequence of polynomials with nonnegative coefficients. Obviously, G is closed for addition, multiplication and forming uniform limits; furthermore, G contains every polynomial with nonnegative coefficients.
First, we prove that the function -x is in G. This immediately follows from the fact that the polynomials x(1 + x)n - x have nonnegative coefficients and Ix(1 + x)nI < (1/n) for every -1 < x < 0 (this can be easily verified by differentiation). However, then -xn = (-x)xn-1 also belongs to G for every n > 1, and so G contains every polynomial whose constant term is nonnegative. Now let g be an arbitrary continuous function on [-1, 0] with g(0) > 0. By Weierstrass's theorem, there is a sequence of polynomials Pn converging
uniformly to f on [-1, 0]. But then the same is true of the sequence with terms Pn (x) = pn (x) - pn (0) + 9(0)
and these polynomials are from G. Hence, g also belongs to G, which proves the sufficiency part in our claim.
3. SOLUTIONS TO THE PROBLEMS
198
Problem F.30. Prove that if ai (i = 1, 2, 3,4) are positive constants, a2 - a4 > 2, and a1a3 - a2 > 2, then the solution (x(t), y(t)) of the system of differential equations zb = a1 - a2x + a3xy,
y = a4x - y - a3xy
(x, y E R)
with the initial conditions x(O) = 0, y(O) > a1 is such that the function x(t) has exactly one strict local maximum on the interval [0, oo).
Solution. x strictly increases in a neighborhood of t = 0, hence it is enough to prove that ± changes sign exactly once. Let us draw the pieces of the hyperbolas ± = 0 and y = 0 lying in the first quadrant of the phase plane (x, y) (see Figure F.1):
\
x=0:a3x y- a2a 1+a1=0 3
y=0:-a3
\x+
3)
\y
a3)
=0.
a3
Y II
Figure F.1.
Denote (xo, yo) the intersection of these. The trajectory of the solution
cannot leave the region I bounded by the curves ± = 0, y = 0, y > yo. Therefore, it is enough to prove that y(t) > yo, t > 0, and the trajectory intersects the curve ± = 0. If the trajectory intersecting the half-line x =
x0, y > yo leaves the region II bounded by the lines x = 0, x = xo, e = (a4 - a2)x - y + a1 = 0, then it must intersect the curve x = 0, namely in the opposite case y(t) J. z1 i x(t) + y(t)
.
const > 0,
and so
x(t) / xl
(t -> co).
3.3 THEORY OF FUNCTIONS
199
From this, it follows that x1 = x0 and yl = yo because the point (x1, Yi) must lie on the curves i = 0, y = 0. But this is a contradiction for x(t) > xo for large t. Next, we show that the trajectory must indeed intersect the half-line x = x0, y > yo. Write the function in the form
i
-a3xy+a4x-y a3xy - a2x + al
-1+ (a4-a2)x-y+ai a3xy - a2x + a1
From this, it follows that the values of y/i on the line e would be equal to -1, which is bigger than a4 - a2. Hence the trajectory does not intersect the line e. We can also see that the function y/i is a decreasing function
of x in the region IF = {(x, y) E II : y < a2/a3} along the lines (a4 a2)x - y + C = 0 (al < C = const). Since _
y
x
-a3xoy + a4xo - y
a3xoy -a2xo + al
X = xo
_ -a3xo(y - yo) - (y - yo) = - (1 +
\
a3xo(y - yo)
1 a3xo
/
it follows that y i
1
1- a3xo
(x,y)EII'
Now we show that
1+
1
a2xo
Simple calculation gives the formula (alai - a2) +
xo =
(alai - a2)2 + 4a1a3(a2 - a4) 2a3(a2 - a4)
,
from which a3xo > (aia2 - a2)/(a2 - a4) follows. Thus, to conclude the required inequality, it is enough to verify 1
a1a3 - a2 >
T-
1
,
a2 -d4
which is obvious from our assumptions.
Now let e' be a line through the point P, the tangent of which lies in between -(a2 - a4) and -(1 + 1/a3xo). Since x
1+ (x, y) Eel f1II'
1 ,
a3xo
the trajectory coming from the region II cannot intersect e' from above. If it does not intersect the line x = xo either, then (x(t), y(t)) -> (x1, yl) 0 (xo, yo), (t -> oo) would follow, which is a contradiction. With this, the proof is done.
3. SOLUTIONS TO THE PROBLEMS
200
Problem F.31. Let us call a continuous function f : [a, b] -* R2 reducible if it has a double arc (that is, if there are a < a < ,3 < ry < 6 < b such that there exists a strictly monotone and continuous h : [a,,3] -> [y, 6] for which f (t) = f (h(t)) is satisfied for every a < t < ,3); otherwise f is irreducible. Construct irreducible f : [a, b] -* R2 and g : [c, d] - R2 such that f ([a, b]) = g ([c, d]) and
(a) both f and g are rectifiable but their lengths are different; (b) f is rectifiable but g is not. Solution. In both a) and b) the curve f will be the following: Let A C_ [0, 1]
be Cantor's ternary set, and let E be the unit interval [0, 1] considered as a curve on the plane. It is known that there is a continuous and increasing surjection cp : A -+ E (write x E A in the ternary form x = 0.e1E2...
where each ej is 0 or 2, and then let V(x) be defined by the binary form 0.(e1/2)(E2/2) ... ). Let cp be the restriction of f to A. In the complementary intervals of A of lengths 3-'l, let f run through circles of radius 3-" in a continuous fashion. It can be easily seen that f is continuous, irreducible, and rectifiable.
In case (a), let g : [-1, 1] -* R2 be defined on [-1, 0] by g(t) = E(-t), and on [0, 1] let g coincide with f. Then g is irreducible, but its length is longer than that of f by 1. In case (b), let us divide the set of dyadic rational numbers of [-1, 1], from which there exists a monotone correspondence with the complementary intervals of A and hence with the circles that f runs through, into ... , each of which is dense in [-1, 1]. Let us also divide the the sets So, Si,..., parameter interval of g : [-1, 1] -* R2 into the consecutive intervals I,,, of
length 2-", n = 0, 1, .... On I,, let g run from E(0) to E(1/2) running also through the circles corresponding to the points of S1 fl I. On 12, let g run from E(1/2) to E(0) running also through the circles corresponding to the points of S2 fl I2. On 13, let g run from E(0) to E(1/3) and so on. Finally, on 1o let us define g so that it goes from E(1) to E(0), and besides the circles corresponding to the points of So, it should also run through every other circle through which it has not yet run. Obviously, the continuity of g has to be checked only at the point 1, which can be easily done. Since each set Si is dense, g is irreducible. But it is not rectifiable, because for every n we can write into it a polygon (with vertices E(0), E(1/2), E(0), E(1/3), ..., E(1/n), E(0)) of length 2 (1/2 + 1/3 + + 1/n). Remark. If we like, we can also assume that [a, b] _ [c, d].
Problem F.32. Let n > 2 be a natural number and p(x) a real polynomial of degree at most n for which
max lp(x)l< 1,
-1<x<1
p(-1) = p(1) = 0.
Prove that then Ip(x)I <
n cos
2
1 - x2 COS2 2n
1
cos -E-n
<x<
1
cos 1nn
3.3 THEORY OF FUNCTIONS
201
Solution. Let c = cos(-7r/2n). It is enough to prove that for xj < 1/c we have lp(x) I < 1, because then by applying the classical Bernstein inequality
on the interval [-1/c, 1/c], it follows that p(X)1C
n
=
1
(IxI<).
1nc
Let T, In > 1 be the number outside [-1, 1] with the smallest absolute value for which Ip(n)I = 1. We have to show that n > 1/c. If we apply a linear transformation, this amounts to the same as proving that if lp(x) 1 < 1
for -1 < x < 1 and lp(l)l = 1, then for c < x < 1 the polynomial p(x) cannot vanish, that is, p(x) 0. Suppose that this is not true, and let Tom, (x) = cos(n arccos x) be the Chebyshev polynomial of degree n. We know that for some sequence 1 = xo > x1 > > xn, = -1, we have Ta(xi) = (-1)i. Furthermore, TT,(c) = 0, and Tn is positive to the right of c. We shall arrive at a contradiction by proving that the polynomial p - TT, has n + 1 zeros. One zero is x = 1. If i > 0, then there is a zero in the interval [xi, xi+1] . If this happens to be one of the endpoints of this interval, then it is a double zero, and hence there are at least n - 1 zeros in [-1, x1] (counting the possible root
xl only once). By our assumption, there is an x1 < x < 1 for which 0 = p(x) < TT,(x), so p - TT, must vanish in at least one point of [x1, x] (recall that p(xl) - T,T(xl) > 0). Thus, p - TT, has at least n + 1 zeros. But since its degree is at most n, this can only happen if p =_ TT,, which is not possible, because p(x) < T,,(x). The obtained contradiction proves our claim.
Remark. It can be proven that equality occurs if and only if p(x) _ ±TT, (x/c).
Problem F.33. Let f be a strictly increasing, continuous function mapping I = [0, 1] onto itself. Prove that the following inequality holds for all pairs x, y E I:
1 - cos(xy) < f f (t) sin(tf (t))dt + I f-1(t) sin(t f-1(t))dt 0
.
o
Solution. Let
T1={(u,v):0
0 < v < f(u)},
T3={(u,v):0
J0 J0
g(u,v)dudv
f 1 0
0
f(u)
/V g(u,v)dvdu + J 0
f
fg(u,v)dudv.
3. SOLUTIONS TO THE PROBLEMS
202
Thus, with the choice
_ 82(1 - cos(xy)) g(x, y)
8y8x
= sin xy + xy cos xy > 0,
we get
1 - cos(xy) < fox f (u) sin(u f (u))du + J y f -1(v) sin(v f -1(v))dv . 0
o
Problem F.34. Let U be a real normed space such that, for any finitedimensional, real normed space X, U contains a subspace isometrically isomorphic to X. Prove that every (not necessarily closed) subspace V of U of finite codimension has the same property. (We call V of finite codimension if there exists a finite-dimensional subspace N of U such that
V+N= U.) Solution. By applying induction, we may assume that V has codimension 1. Let W be a finite-dimensional normed space, and consider W ® W with the norm II(w1,w2)II = max{IIw1II, IIw211}. Let Wl ® W2 be the isometric isomorphic image of W ® W in U. It is enough to show that v n (W1® W2 ) contains a subspace that is isometrically isomorphic to W.
If V n (W1 ®W2) = W l E) W2, then we are done. If v n (Wl ®W2) has codimension 1 in Wl ® W2, then there is an 0 f : Wl ® W2 -+ R linear functional with kernel V n (W1ED W2). With the notations fl(wl) = f (wl ® 0), f2(w2) = f (W2 ® 0) we get two linear functionals on Wl and W2, respectively, and
f (w1 ® W2) = fl(wl) + f2(w2).
I
Without loss of generality, we can assume that 111111 > IIf2II and 0 < I f1 II . Consider the linear functional
\f1-> -a1If2II By the Hahn-Banach theorem, there is a cp E Wl * for which IIf2II/IIf1II and W(fl) = -IIf2II. If 11f211 54 0, then let B(g) _,(g) . f2/IIf2II
(g E Wl ), while in the case IIf2II = 0 let B =- 0. This B : Wi -> W2 is a linear mapping of norm at most 1. Since our spaces are of finite dimensions, we have B = A* for some linear operator A : W2 -* W1. Since I A I I= I I A* I I= I B I I< 1, the mapping w2 Aw2 ® w2 is an isometry of W2 into Wl ® W2. Furthermore, the range of this mapping is part of v n (W1 ® W2) because I
fl(Aw2) + f2(w2) = (A*fl)(w2) + f2(w2) = -f2(w2) + f2(w2) = 0. Thus, we have found an isometry from W 2 into v n (Wl ®W2), by which
we verified that v n (W1 ® W2) contains a subspace that is isometrically isomorphic to W.
3.3 THEORY OF FUNCTIONS
203
Problem F.35. For every positive a, natural number n, and at most an points xi, construct a trigonometric polynomial P(x) of degree at most n for which f27r
P(xi) < 1,
P(x)dx = 0,
J0
maxP(x) > cn,
and
where the constant c depends only on a.
Solution. We shall construct a P of degree 2n instead of n, but this is the same as if we solved the problem with 2a instead of a.
Without loss of generality, we can assume that the midpoint of the largest contiguous interval determined by the points xi is 0 (if this was not the case, then we would have to translate the trigonometric polynomial
to be constructed below). This means that there is no xi in the interval (-7r/an, 7r/an). Let
Pl (x) = 1 + > cos vx = 2
sin (n +
x = Dn(x)
2 sin 2
v= 1
be the nth Dirichlet kernel. For this P1(0) = n + (1/2). Furthermore, P1 monotonically decreases on (0, 7r/n), for 7r > IxI > 7r/n, its absolute value is smaller than 1/2 sin(7r/2n) n/7r, and at the point 7r/an its value is sin (n +
sin
n 2
2 sin 2an
where the coefficient of n is smaller than 1 and where A - B means that the ratio A/B tends to 1 as n -* oo. Hence, there exists a c < 1 such that Pl (x) - cn is negative at every xi but P1(0) - cn > n(1 - c) is positive. Let
P2(x)
_
1
n
_
2sin2
1
be the Fejer kernel. We have P2(0) = n 2 1
{sin(n+1)x)2
2
1>Dv(x)
f and
P2(x)dx = 1.
1
Therefore, if P3(x) = P2(x)(Pi(x) - cn), then
P3(0) > c'n2,
P3(xi) < 0,
and f
27r
2
P3(x)dx _< c"n fo
27r
P2(x)dx = c"n.
J0
1
3. SOLUTIONS TO THE PROBLEMS
204
Finally, let / P(x) = I P3 (x)
Then
j2, 2-
)dx\
P3(x
I
f2
f 0
because of the previous estimate, and
P(0) >"n - 1. Problem F.36. Let f :
El
IR ---> IR be a twice differentiable, 27r-periodic
even function. Prove that if 1
f"(x) + f(x) _
f(x + 37r/2)
holds for every x, then f is 7r/2-periodic.
Solution. Since 11f (x + 3ir/2) is defined and f is continuous for all x, f
must have the same sign on R. From the parity of f, we get f"(-x) _ f"(x). On applying the identity of the problem at -x, we get + f (x) =
1
f (-x + 37r/2)'
therefore
f(x+
37r
37r
2)=f(-x+ 2)=f(x-
37r
2
)
holds for every x. Thus, f is 37r-periodic, and since it is also 27r-periodic, it follows that 7r is a period of f. Hence, the condition in the problem can be written in the form 1
fr(x) + f(x) =
f (x + -7r/2) .
Let us consider the function g(x) = f (x + 7r/2). This is also even:
g(-x) = f (-x + 2) = f (x - 2) = f (x + 2) = g(x) In view of g'(x) = f'(x + 7r/2) and g"(x) = f"(x + 7r/2), we can write f (x) + f(x) _
1
g(x)' 1
g"(x) + g(x) =
f(x)
3.3 THEORY OF FUNCTIONS
205
Multiplying here by g and f and subtracting the second equality from the first, we get
0 = f"9 - f9" = (f'9 - f9')" and so the function c = f'g - f g' is constant. Furthermore, since the derivative of an even function is odd, we find that c must be odd, hence c = 0. Since g does not have a zero, we can conclude that (f /g)' = c/g2 = 0, and therefore f /g is constant. f is continuous and periodic, so it takes its maximum and minimum at some points xl and x0. Hence g(xo) = f (xo + 7r/2) > f (xo) and g(xi) = f (xl + 7r/2) < f (xl). These inequalities and the fact that the ratio f /g is constant show that this constant is equal to 1, which means that f (x) _ g(x) for every x, and this is exactly what we had to prove.
Problem F.37. Let g : JR -+ R be a continuous function such that x+g(x) is strictly monotone (increasing or decreasing), and let u : [0, 00) R be a bounded and continuous function such that
u(t) + J
ti g(u(s))ds
is constant on [1, oo). Prove that the limit limt_cc, u(t) exists.
Solution. It is enough to prove the existence of limt_,,.(u(t) + g(u(t))) because x + g(x) is continuous and strictly increasing. Consider the function v(t) = ftt 1 g(u(s))ds on the interval [1, oo). For this,
v'(t) = g(u(t)) - g(u(t - 1))
.
Since u is bounded, say u : [0, oo) -+ [a, b], we get that v' is also bounded. Hence, v is uniformly continuous on [0, oo). But we have assumed that u(t) + v(t) is constant, hence the uniform continuity of u also follows. On the other hand, g is uniformly continuous on [a, b], which yields the uniform continuity of v' via the preceding formula. Next we show that v' has limit zero at infinity. To this end, it is enough to show that the integral f1 (v'(s))Zds is finite, because the integrand is nonnegative and uniformly continuous. We can write 2
f(vi(s))2ds
=
(g(u(s)) - g(u(s - 1))) ds
= 2 ft g(u(s))(g(u(s))
-
g(u(s - 1)))ds
i
=
f
t
t
g(u(s))2ds + f g(u(s - 1))2ds
2 f9(u(s))v'(s)ds t i
- ft-t i
g(u(s))2ds + fo
9(u(s))2ds.
3. SOLUTIONS TO THE PROBLEMS
206
Since g is bounded, the second and third terms are bounded with a bound independent of t. The same is true of the first term, namely, using v' = -u' (which follows from the assumption that u(t) + v(t) is constant), we can get 2
J
t
g(u(s))v'(s)ds = -2
t
g(u(s))u (s)ds = -2 fu(i) u t g(x)dx,
1
and g is bounded in [a, b]. With this, we have verified that the above integral is finite, furthermore
tliu(t) = - tli m v' (t) = 0. Let us now notice that t
u(t) + g(u(t)) = (0) + v(t)) + (g(u(t)) -
J g(u(s))ds) it 1
- g(u(s)))ds. t (0) + V(0) + j(g(u(t)) 1
By our assumption, the first term is constant. For the second term, we get from the mean value theorem that for every s E [t -1, t) there is a X E (s, t) with Iu(t) - u(s)I = 1(t - s)U'(X)I W(X)I. Therefore, lim sup Iu(t) - u(s)j = 0. t- 00.9 E[t-1,t]
Hence, the integrand in the preceding formula, and together with it also the second term, tends to zero as t --> oo. Thus, the limit limt.c(u(t)+g(u(t))) exists, and the proof is complete.
Problem F.38. Prove that if the function f : R2 -> [0, 1] is continuous and its average on every circle of radius 1 equals the function value at the center of the circle, then f is constant.
Solution. It is enough to prove that for every 1 > a > 0 the function a
g- (z) =
9(z) =
4a2
J
a
pa
J
f (z + x + iy)dxdy a
is constant, since then we get the constancy of f for a -> 0. Note that g also satisfies the assumptions of the problem, namely by Fubini's theorem, f2w
2
27r
g(z + ett)dt = 1
a
1
4a2
0
f
a
fa fa f (z + ett + x + iy)dxdydt 2x
a
=4a2- a,f a 27r fO 1
a
a
f(z+x+iy+&t)dtdxdy
a
f(z + x + iy)dxdy = g(z).
4a2
fa fa (1)
3.3 THEORY OF FUNCTIONS
207
Obviously, g is uniformly continuous. Hence there is a positive function 6 such that lim 8(e) = 0,
E-.0
and for every z, z' E R2 = C
Ig(z) - g(z')l < 5(lz - 4).
(2)
Now let g be the set of all g : R2 -+ [0, 1] that satisfy the mean value property (1) and the smoothness property (2) (with the above b, which we consider to be fixed). G consists of uniformly bounded and uniformly equicontinuous functions, hence G is compact with respect to the uniform norm. Thus, the functional g(1)-g(0) attains its supremum a on G for some go. Since G is clearly translation and rotation invariant, we can conclude that for every g in G and for every z, z' with Iz - z'I = 1 the inequality (3) Jg(z) - g(z') I < a holds. But then, on applying the mean value property (1) for go, we get that
a = go(1) - go(0) =
1
2n
f (go(1 + ezt) - go(eit))dt < 0
2ir
f adt = a, 0
which, in view of the continuity of the functions involved, is only possible if we have equality everywhere under the integral sign, that is,
go(1 + ezt) - go(eit) = a. In particular, go(2) - go (1) = a. Applying the same procedure to this equality, we can conclude go(3) -
go(2) = a, and in general go (k + 1) - go (k) = a for every k = 1,2,.... Adding these together, we arrive at go (n) - go (0) = na, which, taking into account the boundedness of go, is only possible if a = 0. Return now to (3). This says that for every g in G we have g(z) = g(z') provided Iz - z'J = 1. But every two points on the plane can be joined by a chain of points with consecutive distance 1; hence, we can conclude that every function in G is constant. Thus, the function ga from the beginning of the proof is also constant because it is a member of G, and this proves our claim. Remarks.
1. The conclusion holds if we assume only the nonnegativity of f. This is a result of H. Shockey, and J. Deny.
2. The statement is a certain strenghtening of the fact that a bounded and harmonic function on the plane is constant; namely, the harmonic functions are exactly the functions that have the mean value property on every circle (not just on those with radius one). 3. The statement is the continuous variant of the following well-known problem, which can be solved along the lines discussed above: If we write numbers from [0, 1] into the lattice points of the plane in such a way that every number equals the average of the four neighboring ones, then all the numbers are the same.
3. SOLUTIONS TO THE PROBLEMS
208
Problem F.39. Let V be a finite-dimensional subspace of C[0,1] such that every nonzero f E V attains positive value at some point. Prove that there exists a polynomial P that is strictly positive on [0, 1] and orthogonal to V, that is, for every f E V, 1
f
f (x)P(x)dx = 0.
Solution. Let fl,... , fk be a basis in V, and let us define a mapping f : [0,1] - '
Rk
by f (x) = (fi(x),... , fk(x)). If A = (Al, ... , Ak), then
ATf(x) = >aifi(x) E V, and hence there is an x E [0, 1] for which
ATf(x)>0. That is, if L is any half-space containing the origin, then f (x) is an inner point of L for some x. Hence, the origin is contained in the convex hull of if (x)jx E [0, 1]}.
But then there exists an E > 0 such that this set contains the ball with center at the origin and of radius E. Let us consider the points of the form (Eel,...,eek),
where ei = f1. For any choice of the values e1, ... , ek, there are numbers c1 > 0, ... , ck+1 > 0, ci = 1, xl, ... , xk+l E [0,1] for which
1: cifi(x.i) =Eei. Approximating first the distribution >j cab.,, by positive functions, and then these functions by polynomials, we get the existence of a strictly positive polynomial p for which
c;fi(x;) Now let J C {1,
.
ei=1
,
101
f (x)p(x)dx
E
< 2.
k}. Let us choose the numbers ej as follows: if
ei=-1
iEJ;
if
iOJ,
and let pj denote the polynomial p corresponding to this choice. Then
fpJfi>, E
i E J,
3.3 THEORY OF FUNCTIONS
209
and
fpJfi<_, Hence, in every quadrant of the space IRk, there is a point of the form
(fPJf1...fPJfk). Thus, the origin belongs to the convex hull of these points, that is, with some dJ > 0,
f (dJPJ)f=o
,
i = 1, ... , k,
and at the same time the polynomial EJ djpj is strictly positive on [0, 1].
Remark. For more information see A. Pinkus, and V. Totik, One-sided L'-approximation, Can. Bull. Math., 25 (1986), 84-90.
Problem F.40. Let D = {z E C: Izl < 1} and D = {w E C: Iwl = 1}. Prove that if, for a function f : D x B --+ C, the equality
az+b aw+ b
f
b aw+b f(z'w)+f (a'bw+a)
&+-d' bw+a)
1
( )
holds for all z E D, w E B, and a, b E C, lal2 = 1 + Ib12, then there is a function L : ] 0, oo[- C satisfying
L(pq) = L(p) + L(q),
for all p, q > 0,
such that f can be represented as 2 f(z,w)=L(1-Izl2
l1w-z$
,
forall
zED,wEB.
Solution. First, we prove that if f satisfies equality (1), then there exists
acp:D --pCsuchthat f (z, w) = cp(zw), t(1 - S-)
s(1 - s)
( ts(1-s)+1-s S)+
(z, w) E D x B,
= cp(t) + cp(s),
t,sED.
(2)
(3)
Let a2 =w hold for (z, w) E D x B, b = 0, and a E C. Then, from equality (1) it follows that
f(zw,1) = f(z,w)+f(0,1).
(4)
3. SOLUTIONS TO THE PROBLEMS
210
By substituting z = 0, w = 1, we get f (0, 1) = 0, and so, by (4), (2) holds for the following function cp : D -> C:
z E D.
O(z) = f (z, 1), From (1) by (2), we get
(b aw+b
az+b aw+b _ (bz+a bw+a) (z, w) E D x B,
bw+a)
a
\
Ia12 = 1 + 1 b 1 2 .
(5)
Let t, s E D. If we substitute
a=
1
1 -Isle, b =as, w =
1-s 1-s
z = tw
for (5), then (3) follows from the fact (a w + b)/(bw + a) = 1.
LetA={uEO: Reu>0}and u E A.
0(u) = 0(u + 1)
(6)
Substituting for u, v E A: t = (u - 1)/(u + 1), s = (v - 1)/(v + 1), from (3), we obtain
0(uRev+i Imv) =0(u)+'c,(v),
u,v c A.
(7)
We are going to show that
uEA.
(Reu),
(8)
Let x E (0, +oo), y E R. Then from (7) it follows that
0(1 + 2iy) = 0((1 + iy)1 + iy) = 0(1 + iy) + ?(1 + iy) = 20(1 + iy) and
(1 + 2iy) =(2) + 0(1 + 2iy) - 0(2) = 0(2 + 2iy) - 0(2) = ,O((1 + iy)2) - 0(2) = Vi(1 + iy) + V)(2) - 0(2) = 1(1 + iy). Hence, 0(1 + iy) = 0, and from (7) we get '0(x + iy) = z/i(x 1 + iy) = O(x) + 0(1 + iy)
Finally, let L(p) = zi(p), p E (0, +oo). Then from (6), it obviously follows that
L(p4) = L(p) + L(4'), p, 4 E (0, oo) On the other hand, from (2), (6), and (8), we obtain
1-zw 1
- IzI2l
(1
l IZI2
Iw-z121 =L\Iw-zi2), for all (z, w) E D x B.
1-zwJ
3.3 THEORY OF FUNCTIONS
211
Problem F.41. Prove that the series EP cP f (px), where the summation is over all primes, unconditionally converges in L2[0,1] for every 1periodic function f whose restriction to [0, 1] is in L2[0,1] if and only if
uP Icpl < oo. (Unconditional convergence means convergence for all rearrangements.)
Solution. Let ff(x) = f (px). Suppose first that the sum in question is finite. Since we have II fPII = II f II, where II . II denotes the L2[0,1]-norm, we
get IIcPfPII <
IcPIIfPII = IIf11
IcPI < 00.
Hence the completeness of L2 [0, 1] implies the convergence of > cP fP in any rearrangement. Conversely, suppose that E Ic,I is divergent. Let AP denote the norm-
preserving operation f -+ fP. If 6 is an arbitrary number, there is a
finite subset pi, i E I, of the primes such that
I EiEI cPi I
> 6. Let
= EiEI cp,APi and H = {fJ2Ed pi : J C I}. If n E H, then let J n = {i E I : pile}, n = 11{pi : i E I \ Jn}, vn = >{cp : p E Jn}. Consider the functions e27rint c L2 [0, 1), n E H. These form an orSI
thonormal system, and Ap(e2'rint) = e27rinPt. The function >nEH e27rant has norm 21H112. Apply the operator SI to this function. Then for n E H, the coefficient of e2nint in the resulting function is
E cPi piln
if lpi = n for some i E I, and 0 otherwise. Hence, the coefficient is exactly vn. Thus,
IISI(j: e27rint)I12 > > vn = 2 nEH
>1
nEH
(vn,+vn)
nEH
1:(vn+ Vn-)'
CPi)2'
nEI
nEH
that is, Pill >_ 26.
Since here 6 is arbitrary, we can form a sequence of operators SIk with of subsets of the natural numbers, some increasing sequence Il C I2 C
the union of which contains every natural number such that the norms II SIk 11 tend to infinity as k ---> oo. Hence by the Banach-Steinhaus theorem,
there is an f for which the sequence SI,, (f ), k = 1, 2.... is not convergent in L2 [0, 1], and this proves the necessity of the condition.
3. SOLUTIONS TO THE PROBLEMS
212
Problem F.42. Let ao = 0, a1,.. , ak and b0 = 0, b1,. .. , bk be arbitrary real numbers. (i) Show that for all sufficiently large n there exist polynomials Pn of degree at most n for which .
i = 0,1, ... , k,
p (x ) (1) = bi,
Pn' ) (-1) = ai,
(1)
and
maX jPn(x)I
n2,
(2)
where the constant c depends only on the numbers ai, bi. (ii) Prove that, in general, (2) cannot be replaced by the relation limn n2 max Ipn(x)I= 0.
(3)
Solution. Let us search pn in the form 1 2k+1
pn(X) = 12
Cs,nT(2m+1)s(x),
(4)
S=0
where
m
- 2]
= [4k+2
(n > 4k - 1),
(5)
and Tj (x) = cos(j arccos x) is the Chebyshev polynomial of degree j. Then
the degree of pn is at most n, and conditions (1) lead to the system of equations 2k+1
2k+1
cs,nT(2)m+l)s(1) = bin2, i = 0, ... , k.
Cs,nT(2m+1)s(-1) = ain2,
s=0
s=0
(6)
We are going to prove that for large enough n this system can be uniquely solved.
First of all, because Tx)(-1) _
we get from (6) by
addition that k
(i) c2s,nT2(2 ,+1)s (1) -
+bi 2 (- 1)'ai o n,
i = 0, ... , k.
(7)
s=0
If we differentiate the well-known differential equation
(1-x2)T'(x)-xT(x)+j2T(x)=0 for the Chebyshev polynomials (i -1) times and substitute x = 1, we arrive at
T(i)(1) = j2 - (i - 1)2T(2_1)(1)
j2(j2 - 1) ... (j2 - (i - 1)2)
2i - 1
(2i - 1)!!
2i
= ( 2i - 1)!! .. +0
(j2x_2),
i = 1, ... , k; j - oo,
3.3 THEORY OF FUNCTIONS
213
whence from the above equations and from (5) the equations in (7) take the form k
E s=o
c2s,n{S2a +O(n 2)}
1)ai+bi
= 22i(2rn + 1)2i ( - 1)!!n2,
i = 0,...,k. (8)
Now, if n tends to infinity, then in the limit this takes the form (recall that
ao=bo=0) k c23
s=0
s2i =
bl - a1
(2k + 1)51
,
i = 0...k ,
,
,
2
which has a unique solution, for its determinant is the Vandermonde determinant formed of the elements 12, 22, ... , k2, and so it is different from zero. But then (8) has a unique solution for sufficiently large n, and c29,n = 0(1), s = 0, ... , k. Similarly, it follows from (6) that c2s+1,n = 0(1), s = 0, ... , k. c Hence, (4) satisfies (2) whenever we choose c so large that c > Y: 2k+1 i, is satisfied. The fact that in general (3) cannot hold follows from Markov's inequality:
max p 15 n2 max Ipn l .
xE[-1,1]
xE[-1,1]
Indeed, using this we can see that (3) would imply lim max lp' (x) 1 = 0,
n-- Ixl<1
which contradicts (1) unless a1 = 0 and b1 = 0.
Problem F.43. Let f and g be continuous real functions, and let g 0- 0 be of compact support. Prove that there is a sequence of linear combinations of translates of g that converges to f uniformly on compact subsets of R.
Solution. If {g,,} is a sequence of linear combinations of translates of g for which Ig,,,(x) - f (x) I < 1/n for every x in the interval [-n, n], then this sequence uniformly converges to f on every compact subset of the real line. Hence, it is enough to verify that any f can be arbitrarily well approximated on any finite closed interval I by linear combinations of translates of g.
Let us suppose for an indirect proof that this is not the case, and I is such an interval, that the linear hull of translates of g is not dense in C(I). Then, by the Hahn-Banach theorem, there is a linear functional L E C* (I) that vanishes on every translate of g. By the Riesz representation theorem, L can be identified by integration with respect to a signed measure p with
support in I. Let h(x) = g(-x). Then f00
(h*d)(t)=J
00
0o
h(t-x)dµ(x)=J
g(x-t)d(x) = 0o
f g(x-t)dp(x)=0. n
3. SOLUTIONS TO THE PROBLEMS
214
Taking the Fourier transform here (which is possible because h and p have compact support), we find .F(h) F(p) - 0. However, the functions 1(h) and 1(p) are analytic, hence one of them must be identically 0. But this is a contradiction since neither h nor p is identically zero. This proves the statement of the problem.
Problem F.44. Let x : [0, oc) - R be a differentiable function satisfying the identity x' (t) _ -2x(t) sin2 t + (2 - I cos tj + cos t)
j
x(s) sin2 s ds -
on [1, oc). Prove that x is bounded on [0, oo) and that limt . x(t) = 0. Does the conclusion remain true for functions satisfying the identity
(t) _ -2x(t)t + (2 - cos t+ cos t)
f
x(s)s ds ? -1
Solution. Let us consider the equation t
x'(t) = -2a(t)x(t) + (2 - I cost) + cost) it
a(s)x(s)ds,
t-i
where a(t) > 0 is a continuous function. If x(t) is a solution, then let us estimate the upper right derivative D + I x(t) I of Ix(t) I : t
D+Ix(t)I < -2a(t)Ix(t)I + (2 - I cost) +cost)
Jta(s)Ix(s)Ids -
t
-2a(t)Ix(t)I +2 it
-i
a(s)Ix(s)Ids
t
- (I cost[ - cost) it
t-i
a(s)Ix(s)Ids
-2 (f° a(t)jx(t) Ids -
f
o
a(t + s)lx(t + s)Ids )
t
- (I cost) - cost) it
-2 dtd
fj ro
i
t-i
a(s)Ix(s)Ids
t
a(u) I x(u) I duds +9
-(I cost[ - cost)
f
a(s)Ix(s)Ids.
Hence,
D(Ix(t)I+2f oi ft t
+9
a(u)Ix(u)Iduds)
-(IcostI-cost)f a(s)x(s)ds. t t-i (1)
3.3 THEORY OF FUNCTIONS
215
Since the right-hand side of (1) is nonpositive, we get that the function (2)
Ix(t) I + 2 f o ft+s t a(u) I x(u) I duds 1
is decreasing, and so I x(t) I is bounded.
For k = 0,1, ... , let Hk = [(2k + 1)ir - 1, (2k + 1)7r + 1]. We show that
f
t
max
a(s)Ix(s)jds -> 0
tEHk t-1
(k -> oo).
(3)
If we suppose that on the contrary, (3) does not hold, then there is an
a > 0 and a sequence {t,,,} such that t,,, E U' oHk, t,,,, - 00, and t,y
ftn_1 a(s )Ix(s)Ids > a. Then at least one of the equalities t n-1/2
a(s)Ix(s)Ids > a/2,
t
a(s)Ix(s)Ids > a/2
ftn_1
holds. Since I cost I - cost > 0 on the interval Hk = [(2k + 1)ir - 3/2, (2k + 1)ir + 3/2], there exists a 3 > 0 such that at every point of an interval of length 1/2 and containing tn, we have t -Cost) f
(I cost)
a(s)Ix(s)Ids > /3.
t-1
However, this and (1) imply Ix(t) I+ 2
f ft o
1
t +s
a(u) I x(u) I duds -> -oc
(t ---> 00),
which is impossible. Hence, (3) holds. From (3) it follows that
ff o
max tEHk
1
t
t+s
a(u)jx(u)jduds -* 0
(k - oo).
(4)
Using (4) and the monotonicity and nonnegativity of the function in (2), we can see that to prove the existence of the limit of x(t) at infinity it is enough to show that there is a sequence {tn} of points of the set UkoHk for which to -> 00, x(tn) -* 0 as n -* oc. If there was not such a sequence, then there would be a ko E N and a -y > 0 with the property that Ix(t) I > ry for every
t E Hk and k > ko. This, however, contradicts (3), whether a(t) = sin2 t or a(t) = t. This proves the claim of the problem, and we have also obtained that the same conclusion holds if we use the equation in the second half of the problem.
3. SOLUTIONS TO THE PROBLEMS
216
Problem F.45. Let c > 0, c # 1 be a real number, and for x E (0, 1) let us define the function 00 cx2k).
f(x) = fJ(1 + k=O
Prove that the limit f (x3)
lim
x-.1-0 P X) does not exist.
Solution. We argue indirectly, hence let us assume that the above limit exists. In general, let S be the set of all positive real numbers q for which the limit lim
f(x4)
g(q)
x-1-0 f(x)
exists. Thus, the indirect assumption is that 3 E S. First, we are going to show that this implies S = R+, and we also get an explicit representation on the function g. Note that 2 E S because x
lim o f(x2) f(x)
X-1-0 Jim
=
1
1+cx
_
1
1+c*
(1)
Furthermore, by substitution it is easy to verify that S is a multiplicative subgroup of the real field, and g(ql)g(q2) = g(qlq2)
for
gl,g2 E S.
(2)
The numbers 2m31, in, l = 0,±1,... form a subgroup S1 of S that is dense on the positive real line (use the fact that by the prime factorization theorem the number log 2/ log 3 is irrational, hence numbers of the form m log 2 + l log 3 form a dense set on R).
Using the monotonicity of f, we can see that g(q) _< 1 if q > 1 and g(q) > 1 if q < 1. This implies g(q) = q7 for q E Sl for some ry < 0 In fact, let g(2) = a, g(3) = b, and let us choose ry to satisfy a = 27. If b 0 3-(, say b < 37, then by choosing a number q = 2m31, 1 < 0, in the interval [1, 2] such that (b/37)1 > 1, we get a q E Sl with q > 1 and g(q) > 1, which is not possible. Hence, g(q) = q'Y for q = 2, 3, from which the same follows for all q E Sl by the group property (2). Now it is easy to show that S = R+ and that for every q E ][8+ we have g(q) = q7.
(3)
In fact, if q E ][8+ is arbitrary, then for every e > 0 there are ql, q2 E S1 such that ql < q < q2,
1<<1+E. g(q2)
3.3 THEORY OF FUNCTIONS
217
But the monotonicity of f implies that
ff((x ))
9(ql) ? liimsup
> liminof f(x)) > 9(q2),
and here the left- and right-hand sides can be arbitrarily close to qry if e is chosen sufficiently small. This verifies (3) for all q. We will not use it, but it is clear from (1) that y = - log2(1 + c). Since
1-x4'Y -=q7,
lim x-.1-0 (1 - x)ry
we can conclude from (3) that f is of the form f(x) = (1 - x)-PL(x)
(4)
(with p = -y), where L is slowly varying inthe sense that x lim
L(x4)
o L(x) -- 1
for every q > 0.
Let f(x)
00 akxk,
0 < x < 1,
k=0
and sn = k-0 ak. By a well-known Tauberian theorem (see G. H. Hardy, Divergent Series, Clarendon Press, Oxford, 1989, Ch. VII, Theorem 108), we can derive from (4) with any fixed q E (0, 1) and some p > 0 that sn ,.. p.f (ql/n)+
where an - bn means that the ratio an/bn tends to 1. However, m-1
52m = 1 (1+c)+c= (1+c)m+c 0
and C)m-1 + C2,
82-+2--1 = (1 + C)m + c(1 + from which
f
52m+2m-1 c 1+1+c 82-
(1 (1 -
ql/(2--1,3))-P
ql/(2m-1.2))-P
(ql/(2-_1.3)) (gl/(2m-1.2))
.f
2 .2m-1 1 -P
(3
. 2m-1
P
I
(3)p 2
3. SOLUTIONS TO THE PROBLEMS
218
and by (1) and (4),
1+c=2P. Thus, 1 + 2c = 3P, and p must satisfy the equation
3P-2.2P+1=0, which certainly holds for p = 0 and p = 1, but does not hold for any other p, because the function 3' - 2 2t + 1 is convex on [0, co). If p = 0, then c = 0; while if p = 1, then c = 1, and these values for c were not allowed. The obtained contradiction proves the claim.
Remark. Note that if c = 1, then f (x) = 2/(1 - x), and the limit lim f(X9)
x-,1-o f (X) exists for every q > 0.
Problem F.46. Let f and g be holomorphic functions on the open unit disc D, and suppose that If I2 + IgI2 E Lipl. Prove that then f, g E Lip'2* A function h : D -+ C is in the Lipa class if there is a constant K such that jh(z) - h(w)I < Kjz - w1a for every z, w E D.
Solution. We shall use the following Hardy-Littlewood theorem: if a function h : D --> C satisfies
h'(z)I <
M
V
(1)
Ixl
with some constant M, then h E Lip!. In fact, in order to verify the Hardy-Littlewood theorem, it is enough to show that if Iz - z'I < 6 < 1/2, then
If (Z) - f(z')I < C/.
(2)
But if w = (1 - 6)z, w' = (1 - 5)z', then (1) easily implies that zI f(I,-b)lzl
If(z) -f(w)d
M 1-
= 2M(
-zl-
tdt
I ( 1 - 6)Izl) <
M6 < Mme. 1 - (1 - 6)IzI
A similar argument shows that
If W) - f(w')I < C/
and
If(w) - fW) I <- CVi,
3.3 THEORY OF FUNCTIONS
219
from which (2) follows. We shall prove that
IfF(z)I2+ l9
(z)I2 <
1
KIzI,
where K is the Lipschitz constant corresponding to If I2 + IgI2. From here our statement follows by the Hardy-Littlewood theorem. Let z E D and r > 0 be such that the circle with radius r and center at z is contained in D. Let us apply Parseval's identity on this circle:
1 j27r 27r
f (n) (z)
If (z + ei9r) I2 d8 = n=O
2
r2n > If(z)I2 + If'(z)I2r2,
n!
or in rearranged form, 1
2lr
f (If(z+exer)I2 - If(z)I2)d8 > 27r
If'(z)12r2.
Applying the same inequality with f replaced by g, and adding the two together we, get
2f 1
27r
((I f ( z
+ eier)I2 + Ig(z + eier)I2) - (If(z)12 + Ig(z) I2 )) de
> (If'(z)I2 + Ig'(z)I2)r2
But because If I2 + IgI2 E Lipl, the modulus of the integrand on the lefthand side is at most KI eierl = Kr, hence the integral is at most 27rKr. Thus,
1 27rKr = Kr > (Ifl(z)I2 + I9
(z)I2)r2,
that is, Ifl(z)I2 + Ig'(z)I2 < K r
Since r can be any number smaller than 1 - IzI, by letting r tend to 1 - IzI we obtain
If ' (z)I 2 +g ' (z)I 2 <
K
1-IzI,
and this is what we had to prove.
Remark. The example f - 0, g(z) _ (1-z)1/2 shows that the conclusion is sharp.
3. SOLUTIONS TO THE PROBLEMS
220
Problem F.47. Find all functions f : ][83 -> R that satisfy the parallelogram rule
f(x+y)+ f(x - y) = 2f (x) + 2f (y),
x, y E I[83
and that are constant on the unit sphere of R3.
Solution. Suppose that the function f : 13
R satisfies the functional
equation
f (x + y) + f (x - y) = 2f (x) + 2f (y),
x,y E R3,
(1)
and is constant on the unit sphere f (u) = c,
u E R3,
(2)
lull = 1.
We shall prove that, with the help of an additive function a : ][8 --+ R, f can be written in the following form:
f(x)=a(IIxII2),
xER3.
(3)
We begin by showing that f takes equal values on vectors of equal norms. So first let x, y E R3, IIxII = IIyII < 1. Then there is a vector z c R3 such that IIx1I2 + IIzII2 = 1 = IIyll2 + IIzII2 and z1.x, zly. By the Pythagorean theorem, IHx ± zil = 1 = Ily ± zil, and therefore by (1)-(2),
2f(x) = f (x + z) + f (x - z) - 2f (z) =c+c-2f(z) = f(y + z) + f(y - z) - 2f(z) = 2f(y), that is, f (x) = f (y). On the other hand, setting y = 0 in (1),
2f(x) = f(x + 0) + f(x - 0) = 2f(x) + 2f(0), whence f (0) = 0. Thus the substitution y = x leads to the relation f(2x) = f (x + x) + f (x - x) = 2 f (x) + 2 f (x) = 4 f (x),
x E R3 .
Consequently, if for some positive number r and all x, y E R3 satisfying I I x I I = I I y I I < r we have f (x) = f (y), then for any x', y' E ll
satisfying
IIxII = IIyII < 2r, with the notation x = x'/2, y = y'/2, we obtain IIxII = y < r and therefore also f (x') = f (2x) = 4f (x) = 4f (y) = f (2y) = f (y) So, by induction, f (x) = f (y) whenever x, y E R3 and I I x I = IIyII. I
3.3 THEORY OF FUNCTIONS
221
By what we have proved, f (x) depends only on I1 x I1 or equivalently, same, on IIx112; in other words, there exists a function a : JR -> R such that f(x) = a (11XI12) ,
xEJ3.
It remains to verify the additiveness of the function a. To this end, fix ), p E R+ arbitrarily, and choose vectors x, y E JR3 so that A = IIx112' P = IIy112, and xly. Then, in view of (1) and the Pythagorean theorem, 2a(.\) + 2a(p) = 2a (11x112) + 2a (IIy112) = 2f (x) + 2f(y)
=f(x+y)+f(x-y)=a(Ilx+y112)+a(IIx-yl12) = a (11x112 + I1y112)
+a
(11x112 + I1y112) = 2a(A
+ p).
Thus a is additive on R+ and, as we may choose its value on JR_ arbitrarily,
setting a(-a) = -a(A), A E R+, we get the additive function desired. Conversely, simple substitution shows that the functions of the form (3) are solutions of the problem (1)-(2). Remarks. 1. We obtain the same solution if we assume (1) only for y satisfying Ilyll = 1 and, in a more general way, consider vectors of a real inner product space of dimension at least three. Of course, in this case we encounter further, essential difficulties.
2. The two-dimensional case has proved to be still harder. It is an open question whether in this case the problem (1)-(2) admits solutions different from (3).
Problem F.48. For any fixed positive integer n, find all infinitely differentiable functions f : I[8n -+ JR satisfying the following system of partial differential equations: n
>aikf
0,
k=1,2,....
i=1
Solution. Let n E N, and consider the system of partial differential equations n a2kf
= 0,
k = 1, 2, ...
(1/n)
i=1
for the infinitely differentiable unknown function f : Rn -+ R. Obviously, linear combinations of any partial derivatives of solutions are solutions
again. We show that there is a universal solution in the sense that all solutions are linear combinations of partial derivatives of this universal solution. We begin with an important observation.
Lemma 1. If f is a solution of system (1/n), then 3f = 0 (i = 1, 2, ... , n), so f is a polynomial of degree not greater than 2n - 1 in each variable.
222
3. SOLUTIONS TO THE PROBLEMS
Proof. For the commuting partial differential operators al, a2, an, consider the power sums
Pk = aik + 022k + ... + a2k,
k = 1,2,...,
and the elementary symmetric polynomials
S1=a1+a2+...+an S2=c1 S3 = a1 a2a3 + al a2a4 + ... + a.2-2an-1an
Sn = al a2 ... an.
Then the differential equations can be written in the form
k=1,2,....
Pkf =0, Further, by the Newton formulas,
P1 - S1 = 0,
P2-SIP, +2S2=0,
........................
Pn-SIPn_1+...+(-1)"-1Sn_1P1+(-1)nnSn=0.
Hence, it follows that
Slf =Plf = 0, S2f = 1(S1Plf - P2f) = 0,
1 Snf = -(Sn-1P1f - Sn-2P2f + ... + (-1)n-lpnf) = 0. n
Now, for fixed i E {1, 2, ... , n}, consider the differential operator
P(ai) = (ai a2) ... (a1 - an) =a2n-S a2n-2+S a2n-4-...+ ( - 1 ) ' S
.
We see that P(aa) - 0 and, therefore, at n f = Slain-2 f - S2a2n-4 f
+... + (-1)n+lSnf = 0.
Lemma 2. Let n E N, and let the function Qn : 1[Sn - JR be defined by the following determinant: 2n-3 xl \ /X in-1
/Qn(x1, x27 ... , xn) = det
x22n-1
x22n-3
X2
k x2n-1
x2n-3
xn
I
n
n
/
3.3 THEORY OF FUNCTIONS
223
Then f = Qn is a solution of the system (1/n). Proof. We proceed by induction for n. If n = 1, then the assertion is obvious: Q1(X1)=X1'
Q1 (xl) = 0.
Suppose the assertion is true for n - 1, that is, n-1
0ikQn-l(xlIx2,...,xn-1) = 0,
k = 1,2,... .
i=1
Expand the determinant Qn according to the first column: n
Qn(xlex 2
o
Xn) n) = -
j=1
(-)j-1x2n-1 Qn-1(X1, ..., x
1,x{lo
x n)
It follows that n
Ea2kQn(xl, x21... , xn) ii=1 n
n
W 1)
_
2k [2n-1 xj Qn-1(xl, ... , xj-1i xj+1i ... , xn)]
C7i
i=1 j=1 n
_
-1)j-1
j
d2k
2n-1 &x3 kxj
Qn-1(X1.... ,xj-1ixj+li...,xn)
n
(-1) j-1xj2n-1 j=1
ai2kQn-1(x1i...,xj-1,xj+1,...,xn) i0j
The second sum vanishes by the induction hypothesis. Obviously, the first sum also vanishes if k > n, whereas for k < n it is the expansion according to the first column of the following determinant: / x12n-2k-1
xl2n-3 ... xl \
x22n-2k-1
2n-3 x2
\X 2n-2k-1 n
x2n-3
x2
(2n - 1) (2n - 2)...(2n -
n
...
xn
This, too, is zero since the first and (k + 1)th columns of the determinant coincide.
Theorem. The function f : IEBn -> R satisfies the system of partial differential equations (1/n) if and only if it is a linear combination of partial derivatives of Q.
3. SOLUTIONS TO THE PROBLEMS
224
Proof. Lemma 2 shows that Qn satisfies the system (1/n). Therefore, each of its partial derivatives and all linear combinations of them will also be solutions. The converse can be proved by induction for n. For n = 1, it is obvious, since f" = 0 implies f(xl) = axl + b = a Q1(xl) + b a1Q1(x1)
Suppose that the assertion is true for all natural numbers not greater than n. Let f be a solution of the system (1/n+ 1). By Lemma 1, an+1 f = 0, and so a2+ll f is independent of xn+l: 2n+1 an+1 f (xl, x2, ... , xn, xn+1) _ 00(x1, x21... , xn)
Since together with any solution its partial derivatives are also solutions, 00 satisfies (1/n + 1) and, since it does not depend on xn+l, it satisfies (1/n) as well. Thus, by the induction hypothesis, we have the representation
00(xl,x2i.... xn) -
C0
all1
a2212 ...ann1n
Qn(xl,x2,...,xn)
with suitable constants co E R (a E N` o) among which only finitely many are different from 0. On the other hand, we note that Qn(x1i x2, ... , xn) =
(-1)n
2n+1
(2n + 1)! an+1
Qn+1(x1, x2, ... , xn+1)i
which can be seen from the expansion of Qn+1 according to its last row:
E x,n+1 1-1) Dk(n). n
2k+1//
Qn+1(x11 x21... , xn+1)
k
k=0
Here Dk(n)
is
the respective minor and,
obviously,
Dn(n)
_
Qn(xl,x2,...,xn). Comparing the above results it follows that (_l )n
00=1: c0 (2n + 1)! ala1a2
2a2
... annn 2n+1 an+1 Qn+l
Next, define the function go : ]!8n+1 -* R by the relation 90 =
-1
n
0 C11 a22 ... annnan+lQn+1
co (2n + 1)! al a
Clearly, go is a solution of (1/n + 1), and (f - go) = o - to = 0. Thus fl = f -go is a solution of (1/n+1) and an+il fl = 0, that is, an,+1f1 does not depend on xn+1: a2n
n+lfl (xl, x2i ... xn+l) = 01(x1, x21 ... , xn).
3.3 THEORY OF FUNCTIONS
225
Pursuing the process, we obtain the functions 00, 01,
, 02n, 90, 91, . .
.
,
92n, and fl, f2, ... , f2n+1. Here each of the functions go, gl, ... , g2,,, is a linear combination of partial derivatives of Qn+i So with the help of Lemma 2, we see by induction that for k = 1, 2, ... , 2n the function fk+l = fk-9k is a solution of the system (1/n+ 1), and 82+ik+1 fk+1 = 82+1k+1(fk -9k) _ Ok - q5k = 0. Consequently, f2n+i does not already depend on xn+l, that is,
f2n+1(x1, x2, ... , xn, xn+l) = Y'2n+1(x1, x2, ... , xn)
where, similar to the arguments above, it can be proved that ¢2n+1 is a linear combination of partial derivatives of Qn+l. Finally, the representation
f =9o+91+ +92n+f2n+1 proves the theorem.
Problem F.49. Let P be a polynomial with all real roots that satisfies the condition P(0) > 0. Prove that if m is a positive odd integer, then f(k)(0)xk > 0
k=0
k!
for all real numbers x, where f = P--.
Solution. Let
f(k)
F(x) _
k
(0)
xk
k=0
and consider the polynomial Q(x) = P'r(x) F(x). Denoting by n the degree of P, from the factor Ptm (counting multiplicities) we obtain n m real roots of Q. Assume, contrary to the assertion of the problem, that F also has real roots. We distinguish between two cases: 1. F has degree m - 1; then F has at least two real roots since its degree is even.
2. F has degree not greater than m - 2. According to Rolle's theorem, we can count at least mn + 1 roots (with multiplicities) of Q' in the first case, and at least mn roots in the second case. We note that because of the relations P(0) # 0 and F(0) # 0, the root 0 could be counted at most once. Zero is actually a root of Q', of multiplicity not less than m - 1 at that, since for j = 1 , 2, ... , m - 1 we have
Q(i)(0)
(j) [(pm)(1)(0)] l
.
rll [(Pm)(1)(o)] .
[F(u-1)(0)]
[f('-1)(o
)]
=
(PM. f)(i)(o) = 0,
3. SOLUTIONS TO THE PROBLEMS
226
the function PI f being equal to the constant 1. Consequently, at least m - 2 roots can be joined to those counted so far.
Thus Q' has at least nm + m - 1 roots in case 1, and at least nm + m - 2 roots in case 2. Since, however, the degree of Q' is equal to nm + m - 2 in case 1 and is not greater than nm + m - 3 in case 2, we obtain that Q' is identically 0. Therefore, Q is a constant polynomial, and hence it is identically 0, contrary to the relation Q(0) # 0.
Thus, F has no real roots and, since F(0) = P-m(0) > 0, we have F(x) > 0 for all real values of x.
Problem F.50. We say that the real numbers x and y can be connected by a 6-chain of length k (where 6 : ]I8 -> (0, oo) is a given function) if there exist real numbers x0, x1, ... , xk such that x0 = x, xk = y, and xi
Ixi - xi_l < S
C xz-1 2+
/
,
i = 1,...,k.
Prove that for every function 6 : R -* (0, oo) there is an interval in which any two elements can be connected by a 6-chain of length 4. Also, prove that we cannot always find an interval in which any two elements could be connected by a 6-chain of length 2.
Solution. To prove the first assertion, using the Baire category theorem, choose a positive integer n and an interval I such that the set HH, = {x: 6(x) > 1/n} is dense in I; we may also assume that the length III < 1/10n. Let K be the middle third of I; we show that in it, any two elements can be connected by a 6-chain of length 4. So let x.y E K. If c E K fl H,,, and b = c + (y - x) /2, then reflection in b followed by reflection
in c give translation by x - y. If we insert this between two reflections in a, we obtain a translation by y - x, which carries x into y. In order that in this way we obtain a 6-chain connecting x and y, we have to care only about the following two things: first, a E H,,, should be satisfied and, second, 2a - x should fall in the 6(b)/2-neighborhood of b. Since, however, H,, is dense in I, this can easily be achieved. Proving the second half of the problem requires a bit longer argument.
Let B be the set of all real numbers that can be written using a finite number of binary digits. We specify certain pairs of points of B so that for suitable 6 they cannot be connected by a 6-chain of length at most 2, and every interval contain such a pair of points. To this end, let P = U°__1P2, where Pi
(-2 2i -22i +2) 22i
22i
... ,
22i - 2 22i 22i ' 22i) }
We first define 6 on elements of B in the following obvious manner: if
x c B ends in 1/2i, then let 6(x) = 1/2i+1 (if x is an integer, we take i = 0). Then, evidently, pairs of points of P cannot be connected by 6chains of length 1, nor by those of length 2 and passing through an element of B.
3.3 THEORY OF FUNCTIONS
227
We extend 8 to R in the following way. B is an additive subgroup of so R is a disjoint union of residue classes of the form a + B. If a V B, we define 8 on a + B so that for b E a + Bi the value 8(b) be smaller than min{2Jx - bl, 21y - bJ} for all (x, y) E where B = U°0Bi, and Bi consists of those numbers whose fractional part can be written using exactly
i digits. It is easy to see that 6 may be defined in this way. To finish the proof, we verify that for any (x, y) E P and z c a + B the numbers x, 2z, y cannot form a 8-chain. Suppose that (x, y) E Pi. Then (y - x)/2 = 1/2i. If x, 2z, y form a 6-chain, then
8(2+z)>Ix-2zl =2 x-(2+z)
,
whence by the definition of 6 for j > i we obtain x/2 + z ¢ a + Bj. We similarly obtain y/2 + z V a + Bl. Then, however, the fractional part of (y - x)/2 can be written using at most i - 1 digits, which contradicts the relation (y - x)/2 = 1/2i. Remark. It is not known what happens if we require the existence of 8-chains of length not greater than 3. Problem F.51. Find meromorphic functions 0 and V) in the unit disc such that, for any function f regular in the unit disc, at least one of the functions f - 0 and f -,b has a root.
Solution. Introduce the notation i(z) _ ''(z) - O(z) 0 0,
g(z) = f(z) - O(z),
h(z) _ k(z)
Examine the functions that satisfy the following conditions: 1. f = hic + 0 is regular; 2. hic is nowhere 0; 3. (h - 1)ic is nowhere 0. We claim that there exist functions 0 and ic, meromorphic in the unit
disc, with r,(z) # 0 if JzJ < 1 and such that the above set of conditions cannot be satisfied by any h. This immediately gives the assertion of the problem.
If we choose 0 and is so that the locations and orders of their poles coincide, then the fulfillment of condition 1 implies the regularity of h. If
all three conditions are satisfied, then the values of h(z) can be 1 only at the poles of ic, and even this can be executed by choosing K and 0 properly. Assuming first-order poles, we only have to take care that, denoting by A(zo) and B(zo) the residues of is and 0 at the pole z0i respectively, the value B(zo)/A(zo) is different from -1. Actually, then h(zo) = -B(zo)/A(zo) 54 1 by condition 1.
3. SOLUTIONS TO THE PROBLEMS
228
Thus, the regular function h does not take on values 0 and 1 in the unit disc. By Schottky's theorem, it follows that I h(z) I remains below a bound depending on Ih(0)I and IzI only. We show that this is impossible. Consider the following sequences of functions:
On =0= fin=
=
1
z
z- 2
(1 - z)n
z(z- 2)
Calculate the residues of the two poles:
An(0) _ -2,
Bn(0) = 1,
An C) = 2
2-n+1
Bn (2 I = 1-
Hence, it is easy to see that the functions meet the requirements stated so far. If, however, the three conditions above are fulfilled, then hn(0) = 1/2 and hn(1/2) = -2n-1, contrary to the conclusion of the Schottky theorem. Therefore, if n is sufficiently large, the statement of the problem holds for icn, On and the function On = icn + On obtained from them.
Problem F.52. To divide a heritage, n brothers turn to an impartial judge (that is, if not bribed, the judge decides correctly, so each brother receives (1/n)th of the heritage). However, in order to make the decision more favorable for himself, each brother wants to influence the judge by offering an amount of money. The heritage of an individual brother will then be described by a continuous function of n variables strictly monotone in the following sense: it is a monotone increasing function of the amount offered by him and a monotone decreasing function of the amount offered by any of the remaining brothers. Prove that if the eldest brother does not offer the judge too much, then the others can choose their bribes so that the decision will be correct.
Solution 1. In terms of functions, the problem can be expressed in the following way. We are given the continuous functions
gi(x1,...,xn),...,g2(x1,...,xn),...,gn(xi,...,xT,,) defined on [0, oo)n (here g j (x1, ... , xn) is the deviation of the heritage of the jth brother from 1/n times the whole heritage, provided that the judge is offered x1 units of currency by the first brother, x2 units by the second, etc.), the sum of the functions being 0, and the function gj(xl,... , xn)
being strictly increasing in the variable xj but strictly decreasing in all other variables xk, k j. Further, we know that the judge is originally impartial, that is, gj (0, . . . , 0) = 0 for all j. We have to show that there exists an > 0 such that for any 0 < xn < an there are values x1 = xl(xn), ... 1xn_1 = xn_1(xn) with gi(xl. , xn) = 0 for all j. We prove more (to
3.3 THEORY OF FUNCTIONS
229
be exact, we must prove more in order that the proof below remain valid), namely, that there even exist x1 = x1(xn), ... ,xn_1 = xn_1(xn), which, besides satisfying the relations just mentioned, tend to 0 as xn -> 0. We apply induction for n. If n = 1, there is nothing to prove. Assume that the assertion holds for n-1 functions. We claim that there is a number b > 0 such that if 0 < X 2 ,-- . , xn < b are arbitrary, then there exists one and only one y = y(x2i... , xn) satisfying g1(y, x2i ... , xn) = 0, whereas y is a continuous, strictly increasing function of the variables X 2 ,- .. , xn. Really, the uniqueness of y follows from g1 being strictly increasing in its first variable. If x2 = x3 = . . . = xn = 0, then we may choose y = 0, while for other values the existence of y can be seen as follows. Since g1(1, 0, .... 0) > 0, therefore the continuity assumed ensures the existence of b > O such that for 0 < x2, ... , xn < b we have
91(1,x2i...,xn) > 0. On the other hand, 91(0, x2i ... , xn) < 0.
So, again by continuity, the y above must exist. If x'j > xj, j # 1, then
91(y(x2i...,xj1 ...,xn),x2,...,xj,...,xn) = 0 = 91(y(x2,...,xn),x2i...) xn) >
91(y(x2i...,xn),x2,...,xji ,...,xn),
and this shows that y(x2i ... , xn) is an increasing function of xj. We next verify the continuity of y. Since 91(y(x2i...,xn)-E, x2,..,xn) <0<91MX2,...,xn)+E',x2,...,xn)I
there is a b > 0 such that for I x j - x''. I < 6, j = 2, 3, ... , n, we have
91(y(x2, ... , xn) - -,X'2, ... , xn) < 0 < 91(y(x2, ... , xn) + -,x'2,
.
,
xn).
By the foregoing, this proves the inequalities y(x2i ... , xn) - E < y(x2, ... , xn) < y(x2i ... , xn) + E.
After these preparations, consider the n - 1 functions hi (x2, ... , xn) = 9j (y(x2, ... , xn), x2, ... , xn),
j = 2, ... , n.
By the properties of y, they are continuous, have sum 0, and satisfy the relations hj (0, ... , 0) = 0 for all j. We claim that they are strictly monotone in the required sense. Actually, if k j4 j and xk increases, then y(x21 ... , xn)
also increases and therefore hj decreases. Thus, if xj increases, then all hk, k # j, will decrease, so hj must increase since the sum of the h1 is zero. Consequently, the n - 1 functions hj satisfy the hypotheses of the
3. SOLUTIONS TO THE PROBLEMS
230
problem. So, by the induction hypothesis, there exists a' > 0 such that for
any 0 < xn < a' there are x2 = x2(xn), ... xn-1 = xn-l(xn) satisfying . , xn) = 0 for all j > 2, and here each xj (xn) tends to 0 as xn --p 0. b} > an > 0, where b is the constant used Hence, there exists during the preparation, such that if 0 < xn < an, then 0 < xj (xn) < b for j = 2, ... , n - 1. Then, however, with the values
hJ (x2i
x1 =
y(x2(xn),...,xn-1(xn),xn),
x2 = x2 (xn), ... , xn-1 = xn-1(xn),
the relation gj (xl, ... , xn) = 0 is valid for all j (for j = 1 this follows from the definition of y), and here each xj(xn) tends to 0 as xn -+ 0 (we again make use of the continuity of y). Thus, we have proved the statement also for n functions.
Solution 2. Let gi (i = 1, 2, ... , n) be the functions defined in Solution 1. Let ei = (0, ... , 0, 1, 0, ... , 0) be the ith unit vector in Rn (the ith coordinate is 1, the others are 0). By the assumptions, gl(ei) < 0 if i > 1. In view of the continuity of gl, there is a number ei > 0 such that 0 < xl < Ei implies gl (xie1 + ei) < 0. Set E = min{Ei : 2 < i < n}. We show that if 0 < xl < E, then there exist nonnegative numbers x2, x3, ... , xn such that 9i (X1, x2, ... , xn) = 0 for 1 < i < n. To this end, it is sufficient to prove that gi(xl, x2, ... , xn) = 0 for i > 2 (recall that the sum of all functions gi is zero).
Let 0 < xl < E be an arbitrary number, which will be fixed in what follows. Set G(x2, ... , xn) = max gi(x1, x2, ... , xn) and
H = {(x2, ... , xn) E
]fin-1 > 0: G(xl, x2, ... , xn) < 0}.
Then
a. H C [0,1]n-1, since if xi > 1 for some i > 2 then 0 > g1(xlei + ei) 91(x1, x2, ... , xn). Therefore >i gi - 0 gives G(x2i ... , xn) > 0b. H is closed, being a level set of a continuous function. c. H is nonempty, since (0, 0, ... , 0) E H. It follows that 91(x1, x2i ... , xn) takes its minimum on Hat some point (x2, x3, ... , x° ). We prove that
gi(xl,x2ix3,...,xn) = 0 for all i>2. By the definition of H, it is clear that 9i(xl, x02, ... , x0n) < 0
(i = 2, ... , n).
Suppose that for some i > 2 we have 0 , x2, ... , xn) < 0. 9d(xl0
3.3 THEORY OF FUNCTIONS
231
Then making x° a little larger, we remain in H, but the value of g1 becomes smaller, which contradicts the choice of (x2i ... , xn). This contradiction proves the statement. Remarks.
1. It is easy to show by examples that the assertion becomes false if we replace the condition of strict monotonicity by monotonicity in the wide sense.
2. Here is a simple example that if a > 0 is arbitrary then a correct decision cannot already be achieved in case the eldest (say, the nth) brother offers the judge at least a units of currency: let
9.j (xl, ... , xn) = (n - 2)xj +
axe
xj +
- x.j-1
- xl
x.j+l
- xn
if j = 1,2,...,n- 1, and let 9n(x1,...,xn) _ (n - 1)xn -
ax1
axe
x1+1
x2+1
axn_1
xn_1+
Problem F.53. Construct an infinite set H C C[0,1] such that the linear hull of any infinite subset of H is dense in C[0,1].
Solution. Let {9k}
1 be dense in C[O,1] (9k 0 0). We show that the set
H = {hn}°°_2, where
hn=[ k=1
9k
_
119kMIoo
nk
1
meets the requirements. If L is any bounded linear functional on C[O,1], set
hL(z) =
00
k
Lgk
k=1 \119k1loo/ z
Then hL is analytic in the unit disc, and Lhn = hL (1/n). Let
IF
- {hnl, hn2,
hn3,...
}
be an infinite subset of H, and let L E C[O,1]* be any functional that annihilates H'. Then hL (1/nn) = Lhn,,, = 0 for all m, so the analyticity of hL yields hL - 0. Hence Lgk 0 (k = 1, 2, ... ), that is, L 0. This, however, means exactly that the linear hull of H' is dense in C[O, 1].
Remark. Of course, the proof works in any separable Banach space.
3. SOLUTIONS TO THE PROBLEMS
232
Problem F.54. Let a > 0 be irrational. (a) Prove that there exist real numbers al, a2i a3, a4 such that the function f : R -> R, f (x) = ex [al + a2 sin x + a3 cos x + a4 cos(ax)]
is positive for all sufficiently large x, and lim inf f (x) = 0.
(b) Is the above statement true if a2 = 0?
Solution. (a) Put f (x) = ex (2 - cos(x - 2-7ra) - cos ax),
where a is to be defined later. Then f (x) > 0, while f (x) = 0 if x = 2k-7r and x - 27ra = 2n7r for some integers k and n. Hence a(n + a) = k. Assuming
that also f (x') = 0 for some x' # x, we obtain a(n' + a) = k' for some integers n' # n and k' 0 k. The relations a(n+a) = k, a(n'+a) = k' yield a = (k - k')/(n - n'), a contradiction. Thus f (x) > 0 for all sufficiently large values of x. Choose a E [0, 1) so that, for some sequence Ink} of natural numbers, nk
- a I< a
e -nka/a
(mod 1).
nk
The existence of an a of this kind can be established as follows. Let K be the circle of perimeter 1, and Po E K. Starting from po, lay 1/a on K (in a given direction) n times to obtain pn. Denote by In the interval r [Pn
a-n7r/a
e-n7r/a
n
Pn+
n
on K. Let no = 1, and suppose that {nk}m o are given. Define nm+1 so that nm+1 > nm and Inm+l c Inm . There is an nm+1 of this kind since {Pn}°O_k is dense in K and e-""r/a/n -> 0 (n -+ oo). We have Inl D ..., and IInI -+0 (n -4 oo). Let a = fl° oInk. Then Ino
f (--) = e2nk7r/a I 2 - cos l
ak7r
2
- 27ra I - cos 2nkir]
-a -a-mk =e 2nk7r/a [1-cos21r nk where Mk is an integer satisfying the relation
--a-mk a nk
e-nk7r/a nk
3.3 THEORY OF FUNCTIONS
233
Therefore, using the inequality cos u > 1 - u2/2, we obtain
f (?:)
< <
ez« 2 [27r (a - a - mk 2r 2 nk
)]2
(k-*oo).
-40
Hence, lim inf.,,+,,. f (x) = 0. (b) We show that in the linear hull of the functions ex, ex cos x, ex cos ax there is a function with the required properties if and only if for any e > 0 the inequality e-7rn/2 m
a-- nI <e
(*)
n
holds for infinitely many rational numbers
m (m, n E N). Let n
g(x) = ex (al + a3 cos x + a4 cos ax)
be such a function. Then, necessarily, al = la3l + la4l, and ai # 0 for i = 1, 3, 4. Let {xk} be a sequence that tends to oo and satisfies g(xk) --+ 0 as k --+ oo. Since al = la3l + la4l, we have Xk = 7rnk + bk,
axk = 7rmk + Ok,
where nk and Mk are natural numbers, nk --+ 00, 'Mk --+ 00, bk - 0 and Ok - 0 as k --> oo. Let,3 E (0,1) and b = min{la3l, la4l}. If k is sufficiently large, then Q > 9(xk) = ennk+6k (al + a3 cos(7rnk + bk) + a4 cos(7rmk + Ok)) = e7rnk+6k (al - Ia3l cos bk - Ia4l cos A = e7rnk+6k
>
(al - la3l + la3l 2k k2 + O(A2) ) 2 + o(bk) - la4l + la4I
e"nk-13 (la316k
+ la4lAk)
e"nk
3e
(bk + Ok) ,
since cos u = 1 - u2/2 + o(u2) for u -* 0. Hence, for sufficiently large k, 141, lokl
3eQ _7rnk/2 be
and, therefore, Mk
7rmk + Ok
Mk
Ok - nk bk
nk
7rnk + bk
nk
(7 +
< IOkI+2albkI < 2a+1
-
(7r - 1)nk
- 7r - 1
) nk
3e,3 e-7rnk/2 b
nk
3. SOLUTIONS TO THE PROBLEMS
234
Since a E (0, 1) is arbitrary, we obtain that for any e > 0 the relation (*) is valid for infinitely many rational numbers M,. Next let E > 0 be fixed, and suppose that (*) has infinitely many rational
solutions mk/nk, Mk E N, nk E N, k = 1, 2, .... Passing to subsequences if necessary, we may assume that all the nk have the same parity, all the Mk have the same parity, and <e
e-7rnk/2 (k = 1,2,...).
nk
Let al = 2, and furthermore, if the members of Ink} are odd, a3 =
-1 if the members of Ink} are even
and a4 =
-1
if the members of {Ink} are odd, if the members of {Ink} are even.
Then g(x) = ex (al + a3 cos x + a4 cos ax) > 0 for x > 0, and g(nk7r) = enk'r [2 + a3 Cos nk7r + a4 Cos ank7r]
= enk" [1 - cos 7r(ank - Mk)]
= e"k" [12(ank <
- mk)2 + o ((ank - mk)2)
enk'r7r2E2e-ank = E27r2
for sufficiently large k. Consequently, lim inf g(x) < E27r2. X-+00
Thus, if for any e > 0 the relation (*) has infinitely many solutions, then lim inf g(x) = 0. x-.+oo
If a is an algebraic number then, by the Thue-Siegel-Roth theorem, for any b > 0 there are only finitely many m/n such that m
1
n < n2+s ' So, if (*) for any E > 0 has infinitely many rational solutions m/n, then a must be transcendent. There is an a of this kind. For instance, let
N > e'/2 be an integer, nl = N, and nk+1 = Nnk, k = 1, 2, ... , and further let a = E', 1/ni. Then with the notation k
Mk aknZ=nk i=1
1
3.3 THEORY OF FUNCTIONS
235
we have
a - ak <
2
nk+l
and, therefore,
a-- < Nnk _ Mk
-0
2nk
Nnk
,, _
e-1rnk/2
(Ne_ir/2)nk
nk
(Ne-7r/2)nk
2
2nk
2
Since
=
it follows for any e > 0 that
a-- <e Mk
nk
k ->oo
e-7rnk/2
nk
nk
for all sufficiently large indices k.
Problem F.55. Prove that if {ak} is a sequence of real numbers such that 00
EIakl/k = oo
2"-1
00
E k(ak - ak+1)2
and
(k1
n=1
k=1
1/2 < oo,
then f7r
00
ak sin(kx) dx = oo. k=1
Solution. Although the condition
lim ak = 0
k-.o0
(1)
does not appear in the statement of the problem, by the well-known CantorLebesgue theorem, this follows from the fact that the series 00
E ak sin kx
(2)
k=1
is convergent almost everywhere. We note that for the application of this theorem it would be sufficient if the series (2) were convergent on a set of positive measure. To make reference easier, we list the remaining conditions: Iakl b k=1
=°°,
(3)
3. SOLUTIONS TO THE PROBLEMS
236
1/2
2n-1
00
E 2 kl1akl2
< 00,
(4)
n=1 (k=2n-1
where
Dak := ak - ak+1
(k = 1, 2, ... ).
From (4) it follows that the sequence {ak} has bounded variation, that is, 00
EIDakI<00.
(5)
k=1
Really, by the Cauchy inequality, 00
00
E IDakI = 00
2'-1 IDakj
L:
n=1k=2'-1
k=1
1/2
2' -1
00
E (2'' k=2n-1 IDakI2 n=1 00
2"-1
n=1
k=2^-1
1
/2
E ' kIDakl2 Consider the nth partial sum of (2). By Abel's rearrangement, we obtain n
n
E ak sin kx = k=1
Dk (x)Aak + an+1 Dn (X),
(6)
k=1
where Dn(x) is the conjugate Dirichlet kernel: s in kx
Dn (x) :=
=
cos 2 - cos (n + 2) x
k=1
2 sin 2
(n = 1, 2,... ).
Introduce the notation
Dn(x) :_ -
cos (n + 1) x 2 2sin ax
(n = 0, 1, ... ).
Then Dn (x) = Dn (x) - Do (x) and from (6) we derive n
n
E ak sin kx = k=1
(n = 0,1, ... ; Do (x) = 0), n
Dk(x)Dak - Do(x) k=1
Dak + an+1Dn(x) - an+1Do(x) k=1
n
E Dk(x)Dak - a1Do(x) + an+1Dn(x) k=1 n
_ E Dk(x)Dak + an+1Dn(x), k=O
3.3 THEORY OF FUNCTIONS
237
where the convention ao := 0 and its consequence Da0 = -a1 are used. It follows that the series (2) is convergent: 00
00
E ak sin kx = E Dk (x) Dak =: f (X) k=1
(7)
k=0
for every x with the possible exception of the case x = 0 (mod 27r). In the following, we make use of an inequality of the Sidon type: for any integer n > 2 and numerical sequence {bk}, it
1/2
2n-1
2n-1
E bkDk(x) dx < C > kbk
ir/n
k=n
(8)
k=n
where C is a positive constant. To see this, we first apply the CauchySchwarz inequality and then exploit the orthogonality of the system {cos (k +
2)
x}:
2n-1
it
rir
2n-1
E bkDk(x) dx=1/n k=n 1/2
dx
T'n
2)2 1/2
2
bkcos l k+ 2 l x k=n
dx
\\\
1/2
2n-1
k=n
cos (k + 2) x dx 2 sin 2
2n-1
J0
Cn- > bk
<
k=n
it
(2 sin
(2n_1
bk
kbk
k=n
Let s > 1 be an integer. According to (7),
f
if
j-1 > Dk (x)Dak dx
2'-1
(x) dx >
a2
j=1
J'r/(7+1)
2'-1
k=0
00
E Dk(x)Dak dx := I1 - I2 . 7=1
3T k=7
Using the inequality
I1 >
it/J
9-1
l(7+1) k=0 = I11 - 112 .
dx
Dak X- -
2'-1
1
9=1
ir/7
7-1
> (k + 1) jakl dx
fit/(j+1) k=0
(9)
3: SOLUTIONS TO THE PROBLEMS
238
Since
1+1)'
ln(1+i J -
i(i
by the foregoing 2' Ill >
7(7a1 + 1)/
(J
It is easy to see that 2°-1
00
00
sl0akl,
1
aj
j=1.7(7+1)
j=1
whence
2°-1
ao
i k=1 lAakl
Ill >
ld l j=1
Similarly, 2"-1j-1 V-1
00
y(7
112 = 7r j=1
k= =0
Therefore
00
l0ak1 < 7rE lLak1 < 27r ILakl .
+ 1)
k=0
2°-1
k=1 00
laj l
I1 1 E
_ (1 + 27r) E l0akl .
j=1
(10)
k=1
We turn to estimating 12: 2'-1
a
I2= E f1=1 j=21-1 2''--1
e
<
1=1 a
+
2'-1
2^-1 \
,r/'
2'-1
, /(7+1)
k=j +n=l}1,
dx
it/j
2'-1
/(j+1)
k=j
Dk(x)Aak
dx
2^-1
7r/j
00
j=21-1 n=1+1
00
L,+1
Dk(x)Dak dx =: I21 + 122 )
k=2^-'
Making use of the elementary inequality
sinx>
-x
( 0<x< 2)
we obtain 8
I21 !5
2'-1
2'-1
7r/j
1=1 j=21-1 >r9 /(+1) 2 sin 2 k=21-1 g
E !=1
2'-1
7r2-1+1 *7r
2-- -
2x
1:
k=z'-1
lAakldx
l0ak l dx 7r 1n2 9
2"-1
1: lAakl k=1
.
3.3 THEORY OF FUNCTIONS
239
We now apply (8): 2-t+1
00
s
E
122 = E
Dk(x)Aak dx
L_1
1=1 n=1+1
0o n -1
1r 2
2n-1 k=2n-1 2n-1
-1+1
Dk(X)Dak dx
n=21=1 2 i
k=2n-1 2" -1
it
00
- n2
k-2n 1
00
2n-1
n=2
k=2"-1
Dk(x)Lak dx 1/2
E kILakI2
Consequently, ir In 2 12 <-
2°-1
2
1/2
2n-1
00
kkakI+C> E klDakI2
(11)
k=2n-1
n=2
k=1
Relying on (9), (10), and (11), we find that 2°-1
00
ir2--
jaki
E ak sin kx k=1
k=1 o0
\1
00
2-1
n=2
k=2n-1
-CE
k=1
1/2
kIDakI2
In view of conditions (3), (4), and (5), this already yields the desired conclusion.
Problem F.56. Let h : [0, oc) --> [0, oo) be a measurable, locally integrable function, and write
H(t) :=
J0
t h(s)ds
(t >_ 0).
Prove that if there is afconstant Bfo with H(t) < Bt2 for all t, then
r
e-H(t)
t eH(u)du
dt = oo.
J0
Solution 1. Interchanging the order of (integration, we obtain oo
e-H(t) J0
t eH(u) du dt =
m
t
J J 0o 0
= L°° Ju f00
H(t)-H(u)] du dt
0
f
e[-H(t)-H(u)] dt du
fexp L- J h(s) dsI
r
u
3. SOLUTIONS TO THE PROBLEMS
240
It is sufficient to prove the existence of a constant c > 0 such that
ft h(s)ds] dtdu>c I(T):=exp[-J f T
u+1/T
1
u
IT,
for large T. We have u+1/T
2T
I(T) >
eXp - u
Tu
IT,
=
u+1/T
j2T
u+1/T
h(s) ds du.
exp
1
T
h(s)ds dtdu
U
Introduce the notation u+1/T
h(s) ds > 10B
u E [T, 2T] : f
Qt :=
u
and find an upper estimate for the Lebesgue measure µ(QT) of the set QT: T
T+1/T
u+1/T
10Bp(T) < IT, f
8
-1/T h(s) duds
h(s) ds du < IT,
u
s
=Th(s)ds
2
whence µ(QT) !5
10
C2T +
T+
T3
I
.
Therefore,
1(T)>
f
1
e-los du >
e-10B
T
T J[T,2T]\QT
> e- 10B
(T - µ(QT))
1--) 1=: C>0'
provided that T > 1.
Solution 2. We first show that if g : [1, oo) -* (0, oo) is measurable and locally integrable, further t
f g(s) ds < B, t2
(B1 = constant)
1
for every t, then
11 ds = oo. 1
g(s)
3.3 THEORY OF FUNCTIONS
241
Indeed, let T > 1 be any real number. By the Bunyakovski-Schwarz inequality
(12T T2 =
(j27' 1 dt
=
'1g (t)
9(t) dt
(j2Tg(t)dt) (12T1) g(t)
<
dt
< B1 4T2 fT
g(t)
dt .
So 2T
T
dt >
g(t)
which yields the assertion. Let g(t)
(T > 0),
4B1 1
eH(t) fot
(t > 1).
eH(u) du
According to the problem, we have to prove that r°°
1
g(t) dt = oo.
J1
To this end, by the above remark, it is sufficient to verify the inequality t
(t > 1)
g(s) ds < Blt2 1
with a suitable B1. This is simple:
ft g(s) = J 1
eH(e)
ft
ds
f0 eH(u) du t
eH(u) du.
= In f eH(") du - In 0
o
Since H is monotone increasing,
f with a suitable B1.
g(s)
In [teH(t)] = lnt + < Bt2 + In t < B1 t2
3. SOLUTIONS TO THE PROBLEMS
242
Problem F.57. Consider the equation f'(x) = f (x + 1). Prove that (a) each solution f : [0, oo) -> (0, oo) has an exponential order of growth, that is, there exist numbers a > 0, b > 0 satisfying If (x) I < a ebx,
x>0;
(b) there are solutions f : [0, oo) -+ (-oo, oo) of nonexponential order of growth.
Solution. In the statement of the problem, a constant c has been omitted. The statement is correctly, f'(x) = cf (x + 1), where 0 < c < 1/e. Then there exists a A > 0 such that A = ce' and so eAx is a positive solution. On the other hand, the equation f'(x) = f (x + 1) has no positive solution f. If it had one, then f would be strictly increasing, and the Lagrange mean value theorem would give f (1) > f (1) - f (0) = f'(i;) = f (e+1) > f (1), a contradiction. The equation f'(x) = cf (x+l) has a positive solution on [0, oo) if and only if c < 1/e (see T. Krisztin, Exponential bound for positive solutions of functional differential equations, unpublished manuscript). (a) Proof of the exponential order of growth. If f : [0, oo) -> (0, oo) sat-
isfies the equation f'(x) = cf (x + 1), put a(x) = f'(x)/ f (x). Then Ux
f (x) = f (0) exp
a(s) ds
,
and
a(x) = cexp
ra+1
(J
a(s) ds) > 0;
a
that is,
Ina(x) =JX+1a(s)ds, x>0. c
Choose k so that k > a(0) and k > ln(k/c). Begin to define the sequence {xn}° o in the following way: xo = 0,
x1 = max{x E (0,1]: a(x) < k}.
Since
/1
x1 is well defined. If x0, ... , xn are given, put xn+1 = max{x E (xn,, xn + 11: a(x) < k} . Since
JXn+1 n
a(s)ds=Ina(x")
c
3.3 THEORY OF FUNCTIONS
243
xn+l is well defined. Since a(x) > k on (xn+1, xn + 1], it follows that Xn+2 > xn + 1. Hence 2n
[0,n] C U[xl-1,x1] l=1
Therefore, if x E In - 1, n), then 2n
x
j
a
a(s) ds < j a(s) ds <
a(s) ds 1=1
2n
< 1=1
f
x,-1
y1_1+1
2n
a(s) as =
x,_
in
C
< 2nk < 2xk + k
< 2n In
a(xl-1)
.
Consequently,
(JX
f (0)e 2k a 2(x > 0).
a(s ) ds
f (x) = f (0) exp
(b) There is a function ¢ E Coo [0, 1], not identically zero, such that 0(n)(0) = 0(n)(1) = 0; n = 0, 1, 2, .... For example, O(x) = exp (1(x(x - 1))) if x E (0,1), and f (0) = f (1) = 0. Then the formula n = 0, 1.... ;
f (x + n) = nC q(n) (x),
x E [0,1] ,
defines a solution of the equation f'(x) = cf (x + 1). Let X E (0, 1) be such that ¢(x) # 0. By Taylor's theorem, there exists an 77 E (0, x) such that
=
0(l) (0) xl
O(x) 1=0
/
+
0(n)
xn =
ml
_(n) x++
ml
Hence
If (n + n) I = C
0(n)
(,) = 10(X) l
(cx)n
which increases faster than any a ebn as n -4 oo. Thus f has a nonexponential order of growth.
3. SOLUTIONS TO THE PROBLEMS
244
3.4 GEOMETRY Problem G.1. Find the minimum possible sum of lengths of edges of a prism all of whose edges are tangent to a unit sphere. Solution. If the edges of a prism are tangent to a sphere, then each face of the prism intersects the sphere in a circle, tangent to the sides of the face. Hence, each face is a circumscribed polygon.
The translation along the generators of the prism that takes one base polygon to the other moves the inscribed circle of the first polygon to the inscribed circle of the second. These circles are congruent and have parallel planes since the bases of the prism are congruent and parallel. Since the translation that moves one of two congruent and parallel circles on the sphere to the other is perpendicular to the plane of the circles, the generators of the prism are perpendicular to the base, thus, the prism in question is a right prism. Sides of a right prism are rectangles. Since they are also circumscribed, and only squares are circumscribed among rectangles, each side of the prism is a square. Consequently, the edges of the bases and the generators have the same length, the bases are regular, and the sum of the lengths of the edges of the prism is three times the perimeter of the base. Let us cut the prism and the sphere by a plane through the center of the sphere, parallel to the base. The prism is cut in a polygon congruent to the base polygons; the sphere is cut in a great circle. The great circle contains the point of contact of every tangents of the sphere orthogonal to the plane of the great circle. Therefore, it contains a point from each generator. We get that the base polygon is inscribed in a unit circle. Since the base polygon has equal sides, it must be a regular polygon. We conclude that the prisms that satisfy the prescribed condition are regular prisms, with the length of the generators equal to the length of the sides of the base polygon scaled in such a way that the circumcircle of the base polygon has radius 1. Conversely, every prism of this type satisfies the condition of the problem, since the center of such a prism is located at distance 1 from each edge. Indeed, the distance to the center from a generator is clearly equal
to 1. On the other hand, the orthogonal projection of the center onto a lateral face, which is a square, is the center of the face. Thus the distances to the center from the sides of this square are equal.
To get the sum of the lengths of the edges of such a prism, we have to multiply by 3 the perimeter of a regular polygon inscribed in a unit circle. It remains to decide which of these regular polygons has the shortest perimeter. The perimeter of a regular n-gon inscribed in a unit circle is equal to 71
sin a
n
a
2n sin - = 2Tr
,
3.4 GEOMETRY
245
where a = it/n. Since sin x is concave on the interval (0, 7r), the slope of the chord connecting the origin to the point (x, sin x) is decreasing as x is increasing. Therefore, sin a/a is minimal when a = 7r/n is maximal, that is, n is minimal. Thus, the minimum is attained by a regular triangle. As for the prisms, the minimum of the sum of the edge lengths is obtained for a regular prism over a triangular base, with lateral faces being squares, and the value of the minimum is 3. 6 sin 60° =9i.
Problem G.2. Show that the perimeter of an arbitrary planar section of a tetrahedron is less than the perimeter of one of the faces of the tetrahedron.
Solution. Obviously, planar sections of a tetrahedron are triangles or quadrangles, since each face of the tetrahedron contains at most one edge of the intersection provided that the intersection does not coincide with the face. The case of quadrangles can be reduced to the case of triangles. Namely, we show that if we translate the plane of a quadrangle intersection in both directions until it passes through the nearest vertex, then one of the translated planes intersects the tetrahedron in a triangle (possibly degenerated to an edge), the perimeter of which is not shorter than the perimeter of the quadrangle. If both translated planes intersect the tetrahedron in a triangle, then introduce the notation shown in Figure G.1 and express the sides of the quadrangle with the help of the sides of the two triangles. Since the planes of the three sections are parallel, the ratio of the segments of a straight line cut off by these planes is the same for all straight lines. Set
.A=BP: AB=BQ: BT =FR: OF=GS: AG. Then for the corresponding sides we have p = .dal,
q = bl + (1 - A)(b2 - bl) = .tbl + (1 - .)b2i
r=c2+)1(cl-c2)=Acl+(1-)%)c2i s=(1-.))a2. From these, p + q + r + s = A(al + bl + cl) + (1 - A) (a2 + b2 + C2), that is,
K = AK, + (1 - A)K2, where K = p + q + r + s, Kl = al + bl + cl, and K2 = a2 + b2 + c2. This implies K < max{K1, K2} .
If one of the triangles AAUT, or LBGF, or both degeneratse to the edges AD or BC, respectively, then this inequality remains valid, putting Kl = 2AD or K2 = 2BC, respectively. Then we get that the perimeter of one of
246
3. SOLUTIONS TO THE PROBLEMS
Figure G.1.
the faces of the tetrahedron is greater than the perimeter of the intersection quadrangle. Hence, it is enough to consider the case of triangle intersections. We may suppose that the intersection triangle has a vertex in common with
the tetrahedron. Otherwise we may translate the plane of the triangle to reach this situation (see Figure G.2). The translated plane meets the tetrahedron in a triangle similar to the original one, with ratio of similarity greater than 1 and it passes through that vertex of the nonintersected face of the tetrahedron that is nearest to the intersecting plane.
If the intersecting plane is parallel to the face not met by it, the assertion is trivial. Otherwise, we show that the perimeter of AKBM is smaller than
3.4 GEOMETRY
247
the greater of the perimeters of OKBC and LKBD. For this purpose, consider the smallest ellipsoid of revolution with foci K and B that contains
both C and D. Since M is an internal point of it, we get KM + MB < max{KC + CB, KD + DB} . But then, either kKBM < kKBD < kABD or kKBM < kKBC, where kPQR denotes the perimeter of OPQR. In the first case, the assertion is proved, the second case can be finished by a repeated application of the previous arguments.
Problem G.3. Show that the center of gravity of a convex region in the plane halves at least three chords of the region.
Solution. Let us denote the disc by T and its barycenter by S. If X is a point of the boundary curve G of T, then denote by Y(X) the second intersection point of the straight line XS and the curve G. Let f (X) = )FS - Y(X)S. We have f (X) = -f (Y(X)) for any X E G. Since f (X) changes continuously as X runs over the are XI (X ), it attains any value
between f (X) and -f (X). This implies the existence of a point X1 E G such that X1S = Y(X1)S. If X1S were the only straight line whose intersection with T is halved by S, then f would be positive along one of the arcs Y(X1). Reflecting this arc through S, the reflected are together with the other arc Xfj-Y(XI) of G would bound a domain T1, the barycenter
of which is different from S, since T1 lies in a half-plane bounded by a straight line passing through S. Therefore, the barycenter of the union T2 = T1 U T should be different from S. On the other hand, T2 is centrally symmetric with respect to S, so its barycenter should be invariant under the reflection in S, which means that its barycenter should be S, which is a contradiction. The same arguments show that if f has finite zeros, then it can not be nonnegative on a half-arc of G. Suppose that there are only two straight
lines X1S and X2S for which the segment of the straight line cut off by the figure is halved by S (X1, X2 E G). By the previous remark, f must be negative (or positive) on the arcs '(X1) and 11'(X2), which is a contradiction, since the sign of f is opposite on arcs opposite to one another. We conclude that at least three chords of the figure are halved by S. The existence of four such chords cannot be proved in general, as it is shown by the case of an arbitrary triangle. Problem G.4. Let A1i A2, ... , A,,, be the vertices of a closed convex n-gon K numbered consecutively. Show that at least n - 3 vertices Ai have the property that the reflection of Ai with respect to the midpoint of Ai_1Ai+1 is contained in K. (Indices are meant mod n.)
Solution. We shall call a vertex P of K reflectible with respect to K if its reflection in the midpoint of the segment connecting the two neighbors of P belongs to K.
3. SOLUTIONS TO THE PROBLEMS
248
1. Let us begin with the first nontrivial case, with the case of quadrangles. For both pairs of opposite sides, the sum of angles lying on one of the two sides is at least -7r; let us denote by A a common vertex of two such sides and denote the other vertices in a cyclic order by B, C, D (see Figure G.3). We show that A is reflectible with respect to the quadrangle. We can construct the reflection A' of A by taking a parallel to AB through D, which intersects side BC in a point E because 7r,
and then taking the parallel to AD through B, which intersects the segment DE at A' because 7r. Therefore, A' belongs to the quadrangle ABCD.
Figure G.3.
Figure G.4.
2. If we show that among any four vertices of the convex polygon K, at least one is reflectible with respect to K, then the assertion of the problem is proved.
We shall say that a vertex P of the polygon precedes the neighboring
vertex R if going around K starting from P in the direction of R is a movement of positive orientation. Consider four arbitrary vertices of K and apply for them the previous notation, that is, let 7r, 7r (see Figure G.4). If the vertex E of K that follows A is not B, then E is an internal point of the intersection S of three halfplanes S1, S2, S3, where S1 is bounded by the straight line AD and contains ABCD, S2 is bounded by BC and contains ABCD, and S3 is bounded by
AB and does not contain ABCD. Similarly, if the vertex F that precedes A is not D, then F is an internal point of the domain T = T1 fl T2 fl T3, where the half-plane T1 is bounded by the straight line AB and contains ABCD, T2 is bounded by DC and contains ABCD, and T3 is bounded
by AD and does not contain ABCD. It is enough to prove that A is reflectible with respect to the quadrangle AECF, since then it is obviously reflectible with respect to K as well. For this purpose, it suffices to show that <EAF + 7r and Tr. It is enough to see
3.4 GEOMETRY
249
the first inequality; the second can be shown in a similar way. But we have
<EAF + QBAC + it - QACD = QBAC + ( 7r, which proves the assertion of the problem. Remarks. 1. The assertion cannot be improved; that is, there exists for all n a convex n-gon having exactly n - 3 reflectible vertices. Indeed, consider a convex quadrangle ABCD with no parallel sides. It is easy to see that such a quadrangle has only one reflectible vertex; denote it by A. Now consider
a convex n-gon K such that B, C, and D are vertices of K while the other vertices of K lie in a neighborhood of radius e of A (obviously, such a convex n-gon exists for any n). It can be easily seen that if a is sufficiently small, then B, C, and D are not reflectible with respect to K. 2. The assertion does not hold for concave polygons. Problem G.S. Is it true that on any surface homeomorphic to an open disc there exist two congruent curves homeomorphic to a circle ?
Solution. The answer is no: There exists a surface homeomorphic to a disc which does not contain two congruent curves homeomorphic to a circle. We show this by giving an example.
We start from a minimal surface, that is, from a surface for which H = 1/2(gl + 92) = 0. (Here H denotes the Minkowski curvature, and gi and 92 are the principal curvatures of the surface.) Although unnecessary, we explain why we are looking for counterexamples among minimal surfaces. If two congruent copies of a surface intersect one another in a closed curve, then by covering one copy with the other, the two positions of the intersection curve give two congruent curves on the surface, which are different in general. If two copies are tangent to one another at an isolated common point, then by a slight movement of one of them, they can be made to intersect one another along a curve. The choice of minimal surfaces is justified by the fact that if H > 0 at a point of a surface, then we can always find a congruent surface so that this point is an isolated common point of them. We can get such a surface by reflecting the original surface in the tangent plane at the point in question and then rotating it 90° about the normal of the surface. This follows from Euler's theorem, since by the inequality gl cos2 (p + 92 sin2 (p > -g2 cos2 (p - gl sin2 (p ,
the curvatures of the normal sections of the original surface are always greater than the curvatures of the corresponding normal sections of the
250
3. SOLUTIONS TO THE PROBLEMS
transformed surface, and therefore the first surface is above the second one in a small neighborhood (looking at them from the direction of the normal vector).
Now we show that in a sufficiently small domain of a minimal surface, any two congruent closed curves bound congruent domains of the sur-
For this purpose, take two copies of the surface and move one of them so that the congruent curves cover one another. Suppose that the piece of the minimal surface is small enough to ensure that both surfaces can be obtained as the graph of the single-valued functions z = f 1(x, y) and z = f2 (x, y). The areas of the domains bounded by the closed curve are obtained as integrals of the lengths of the vectors ml(-1,p1,Q1) and
face.
m2 (-1, p2) q2), respectively, over the same domain. (Here pi and qi denote the partial derivatives of fi.)
We use the fact that the variation of the surface area of a minimal surface is 0, that is, if a minimal surface is embedded into a one-parameter
family of surfaces all having the same boundary, then the derivative of the area of these surfaces with respect to the parameter is zero at the minimal surface. Consider the one-parameter family of surfaces defined by z = Af, 1 + (1- A) f2. The area of a member of this family is obtained as the integral of the magnitude of the vector amt + (1 - A)m2. This magnitude is a strictly convex function of the parameter A if m1 54 m2 and does not depend on A if m1 = m2. Consequently, the area, obtained by integrating convex functions of A, is itself a convex function of A. Its derivatives at A = 0 and 1 can vanish only if it is a constant function, and this happens
only if m1 = m2 at any point of the domain, that is, if the two pieces of surfaces coincide.
According to this observation, it is enough to find a minimal surface that does not contain different congruent pieces. Since two irreducible algebraic surfaces that have a domain in common coincide, it is enough to present an algebraic minimal surface that has no or only a finite number of automorphisms different from the identity. (An automorphism of a surface is a bijective mapping of the surface onto itself that can be extended to a congruence of the space.) If the surface has finite automorphisms, then a small neighborhood of a point different from its automorphic images has no automorphisms different from the identity. Actually, any algebraic minimal surface would do, provided it is not a surface of revolution. However, we need to find only one of them. A simple computation shows that for the surface
x = u3 - 3uv2 + 3u,
y=v3-3u2v+3v, z = 6uv,
H = 0 and K = -2(U2 (u2 + v2 + 1)2 (K is the Gauss curvature). The latter attains its minimum only at it = v = 0; thus this point and the Dupin indicatrix at this point must be fixed by any automorphism of the surface.
3.4 GEOMETRY
251
There may be only a finite number of such automorphisms, however, since the Dupin indicatrix is a hyperbola by K < 0.
Problem G.6. The plane is divided into domains by n straight lines in general position, where n > 3. Determine the maximum and minimum possible number of angular domains among them. (We say that n lines are in general position if no two are parallel and no three are concurrent.)
Solution. The minimal number of angular domains is three. Indeed, the convex hull of the intersection points of the straight lines is a convex polygon having at least three vertices. Each vertex is an intersection point of two straight lines and those half-lines of these lines, which go outside the convex hull, bound an angular domain of the considered subdivision, since there are no intersection points on them.
On the other hand, one can always position n _> 3 straight lines in the plane so that the number of angular domains is exactly 3. Such a construction is given by n tangents to a quadrant (see Figure G.5).
Figure G.5.
The maximal number of angular domains is (-1)n+l
n+
-1
2
that is, n if n is odd and n - 1 if n is even. Indeed, each straight line is divided by the others into two half-lines and some segments. Angular domains can be bounded only by these half-lines, and each half-line bounds at most one angular domain (otherwise there would be a point that lies on three of the straight lines). It follows that 2n half-lines can bound no more than n angular domains.
For odd n, we can present an (essentially unique) construction with n straight lines and n angular domains: Consider the longest diagonals of a regular n-gon. These n straight lines are in general position. There is an
252
3. SOLUTIONS TO THE PROBLEMS
Figure G.6.
angular domain at each vertex of the polygon. Thus the number of angular domains is n (see Figure G.6).
We prove that if n straight lines in general position bound n angular domains, then n is odd. Removing the two half-lines considered above from each straight line gives n segments. Any endpoint of such a segment is shared by exactly two segments. If two of the segments have different endpoints, then they cross one another since the intersection point of their straight lines can lie neither outside nor at the end of the segments. Thus, the union of the segments yields a (number of) self-intersecting closed broken line(s). Fix an orientation on each of these broken lines, and consider three consecutive segments a, b, and c on one of them. Since a and c intersect, they lie on the same side of b. Omitting the endpoints of a and b
(that is, three points), we can couple the remaining vertices, saying that the points P, Q form a pair if P is on the same side of b as the segments a and c, and Q is the neighboring vertex that comes after P according to the fixed orientation on the broken line containing P. Hence, the number, n, of vertices is odd. In the case of even n, we can attain the maximal (n-1) angular domains. This follows directly from Figure G.6; if we remove one straight line, the number of angular domains decreases by 2.
Problem G.7. Let A = A1A2A3A4 be a tetrahedron, and suppose that for each j $ k, [Aj, Ask] is a segment of length p extending from A3 in the direction of Ak. Let pj be the intersection line of the planes and [AkA1A.,,,,].
Show that there are infinitely many straight lines that
intersect the straight lines P1, P2, P3, P4 simultaneously.
Solution. It is natural to assume that we are in the projective space obtained from the Euclidean space by joining ideal elements.
3.4 GEOMETRY
253
Exclude first the singular cases, and assume the tetrahedron has no edges of length p. Let {j, k,1, m} = {1, 2, 3, 4}. Denote the intersection point of the straight lines AkAl and AikA;l by Bj,,,,,. Obviously, Bum, is the intersection point of the plane S,,,, _ [A;AkAI] opposite to A,, and the straight line pj. Consider now the plane S3 (see Figure G.7). The points Bkj, B1 5 B.j are adjacent to a straight line ej, as is easy to see by a multiple application of Menelaos' theorem. This implies that ej intersects P1, P2, P3, P4 simultaneously. (Indeed, the points Bki, Bid, Bmj belong to pk, pi, p,,,,,, respectively, while pj and ej are coplanar.) The line ej is not an edge of the tetrahedron, since it would result, the excluded case. The lines el, e2, e3, e4 are all different since they lie in different planes of the tetrahedron, but none of them coincides with an edge.
Figure G.7. Therefore, the lines pi, P2, P3, P4 are intersected by four different straight lines simultaneously. If two of the lines pi (i = 1, 2, 3, 4) are intersecting, then their point in common must be contained in the plane spanned by the two other straight lines; hence we can easily find infinitely many straight lines intersecting all four of them. If P1, P2, p3, p4 are mutually skew, then take P1, P2, P3. It is known that the straight lines that intersect all of these three lines sweep out a doubly ruled second-order surface. One family of straight lines on this surface is given by the straight lines intersecting pl, P2, P3, while pl, P2, P3 belong to the other family. The line p4 has more than two points in common with the surface, hence it is also a generator of it and intersects every straight line in the first family. Now let us discuss the singular cases. (a) If the three edges starting from one of the vertices Ak have length
p, then Pk is not properly defined, while the three other lines are adjacent to Ak. As Pk is not determined, the statement cannot be applied to this case. If, however, we define Pk as an arbitrary straight
3. SOLUTIONS TO THE PROBLEMS
254
line in the two coinciding planes that should define it, the statement is clearly true.
(b) If there are two edges of length p starting from a vertex Ak, then denote by A, the fourth vertex, not lying on these edges. It is easy to see that in this case pk is the intersection of the planes Sk and S1, pi c S1, while the two other lines, pj, p,,,,, go through Ak. This means that any straight line e such that Ak E e C S1 intersects all the lines A. (c) Assume now that the tetrahedron has exactly one edge of length p. In this case, one of the straight lines ei coincides with this edge, and therefore we can only say that there are three different straight lines
among el, e2, e3, e4. However, to apply the arguments used in the generic case, it suffices to have three straight lines that intersect all of P1, P2, P3, P4. Thus, P1, P2, P3, p4 are intersected by infinitely many straight lines simultaneously.
Figure G.8.
(d) Assume, finally, that the tetrahedron has two edges of length p, opposite to one another, say A1A3 = A2A4 = p = a. Introduce the notation A1A2 = b, A2A3 = c, A3A4 = d, A1A4 = f (see Figure G.8). Applying Menelaos' theorem, we obtain the following divided ratios: (A1A3B24) _ (A2A4B31) _
b-a a-c'
a-d '
(A2A4B13)
a-frn
c -a
a-d'
a
(A1A3B42)=
b
-a
Turning to cross ratios,
(AlA3B24B42) _
(b
- a)la - d) = (A2A4B13B31),
(a - c)(f -a)
3.4 GEOMETRY
255
but then (A1A3B24B42) = (B31B13A4A2)
Let us take into consideration that Al and B31 are on p3i A3 and B13 are on pi; B24 and A4 are on p2i B42 and A2 are on p4. The equality we have just obtained means that p1, P2, P3, p4 cut the skew lines A1A3 and A2A4 in two 4-tuples having the same cross ratio. Thus these lines belong to the same family of generators on a doubly ruled second-order surface. This implies the statement of the problem.
Problem G.8. Consider the radii of normal curvature of a surface at one of its points P0 in two conjugate directions (with respect to the Dupin
indicatrix). Show that their sum does not depend on the choice of the conjugate directions. (We exclude the choice of asymptotic directions in the case of a hyperbolic point.)
Solution. If P0 is a parabolic point, then the indicatrix is a couple of parallel lines, and a direction not parallel to them is conjugate only to their (that is, the asymptotic) direction. The radius of normal curvature is infinite in the asymptotic direction and finite in any other direction, so the sum of the radii of normal curvature in two conjugate directions is always infinite.
It is known that the absolute value of the radius of normal curvature in a given direction is the square of the length of the segment from Po to the indicatrix point in the given direction. According to this, we need only to show that the sum (for the elliptic case) or the difference (for the hyperbolic case) of the square of the length of conjugate half-diameters of the Dupin indicatrix does not depend on the choice of the conjugate diameters.
A theorem of Apollonius states that the indicatrix is an ellipse at an elliptic point.
Now suppose that P0 is a hyperbolic point, that is, the indicatrix is a pair of conjugate hyperbolas whose equation is of the form X2
a2
y2
- b2 = ±1
in a properly chosen coordinate system. Let us denote by Si and s2 the common asymptotes of them (see Figure G.9). Let S be a point on one of the hyperbolas and e1 be the tangent of the hyperbola at S. This tangent intersects s1 at the point L, s2 at the point M. Draw a tangent (different from s2) from M to the other hyperbola. This tangent touches the hyperbola at the point T and crosses s1 at N. It is known that the area of the triangle enclosed by the asymptotes and a tangent of the hyperbolas is equal to ab. Thus, LPOLM and LPOMN are of the same area, which, due to their common altitude MM*, yields POL = PON.
It is also known that S halves LM and T halves MN. Hence,
L POST is the median triangle of ILMN; consequently, POS 11 e2 and
3. SOLUTIONS TO THE PROBLEMS
256
Figure G.9. POT II el.
This means, however, that POS and POT are conjugate half-
diameters. Applying the cosine law for the triangles OPOLM and OPOMN, and using the equality P0L = PON shown above, we get
4 (PoT2 - P-S2) =LM2 - NM2 =LPo2 + PoM2 - 2LPOPoM cos a[NPo2 + PO M2 - 2NPOPOM cos(7r - a)]
=-4LPOPOMcosa=-16
22
LPo PoM
cosa,
where LPo/2 and POM/2 are the contravariant coordinates of the vector P with respect to a basis consisting of unit vectors pointing in the directions of the asymptotes. The product of them (as it is known) does not depend on the choice of the point S of the hyperbola. In such a way, 2 ( 2 - AO-S) does not depend on S, that is, on the choice of the conjugate pair of diameters.
Problem G.9. Show that a segment of length h can go through or be tangent to at most 2[h/../) + 2 nonoverlapping unit spheres. ([.] is integer part.)
Solution. Let us denote by e the supporting straight line of the given segment of length h, project the centers of the spheres onto this straight line, and consider the open intervals of length on e centered at the projections of the centers. We claim that no point of e belongs to more than two such intervals. If not, there would exist a cylinder of radius 1 and altitude m < f containing three points A, B, and C (the centers of the spheres) that are at least 2 units away from one another. We shall show that this is impossible.
r
3.4 GEOMETRY
257
0
Figure G.10.
In the following, we shall use the term "the plane of a point" for the plane that is orthogonal to e and passes through the point. The plane of A intersects the cylinder in a circle centered at K. Let us project B and C onto the plane of A. Denoting the projections by B' and C', respectively (see Figure G.10), we have
m2 +B'A2 > 22, that is,
B'A2>22-m2>2, hence, LB'KA is obtuse angled at K, since the radius of the circle equals 1. It follows that if we move A to the perimeter of the circle along the radius KA, AB' and hence AB increase (and similarly, so do AC' and
hence AC). The same is true for the other points, so we may suppose that A, B, and C are on the surface of the cylinder, at distance 1 from e. for the height of the cylinder, we may also Since we required only m < suppose that one of the points, say A, lies in the upper base of the cylinder and one of the points, say C, lies in the lower base.
Figure G.11.
258
3. SOLUTIONS TO THE PROBLEMS
Consider the plane of B, and let B1 be the reflection of B in e, A' and C' be the projections of A and C onto the plane of B, respectively (see Figure G.11). Since the distances between the planes of A, B and C are smaller
than f , while the distances between the points are not less than 2, the distances between the projections of these points are greater than 0. This implies that the endpoints A' and C' lie on different arcs determined by B and B1. If we move both A' and C' toward B along the arc, and move A and C in a corresponding way, we can manage to have AFB- = CB = 2. But in this case A'B1 is just the distance of the planes of A and B, and C'B1 is just the distance of the planes of C and B. Therefore
A'B1+C'B1<m
A'C' < A'B1 + CBl < v f2. Thus,
AC2=m2+A'C'2<2+2,
and so AC < 2.
This is a contradiction that proves the propo-
sition. Let us elongate the segment of length h in both directions by a segment
of length /. Since the segment goes through or is tangent to the spheres, the intervals assigned to the spheres are all contained in this enlarged segment. Since every point of the enlarged segment is covered by at most two intervals, we get that the number of spheres is at most
h+2 2
2
1
vf2-
1
2[VJ+2.
To see this, we put the intervals in an increasing order by their left endpoints (this ordering is not always unique). Then the intervals standing at odd (or even) positions are necessarily disjoint. Thus, the number of all intervals is at most
[h]+,+[h]+,.
It is easy to see that the given estimation is exact.
Problem G.10. Characterize those configurations of n coplanar straight lines for which the sum of angles between all pairs of lines is maximum.
Solution. We may suppose that all straight lines go through one point. Starting from an arbitrary straight line, let us number the straight lines in counterclockwise orientation (the order of coinciding straight lines is
3.4 GEOMETRY
259
arbitrary): el, ... , en. For 1 < i < [n/2], we say that el is the ith neighbor of ek if
l - k + i (mod n) . We shall denote the ith neighbor of ek by ek. Let us denote by (e, f) the angle between the straight lines e and f and denote by < e, f > the angle of the counterclockwise rotation that takes e to f. For i fixed,
=i(<e1,e2>+<e2ie3>+...+<en,e1>)=i7r, and equality holds if and only if < ek, ek >< 2
(1)
for all k. Thus, for odd n, the sum of angles between all pairs of straight lines can be estimated as follows:
a
n-1
a i7r = n2-17r n-1
[(ei, ei) + ....+ (en, e')] <
4
(2)
2'
and equality holds if and only if (1) holds for all k = 1, 2, ... , n and i = 1, 2, ... , (n - 1)/2. For this, it is enough to require that ek,ekn-i)/2
<
>< 2
(3)
hold for k = 1, 2, ... , n. For even n, the sum of angles is 2-1
[(ei, ei) + ... + (en, a ,)] + z=i
2
n7r
[(el, ei/2) + ....+ (en, e; /2)J
n27r
42
and equality holds if and only if n/2
<
><
2
holds f o r k = 1, 2, ... , n. For even n, these inequalities imply the requested
characterization, since the inequalities < ek, ek/2 ><
2
and
< ek+n/2,
can be satisfied simultaneously only if n/2 < ek, ek
7r
>= 2 ,
n/2
Ir
>< 2
260
3. SOLUTIONS TO THE PROBLEMS
that is the (n/2)th neighbors are perpendicular to one another. To summarize, if n is even, the maximum is attained if the family of straight lines is composed of n/2 orthogonal pairs and any family of orthogonal pairs of straight lines yields maximal sum of angles. Returning to the case of odd n, let us observe that orthogonal couples of straight lines can be removed from the family, since the omission of an orthogonal pair there decreases the sum of angles by (n - 1) 2 and n2
(n - 2)2
4
4
=n-1.
Suppose that after the removal of orthogonal pairs there remain 2m + 1 straight lines. These straight lines must be different, because if e were a double line then the mth neighbor of the mth neighbor f of e would be e, which would mean that e and f are perpendicular. If e is an arbitrary member of the 2m + 1 straight lines, then conditions (3) imply that each quadrant of the plane determined by e and the straight line perpendicular to e must meet exactly m straight lines. Adding to such a family of an odd number of straight lines an arbitrary number of orthogonal pairs, we get all the configurations that yield the maximum, and the maximum is equal to [(n2 - 1)/4] (a/2) by (2). Finally, if we do not suppose that the straight lines go through one point, then taking a configuration characterized above and replacing each straight line by a parallel one, we can get all the requested configurations.
Problem G.11. Let f (n) denote the maximum possible number of right triangles determined by n coplanar points. Show that lim
f n) = oc and
n-.oo n2
n
lim f () = 0. n-+oo n3
Solution. Lower estimation. We shall show the following sharper result. There exists a constant c > 0 such that f (n) > cn2 log n
(n > 3).
Suppose first that n = (3k + 1)2, where k is a natural number. Consider those points of the lattice of points with integer coordinates, which lie in the square spanned by the vertices (0, 0), (0, 3k), (3k, 3k), (3k, 0). The number of these points is obviously (3k + 1)2. We shall count only those right triangles for which the right-angled vertex (q, r) satisfies k < q < 2k, k < r < 2k and the two other vertices of which lie in the square of sides 2k centered at (q, r). Obviously, all these right triangles are good for us. We have to determine the number of right triangles spanned by the
lattice points of the square (-k, -k), (-k, k), (k, k), (k, -k) so that the right-angled vertex lies in the origin (0, 0).
3.4 GEOMETRY
261
The perpendicular sides of such a right-angled triangle lie on straight lines given by equations of the form ix = jy and jx = -iy, where (i, j) = 1, -k < i, j < k. Obviously, we estimate the number of right triangles from below if we take into consideration only those straight lines that satisfy
0 < i < j as well. In this case, each of the straight lines ix = jy and jx = -iy contains 2 [k/j] lattice points different from (0, 0), so the number of right-angled triangles determined by them is 2
4
[k]
>.T 2
k2
j
For a fixed j, we find at least
right triangles (cp denotes Euler's phi function), which gives k2
E 0(j)
j=1
72
right-angled triangles altogether. Now returning to the original problem, the (3k + 1)2 points we constructed yield at least k
(2k + 1)2k2
L.
P j2
3=1
right triangles. Introducing the function
fi(x) = 1:W(7) j<x
and summing up partially, we get
t
f ((3k + 1)2) > (2k + 1)2k2 j=1
(2j
1)
+2 j (2) 2U + 1)2
Combining this with the well-known inequality 'D(x) > clx2,
we get
(1)
k-1 > c3k4log k.
f((3k + 1)2) > c2(2k + 1)2k2 j=1 j
3. SOLUTIONS TO THE PROBLEMS
262
Now let n be an arbitrary number. Set k = [(f - 1)/3] > V/n-/4. Then
f(n) > f((3k+1)2) > c3k4logk> cn2logn. Upper estimation. We shall prove a sharper result again: using the "descente infinie" method we show that
f(n) < n2/.
(2)
The statement is trivial for n = 1, 2, 3, 4, 5 (for these values n2 Vjn- > (3) ). If (2) were not true, then there would be a smallest natural number n such
that f(n) > n2V/n.
Let us take n points P1i ... , PP, on the plane that span f (n) right-angled triangles. We claim that there must be a straight line that contains from the n points at least 2/ ones. Consider all possible ordered couples (Pi, Pj), and assign to each of them the number of right triangles PPP3Pk such that the right angle is at Pi. This number is the number of points different from Pi on the straight line perpendicular to PiPj at Pi. The sum of these numbers is twice the number of right triangles and so it is at least 2n2 Vrn-. Since the number of summands is n(n - 1)(< n2), one of the summands is at least 2 f, and thus one of the perpendicular lines contains at least 2Vrn_ points.
Let us remove the points of this straight line and examine how many right triangles are destroyed by this at most. We divide the triangles destroyed into two groups: (A) triangles for which the right-angled vertex is omitted; (B) triangles with right-angled vertex lying off and one of the other vertices lying on the critical straight line. We can give an upper bound for the number of triangles of type A in such a way that we have n(n - 1)/2 choices for the hypotenuse of such a triangle, and a segment PPP may serve as hypotenuse for at most two right triangles of type A (Thales' circle!), so the number of triangles of type A is at most n(n - 1) < n2. In the case of triangles of type B, we have n(n - 1)/2 choices for those vertices that are not required to lie on the critical straight line, and we can choose at most two points on the critical straight line which form a triangle of type B together with a fixed couple of points P2P3 so that the right-angled vertex is either at Pi or at Pj. It follows that the number of triangles of type B is also at most n(n - 1) < n2. In such a way, the removal of the points of the critical straight line destroys at most 2n2 right-angled triangles; consequently, the remaining,
at most, n - 2 points span at least n2
- 2n2 right triangles. By the
induction hypothesis, n2Vn-
- 2n2 < (n - 2 V/n) 2
n-2 n<(n-2/ )2/i,
3.4 GEOMETRY
263
from which
2n2 < 4nj, that is,
n<4,
for which cases we have already seen that the assertion is true. Remarks. 1. We could have avoided the use of (1) using only the elementary fact that cp(p) = p - 1 for prime numbers. However, this would have given us only the estimation f (n) > c4 n2 log log n,
where c4 is a, suitable positive constant. 2. Laszlo Lovasz remarks that there is a positive constant c5 such that the number of right triangles in the lattice construction described in the first part of the solution is at most c5 n2log n.
3. Bela Bollobas proves the statement of the problem in the following more
general form: Let us denote by f (n, k) the maximal number of rightangled triangles spanned by n points of the k(> 2)-dimensional space, where the maximum is taken for all possible configurations of the points. Then dkn3_2'-k , cn2 log n < f (n, k) <
where c and dk are positive constants. The lower bound comes from our previous arguments because of f(n,k) _> f(n,2). We prove the upper bound for k = 3. Denote by R a collection of n points in the three-dimensional space, and denote by p the number of those straight lines that contain more than 3/ points of R. Let el, e2, ... , e, stand for these straight lines. el contains at least 3V,`n points of R, el and e2 together contain at least 3/i(3 / - 1) points, because two different straight lines may have at most one common point, and so on, el, ... , en' cover at least p'-1
E(3/- 1)>p3 ' -P2 i=1
points of R for every p' < p. Consequently, p < V/n. We prove that if "big" straight lines, the number of right triangles decreases at most by 4n512. Let e be a "big" straight line. Those right we remove at most
triangles that have at least one vertex on e can be divided into two classes: i.
Triangles that have two vertices on e. The third vertex of such a triangle can be chosen out of at most n points. This point, together with a point on e, leaves only two possibilities, and so the number of right triangles of this class is at most 2n2.
264
ii.
3. SOLUTIONS TO THE PROBLEMS
Triangles that have only one vertex on e. We have n(n - 1)/2 < n2/2 choices for the two other vertices, which form a right-angled triangle together with at most 4 points on e, thus the number of right triangles of this type is less than 4n2/2 = 2n2. Since we have at most V/n "big" straight lines, their removal causes the vanishing of not more than v 4n2 = 4n5/2 right triangles. After the removal, each straight
line contains at most 3V/'n points. Let us denote by q the number of planes passing through the point P and least 6n3/4 points of the remaining family. By the idea used above, we get q'6n3/4
- gj23v < n
for all q' < q, since two planes have at most 3 f common points. From here, q < n1/4. There are less than n1/43v/n triangles satisfying
that its right-angled vertex is at P, one of its perpendicular sides lies in one of these "big" planes, while the other is perpendicular to the plane. The number of right triangles with right-angled vertex at P, not satisfying these conditions, can be estimated by the sum i-1 fi s:, where l denotes the number of orthogonal pairs of a plane and a straight line passing through P such that both the plane and the straight line contains at least one point different from P and fi, si denote the number of points different from P lying on the plane and the straight line of the ith pair, respectively. But then fi < 6n7/4 . fisi < 6n3/4 Thus, the number of right triangles in R is less than 4n512 + 6n714n < 10n1114
We remark that the case k > 3 requires the consideration of not only straight lines and 1 (1 < k)-dimensional linear subspaces but also that of circles and spheres.
Problem G.12. Suppose that a bounded subset S of the plane is a union of congruent, homothetic, closed triangles. Show that the boundary of S can be covered by a finite number of rectifiable arcs.
Solution 1. Let us denote the boundary of S by H and let P E H be an arbitrary point. Then one can find a sequence of points {P,} converging to P such that each point P,,, lies on the boundary of a triangle A,,, constituting S.
We may suppose that each point P,, lies on the boundary of A at the same position (that is, the translation that takes Auto A,,,, takes P,,, to P,,,.). (This can be shown by our assumptions. Translate the triangles A,,,
to a fixed triangle A; denote by Q the point corresponding to P. The sequence {Q} is bounded, and it thus contains a convergent subsequence
3.4 GEOMETRY
265
{Qnk}, which tends to Q. Translating A back to Ank, Q corresponds to Pnk . Obviously, Pnk -> P and the points P"'k lie at the same position on the boundary of Ank ). Obviously, each triangle An contains a homothetic triangle A of half size such that Pn is a vertex of A lying at the same position for all n. These triangles converge to a homothetic triangle Ap with vertex P such that the interior of A contains no point of H. Let us number the three vertex positions of the triangles homothetic to An. This gives rise to a natural ordering of the vertices of triangles homothetic to An. Let us define the sets Hi C H (i = 1, 2,3) in such a way that a point P E H belongs to Hi if and only if P is the ith vertex of A ,. Obviously, H = H1 U H2 U H3. Let us study H1. Denote by a the angle lying at the first vertex of OP. Let us introduce a coordinate system on the plane so that the y-axis shows in the direction of the bisectrix of the angle a, the direction of the x-axis coincides with the direction of the supplementary angle of a. Let us divide the plane into bands of equal width parallel to the x-axis. If the width of the bands is chosen to be small enough (for example, if it is smaller than
the bisectrix of Ap that passes through P), then the intersection of H1 and one of the closed bands is the graph of a function y = f (x), where f is defined on a bounded subset of the x-axis and satisfies Lipschitz condition
with constant k = cot (a/2), which gives the possibility to extend it to the closure of its original domain. The extension of f is also k-Lipschitz. After this, by extending f to the components of the open complement of its domain in a linear way, we can obtain a k-Lipschitz extension of f, defined on a sufficiently large bounded interval. The graph of this extension is a rectifiable curve that covers the intersection of H1 and the considered band. Since S is bounded, it crosses only a finite number of bands; thus, Hl can be covered by a finite number of rectifiable arcs. We can proceed similarly for H2 and H3, so the theorem is proved. Solution 2. Let 1[8` be the two-dimensional Euclidean vector space, and let H C 1E82 be a nonsingular closed triangle whose angles are the regions A1, A2, A3, that is,
H=A1nA2nA3.
(1)
The statement of the problem can be reformulated as follows.
Statement. If M (C R2) is bounded, than the boundary of M + H can be covered by a finite number of rectifiable arcs. (From now on, we use operations on complexes.)
Since M is bounded, there exists a finite set V C 1[82 such that M C V + 12 H. Setting M(v) = (v + 12 H) n H, we get
M=
M(V);
(2)
VU
hence,
M+H=
vE
(M(v) + H).
(3)
3. SOLUTIONS TO THE PROBLEMS
266
First we show that M(v) + H = xf11(M(v) + Ai)
(4)
for all v c V. It is clear from (1) that M(v) + H C iA1(M(v) + Ai). Suppose now that
p E inl(M(v) + Ai). Then p E M(v) + Ai (i = 1, 2, 3), but
(p - M(v)) n (Ai \ H) C (p-v- 1H)n(Ai\H) =0 must be satisfied at least for one i. Thus
(p-M(v))nH
0;
and this means that
p E M(v) + H. Hence,
M(v) + H D inl(M(v) + Ai) holds as well, and this proves (3). Let us denote by fr (X) the boundary of a set X. Using that V is finite, (3) and (4) yield
fr (M + H) C U U fr (M(v) + Ai) vEV i=1
Now it is enough to show the following proposition.
Proposition. Let A be the angular domain of angle 2a (0 < 2a < 7r), the vertex of which is the origin and the bisectrix of which is the negative y-axis. Let M C R2 be a bounded set; furthermore,
T={(x,y) ER2:0<x<1}. Then
G=fr(M+A)nT is a rectifiable curve.
Proof. To show this, let (t, g(t)) denote for 0 < t < 1 the point of G the whose abscissa is t, that is, set
g(t) = sup{y - x - tj cot a: (x, y) E M}. Since g satisfies the Lipschitz condition
Ig(t) -g(t')I < It - t'Icota, the latter proposition is trivial.
3.4 GEOMETRY
267
Problem G.13. Let F be a surface of nonzero curvature that can be represented around one of its points P by a power series and is symmetric around the normal planes parallel to the principal directions at P. Show that the derivative with respect to the arc length of the curvature of an arbitrary normal section at P vanishes at P. Is it possible to replace the above symmetry condition by a weaker one?
Solution 1. Since the normal planes parallel to the principal directions are perpendicular, the symmetry of F around them implies the symmetry around their intersection, that is, around the normal of F at P. Since the plane of normal sections contains the normal line, normal sections at P are also symmetric around the normal. The analyticity of the surface ensures that the normal sections have arc length and curvature at each point, which is a differentiable function of the are length. Let g(s) be the curvature expressed as a function of the arc length. If P corresponds to the parameter so, then the symmetry of the normal section around the normal at P gives the identity g(so + s) = g(so - s). Differentiating with respect to s at s = 0, we get g'(so) = -g'(so), that is, g'(so) = 0, which proves the proposition. As we see, symmetry around the normal planes parallel to the principal
directions can be substituted by the symmetry around the normal of F at P.
Solution 2. By our assumption, introducing a coordinate system with origin at P, the surface is given by an equation of the form i k
z
aikx y
,
i,k=0
where aoo = 0. We may also suppose that the x- and y-axes show in the principal directions of F. In this case, alo = a01 = 0. With such a choice of coordinate system, according to our assumptions, z(x, y) = z(-x, y) =
z(-x, -y) = z(x, -y), which can be the case only if aik = 0 whenever either i or k is odd. Then the equation of the surface has the form 00
z=
a2i,2kx2iy2k
i,k=0
A normal section at P can be parameterized as follows:
r=r(t): x=clt,y=c2t, cc
00
z=
a2i,2k(Clt)2i i, k=0
(C2t)2k ==
bnt2n
n=1
where the constants c1, c2 (ci +c2 # 0) depend on the plane of the normal section, and the coefficients bn are functions of a2i,2k.
3. SOLUTIONS TO THE PROBLEMS
268
The curvature of the curve r = r(t) can be computed by the formula (rl r")2
9(t) =
Ir/l3/2
By our assumption, g(O) # 0, and instead of needing to prove the equation dg/ds = 0, it is enough to show the relation g'(0) = 0, since
dg 1
dg dt dt ds
dg
ds
dt Ir'I
Taking into consideration that
r ' = (Cl, r" =
0" n=1
bn
t2n-1
2n(2n - 1)
0, 0,
,
bnt2n-2
n=1
r /l/ =
2n(2n - 1) (2n - 2)
0, 0,
bnt2n-3
n=1
the relation g'(0) = 0 is obtained by direct computation. The above solution shows that in order to weaken the symmetry con-
dition, it is enough to require that the equation z = E°k=o aikxiyk of the surface in the coordinate system attached to the normal and principal directions does not contain terms of order three.
Problem G.14. Let a(Sn, k) denote the sum of the kth powers of the lengths of the sides of the convex n-gon Sn inscribed in a unit circle. Show that for any natural number greater than 2 there exists a real number ko between 1 and 2 such that a(Sn, ko) attains its maximum for the regular n-gon.
Solution. We shall show the following sharper result. If 1 < k < (tan(7r/n))/(ir/n), then a (Sn, k) < a(SS,, k) for any convex n-gon inscribed in a unit circle, where Sam, is the regular n-gon inscribed in a unit circle. P r o o f . T o show this statement, it is enough to consider only those poly-
gons Sn that contain the center of the circle on the boundary or inside. Indeed, let S n be a polygon not satisfying this condition, and let A1, ... , An be the vertices of Sn. Suppose that A1A2 is the closest side of Sn to the center of the circle. Denote by A'2 the antipodal pair of Al, and consider the polygon S'n with vertices Al, A'2, A3, ... , An. Then AlA2 > A1A2, and A2A3 > A2A3 (since the angle opposite to the side A3A2 in DA2A2A3 is obtuse). Fork > 1, obviously U(SS,, k) > Q(Sn, k).
3.4 GEOMETRY
269
Now let Sn be a convex n-gon inscribed in a unit circle, such that S,,, contains the center of the circle on the boundary or inside. Let x1 ,. .. , xn denote half of the central angles corresponding to the sides of Sn. Then n
u(Sn, k) = 2k > sink xi, where 0 < xi < 2 ,
xi = 7r. i=1
i=1
Our proposition is well known for k = 1, so it suffices to deal with the case k > 1. Let us fix the exponent 1 < k < (tan(7r/n)/(ir/n). Since the function (tan x)/x is continuous and strictly increasing in the interval (0, it/2, furthermore, lima-o(tan x)/x = 1, there exists a unique real number a = a(k)
such that 0 < a < ir/n and k = (tan a)/a. Consider the function f (x) = sink x. Since f"(x) = k(k - 1) sink-2 x cos2 x - k sink x = k sink-2 x(k cos2 x - 1) , one can easily check that f (x) is convex on the interval and concave on the interval (arc cos (1//), 7r/2). There(0, arc cos fore, it is concave on the interval (a, -7r/2) because
f" (a) = Set
x
gO
t a
a sink-2 a
( tan a
cos2 a-1 ) <0. if 0 < x < a, if a < x < 7r/2.
k sink-1 a cos ax, sink x,
Obviously, g is continuous in the interval [0, 7r/2]. One can see easily by a study of the derivative that the function (sink x)lx is strictly increasing on the interval [0, a] and strictly decreasing on the interval [a, 7r/21. One can derive from this that g is convex on [0, 7r/2] and also that f (x) < g(x) holds for x E [0, 7r/2]. These facts, together with the inequality n
Eg(xi) < ng (n)l i=1
obtained by an application of Jensen's inequality, yield n
a(Sn, k) = 2k
f(xi) :5 n2kg (n/l
i=1
Since a < 7r/n, we have f (ir/n) = g(ir/n), and thus a(Sn, k) _< n2k sink The proposition is proved.
7r
n
= a(S*, k).
3. SOLUTIONS TO THE PROBLEMS
270
Remark. Many contestants observed that if 3 < n' < n and 1 < k < (tan(ir/n))/(ir/n), then we also have the inequality u(Sn,, k) < o(Sn, k). Indeed, considering the derivative of the function y = issink(7r/x), we see that
y' = sink - is
k7r is
k7r
sink-1 - cos - = sink-1 - 1 sin - - - cos -J > 0 is
x
is
provided that is > 2 and k < (tan(ir/x))/(ir/x) 7r/n' > it/n and
tan, n'
>
is
.
is
x
Using the relations
tan n n
we obtain
v(S,, k) = 2kn sink - > 2kn'sink = a(S,, k) > a(Sn, k). n n'
-
Problem G.15. Let h be a triangle of perimeter 1, and let H be a triangle of perimeter A homothetic to h. Let hl, h2i ... be translates of h such that, for all i, hi is different from hi+2 and touches H and ht+i (that is, intersects without overlapping). For which values of A can these triangles be chosen so that the sequence hl, h2i ... is periodic? If A > 1 is such a value, then determine the number of different triangles in a periodic chain hl, h2i ... and also the number of times such a chain goes around the triangle H.
Solution. We may restrict ourselves to the case of regular triangles since any triangle can be transformed into a regular one by an affine transformation and affinities preserve homothetic and touching position of triangles and also the ratio of perimeters of homothetic triangles. Let A, B, C be the vertices of the triangle H; let Pi, Qi, Ri be the vertices of the triangle hi; and let a, b, c and pi, qj, ri be the sides of the triangles H and hi opposite to the vertices A, B, C and Pi, Qi, R, respectively. These sides are considered to be half-open segments excluding the endpoint C
from side a, the endpoint Ri from side pi, etc. It is assumed that the vertices A, B, and C correspond homothetically to the vertices Pi, Qi, and Ri, respectively (see Figure G.12). There are two ways a triangle hi can touch the triangle H: either a vertex of H lies on the side of hi opposite to the homothetically corresponding vertex of hi, or a vertex of hi lies on the side of H opposite to the homothetically corresponding vertex. According to our assumptions, we always have exactly one of the two cases. In the first case, we say that hi touches
a vertex of H; in the second, we say that hi touches a side of H. The common point of the two triangles will be called the point of contact. We fix the orientations (Pi, Qi, Ri) and (A, B, C) of the triangles hi and H.
3.4 GEOMETRY
271
Figure G.12.
If hi touches a side of H, then denote by f (hi) the distance between the point of contact and the vertex of H that comes next to the contact point according to the fixed orientation. If hi touches a vertex of H, then denote by f (hi) the distance between the point of contact and the vertex of hi that comes next to the contact point according to the orientation opposite to the fixed one. Obviously, 0 < f (hi) < A/3 in the first case, 0 < f (hi) < 1/3 in the second (see Figure G.13).
Figure G.13.
It is clear that the contact points of hi and hi+i are contained in one (closed) side of H. Thus, the fixed orientation of H gives rise to an ordering of the triangles hi and hi+i We may establish the following facts. Let h' denote an arbitrary translate of h touching H. We want to describe the translates h" of h that touch both h' and H and come after h' (according to the fixed orientation). (See Figures G.14.a-e). (a) If h' touches a side of H and f (h') > 1/3, then h" is unique, it touches
the same side of H, and the distance of contact points is 1/3 (thus f(h") = f(h') - 1/3)) (Figure G.14.a). (b) If h' touches a side of H and f (h') < 1/3, then h" is unique, it touches the vertex of H next to the contact point of h' with respect to the
272
3. SOLUTIONS TO THE PROBLEMS
f(h')>1/3 H
Figure G.14.a
Figure G.14.b
fixed orientation, and f (h") = 1/3 - f (h') (Figure G.14.b).
(c) If h' touches a vertex of H and f (h') < A/3, then h" is unique, it touches the side of H that contains the contact point of h', and f (h") = A/3 - f (h') (Figure G.14.c). (d) If h' touches a vertex of H and f (h') = A/3, then we can choose for h" any triangle that touches the vertex of H next to the contact point of h' satisfying f (h") < (1 - ))/3 (Figure G.14.d). (e) Finally, if h' touches a vertex of H and f (h') > A/3, then h" is unique again, it touches the vertex of H next to the contact point of h', and it satisfies f (h") = (1 - A) /3 (Figure G.14.e). I. Suppose first that A < 1. We show that in this case one can always find a periodic chain of triangles.
Let h1 touch the vertex A of H so that f(h') = 1 - A. Set k = - (-A/(1 - A)). Starting from h1 and advancing according to the fixed orientation, one can construct the sequence h2, ... , h2k+1 uniquely by the
3.4 GEOMETRY
273
h'
f(h')< 3
h"
f(h")= 3 -f(h') Figure G.14.c
Figure G.14. d
previous observations (a)-(e), so that the even triangles h 2 ,-- . , h2k touch a side of H, hl, ... , h2k+1 touch a vertex of H, and f (h2k+1) >- A. Then we can choose h2k+2 according to (d) so that f (h2k+2) = 1 - A is fulfilled. The triangle h2k+2 can be obtained from h1 by a rotation of angle 2iri/3 about the center of H. Therefore, continuing the construction from h2k+2, choosing f (h4k+3) = 1 - A, we can close the chain by setting h6k+4 = hi, which finishes the proof.
274
3. SOLUTIONS TO THE PROBLEMS
Figure G.14.e
II. Now let us examine the case A > 1. Then only the cases (a), (b), and (c) are possible, and each h" is uniquely determined by h'. The same would be true if we changed the orientation; hence, there are exactly two translates of h that touch H and a given h'. Consequently, given hl and h2 that touches hl and H, the sequence hl, h2i ... can be continued uniquely. Suppose, we are given a periodic chain hl, h2i ... with period n(hn+l = hl) winding around H k times. The phrase "the chain winds around H k times" makes sense since the contact points of the triangles hi follow one another according to a fixed orientation of the boundary of H. We may suppose that this orientation coincides with the orientation (A, B, C). Observations (a), (b), and (c) make clear that the contact points of the triangles hi that touch a side of H form a cyclic sequence of n - 3k elements running around H k times such that the peripheral distance between two consecutive contact points is 1/3. Thus,
3(n-3k)=Ak, that is,
A = n 33 k,
n = 3k(A + 1).
(1)
One obtains from this equation that in the case A > 1, the existence of a periodic chain implies the rationality of A. In this case, the number of times
3.4 GEOMETRY
275
this chain goes around the triangle H is the denominator in the reduced simple fraction form of 3A, and the number of triangles contained in the chain is obtained as the product of this winding number and 3(A + 1) (of course, there is a periodic chain for any integer multiple of n and k with period n and winding number k). We still have to show that if (1) holds for some natural numbers k and n, then there exists a periodic chain with period n. Then it will also prove that the rationality of A is sufficient for the existence of a periodic chain. In addition to this proposition, we show that no matter where we put the starting triangle hl, the sequence of his will get closed at the nth step after k rounds around H. We may restrict ourselves to the case when h1 touches a side of H, since either h1 or h2 touches a side of H, and if the chain starting from h2 gets closed after the nth step, then so does the chain starting from h1 because of the unique reconstructability of the sequence in both directions. Let us construct the beginning part hl, ... , h,,,+1 of the sequence until it goes around H k times and the distance between the contact points of h,,,,+1 and h1 is less than 1/3. This situation is surely arrived at sooner or later, because at most A consecutive elements of the chain can touch the same side of H. Then, as above, we have
0< kA -
1 m-3k <3. 3
(2)
Expressing A with the help of (1) and multiplying by 3, we get
0
Problem G.16. The traffic rules in a regular triangle allow one to move only along segments parallel to one of the altitudes of the triangle. We define the distance between two points of the triangle to be the length of
the shortest such path between them. Put (n21) points into the triangle in such a way that the minimum distance between pairs of points is maximal. Solution. In the course of the solution, "distance" will refer to the ordinary distance of points while the italic "distance" will mean distance in the new
metric. Let ABC be the given triangle, AB = 1. In order not to lose the idea of the solution among the precise verification of simple geometric facts, we shall omit technical details. (a) Construction for lower bound. Let us divide each side of the triangle into n - 1 equal parts, connect each node of BC to a corresponding node
of CA by a segment parallel to AB, and divide these segments into n 2, n - 3, ... , 2,1 equal parts, respectively, in the order of their distances
276
3. SOLUTIONS TO THE PROBLEMS
from the side AB. We obtain the same set of points if we take through all nodes on the sides AB, BC, and CA straight lines parallel to the two other sides and consider all possible intersections of these straight lines. One may expect that this system of (n21) points is optimal. Let us denote by rn the minimum of distances for this system. (b) Theorem. Let (2) < N < (n21). If we are given N points in LABC, then one can always find two among them with distance < rn. If N = (n2 1) and the distance of any two points is at least rn, then the system of points coincides with the one constructed in (a). (c) For the proof, we need the proposition that a ball of radius r with center R in the metric space described in the problem is a regular hexagon centered at R, the vertices of which are obtained by moving R off to distance r parallel with one of the altitudes. The proof of this proposition is left to the reader. (d) We show here that if the distance of any two points of a certain subset
of the AABC is greater than rn = 2c = //n -1, then the subset contains at most (2) points. Consider triangles of the triangular lattice constructed in (a). Triangles homothetic to LABC with ratio 1/(n - 1) will be called
recumbent, and those homothetic to ABC with ratio -1/(n - 1) will be called standing. There are (2) recumbent triangles. Let us place a ball of radius c that is a regular hexagon of side c around the centers of recumbent triangles. They cover the recumbent triangles and the standing ones as well,
since each standing triangle is surrounded by three recumbent ones and the hexagons put around the centers of the neighboring triangles cover the standing triangle obviously. Thus, the given (2) balls cover AABC, that is, every point of the considered subset belongs to one of these hexagons. Because the diameter of such a hexagon is 2c = rn, a system of points for which the distance between any two points is greater than rn meets each hexagon in at most one point; hence, it contains at most (2) points. This proves the first part of the theorem. (e) We can show the uniqueness for N = (n21) points in the following way. If q < 1 and we shrink AABC from A with ratio q, then we get a triangle AB'C' that contains at most (2) points. If q -> 1, we obtain that side BC contains at least n points. Of course, side BC cannot contain
more than n points, and the points have to divide side BC into n - 1 equal parts. Let AB+C+ be the dilatation of ABC with respect to A with ratio (1 - 1/(n - 1)). The hexagons of side c around the points lying on BC cover the difference of the triangles ABC and AB+C+. Consequently, tAB+C+ contains (2) points, so that the distance of any two points is at
least rn = rn_1 (1 - 1/(n - 1)) = rn_1AB+. Since the theorem is trivial for n = 2, we may complete the proof by induction.
Problem G.17. Let C be a simple arc with monotone curvature such that C is congruent to its evolute. Show that under appropriate differentiability conditions, C is a part of a cycloid or a logarithmic spiral with polar equation r = ae''9.
3.4 GEOMETRY
277
Solution. Let rl = r(s), s E [0j] be the parameterization of C by arc length for which the radius of curvature p(s) is a monotone increasing function. The evolute E is parameterized then by r2 = r(s) + p(s)n(s), where n is the principal normal vector. The length of the arc of the evolute bounded by r2(0) and r2(s) is Q(s) = p(s) - po, where po = p(0). Let 4) be the congruence that superimposes C on E. There are two possibilities: 1. 4) takes r(O) to r2(0);
2. 4) takes r(O) to r2($). The radius of curvature of the evolute at the point r2 (s) is p(p(s) - po) in the first case and p(t - p(s) + po) in the second. Since, by Frenet equations, dr2
d2r2
=
da - n'
dal
P(s)P'(s)
t
we have p(p(s) - po) = p(s)p'(s) in the first case and p(B - p(s) + po) _ p(s)p'(s) in the second. We show that s = p(s) - po for all s E [0, .£] in the first case. Otherwise, we could find an interval [a, b] by the suitable differentiability conditions
such that a = p(a) - po, b = p(b) - po, and s > p(s) - po or s < p(s) - po holds everywhere inside [a, b]. However, in this case, we would have
b - a = P(b) - P(a) = f b P (s)ds = b P(P(s) - Po) ds j4 b - a a
a
P(s)
since p(s) - po > s implies p(p(s) - po) > p(s) and p(s) - po < s implies p(p(s) - po) < p(s). We conclude that the solutions belonging to the first case have a natural equation of the form s = p(s) - po. These curves are logarithmic spirals the polar, equation of which has the form r = ae'9, and one easily checks that any arc of such a spiral is a solution. In the second case, .£ - s = p(t - p(s) + po) holds for all s E [0, f]. To see this, we have to show first that 4) is an orientation-reversing transformation of the plane. We may suppose without loss of generality that
while r(s) is running along the curve C, the unit tangent vector t(s) is rotating counterclockwise. Meanwhile r2(8) is running along the evolute E, and its unit tangent vector n(s) is also rotating counterclockwise. Since 4) takes r(0) to r2(P), the point 4)(r(s)) is running along E in the opposite direction and its unit tangent vector is rotating clockwise. Therefore, 4) reverses orientation. Orientation-reversing isometries of the plane are glide reflections, so the evolute F of the evolute of C is a translate of C. Let r3(s) denote the center of curvature of E at r2(s). The length of the arc of F lying between r3 (f) and r3(s) is equal to p(2 - p(s) + po) - Po Monotonicity of the curvature implies that the direction of the tangent vector uniquely characterizes the point at which the vector is tangent to the curve. As a consequence, we get that the translation above takes r(s) to r3(s).
For this reason, p(f - p(s) + po) = f - s holds, as we wanted to show. From here, we obtain p(s)p'(s) = f-s+po by our previous considerations. Thus, in the second case, the solutions are given by arcs of the curves
3. SOLUTIONS TO THE PROBLEMS
278
with natural equation p2(s) + y - s + po)2 = c. These curves are cycloids given by x = a(u - sin u), y = a(l - cos u), where a = 4 po + (t + po)2. We may conclude that, in the second case, the solutions are those subarcs i'2 of the arc corresponding to the parameter domain u E [0,7r] for which a = p2(Pl) + [iP2 + p(Pl)]2 (for example, the whole arc fits the requirement). Remark. We may pose the more general problem of finding curves that are similar to their evolute. The solution of this problem involves epi- and hypocycloids and logarithmic spirals of the general polar equation r = ae°o as well.
Problem G.18. Given four points A1, A2, A3, A4 in the plane in such a way that A4 is the centroid of the AA1A2A3i find a point A5 in the plane that maximizes the ratio mine<j<j
(T (ABC) denotes the area of the triangle LXABC.)
Solution. The medians, which go through the centroid of the triangle, divide the plane into six angular regions. We claim that there is exactly one point in each of these regions for which the ratio in question attains its maximal value, namely 1/4. Let F be the midpoint of A2A3, let E denote the reflection of A in F, let R denote the trisecting point of the segment A3E that is closer to A3, and finally, let A5 denote the point on the halfline A4R for which A4A5 = 3/2. A4R. A5 is the point in the angular region A3A4E, that yields the maximum (see Figure G.15).
Figure G.15.
Suppose for a while that A5 is an arbitrary point in the angle A3A4E.
If this point is not in tA3A4E, then using the notation T(A4A5E) =
3.4 GEOMETRY
279
2T(AlA5A4) = 2b, T(A3A5A4) = a, T(A1A2A4) = c, T(A1A2A5) = d, we can express the area of the hatched domain as a + 2b + c = d,
(1)
since this area is the sum of the area of three triangles each having a side of length A1A2 = A3E, and the sum of the corresponding altitudes equals the altitude of LA1A2A5 corresponding to the side AlA2. Formula (1) remains true also when A5 is inside AA3A4E, only we have to take signed areas. In both cases, the areas with no sign satisfy
a+2b+c
4d.
(2)
Therefore, the ratio in question is < 1/4. The necessary and sufficient condition of having equality in (2) is a =
b = c = 1/4. On one hand, this means that A5 is four times farther from the straight line A1A2 than A4 is; on the other hand, it implies that T(A3A4A5) = T(A1A4A5), that is, A4A5 1 A1A3. These two conditions together are satisfied only by the point A5 constructed above. One can show by a direct computation that in this case AA1A2A5 has the largest area among all DAjAjAk and that T(AiA;Ak) > d/4 for all i,7,k. Problem G.19. Let K be a compact convex body in the n-dimensional Euclidean space. Let P1, P2,. .., Pn+1 be the vertices of a simplex having maximal volume among all simplices inscribed in K. Define the points
Pn+2, Pn+3, ... successively so that Pk (k > n + 1) is a point of K for which the volume of the convex hull of Pi,. .. , Pk is maximal. Denote this volume by Vk. Decide, for different values of n, about the truth of the statement "the sequence Vn+l, Vn+2,... is concave."
Solution. The statement makes no sense for n = 0 and is obviously true for n = 1. Consider the case n = 2. It is clear that the vertices of the starting simplex lie on the boundary of K, otherwise we could move the point lying inside K farther from the opposite hyperplane. We show by general induction for the two-dimensional case that further points are also chosen from the boundary.
The base clue has already been proved. Suppose that the first n points lie on the boundary. In this case, they are vertices of a convex n-gon, each side of which cuts off a (possibly trivial) part of the convex figure. The new point is chosen from one of these parts, which results in the polygon "growing" by a triangle. Because of maximality, the new point must come from the boundary.
280
3. SOLUTIONS TO THE PROBLEMS
To prove concavity, we have to show that when taking two consecutive differences of the sequence, the second does not exceed the first. If the new points come from arcs belonging to different sides, then this is obvious, since otherwise we would have to add these points to the sequence in the opposite order. If two consecutive points are chosen from the same are over a side of the convex hull of preceding points, then let A and B
denote the endpoints of this side and C and D be the points added to the sequence. Since C is taken first, the area of AABC is greater than or equal to the area of LABD. Denoting by E the intersections of the diagonals of the quadrangle spanned by the points ABCD, the statement follows from the inequality T(EAB) > T(ECD), since the first difference is T(ABC), while the second is T(ABD) + T(ECD) - T(EAB). Observe that the supporting half-lines of the convex figure taken at A and B in the half-plane bounded by AB containing C (see Figure G.16) are parallel or intersect each other; otherwise the one from A and B that was chosen later into the sequence would not have the maximality property. Let us draw a straight line through C parallel to AB. This intersects the straight line BD at a point D' lying outside the segment BD, so it suffices to prove the inequality with D' instead of D. The triangles ABE' and CD'E' are similar, and by the above observation, the segment AB is not shorter than the segment CD', and thus the statement is implied.
Figure G.16.
Now we are going to prove that the sequence is not necessarily concave for n > 3. For n = 3, the regular octahedron serves as a counterexample. Let us choose four arbitrary, noncoplanar vertices of the octahedron for the vertices of the first inscribed simplex. The volume of this simplex is maximal. Indeed, suppose that ABCD is an inscribed simplex of maximal volume having as many common vertices with the octahedron as possible. If one of the vertices, say A, is not a vertex of the octahedron, then consider
the plane through A parallel to the plane BCD. By the maximality assumption, this plane may not contain an internal point of the octahedron, but it has a nonempty intersection with the octahedron and thus contains at least one vertex of the octahedron, which could have been chosen instead of A. The contradiction proves our proposition. One can similarly
3.4 GEOMETRY
281
show, that the fifth point can also be taken from the vertices of the octahedron. The point is that opposite faces of the octahedron are parallel, so the points lying at maximal distance from two neighboring faces form an edge of the octahedron. The sixth point then must be the sixth vertex of the octahedron. The second difference of this sequence is twice as much as the first one; hence the sequence is not concave. Turning to the general case, consider a prism over the octahedron, that
is, the convex hull of a three-dimensional octahedron and an (n - 3)dimensional simplex lying in a complementary subspace of the octahedron.
We can show that this is a suitable counterexample in the same way as above, since due to maximality, the starting simplex must contain all the vertices not belonging to the octahedron and the further points must be chosen from the three-dimensional plane generated by the octahedron.
Problem G.20. Let us connect consecutive vertices of a regular heptagon inscribed in a unit circle by connected subsets (of the plane of the circle) of diameter less than 1. Show that every continuum (in the plane of the circle) of diameter greater than 4, containing the center of the circle, intersects one of these connected sets.
Solution. Let A1, &..., A7 be the vertices of the heptagon, let 0 be the center of the circumscribed circle, and let H1, H2,. . . , H7 denote the connected sets that connects the pairs A1, A2, A2, A3, ... ,A7A1, respectively. Denote by C a continuum in question, by D the circle of radius 2 centered at O.
By the assumption d(Hi) < 1 (i = 1, 2, ... , 7), Hi is inside D and 0 ¢ Hi. On the other hand, conditions d(C) > 4 and 0 E C imply that there exists a point Q E C which is outside D. Suppose to the contrary that UiHi and C are disjoint. In this case, the distance 6(x) of a point x E Hi from the closed set C U D not covering x is positive. Consider the open disc U(x, 6(x)) of radius 6(x) centered at x for all x E Hi. The union Fi = UXEH; U(x, 6(x)) of these discs is open and connected, and the
same holds for the union r = UiI'i. The set IF is inside D; furthermore, IF n (C U D) = 0. Since IF is open and connected, one can draw a closed broken line T* inside P that connects the points A1, A2i ... , A7. Let T be the boundary of the connected component of R2 \ T* containing O. T is a simple, closed, broken line inside D such that T and C are disjoint, 0 E C is inside T, and Q E C is outside T. We are going to show that the existence of such a T contradicts the fact that C is a continuum. Denote by B the set of points that lie inside T, by K the set of points lying outside T. By C n D = 0, we have
C= (CnB)U(CnK), where
(a) both C n B and C n K are open-closed in the subspace topology of C; furthermore,
282
3. SOLUTIONS TO THE PROBLEMS
b) none of them is empty because 0 E C fl B and Q E C fl K. However, these two facts together contradict the connectedness of the continuum C. Remark. One can give a counterexample for the proposition of the prob-
lem if C is required only to be connected, dropping the assumption that it is also closed. Indeed, draw two disjoint, infinite, broken lines into a square EFLJ so that one of them starts from E, the other starts from F, and both have closure that cover the segment U. Thus, M1 = T1 U J and M2 = T2 U J are two disjoint connected sets. Let us shrink and move the square and the figures contained in it in such a way that the points E, F that correspond to the vertices E and F get onto the segment A7A1 in the order A7, E, F, Al while L gets inside the heptagon. (We denote the image of a figure under this similarity by putting a_bar over the corresponding sign.) Let Q* be a point on the half-line OF whose distance from 0 is greater than 4. Then the claim of the proposition does not hold for the connected sets Hj = AjAj+1 (j = 1,2,... , 6), H7 = E7E1 U M1 U LA1, and the bounded and connected set C = OJ U M2 U FQ*. (This remark is due to LAszlo Babai.)
Problem G.21. What is the radius of the largest disc that can be covered by a finite number of closed discs of radius 1 in such a way that each disc intersects at most three others?
Solution. Draw a closed disc of radius 1 around each vertex of a square of side V r2-. These four discs cover the disc of radius drawn around the center of the square. We are going to show that a circle of radius greater than i does not have a covering having the prescribed properties. In the following, we shall denote by' the boundary of a set (the boundary of A is A') and "disc" will always mean a closed disc.
Let K be a disc of radius greater than vf2-, and suppose that it has a covering with the prescribed properties. Let us consider a subsystem A1, A2, ... , A,,,, of the covering in such a way that (i) the system A1, A2, ... , A,,,. covers K', (ii) if any of the discs A1, A2i ... , are removed, the remaining discs fail to cover K'. Such a subsystem can be produced obviously by omitting unnecessary discs one by one. There are no single points and empty sets among the intersections K' n Ai, since the discs different from Ai cover a closed subset of K', so if K' n Ai is empty or consists of one point, then Ai can be omitted. We claim that m > 5. Obviously, K' n Ai is an arc with diameter not greater than 2; however, the central angle of such an arc is less than 7r/2 in a circle of radius greater than f . Hence one needs at least five discs to cover K'. Now let X1 and X2 be the endpoints of the arc K'nA1. We may suppose without loss of generality that A2 covers X1 and A3 covers X2 since if both
3.4 GEOMETRY
endpoints of an arc K' n Ai were contained in the disc Aj (i whole arc would lie in Aj and Ai could be omitted.
283
j), then the
Now if we suppose that ((Al n K) \ A2) \ A3 is not empty, then it is an arc of positive length. Observe that this arc must be covered by one disc, otherwise Al would intersect more then three discs. Denote by A this
disc, and consider the arc A3 n K. This arc must be covered by A and Al together. Otherwise, A would also have a point in common with the disc that covers the remaining part, that is, A would intersect at least four other discs. Therefore, A U Al D A3 n K. But then A covers the endpoint of K' n A3 not contained in A1, and since four circles can not cover K', A must intersect at least one more disc, which would be the fourth disc intersected by A, and this is impossible. We conclude that ((A' n K) \ A2) \ A3 = 0, that is, A2 u A3 D A'1 n K. Similarly, we can find further two discs, A4 and A5, such that Al U A4 A3 n K, Al u A5 D A1n K, and A1,... , A5 are different. However, this is a contradiction, since in this case Al would intersect more than four other circles.
Thus, a disc of radius v admits a covering with the prescribed properties, but larger discs do not. On the basis of the above proof, we can also see that the covering of the disc of radius f is "essentially" unique (that is, unique up to rotations).
Problem G.22. Assume that a face of a convex polyhedron P has a common edge with every other face. Show that there exists a simple closed polygon that consists of edges of P and passes through all vertices.
Solution 1. Let S be the distinguished face of P, and suppose that P has n vertices off S. The proof goes by general induction with respect to n. The base clause is obviously true for n = 1; now let n > 1. It is easy to see that by omitting the edges of S from the graph of edges of P, we obtain a tree. (Indeed, imagine that P is a planet and its edges are dams, S is filled with water, while the other faces are empty basins and explode the dams that bound S.) For this reason, P has a vertex B not belonging to S such that the edges starting from B end on S with the exception of one edge. Let us denote by C the endpoint of the exceptional edge and by A1, A2, ... , A,,, the endpoints of the other edges ordered in correspondence with an orientation of S. Finally, let A0 be the vertex of S that precedes the vertex A1, andletAm+1 be the vertex that follows the vertex A,,,, according to the above orientation of S. (Figure G.17 shows a case with m = 3.)
Now if the half-line CB intersects the plane of S in a point A, then construct a polyhedron P' by gluing the prism AA1 ... AmB to P. P' is obviously convex itself and satisfies the conditions of the proposition. However, B is not a vertex of P, and thus P' has only n - 1 vertices off the plane of S. By the induction hypothesis, the edge skeleton of P' contains a simple closed polygon that passes through each vertex of P'. This polygon goes through A in one of the orders AOAA77.+1, AOAC, CAA,,,+1.
284
3. SOLUTIONS TO THE PROBLEMS
Figure G.17.
According to these three cases, replace vertex A of this polygon by the paths A1... A,,,._1BA,,,,, A1A2 ... A,..B, BA1A2 ... A,,,., respectively. The simple closed polygon we obtain this way passes through each vertex of P.
If CB is parallel to the plane of S, then choose a plane -y such that but has empty intersection with P. Now if we take the projective augmentation of the Euclidean space by adding a -y crosses the half-line BB
plane w of points at infinity and apply a projectivity II that takes -y to w, then the polyhedron IIP will completely consist of proper points, and the intersection point of the half-line (IIC) (IIB) and the plane of HS will also be proper; consequently, we may apply the previous consideration. We can proceed similarly if the half-line BAG intersects the plane of S.
Solution 2. Let S be the distinguished face, and let G be the graph obtained by removing the edges of S from the graph formed by the vertices and edges of the polyhedron. G is a tree, and the degree of a vertex of G not belonging to S is at least three. Let us choose an arbitrary face, B1, of P, and color the edges of G that lie on the boundary of B1. Suppose that we have already chosen the faces B1i B2, ... , Br in such a way that (1) Bi and Bj have no vertex in common if i # j, (i, j = 1, 2, ... , r), and
(2) for all i (i = 1, 2, ... , r), there exists an edge of G that connects Bi to one of the faces B1, B2, ... , Bi_1. Using that G is connected, we get that if P has a vertex not belonging to S that is not covered by the faces B1, B2, ... , Br, then there is a vertex x among these that is connected to some Bi (i E {1, 2, ... , r}) by an edge xy. Since the degree of x is at least three, x is covered by a face B,.+1 such that xy is not an edge of B,.+1. If some vertices of B,.+1 were covered by the faces B j (j = 1, 2, ... , r), then one of these vertices, say z, could be connected to x along edges of G without passing through points of the faces Bp On the other hand, according to the induction hypothesis, we
3.4 GEOMETRY
285
can join x and z by a path in G in such a way that all the vertices we go through belong to some Bj, which contradicts the fact that G is a tree. Consequently, the choice of faces B1, B2, ... , B,.+1 meets requirements (1) and (2). Now color the edges of G that lie on the boundary of Br+1. At the end of this process also color the edges of S that do not belong to any of the faces Bj. Thus, we obtain a system of colored edges that cover all vertices of P and form a polygon with the required properties.
Problem G.23. Let D be a convex subset of the n-dimensional space, and suppose that D' is obtained from D by applying a positive central dilatation and then a translation. Suppose also that the sum of the volumes of D and D' is 1, and D fl D' $ 0. Determine the supremum of the volume of the convex hull of DUD' taken for all such pairs of sets D, D'.
Solution. The conditions imply that D and D' are bounded convex sets with nonempty interior. The problem is equivalent to the determination of the supremum of the quantity V(co(D U D'))//(V(D) + V(D')) , where besides, having the properties just mentioned, D and D' have nonempty intersection and D' is obtained from D by a positive central dilatation and a translation. Let the ratio of the dilatation be a fixed positive number A. We show that in the case of a fixed A, the supremum in question is (1 + A + ... + A'')/(1 + An) = f (A), and this supremum is attained. The case n = 1 is trivial; let n > 1 in the following. First, we show that the case A = 1 can be derived from the case A $ 1. Indeed, if A = 1, then stretch D' from a point p E D fl D' with ratio 1 + E. Since D and the obtained D meet each other at p, we have
V(co (D U D'))
P c D. It is easy to see that V(P) tends to V (D) if and only if r(P, D) = supXED d(x, P) tends to 0, where d(x, P) denotes the distance between x and P. It is also not difficult to see that if r(P, D) _< E, then r(co (P U P'), co (D U D')) < E (P' denotes the image of P in D'). Suppose that P contains a common point of D and D' and its preimage with respect to the similarity; then P fl P' $ 0. Taking such polyhedrons P, if V(P) -* V (D), then V(co (PUP')) -> V(co (DUD')), and thus to prove that the supremum is < f (A), we may restrict ourselves to polyhedrons.
Now let D be a (closed) polyhedron, D' = AD, p E D fl D'. We may assume that 0 ¢ D. It is easy to see that co (D U AD) = UA
U uD=DU(E\AE), a<µ<1
3. SOLUTIONS TO THE PROBLEMS
286
and thus,
U AD =V(D)+(1-a")V(E).
V
a<µ<1
We show that V (E) < V (D) A/(1 - A) . Let F1, F2, .. ., Fr be those (n - 1)-dimensional faces of D whose hyperplane separates 0 strictly from D. Then the segment that connects 0 to p intersects the hyperplane iri at a point pi. Therefore, using the notation q = (1/A) p, we have
d(0,7i) =
epid(4,ii) <
4pi
pd(4,7ri)
4p
_ 1-
d(4,7r )
Since E is the union of pyramids over the faces Fi with vertex 0, and since the pyramids over the faces Fi with vertex q are contained in D, the previous inequality gives
Ed(O,iri)VV_i(Fi)
V(E) =
n
i=1 A
<
A
r
1-a i=1
>d(4,7i)V _i(Fi)
1-AV(D)
(V,i_1 denotes the (n - 1)-dimensional volume). From this,
V(co (DUD'))
V(D)+V(D')
- V (A<µ
V(D)(1+A' )
1 + (A + A2 + . .. + An)
1+A
.f (A)
Now we show that f (A) is an exact maximum. Let D be a simplex with vertices p1, p2, ... , pn+l; the corresponding vertices of D' are pi, p2, ... , p;+1+
and let p1 = p2. (For A = 1, we choose for D' the translate of D for which p'1 = P2.) In this case, co (D U D') is made up of the simplex D' and a truncated pyramid (or a prism if A = 1), so its volume is AnV(D) + (1 + A + + An-1)V(D) = f (A) (V(D) + V(D')). It remains to determine f (A). We claim that f (A) < f (l) (n + 1)/2, that is, (n+1)(1+A"`)
O.
However, this follows from summing up the inequalities
1+An-A1-An-2=(1-AZ)(1-An-z)>0 for 0
3.4 GEOMETRY
287
Problem G.24. Consider the intersection of an ellipsoid with a plane or passing through its center O. On the line through the point 0 perpendicular to a, mark the two points at a distance from 0 equal to the area of the intersection. Determine the loci of the marked points as or runs through all such planes.
Solution. Let B be an arbitrary body, 0 a point of it. Denote by F(a; B, 0) the pair of points P = {P+, P- } that lie on the straight line perpendicular to a at 0 and whose distance from 0 is equal to the area of the intersection of or and B: Area (a fl B). Let 4)(B) stand for the figure formed by the points P as or is varied: 4) (B) = U F(a; B, O) . a
Let E and t be a plane and a straight line intersecting one another orthog-
onally at 0, U the stretching in the direction of t with ratio A, and U* the stretching in the direction of E with ratio A. Obviously, U and U* are affinities.
Lemma.
4)(Lf(B)).
Proof. Set a' = U(a),
P = .P(a; B, O),
P' = F(u';U(B), 0) =.F(U(a);U(B),U(O)) The planes a and a' intersect each other in a straight line m, lying in the
plane E. Thus P' belongs to the plane A spanned by t and P. (Figure G.18 shows the trace of the mentioned figures in the plane A viewed from the direction of m.) Consider the points U E or and U' E a' that lie over a given point N E E n A (that is, UN, U'N I E). They satisfy NU' = ANU, and by elementary properties of affinities,
OP' _ Area (a' fl U (B)) _ OU' QP Area (a fl B) OU This equation shows that AOPP' can be obtained from tODU' by a dilatation and a right-angled rotation about 0 in the plane A. This similarity takes E n A to t, the point N to the intersection M oft and PP', and since similarities preserve angles and the ratio between corresponding segments,
we get that PP' I t and MP' = MP, which means that U* (P) = P', and this proves the lemma. Now let a, b, c be the half-axes of the ellipsoid E (E(a, b, c)). The equation of E takes the canonical form x2
2
z2
a2+b2+C2=1
3. SOLUTIONS TO THE PROBLEMS
288
Figure G.18.
in a suitable coordinate system. Consider the sphere S2 (a) of radius a around the origin. Then ID(S2(a)) is also a sphere: S2(a27r). Now let 111 be the stretching with ratio A = b/a in the direction of the y-axis. Ul (S2 (a) ) is the ellipsoid El (a, b, a), and by the lemma, -D(El) = D(Ul(S2(a))) U, (.1, (S2(a))) = U (S2(a27r)) = E2(abir, a27r, ab7r) .
Now let U2 be the stretching with ratio c/a in the direction of the z-axis. Then U2 (El) is the given ellipsoid E(a, b, c), and a repeated application of the lemma gives ,D (E) = D(U2(El)) =1d2 (f'(E1)) = U2 (E2) = E3(bc7r, ac7r, ab7r),
which was to be determined. Remarks.
1. We remark that the proposition of the lemma is true also for n-dimensional bodies B and (n - 1)-dimensional hyperplanes. Thus, an application of the lemma gives an immediate answer to the n-dimensional version of the problem as well. 2. We also remark that the lemma, which lies in the base of the solution, concerns an arbitrary body B and an arbitrary point 0 of it. Hence, it is applicable to a much wider class of problems.
Problem G.25. Construct on the real projective plane a continuous curve, consisting of simple points, which is not a straight line and is intersected in a single point by every tangent and every secant of a given conic.
Solution. Removing a straight line from the projective plane we obtain a Euclidean plane. Let us introduce Cartesian coordinates (x, y) on it and
3.4 GEOMETRY
289
denote by K the circle of radius 1 centered at (0, -2). K is a conic of the projective plane. All conics of the projective plane are equivalent, that is, one can, find a projective transformation P for any conic C such that P takes C to K. Therefore, if r is a solution of the problem with respect
to the circle K, then P-1(F) is a solution with respect to C since the projectivity P and its inverse P-1 are bijections that take straight lines into straight lines, and secants and tangents of a conic into a secant or tangent of the image conic, respectively. Thus if F meets the requirements of the problem with the conic K, then so does P-1 (F) with the conic C. First, we construct on the Euclidean plane a continuous curve r consisting solely of simple points that is not a straight line and is intersected in a single point by every secant and tangent of K that is not parallel to the x-axis. Let y = f (x) be a function defined on the whole x-axis such that (a) f is continuous and differentiable, (b) 0 < f(x) < 1,
(c) the absolute value of the derivative of f is less than 1, (d) f is increasing on the left of x = 0 and decreasing on the right, (e) the limits lima-,,,, f (x) and lima-_,,. f (x) exist. We show that secants of the graph r of f do not intersect K. Indeed, suppose that £(x) is a linear function such that
e(X1) = f(xl), £(x2) = f(x2);
x1 < x2.
(1)
The graph Q of £(x) is a secant of I'. We have
It'(x)I = ItanaI < 1 because of (c), where a is the direction angle of £. If a = 0, then Q either coincides with the x-axis or lies above it because of (b); hence it does not
meet K. If a > 0, then the zero x0 of t satisfies x0 < xl < x2 by (b) and (1). In this case, the assumption 0 < x0 together with a > 0 and (1) implies f (xi) < P X2), 0 < x1 < x2i which is in contradiction with (d). If, however, x0 < 0 and I tan aI < 1, then Q does not meet K. We can prove similarly that £ has empty intersection with K also in the case
-1 < tana < 0. We conclude that the secants of F do not meet K, or, in other words, the tangents and secants of K have at most one point in common with F. Those secants and tangents of K that are parallel to the x-axis (denote the set of them by x) obviously have empty intersection with V. Other tangents and secants with nonzero direction angle intersect r since F divides the plane into two connected parts, one of which contains
the half-plane y < 0 while the other contains the half-plane y > 1. A straight line with nonzero direction angle has points in both half-planes, hence it must cross the graph F of f. Thus, the curve r of the Euclidean plane has the required properties. We augment the Euclidean plane to a projective plane by adding an ideal point to each family of parallel lines. Attaching the ideal point P,,. of the x-axis to IT, we get a closed curve F on the projective plane, which
3. SOLUTIONS TO THE PROBLEMS
290
is continuous by (e) and is intersected in a single point, namely in Pte, by the straight lines of X. The points of r are simple since the graph r of the continuous function f consists of simple points and we have closed r by adding one single boundary point. Finally, we have to present a nonlinear function that has properties (a)(e). For example, y = e_x2/2. This function gives a solution according to the previous considerations. But one can easily construct other suitable functions. The simple, though not differentiable, function whose graph is obtained from the x-axis by replacing the segment between the points A(-2, 0), C(2, 0) by the broken line ABC, where B has coordinates (0, 1), also does the job.
Problem G.26. Let T be a surjective mapping of the hyperbolic plane onto itself which maps collinear points into collinear points. Prove that T must be an isometry.
Solution. (A) First we show that T is injective. The proof (based on the idea of Nandor Simanyi) consists of three steps.
1. We claim that the inverse image of a point is convex. For this
purpose, we need to show that if T(A) = T(B), then every point C of the segment AB is mapped to T(A). Suppose it is not so: T(C) # T(A) for some C E AB. Let e and f be the straight lines perpendicular to T(A)T(C) at T (A) and T (C), re-
Let us choose the points P and Q in such a way that their images are points of e and f, respectively, different from T(A) and T(C) (see Figure G.19). There are no three collinear points among A, B, P, Q since there are obviously no three collinear points among their images T(A), T(B), T(P), spectively.
T(Q). According to this, the straight line CQ intersects side AB of DABP in an inner point and thus must cross one of the two other sides. The image of this second intersection point must lie
on both e and f, which is a contradiction since e and f do not intersect each other.
T(Q)
Figure G.19.
T(C)
3.4 GEOMETRY
291
2. Now we show that the inverse image of a point X contains no interior point. Suppose to the contrary that the interior of T-1(X)
is not empty. Draw a straight line f through X. We claim that int (T-1(f \ {X})) # 0. Indeed, consider a point Z of f different from X and denote by P a point of its pre-image (see Figure G.20). Then the images of straight lines that go through P and intersect T-1(X) lie in f . In other words, T-1(f) contains a pair of opposite angular domains with common vertex P. Since T-1(X) is completely contained in one of these angles, it is disjoint from the other; therefore, int(T-1(f \ {X})) # 0, as we claimed. Now rotate f about X and consider the sets T-1(f \ {X }). They form an uncountable family of disjoint subsets of the hyperbolic plane such that each set has an interior point, which is a contradiction.
f X
Z=T(P)
Figure G.20.
3. Now we are in the position to prove that the preimage T-1(X) of every point X consists of one single point. Again we use the indirect method. If T-1 (X) is not a point, then it is a segment, half-line, or a straight line according to the previous considerations. In any case, T-1(X) is contained in a straight
line, which will be denoted by e in the following. Let A and B be two points of e that are mapped to X (see Figure G.21). It is easy to see that T(e) is a part of a straight line g. Draw a straight line f through X other than g. One can show as above that the interior of the set T-1(f \ {X}) is not empty. Rotating f about X, we can again get a contradiction.
g DT(e)
Z=T(P) T(A)=T(B) f
Figure G.21.
292
3. SOLUTIONS TO THE PROBLEMS
(B) The major difficulties of the proof are over. As a next step, we show that the images of noncollinear points are noncollinear. Indeed, if the noncollinear points P, Q, and R were mapped to a straight line e, then the images of the straight lines PQ, QR, and RP would also be contained in e. Consequently, every straight line that twice crosses the boundary of LPQR twice would be mapped into e. However, such a straight line passes through any point of the plane; therefore, the image of the whole plane should be contained in e, which contradicts the fact that T is surjective. (C) From this it follows that T-1 also takes collinear points to collinear points; the image of a straight line (under T) is a straight line, and the images of intersecting straight lines are intersecting, the images of nonintersecting straight lines are nonintersecting. (D) T preserves the ordering on straight lines. Indeed, if A, B, and C are three different points of a straight line e, then, as is known, C does not separate A and B if and only if one can find straight lines a, b, and c through A, B, and C, respectively, such that a and b intersect each other but c intersects neither a nor b (see Figure G.22). T preserves the latter property by (C), hence preserves ordering.
Figure G.22.
(E) Therefore, T preserves half-planes, half-lines, and segments. Furthermore, it preserves the ordering of pencils. Thus, the images of asymptotic half-lines are asymptotic. (F) Now we show the existence of a length d such that T maps segments of length d onto segments of length d. Let us perform the following construction (see Figure G.23.a). We denoted the two half-lines of e determined by P by f and g. Let us take a point X not lying on e and draw the half-lines h and k starting from X and asymptotic to f and g, respectively. Take a point Y on the elongation of k beyond X; let the segment YP intersect h at U. Draw an asymptotic half-line to g from U, and suppose that it intersects the segment XP in V. Denote by Q the intersection point of YV and g.
3.4 GEOMETRY
293
Figure G.23.a
We claim that Q does not depend on the choice of X and Y.
It is very easy to check this statement in the Cayley-Klein model of hyperbolic geometry (and it is also sufficient, since we know that
if a proposition is true in every Cayley-Klein model, then it is a true theorem of hyperbolic geometry). The above construction is a construction of a complete quadrangle in the model (see Figure G.23.b); consequently, denoting by I and J the horizontal points of f and g, respectively, the pairs JP and I, Q are conjugate, and the points P, J, and I determine the fourth harmonic Q uniquely.
Figure G.23.b
It is clear that the length d of the segment PQ does not depend on the choice of P either. Since T transforms Figure G.23.a into a similar one, it moves segments of length d into segments of length d, as we stated.
(G) Of course, the segments of length nd are also invariant under T, as
3. SOLUTIONS TO THE PROBLEMS
294
are the angles of regular triangles of side nd. Because the angle of these triangles tends to zero as n tends to infinity, there exist arbitrarily small invariant angles. As the sum of invariant angles is also invariant, a dense set of angles will be invariant under T. Taking into consideration that T preserves the ordering on pencils, it follows that T preserves angles. However, in hyperbolic geometry, a triangle is determined up to isometries by its angles, so the image of every triangle
under T is a triangle congruent to the original one, hence T is an isometry.
Problem G.27. Let X1,.. , X,, be n points in the unit square (n > 1). Let ri be the distance of Xi from the nearest point (other than Xi). Prove the inequality .
r2+...+r2 <4 Solution. We shall prove by general induction. The base clause is trivial
for n = 2. Let n > 2, and suppose that the proposition is true for any system of points consisting of less than n points. This assumption means also that if the points Y1i ... , Yk are enclosed in a square of side a and qj denotes the distance of Yj from the other points, then for 1 < k < n we have
q < 4a2.
Let us divide the square into four congruent squares by the medians. Denote the small squares by N1, . . . , N4 (see Figure G.24). We shall separate some cases according to the possible distributions of the points Xi in the small squares and then study each case individually. If a point Xi lies on the common boundary of two squares, then decide which square it belongs to; it does not matter how.
N1
N2
N,
N3
Figure G.24.
Case A. None of the small squares contains exactly one point. Here we may distinguish two further subcases: Either all points are in one small
3.4 GEOMETRY
295
square or the points are distributed in more than one square. In the first case, applying the induction hypothesis f o r X 1 ,.. . , X,,,_1i we get
Since r2 < 2, the proposition of the problem is true in this case. In the second subcase, however, each small square contains less than n points (and if it contains a point at all, then it certainly contains more than one). Therefore, by the induction hypothesis, 2
<4(1) =1
(1)
X;EN
for 1 < v < 4, from which the proposition of the problem follows. Case B. There is exactly one small square, say N1, that contains one point. For example, let X1 E Ni. Inequality (1) holds also in this case for
v = 2,3,4. Therefore, if ri < 1, then we are ready. We may suppose for this reason that r1 > 1. Of course, r2 < 2, and this estimation would be enough if one of the small squares were empty. Thus, we may suppose in the following that each small square contains a point; from this it follows that r1 < 5/4 because of the point in N4.
. X, N2
N3 N43
N44
Figure G.25.
Let us divide N4 into four smaller squares with the help of its medians, and we will denote these squares by N41, N42, N43, N44 (see Figure G.25).
As r1 > 1, the shaded domain contains no point. We shall show that 5 r2
XiEN4j
<
5 16
(j _ 3,4).
(2)
This follows from the induction hypothesis applied for N4a if the number of points contained in N4j is not exactly one. If, however, N4a
3. SOLUTIONS TO THE PROBLEMS
296
contains exactly one point, then its distance from another point in N4 is
at most '/4, that is, (2) holds in any case. Then n
r?=r2+ E r?+ > r?+
i=1
XiEN2
XiEN3
XiEN4
< 4+1+1+216 <4. Case C. The number of points is equal to one in exactly two small squares. In this case, the squares can be coupled in such a way that squares in a couple are neighbors and one of the squares in a couple contains exactly one point while the number of the points in the other is not one (see Figure G.26). Assume, for example, that N1 and N4 are in one couple and N1 is the one that contains exactly one point. Obviously, it suffices to show
that
E ri < 2. XiEN1UN4
This can be done in the same way as in case B.
1 point
1 point
1 point i
1 point Figure G.26.
Case D. Exactly three squares, say N1, N2, and N3, contain one point. Call these points X1, X2, and X3, respectively. Let us divide N4 into four small squares, as shown in Figure G.27. First of all, we remark that r2 < 5/4. We distinguish three subcases. Assume first that the rectangles N41 UN42 and N43 U N42 both contain a point. Then r3
ri
4
16
2
Consequently,
1 i=1
2
=
1
2 3
+ E r? < Xi EN4
13 16
+
5 4
+
13 16
+1<4
.
3.4 GEOMETRY
297
. x1
.x2
N2
N,
N41
N42
N3
N.
N43
x3
Figure G.27. Second, suppose that, for example, N41 U N42 contains no point but N43
and N44 do contain a point. In that case, r3 < 13/16, ri < 1 + 1/16, and furthermore, (2) is now valid. Hence, n
r?=ri+r2+r3+ i=1
is+2i <4.
rZ < XiEN4
Third, it remains to study the case when neither N41 nor N42 contains a point and one of N43 and N44 is also empty. Applying the induction hypothesis to the other (nonempty) square we get n
i=1
ri=ri+r2+r3+ E ri- -
A
+
A
+
A
+
A
=4.
XiEN4
Case E. The last case that remains is when each of the small squares contains one point. Let Xi E Ni. We prove that d(X1, X2)2 + d(X2, X3) 2 + d(X3, X4) 2 + d(X4, X1)2 < 4.
(3)
((d(P, Q) denotes the distance between the points P and Q.) This already implies the inequality of the problem.
Let us think of (3) as a strictly convex function defined on the set
xv=1N C R8. Such a function can attain its maximum only at the extremal points of the boundary of its domain. An extremal point corresponds to such a configuration of the points Xi in which every Xi is positioned at a vertex of the corresponding square Ni. Since for these cases (3) can be easily verified, (3) holds in general.
Remark. Contestants B. Brindza, V. Komornik, P.P. Palfy, V. Totik, and Zs. Tuza described those configurations of the points for which the sum of squares in question is equal to 4. They found that there are only
298
3. SOLUTIONS TO THE PROBLEMS
two possibilities: n = 2 and the points are positioned at opposite vertices of the square, or n = 4 and the points are put to the vertices of the square.
Problem G.28. Give an example of ten different noncoplanar points
Pi, ... , P5, Q1, ... , Q5 in 3-space such that connecting each Pi to each Qj by a rigid rod results in a rigid system.
Solution. First, we describe two constructions. 1. Let g be a rigid structure. Take a further point P, and connect it by a rigid rod to three points Q1, Q2, Q3 of g, for which P, Q1, Q2, Q3 are not coplanar. Then a rigid system is obtained. II. Let g be a rigid system and PQ a rod in it. Take a point T on this rod and connect it to two further points R, S of g, for which P, Q, R, S are not coplanar. The system obtained will also be rigid. Putting it the other way around, these two statements mean that to get a rigid realization of a graph g, it is enough to have (a) a rigid realization of a graph obtained from g by deleting a point of degree 3, or (b) a rigid realization of a graph obtained from g by deleting a point of degree four and connecting two of its neighbors by an edge. In our case, the graph to be realized is as shown in Figure G.28.
We omit the edge P1Q1. Without drawing edges between Pi's and Qt's it is sufficient to find a realization of the following graphs: by (b),
again by (b),
again by (b),
P2
P3
P4
P5
0
0
0
0
0
0
0
0
0
Q1
Q2
Q3
Q4
Q5
P2
P3
P4
P5
0
0
0
0
0
0
0
0
Q2
Q3
Q4
Q5
P3
P4
P5
0
0
0
3.4 GEOMETRY
299
0
/0
0
0
Q2
Q3
Q4
Q5
by (a),
P3
P4
P5
0
0
0
0
0
0
Q3
Q4
Q5
P4
P5
0
0
by (a),
11
0
0
0
Q3
Q4
Q5
by (a)
P4
P5
0
0
11
0
0
Q4
Q5
The last graph is a tetrahedron, which is obviously rigid.
Problem G.29. Let us define a pseudo-Riemannian metric on the set of points of the Euclidean space 1E3 not lying on the z-axis by the metric tensor 1
0
to
1
0
0
0
0
,
- x2 + y2
where (x, y, z) is a Cartesian coordinate system in V. Show that the orthogonal projections of the geodesic curves of this Riemannian space onto the (x, y) -plane are straight lines or conic sections with focus at the origin.
Solution 1. Let a, 3, and y run through the indices 1,2, and denote by (x1, x2, x3) the coordinates (x, y, z). We compute the differential equation of the geodesics. We can write the metric tensor and its inverse (gik) in the following form ga(3 = bap,
g = -
gao = bao, (x1)2
ga3 = 0,
ga3 = 0, 1
+ (x2)2,
g33 = -
(x1)2 + (x2)2
Using the formula
E 3
2
h=1
gih
09ih + aghk axk axi
_ 89jk aXh
300
3. SOLUTIONS TO THE PROBLEMS
for the Christoffel symbols, we obtain
rap =Pap = r30 = r.3 = 0; x-"
1
F33 = 2 Nl_(Xl)2 + (x2)2 ' r3 73 =
r3 3ry _-
x7
1
2 (x1)2 + (x2)2
By these, the equation of the geodesics,
i,j
can be written in the form
+ 2 2-zj2 =0; x +y
x//
(1)
2
1
x2+ y2
+2
yf
z'20
z"+x2+ Y2 x'z'+x2+ 2y'z'=0.
(2)
(3)
y
This system is satisfied by the curves
x=as+b, y=as+b, z=c, the projections of which onto the (x, y)-plane are straight lines. Furthermore, looking at the equations, we see that if z' = 0 at the initial point of an integral curve, then z' - 0 along the curve. Assume z' # 0. Then the third equation can be written in the form
2 ds
z'/
ln(x2 + y2)
Therefore,
c
lnz = In
z
x2 + y2
.
,_
c x2 -+y2
(where c is an arbitrary positive constant). Substituting this into equations (1) and (2), we get 2
x' + 2 (x2 + y2)3/2 = 0 ; c2
y =0. Y + 2 (x2 + y2)3/2
3.4 GEOMETRY
301
This system is non other than the well-known system of differential equations 8
x" _
aTX
c2 2
1
x2 _+V 2
e (C2
y=
ay
2
1
x2 + y2
of the Kepler problem, the solutions of which are conic sections with focus at the origin.
Solution 2. The problem of finding geodesics of the given space is equivalent to the problem of finding stationary curves of the variational problem corresponding to the integral VX72
J
+ y2 -
x2 + y2z2 dt
Since the variational function is positively homogeneous of degree one in the variables x, y, z, it is well known that those stationary curves of it that satisfy the condition x2
+ y2 - v/x2 + y2z2 = 1
(4)
coincide with the stationary curves of the problem corresponding to the integral x2
J
+ y2 -
x2 + y2z2 dt.
Changing over to the coordinate system (r, cp, z) with the coordinate transformation x = r cos cp, y = r, z = z, we obtain the variational problem
f
(r2 + r2cp2 + rz2)dt.
The integrand function is independent of cp and z. Therefore, the EulerLagrange equations OF
d aF
8F
d aF
aW
dt acp'
az
dt az
imply that OF = A,
a
aF = B az
(A and B are constant),
that is, we obtain the first integrals 2r2cp = A,
-2rz = B.
3. SOLUTIONS TO THE PROBLEMS
302
We introduce the new unknown function u(cp) = 1/r(cp). For this, we have
r
_
1
1 du . 2 dcp
.
uu
1 du
A du
A 2r2
u2 dcp
2 dcp
Therefore, by the relation (4), 1 = T + r2cp2 - rz2 22
(A
d
\ u/2 +
=
(du 12
z u4I
2
2
4-
42
-
B2 u2
1
B2 4
Let us differentiate with respect to cp: z
2U
1l
zl
{ 2 L2+uJ-2 }dcp=0. The equation du/dcp = 0 characterizes circles centered at the origin. Dividing by du/dcp, we get the equality d2u d
B2
+u = 2A2
the general solution of which can be written in the form B2 u(cp) = 2A2 (1 + e cos(cp + w))
that is, r(cP)
Bz
p 1 + e cos(cp + w)
(p
2A2)
'
and this is the focal equation of a conic section with focus at the origin.
Remark. Analogously, the trajectory of a point moving in a conservative force field can be obtained as the spacelike projection of a geodesic curve of a suitable Riemannian metric on the four-dimensional space-time plane.
Problem G.30. Let us divide by straight lines a quadrangle of unit area into n subpolygons and draw a circle into each subpolygon. Show that the sum of the perimeters of the circles is at most 7r ,,fn- (the lines are not allowed to cut the interior of a subpolygon).
Solution. It is known that the area of an n-gon circumscribed about a circle of radius R is at least R2n tan(ir/n) . Cutting the quadrangle by one straight line after the other, we determine the number of the domains
produced and the sum of the angles of the domains. When we draw a
3.4 GEOMETRY
303
new straight line, the number of polygons increases by k, and the sum of the angles increases by at most 2k7r, depending on how many nodes of the previous decomposition the new straight line goes through. This way, if the number of sides of the polygons in the resulting decomposition: k1, k2, ... , kn, then n
>(ki-4) <0. i=1
Denote by Ti the area of the ith domain and by Ki the perimeter of the inscribed circle. We know that 2
K? <
ki tan k i
Ti
Then n
n
KK <7r i=1
i=1
n
4
ki tank i
Ti < 1r N
i=1
4
ki tan : i
i
V
by the Cauchy-Schwarz inequality. Therefore, it suffices to show that n i=1
4
ki tan'.i
> n.
If ki = 4, then the corresponding term in the sum is just equal to 1. If ki > 5, then by >,(ki - 4) < 0, this term comes together with (ki - 4) triangles, and for this reason, it is enough to see that for k > 5,
+(k-4)3tan3
ktan2 Since
4
3 tan 3
it suffices to show that for k > 5, 4
k tan
< 0.23k + 0.079.
The left-hand side is less than 4/-7r, and this is enough for k > 6. When k = 5, the left-hand side is less than 1.11 and the right-hand side is greater than 1.2.
3. SOLUTIONS TO THE PROBLEMS
304
Problem G.31. Let K be a convex cone in the n-dimensional real vector space Rt, and consider the sets A = KU(-K) and B = (R'" \A)U{0} (0 is the origin). Show that one can find two subspaces in Rtm such that together they span RT, and one of them lies in A and the other lies in B.
Solution. First, we show that if x c int K then x E K. Take a simplex with vertices in K that has x in its interior. The vertices can obviously be moved to K by a small perturbation in such a way that x remains in the simplex; thus x E K. Now we show that if x K, then there exists u E RT such that (u, x) < 0, but (u, y) > 0 for every y E K. Let k be the point of K nearest to x. The perpendicular bisector hyperplane of the segment connecting x to k separates x from K, that is,
there exist u E Rn and b E R such that (u, y) > b for every y E K, but (u, x) 0. Thus, (u, y) > 0 for every y E K. We prove the proposition of the problem using induction. The statement
is obvious for n = 0. Assume that n > 1 and the statement is true for W provided that 0 < s < n. If the maximal number of linearly independent elements of K is s and s < n, then K is contained in an (n-1)-dimensional subspace H. In this case, let E0 C K U (-K) and Fo C (H \ A) U {O} be subspaces with which the proposition holds in H. Taking E0 for E and the span of Fo and a vector f E R' \ H for F, we find that the proposition is true. We may thus assume that K contains n linearly independent vectors and, as a consequence, it has an interior point. Set
N = {u: (u, x) > 0 for every x E K).
If N = {O}, then K = Rn, and thus K = R. Therefore, we may assume
that N
{O}. Choose a maximal linearly independent system u1,. .. , uk consisting of vectors from N. Let H be the hyperplane
H={x: If H n K is a cone,
(HnK)U(-(HnK))=HnA, (H\(HnA))U{O}=HnB; therefore, the proposition can be applied to H n A and H n B. We obtain subspaces E0 C H n A and F0 C H n B, which span H. Let us observe
that E0 C {x: (ul, x) = (u2, x) = ... = (uk, x) = 0}. Let l1i ... , l,, be a basis of E0, and let fl, ... , f, be a basis of F0. We may E K; furthermore, m + 1 > n - 1 holds. obviously assume that 1 1 ,. .. (K \ H) has at least one interior point, call any of them e, and let E be
3.4 GEOMETRY
305
the subspace generated by E0 and e. If we show that E C A, then we are done with the choice F = F0.
First, we show that Eo C K. If x E E0 and u E N, then u = )lul + + Akuk, and thus (u, x) = Al (ul, x) + ... + Ak (uk, x)
,
from which x E K. Now, if to E E0, then to E K and hence to + e c K; moreover, to + e E int K C K. From this, )(1o + e) E A for every A E that is, E C A.
Problem G.32. Let V be a bounded, closed, convex set in R', and denote by r the radius of its circumscribed sphere (that is, the radius of the smallest sphere that contains V). Show that r is the only real number with the following property: for any finite number of points in V, there exists a point in V such that the arithmetic mean of its distances from the other points is equal to r. Solution. (a) Since V is bounded, it is easy to see that there exists a minimal sphere
of radius r that contains V. Call this sphere G, and denote its boundary by S, its center by 0 (the origin). We show that 0 E co (S fl v). Suppose that this is not true. Then there exists a hyperplane H that separates 0 from s n v, since s fl v is compact. We may assume that
H = {(xl, ... , x",) : X. = c}, where 0 < c < r.
Since r is minimal, the spheres of radius less then r centered at Ok = (0, . . . , 0,1/k) do not cover V. Therefore, we can find points ak E V such that Iak - OI > r. The sequence {ak} possesses a convergent subsequence
ak, - a*. Obviously, a* E V and Ia*I = r; therefore, a* E S fl V and thus a* = (a*, ... , a* ), where a* > c. Then, for i > io, aki = (at, ... , an), r2 since ak E V. From this, we get where an > c and E n-
Iak, - Ok;I =
(:(a2 + (a- 1/k)2
< (r2-2c/ki+1/ki2)1/2
0 and the points ai is < r. On the other hand, s = (1/m) Em l 1 ai E V, and by the minimality of r we can find a point y in V such that I s - yI > r. Thus,
--
Iai - yI
--
(ai - y)I = Is-yI>r.
3. SOLUTIONS TO THE PROBLEMS
306
By the convexity of V, the segment [0, y] lies in V, and one can obviously
find a point z on this segment for which (1/m) E jai - zl = r. (b) If u > r, then for a1 = 0, there is no such point x E V for which Ia1 - xI = u. (c) Now let u < r. Since 0 E co (S fl v), there exist points a1, ... , ak E
S fl V and numbers t1i ... , tk such that for 0 < ti, E
ti = 1 and
Ek 1 tiai = 0. Set 6 = (r - u)/kr > 0, and choose the natural numbers q 1
and pi in such a way that k
(i=1,...,k).
pi =q, and Ipilq-til <E i=1
Let the system of points b1, . . . , b9 consist of the points ai, so that ai is taken pi times for each i. Then Ib; I = r (since bj E S) and k
k
E(pi/q - ti)ai < CE lai I = krE. i=1
i=1
We show that for any x E V, (1/q) >9=1 Ibi - xI > u. Indeed, denoting by (a, b) the dot product, for x E V, we have gEIbi-xI> i=1
=1 r =1
E iE1(bi-x,bi)=Q (q T
r2_ ( x,bi q
i=1
=r- 1qr x,bi >r - it qr 9
i=1
>r-krE>u.
i-1
Problem G.33. Show that for any natural number n and any real number d > 31/(3n - 1), one can find a covering of the unit square with n homothetic triangles with area of the union less than d.
Solution. Let us circumscribe congruent homothetic triangles about the faces of a regular hexagonal tessellation such that the area of the triangles is 3/2 times larger than the area of the hexagonal face. These triangles cover the plane with density 3/2. Let the radius of the circle inscribed in one of the triangles be equal to 1. Let us translate the sides of every triangle toward its center by e. The density of the new triangles is equal to (3/2) . (1 - E)2. There are holes produced in the covering: upside-down triangular holes circumscribed
about a circle of radius e. The relative number of the holes is twice the number of the triangles, so the density of the holes is (3/2) 262 = 362. Let us cover each hole with some density d > 1. By this we mean that we cover each hole with some plates, whose total area is equal to d times the area of the hole. The joint density of the triangles and the plates is
3.4 GEOMETRY
307
D = (3/2) (1- e)2 +3de2. D is minimal ate = 1/(2d - 1) . The minimum is D = 3d/(2d+ 1) . Let D,,, denote the infimum of the densities of the coverings with homothetic triangles of n different sizes. As we have seen, D1 < 3/2 . Since the holes in the above construction can be covered by homothetic triangles of (n - 1) different sizes with density arbitrarily close to Dn_1i we have
D2 <
3D1
2D1+1'
D3 < 2D2+1'...' 3D2
that is, Dn<33n1.
El
Problem G.34. Let R be a bounded domain of area t in the plane, and let C be its center of gravity. Denoting by TAB the circle drawn with the diameter AB, let K be a circle that contains each of the circles TAB (A, B E R). Is it true in general that K contains the circle of area 2t centered at C?
Solution. We shall show that the answer to the question is negative. Let R be the semicircle in the plane with Cartesian coordinates (x, y) defined
by the inequalities x2 + y2 < 1, x > 0. The area of R is t = it/2, and the radius of the circle of area 2t is equal to 1. Assume that P E TAB for some pair of points A, B E R. Denote the position vectors of the points P, A, B by p, a, b, respectively. The condition P E TAB is equivalent to the inequality Ip - (a - b)/2I < Ia - bI /2. Since
Ip- (a+ b)/21 ? IpI - I(a+b)/2I, we have
IpI < Ia + bI /2 + Ia - bI /2 < ((Ia + b12 + la= ((Ia12 + Ib12 + 2(a, b) + Ia12 + Ib12 - 2(a, b))/2) 1/2 _ (Ia12 + Jb12)1/2 <
Therefore, if we choose for K the circle of radius' centered at the origin, then K will contain all the circles TAB (A, B E R). Let us compute the coordinates (xC,yC) of the center of gravity C of the figure R. By the symmetry of the figure, yC = 0. To compute xc, we use the well-known method
xC = t
1x2
1 - x2 dx =
t [_(1_X2)3/21' 1= 10
2
3t
- 34
3. SOLUTIONS TO THE PROBLEMS
308
We want to show that the circle of radius 1 centered at (4/3ir, 0) does not contain the circle K. For this purpose, we have only to show that 4/3ir+ 1 > v/-2-. Using it < 3.2, 2.4.0.416 < 1, and (1.416)2 > 2, we obtain 4
4
4 +1>
37r
3
.
_
3.2+I
1
.4+1>1.416>V2.
2
Therefore, R is indeed a counterexample to the proposition in question.
Remark. Competitors disproved the proposition with many different counterexamples. Most constructions resembled the set-theoretic union of the circle x2 + y2 < 1 and the ellipse x2/(1 + e)2 + y2/E2 < 1 for some suitably small positive E.
Problem G.35. Let M' C R"+1 be a complete, connected hypersurface embedded into the Euclidean space. Show that M" as a Riemannian manifold decomposes to a nontrivial global metric direct product if and only if it is a real cylinder, that is, M'n can be decomposed to a direct
product of the form M' = Mk x R"-k (k < n) as well, where Mk is a hypersurface in some (k + 1)-dimensional subspace Ek+l C Rn+1 the orthogonal complement of Ek+1
Rn-k is
Solution. The solution rests upon the two fundamental equations of classical surface theory, namely, the Gauss equation
R(X, Y)Z = g(Z, A(X))A(Y) - g(Z, A(Y))A(X) and the Codazzi-Mainardi equation
(VxA)(Y) = (VyA)(X) In these formulas, X, Y, Z are vector fields tangential to the hypersurface,
g(X, Y) is the first fundamental form, A(X) = dxm is the Weingarten map (m is the unit normal vector field of the hypersurface, dx is the directional derivative in R"+1), VxY is the covariant derivative, and, finally, R(X, Y)Z is the Riemannian curvature tensor. A(X) is a self-adjoint linear mapping on each tangent space. It is also called the extrinsic geometrical fundamental form, since with its help one can describe the shape of the surface in the space. For example, the following statements are known and can simply be derived from the CodazziMainardi equations. Let WP be the kernel of Ar at the point p E M", and assume dim Wr°
= constant = k on some open subset U C M'. Then the distribution of the subspaces Wr° is integrable, and the integral manifolds are open subsets of a subspace Rk C R'n+1 a The Riemannian curvature of the hypersurface M" is 0 if and only if the rank of A is at most 1 at any point. For such a surface let U denote the open set of those points at which A 0 0, and let n be a unit vector field on
3.4 GEOMETRY
309
U consisting of eigenvectors of A such that A(n) = An holds with A Y1- 0. Then, by the previous proposition, integral manifolds perpendicular to n are open subsets of some subspace Jn-1 in JRn+1 Let c(s) be a curve on such an integral manifold parameterized by arc length, and consider the
function A(s) = A(c(s)). Codazzi-Mainardi equations yield by a simple computation that A' = V(s)A, where V(s) := -g(c(s), Vnn), and thus A(s) = A(0)efo (t) dl Consequently, if A 0 at one of the points of the integral manifold perpendicular to n, then A 54 0 along the whole integral manifold and also at
its boundary points. For this reason, if M' is complete, then the maximal integral manifolds perpendicular to n must fill the whole subspace Rn-1. Consider a bending of such a complete hypersurface having Riemannian curvature 0 onto Rn. (If Mn is not simply connected, then we bend its universal covering space.) The above integral manifolds will be bent onto parallel hyperplanes of Rn; Thus the distance between the points of an integral manifold and another integral manifold is constant. This means, that the integral manifolds themselves are parallel, (n-1)-dimensional subspaces JRn-1 in Rn+1 since otherwise the above distance would not be constant. The orthogonal complement JR2 of the subspaces Rn-1 cuts Mn in a curve M1, and we obviously have the cylinder decomposition M' = M1 X Rn-1 To sum up, we have the following propositions. Proposition 1. A complete hypersurface of Riemannian curvature 0 is always a cylinder of the form Mn = M1 x Rn-1, and M1 is a curve in the orthogonal complement 1[82 of
Rn-1
Proposition 2. If a complete hypersurface Mn decomposes into a metric
direct product in the form Mn = Mk X Mn-k, then either Mn has Riemannian curvature 0, or the Weingarten map vanishes at each point on the tangent spaces of one of the manifolds.
Proof. Denote by TT the tangent space of Mn at p E Mn and by Tp = T1 x T the decomposition of the tangent space corresponding to the direct product decomposition of Mn. Since for a metric direct product the Riemannian curvatures are also multiplied in a proper way, R(T I , T,2)X = 0 holds. Combining this with the Gauss equation we get
g(X, A(Tr1))A(TP) = g(X, A(Tr2))A(T')
.
This identity can hold for a self-adjoint mapping A only if A has rank 1 or A vanishes along one of the subspaces T. We have to show that if the Riemannian curvature of the surface Mn is not identically equal to zero, then A vanishes along Ti at each point of the manifold.
Let p E Mn be a point at which R(X, Y)Z 54 0. Then, as we have seen above, A vanishes on one of the subspaces T. Suppose that A(T2) = 0 holds. By Gauss's equation, the Riemannian curvature R(2) (X, Y) Z of the
3. SOLUTIONS TO THE PROBLEMS
310
manifold Mn-k vanishes, while the curvature RM (X, Y) Z of the manifold Mk is different from 0 at p. Since the tensor R(1)(X,Y)Z is constant along
each copy of M,-k in the direct product, A(T2) = 0 along the copy of M,-k that goes through p. This means that the Riemannian curvature of Mn-k is equal to 0 identically. It follows also that A(T9) = 0 in every point q = (ql, q2) E Mk X M,-k such that the tensor RM is different from
oatg1. It remains to show that A(Tg2) = 0 also at points q = (ql, q2) for which R(1}
1q, $ 0.
Let Vk C Mk be a maximal connected open subset such that RM = 0 at each point of Vk. The Riemannian curvature vanishes on the part U = Vk X M,-k C Mn of the hypersurface, and for this reason, the rank of A is at most 1 here. We have to show that the only eigenvector n of A corresponding to the nonzero eigenvalue lies in the tangent space V. Suppose to the contrary that n VT1 at some point p. Let N be the maximal integral manifold perpendicular to n and going through p. Since the submanifolds N, Vk, and M,-k are subspaces (totally geodesic submanifolds with 0 curvature) of the locally Euclidean space U, the angle between n and the subspace T1 is constant and nonzero along N. Since the eigenvalue A of n does not vanish on the boundary of N either, A $ 0 on the boundary of N and n VT1 holds at these points as well. In any neighborhood of such a boundary point, one can find a point q = (ql, q2) such that Rill Iql $ 0, and thus the image A(Tq) lies in the subspace TQ while at the boundary points A(TP) is not contained in T 1. For this reason, the existence of such a boundary point contradicts the continuity of A. However, such a boundary point does exist obviously, otherwise N would be complete and its orthogonal projection onto the manifold Vk would give a complete open subset of Vk. This is impossible because of the definition of Vk. This proves Proposition 2. The theorem follows immediately from the two propositions. Indeed, it is clear that cylinders decompose into a metric direct product, and conversely,
if Mn decomposes into a metric direct product, then either R(X,Y)Z - 0, and then by Proposition 1 Mn is a cylinder, or A(T2) = 0, and then the copies of M"-k are parallel (n - k)-dimensional subspaces. Mn is obviously a cylinder in the latter case as well.
Problem G.36. Among all point lattices on the plane intersecting every closed convex region of unit width, which one's fundamental parallelogram has the largest area?
Solution. Let r* be the lattice generated by the vertices of a regular triangle of altitude 2. The area of the fundamental parallelogram is equal
to 1/(2f). Let r be another lattice such that r is not congruent to r*, but the area of the fundamental parallelogram of r is also 1/(2"). We shall show that
3.4 GEOMETRY
311
in this case, there exists a regular triangle of altitude 1 which shares no point in common with r. Let P*, Q* and P, Q be the two nearest points in the lattices r* and r, respectively. Obviously,
1
a=PQ
.
Let the straight line PQ be horizontal. Place a regular triangle with altitude 1 onto the plane in such a way that one of its sides lies horizontally below PQ, while the other two sides go through P and Q, respectively. The distance of the straight line PQ from the upper vertex of the triangle is (av/J/2), and its distance from the base side is (1 - av/3-/2). The distance between two horizontal ranges of points in r is equal to 1/(2va). Since
1-a2
0
the triangle contains no lattice points other than P and Q. Translating the triangle a little downward, we obtain a triangle of unit width that contains no lattice points. It remains to show that r* has a point on every closed convex figure k of width 1. For this purpose, let us remark that the radius of the maximal inscribed circle of a convex figure of width 1 is at least 1/3. This can be proved from the facts that every convex figure is contained in a triangle or a band circumscribed about the inscribed circle of the figure and that among the triangles circumscribed about a given circle, the width of the regular triangle is maximal. As a consequence, the width of k is at most three times the radius of its inscribed circle. Now we need only remark that every circle of radius 1/3 contains a lattice point from r*. Therefore, r* has a point in common not only with every convex figure of unit width, but also with the inscribed circle of such a figure. 0 Problem G.37. Let S be a given finite set of hyperplanes in RT, and let O be a point. Show that there exists a compact set K C_ W containing 0 such that the orthogonal projection of any point of K onto any hyperplane in S is also in K.
Solution. We may assume without loss of generality that 0 is the origin of the space R' and that the normal vectors of the hyperplanes in S generate R' (if this latter condition is not fulfilled, then we may add further hyperplanes to S until we reach the desired property). Denote b y P1 the set of orthogonal projections onto the (n - 1)-dimensional linear subspaces of R" parallel to a hyperplane in S. F o r i = 2, ... , n,
let Pi be the collection of all projection operators, the image of which is (n - i)-dimensional and can be obtained as the intersection of the images of some operators from P. Then, by the finiteness of S and the assumption we made for the normal vectors of the hyperplanes, Pi is finite and nonempty
312
3. SOLUTIONS TO THE PROBLEMS
for every 1 < i < n. Thus, P,, has one element, P,1 = {Pn}, where P,, is the zero operator. For the sake of unified notation, set Po = {P0}, where Po is the identical transformation. If Pi E Pi and P3 E Ps, and the image of Pi contains the image of Pj, or equivalently, if the kernel of Pi is contained in the kernel of Pj, then we write Pi > P3. The orthogonal projection onto a plane s from S can be given in the form x F-, Px + p3, where P8 E P1 is the operator of the orthogonal projection
onto the (n - 1)-dimensional linear subspace parallel to s, and p8 is the vector drawn from 0 perpendicular to s. Obviously, P8p8 = 0 for every s E S.
Lemma. One can find a sequence of numbers 0 = Ro < R1 < such that for any 1 < j < n, Pj E P;, and R > R3, the set
. < R,,,
G(Pj, R) := {x E W'': Pox = 0 and < R2 - Ri for every 0 < i < j - 1, P3
is mapped into itself by any projection x H P8x + p8 onto a plane s E S for which P8 > Pj . Since P,,, = 0, applying the lemma for j = n and R = R,,, we obtain that the closed set K = G(0, R,,,) is mapped into itself by any projection onto a plane in S. On the other hand, this set is bounded, for it is contained in the sphere of radius Rn centered at the origin. Thus, it satisfies the requirements formulated in the problem.
Proof. We prove the lemma by induction on j. For j = 1, let P1 E P1 be an arbitrary projection. Then
G(P1iR)={xE1Rn: Plx=0,IIxII P1, then, since the images of the operators P1 and P, are (n - 1)-dimensional, P8 = P1. Therefore, P8x0 + p., = p8 for any x0 E G(P1, R). Thus, if we take supSES Ilp,, II for R1, the claim of the lemma holds.
Now assume that the lemma has been proved for the values 1,2,. .. , j - 1 and that the numbers Ro < Rl < . . < Rj_1 have al.
ready been constructed. Let P3 E P; be an arbitrary projection and s E S such that P8 > Pj. If Pjxo = 0, then Pj(P8xo + pe) = Pjxo + Pjp8 = 0. Thus, projection onto s preserves the first defining property of the elements of G(Pj, R). Second, we show that if R > Rj_1 and x0 E G(Pj, R), then IIP (P8xo +ps)112 < R2 - Rz
-
(1)
holds for0 Pi, then Pi(P8xo + p8) = Pixo, and there is nothing to prove. If P8 > Pi, then let us denote by Pi+1 the projection operator whose image is the intersection of the images of P8 and Pi. Consider the set G(Pi+1, R*) = G(Pi+1,
R2 - IIPi_1x0112)
3.4 GEOMETRY
313
Then
R*2=R2-IIPi+lxol12>R2-(R2-R+1)=R+1 and therefore, R* > Ri+1. Thus, i + 1 < j - 1, and by the induction hypothesis, the projection x H P,x+p, maps G(Pi+1, R*) into itself (since P, >- Pi+1). However, (Po - Pi+1)xo E G(Pi+1,R*), since for Pi+1(Po Pi+1)xo = 0 and Pi+1 < Pk E Pk (k = 0, ... , i), we have IIPk(Po - Pi+1)xo112 =11(Pk - Pi+1)xo112 = IIPkxo 112 -IIPi+1xo112
[(p.'
- Pi+1)xo +po]112 < R*2 - R?,
that is, IIPi(P9xo +ps) - Pi+1xo112 < R2 - IIPi+1xoII2 - R2
from which we get (1). To prove the Lemma we have only to show that if R > Rj
is sufficiently
large, then (1) holds for i = j - 1 and any x0 E G(Pj, R) and any Pj_1 E P;_1 such that Pj_1 > Pj. If P, > Pj _1i then we are done because Pj_1(P,xo +p,) = P3_1xo.
If P, > P3, then the intersection of the images of P, and Pj_ 1
is
(n - j)-dimensional and consequently coincides with the image of Pj.
We show that Pj_1P8 is a contraction on the kernel of the operator Pj. For this purpose, it suffices to show that if x # 0, Pox = 0, then IIPi _ 1 Psx I I < I I x I I The assumption that the two sides of the latter inequality are equal yields IIP,xlI = IIxII, from which we obtain P,x = x. Repeating
the same argumentat yields Pj_1x = x. Therefore, x is in the image of P, and Pj_ 1i that is, in the image of Pj. However, this contradicts x # 0 and Pox = 0. This means that Pj_1P, is indeed a contraction, that is, there exists a number 0 < q < 1 such that Pox = 0 implies I I Pj_ 1 P, x l l< g l l x l l Now let us choose the numbers rj = rj (P Pj_1i Pj) in such a way that
R > rj.
qR + R1 < JR2 -
(2)
(This is possible since if both sides are divided by R, the left-hand side tends to q and the right-hand side tends to 1 as R -+ oo.) Now, if R > rj and x0 E G(P;, R), then IIPj_1(Psxo+p,)11
gllxo11+11p9II
therefore, by (2), (1) holds for i = j - 1. Now let Rj be the maximum of the numbers rj _1iPj) for all P8,Pj_1iPj. Then we obtain that (1) holds for R > R, with i = j - 1, if Pj E P xo c G(Pj, R), Pj < Pj_1 E P,_1,
sES
and P,>P3. This completes the proof of the lemma.
314
3. SOLUTIONS TO THE PROBLEMS
Problem G.38. Let k and K be concentric circles on the plane, and let k be contained inside K. Assume that k is covered by a finite system of convex angular domains with vertices on K. Prove that the sum of the angles of the domains is not less than the angle under which k can be seen from a point of K.
Solution. Let 0 be the common center of the circles, and r and R the radius of k and K, respectively. Let us define a function F on the interior of k as follows. If the distance from 0 to P is p, then set
F(P) = f(P) =
R2 - r2
7r(R21
P2)
r2 - p2
If T is a measurable set in the interior of k, then let
m(T) = fF(P)dP. The function m is obviously a measure on the interior of k. We show that m has the following property:
If g and h are half-lines starting from a point A of K such that g is tangent to k, h intersects k, and the angle between h and g is a, then the angular domain bounded by g and h cuts the interior of k in a domain T, the measure m of which is a.
Figure G.29.
Let us compute m(T) by successive integration. Let e be the half-line starting from 0 perpendicular to h. Let us introduce a polar coordinate system on the plane, with polar axis e. Then a point P(p, cp) belongs to T
3.4 GEOMETRY
315
if and only if d < p < r, cp < 0, where d = p cos ?9 = R sin(E - a) is the distance of 0 from h, E is the angle between the half-lines AO, and g (see Figure G.29). Based on these, we get
m(T) = JT F(P) dP = =
fr
parccos(d/p)
f (p)pdcpdp
JJ d
Ir
arccos(d/p)
2f (p) p arccos(d) dp
r R sin(e-a)
2 (P)P arccos
Rsin(E - a) P
d Pp
Therefore, we have to show that, for 0 < a < 2e, we have IT
2f (P)P arccos
R- sin (E
Rsin(e-a)
P
-a))
dP= a.
Substituting R sinn(e - a) = t, this means that fr
2f (p) p arccos()
dp = E- aresin(R )
if -r < t < r. This relationship is obviously true for t = r. Thus, it is enough to show that the derivatives of the two sides with respect to t are the same: 2f(p)p dP =
Jt Jr
p2 - t2
2f (P)P
Jr it
P2
_t2dp
1 t2 ' R2_
-r < t < r
R2 - r2
p
2
r 7r R2-p2
dp
(r2 - p2)(p2 - t2)
p=r 2 Tr
1
R2-t2
tan
R2 - r2 2 - t2
1
p
R2- t2 r2-p2
vfR2 -
p=t
The proposition proved, combined with the additivity of the measure m, yields immediately that if both half-lines bounding a convex angular domain of angular measure a intersect k, then the m measure of the intersection of the angle and k is equal to a (since the intersection can be obtained as the difference of two segments of k). Now consider a covering of k with a finite number of convex angular domains. Let us denote by a1, ... , an the angular measure of the angles, and by T1i . . . , Tn the intersections of the domains with k. Then, by the proposition just proved, ai > m(Ti), i = 1, . . . , n. (Equality holds here if and only if the both boundary lines of the ith angle have a point in common with k.) Therefore, by the subadditivity of measures,
al+
+On >m(TI)+
+m(Tn)>U -
1
TO =2E
(since U' l Ti fills the interior of k and the m measure of k is exactly 2E). This completes the proof.
3. SOLUTIONS TO THE PROBLEMS
316
Problem G.39. Let (E, 7r, B) (7r : E -> B) be a real vector bundle of finite rank, and let TrE =VC®HC
(*)
be the tangent bundle of E, where V = Ker d7r is the vertical subbundle ofTE. Let us denote the projection operators corresponding to the splitting (*) by v and h. Construct a linear connection V on VC such that
Vx V Y - Vy V X =v[X,Y] -v[hX,hY]. (X and Y are vector fields on E, [.,.] is the Lie bracket, and all data are of class C°°).
Solution. Let CO°(E) be the ring of differentiable (of class C°°) functions E ---* R, and denote by X (E) and Xv(E) the C°° (E)-module of all vector fields on E and the submodule of vertical vector fields, respectively. A mapping
D: Xv(E) x Xv(E) -.-4 Xv(E),
(Z,W) F-> DZW
will be called a pseudoconnection on V if it has the formal properties of linear connections, that is, it is C°° (E)-linear in Z and an R-linear derivation in W (DZ f W = (Z f )W + f DZW, f E C°° (E)). Assume that D is a pseudoconnection such that DZW
- DWZ = [Z, W] for every Z, W E Xv(E),
and define the mapping
V: X(E) x Xv(E) --> Xv(E),
(X,Y) --*VxY
by the formula
VxY :=
v[hX, Y].
Then V is a linear connection. It can be seen immediately that V is linear in X and Y and that for f E C°°(E) we have (vX) fY + fv[hX,Y] + v(hX)fY VxfY = = fVxY + (vX + hX)fY = fVxY+(Xf)Y (applying linearity at each point and vY = Y). Finally, a similarly simple
computation shows that V fxY = fVxY. We show that V satisfies the required relationship. Indeed, v[hX, vY] - v[hY, vX] VxVY - VYVX = = [vX, vY] + v[hX, vY] - v[hY, vX] = v([vX, vY] + [hX, vY] - [hY, vX]) = v([X,Y] - [hX, hY]) = v[X,Y] - v[hX, hY] .
3.4 GEOMETRY
317
It remains to show that an "auxiliary" pseudoconnection D does exist. We shall prove this by a simple local construction. Assume dim B = n, rank = r. By the local triviality of the vector bundle, each point of B has a neighborhood U such that 7r-1(U) -+ U is diffeomorphic to the trivial bundle U x R' --+ R'. If V): x-1(U) - U x RT is a vector bundle isomorphism, (u') 1 is a local coordinate system on U, and (la)a=1 is the dual basis of the canonical basis of R'', then
xi:=uio7r
(1
y:=laopr2o0 (1
(1 < a,,3 < r)
.
(It follows from standard methods of the theory of linear connections that 17 can be defined this way.) Then for any two vector fields Z
a Za ay a,
(9 W W,3 ayo
over it-1(U), we have o
L5-,w = Za aya ayo
and consequently, DzW - DwZ = [Z, W], which completes the proof.
Problem G.40. Consider a latticelike packing of translates of a convex region K. Let t be the area of the fundamental parallelogram of the lattice defining the packing, and let denote the minimal value oft taken for all latticelike packings. Is there a natural number N such that for any n > N and for any K different from a parallelogram, ntmli,(K) is smaller
than the area of any convex domain in which n translates of K can be placed without overlapping ? (By a latticelike packing of K we mean a set of nonoverlapping translates of K obtained from K by translations with all vectors of a lattice.)
Solution. We show that such an N does not exist. Let h be the branch of the hyperbola defined by the equation xy = 1 that lies in the first quadrant of the plane. Let A be the intersection point of the x-axis and the tangent of h at (a, 1/a), and let B be the intersection point of the y-axis and the tangent of h at (1/a, a). Let us replace the arcs of the hyperbola "beyond" (a, 1/a) and (1/a, a) with the segments of the tangents between A, (a, 1/a)
318
3. SOLUTIONS TO THE PROBLEMS
and B, (1/a, a), respectively. Denote by g the curve consisting of the two segments and the arc of the hyperbola between them. Then the area of any triangle bounded by the coordinate axes and a tangent of h equals 2. On the other hand, if x -* oo, then the area of the domain between the coordinate axes and g tends to infinity as well. Therefore, we can choose a in such a way that the ratio of the two areas is equal to an arbitrarily small e. Let us round two opposite corners of the unit square with the help of two suitably scaled copies of the arc g cutting off the square two domains of area 6. The area of the rounded square q equals 1 - 26. Since q is central symmetric, tmin(q) is the area of the circumscribed hexagon of q having minimal area, that is, tmjn(q) = 1 - 2eb. On the other hand, n translates of q can be packed into a domain of area n - 26. Thus, for any n, we can choose an E > 0 such that en < 1, and then ntmin(q) = n-2e6 > n-26.
Problem G.41. Show that there exists a constant ck such that for any finite subset V of the k-dimensional unit sphere there is a connected graph G such that the set of vertices of G coincides with V, the edges of G are straight line segments, and the sum of the kth powers of the lengths of the edges is less than ck. Solution 1. Let G be a connected graph such that the set of vertices of G is V, the edges of G are straight line segments, and the sum of the length of the edges is minimal. By the characterization of G, G is a tree. Take an open ball about the midpoint of every edge of G with radius r(V/3- - 1)/4,
where r is the length of the edge. We prove that these spheres do not intersect one another. Assume, to the contrary, that the segments AB and CD are edges in G, p(A, B) = R, p(C, D) = r, and
S((A+ B)/2, R(V' - 1)/4) n S((C + D)/2,
r(V - 1)/4)
0
,
where p is the distance and S(a, r) is the open ball of radius r centered at a. By symmetry, we may assume R > r. Then
p(A, (C + D)/2) < p(A, (A + B)/2) + p((A + B)/2, (C + D)/2)
< R/2 + R(vf3- - 1)/2 = Rv /2 . Since at least one of the angles (A, (C + D)/2, C) and (A, (C + D)/2, D) is not obtuse if A :A (C + D)/2, we get min{p(A, C), p(A, D)} < (p2(A, (C + D)/2) + r2/4)112
< (3R2/4+R2/4)1/2 = R. If the edge AB is removed from the graph G, then the subgraph is the union of two trees. By symmetry, we may assume that B is contained in the same component as C (and D as well). But then removing from G the edge AB and adding to it the shorter of AC and AD, the length of which is less then R, we obtain a connected graph G' satisfying all the requirements for G, but the sum of the length of its edges is less then the sum for G, and this is a contradiction.
3.4 GEOMETRY
319
Since the balls we constructed are disjoint and are contained in a sphere of radius 2, their total volume is less than the volume of the ball of radius 2. Thus, k
2kwk ?
E
41
Pk(A, B)
AB is an edge in G
Wk
where Wk is the volume of the k-dimensional unit ball. Therefore, Pk(A, B) < AB is an edge in G
(
L
4
V3- -1
) 2k = ek
as was to be proved.
Solution 2. It suffices to show the following claim:
Claim. If T is a box such that the ratio of any two edges of T is less than or equal to 3, and V C T is a finite set of points, then there exists a connected graph G = G(V) set of vertices V such that the sum of the kth power of the edges of G satisfies S(G) < Ck Vol (T),
where ck is a constant depending only on k, Vol (T) is the volume of T.
Using induction on the number of points in V, we shall show that one may choose ck = 3k+lkk/2. Let IVl = n. The statement is true for n = 1, 2, and assume it has already been proved for 1, 2, ... , n - 1. Let us shrink T,
that is, move the faces of T inside as long as the assumptions on T are not violated and T contains V. This way, we make the inequality to prove sharper. If T cannot be shrunk further and e is one of the longest edges of T, and furthermore, Q1 and Q2 are the faces perpendicular to e, then V has a point on both Q1 and Q2 (otherwise, we could shrink T further). Let T1, T2, and T3 be the consecutive boxes obtained by cutting T into three equal parts by hyperplanes perpendicular to e. We distinguish two cases depending on whether T2 contains a point of V or not. Case 1. V has no point in T2. Let V1 = V fl T2, V3 = V fl T3. Since Qi fl v 0 for i = 1, 2, we have l Vi l < n. Applying the induction hypothesis for T1, V1 and T3, V3, we obtain connected graphs G1 = G(V1) and G3 = G(V3), for which S(Gi) < ckVol(Ti) = (1/3)Vol(T), i = 1, 3. Connecting a point of Gl to a point of G3i we obtain a connected graph G such that S(G) < S(G1) + S(G3) + (V/k-)k jell
< 2ckVol(T) + (j)k3kVol(T) < ckVol(T) by the choice of ck.
Case 2. V has a point in T2. Choose x E V fl T2 and cut T into two closed boxes T1 and T2 by a hyperplane passing through x perpendicular
320
3. SOLUTIONS TO THE PROBLEMS
to e. Then T* also has the property that the ratio of any two of its edges is not more than 3. Thus, the induction hypothesis can be applied again to Tl , Tl fl V and T2*, T2 fl V since the sets Tl fl V and TT fl V have less than n points. Therefore, one obtains the connected graphs Gi = G(V fl Ti ) and G2 = G(V fl T2*) such that S(G2) < ckVol(T*)
(i = 1, 2).
Setting G* = Gi U GZ yields a connected graph on the set of vertices V and
S(G*) = S(Gf) + S(G*) < ckVol(T1) + ckVol(T2) = ckVol(T).
El
Remarks. 1. We may choose for G a Hamiltonian circuit as well, but the proof of this fact is more difficult.
2. One can prove that c2 = 4 is the smallest suitable constant and that it is good for the Hamiltonian circuit problem also. It is an interesting problem to find the exact constants for k > 2.
Problem G.42. Let us draw a circular disc of radius r around every integer point in the plane different from the origin. Let Er be the union of these discs, and denote by d, the length of the longest segment starting from the origin and not intersecting Er. Show that 1
lim (d,. - 1) = 0.
r--.o
r
Solution. Let the second endpoint (x, y) of one of the longest segments be on the circumference of the disc centered at (k, n). Reflecting the segment in the point (k/2, n/2), we obtain that the segment connecting the points (k - x, n - y) to (k, n) has a point in common only with the circle centered
at (k, n). Since the points (0, 0), (k - x, n - y), (x, y), and (k, n) are vertices of a parallelogram, the two other sides of which have length r, this parallelogram contains no lattice points in its interior. The segment connecting the origin to (k, n) has a point in common only with the disc about its endpoint. The length of this segment differs from dr by at most r.
Let Dr be the length of the longest segment connecting the origin to a lattice point and having a point in common only with the disc about its endpoint. As we observed, I Dr - dr 1< r. According to this, it suffices to show that
lim(Dr-1)=0. r
r 0+
For this, we shall apply the well-known theorem of lattice geometry saying that a triangle that has lattice points for its vertices but contains no further
3.4 GEOMETRY
321
lattice points in its interior and on the boundary (except for the vertices) has area 1/2.
Let V = (n, k) be the endpoint of a segment of length Dr, and let H be the domain containing those points of the plane whose orthogonal projection onto the straight line OV lies in the closed segment OV. Since translates of H with integer multiples of the vector (n, k) cover the whole plane, the distance of any lattice point P E H from the straight line OV is at least r if P is different from 0 and V. Let Q be a lattice point in H different from 0 and V such that the distance of Q from the straight line OV is minimal. Then there are no further lattice points on the segments OQ and OV, nor in the interior of triangle OVQ, because these points are closer to OV than Q. Furthermore, there are no lattice points in the interior of the segment OQ since every disc of radius r centered at a lattice point different from 0 and V is disjoint from the segment OV. Therefore, the area of triangle OVQ is equal to 1/2. Consequently, the distance of
Q from the straight line OV, that is, the altitude of triangle OVQ, is 1/ 1 OV 1. We find that the distance of Q from OV is at least r if and only if I OV 1< 1/r.
This yields another characterization of Dr: Dr is the length of the longest segment connecting 0 to a lattice point without going through further lattice points (that is, the coordinates of the endpoint are relative primes) and having length at most 1/r. By this, Dr < 1/r, of course. It is well known that for any e > 0, if n is large enough, then there is a prime number between n and n + en. Let 0 < e < - 1, and choose a prime p satisfying 1
1
(1+e)r
Let q be the largest nonnegative integer such that p2 + q2 < r2 Then q < p since if we had q > p, then by p2
2p2
r
+ q2 > 2r < (1 + E) r < would hold. On the other hand, q > 1, as for r < 1, p2+12 < (1/r-1)2+1 < 1/r2. Thus, 1 < q < p, and since p is a prime, p and q are relative primes.
"
This implies that the segment connecting the origin to (p, q) is suitable. Therefore 2
D2 > p2 + q2 > p2 +
(
r2 - p2 - 1
)
- 2
p2 + 1
T2
r2 2
1 - (1+6)2 +
1 > (r - 1 -
1
(1 + E)2
)
322
3. SOLUTIONS TO THE PROBLEMS
that is,
Dr >
r - 1'1 -
1
(1 +
E)2
- 1, and for any sufficiently small r,
We conclude that for any 0 < E < we have 1
1
1
(1 + E)2
r
Since
1-
1
(1+E)2
-'0
as
e -+ 0,
the proposition follows immediately.
Problem G.43. We say that the point (al, a2, a3) is above (below) the point (b1, b2, b3) if a1 = b1, a2 = b2 and a3 > b3 (a3 < b3). Let e1i e2, ... , elk (k > 2) be pairwise skew lines not parallel with the z-axis, and assume that among their orthogonal projections to the (x, y)-plane no two are parallel and no three are concurrent. Is it possible that going along any of the lines the points that are below or above a point of some other line ei alternately follow one another? Solution. Assume that it is possible. From this, we derive a contradiction.
Let Pi,j be the point of ei, which is above or below the line ej (1 < i, j < 2k, i # j). In the first case, we call Pi,3 an upper point, and in the latter case we call it a lower point. In the text below, by a point of ei we mean one of the points Pi,j. Those two points of ei that have all points of ei on one side will be called the "ends" of ei. Two points of a line will be called neighbors if there is no point of the line between them.
Lemma. Every straight line contains at least two points that are above or below an end of another line. Proof. Obviously, it is enough to consider the straight line el. We may also assume that el is horizontal (that is, parallel with the (x, y)-plane), since this can always be managed applying a transformation of the form (x, y, z) --> (x, y, z + ay +,3z), which leaves the location of upper and lower
points unchanged. Let S be a plane perpendicular to e1, and denote by f2,f3,...,f2k the orthogonal projections of the lines e2ie3,...,e2k onto S (see Figure G.30). Let us introduce a coordinate system on S having horizontal and vertical axes and take from among f2, f3, ... , f2k the ones, say f2 and f3i whose slope is minimal and maximal, respectively. We claim that if e2 is below (above) e1, then each point of it lies on the negative (positive) side of el with respect to the positive direction of the horizontal coordinate axis. Assume that this is not so and, for example, e2 is below el and has a point on the positive side of el (the other cases can be treated analogously). Let Pi,,,2 be the nearest to P2,1 among these points. This
3.4 GEOMETRY
323
is an upper point since it is a neighbor of a lower point. The point P2,i0 below it lies on the straight line ei,,. It is easy to see that one can find a sequence of indices i1 , . , i,, such that 1. in = 1;
2. the points and are neighbors on ei, (j = 1, 2, ... , n - 1); is in the negative direction from 3. (j = 1, 2, ... , n - 1).
Figure G.30.
It is also easy to see, however, that in this case, the slope of one of the straight lines fi...... is less than the slope of f2, and this is a contradiction. This way, we proved that one of the ends of e2 and e3 is above or below el. Therefore, the lemma holds.
We remark that there cannot be more than two ends above or below a straight line, because it would mean that the 2k straight lines would have more than 4k ends. Therefore, there are exactly two points on every straight line that are above or below an end of another line, and these latter lines are the ones giving the minimal and maximal slopes after the above transformation and projection. Let the ends of el be above or below e2 and e3, respectively. Since the straight line contains 2k - 1, that is, an odd number of points, either both
e2 and e3 are above el or both of them are below el. We may assume that they are above el. We also suppose that e2 and e3 are horizontal since this can be reached by applying a suitable transformation of the form (x, y, z) H (x, y, z + ax +,3y). Let S be a plane perpendicular to e2, and introduce a coordinate system on S as shown in Figure G.31. By the previous remark, among the orthogonal projections of the straight lines ej(j # 2) onto S, the projection of el has maximal slope, el has maximal slope, since el goes below e2 and has a point in the positive direction from P1,2, and since the projection of e3 is horizontal, that of el has positive slope. This implies that P1,3 is higher than P1,2 (its z-coordinate is
3. SOLUTIONS TO THE PROBLEMS
324
el
P3 Figure G.31.
greater). Repeating the same argument but changing the role of e2 and e3, we obtain that P1,2 is higher than P1,3. However, these two implications contradict one another. We conclude that the indirect assumption is false: upper and lower points cannot follow one another alternating on each straight line.
Problem G.44. Suppose that HTM is a direct complement to a vertical bundle VTM over the total space of the tangent bundle TM of the manifold M. Let v and h denote the projections corresponding to the decomposition TTM = VTM ® HTM. Construct a bundle involution P : TTM -> TTM such that P o h = v o P and prove that, for any pseudoRiemannian metric given on the bundle VTM, there exists a unique metric connection V such that
VXPY-VYPX=Poh[X,Y] if X and Y are sections of the bundle HTM, and
VXY-VyX = [X, Y] if X and Y are sections of the bundle VTM.
Solution. As usual, call the sections of the bundles VTM --> TM and HTM -> TM vertical and horizontal vector fields on TM, and denote their modules by XvTM and . HTM, respectively. If X(TM) denotes the module of vector fields in TTM -+ TM, then
X(TM) = XvTM ® XHTM, and, for simplicity, the projections corresponding to this decomposition are also denoted by v and h.
3.4 GEOMETRY
325
Let (U, (ul, ... , un )) be a chart for M, and consider the induced chart
for TM, where 7r : TM --+ M is the projection of the tangent bundle, and
xi=uio7r, y'(v)=v(u') (vETM, 1
Nk:ir-1(U)--+R (1
(1 < i < n)
Nk aa
Sri
y
form a local basis in XHTM. Then a
6)
ay i' axi
(1
is a local basis in .A(TM), and on the coordinate patch 7r-1(U) every vector field X E .A(TM) can be written in the form
X Xi 6xi +Yi
y
i
where the first term is horizontal and the second is vertical. Define the map P by the formula Z
X
S2i
+ YZ ayi H YZ
Ski
+ Xx yi
.
An easy calculation using the above formulas shows that this map is independent of the choice of the chart. From the formula, it is immediate that p2 = 1 and P o h = v o P. It is also clear that P is a bundle involution.
Since .A(TM) _ XvTM ® . HTM, and a connection V :.A(TM) x . vTM ---+ XvTM is linear over the function ring in its first variable, it suffices to define V on the summands .XvTM x . vTM and XHTM x
.vTM.
Case 1. Consider the map F : (3vTM)3 - XvTM defined by F(X, Y, Z) = X (Y, Z) + Y(Z, X) - Z(X, Y)
- (X,[1',Z])+(Y,[Z,X])+(Z,[X,Y]), where (., .) denotes the pseudo-Riemannian metric given on the vertical bundle. For fixed X and Y, the map Z -4 F(X, Y, Z) is linear over the
326
3. SOLUTIONS TO THE PROBLEMS
function ring, and, since the form (. , .) is nondegenerate, there exists a unique VxY E XvTM such that (2VxY, Z) = F(X, Y, Z)
for all Z E .vTM. An argument analogous to the proof of the theorem on Levi-Civita connections shows that the map (X, Y) ---> VxY satisfies all requirements. Case 2. The restriction of the connection V to XHTM x XvTM is defined by the requirement that
(2VuY, Z) = U(Y, Z) + PY(Z, PU) - PZ(PU, Y) - (PU,Po h[PY,PZ]) + (Y,Po h[PZ, U])
+(Z,Poh[U,PY]) hold for all U E tHTM, Y, Z E XvTM. A simple but lengthy calculation shows that this restriction of the connection also has all required properties. In order to prove its uniqueness, suppose that the map
V' :.CHTM x XvTM -4 XvTM also fulfills the requirements of the problem. Then, for all U E .CHTM, Y, Z E XvTM, we have
U(Y, Z) + PY(Z, PU) - PZ(PU, Y) (1) (VUY, Z) + (Y, VUZ) + (V' Z, PU) + (Z, V PYPU) - (VZZPU, Y) - (PU, VPZY) =2(V' Y, Z) + (V' PU - VUY, Z) + (VUZ - VPZPU, Y) + (VPYZ - VPZY, PU) (?)
2(VUY, Z) - (P o h[U, PY], Z) - (P o h[PZ, U], Y) + (P o h[PY, PZ], PU),
where at (1) compatibility of V with the metric is used, and at (2) the assumption V'XPY - V',PX = P o h[X, Y] is used. Therefore, it follows that (V UY, Z) = (V UY,
Z),
that is, (VUY - VuY, Z) = 0
for all
Z E . vTM.
Since the form (., .) is nondegenerate, this implies that V = V, which completes the proof.
3.4 GEOMETRY
327
Problem G.45. Let A be a finite set of points in the Euclidean space of dimension d > 2. For j = 1, 2, ... , d, let Bj denote the orthogonal projection of A onto the (d-1)-dimensional subspace given by the equation
xj = 0. Prove that
d d
I Bj I> I AI d-1.
j=1
Solution. We use induction on the cardinality of the set A. If JAI = 0 or 1, then the statement is obvious. If JAI > 1, then the projection of A onto a suitable coordinate axis contains at least two distinct points. Without loss of generality, we can assume that this coordinate is the dth. This means that a suitable hyperplane with equation Xd = a divides A into two disjoint nonempty subsets: A = X U Y.
Let Cj and Dj denote the projections of X and Y, respectively, onto the hyperplane xj = 0; then ICj I + JDj I = IBj I for each j E {1, ... , d - 1}, and ICdI < IBdI and IDdI < JBdI. By the induction hypothesis, we have d-1
Ix Id-1
and
j=1
d-1
IYId-1 < IBdI
f IDj I.
j=1
Using the inequality 1/k
k
(al ..... ak) k + (bl ..... bk) k <
(H(ai+bi)) %=1
which is an equivalent form of the usual inequality between arithmetic and geometric means, we have
f d
d-1 IBjI11(d-1)
=
IBdI11(d-1)
II (ICj I + IDj
I)1/(d-1)
j=1
j=1
>
BdII/(d-1)
d-1
d-1
j=1
j=1
. r ICj I1/(d-1) + H IDj I d111
>- IXI + IYI = JAI,
that is, d
IB,I > IAId-1, j=1
which is what we wanted to prove.
3. SOLUTIONS TO THE PROBLEMS
328
Problem G.46. Let Al , ... , An be a sequence of n > 3 points in the Euclidean plane R2. Define the sequence A('),.. . An(') (i = 1, 2.... ) by ,
induction as follows: let A(z) be the midpoint of the segment A(i-)
=
A12-1).
where n+l Show that, with the exception of a set of zero Lebesgue measure, for every initial sequence (A(O), ... , A(°)) E (R 2)n, there
exists a natural number N such that the points A(lN), ... , An) are consecutive vertices of a convex n-gon.
Solution. If R' is naturally identified with the complex plane C, then our sequences can be considered as vectors in the space Cn, and the induction step corresponds to the linear transformation L(z1,...,zn) =
zl + z2 zn + zl 2 ,..., 2
In Cn there exists a basis consisting of eigenvectors of L: we have
L(ej)_Ajej for the vectors ej =
(j=1,2,...,n) (1,ej,e2j,...,e(n-,)j) and scalars Aj
= (1 + ej)/2
1,.. n) where e = e2"i/n is a primitive nth root of unity. All vectors v in Cn can be written in the form v = E; 1 ajej. Since en = (1, 1, ... ,1), adding the vector an en means a translation of our system of points in the original plane R2 by a fixed vector. Therefore, it suffices to consider the problem restricted to the subspace generated by the vectors e1,. .. , en-l- Suppose that n-1
(A10),
... , 4 0 ) ) = v = ajej. j=1
Applying the induction step N times, we obtain n-1 A1N),...,A,NI)
= LN(v) _ a3A ej. j=l
An elementary calculation shows that I AjI < IA' for 1 < j < n - 1, and IAn_1I = IA, I. So, for 1 < j < n - 1, we have IAjI N/IA,IN -> 0 when N -> 0. Therefore, it suffices to show that for almost all v the vector
ale, +an_len_, corresponds to successive vertices of a convex n-gon. This is indeed the case if Iai l : Jan-11, because e1 is a vector corresponding to successive vertices of a regular n-gon, and the vector ale1 + an_1en_1 corresponds to the sequence of points obtained from successive vertices of this regular n-gon by applying the linear map z - a1 z + an _ 1 z in R2. This transformation is nondegenerate, since a,z+an_12 = 0 can only hold when z = 0 or jai I = Jan-11, and thus it takes convex n-gons into convex n-gons. To summarize the property required in the problem is true for all vectors Enj=1 ajej of Cn except when la1I = lan-,I, and this exceptional set does indeed have zero Lebesgue measure.
3.4 GEOMETRY
329
Problem G.47. Suppose that n points are given on the unit circle so that the product of the distances of any point of the circle from these points is not greater than two. Prove that the points are the vertices of a regular n-gon.
Solution. Let the circle be the unit circle around the origin in the complex plane. Let z1, z2, ... , zn be the complex numbers that represent the points. By a suitable rotation, we may assume that z1 Z2 Zn = 1. Consider the following polynomial: P(W) _
(W-z1)(W-z2)...(W-Zn)
=wn+Q(w)+1. Then IP(z)I is the product of the distances of the point represented by the complex number z from the given points. So, if z is a complex number of absolute value 1, then IP(z) I < 2. Let w1i w2, ... , wn denote the nth roots of unity. It is well known that w1 + w2 + + Wk = 0 for all k = 1, 2, ... , n - 1. This implies that Q(wl) + + Q(Wn) = 0. If Q(w) is not identically zero, then, for some j, the complex number Q(wj) is different from zero and has nonnegative real part. Then P(wj) = 12 + Q(wj) l > 2, which contradicts the assumption. Therefore, the polynomial Q is identically zero, that is, P(Z) = zn+1. Then the roots z1i z2, ... , zn of the polynomial P(z) form a regular n-gon.
3. SOLUTIONS TO THE PROBLEMS
330
3.5 MEASURE THEORY Problem M.1. Let f be a finite real function of one variable. Let D f and D f be its upper and lower derivatives, respectively, that is, D f (x) = limsup h,k-.0
f (x+h) - f (x-k) f (x+h) - f (x-k) D f (x) = liminf h+k h+k h,k-.0 h, k> 0
h, k> 0
h+k>O
h+k>O
Show that D f and D f are Bore]-measurable functions. Solution. It is obviously sufficient to prove the assertion for D f . It is also clear that 00
00
{x: Df(x)>c}=.U n Aik Z=1 k_1 for all real values c, where Aik denotes the union of intervals [a, b] shorter
than 1/k satisfying (f (b) - f (a))/(b - a) > c + 1/i. We show that Aik is a set of type FQ which implies the assertion. More generally let {I.y} (ry c F) be a system of open intervals, and let A = U.yErL (where 7, stands for the closure of Ly). Obviously,
A=BuCuD, where B = U yErly, C is the set of all points that are left-hand endpoints of at least one Ly but are not interior points of any of them, and D is the set of those points that are right-hand endpoints of at least one I. but do not lie in the interior of any Ly. Clearly, B is an open set. If x E C, let rx be a rational point in the interior of the interval Ly that has the left-hand endpoint x. Points rx corresponding to distinct points x are distinct since
in the case x < y, rX = ry, the point y would lie in the interior of the interval I.y starting at x. Consequently, C, and similarly D, is countable. So each of the sets B, C, D, and therefore A as well is of type Fo.
Remark. It is interesting that for the left-hand and right-hand derivatives a similar theorem does not hold. For instance, if E is a nonmeasurable
set that along with its complementary set is dense, and f (x) is the char-
acteristic function of E, then D+ f (x) = 0 or +oo if x E E or x V E, respectively, and so D+ f (x) is nonmeasurable in this case. It should be noted that in the book by S. Saks (Theory of the Integral, Hafner, New York, 1937, Vitali's covering theorem, pp. 112-113), the author proves only Lebesgue measurability of D f and D f with much more powerful tools, although generalized to arbitrary dimension.
3.5 MEASURE THEORY
331
Problem M.2. Let E be a bounded subset of the real line, and let Sl be a system of (nondegenerate) closed intervals such that for each x E E there exists an I E 1 with left endpoint x. Show that for every e > 0 there exist a finite number of pairwise nonoverlapping intervals belonging to Sl that cover E with the exception of a subset of outer measure less than e. Solution. Let A be a bounded open interval containing E. Denote by Sn the set of all points of A having neighborhoods not containing points of E that are starting points of intervals belonging to Sl and longer than 1/n. Obviously Sn is an open set, Sn+1 C Sn and, furthermore, (n',Sn)fE = 0 since at each point of E there begins at least one nondegenerate interval belonging to I and if 1 is already less than the length of the latter, then n the point cannot belong to S,,. It is a well-known property of the outer measure that for any set A the outer measure A* (X fl A) as a function of X is a measure on the Lebesgue-measurable sets. Thus, taking A = E and using the relation Sn+1 C Sn as well as the finiteness of the outer measure of E, we obtain n
limo A* (Sn fl E) = A'(( _ 1 Sn) n E) = A* (0) = 0.
Fix no so that A* (Sno fl E) < e/2. If the set El = E - (Sno fl E) is nonempty, which we may assume, then in each neighborhoood of any of its points there is a point of E that is also a point of El and at which an interval longer than 1/no begins. There is a point a1 E El such that A*((-oo, al) n El) < 2(noA(A) + 1)' and by the former arguments we may assume at once that some interval [a1, b1] starting from al and belonging to I is longer than 1/no. Then we choose (if still possible) a point a2 E El such that a2 > b1, *((bl, a2) fl El) < 2(no\(0) + 1)' and
[a2, b2] E Il,
b2-a2>
1
no
and so on. The procedure ends after a finite number of steps, namely we can construct at most no.\(A) + 1 intervals [ak, bk] in this way (since bk - ak > 1/no). What is left out from E by these intervals is covered by Sno nE and the not more than no\(0)+1 sets (bk_1i ak)flEl (bo = -oo) of outer measure not exceeding e/(2(no\(0) + 1)). Their total outer measure is less than e, which completes the proof.
3. SOLUTIONS TO THE PROBLEMS
332
Remarks.
1. In addition to being nonoverlapping, the intervals we have chosen are disjoint.
2. The assertion cannot be replaced by the stronger statement that it is possible to choose an at most countable number of nonoverlapping intervals such that the part not covered by them has measure 0. This is shown by the following simple example: E = (0, 1), S2 = {[x,1] : x E (0, 1)}.
Problem M.3. Let f (t) be a continuous function on the interval 0 < t < 1, and define the two sets of points At = {(t,0): t E [0,1]},
Bt = {(f(t),1): t E [0,1]}.
Show that the union of all segments AtBt is Lebesgue-measurable, and find the minimum of its measure with respect to all functions f.
Solution. We first show that the set A = Uo Po. By the definition of A, to every n there is a to such that Pn E According to the Bolzano-Weierstrass theorem, the sequence to contains a convergent subsequence: tnk -> to. The distance from the point Pnk to the segment AtoBto is not greater than max{lf (tnk) - f (to)l, Itnk Atn Btn .
toI}, which tends to 0 by the continuity of f (t). It follows that Po E AtoBto C A. Since A contains the limit point of any convergent sequence of its points, A is closed.
A simple calculation shows that the point of the segment AtBt that lies on the straight line y = c has abscissa (1 - c)t + cf(t), which, by the continuity of f (t), is a continuous function of t. Consequently, if two points
of the set A lie on the line y = c, then A contains the segment that joins these points. Now we can determine the minimum of the measure of A. If f (t) is
constant, then A is a triangle of unit base and unit altitude, so it has measure 1/2. If the segments AoBo and A1B1 do not intersect, then the trapezoid with vertices Ao, Bo, A1, and B1 is a subset of A, so the measure of A is not less than 1/2. If the segments AoBo and A1B1 intersect at some
point C, then the triangles AOCA1 and B0CB1 are subsets of A, so the measure of A is not smaller than 1 1 d _ 11+d2 t(d) =
2 1+d'
where d denotes the distance from Bo to B1. By a simple calculation, we - 1. Thus, obtain that the minimum of t(d) on the positive half-axis is
the measure of A cannot be less than - 1. On the other hand, for - 1. f (t) = (v/2-- 1)(1 - t), the measure of A is exactly Remark. I. N. Berstein (Doklady Acad. Nauk. SSSR 146 (1962), 1113) refers to the result of the problem but he states erroneously that the minimum is 1 .
3.5 MEASURE THEORY
333
Problem M.4. A "letter T" erected at point A of the x-axis in the xyplane is the union of a segment AB in the upper half-plane perpendicular to the x-axis and a segment CD containing B in its interior and parallel to the x-axis. Show that it is impossible to erect a letter T at every point of the x-axis so that the union of those erected at rational points is disjoint from the union of those erected at irrational points.
Solution 1. We call width of the letter T erected at A the minimum of the lengths CB and BD. We devide the irrational points into countably many classes H1, ... , H,, .... The class Hk consists of all irrational points for which the widths of the letters T erected at them are greater than 1/k. Baire's theorem implies that the set of all irrational points cannot be represented as the countable union of nowhere-dense sets, so some Hk is dense in a suitable interval I. Selecting an arbitrary rational point A of the interval I, denote by b the width of the letter T erected at it. The class Hk contains an element Al whose distance from A is less than min{b,1/k}. The letters T erected at A and Al obviously intersect each other since they
both have widths greater than the distance from A to A1. The proof is complete.
Solution 2. Project on the x-axis the segment CD parallel to the x-axis of each letter T, and denote the projection by I. The projection II is an interval containing x in its interior. We prove the existence of an irrational
a and a rational r such that I,, and Ir contain r and a, respectively, in their interior. This already implies that the letters T erected at a and r intersect. For each rational number r = p/q (we shall write all rational numbers in irreducible form), we choose a closed subinterval Jr of I, containing r in its
interior and having length not greater than 1/q2. Consider a sequence of rational numbers where the denominators are strictly monotone increasing and each number lies in the intervals J, corresponding to the previous terms: P1 rl = -, r2 =
P2
ql
q2
,. .. , rn, _ Pn , ... ; qn
q1 < q2 < ... < qn, < ...; r,nEJrinJr2n...nJrn_1.
Such a sequence obviously exists and, moreover, it is a Cauchy sequence, so it has a limit limes-,* rn, = a. Since Jr, contains the points r7,+1, rn,+2, . . and is closed, it follows that a E Jr,a and, in particular,
Ia - r,I <
1
qn
From the theory of numbers it is well known that in this case a can only be irrational. I, contains a in its interior, so for sufficiently large n we have
334
3. SOLUTIONS TO THE PROBLEMS
rn E I,,. On the other hand, a E Jr, C Ir,. Thus, for sufficiently large n, the letters T erected at the points a and rn intersect each other. Solution 3. Assume that the assertion is false, that is, we have succeeded
in erecting a letter T at each point of the real line so that none of the letters T erected at rational points intersects a T erected at an irrational point. We may assume that the letters T are symmetric, that is, B is the midpoint of the segment CD (let C always have the smaller abscissa among
C and D). We define a monotonically increasing sequence of numbers al, a2.... an, ... by recursion. Let al be any real number. If an is already defined as a rational number, then let an+1 be irrational, while if an is irrational,
then let an+1 be rational, and choose an+l(> an) so that an+1 - an is smaller than one-quarter of the length of BnDn. Denote the abscissa of Dn
by bn. Making use of the fact that the letters T erected at an and an+1 do not intersect, we obtain the following sequence of inequalities:
an+1 < an+1 < an + 2(an+1 - an) < an + 2
bn - an 4
=
an + an 2
< 6n.
It follows that {an} is bounded, and so a = liman exists, and also that a < 6n for every n. Then, however, for sufficiently large n, the letter T erected at an intersects the T erected at a, which contradicts the initial assumption. Remarks.
1. In addition to the assertion of the problem, a contestant has proved that in any interval there is a continuum of irrational numbers a such that the letter T erected at a intersects a T erected at a rational point. Really, from the reasoning of Solution 2 it turns out that if r1, r2.... is a sequence of rational numbers such that Jrl 3 Jr2 D ... and ql < Q2 < .... then {rn} is obviously convergent and its limit is a "suitable" irrational number. Let r = p/q be a rational number belonging to the
interval (a, b) such that Jr C (a, b). Let ro = Po/qo and r1 = pl/ql be rational numbers satisfying qo > q, ql > q, Jro C Jr, Jr1 C Jr, and Jro n Jr1 = 0. The existence of ro and r1 of this kind is evident. Once we have defined the rational numbers ri1,,..,ik (i1 = 0,1; ... ; ik = 0,1), let ril,,..,ik,o and ril,,,,,ik,1 be two rational numbers such that gil,...,ik,o > qil,..., ik, JT'il...... k,0 C Jril...... k ,
gi1,...,ik,1 > gi1,...,ikI Jri1,..... k,1 C Jril...... k
Jril....,ik,0 n Jri1...... k'1 = 0.
With each number 0 < x < 1, we associate an irrational number a E (a, b) such that the letter T erected at a surely intersects a T erected at some rational point. If the infinite dyadic form of x is
x = 0.iJ2 ....
3.5 MEASURE THEORY
335
we associate with x the limit of the sequence
This will be a "suitable" irrational number, and the construction guarantees that a1 and a2 associated with distinct numbers x1 and x2 are distinct.
2. The union of a segment AB in the upper half-plane that is erected perpendicularly at point A of the x-axis and a segment BC that is parallel to the x-axis will be called a [ or a I depending on whether the
abscissa of C is greater than or less than that of B. Two contestants proved that if we erect a [ at each rational point and a I at each irrational point, then the union of the [ intersects the union of the 1. This assertion
is obviously sharper than that of the problem. Two other contestants showed by examples that it is possible to erect a [ at each point so that the I are pairwise disjoint. Indeed, if the points B of the [ lie on the graph of a strictly monotone decreasing positive function and the length of BC is arbitrary, then clearly any two [ are disjoint. 3. It is easy to see that the next theorem generalizes the statement of the problem: Let X be a complete metric space of second category, and let
X = P U Q, where P and Q are disjoint dense sets. Then it is impossible to define a real function f on X such that each point p E P has a neighborhood Vp for which x E Vp fl Q implies f (x) < f (p) and such that each point q E Q has a neighborhood Vq for which x E Vq fl P implies
f(x) < f(q)
This was proved by some contestants for complete metric spaces and compact metric spaces, respectively. It was also shown that the hypothesis of completeness cannot be dropped.
Problem M.S. Let f and g be continuous positive functions defined on the interval [0, oo), and let E C [0, oo) be a set of positive measure. Prove that the range of the function defined on E x E by the relation X
F(x, y) =
J0
y
f (t)dt + fo g(t)dt
h as a nonvoid interior.
Solution. Put O(x) =
LX
f(t)dt,
fi(x) =
J0
g(t)dt.
The function 0 and its inverse 0-1 (which obviously exists) are absolutely continuous, so they map sets of measure 0 onto sets of measure 0. Therefore, the image O(E) = A of the set E of positive measure is measurable and has positive measure. Similarly, the set O(E) = B has positive measure. Since the range of the function F(x, y) considered on E x E is exactly
336
3. SOLUTIONS TO THE PROBLEMS
the set A+ B = {a + b : a E A, b E B}, it is sufficient to prove that for sets A and B of positive measure, the sum A + B contains some interval. Let u and v be points of density 1 of A and B, respectively, and let e > 0 be chosen so that the relations 0 < 6 < 2e, 0 < 6' < 2E, 6 + 6' > 0 imply
p(An [u-6,u+6']) > 2(b+6'), p(Bn [v -6,v+6']) > 2(6+6'), where p stands for Lebesgue measure. We show that (u+v-E,u+v+E) C
A+B. Let t E (u+v-E,u+v+E) and
A*=t-A={t-a: aEA}. Then A* n [t - u, v + E] is congruent with A n [t - v - E, u], so
p(A* n [t-u,v+E]) > 2(u+v+E-t)
and
p(Bn [t-u,v+E]) > 2(u+v+E-t), from which it follows that A* n B # 0. Let x E A* n B; then t - x E A, x E B, and, consequently, t = (t - x) + x c A + B. Remarks.
1. Two contestants showed that it is sufficient to assume that f and g are positive functions integrable in every finite interval. 2. A contestant remarked that if we know of E only that it has positive outer measure, then the statement becomes false.
3. Denote by Al and B1 the set of all points of density 1 of A and B, respectively. If u c Al and v E B1, then by the density theorem of Lebesgue u and v have density 1 also in Al and B1, respectively, so by the solution above, Al + B1 contains some neighborhood of u + v. This means exactly that Al + B1 is an open set (see J. B. H. Kemperman, A general functional equation, Trans. Am. Math. Soc. 86 (1957), 28-56, Theorem 2.2).
Problem M.6. In n-dimensional Euclidean space, the union of any set of closed balls (of positive radii) is measurable in the sense of Lebesgue. Solution. Let {Ga} (a E A) be an arbitrary set of closed balls of positive radii. Put H = UQEAGO. Denote by G the set of all closed balls (of positive radii) that are contained in some Ga. Clearly, the elements of G constitute a cover of H in the sense of Vitali. Therefore, by Vitali's covering theorem (see for example S. Saks, Theory of the Integral, Hafner, New York, 1937,
p. 109), we can choose an at most countable number of pairwise disjoint balls S1, S2,... belonging to G such that H \ UiSi has Lebesgue measure 0. Since Si C H, the relation
H=
(H\us)u(us)
implies that H is Lebesgue-measurable.
3.5 MEASURE THEORY
337
Remarks.
1. Most contestants solved the problem with the help of Vitali's covering theorem.
2. Several contestants remarked that H can be obtained as the union of at most countably many Jordan measurable sets. 3. One contestant proved the following generalization of the problem: The union of any set of convex sets with nonvoid interiors is measurable in the sense of Lebesgue.
Problem M.7. In n-dimensional Euclidean space, the square of the two-dimensional Lebesgue measure of a bounded, closed, (two-dimensional) planar set is equal to the sum of the squares of the measures of the orthogonal projections of the given set on the n-coordinate hyperplanes.
Solution 1. We solve the problem in the generalized form where a bounded, closed subset of H or an r-dimensional subspace L is projected to the r-dimensional coordinate subspaces. H is the difference of two bounded open sets N2 E N1. Each of them is the union of an at most countably infinite number of disjoint parallelepipeds t1i and t2j. Their images under a linear transformation T (the orthogonal projection) are parallelepipeds whose volume (Lebesgue measures, denoted by A) satisfies
A(Ttli) = c)1(tli)
and
A(Tt27) = cA(t2.7),
where c is a constant depending on the position of the two subspaces and the direction of projecting only. Further, since a projection takes the union and the difference of sets into the union and the difference, respectively, of the image sets, from the additivity of the measure it follows that TH is
measurable and .(TH) = c.(H). If Til...i,, is the orthogonal projection to the coordinate subspace [xi1, ... ,
xi,.] and cil...i,, denotes the corresponding constant, then the sum of the squared measures of the projections is A2(Ti1...i,.H)
_
ci1...i,.A2(H) = \2(H)
Ci1...i,.,
(1)
where the summation is extended to all combinations i 1, ... , i, of order r of the numbers 1, 2, ... , n. If e1,.. . , en are unit vectors of a Cartesian system of coordinates in the n-dimensional Euclidean space and H is the r-dimensional cube spanned by the orthogonal unit vectors n
uk = E ai k)ei i=1
(k = 1, ... , r)
3. SOLUTIONS TO THE PROBLEMS
338
belonging to L, then ).(H) = 1, and the value cil...i,. _ A(Ti1...irH) is equal to the determinant IAi1,..ir I formed of the columns i1 i . . . , it of the matrix A = {ack)}. Then, however, by the Binet-Cauchy formula, c21... it
j
=
IAil... it I2 = IAA* I
= 1,
which, owing to (1), verifies our assertion.
Solution 2. We prove the assertion in the previous generalized form where a (not necessarily compact) Lebesgue-measurable set H lying in the r-dimensional subspace L is projected to the r-dimensional coordinate subspaces.
Let e1, ... , en be the orthonormal basis vectors of a Cartesian system of coordinates in the n-dimensional Euclidean space En, and u1,. .. , ur be the orthonormal basis vectors of a system of coordinates in the subspace L. We may assume that L contains the origin since this can be achieved
by translation, and translation means translation of the projections, so the measure of H and the measures of its projections remain unchanged. Consequently, we may choose n
ai
uk =
k
(k = 1,...,r),
ei
i=1
satisfies IAA* I = 1. where the matrix A = Denote by Ail...ir the operator of orthogonal projection from L to the coordinate subspace [xil, ... , xir] of En. The operator Ail...i,, sends the basis vectors Uk to the vectors of the latter subspace with components Let (As,...i,,) be the matrix of this projection operator (that
is, the square matrix formed of the columns i1 i ... , i, of A), and let IAi1...ir I
be the determinant of this matrix. The Lebesgue measure of H is given by
IL = fXHdA, L
where XH denotes the characteristic function of H. By the formula for transformation of integrals, the measure of Ail ...i,.H is
f
XAi1...i,.H dA = IAil...ir I I XH dA = IAil...ir 11L-
(This is valid also when IAil...ir1 = 0, since in this case Ai,...i,.H lies in an at most (r -1)-dimensional subspace, that is, a subspace of measure 0 with
respect to A.) Summing for all combinations i1, ... , i,. of order r, by the Binet-Cauchy formula we obtain {
fXA1...I,.HdA}2 = /L2
[x,l.... ,xirl
which proves the assertion.
IAil...ir
I2
= µ2 IAA.I = /U2,
3.5 MEASURE THEORY
339
Problem M.B. Let us use the word N-measure for nonnegative, finitely additive set functions defined on all subsets of the positive integers, equal to 0 on finite sets, and equal to 1 on the whole set. We say that the system 2l of sets determines the N-measure p if any N-measure coinciding with p on all elements of '21 is necessarily identical with p. Prove the existence of an N-measure p that cannot be determined by a system of cardinality less than continuum.
Solution. We shall need a definition and a lemma.
Definition. We say that the system A consisting of certain subsets of a set is independent, if for any distinct members X1i ... , Xj, Y1, ... , Y,, of A, the set is infinite (here U denotes the complement of U).
Lemma. Any countable set has an independent system of subsets that is of power continuum.
Proof. Indeed, take the following set for the countable basic set. Let
M be the set of all sets of real numbers that can be obtained as finite unions of intervals belonging to any type of closedness but having rational endpoints. Clearly, there is only a countable number of such sets. Denote by Ma the set of those sets of real numbers described above that contain the real number a. Then the system {Ma} of subsets of M, for a running
through the set of all real numbers, has power continuum. It is easy to see that {Ma} is independent. Really, for given real numbers al,... , aj, bl,... , bk of A, the set Ma,
nMb, n...nMbk
consists of those elements of M, that is, those sets of real numbers described above, that contain a1, . . . , aj but do not contain b1, ... , bk. The set of these sets, however, is countably infinite.
Next we construct an N-measure that takes the values 0 and 1 only. To this end, we shall need Zorn's lemma, so our proof is not purely constructive.
A system of subsets of the set N of all real numbers is called a filter if it is closed for taking finite intersections and if together with any set it contains all larger sets as well. To exclude the filter consisting of all subsets,
we also require that a filter should not contain the empty set. Since the union of an increasing chain of filters is again a filter, to any filter with the help of Zorn's lemma we can find a maximal filter containing it. This is called an ultrafilter. If a filter contains neither the set A nor q, then it can be extended by A. If it contained both, then it would also contain their intersection, that is, the empty set. Therefore, an ultrafilter contains exactly one of the sets A and A. If we now associate with an ultrafilter
340
3. SOLUTIONS TO THE PROBLEMS
M the N-measure p that takes the value 1 on elements of M and 0 on their complements, then we obtain a set function with values 0 and 1 that is defined on all subsets of N and is additive. Additiveness could fail only if neither of the sets has measure 0, but this is impossible: as the empty set does not belong to our ultrafilter, two disjoint sets of measure 1 do
not exist. We shall prove for a measure of this kind that it cannot be determined by less than a continuum number of sets. From the subsets of N, in the manner described in the lemma, we construct an independent system of power continuum, and denote its elements by Na,. If c is an infinite sequence consisting of distinct real numbers, we denote by S, the union of the complements of the corresponding sets Na,. The filter generated by the Na,, the S,, and the complements of finite sets consists of all sets containing finite intersections of these. To see that the latter is really a filter, we have to show that it does not contain the empty set. In the opposite case, we could find al, . . . , am and cl, . . . , c,,, so that the set
x =Nal n...nNam nscl n...nSc is finite. Here ci is an infinite sequence of distinct real numbers, and consequently we can choose real numbers bl, . . . , b,. from the elements of cl, ... , c,, so that al, ... , am, and bl, . . . , b,, are distinct. Independence implies that the set
Nal n...nNam nNbl n...nNb =Y is infinite. On the other hand, the definition of the Sc yields Y C X. This contradicts our assumption. It follows that the system of sets considered does not contain the empty set; hence it is a filter. We extend this filter to an ultrafilter M and define the N-measure p with the help of the latter. We have to prove that /.c cannot be determined by a system 2i of power less than continuum. If % is a determining system containing, among others, sets of measure 0, then these can be replaced by their complements: it is sufficient to consider determining systems 2i all of whose elements belong to M. Taking a % of this kind, we assume that it contains less than a continuum of sets. If we consider the finite intersections of the sets in 2i, their power is still less than continuum. Therefore, we may assume from the beginning that 2i is closed for taking finite intersections. If there were a Z E M that contained no element of 2i, then Z would not be disjoint from any set in % and, consequently, the sets containing sets of the form ZnA (A E 2() would constitute a filter. This filter could be extended to an ultrafilter, and the latter would define an N-measure v. Then µ and v would coincide on the elements of %, whereas µ(Z) = 0 and v(Z) = 1. This is impossible since 2( is a determining system. So each element of M contains a set from 2i. Then at least one element W of 2t is contained in infinitely many elements of {Na}; let c be a countable sequence of the corresponding subscripts. The set Na is disjoint from W for each element of c, from which it follows that Sc = UNa is also disjoint from W. But this means that M contains
3.5 MEASURE THEORY
341
two disjoint sets, which contradicts our assumption. The contradiction has
been caused by the assumption that p can be determined by less than a continuum of sets. For the completion of the proof, it remains to show that p vanishes on every finite set. This is evident since p(D) = 1 whenever D is the complement of a finite set. Remarks.
1. If the continuum hypothesis is assumed, the problem becomes trivial. In fact, it is easy to prove that a countable set cannot determine an N-measure. 2. It would be a sharpening of the problem to prove that no N-measure can be determined by less than a continuum number of sets. This, however, is undecidable within the usual Zermelo-Fraenkel system of axioms for set theory.
3. The problem can be generalized to other cardinal numbers. For any cardinal m, there exists a finitely additive 0-1 measure that vanishes on all sets of power less than m and cannot be generated by less than 2' subsets. To prove this, it is sufficient to give an adequate generalization of the lemma. One contestant proved this generalization.
4. One may raise the question of whether there are relatively many or relatively few N-measures for22m the determination of which 2' subsets are N-measures, and there are equally so needed. There are altogether
many N-measures that cannot be determined by less than 2' sets. In the original case where the basic set is countable, one contestant proved this.
Problem M.9. Let {q5 (x)} be a sequence of functions belonging to L2(0,1) and having norm less than 1 such that for any subsequence {0nk (x)} the measure of the set {x E (0,1) :
1
N
> 0nk (x)
y}
tends to 0 as y and N tend to infinity. Prove that ¢n tends to 0 weakly i the function space L2(0,1).
Solution. Suppose there is a function f E L2(0,1) such that II
0.(x)f(x)dx
J
does not tend to 0 as n -> oo. Then in view of the relations
10.(x)f(x) dxl J1 0
Il0nll llf II
<_ 11111
(n =1, 2, ... )
3. SOLUTIONS TO THE PROBLEMS
342
(replacing, if necessary, f (x) by e"" f (x) with a suitable 0), there is an a > 0 and a subsequence {¢flk(x)}k 1 such that r1
ReJ 0lk(x)f(x)dx>a 0
for every k (I If II stands for the norm of the function f (x) in L2(0, 1)). Obviously, to every e > 0 there is 6 > 0 such that the relations m(A) < 6, A C (0, 1) imply IIXAf II2 < E (here m(A) denotes the measure of the set A, and XA(x) denotes the characteristic function of A). Let E be a positive
number less than a, let 6 be a positive value corresponding to e in the sense just indicated, and choose y(E) and N(E) so that for N > N(E) the measure of the set
E = xE(0,1):
N
N E 0-k (x) k=1
is smaller than 6. Then, for N > N(E), it follows that N
1
Ref
Na <
f (x)0.,, (x) dx
0
k=1 r
N
E
k=1
N
r
=Ref f(x)E0..k(x)dx+Re>J f(x)0.k(x)dx J If (x) I
k=1 E
E
0.k
xdx +
< vy(E)IIf II2 + NIIXEf II2 <
1: fE I f(x)Onk (x) I dx
VrN-y(e)IIf
II2 + NE,
that is, V-N-(a - e) < y(e) IIf 112, which is impossible for sufficiently large N
since a - e > 0. Consequently, 1
I
0.(x) f (x) dx - 0
(n
00)
for all functions f E L2(0,1), as stated. Remarks. Essentially all acceptable solutions have followed this way. 1. Two contestants noted that it is sufficient to require the following: for any subsequence {(pf,k (x)} and any number c > 0, the measure of the set
N
xE(0,1):
0-k (x) k=1
tends to0asN--roc. 2. One contestant showed by a counterexample that without boundedness
of the sequence {0.(x)} in the norm of L2(0,1) the assertion is not
3.5 MEASURE THEORY
343
always true. He also mentioned that the hypotheses of the problem do not imply strong convergence. 3. Another contestant pointed out that it is sufficient to make the following assumption: to any subsequence {nk} of the natural numbers, there is a function f (N) 54 0 for which f (N)/N -> 0 as N -> oo and the measure of the set N
X E (0,1) :
1 k-i E 0nk (x) > y
Al
)
tends to 0 as y -* oo, N --> oc. He also noted that the sequence {0",(x)} does not necessarily converge either in measure or in mean. .4. A contestant observed that the hypothesis may be weakened as follows: from any subsequence of the sequence {0,,,(x)} one can choose a subsequence {cbl,.(x)} for which the measure of set (1) tends to 0 as y --> oc, N -* oo. 5. One contestant called attention to the fact that if one assumes boundedness of the sequence {0, (x)} in the norm of LP(0,1), where 1 < p < oo, then it can be shown that, under the hypothesis of the problem,
i
1
0,,.(x) f (x) dx --* 0
(n -> oc)
follows for all functions f (x) E L9(0,1), where 1/p + 1/q = 1. 6. Four contestants treated the problem in the complex case.
Problem M.10. We say that the real-valued function f (x) defined on the interval (0, 1) is approximately continuous on (0, 1) if for any xo E (0, 1)
and e > 0 the point xo is a point of interior density 1 of the set
H= {x : If (x) - f (xo) I< E}. Let F C (0, 1) be a countable closed set, and g(x) a real-valued function defined on F. Prove the existence of an approximately continuous function f (x) defined on (0, 1) such that f (x) = g(x)
for all x E F.
Solution 1. Let F = {c1, C2.... }. Denote by I(x, n) the closed interval of center x and radius 1/n. By the "central p-multiple" (0 <_ p < 1) of the interval (a, b) we shall mean the closed interval
[a+(b_a)(1_P),b_ 2(b-a)(1-p)] We shall define a system Jn of intervals by recursion so that it has the following properties. J,, consists of countably many disjoint closed intervals, none of which intersects F. Further, c, is the only accumulation point of
344
3. SOLUTIONS TO THE PROBLEMS
the set of endpoints of the intervals forming Jn and, finally, K,,,, and Kn are
disjoint if m # n, where Kn stands for the union of the intervals forming Jn. Suppose that for n' < n we have already defined J,,, so that they have the properties required. Then the closure of the set U'-i Ki = Ln is Ln U {cl,...,cn_l},
and it does not contain cn. Therefore, it does not intersect a suitable neighborhood I (cn, rn) of cn either. We may assume that rn > n. For each
j (j = 1, 2, ... ), consider the interior of the set I (cn, rn + j) \ I (cn, rn + j + 1) \ F. This is the union of a (finite or) countable number of open intervals. Obviously, the total length of these open intervals is equal to the measure of the set I (cn, rn +j) \ I (cn, rn +i + 1) = M (n, j ), so the union of a finite number of them has at least (1-1/ j) times this measure. Take the
central (1 - 1/j)-multiples of these finitely many intervals, and let these new intervals constitute the system J,3. Put Jn = U _ l Jn, j . Clearly, the systems Jn of intervals so obtained possess the required properties (in particular, the recursion is correct). Further, the intervals
forming Jn,j fill at least (1 - 1/j)2 times the measure of M(n,j) and, therefore, Kn, the union of the intervals forming Jn, has density 1 at cn. Define f in the following way. Let f (x) = g(cn) if x E Kn or x = cn. If we prove that the set U°°_1KnUF is closed, then f can be defined at points not belonging to this set so that on the countable number of disjoint open intervals forming the complement of this closed set, f is linear, and at the endpoints of the intervals it takes the values already fixed. It is obvious that the function f defined in this way is approximately continuous in points of F and continuous in all other points. To see the latter, observe that for x V F the distance from x to F is positive, say greater than 1/m, where
m is a positive integer. Thus, by the assumption rn > n, Kn does not intersect the 1/2m-neighborhood of x if n > 2m. Consequently,
I(x, 2m) fl (U Kn U F) = I(x, 2m) fl (U Kn). n=1
n=1
The right-hand side, however, is the union of a finite number of closed intervals. This proves, on the one hand, that U°_iKn U F is closed and, on the other hand, that the restriction to I(x, 2m) of the function f constructed with the help of this set is a continuous polygon function.
Solution 2. Let F = {rl, r2.... }. We start from the idea that the closer x is to ri, the more f (x) should "feel" the value taken at rti. We try to achieve this by a definition of the following type:
f(x) -_
u( )lid *:
Lti=1 u(i)Ix-r;
if x
F.
(1)
3.5 MEASURE THEORY
345
We now impose various conditions on the order of u(i) to ensure convergence of this series and approximate continuity of the function obtained, and finally we show that these conditions can be fulfilled. The series will always be convergent if, for instance, (2) u(i) > i2lg(ri)I and u(i) > i2 (i = 1, 2, ... ). Really, for x 0 F, the closedness of F gives mini I x - riI > 0. f will be continuous in all points of (0, 1) \ F (in a small neighborhood, both the numerator and the denominator are sums of uniformly convergent series of continuous functions), hence it is also approximatively continuous there.
Consider therefore an ri. Let d = minj
E
9(rj)
= 0(1) (3)
j
j d - 6/2 > d/2). We divide the values x into two classes. One consists of those x that satisfy the relation 1 + I9(rj) I <
u(j)Ix - rjI
(4)
1
i2
for all j > i. For these x, by (3) and (4), u()Ix-r:l +O(1) g(ri)+O(Ix-riI) 1+0(Ix-riI) u(05-r+l +O(1) which lies in the e-neighborhood of g(ri) if 6 is sufficiently small. Thus, we have to achieve that the measure of such x in (ri - 6, ri + 6) be (2 + o(1))b. If, however, (4) does not hold, then
f(x) =
ix - rj l < (j) (1 + I9(rJ)I) = Ej. The measure of such "bad" x in (ri - 6, ri + 6) is <2
Ej. Ir i -r9 I
If now Ej < Iri - rjl2/j2, then this measure is j2 I2 Iri -rj < 2(46)2
<2 z
I r; -rj l <6+
3
j2 1
= 0(62) = o(6)
as 6 - 0 (since Iri - rj l2/j2 < Iri - rj I/4 and so (3/4). Iri - rj l < 6). Consequently, together with (2), we have only a finite number of conditions for each j, so they can be realized.
3. SOLUTIONS TO THE PROBLEMS
346
Problem M.11. Let {fn}°D 0 be a uniformly bounded sequence of realvalued measurable functions defined on [0, 1] satisfying
ff2 l
Further, let {cn} be a sequence of real numbers with 00
cn = +00. n=0
Prove that some re-arrangement of the series E°° 0 cn fn is divergent on a set of positive measure.
Solution. We first show that 00
2 f2
(1)
Cn n
n=0
is divergent on a set of positive measure. Suppose that (1) yields a function g that is finite a.e. By Egorov's theorem there is a set A C_ [0, 1] on which
the series (1) is uniformly convergent and for which .\(A) > 1 - (/2K), where K is a common upper bound of the functions fn. Then, by the uniform convergence, the limiting function g(t) is integrable on A, and Co
c cn n=0
JA
fn2(t) dt =
JA
g(t) dt < +oo.
(2)
On the other hand,
JAfn(t) dt = Jo 1 f,2, (t) dt - 1,0,11\A fi(t) dt > 1 - K2K = 2 which, in view of (2), yields 1
°O
21: cn
a contradiction.
As a second step we show that the series E° 0 cn fn has a subseries divergent on a set of positive measure. Let e1, ... , En.... be independent random variables on some field (1, A, P) taking the values 0, 1 with probability 1/2, and let 0 < t < 1 be a number for which (1) is divergent. Then for the random variables Encn fn(t) we have Var (Encnfn(t)) = n=0
E n=0
+oo
3.5 MEASURE THEORY
347
and therefore, by Kolmogorov's three-series theorem, E°° o encn fn(t) is divergent with positive probability. Thus, the set of all elements (w, t) in the probability field (S2, A, P) x [0, 1] for which 00
1: (en(w)Cnfn(t)
(3)
n=0
is divergent has positive measure. Consequently, there exists w E f for which the series (3) is divergent on a set of positive measure. It follows that there is a set D of positive measure and an infinite sequence n1 < n2 < ... such that cc
1: Cnk fnk (t)
(4)
n=0
is divergent for all t E D; so there exists b(t) > 0 such that for arbitrarily large indices the partial sums of (4) have oscillation not less than b(t). Define next the numbers N1 = 1 < N2 < ... as follows. Let the number Nk+1 be such that the measure of the set FL
Dk
f ED:
max
Nk
Cnifni(t)
8=v
is greater than (1 - (1/2k+1)) , A(D). Such an Nk+1 exists since for sufficiently large Nk+l, all values t E D belong to Dk. The set
D* = fl Dk k=1
has a positive measure, and >
max
Nk
Cnifni(t)
b(t)
2
i=v
if t E D* and k > 1. Therefore, any rearrangement of the original series >°°=o cn fn, for which the sums r_NNk cni f,,, consist of consecutive terms, is divergent at all points t E D*.
Problem M.12. Let { fn} be a sequence of Lebesgue-integrable functions on [0, 1] such that for any Lebesgue-measurable subset E of [0, 1] the sequence fE fn is convergent. Assume also that limn fn = f exists almost
everywhere. Prove that f is integrable and fE f = limn fE fn. Is the assertion also true if E runs only over intervals but we also assume fn > 0? What happens if [0, 1] is replaced by [0, oo)?
Solution. Let E _ [0, 1] or E = [0, oo), and E = U' 1Ek, where Ek is a measurable set of finite measure and on Ek the convergence of the sequence
348
3. SOLUTIONS TO THE PROBLEMS
{fn} is uniform. By Egorov's theorem, such a sequence {Ek} obviously exists. It is clear that fEk f exists for every k, and fF f = limn fF fn if F is a measurable subset of Ek. The formula p(G) = limn fG fn defines a finitely additive signed measure
on measurable subsets G of E. The theorem of Beppo-Levi shows that f together with the relation µ(G) = IG f for each the existence of JE measurable subset G of E are equivalent to the a-additivity of P. But the latter follows from the well-known fact that the limit of a pointwise convergent sequence of finite signed measures is a finite signed measure (in
particular, it is a-additive; see P. R. Halmos, Measure Theory, Springer, New York, 1974, p. 170, relation (14)). It can also be seen that fn -
JE
--+ 0 (n - oo). Thus, we have answered the first and third questions of the problem. The answer to the second question is negative, as shown by the example fn(x) = n if 0 < x < 1, and fn(x) = 0 if 1 <X<1. fI
n
n
Problem M.13. Let 0 < c < 1, and let q denote the order type of the set of rational numbers. Assume that with every rational number r we associate a Lebesgue-measurable subset H,. of measure c of the interval [0, 1]. Prove the existence of a Lebesgue-measurable set H C [0, 1] of measure c such that for every x E H the set
{r:xEHr} contains a subset of type q.
Solution. We give two solutions. Both make use of the following simple lemma:
Lemma. Let M be a system of sets that consists of certain subsets of the interval (0, 1), and suppose that for each A E M there is an x E A such that A n (0, x) E M and A n (x, 1) E M. Then every element of M has a subset of type q. Proof. Arrange the rational points of (0, 1) in a sequence Ti, r2, .... Let A E M be arbitrary. We shall define by recursion a sequence of points xn E A having the following properties: 1. It is ordered in the same way as the sequence {rn}; 2. A n (0, xn) E M, A n (xn, 1) E M for every n, and A n (xi, xj) E M for every pair xi < xj.
By assumption, there is an x1 E A such that A n (0,x1) E M and An(xi,1) E M. Next suppose that X1, X 2 ,-- . , xn have already been chosen, and let the immediate neighbors of rn+1 from among r1i r2, ... , rn be ri and
rj, ri < ri. By the induction hypothesis, we have xi < x; and An(xi, xj) E M , so there is an xn+1 E A n (xi, xj) such that A n (xi, xn+1) E M and A n (xn+1, xi) E M. A similar choice of xn+1 is possible in the cases where rn+1 > ri (i < n) or rn+1 < ri (i < n). The proof of the lemma is complete.
3.5 MEASURE THEORY
349
Solution 1. We need the following auxiliary theorem:
Auxiliary theorem. Let the sets Hn E [0, 1] be measurable and have measure not less than c (n = 1, 2.... ). Then the set H of all points x for which the sequence In : x E Hn} has positive upper density is also measurable and has measure not less than c.
Proof. Denote by fn the characteristic function of Hn, and put 1
gn = 1 (f1 + f 2 + ... + fn).
Obviously, x E H if and only if g, ,(x) does not tend to 0 as n -> oo, which yields the measurability of H. Since 0 < gn(x) < 1 for every n, we have
A(H) >
f
f,1]\H
1
g. (x) dx = fo g. (x) dx -
H
=c-
0
f
[0,1]\H
J[0
gn(x) dx
gn(x)dx.
Now x E [0, 1] \ H implies limn-oo gn(x) = 0, so by Lebesgue's theorem on the passage to the limit under the integral sign, lim
f
g (x) dx = 0.
o,1]\H
Thus, from the previous inequality, we obtain .(H) > c, which proves the auxiliary theorem. Let {rn} be an arrangement of the rational points of (0, 1) in a sequence
of uniform distribution, that is, for any subinterval (a, b) E (0, 1), the density of the sequence In : rn E (a, b)} is b - a. (A simple example of an arrangement of this kind is the following. We first fix an arbitrary arrangement in a sequence {sn}, then from {sn} we consecutively choose the elements of minimal index lying respectively in the intervals [0,1], 0,
1
12
,
2,1
[0,1
n
1
,
2
n, 1n
,...,
n-1 ,1 ,..., n
meanwhile, we should make sure that each element is selected only once.) Let A E (0, 1), and assume that the sequence In : rn E Al has positive upper measure. We show that A contains a subset of type g. By the lemma we have presented in advance, it is sufficient to prove the existence of an
x E A such that both In : rn E A fl (0, x)} and In : rn E A n (x,1)} have positive upper density.
Denote by d the upper density of the sequence in : rn E Al, and let < xm = 1 be a subdivision of [0, 1] finer than d/3. Since {rn} is uniformly distributed, the upper densities of the sequences in : rn E A n (xi_1i xi)} are smaller than d/3 for every i. Consequently,
0 = xo < x1 <
350
3. SOLUTIONS TO THE PROBLEMS
denoting by di the upper density of the sequence in : rn E A n (0, xi)}, we
have 0 < di - di_1 < d/3 for i = 1, 2, ... , m. Since d,n = d, there is an i with 0 < di_1 < di < d. Then di_1 < di implies that the set A n (xi_1, xi) is infinite, and it is easy to see that for any element x E An(xi_1, xi)}, the sequences in : rn E A n (0, x)} and In : rn E A n (x, 1)} also have positive upper measures.
Now consider the sets Hr appearing in the statement, and let H1 be the set of those x for which the sequence in : x E Hr,, } has positive upper measure. By the previous observation, if x E H1, then {r x E Hr} contains a subset of type q. Applying our auxiliary theorem to the :
sequence of the sets Hr,,, it follows that H1 is measurable and ) (H1) > c, as required. Solution 2. If the closure of a set A has positive measure, then A contains a subset of type i since the system M = {A C (0, 1) : \(A) > 0} obviously satisfies the condition of the lemma. It is therefore sufficient to prove that the set
H2={x: A({r:xEHr})>0} is measurable and \(H2) > c. Consider the function f (x) = A({r : x E Hr}). Clearly, 0 < f (x) < 1; if we show that f (x) is Lebesgue-measurable and fo f (x) dx > c, then we shall be done. For any A E [0,1], put
An =
U 2
2-1
i
2n
, 2n
2W
Obviously, Al D A2 D ... and n= nlAn = A; hence limn . a(An) = A(A)So let fn(x) = \({r : x E H,.}n); then fn f, and therefore we need only to show that fn (x) is Lebesgue-integrable with fo fn (x) dx > c. We have the relation 1
,\(An) =
2^
9n,i (A), 2n E i=1
where gn,i(A) = 1 or 0 depending on whether A has or does not have a point in [(i - 1)/2n, i/2n]. Clearly,
fn,i(x):=gn,i({r: xEHr})= sup{k,.: rE I
i-1
i 11
2n
2n J 1
,
where k, is the characteristic function of Hr. The function k, is integrable and fo kr(x) dx = c, so fn,i is integrable and ff fn,i(x) dx > c. Since fn = (1/2n) Ei= fn i, it follows that fn is integrable and ff fn(x) dx > c. Remark. Denote by Ho the set of those x for which the set {r : x E Hr} contains a subset of type 77. Let {rn} be a fixed arrangement of the rational
3.5 MEASURE THEORY
351
points of (0, 1) in a uniformly distributed sequence, and denote by H1 the set of those x for which the sequence {n : x E Hr,, } has positive upper density. Let H2 be the set of those x for which the closure of the set
{r : x E H,. } has positive measure. Finally, let H3 be the set of those x for which {r : x E Hr} is dense in a subinterval of (0, 1). It is easy to see that for any set A of rational numbers, the upper density of the sequence in : r,,, E Al is at most .(A). Therefore, H3 C H1 C H2 C Ho holds for any system Hr (thus Solution 1 proves a deeper assertion than Solution 2, since it shows from a narrower set that its measure is at least c). On the other hand, the statement .(H3) > c is false: it can be shown that, for any 0 < c < 1, there exists a system Hr satisfying the conditions of the problem and such that the set {r : x E H,.} is nowhere dense for every x E [0, 1], in particular, H3 = 0. We also note that if the sets Hr are measurable, then so is Ho. Actually, it can be proved that if the Hr are Borel (or, more generally, analytic) sets, then Ho is analytic, hence measurable. From this it follows easily that Ho is also measurable in this general case.
Problem M.14. Find a perfect set H C [0, 1] of positive measure and a continuous function f defined on [0, 1] such that for any twice differentiable function g defined on [0, 1], the set {x E H : f (x) = g(x)} is finite. Solution. Delete an interval of length 1/6 from the center of [0, 1], intervals of length 1/18 from the centers of the remaining two intervals, and, following the procedure, intervals of length 1/2 3n+1 from the centers of the 2' intervals remaining after the nth step. This is possible because the length of the remaining intervals is -1
2-n 1- 1: 2k
1
1
2 3n+1
>2'
1
°°
6
=2 n(1
2
k
3
2)=2
1>2.3n+1
Denote by In,i (n = 0, 1.... ; i = 1, ... , 2") the open intervals of length 1/2.3n+1 deleted in the nth step. Put H = [0, 1] \ UIn,i. Then H is perfect and has measure 1/2. Denote by h the function that vanishes on H and whose graph on In,i is an isosceles triangle of altitude 1/n for n = 0, 1, .. . and i = 1, . . . , 2n. Then h is continuous on [0, 1], and so the function
AX) = f h(t) dt 0
is continuously differentiable on [0, 1] and satisfies the relation f'(x) = 0 for all x E H. Let g be twice differentiable on [0, 1]; we establish that the
set A = {x E H : f (x) = g(x)} is finite. Suppose A is infinite. Then
3. SOLUTIONS TO THE PROBLEMS
352
there is a convergent sequence xk -> x0 in A. Since f'(xo) = 0, we have g'(xo) = 0; hence by l'Hopital's rule, lim 9(x) - 9(xo) = lim
x-,xo
(x - x0)2
g'(x)
x--.xo 2(x - x0)
g" (X0)
9'(x) - g' (X0) - 1'm x-,xo
2(x - xo)
It follows that lim f (xk) - f (xo) k-oo (xk - xo)2
2
- 9"(xo) 2
The contradiction will arise from the fact that f(x) - f(xo)
lim
x-.xo,xEH
=oo.
(x - xo)2
Indeed, let x c H, x # xo, be arbitrary. Then there is an interval I,',i that separates x and x0. Let n be the smallest subscript for which there exists an In,i of this kind. It is easy to see that Ix - xoI < 2 On the other hand, if, say, x > x0, then f (X) - f (xo) =
x
JxQ 1
h(t) dt > 1
n
Jr 1
2'2.3'+' *n
h(t) dt
_
1
12.3n .n.
Hence 1 f(x) - f(xo) > (4) n (x-xo)2 - 12.3" .n.4-'t - (3) I
1
12n
If x -> x0, then n -4 oo and (4/3)n (1/n) -+ oo, which proves the assertion.
Problem M.15. Prove that if E C R is a bounded set of positive Lebesgue measure, then for every u < 1/2, a point x = x(u) can be found so that
I(x-h,x+h)nEI>u,h
and
I(x-h,x+h)n(R\E)l>uh
for all sufficiently small positive values of h.
Solution. We shall prove a stronger statement, namely, we verify the conclusion for u = 1/2. Let
F(x) = m ((-oo, x) n E), where m denotes the Lebesgue measure. Then F is continuous, increasing,
and satisfies IF(x) - F(y)l < Ix - yI for all x, y C R. By Lebesgue's
3.5 MEASURE THEORY
353
density theorem, there exists x E R, which is a point of density 1 of E, and hence there is a positive k E R with F(x + k) - F(x) > k/2. Since for sufficiently small x we have F(x + k) - F(x) = 0, continuity ensures the existence of an x0 satisfying F(xo + k) - F(xo) = k/2. Consider the function G(x) = F(x) - F(xo) - (x - xo)/2. Then G(xo) = G(xo + k) = 0. As almost every point of R is a point of density 1 of either E or ][8 \ E, we have I G'(x) = 1/2 a.e. Thus, G is not identically 0 on [xo,xo + k]. Consequently,
either
max
tE[xo,xo+k]
G(t) > 0
or
min
tE[xo,xo+k]
G(t) < 0.
It is sufficient to consider the first possibility since the second is similar to it. Let x1 E (xo, x0 + k) be a point at which G assumes its maximum just mentioned. Then for every
0 < h < min{xl - xo, xo + k - xl},
we have G(xl -h) < G(xl) > G(xl+h), that is, F(xl -h)+h/2 < F(xl) >
F(xl+h)-h/2. Thus, F(xl+h)-F(xl-h) < F(xl)+h/2-(F(xl)-h) _
3h/2 and F(xi + h) - F(xl - h) > F(xi) - (F(xl) - h/2) = h/2. From these two inequalities, we obtain
m ((x1 - h, x1 + h) fl E) > 2 and
m((xi-h,xl+h)fl (]I8\E))> 2 The proof is complete.
Remark. One contestant did not make use of the boundedness of E and
for u = 1/2 proved the assertion in the case where E is measurable and
neither E nor R \ E is a null set, and also in the case where E C m* (E) 0, and m* (E) # 0 (here m* and m* denote the exterior and interior Lebesgue measures, respectively).
Problem M.16. Show that there exist a compact set K C ][8 and a set A C R of type FF such that the set
{xER:K+xCA} is not Bore]-measurable (here K + x = {y + x : y c K}). Show that there exist a compact set K C ][8 and a set A C ][8 of type FQ such that the set
{xEJ:K+xCA} is not Bore]-measurable (here K + x = {y + x : y E K}).
Solution. Let P and K be bounded perfect sets such that the sums x + y (x E P, y E K) are pairwise distinct, that is, the relations x1i x2 E P,
3. SOLUTIONS TO THE PROBLEMS
354
y1, y2 E K, xl + yi = x2 + y2 imply x1 = x2 and yl =Y2- It is easy to see that the sets
P=
i;
a2=O,1;
i = 1, 2, .. .
lO;
a,=O,2;
i = 1, 2, .. .
l i=1
and
K i=1
are of this kind. It is well known that there exists a set U C P x K of type G5 such that the set
B = {x E P : there exists y E K with (x, y) E U} (which is the projection of U onto P) is not Borel-measurable.
Put V = (P x K) \ U and A = {x + y : (x, y) E V}. The mapping O(x, y) = x + y (x E P, y E K) is continuous and, by the choice of P and K, one-to-one. Therefore, 0 is a homeomorphism of the compact set P x K onto {x + y : x E P, y E K}. Since V is of type FQ, it follows that A = O(V) is an Ftype subset of O(P x K) = P + K. Since the latter set is closed in R, the set A is a subset of type FQ of R as well.
It is easy to see that K + x c A if and only if 0-1(K + x) C(A), that is, if x E P and
({x}xK)nU=0, (that is, if x E P \ B). So
{xER: K+xcA}=P\B, which is not Borel-measurable.
Problem M.17. For which Lebesgue-measurable subsets E of the real line does a positive constant c exist for which sup
< c
-00
sup
I
eanx f(x)dx
for all integrable functions f on E ?
Solution. We show that (*) holds for those and only those Lebesguemeasurable sets that - apart from a subset of measure 0 - can be covered by a set consisting of the finite union of closed intervals and containing no pair of congruent points modulo 2ir. Denote by E the system of all sets satisfying the requirements of the
problem and by F the system of all sets having the properties just described.
3.5 MEASURE THEORY
355
If E E E, then those points of E for which there is a congruent point in E (mod 27r) form a set of measure 0. Indeed, if, for example, the set E' = E n (E + 2j7r) n (2k7r, 2(k + 1)7r)
had positive measure (j # 0 integer, E + u denotes the translate by u of E), then with
11
ifxEE',
f(X) -1 ifxEE'-2j, 0
otherwise,
it would follow that
L eZnx f (x) dx = 0 while, as f
for every n,
0 a.e., sup
I
eitx f(x) dx >0.
The same property trivially holds also for every E E F. Denote by * the reduction modulo 2 with range space (-7r, 7r]. That is, u* is the (signed) deviation of u from the nearest multiple of 2-7r (if there are two such multiples, then u* = 7r). Furthermore, put
E*={u*: nEE}, h*(x) = h(u), where x E E*, x = u*, u E E. By the property above, the latter definition is correct for a.e. x E E* since there is only one such u. Therefore, we may write
(eitx) f* (x) dx,
0(t) =
E
eitx f(x) dx =
0(n) =
E
einx f(x) dx = E (einx)*f* (x) dx.
E
Here f *(x) E £ l (E* ), and every element of Gl (E*) occurs among the f *(x),
where f(x) E £1(E). We show that the relation JE a (x)f*(x)dx
< cmax 1¢(n)I = cmaxJ n
einx f*(x)
dx
E.
is valid for every function f * E G1(E*) if and only if there exists a function b(x) = E aneinx, EIan I < +oo, such that b(x) = a(x) for a.e. point
x E E*. Then the choice c = E _. Ian1 is possible.
3. SOLUTIONS TO THE PROBLEMS
356
Suppose first that there exists a function b(x) of this kind. Then a(x)f*(x)dx
JE' 00
n=-oo
einx f* (x) dx
an I *
00
cc
E ano(n)
n=-oo IanI mnax I0(n)I.
n=-oo
Conversely, consider the sequences {jn}n+_° 0o that tend to 0 as Inl -> oo; they form a Banach space with respect to the norm max l in I . The set of all sequences of the form {0(n)} is a subspace of this space, and the
functional fE* a(x) f * (x) dx is linear and, by our assumption, bounded on it. According to the Hahn-Banach theorem, this functional can be boundedly extended to all sequences { jn }. But the bounded linear functionals on the latter have the form E an jn, where E I an I < +00. Consequently, restricted to the sequences {0(n)},
J a(x)f* (x) dx = > anq5(n) _ *
=
00
n=-oo
an
f einx f* (x) dx E*
b(x)f*(x) dx fE*
(where b(x) _ aneinx), for every f* E £1(E*). Hence, a(x) = b(x) for a.e. x E E*. Apply the proposition just proved to the function a(x) = eitx with fixed t. Let E E F. If the finite number of intervals covering E is somewhat augmented, then the set obtained will not contain any pair of points congruent modulo 27r either. Let k(x) be a "sufficiently smooth" function that equals 1 on the original intervals, and 0 outside the augmented intervals. Finally, put 00
b(x) _
k(x +
2j)eit(x+2jn).
j=-00 Since at each point at most one term of the series is different from 0, the definition of b(x) makes sense. Furthermore, b(x) = a(x)* = (eitx)*
if x E E*.
The function b(x) is periodic with period 27r and Fourier coefficients bn = 27r
foo
°°
27r
00
e-"x 1` k(x + 2j)eit(x+2i7r ) dx j=-00
°°
27r
e-inxb(x) dx
-oo
eitx-inxk(x) dx.
3.5 MEASURE THEORY
Then
357
'00
Ibnl <
Ik(x)I dx < cl,
27r
00
or, after integrating by parts twice, Ibnl C
1
27r(n - t)2
f
Ik"(x)I dx < 00
c2
(n - t) 2
(for a suitable choice of k(x)). Hence, 00
1
Y:Ibnl
n=-oo (n --t)2
= c3.
Applying the statement we have proved previously, it follows that E E E. Conversely, suppose that E E £. We establish the existence of 6 > 0 such that for any two Lebesgue density points u, v of E, the relation I u - v I > 7r implies I(u - v)*I > S. Otherwise, indeed, there are sequences {un}, {vn} of density points with Iun - vnI > 7r but (un - vn)* -+ 0. It can be shown
that there is a number t such that [t(un - vn)]* 74 0. To see this, first let un - vn = wn be a bounded sequence; then t = 1/ sup I un - vnI is an appropriate value. If wn is unbounded, then we may assume that wn -p 00 and, what is more, wn+l/wn - oo. We form a sequence to as follows:
to=0,
to=tn_1-
(tn_lwn)* - 7r/2 ,
wn
n=1,2,...
Here to - tn_1 = O(1/wn), and by the fast increase of wn the limit of the sequence to exists. Put 00
t = limtn = E(tn - tn_1). n=1
Then
t-tn=0 Consequently, wn
twn = (t - tn)wn + tnwn = 0
wn+1
+ tn_lwn - (tn_lwn)* + ', 2
and, since tn_lwn - (tn_lwn)* is a multiple of 27r,
(twn)*=01 wn 1+W=7r +0(1)740. wn+l /
So, this t is suitable.
2
2
358
3. SOLUTIONS TO THE PROBLEMS
Using this t, we form the function a(x) = (ettx)*,
x E E*,
and apply the statement proved earlier. Omitting a set of measure 0, a(x) can be extended to a 27r-periodic continuous function (even a function with absolutely convergent Fourier series), to be denoted by b(x). Since 1
1-
\
un --,un+ J flE, n n
I
as well as its *, has positive measure, and similar statements hold for vn, after the omission of the set of measure 0 mentioned above, there remain some points in these sets. Let u'' , v' be such points. Then u'n - un -> 0, 0, hence 0, and a (u'*n) = b (u';,), a b (v'n). Therefore, by the continuity of b(x), it follows that a (u' ,) - a (v'*) , 0. But a (u'*) = eitun, a (v'*) = eztv;, n
n
,
whence
0 = lim[t(un -
lim[t(un - vn)]*,
and this contradicts the construction of t. Thus 6 exists. For each density point of E, consider its neighborhood of radius 6/4. Their union S is open and almost covers E. For any two points ul and vl of S, the relation I (ul - vl)*I < 6/4 implies Jul - v1I < 6/4; otherwise (assuming that 6 < 7r), Jul -vi > (3/2)7r would follow, and by the construction of S there would exist density points u and v in E with I u - ul I < 6/4, 1v-viI < 6/4, hence iu - vJ > 7r, while I(u-v)*I < I(ul -vl)*I+2.6/4 < b, which contradicts the property of S. Thus, the closure of S cannot have two congruent points modulo 27r. Hence, it also follows that the measure of S does not exceed 27r, and since each component of the open set S has length greater than or equal to 6/2, the number of the components must be finite, that is, the closure of S is a finite union of closed intervals. Consequently, E E F.
Problem M.18. Show that any two intervals A, B C R of positive lengths can be countably disected into each other, that is, they can be written as countable unions A = Al U A2 U ... and B = Bl U B2 U ... Of pairwise disjoint sets, where Az and Bz are congruent for every i c N.
Solution 1. In R, consider the equivalence relation
{(x,y)ER2: x - yEQ}, which induces the disjoint classification R = U.yErQ.y. Obviously, each equivalence class Q.y is dense in R, so A fl Q7 and B fl Q.y are countably
3.5 MEASURE THEORY
359
infinite sets. Consequently, assuming the axiom of choice, there exists a bijection
O.y: AnQ., between them for each -y. Then O.,(x) - x for every -y E P, x E A n Q. Let Q = {q1, q2, ... } be a numbering of the rationals, and consider the sets
YEP{xEAn Q.,: q.y(x)-x=qi},
Ai
Bi= U {(a-'(x): xEAnQ.y,0ry(x)-x=qi}, ryEP
for i E N. From the definitions, we see that
B
and B =
A = U Ai i=1
i
i=1
are partitions, and Bi = Ai + qi for every i E N.
Solution 2. Denote by a the relation that holds between the subsets of if and only if they can be countably dissected into each other. It is easy to verify that a is a translation-invariant equivalence relation satisfying
\n
1'S"/
n1T")
if Sn and Tn, respectively, are pairwise disjoint sets (n E N), and SnaTn for every n.
We now observe that any interval is the disjoint union of a countably infinite number of open intervals and a countably infinite set. So, by the above remarks, we may assume that A =]0, a[, B =]0, b[, 0 < a < b. In R, we now consider the equivalence relation
- { (x, y) E R2 : x - y E Q1. Its equivalence classes are dense in R, so - assuming the axiom of choice - there exists a set X C ] 0, a/3[ containing exactly one element of each class. Put a 2a
QA=Qnlo,
3 [,
QB=Qn]O,b-
3
[
and
A* = U (X + q), gEQA
B* = U (X + q), qEQB
where X + q stands for the set X translated by the number q E R. Obviously, A* and B* are countably infinite, disjoint unions of sets congruent with X, so A*aB*. One also easily verifies the inclusions
1a 2a 3' 3
c A C A,
a
[3,b
-
a
3]
C B
B.
360
3. SOLUTIONS TO THE PROBLEMS
Put
A-=]0,a[\A*, A+_] 3,a[\A*, B- =]0, a [\B*,
B+ =]b- 3,b[\B*.
A simple calculation shows that B- = A- and B+ = A+ + (b - a). Consequently, the partitions
A=A-UA*UA+ and B=B-UB*UB+ prove the statement. Remarks.
1. A direct consequence of the statement is the significant fact that there is no o-additive, translation-invariant measure defined on all subsets of R for which the unit interval has measure 1. 2. The first proof can immediately be applied to subsets with nonvoid interior of nondiscrete, separable, Hausdorff topological Abelian groups. 3. By the usual tools of measure theory, the statement can be extended to Lebesgue-measurable sets as follows. Any sets A, B C_ R of positive Lebesgue measures can almost be countably dissected into each other, that is, they have partitions 00 A= i=o U Ai
4.
and B = U Bi i-o
such that A0 and Bo have measure 0, while Ai is congruent with Bi, for every i E N. The assertion of the previous remark cannot be sharpened by deleting the sets of measure 0, since - as noticed by a contestant - a set of first Baire category cannot be countably dissected into a set of second Baire category; however, among sets of positive Lebesgue measure, there are examples of this kind.
Problem M.19. Let H C R be a bounded, measurable set of positive Lebesgue measure. Prove that
limofa((H
t)
>0,
where H + t = {x + t : x E H} and A is the Lebesgue measure.
Solution. We show that
limofA((H
)\H)>1.
+t
3.5 MEASURE THEORY
361
Lemma. For any 0 < E < 1, there is an interval [a, b) such that A (H n [a,b)) > (1 - E) (b - a). Proof. Assume that the conclusion of the lemma is false. Let us cover H by a countably infinite number of intervals (closed from the left), so that the sum of the lengths of the intervals is less than A(H)/(l - E). By the definition of the Lebesgue measure, this is possible. Let the ith interval be [ai, bi). Then by the indirect assumption, A(Hn [ai,bi)) < (1 - e)(bi - ai) and therefore M
00
A(H) <
A(H n [at, bi)) < (1 - E) E(bi - ai) < A(H), i=1
i=1
which is a contradiction. The proof of the lemma is complete.
Let 0 < e < 1, and let the interval [a, b) be such that
A(Hn [a, b)) > (1 -E)(b-a). Let 0 < t < b - a and n = [(b - a)/t] + 1. Then n
(1- E)(b - a) < A(H n [a, b)) < E A(H n [a + (k - 1)t, a + kt)). k=1
Hence, there exists an integer k, 1 < k < n, such that (1 - E)(b - a) A(H n [a + (k - 1)t,a + kt)) > n (1 - E)(b - a)
bta+1
=
(1 - E)t
+bta
Put
Ai=Hn[a+(k+i-1)t,a+(k+i)t)
(i=0,1,2,...).
Since, H being bounded, Ai is empty if i is sufficiently large, we have 00
0C
A ((H + t) \ H) > E a ((A. + t) \ Ai+1) > E (.A(Ai) i=o
i=o
- \(Ai+1))
- E)t =A(A)=A(Hn[a+(k-1)t a+kt))> (1 0 1+bta
Quite similarly,
\((H-t)\H)>-
(1 - E)t
1+bta.
It follows that o
liminf), ((H tt)\H)>l
e
o l + bta
But this is true for every 0 < E < 1. Thus, limionf
\ ((H Itlt) \ H) > 1.
=1-E.
362
3. SOLUTIONS TO THE PROBLEMS
3.6 NUMBER THEORY Problem N.I. Let f and g be polynomials with rational coefficients, and let F and G denote the sets of values of f and g at rational numbers. Prove that F = G holds if and only if f (x) = g(ax + b) for some suitable rational numbers a # 0 and b. Solution. By a "polynomial" we shall always mean a polynomial with rational coefficients, and by the "range" of a polynomial we shall mean the set of its values assumed at rational numbers. We use two classifications of polynomials. We write f " g if the ranges of these polynomials coincide, and f : g if f (x) = g(ax + b) for all x with suitable rational numbers a # 0, b. These are clearly equivalence relations, and they are compatible with multiplications by a constant, that is, if f - g or f g, then c f " cg or c f cg, respectively (c # 0 rational). Our aim is to show that these relations are identical. One implication is clear: if f g, then f - g holds obviously. The converse will be proved in several steps. 1. It is sufficient to prove the statement for polynomials with integral co-
efficients. Indeed, assume f - g, and let c be an integer such that the coefficients of c f and cg are all integers. We have c f - cg, thus c f assuming the integer case, which yields f g as wanted.
cg,
2. If f - g are polynomials with integral coefficients, we show that they must be of equal degree. Indeed, assume that
f(x)=ax' +...,
g(x) = bx k +
.
. . ,
where a # 0, b # 0, and n < k. Choose a prime p/ab. We have f (1/p) = c/p'', where p/c. By assumption, g also takes on this value at some number u/v with (u, v) = 1. Now if p%v, then p cannot divide the denominator of g(u/v); consequently, p must divide v. Now, by (p, b) = 1, the exponent of p in the denominator of b(u/v)k is higher than in any other term of g(u/v), thus the denominator (in the reduced form) of g(u/v) contains p with an exponent > k. This yields n > k, hence n = k as asserted. 3. It is sufficient to prove the statement for polynomials with integral cofficients where at least one of the leading cefficients is 1. Indeed, let f - g, f (x) = aXn + .... Then we\ have
an-l f (x)
al
-
an-1f
(x) '" a"'-lg(x)'
and here an-l if (x/a) still has integral coefficients while its leading coefficient became 1. By assumption, we conclude an-l f (x/a) an-lg(x), which yields f .: g.
4. Let f - g be polynomials with integral coefficients, f (x) = xn + ... , g(x) = bxn+.... We claim that every prime divisor of b has an exponent
3.6 NUMBER THEORY
363
> n. To show this, take a p1b and consider the polynomials
fl (x) = Pn(n-1) f
=
x
( Pn-1 )J ^'
Pn(n-l) f
( X ) ^' gl (x) pn
Pn(n-1)g ().pn
We know that all coefficients of f1 as well as all but the leading coefficient of g1 are integral, and the leading coefficient of f1 is 1. Hence every value of f1 has the following property: if (in the reduced form) its denominator is divisible by p, it is divisible by pn. By f1 - g1 the same must hold for
g1. Now g1(1) = (b/pn) + c with an integer c, and by pub the exponent of p in the denominator is strictly less than n, thus it must be 0, that is, pn I b.
5. It is sufficient to prove the statement for polynomials with integral cofficients where both leading cefficients are 1. Let f - g be polynomials with integral coefficients, f (x) = xn + ... , g(x) = bxn + .... Let cn be the largest nth power dividing b. Consider the polynomials
fl (x) =
x
Cn(n-1) f
C-1
= xn + ... ,
x 1 = b J Cx+ n ... C-1 We have f1 - g1. By step 4, the exponent of any prime in the leading coefficient of g1 must be either 0 or > n, and by the definition of c this coefficient must be 1 or -1. We can exclude the case -1 for even n, since 91(x) = C n(n-1) g
.
then one domain would be bounded from below and the other from above. Finally, if n is odd, then replacing c by -c we can change a -1 into 1. So if the statement is true under this restriction, then we obtain f1 .^s g1, which yields f g. 6. We can also assume that the coefficient of xn-1 is 0 in both polynomials. Indeed, if f(x) = xn + axn-1 + ...,v g(x) = xn +
bxn-1 +
...,
then the polynomials fi (x) = nnf ( x-7.
- a) - g1(x) = nng (n - al
have this property, and f1 ~ g1 again implies f g. 7. Combining the previous arguments, we find that to solve the problem it is sufficient to prove the following statement:
Statement. If f - g are polynomials of the form f (X)
= xn + axn-2 + ... ,
g(x) = xn +
bxn-2
+ .. .
with integral coefficients, then f Pz g.
Proof. Observe that, since the leading coefficient is 1, the only rational numbers where f and g assume integral values are the integers.
3. SOLUTIONS TO THE PROBLEMS
364
Since f (x + 1) - g (x) = nxn-1 + ... , we have g (x) < f (x + 1) for large x, and similarly we obtain g(x) > f (x - 1). Thus, for a large positive integer
x, the only positive rational y that can satisfy f (y) = g(x) is y = x. If n is even, then there can also be a negative y, and similarly we find that its only possible value is y = -x. This means that one of the equations f (x) = g(x) and f (-x) = g(x) has infinitely many solutions; hence it must be an identity. This concludes the proof.
Problem N.2. Show that
[J
(x2+y2) = (-1)[8]
(mod p)
1<x
for every prime p =_ 3 (mod 4). ([.] is integer part.)
Solution 1. Write p = 4k + 3. Consider all the sums x2 + y2, 1 < x, y < 2k + 1. Let r denote the number of those pairs x, y for which this sum = 1.
First, we show that the number of those pairs for which x2 + y2 - -1 (mod p) is r + 1. Since every quadratic residue has a unique representation
in the form x2, 1 < x < 2k + 1, while every nonresidue has a unique representation in the form -y2, the solutions of x2 + y2 = x2 - (-y2) = 1 count the consecutive numbers in the form (nonresidue, residue) within the sequence 1,2,... , p - 2,p - 1. Similarly, the pairs with x2 + y2 = -1 correspond to pairs of type (residue, nonresidue). Since this sequence starts
with a residue and (recall that p - -1 (mod 4)) ends with a nonresidue, the second case must happen r + 1 times. Next, we prove that the number of solutions of x2 + y2 = a2 (mod p),
1 < x, y < 2k + 1, is r for every a. Indeed, the mapping x - fax1, y - ±ayl, where the signs are chosen so that x, x1i y, y1 are all in [1, 2k+ 1], provides a one-to-one correspondence between solutions of x2 + y2 =- a2
and x2 + yl = 1. Similarly, we obtain that the number of solutions of
x2+y2--1 is r + 1. The above observations mean that these sums x2 + y2 represent every quadratic residue r times and every nonresidue r + 1 times. Since the total number of these pairs is ((p - 1)/2)2 while the number of residues and the number of nonresidues are both equal to (p-1)/2, we find r = (p-3)/4 = k. Now, the product we want to compute is not over all these pairs but only over those with x < y. Consider a quadratic residue a. We know that the total number of solutions of x2 + y2 = a is k. If among them there are
u with x < y, v with x = y, and w with x > y, then u = w by symmetry and v is 0 or 1, hence u = [k/2]. Analogously, for a nonresidue the number of solutions with x < y is [(k + 1)/2]. The product of all quadratic residues is 2)2
12.22
...
((P_ )!)2 2 1
Cp
(mod p),
3.6 NUMBER THEORY
365
and the product of nonresidues is
/
-f
= (-1)
1 /I2
\
- -1
(()
!12
(mod p).
Hence, the original product is 1[k/2](-1)[
2
]
= (-1)[ 8 1
Solution 2. Denote this product by P. In the product, all sums of pairs of quadratic residues are multiplied. Let g be a primitive root modp, and put h = g2. The numbers 1, h, h2, ...,h 2k represent every quadratic residue once; thus we have
P-
fi
(hi + hi)
(mod p).
O
This implies
P fi
(hi-hj)=
O
(h2i-h2j) fi O
(mod p).
Here each product is the value of a nonzero Vandermonde determinant. Thus, this equation can be rewritten as
2,...,h 2k)=V(1,h 2,...,h 4k)
(mod p).
Since h2k+1 = 94k+2 = 1 (mod p), the generators of the second Vandermonde determinant are congruent to 1 h2 ,..., h2k h h3 ,... ,
h2k-1.
Thus, the second determinant can be obtained from the first by k+(k-1)+ + 2 + 1 = k(k + 1)/2 transpositions of rows. By the familiar properties of determinants, this means
P = (-1)
k(
(mod p).
2}1
Problem N.3. Let p be a prime and let lk(X, Y) = akx+bky
(k = 1,...,p2),
be homogeneous linear polynomials with integral coefficients. Suppose that for every pair i7) of integers, not both divisible by p, the values lk( , ri), 1 < k < p2, represent every residue class mod p exactly p times.
3. SOLUTIONS TO THE PROBLEMS
366
Prove that the set of pairs {(ak, bk) : 1 < k < p2} is identical mod p with
the set{(m,n):0<m,n
Solution 1. Assume that the statement does not hold. Then there are numbers i j such that ai - aj and bi - bj (every congruence is meant mod p). Consider the number of those triplets (k, x, y), (x, y) # (0, 0) that satisfy lk(x, y) ° li(x, Y) -
By the assumption, for every fixed pair (x, y) (0, 0), the number of solutions in k is p; thus the total number of these triplets is p(p2 - 1). Now consider the solutions in x, y for a fixed k. If k = i or j, this is an identity, which means 2(p2 - 1) solutions. For any other k it is easy to see that the number of solutions is at least p - 1. This means that the total number of solutions is at least
(p2-2)(p-1)+2(p2-1)=p(p2-1)+p(p-1)>p(p2-1), a contradiction.
Solution 2. It is sufficient to prove that for every u and v there is exactly one k with ak - u, bk - V. The uniformity of representations implies that for every (, rl) 0 (0, 0), we have P
2
P.
eP
0.
k=1
to get
We multiply both sides by P
2 C
e2pt((ak-u)S+(bk-v),7) = 0, k=1
for
rl) 0 (0, 0). For C - rl - 0, the same sum obviously gives p2. Now
summing these sums for all possible values of l; and 77, including (0, 0), we obtain p2 P-1 p-1
S-
U e'
-v)n) = p2.
r;=0 77=0 k=1
Changing the order of the summations, we get P2 P-1 p-1
S =k=1Er;=077=0 E P2
(:e_
(e_71)) 77=0
=J C
3.6 NUMBER THEORY
367
Now, a typical factor in this product vanishes, unless ak - u (for the first) or bk =_ v (for the second). Thus, the whole product is 0, unless ak - U and bk _= v, in which case it is equal to p2. Since the sum of these products is p2, this case must happen for exactly one value of k. Remarks. 1. The method of the second solution can be applied to prove the following
generalization of the problem: Let p be a prime, r a positive integer, and the lk's homogeneous linear polynomials in n variables of the form
lk (x1, ... , xn) = akx1 + ... + akxn
(k = 1, ... , rpn).
Assume that for every k-tuple (1 i ... , W of integers, not all divisible by p, the values of ( 6 ,- ... , n) represent every residue class mod p exactly
rpn-1 times. Then every n-tuple (ml, ... , mn) is represented mod p among the n-tuples (ai, ... , ak) of coefficients for exactly r values of k. 2. If we relax the requirement of uniform representation to those pairs 77) where neither nor q is divisible by p, then the following can be asserted: Assertion. Let f (m, n) denote the number of those pairs (ak, bk) that satisfy ak = m, bk =_ n. Then the values lk (C 77) will be uniformly disif and only if tributed for all of the p2 - 2p + 1 admissible pairs f (m, n) can be represented as f (m, n) = g(m) + h(n).
Problem N.4. Let p be a prime, n a natural number, and S a set of cardinality pn. Let P be a family of partitions of S into nonempty parts of sizes divisible by p such that the intersection of any two parts that occur in any of the partitions has at most one element. How large can I P I be ?
Solution 1. This maximum is (pn - 1)/(p - 1). Let H be the set to be partitioned, and let C1i . . . , Ck be its partitions into sets whose cardinalities are multiples of p, such that any two classes may have at most one common element. Consider an h E H, and let Ci(h) be the class of Ci that contains h. By the assumptions, the sets C1(h)\{h},..., Ck(h)\{h}
are pairwise disjoint, and each has at least p - 1 elements. Consequently,
wehavek(p-1)
Now we prove the corresponding lower estimate. Consider the n-dimensional projective space Pn over a finite field K of p elements, a hyperplane 0' (pn+1-1)/(p-1) in it, and the affine space P, = P,, \o,. Observe that Pn has
points, o, has (pn - 1)/(p - 1) and P,P has pn. For an arbitrary point P E a, consider those lines of Pn that contain P but do not lie in or. If we omit the point P from each such line, the remaining affine lines (as sets of points) form a partition Cp of the affine space P' into sets of cardinality p. The number of these partitions is the same as the number of points of or, that is, (pn - 1)/(p - 1).
3. SOLUTIONS TO THE PROBLEMS
368
Solution 2. We keep the upper estimation from the first proof, and present a different construction to show the lower bound. For H we take the set of all n digit numbers in base p. This set has pn elements. The partitions will be given in forms of integer-valued functions;
f (x) will mean the number of the class containing a given x E H. The functions will be indexed by pairs (j, a), where j = 0, . , n - 1, and for a given value of j, the possible values of a are a = 0, . . , pj - 1. The number . .
.
of possibilities for a is pj; thus the total number of these functions is 1+p+p2+...+pn-1
= pn
p-1
as wanted.
We define the function fja(x) as follows. Let the representations of x and a to base p be
+... + bn-lpn-l ' a = ao +alp+ ... + aj_1Pj -1, Let [m] denote the (smallest nonnegative) residue of an integer m mod p. x = So + Slp
Now we put C [S1 + a1Sj]p + ... +
CC
CC
fja(X) = C[So +
+ Sj+1P' + ... +
CC
[Sj_l +
CC
aj-1Sj]Lj-1
tn-lpn-2.
In the case j = 0, a = 0, we interpret this as foo(x) = S1 + S2P + ... +
C
Sn-1pn-2.
We show that every class contains p elements. Indeed, given the value of
fja(x)
=Y=770+...+17n-2p
n-2
we have G = 71k-1 for k = j + 1,...,n, we can choose the value of j that will give p possibilities, and after fixing j we can uniquely determine 0i ... , j from the congruences z + c
j = q:
(mod p)
(i = 0,...,j - 1).
Next, we have to show that the intersection of two classes has at most one element, that is, the system of equations
fj'a'(X) = y
fja(X) = Y,
has at most one solution. Assume first that j # j', say j' < j. Then we obtain 77'. from the second equation, and we already know that fja (x) and j determine x uniquely.
Finally, consider the case j = j'. We must have a # a', say ak # ak for some digit k < j - 1. Then the congruences Sk + Qkej = 77k
(mod p),
k+
(mod p)
77k
determine j, which together with faj (x) determines x uniquely.
3.6 NUMBER THEORY
369
Problem N.S. Let f be a complex-valued, completely multiplicative, arithmetical function. Assume that there exists an infinite increasing sequence Nk of natural numbers such that f (n) = Ak
provided Nk < n < Nk + 4 Nk.
0
Prove that f is identically 1. Solution. We show a slightly stronger statement, where the 4 is replaced by 2 + e. Assume that f is (nonzero and) constant on the intervals Ik = [Nk, Nk + Mk], where Mk = (2 + e) Nk.
First, we prove that f (n) $ 0 for any n. Indeed, if Mk > n, then nx E Ik for some x; hence f (n) f (x) = f (nx) $ 0. Next, we use an interval
of constancy to create another. Let f be constant on I = [N, N + M]. If for an n we can find an integer x such that nx, (n + 1)x E I, then f(n)f(x) = f(nx) = f((n + 1)x) = f(n + 1)f(x), hence f(n) = f(n + 1). Now this condition can be reformulated as
N<x
N+M-N>1 n+1
n
'
or, after rearranging, n2+n(1-M)+N < 0. This means that n lies between the roots of the corresponding quadratic equation, that is, (M - 1)2 > 4N (which shows that our method does not work with M = (2 - e)vrN-) and
M-1- (M-1)2-4N
I" = [c3i e4M] of constancy. As I runs over all the intervals Ik, these intervals I" cover the whole half-line [c3i oo]; thus f (n) is constant for n > e3. Now take an arbitrary positive integer m, and select an n > c3. We have f (n) = f (mn), hence f (m) = 1 as asserted.
Remark. The statement will not hold if the 4 Nk is reduced to exp (c log Nk flog log log Nk) with a suitably small positive c.
To see this, consider a sequence Ik = [Nk, Nk + Mk] of intervals, Nk >
Mk > Nk_1 + Mk_1i with the property that every number n E Ik has a prime factor p > Mk. Choose the numbers Ak $ 0 arbitrarily. We claim that there is a completely multiplicative function f that is identically Ak on Ik; if not, every Ak is 1, and then our function will not be identically 1.
3. SOLUTIONS TO THE PROBLEMS
370
We construct our function recursively. Assume that f (p) is fixed for every prime p < Nk_1 + Mk_1i we define f (p) up to Nk + Mk. For every n E Ik, let pn > Mk be the largest prime divisor of n. These primes are distinct, since for n n' we have (n, m) < In' - nj < Mk. Choose f (p) arbitrarily for every prime that does not occur among the pn's, and then set f (pn) to achieve f (n) = Ak. Now we have to find such a sequence of intervals. Assume that I1, .. . Ik_I are given. Take a large N, say N > expNk_1i we try to find an Ik in [N/2, N].
Let R(x, y) denote the number of those integers n < x whose prime factors are all less than y. If M is such that M(R(N, M) + 1) < N/2, then the interval [N/2, N] contains a subinterval of length M that is free of these numbers. This interval can serve as our Ik. The feasibility of taking M = exp(c log Nk log log log Nk) follows from Rankin's inequality:
R(x, y) < x exp
(
- logloglogy y log x + O(log log y) J
.
Problem N.6. If c is a positive integer and p is an odd prime, what is the smallest residue (in absolute value) of P-1 2.
C2n1
n=O
nJ
(mod p)?
cn
Solution. Let q = (p-1)/2. With this notation, we have 2j+1 - -2(q-j) for every j. Consequently, we have 1 . 3 ... n!2
2n n!
(2n - 1)
2n
(2n)!
n)
n&2 _ 2n.1.3.....(2n-1) n!
_ n!
= (-4)n l
q
n)
-
Hence,
(q) (-4c)n = (1 -
(2:) cn = 0
This is congruent to 0 if 1 - 4c is divisible by p. Otherwise, it is 1 if 1 - 4c is a quadratic residue mod p and -1 if it is a nonresidue.
3.6 NUMBER THEORY
371
Problem N.7. Find a constant c > 1 with the property that, for arbitrary positive integers n and k such that n > ck, the number of distinct prime factors of (k) is at least k. Solution. Write t = [1, 2, ... , n]. We shall show that the number of prime factors of (k) is at least k for n > t + k. Take a prime p < k, and let p9 < k < pk+1 The exponent of p in the decomposition of k! is E8=1 [k/pi], and in t it is s. Among the numbers n, ... , n - k+ 1, there are at least [k/pi] multiples of pi. For i < s, these also enter the decomposition of the corresponding (n - j, t); thus the exponent of p in the product (n, t) (n - 1 , t) ... (n - k + 1, t) is at least as high as in k!. Since this holds for every p, we conclude that k!I (n, t)
(n - 1, t) ... (n - k + 1, t).
This implies that k-1
n-i
ri (n - i, t)
(n).
The factors on the left side are pairwise coprime, since (n - i, n - j) I (i - j) It for i # j, and if n > t + k, they are all larger than 1. Taking a prime factor of each, we find k different prime factors of (k).
Since t = exp(k + o(k)) by the prime number theorem, every c > e satisfies the requirement of the problem for large k. If we want to exhibit a concrete value of c that works for every k, we can argue as follows. Applying Chebyshev's estimate,
Hp<4x, p<x
we obtain
t=
fl p3< fl k fl pa
p
p
Therefore, k
t + k < k-k4 k = (4k1/-k ) < 9k, since the function x1lvf'- assumes its maximum at e2, and its value is e2/e < 2.1.
Remark. The value c = 9 we gave can be further reduced with a little extra effort; one of the contestants gave c = 4.3. It seems to be difficult to decide the improvability of c = e + E for large k. If the "prime k-tuple conjecture" is true, then there are arbitrary large values of n for which (k) has exactly k prime factors.
3. SOLUTIONS TO THE PROBLEMS
372
Problem N.8. Let f (n) be the largest integer k such that nk divides n!, and let F(n) = max2<m,
Solution. First, we find an upper estimate for f (n). Let n = fk 1 p'i be the prime factorization of n. Since of (n) In!, we have n
n
aif(n) < E
<
7- 1
Pi
-
1'
that is, ai log pi <
n log pi f(n)pi-1'
for all i. Summing these inequalities, we obtain log n =
n E log pi
ai log pi <-
f(n)
pi-1
Let q, < q2 < ... be the sequence of all primes. Since (logp)/(p - 1) is decreasing, we have
E log Pi < r. loggi = logk+O(1).
i=1pi-1
i=1pi-1
Combining the last two inequalities, we get
f(n) <
(1+o(1))nlogk
log n
On the other hand,
n=p"...pk'` >g1...gk>2k Consequently, k << log n, log k < log log n+O(1), and the previous estimate
of f(n) yields f (n) < (1
+o(1))nloglogn
log n
as wanted. Now we construct a number m < n with a large value of f (m). Define m1 by (m1+1)! < n < (m1+2)!. We have m1 - (log n)/(log log n)
as n --+ oc. Let p be the largest prime up to (n/m1!)1/3and put m = p3m1!. We have m < n obviously, and, since n/m1! > m1
p - (n/m1!)1/3, that is, m - n.
oc, we have
3.6 NUMBER THEORY
373
Since (mi!)m/ml gym!, the exponent of any prime other than p is at least m/ml times as high in m! as in m. We have to compare the exponents of p. The exponent of pin ml! is F_°__1 [ml/pi] < ml/(p - 1), hence in m
it is less than ml/(p - 1) + 3. On the other hand, the exponent of p in m! is [m/pi] > [m/p] > (m/p) - 1. Therefore, the quotient of these exponents is at least 1
P
m
p +3m1' because p -> oc, but
p:5 (n/mi!)1/3 < ((m1 + 1)(ml + 2)) 1/3 = o(m). Consequently, we have (1+o(1))nllogloogn
F(n) > f(m) > (1+o(1))ml = (1+o(1))m1 =
gn
as asserted.
Remark. It can be proved that
F(n) =
n(log log n - log log log n) log n
+0
(n log n
Problem N.9. Prove that a necessary and sufficient condition for the
existence of a set S C {1,. .. , n} with the property that the integers 0, 1, ... , n - 1 all have an odd number of representations in the form x y, x,y ES, is that (2n -1) has a multiple of the form 2 -1.
Solution. Let S be a set with this property, and write xa-1
1(x) = aES
g(x)=E xn-a aES
which we regard as polynomials over GF(2). Then we have f (X)
= xn-1g(1),
(1)
and 2n-2
f(x)g(x) _
1 = 1 + x + x2 +
xi
j=0
+
x2n-2,
(2)
a,bES a-1+n-b=j
by our assumptions on S. Conversely, if f and g are polynomials over GF(2) that satisfy (1) and (2), then the set S = {a : the coefficient of xa-1 in f is 1} has the desired property.
3. SOLUTIONS TO THE PROBLEMS
374
Now put
(3) 1 + x + x2 + + x2n-2 = p1(x)...p1 (x), where the factors pi, ... , pj are irreducible over GF(2). With a suitable choice of the suffices, we have f =P1 ... Pr,
9 = pr+1...pt
by (2).
Observe that the roots of 1 + x + x2 + + roots of unity except 1. Furthermore, we have
x2n-2
are the (2n - 1)th
(1+x+x2+...-x2n-2)'=1+x2+...+x2n-4
and
and hence this polynomial has no multiple roots. Since the roots of f are the reciprocals of the roots of g by (1), no pi can have two roots that are reciprocals of each other. Conversely, if no pi has reciprocal roots, then the polynomials p1i ... , pi can be coupled so that the roots of any pi are the reciprocals of the roots of its mate. We now multiply one element from each pair to get the polynomial f and the others to get g; these obviously satisfy (1) and (2). Thus, such a set S exists if and only if for every root a a and 1/a belong to different irreducible factors. of l+x+x2+ We construct this decomposition (3). +x2n-2,
Lemma. Consider the permutation of the roots of the equation x2n-1 1 = 0 given by the operation of squaring. Let C1, ... , C1 be the cycles of this permutation. The irreducible factors of x2n-1 - 1 are the polynomials
i=1 ...,1.
11 SEC;
We prove this lemma at the end of the solution; see also E. R. Berlekamp, Algebraic Coding Theory, McGraw-Hill, New York, 1968, Chapter 6. Hence, our condition for the existence of the set S is satisfied if and only
if the roots a and 1/a cannot be tranformed into each other by repeated squarings for any a # 1, that is, a2k+1 54 1
for all k > 0 and (2n - 1)th roots of unity a as for all k. Since
(x2n-1 - 1, x2k+1
1. This can be reformulated
- 1) = x - 1
(xu-1,x"-1)=x(u,")-1,
the set S exists if and only if
(2n-1,2k+1)=1
(4)
3.6 NUMBER THEORY
375
for all k. Now let d be the smallest exponent for which 2n - 112d - 1. If d is even, then 2n - 11(2d/2 -1)(2
d/2
+ 1),
but 2n - 1X(2d/2 - 1), and thus (2n - 1, 2d/2 + 1) > 1, which contradicts (4). Conversely, if d is odd, then we have
(2n-1,2k+1)=(2d-1,2k+1)=1. Consequently, S exists if and only if d is odd, which is clearly equivalent to the condition given in the problem. Proof of the lemma. Let N akxk
O(x) _
k=0
be a polynomial over any field of characteristic 2. We have N
(O(x))2 =
akx2k k=0
Thus, the equation (4'(x))2 = O(x2)
holds identically if and only if ak = ak for all k, that is, every ak is 0 or 1. On the other hand, the above equality is clearly equivalent to the assumption that for every root a of 0, a2 is a root as well.
Problem N.10. Prove that the set of rational-valued, multiplicative arithmetical functions and the set of complex rational-valued, multiplicative arithmetical functions form isomorphic groups with the convolution operation f o g defined by
f(d)9(d)
(f o9)(n) _ din
(We call a complex number complex rational, if its real and imaginary parts
are both rational.)
Solution. Let Q be either the field of rational numbers or the field of complex rational numbers, and let GQ denote the group of Q-valued multiplicative functions. It is well known that GQ is an Abelian group. Now we prove that for every natural number n and every f E GQ,the equation
gogo...o9=9
`
n times
is solvable and the solution is unique.
3. SOLUTIONS TO THE PROBLEMS
376
Indeed, we have g(1) = 1 by multiplicativity. For every prime power pk, we have
f (PI) _ E
g (pk 1) ... g
i
(pk )
g(pk1) ... g(pkn) + ng(pk)
k;
=k
This yields a recursion for f (pk), which shows the unicity. To show the existence, consider the function g whose values are defined by the above recursion at prime powers and that is extended multiplicatively to the other o g is a multiplicative function that coincides with f at numbers. go prime powers; thus they must be identical. Thus, both groups are divisible, torsion-free Abelian groups. By the fundamental theorem of divisible Abelian groups, both are isomorphic to some discrete direct power of the additive group of rational numbers. Since we can prescribe the values of a multiplicative function at prime powers arbitrarily, we conclude that the cardinality of GQ is 2x° for both choices of Q. This implies that the direct power must have 2x° factors in both cases. This proves the isomorphy of the two groups.
Problem N.11. Let H denote the set of those natural numbers for which -r(n) divides n, where r(n) is the number of divisors of n. Show that (a) n! E H for all sufficiently, large n, (b) H has density 0.
Solution. (a) We show that r(n!)In! for all n # 3, 5. We have 00
n! = 11 pap,
ap = E[n/p'],
p
i=1
and consequently,
-r(n!) = 11(ap + 1). p
To prove rr(n!) In!, it is sufficient to find, for every prime p
n, a
natural number h(p) < n such that ap + 11h(p) and h(p)
h(q)
whenever p # q. If p < V/n, put h(p) = ap + 1. We have 00
h(p) = 1 + ap < 1 +
00
[n/21] < 1 +
n/2' = 1 + n,
hence h(p) < n. Also, if p < q < \, then
n_n=(q-p)n> n >1; p
q
pq
pq
3.6 NUMBER THEORY
377
consequently, [n/p] > [n/q]. Since also [p/p'] > [n/q'], for all j, we can infer that h(p) > h(q). For the primes < p < n, we define h(p) by recursion. Assume that h(q) is already defined for every prime
q < p. We need to find a multiple of ap + 1 that is not among the already distributed numbers h(q), q < p. Observe that [n/pi] = 0, for j > 2; thus ap = [n/p]. Hence, the number of multiples of ap + 1 up to n is
_
n
> n - [n/p] > n - (n/p)
n
1 + [n/p] - 1 + (n/p)
1 + [n/p] n
Lap + 1
n+p>p-1
=(p-1)
2
,
with strict inequality unless p = n. The number of already occupied
values is 7r(p) - 1 < (p - 1)/2, with strict inequality unless p = 3, 5, or 7. Thus there is a free number for h(p) except possibly in the cases n = p = 3, 5, 7. In these cases, the divisibility T(n!)In! holds for n = 7 and does not hold for n = 3 or 5. (b) We split H into two parts. Choose a parameter K > 1. We put those numbers in which the exponent of every prime is < K into H1 and the rest into H2. Consider first an n E H1. If n = f,-1 a < K, then T(n) = fl(aj + 1) consists exclusively of primes < K, and by T(n)In, the exponent of every prime is less than K. Consequently,
T(n) <
q,
p
< KK
a.
On the other hand, we have T(n) > 28, and these inequalities together imply s < [K21og2 K] = r.
It is well known that the sequence of numbers that have at most r prime factors has density 0, hence so does H1. Every element of H2 is divisible by the Kth power of a prime. Hence, the number of elements of H2 up to x is at most 00
>[- -]< x> i-K < x Jf i t-k dt = x E[ K-1 1
p
i=2
Hence, the density of H is at most 1/(K - 1). Since K was arbitrary, H must have density 0.
Remark. Let H(x) denote the number of elements of H up to x. It can be shown that x(logx)-(1/2+E) << H(x) << x(log x)-1/2,
for x > xo(E). We outline a proof of these inequalities.
378
3. SOLUTIONS TO THE PROBLEMS
We start with the upper estimate. Write n in the form n = 21p1p2 ... pkm2,
where pi, ... , pk are odd primes and m is an odd integer. The number of possibilities for m is < Vx-, for lit is O(log x). Hence, the
number of those n for which pl
.
pk <
x1/3
is O(x5/6log x) _
o(x(logx)-1/2). In what follows, we assume that P1 ... pk > x113 The exponent of each pj in n is odd, hence 2k Jr(n), so k < 1 if n E H. For fixed 1 and m we have x P1 ... pk
y = 21m2 ,
and y > x1/3 by the previous assumption. Now by a classical theorem of Hardy and Ramanujan, the number of integers < y that are products of k distinct primes is <<
y
(c+loglogy)k-1
(k - 1)!
logy
with an absolute constant c. Since log x > logy > (1/3) log x in our case, this is <<
y (c + log log x)k log x (k - 1)!
We have to sum this for all possible values of l and m. Write 1 = k + j; we know that j > 0. With this substitution, our sum is x
E 2-k-9m-2 log x
(c + log log X) k-1
(k - 1)!
Here the sum over m and j contributes only a constant factor, while the sum over k is just the power series expansion of log log x = c (log x)1/2. exp c + 2
This concludes the proof of the upper estimate. The lower estimate can be obtained by considering numbers of the form n = 2q-1gP1...pk,
where q, pi, ... , Pk are distinct odd primes. This number has r(n) = 2k+lq divisors, thus n E H if k < q - 2. For a fixed q, q < (1 - e) log log x, the number of such n < x is approximately
q-12-q x log x
(log log x)g-3
(q - 3)!
The lower estimate follows by taking q -
log log X. 2
3.6 NUMBER THEORY
379
Problem N.12. Prove that if a, x, y are p-adic integers different from 0 and pox, paixy, then
1 (1+x)Y-1
log(1+x)
x
y
(mod a).
x
Solution. Every p-adic number a can be uniquely represented in the form
a = P'(')E, where a is a unit and v(a) is an integer; p-adic integers are characterized by v(a) > 0. Hence, a divisibility al,l is equivalent to v(a) < v(0). In particular, the condition palxy means 1 + v(a) < v(x) + v(y), and the statement to be proved means v(A) > v(a), where
A- 1y (1+x)y-1 x
log(1+x) x
By the previous observation, it is sufficient to show that v(A) > v(x) + V(Y) - 1. The series
(1+x)y=1+Il)x+...( )xn+... and
n
2
Xare convergent since pi x, Q) is a p-adic integer, and n - v(n) -+ oc. Substituting these expansions into A, we obtain
A=I:00Bn, n=2
where
Bn = (1 (y) y
-
{(y
1)(y
(-1)n-1
n) xn-1 /J
xn -1 2)...(y
(n
1))
(- 1)n-1(n
1)!}
n-
Here the first factor is a polynomial in y whose constant term vanishes;
thus it is a multiple of y. Consequently, Bn = cyxn-1/n! with a p-adic integer c, thus
v(Bn) > v(y) + (n - 1)v(x) - v(n!). To prove that v(A) > v(x) + v(y) - 1, it is sufficient to show that v(Bn) > v(x) + v(y) - 1 for every n; by the previous inequality, this would follow from v(n!) < (n - 2)v(x) + 1. Since v(x) > 1 by assumption, we need only to show v(n!) < n - 1. But the exponent of p in n! is v(n!)
_
°°
nl P3J
<j=1i'= °° n
n P-1
< n.
Consequently, v(n!) < n - 1 since it must be an integer. This concludes the proof.
3. SOLUTIONS TO THE PROBLEMS
380
Problem N.13. Let 1 < a1 < a2 < < an < x be positive integers such that En 1/ai < 1. Let y denote the number of positive integers smaller than ix not divisible by any of the ai. Prove that cx
Y > logx
with a suitable positive constant c (independent of x and the numbers ai).
Solution. The number of multiples of aj up to x is [x/aj]. Hence, the total number of multiples is
[X1 < a<x. aj We have to improve on this trivial bound. Choose an x0 such that the number of primes between x/2 and x is at
least x/(3logx) for all x > x0; such an x0 exists by the prime number theorem. For x < x0, we use the trivial estimate y > 1. For x > x0, we distinguish two cases.
If the number of aj > x/2 is less than x/(6logx), then there are at least x/(6logx) primes x/2 < p < x that are not contained among the aj. These primes cannot be divisible by any aj; thus in this case we obtain y > x/(6logx). If the number of aj > x/2 is at least x/(6logx), then we have
aj <x/2
a <1- aj>x/2 a <1-x6logx
=1
6logx.
Hence, the number w of integers up to x/2 that are divisible by some aj satisfies w
<x/2 Lx/2J :5ai E
This implies
y?
x 2
-w>
2
121ogx
x
12logx
Problem N.14. Let T E SL(n, Z), let G be a nonsingular n x n matrix with integer elements, and put S = G-'TG. Prove that there is a natural number k such that Sk E SL (n, Z).
Solution. Let d be the absolute value of the determinant of G, a positive integer by the assumption. Every element of G-1 is a fraction with denominator d. Since we have Sk = G-1T kG = G-1(T k - I)G + I,
3.6 NUMBER THEORY
381
it is sufficient to prove that there is a positive integer k for which every element of T'k - I is a multiple of d (the determinant of Sk is automatically 1).
By the box principle, there are two matrices among To, T', ... , TC2 say T' and T', 1 > m, such that every element of TI is congruent to the corresponding element of T. This means that every element of Tl -T' = Tm(Tl-m - I) is a multiple of d. This property is preserved if we multiply it by the matrix T-m, which has integer elements by T E SL(n, Z).
Problem N.15. Let
f(n) = E pa p1n
p°
Prove that
lim sup
f(
n-oo
n
)
log log n = 1. n log n
Solution. It is well known that the number w(n) of different prime divisors of n is at most (1 +o(1)) (log n)/(log log n). Since obviously f (n) < nw(n), we conclude immediately that f (n) log log n < 1. lim sup n-oo nlogn To prove the other inequality, we select a parameter k and try to compose
an n from primes below k. The number of these primes is - k/ log k by the prime number theorem. Now take another number m, which we shall later specify in terms of k. Every prime p < k has a power between m and km. We cover the interval [m, km] with intervals of the type [m(1 + c)'- 1, m(1 + E)t], where i = 1, ... , [c log k], c depends only on e. For a
suitable i, there are at least l = [clk/(logk)2] primes p < k that have a power in this interval. We select 1 of them; let P be the product of these 1 primes. If P < Em, then there is a multiple of P in the interval [m(1 + E)t, m(1 + e)i+1]; let n be such a multiple. We have f(n) > lm(1 + E)i-1 > ln(1 - E)2.
We have to estimate n in terms of 1. We have P < k' anyway; thus the choice m = k'+1 guarantees P < Em if k > 1/E. Further, we have n < m(1 + E)t+i < km(1 + E)2 < k1+2,
hence l > (log n) /(log k) - 2. Also n < kk from the above inequality. Thus k >> (log n) /log log n, and we obtain log n 3 n log n l > (1 - E) f (n) > (1 - e) log log n'
log log n'
for large k.
Remark. It can be shown that max f (n) n<x
n log n log log n
382
3. SOLUTIONS TO THE PROBLEMS
Problem N.16. Let a and b be positive integers such that when dividing them by any prime p, the remainder of a is always less than or equal to the remainder of b. Prove that a = b.
Solution. By taking a p > max(a, b), we immediately see that a < b. Assume that a < b, and write
k = min(a, b - a). By a theorem of Sylvester and Schur, there is a prime p > k that divides (b). We show that this prime contradicts the assumption. k We know plb(b - 1)...(b - k + 1),
and thus the residue of b is a number r < k - 1. Now if a < b/2, then k = a < p, and thus the residue of a is a = k > r, as claimed. If a > b/2, then we have k = b - a. Write b = qp + r. We have
a - (q - 1)p = (b - k) - (q - 1)p = p + r - k > r, while p + r - k < p - 1, and thus this is the residue of a and it was larger than r.
Problem N.17. Call a subset S of the set {1, ... , n} exceptional if any pair of distinct elements of S are coprime. Consider an exceptional set with a maximal sum of elements (among all exceptional sets for a fixed n). Prove that if n is sufficiently large, then each element of S has at most two distinct prime divisors.
Solution. We prove the following slightly more general statement.
Statement. We call a subset S of the set {1, ... , n} k-exceptional if any k distinct elements of S are coprime. Consider a k-exceptional set with a maximal sum of elements (among all exceptional sets for a fixed n). If n is sufficiently large, then each element of S has at most two distinct prime divisors.
By maximality, every prime p < n must have a multiple in S. We divide the primes p < n into two classes: into P we put those for which there is an m > 1 such that pm E S, and into Q those for which the only multiple of p in S is p itself. We can estimate the number of p E P, p > x, as follows. For every
such p, take an m > 1, pm E S, and then a prime q1m. These primes satisfy q < n/x and any prime is represented at most k - 1 times among them; thus the number of our p's is at most (k - 1)7r(n/x). Consequently, if 1 < x < y < n, then the number of primes q E Q, x < q < y, is at least 7r(y) - -7r(x) - (k - 1)7rn. x
(1)
3.6 NUMBER THEORY
383
Now assume that there is a t E S that has at least three distinct prime divisors, say t = u"vQw7r, where u < v < w are primes, and a, /3, ry, and r are positive integers. Our plan is to find two suitable primes p, q E Q and replace t, p, q by up and vq. We must have p < n/u, q < n/v; thus the loss is
n n n n 11 t+p+q
If we can achieve up > cn and vq > cn, where c = 11/12, then the inclusion of up and vq offsets this loss. If there are at least two elements of Q in both (cn/u, n/u] and (cn/v, n/v], then this is possible. (We need two elements to guarantee that p q.) By (1), for this it is sufficient that
7ru-7r-> (k-1)7ru+2 u c U
and
n r--7r->(k-1)-7rv+2. V V c
By the prime number theorem, these inequalities hold if v < b f with a suitable constant b depending on k (say, 6 = 1/(3k) is a good choice). If v > b f , so that the previous procedure fails, we argue as follows. We have
n n U vw<-<-<-. 1
v2
b2
We try to find a prime p E Q and replace t and p by u3 p and vw with a suitable exponent j. The new number is admissible if p < n/ui. The loss is at most p + n, while the sum of the new terms is pug + vw > pug + 62n. Thus the process yields a profit if p(ui - 1) > (1 - 62 )n, that is, we need to find a p E Q satisfying
1-b2 n
u3 - 1
uJ
By (1), such a p exists if 7r ( n U3
)
-
7r
(
bl n) > (k - 1)7r
U3 -
(1,- b2)
.
(2)
Recall that u < 1/b2, hence there is a power of u in the interval (b-4, b-'] With this as uj, the left side of (2) is of order n/ log n, while the right side stays bounded, so (2) will hold for every sufficiently large n.
Problem N.18. (a) Prove that for every natural number k, there are positive integers al < a2 < < ak such that ai - aj divides ai for all 1 < i, j < k,
i# j.
3. SOLUTIONS TO THE PROBLEMS
384
(b) Show that there is an absolute constant C > 0 such that a1 > kck for every sequence al, ... , ak of numbers that satisfy the above divisibility condition.
Solution. (a) For k = 1, we can put a1 = 1. Now we show the inductive step from
k to k + 1. Assume that 0 < a1 < < ak is such a set. Write b = nk1 ai. Then the numbers b < b + al < ... < b + ak form a suitable collection of k + 1 numbers. Indeed, we have
(b + ai) - (b + aj) = ai - aj jai lb + ai, for 1 < i, j < k, i # j, and also (b + ai) - b = ai I b.
(b) Consider any prime p < k. If ai - aj (mod p), then ai - 0 (mod p) by the divisibility condition. Consequently, there are at most p - 1 numbers ai that are not divisible by p. Consider now the divisible ones. Dividing them by p, we again get an admissible system. Thus there can be at most p - 1 numbers not divisible by p among them.
Repeating this argument, we find that the exponent of p in A =
]lk1 ai is at least
k (k - (p - 1)) + (k - 2(p - 1)) + ... + (k - [
k
p-1
(k
p
](p - 1)1 1
> - p21(1 + Lp-l]I)3p
if p < -vfk-. Consequently, we have (with c = 1/3)
A > [ pck2/P, p
ak = max aj > Al/k > 11 pck/p p
!2g P > exp c'k log k = kc'k
= exp ck
p
p
for k > 4. From the relation ak - a1 jai, we infer a1 > ak/2 > 2, hence
a1 =
ai >
2a1 >
ak > kck
with C = c'/2. We have proved the inequality a1 > kck for k > 4. For k = 3 it is easy to see that the minimal possible value of a1 is 2, hence it holds if C < (log 2)/(3log 3). For k = 1 and 2, it fails by the examples {1} and {1, 2}.
3.6 NUMBER THEORY
385
Problem N.19. Let n1 < n2 < ... be an infinite sequence of natural tends to infinity monotone increasingly. Prove numbers such that that Ek_1 1/nk is irrational. Show that this statement is best possible in a sense by giving, for every c > 0, an example of a sequence n1 < n2 < ... k c for all k but >k 1 1/nk is rational. such that
Solution. Assume indirectly that E 1/nk = p/q, where p and q are positive integers. Then, for arbitrary k, we have °°
k
ni
1
p
=q
z-1 ni
Multiplying both sides of this equality by qnl ... nk, we see that 00
1ni
qnl ... nk i=k+1
is a positive integer for all k. Now we show that cc
n1 ... nk
1 --> 0
(k -' oo),
i=k+1 ni
and this contradiction will disprove the indirect assumption. By the monotonicity condition, we have 2-(k+1) 2-1 2-k 2-(k-1) > ... > n1 > nk > nk-1 nk+1
which yields nk+1+...+2
n1 ... nk <
k
On the other hand, by applying the inequalities 2-(k+2) 2-(k+3) 2-(k+1) < ... < nk+3 nk+1 < nk+2
and obvious estimations, we obtain 1
nk+1
+
1
nk+2
+
1
nk+3
+...<
1
1
1
nk+1
nk+1
nk+1
1
1
1
2
+ n3 nk+1 + n2k+1 k+1 1
nk+1-1 These two estimations together imply k
= nk+1 _2-(k+1) n < nk+i nl...nk ) (nk+l nk+1 - 1 nk+1 - 1 i=k+1 i
2 .
3. SOLUTIONS TO THE PROBLEMS
386
Here the first term tends to 1 and the second to 0 as k -4 oo; thus the whole expression tends to 0. To solve the second part of the problem, we are going to construct a sequence such that
nk
k
>c (k=1,2.... )and E 1
is rational.
k=1 nk
Take an arbitrary integer n1 > c2 + 1, and let nk+1 = nk - nk + 1 for k = 1, 2, .... The definition implies
nk+1 - 1 > (nk - 1)2 > ... > (n1 - 1)2k > c2k}1. Consequently,
nk-k
> c for all k. On the other hand, an induction shows
that k
1
nz
_
1
n1 - 1
1
- nk(nk - 1).
Thus, the sum of the series is the rational number 1/(n1 - 1).
Problem N.20. Prove that for every positive number K, there are infinitely many positive integers m and N such that there are at least KN/ log N primes among the integers m + 1, m + 4, ... , m + N2. Solution. The basic idea of the proof is to average for several values of m while excluding divisibility by small primes. Let Q = 3 5 ... pl, the product of the first 1 odd primes, and let c
be a natural number < Q such that -c is a quadratic nonresidue for the moduli 3, 5, ... , pl; such a c exists by the Chinese remainder theorem. We try to find m in the form m = c + kQ, where 1 < k < N2.
Consider an 1 < i < N; by the choice of c, we have (i2 + c, Q) = 1. By the prime number theorem for arithmetical progressions, the number of primes among the first N2 elements of this progression is asymptotically equal to N2Q 1 O(Q) log N2Q' thus for sufficiently large N (for a fixed value of Q), it exceeds
Q
N2
30(Q) log N'
Thus, the total number of primes (counted with multiplicity) among the integers
i2+c+kQ, is at least
1
N3
3¢(Q) log N
3.6 NUMBER THEORY
387
Consequently, for a suitable value of m = c + kQ, there are at least
N
Q
30(Q) log N
primes among the integes m + 1, m + 4, ... , m + N2. Since i
O(Q)
j['1Pi
is
> 1
t
A
and the sum E 1/p is divergent, we can select Q so that Q/30(Q) > K. Then the previous arguments yield that for every sufficently large N, the integer m can be chosen so that there are more than K K. (N/ log N) primes in the sequence m + 1, m + 4, ... , m + N2.
3. SOLUTIONS TO THE PROBLEMS
388
3.7 OPERATORS Problem 0.1. Let a, bo, b1, ... , b,i_1 be complex numbers, A a complex square matrix of order p, and E the unit matrix of order p. Assuming that the eigenvalues of A are given, determine the eigenvalues of the matrix
B=
boE
b1A
b2A2
...
abn_1An-1
boE
biA
... bn-2An-2
abn_2An-2
abn-iAn-1 b0E
\ab1A
ab3A3
ab2A2
bn_1An-1
...
bn_3An-3
...
boE
I
Solution. We show that if the eigenvalues of A are a1 i ... , AP then the eigenvalues of B will be the numbers
0(akAi)
(k = 1,...,n; j = 1,...,p),
where a1 i ... , an denote the roots of the equation Zn - a = 0, and
O(x) = bo + b1x + ... +
bn_1xn-1.
In fact, if AO, A,,-, An_1 are quadratic matrices of order p having complex entries, a is an arbitrary complex number and
Al
A2
... An-1
aAn_1
Ao
Al
...
An-2
aA1
aA2
aA3
...
A0
f Ao
C
I
then
det C = det M(a1) ... det M(an),
(1)
where
M(x) = Ao + A1x + ... + For a = 0 the assertion of this theorem is trivial. For a # 0, the validity of the theorem follows from the simple fact that M(a1) ... (0) An-lxn-1.
C=W (0)
... M(an)
3.7 OPERATORS
389
where W is the Kronecker product with E of the Vandermonde matrix built from the roots of the equation z' = a, that is,
E
f E
a1E
... anE
W =V®E= n-lE ... an-1E/ n 1
Apply (1) to the matrix B - AE (where £ is the unit matrix of order np) :
n
det(B - A£) = 11 det (q(akA) - AE). k=1
On the other hand it is well known that the eigenvalues of the matrix
q(akA) = b0E + blakA + ... +
bn_1ak-1An-1
are
0(akAl), ... W(akAp)' This proves the assertion.
Remark. Another group of solutions is based on the theorem stated, but not proved, by one participant in the following general form:
Theorem. If the eigenvalues of the quadratic matrix A of order p are (i, j = 1, ... , n) are regular functions on the disc Izi < A, then the eigenvalues of the matrix
1 i ... , gyp, further maxi JA2 I = A, and fib (z) f11(A)
... fin (A)
fnl(f1)
"'
fnn(A)
are given by the eigenvalues of the matrices f11(Ai)
"'
fln(Ai)
(i = 1, ... , p) fn1(Ai)
...
fnn(Ai)
One participant proved this theorem in the less general case where the functions fib (z) are polynomials; this proof can be applied in the more general case as well.
3. SOLUTIONS TO THE PROBLEMS
390
Problem 0.2. Let U be an n x n orthogonal matrix. Prove that for any n x n matrix A, the matrices Am
m+
1:
U-jAU'
j=o
converge entrywise as m -> oo.
Solution 1. For an arbitrary n x n matrix A, let IIAII denote the norm of A, that is, IIAII = sup IIAxII, IIxII=1
where x varies in the n-dimensional Euclidean space. If U is orthogonal, then U-1 is also orthogonal, and IIUAII = IIAUII = IIAII Entrywise convergence and convergence in norm of a sequence of matrices are equivalent. Thus, by the Bolzano-Weierstrass theorem, from a norm-convergent sequence of matrices it is possible to select a normconvergent subsequence.
For a given orthogonal matrix U, natural number m, and arbitrary ma-
trix B, put
m
1
Bm
1: U-'BU'.
qp + 1
i=O
We shall need the following lemma:
Lemma. For any natural number m,
II(Am)p-ApII=0. Proof. Let p > m. Then
_
1
p
1
+1m+1
(A"")p
1
1
m
E i=o j=o
U-i-j AUi+j
p+m
s(k)U-kAUk, p +lm+1 E k=0 where
1<m+l (k=0,...,p+m);
s(k) = o
if m < k < p, then s(k) = m + 1. Hence
pE II (Am),, - ApII =
p+lm+1 1
s(k)U-kAUk
1
k=O
M-1 1
1
p+lm+1
- p+lk_o 1
U-kAUk
p+m
(s(k) - m - 1)U-kAUk + k=0
s(k)U-kAUk k=p+1
3.7 OPERATORS
391
Now the assertion of the problem can be proved as follows. Since I I A,,,. II < All for all natural number m, the sequence {A,,,,} contains a subsequence {Am, } that is convergent in norm to a matrix H. Then for k oo we have
11H - U-1HUII < IIH - Amk II + IIAmk - U-1AmkUII
+ IIU-1(Ank - H)Ull <
2IIH-AmkII+mk+IIIAII -'0,
whence H = U-1HU. Thus, Hp = H for every natural number p. We prove that limp-. Ap-HII = 0. Actually, for every natural number k we have IIAp - HII = IIAp - HHII
IIAp - (A-k)pll + II(Arnk)p-HplI
IIAp - (A+m)pll + IIAmk - HII,
whence by the lemma we obtain limsup IIAp - HII < IIAmk - All.
p-oo
For k -* oo, the assertion follows.
Solution 2. Consider the natural embedding of the n-dimensional real space in the n-dimensional complex space. In the latter, the n by n real orthogonal matrices are unitary matrices. Thus, we prove more than required if in the statement of the problem we replace the orthogonal matrix by a unitary matrix, and real matrices A by matrices with complex entries.
So let U = (Uij) be a unitary matrix in the n-dimensional complex space. Then, as we know, in a suitable basis the matrix U has the form (Uij) = (Eibij), where bi.7 is the Kronecker symbol and Ie1I = Obviously,
(U')ik = Eibik,
= IEnI = 1.
(U-7)ik = Ei'bik
Now, if A = (aik) is an arbitrary matrix, then (U 7AU7)ik =
Ei jbirarsEBbsk = Ei 3aikEk = Et 3E3aikr,s
Therefore,
1
(Am)ik = m+1
m
1.` E-3Eaik =
If Ei = Ek, then Ei jEk = 1, so limn .
aik
m+1
m E.-aei
(Am)ik = aik.
3. SOLUTIONS TO THE PROBLEMS
392
On the other hand, if Ei # Ek, then (EiEk)m+1
aik j aik (Am)ik = m + 1 La (EiEk) = m + 1 j=0
- 1 -+0
EiEk - 1
as m -+ oo, since (EiEk)+1
-1
EiEk - 1
2
- (EiEk -1I .
We see that in this case limm_oo (Am)ik = 0.
It remains to note that if for a sequence of matrices (ai, i)
oo
M=1
we
know that for each pair of indices i, k the numbers tend to some aik as m -+ oo then the sequence of matrices, evidently, is entrywise convergent to the matrix (aik).
Solution 3. As a generalization of the problem, we prove the following theorem.
Theorem. Let H be a complex Hilbert space, and U a unitary operator in H. Then, for any compact operator A of H, the sequence of operators (Dm (A)
(m = 0,1, ... )
E U-jAU'
m+1
j=0
is weakly convergent, that is, for every pair of elements f, g in H (Dn(A))f,g) = 0,
lim
m,n-oo
where (f, g) denotes the inner product of the elements f, g E H.
(This statement obviously contains the statement of the problem. Indeed, the n by n real orthogonal matrix appearing in the problem induces a unitary operator of the n-dimensional complex Hilbert space; in this ndimensional space all linear operators, in particular those induced by n by n real matrices, are compact, and in finite-dimensional spaces weak and entrywise - so-called strong - convergences of operators coincide.) Proof. For proving the theorem, we first observe that the mappings A 4Dm(A) (m = 1, 2, ... ), defined for all bounded operators A of H, have the following properties:
a. For any pair of bounded linear operators A, B of H and any pair of complex numbers a and ,Q,
(Dm(aA+,3B) = 0- m(A) +,3(m,(B)
(m = 0,1, ...) .
b. For any bounded linear operator A of H, I I,Dm (A) I <_ IIAII I
where IIAII stands for the norm of A.
(m = 0,1, ...
3.7 OPERATORS
393
c. If the sequence {Ak}k 1 of bounded linear operators is uniformly convergent to a bounded linear operator A, that is, I Ak - A l -> 0 as I
l
k -+ oo, and the sequence is weakly convergent for each k, then the sequence {4'm(A)}m=o is also weakly convergent. From these three properties only c. needs to be verified. So let {Ak} 1
be a sequence of operators with the properties above, and let f and g be two elements of the Hilbert space H. Without loss of generality, we may assume that Ilf II = llgll = 1. Then l
(A)
- m.(Ak) + 4'm(Ak) - ,Dn(Ak) + Ak)f, 9) 1 +
< 211A - Akll +
,Dn(A)] f,9)l f, 9) 1 +
A) f, 9)1
P.(Ak)) f,9)1
(we have made use of properties a and b and applied the Schwarz inequality).
Now let e > 0 be arbitrary. Then, by one of the assumptions on the 1, there is an index k = k(e) such that
sequence {Ak}
IlAk(e) - All < 4.
Next, let N = N(k(E), E) be a positive integer satisfying Pn(Ak(e))) f,9) I <
2
for m, n > N. By the assumptions, such N exists. Then, in view of the foregoing, we obtain -P. (A)) f,9)I < E
for m, n > N. This proves property c. It is well known that every compact operator is the uniform limit of finite rank operators, further every finite rank operator is a finite linear combination of rank 1 operators. Consequently, by properties a. and c., it is sufficient to prove our assertion for operators of rank 1. So let A be an operator of rank 1. Then there are two elements 0 and ii in the space H such that Af = (f, O) for all f in H. Now, if h and g are any two elements of H, then (,Pm(A)h, 9) =
1 E (Ujh, 0) (U-j0, g) . 1 E (AU' h, U39) = m+1 m+1 j=0 j=0
Denote by H 0 H the tensor product of H with itself. (The definition of the Hilbert space H 0 H is the following. Denote by Ho the algebraic
394
3. SOLUTIONS TO THE PROBLEMS
tensor product of H with itself. As is well known, this is the free module generated by the symbols x ® y (x, y E H) and having the complex field for operator domain. By the inner product of two elements m
n
Xi ®yi,
x/
iV = > j=1
i=1
7
®y7
of H0, we mean the number n
m
(Xi,Xj)(yi,yj,)
(1)
i=1 j=1
Two elements', zb' E Ho are considered to be identical if (i - ', - ') = 0. On the factor space Ho that arises after this identification, (1) defines a norm. Completing Ho in the metric induced by this norm, we obtain the Hilbert space H = H ® H.) Let U0 be the operator of H satisfying the relation
Uo(x ®y) = (Ux ®U-1y)
(x, y E H).
U0 is uniquely determined in this way and is unitary in H. On the other hand it is clear that
m+1 j
(Uo(h®g)
.
By a classic theorem of ergodic theory (see for example F. Riesz and B. Sz. Nagy, Functional Analysis, Blackie, London, 1956, §144), the righthand side is convergent for every pair of elements ,, z 1' E H and so, in particular, for the pair = h ® , i' = 0 ® g. The proof is complete.
Problem 0.3. Prove that if a sequence of Mikusiriski operators of the form pe-as (A and µ nonnegative real numbers, s the differentiation operator) is convergent in the sense of Mikusiriski, then its limit is also of this form. Solution. Let µne-A^s be the convergent sequence of operators considered.
We may assume thatµn -ppandA-* A (0
{P} {q}
(1)
3.7 OPERATORS
395
is an almost uniformly (that is, uniformly on each finite interval) convergent sequence of functions in C[0, oo). We may assume that p(0) = 0, since otherwise {p}/{q} in (1) can be replaced by the equal expression ({p}{1})/({q}{1}). Define p to be zero on the negative half-line; then p will be a continuous function on the whole real line and not identically zero on the positive half-line. From (1) it follows that the sequence of functions µne-lns{p}
= {µnp(t - An)}
(2)
is almost uniformly convergent. If A = +oo, then we see that (2) is almost uniformly convergent to the zero function; thus µne-An9 --+ 0 = 0
e_8
(n -ti oo).
If A < +oo, then we show that µ is also finite. Let to be a point with p(to - A) # 0. Then p(to - An) - p(to - A) (n -' oo), so µnp(to - An) can only be convergent if p < +oo. Then, however, µnp(t - An) tends almost uniformly to the function {µp(t - A)} = pe-a9{p}. Consequently, µne-ans
-+
pe-as
(n -+ oo)
.
Problem 0.4. Prove that an idempotent linear operator of a Hilbert space is self-adjoint if and only if it has norm 0 or 1.
Solution 1. If T is self-adjoint, then it is an orthogonal projection operator; it is well known that such operators have norm 1 or 0. Conversely, let IITII < 1. For any element x of the space and any number p, we have
T (µTx - (x - Tx)) = pTx and therefore I,zI2IITxII2 < II uTx - (x -Tx)II2 = -2Re [,a(Tx, x - Tx)] + IPI2IITx1I2 + II x - Tx 112. Consequently,
Re[p(Tx,x-Tx)] <
2IIx-TxII2.
But this is possible for every p only if
(Tx,x-Tx)=0. Hence, it follows in a simple and well-known manner that T* = T. The proof applies to real as well as complex spaces.
3. SOLUTIONS TO THE PROBLEMS
396
Solution 2. The kernel (null space) of T, in view of the property T2 = T, coincides with the range of I - T, that is, with the set of all elements of the form (I - T)x where x runs through all elements of the space. It is a wellknown fact, and easy to verify, that the orthogonal complement of the range of I -T is equal to the kernel of I -T*. By a well-known result of Nagy (see, for example, B. Sz. Nagy and C. Foiaq, Harmonic Analysis of Operators on Hilbert Space, Akademiai Kiado and North-Holland Publ. Co., 1970), from the inequality ITII < 1 it follows that the invariant elements of T and T* are the same. So, the kernel of I - T* coincides with the kernel of I - T, which, in turn, coincides with the range of T (the latter fact follows from
the property (I - T) 2 = I - T). Thus (Tx, x - Tx) = 0 for every x, which implies the statement.
Problem 0.5. Let T be a bounded linear operator on a Hilbert space H, and assume that I ITn II < 1 for some natural number n. Prove the existence of an invertible linear operator A on H such that II ATA-1 II < 1.
Solution. Let (.,.) denote scalar multiplication in H. It is easy to verify that the formula n-1
[x,y] = 1: (T`x,Tiy) i=o
defines a new scalar product on H. Let H denote the space equipped with the scalar product [., .]. Obviously, n-1
(x, x) < [x, x] <
1 II"` 112
(x, x)
i=O
for all x c H, that is, the norm in H is equivalent to that in H. Consequently, H is also a Hilbert space. Since the dimension of a Hilbert space is the minimal power of complete sets, that is, sets whose closed linear span is the whole space, the dimen-
sions of H and H are equal. But Hilbert spaces of equal dimension are isomorphic. So, consider a Hilbert space isomorphism A : H -> H. Since the underlying vector space for H and H is the same, A is an invertible linear operator on H. Let x E H be arbitrary, and put y = A-1x. Then (x, x) - (ATA-1x, ATA-1x) = (Ay, Ay) - (ATy, ATy) = [y, y] - [Ty, Ty] n-1 n-1 _ [ (Tiy Tiy) - E(Ti+ly,Ti+ly) _ (y, y) - (Tny,Tny), i=0
i=ol
which is nonnegative since IITnII < 1. Thus, we have obtained
(x, x) > (ATA-1x, ATA-1x)
3.7 OPERATORS
for all x E H, that is,
397
IIATA-111 < 1. Therefore, the
operator A satisfies
the conditions of the problem.
Remark. One contestant showed that the positive square root of the positive, bounded, self-adjoint, linear operator E o T*''Ti can be chosen for A.
Problem 0.6. Is it true that if A and B are unitarily equivalent, self-adjoint operators in the complex Hilbert space f, and A < B, then A+ < B+? (Here A+ stands for the positive part of A.) Solution. We first show that there are self-adjoint operators A and B in C2 with eigenvalues al, A2 and .3i.\4, respectively, such that A < B but A+ B+. For instance, let A and b have matrices 0
-1
( 1 -1)
and
(0 0)
respectively. Then for (x, y) E C2, we have (B(x, y), (x, y)) - (A(x, y), (x, y)) = Ix12 + xy + ty + Iy12 = Ix + y12 > 0,
that is, A < B. From the matrices, it is clear that A and b are selfadjoint, B positive semidefinite, A indefinite. Thus b+ is equal to b, and A+ is positive semidefinite with A+(0,1) # (0, 0), since (0, 1) is not an eigenvector of A. Therefore,
(A(0, 1), (0,1)) > 0 whereas (B+(0,1), (0, 1)) = 0. B+. Now let f = 12 (complex). For simplicity, we write each element of N as a series E°° 1 anen convergent in 12, where So A+
n.
en = (0,...,0,1,0,...).
Let A
(anen) = A(al, a2) + n=1
Ananen, n=3
where An = .k if n = 4k + s, 1 < s < 4, and A1, A2, )3, A4 are described above. Here el and e2 are identified with (1, 0) and (0, 1). Further, let
= B(ai, a2) + E Ananen. B(anen) E 00
n=1
n=3
3. SOLUTIONS TO THE PROBLEMS
398
Both A and B are sums of two self-adjoint operators, so they are selfadjoint. Since B - A = B - A, we have o0
K (B - A)
anen,
anen n=1
n=1
= \(B -
A)(al, a2), (al, a2) J > 0,
that is, A < B. Similarly, B+ - A+ = B+ - A+ and, consequently, A+¢ B+ We now prove that A and B are unitarily equivalent. Let 00
00
U 1(anen) E
= U1(a,,a2) +
anen, n=3
n=1
where U1 denotes the unitary transformation of coordinates that reduces the matrix of h to diagonal form: ((JiEU1 1)(a, b) = (A3a, A4b).
Ul is obviously unitary, and 00
(U1BU1 1) E anen) = A3alel + A4a2e2 +
oc
Ananen
n=3
n=1
Let
00
U2
(fanen)
= > af(n)en7
n=1
n=1
where f denotes the following permutation of N: f (n) = n-4 for n = 4k+3
or n = 4k + 4 with k > 1. Further, f(3) = 1, f(4) = 2, and f (n) =n+4 for n = 4k + l or n = 4k + 2 with k > 0. Then 00
00
cc
n=1
bnen) _ E af(n)bn n=1
(U2 E anen n=1
00
00
00
_ E anbf-i(n) = (> anen, U2 1 E bnen n=1
n=1
n=1
which implies that U2 is also unitary. Further, we have (U2U1BUj 1U2 1)
(anen) n=1
00
=
Ananen. n=1
Finally, let 00
U3 Eanen =U3(a,,a2)+Eanen, n=3
n=1
where (U3 1AU3)(a, b) = (Ala,.2b), and U3 is unitary. With this third unitary transformation, we have
(U3U2U1BUUU) (anen) =A (anen) 1
and so A and B are unitarily equivalent. Consequently, the assertion is false.
n=1
3.7 OPERATORS
399
Problem 0.7. Let K be a compact subset of the infinite-dimensional, real, normed linear space (X, II . II). Prove that K can be obtained as the set of all left limit points at 1 of a continuous function g : [0,1 [---+ X, that is, x belongs to K if and only if there exists a sequence tn, E [0, 1[ (n = 1, 2.... ) satisfying limn_,,. tn, = 1 and limn. II9(tn) - xII = 0.
Solution. The compact set K C X has, for every E = 1/n (n E N), a finite E-net zl (n), z2 (n), ... , zin (n) E K, that is, if x E K, then I I x - zi (n) 11 < E = 1/n for some 1 < i < in. Denote by xo, x1, X2.... the sequence obtained by writing the finite (1/n)-nets z1(1), z2(1), ... , zjl (1), zl (2), Z2(2).... , ziz (2), zl (3).... one after the other. It would already be possible to define a piecewise linear function g: [0,1[-+ X such that g(1 - 1/k) = xk (k E N), and therefore each point of K is a left-hand limit point at 1 of the function g. The problem is that g may have further limit points that do not belong to K. So, we have to improve the procedure by interpolating a new sequence of function values yo, y1, y2, .... To define the latter, we shall need the following proposition (which is essentially the same as the well-known Riesz lemma on almost orthogonal vectors):
If L C X is a linear subspace of finite dimension, then there exists a vector y E X \ L such that I I y I I = 1 and dist(y, L) = 1, where dist(A, B) = inf { II a - bI I : a E A, b c B}
denotes the distance between the subsets A, B C X, and the singleton {a} is replaced by a. Indeed, since X is infinite dimensional, it contains a vector v E X \ L, and the subset
D={u: uE L,IIv - uII < IIvII} of the finite-dimensional subspace L is bounded and closed. Thus D is compact, so it contains a vector uo E D, which lies closest to v. Then, setting vo = v - uo, IIvoII = min{IIv - u I : u E D} = dist(v, L) = dist(vo, L). I
Consequently, the vector y = vo/IIvoII has the required properties. Next, we define a sequence in X by induction. Let Lo = lin {x0, xl }, and using the proposition just proved, choose a vector yo E X \ Lo such
that IIyo II = 1 = dist(yo, Lo).
(k E N) are already defined, then set
Lk = lin{xo,yo,xl,yl,...,xk-1,yk-1,xk,xk+1}, and using the preceding proposition again, choose the vector so that II yk 11 = 1 = dist(yk, Lk).
ykEX\Lk
3. SOLUTIONS TO THE PROBLEMS
400
For simplicity, we first define a continuous function 0: [0, oo[ -> X that has K for the set of limit points at oo. Fork E No = NU{0} and 0 < r < 1, put
0(3k + r) = (1 - r)xk + r(xk + yk), 0(3k + 1 + r) _ (1 - r)(xk + yk) + r(xk+l + yk),
0(3k + 2 + r) = (1 - r) (xk+l + yk) + rxk+l. Let us verify the desired properties. a. By the construction of the sequence xO, x1i x2, .... for any x E K there is a subsequence xk -* x; setting T,,, = 3k,,, we have
llm 0(7-,,,) = llm xk = x. b. On the other hand, we show that no x ¢ K can be a limit point of at oo. In fact, the closedness of K yields 6 = min{dist (x, K),1} > 0, and if for some To E [3ko, 3ko + 3] (ko E No) we have dist (x, 0(To)) <
6/3, then for every r > 3ko+3 the relation dist (x, 0(r)) > 6/3 is valid. Now there exists a natural number ko < k E N such that bl. either 3k+(26/3) < T < 3k+3- (26/3), in which case the relation P(TO) E Lk and the definition of 0 imply dist (0(To), 0(T)) >
3 26
and therefore
dist (x, (p(T)) > dist (0(ro), (b(T)) - dist (x, 0(TO)) >
37
b2. or 13k - T1 < 26/3, in which case for xk E K, by the definition of ¢, we have dist (xk, (b(T)) < 3
and so dist (x, (a(T)) > dist (x, K) - dist (K, (b(T)) > 3.
Thus, there is indeed no sequence r -* oo with 0(T,y) --+ x. Finally, the required function g : [0,1[--+ X is provided by the transformation
tl
tE[0,1[.
3.7 OPERATORS
401
Problem 0.8. Denote by B[0, 1] and C[0, 1] the Banach space of all bounded functions and all continuous functions, respectively, on the interval [0, 1] with the supremum norm. Is there a bounded linear operator T:B[O,1]->C[O,1]
such that T f= f for all f c C[0,1]? Solution. Suppose the existence of such an operator T. For any 0 < a < 1, consider the function
fa(x)
0
if x < a,
1
ifx>a.
The function ha = fa - T fa is, apart from the jump of size 1 at point a, continuous and satisfies T(ha) = T(fa - T fa) = 0. Replacing ha by -ha is necessary. We may assume that ha > 1/3 in a small, punctured, right-hand or left-hand neighborhood of a. Let a1 = 1/2, and define the sequence a1, a2i ... so that ai E (0, 1) and ai belongs to a corresponding small neighborhood of ai_1 for all i > 2. Then for the function gn = Ii 1 hay, we have gn E B[0, 1], gn is continuous apart from the jumps of size 1 at the points a1, . . . , an, and gn > n/3 in a small, one-sided, punctured neighborhood of an. It is easy to construct a continuous function fn satisfying the relation 1
I1fn-9nII sup Ifn(x)-9n(x)I < _. 2 xE [0,1]
(To speak exactly, the sign of equality is valid.) Then, for the norm of the operator T, we obtain the estimate
IIT(fn -9n)II > 2IITfn -T9nII 1, = 2IIfn!I > 2nIlfn-9nII 3
IITII >
valid for any natural number n. Consequently, T cannot be a bounded operator. It follows that there is no operator T satisfying the requirements of the problem.
Problem 0.9. Does there exist a bounded linear operator T on a
Hilbert space H such that 00
00
n Tn(H) _ {0} but n Tn(H)_ # {0}, n=1
where - denotes closure?
n=1
3. SOLUTIONS TO THE PROBLEMS
402
Solution. There exists an operator T of this kind. Let H be a Hilbert space of countably infinite dimension, and let {en}n 1 be an orthonormal basis in H. We define T on the basis vectors first. Let
ifn=1,
10 Ten =
if n > 2 is not a square, ifn = j2 for an integer j > 2.
en+1
ajel +
en+1
2, we assume that 0 < Iaj I < 1 for each j, and
For the sequence {o
aIaj12
We next extend the definition of T to all of H. For an arbitrary vector x E H, put 00
Tx :_
CnTen,
where Sn = (x)en).
n=1
First, we have to prove that the series that defines Tx is convergent, that is, the sequence of the partial sums is a Cauchy sequence. Let k < 1 be positive integers. Then 2
2 l
SnT en
ISn12 +
n=k
E
I
k<j2
n=k
since, substituting the defining expressions of the vectors Ten in the sum Etn=k SCnTen, in the linear combination of basis vectors so obtained the coefficients of the vectors en with k + 1 < n < 1 + 1 have absolute value less than or equal to 1, the coefficient of e1 is Ek<j2
nTen n=k
l
< n=k
1e3212
IajI2
IenI2 + k<j2<1
k<j2<1
I
<(1+a)1: ICn12. n=k
Since En 1 Ieni2 = IIxI12 < oo, the series E°° 1bnTen is convergent. The linearity of T is now obvious, and if in the preceding inequality we write
k = 1 and let 1 tend to infinity, then we obtain the boundedness of T, namely, IITII < 1+a. We show that e1 E T1(H)
for every positive integer 1. In fact,
ej2 = T2j-2e(j_1)2+1
for
j>2
3.7 OPERATORS
403
and therefore e1 + 1ej2+1
= a Tej2 =
T2j-1
Ca
7
e(j-1)2+1)
E
7
Thus,
el + ej2+1 E T2 1(H) C T1(H) if 2j - 1 > 1, and letting j tend to infinity, we obtain ET1(H)
.
7
It remains to show that n,1T1(H) = {O}. Suppose x E ni_°1Tt(H). From the definition of T it is clear that, for any 1, the vectors e,,, with 2 < n < 1 + 1 are orthogonal to Tl (H). Thus (x, en) = 0 for all integers n > 2. On the other hand, if Ty 0 for some y E H, then there necessarily exists an integer n > 2 with (y, en) # 0. Then, however, (Ty, en+1) :34 0 and, by the foregoing, x = TO = 0.
3. SOLUTIONS TO THE PROBLEMS
404
3.8 PROBABILITY THEORY Problem P.1. From a given triangle of unit area, we choose two points independently with uniform distribution. The straight line connecting these points divides the triangle, with probability one, into a triangle and a quadrilateral. Calculate the expected values of the areas of these two regions.
Solution. First, observe that an affine transformation does not change the ratio of the areas, therefore the expectation of the ratio does not change. We will use this fact by choosing the type of the triangle in the most convenient way for our purposes. Denote the vertices of the triangle by A1, A2, A3, and the chosen points
by X and Y. The probability that the line e connecting the points X and Y crosses one of the points A2 (i = 1, 2, 3) is equal to zero because of the independence of X and Y. So the probability that we will get a triangle and a quadrilateral by dividing the triangle by the line e is equal to one. Denote by ..4,t the event that the point Ai is a vertex of the small triangle. The events .At (i = 1, 2,3) form a complete system of events, so by the theorem of "complete expectations," 3
E(t I A-)P(A.),
E(7,) t_1
where t denotes the area of the small triangle, and T denotes the area of the original triangle. If the original triangle is equilateral, then E(t/TIA,t) is independent of i (so does P(.A1) for i = 1, 2, 3), hence 3
E(T) = E(TIAiP(.A..) = E(1IA1) i=1
Suppose now that we have a right-angled isosceles triangle, Ai is the right-angled vertex, and the equal sides have unit length. Denote by F(a, b) the (conditional) probability, that for the line segments C and q of the sides cut by e, we have < a and q < b, assuming that < 1 and q < 1. That is,
F(a,b)=P(t;
P(t; < 1, 77 < 1)
If X = (x1i x2) and Y = (y1i y2), thenr
P( < a, r7 < b) = c
J(a,b)
dxldx2dyldy2i
3.8 PROBABILITY THEORY
405
where c is a constant, independent of a and b, and 0(a, b) is the domain of the four-dimensional space such that, for the corresponding line, < a and y < b. By substitutions xi = axi and yi = by, the domain L(a, b) is transformed to A(1,1), and therefore
P( < a, y < b) = ca2b2
f
dxtdx2dyldyI
0(1,1)
= a2b2P( < 1, y < 1), that is, F(a, b) = a2b2. So the density function is f (a, b) =
a2F(a, Saab
b)
= 4ab,
and thus the ratio of the areas of the triangles is given by g(a, b) = 1 2 ib 1 = ab. 2
The expectation of the ratio is
jj 1
E(T) = E g(a, b) = 1
f1
=
J
1
g(a, b) f (a, b) da db 0
4a2b2 da db = 4
So the expectations of the areas of the triangle and the quadrilateral are 4/9 and 5/9, respectively.
Problem P.2. Select n points on a circle independently with uniform distribution. Let Pn be the probability that the center of the circle is in the interior of the convex hull of these n points. Calculate the probabilities P3 and P4.
Solution 1. Assume that the circumference of the circle is one. Fix a point on the circle, and introduce the arclength parameter. Let Tl...... n be the parameter values associated with the chosen points. Denote by An the event that the center of the circle is not inside the convex hull of the points. Furthermore, denote by Bi the event that there is no point on the arc (Ti, Ti + 1/2). By the independence and uniform distribution of the points, it is obvious, that, except for an event of probability zero, the events + Bn, P(Bi) = 1/2n-1, and Bi are mutually disjoint and An = B1 +
therefore Pn = 1 - P(An) = 1 - n/2n-1. Hence P3 = 1/4, P4 = 1/2. Solution 2. Denote by TiTj the length of the shorter arc with endpoints Ti and r3 and by Bi3 (1 < i < j < n) the event that all of the points are
3. SOLUTIONS TO THE PROBLEMS
406
on the shorter arc with endpoints ri and rj. (If ri and rj are the endpoints of a half-circle, then Bi, denotes the event that all of the points are on one of the half-circles defined by them.) It is obvious that An = Li,7 Bid If the points are distinct, the events Bi3 are mutually disjoint. Since the points are independent and uniformly distributed, they do not coincide with
probability one, and so P(An) = >i i P(Bij). Furthermore, P(T x) = 2x, so (P(TiTj < x))' = 2 (0 < x < 1/2) and P(BijJ'r < x) = xn-2. Because of the theorem of total probability,
P(Bij) = 2
f
1/2
1
x'L-2dx
= (n
0
-
1)2n-2
Hence,
Pn = 1 - P(An) = 1 - (2)
(n -
1)2n-2
=1-2n-1 .
Problem P.3. Let El, E2,- - -, E2n be independent random variables such that P(ei = 1) = P(Ei = -1) = 1/2 for all i, and define Sk = 'i 1 Ei, 1 < k < 2n. Let N2n denote the number of integers k E [2, 2n] such that either Sk > 0, or Sk = 0 and Sk-1 > 0. Compute the variance of N2n.
Solution. It is known (see, for example, W., Feller, An Introduction to Probability Theory and Its Applications, Wiley, New York, 1957, Vol. I, p. 77) that the distribution of N2n is given as 12k2(n-k))
(k)(
P(N2n = 2k) =
(k = 0, 1, ... , n),
22n-k n
(k = 0,1, ... , n - 1).
P(N2n = 2k + 1) = 0
(1) (2)
The symmetry of the distribution implies that E(N2n) = n, and from (1) and (2) we conclude that 12k) (2(n-k)\
n
E(N22n) _ E(2k)2 k
n-k 1
22n
k=0
Assume that JxJ < 1, and consider the functions 2k)
F(x) _
- (2k)2 2k
x2k and G(x) = (1 -
x2)-1/2.
k=1
It is easy to see that CIO
(2k)
G(x)-= E 92k kx2k k=0
(3)
3.8 PROBABILITY THEORY
407
The generating function of E(N2n) is F(x)G(x), that is, F(x)G(x) = 1: E(N2
)x2k
k=0
Using (3), it is easy to verify that
F(x) = x(xG'(x))', which implies that F(x) = x(x2(1 - x2)-3/2)t = 2x2(1 - x2)-3/2 + 3x4(1 - x2)-5/2.
Therefore, F(x)G(x) = 2x2(1 - x2)-2 + 3x4(1 - x2)-3.
(4)
Simple calculations show that x2)-2 =
(1 -
0" kx2k-2 k=1
and
x2)-3 = [ k(k2 1)x2k-4
(1 -
k+1
So from (4), we get
F(x)G(x) =
2kx2k + k=1
00 3 k=1
k(k - 1)x2k =
2
oc'
k=1
(2
k2 + 2 k I x2k
Therefore,
E(N2n) = 3n2 + 2n . Hence,
Var (N2n) = E(N2n)
- E2(N2n) = 12n2 + 2n J - n2 = n(n2+ 1)
Problem P.4. A gambler plays the following coin-tossing game. He can bet an arbitrary positive amount of money. Then a fair coin is tossed, and the gambler wins or loses the amount he bet depending on the outcome. Our gambler, who starts playing with x forints, where 0 < x < 2C, uses the following strategy: if at a given time his capital is y < C, he risks all of it; and if he has y > C, he only bets 2C - y. If he has exactly 2C forints,
3. SOLUTIONS TO THE PROBLEMS
408
he stops playing. Let f (x) be the probability that he reaches 2C (before going bankrupt). Determine the value of f (x).
Solution 1. We are going to prove that f (x) = x/2C (0 < x < 2C). First, we show that f (x) is nondecreasing. Let 0 < x1 < x2 < 2C, and suppose there is a sequence of tossings such that he can reach 2C from x1. We show that by the same sequence of tossings he can reach 2C from x2 (maybe earlier), so f (X2) > f (xl) . Let us see how the amount of the player's money changes after one toss: (a) if x1 < x2 < C, then 2x1 < 2x2 for a win, and 0 < 0 for a loss; (b) if x1 < C < x2i then 2x1 < 2C for a win, and 0 < 2(x2 - C) for a loss;
(c) if C < x1 < x2, then 2C < 2C for a win, and 2(x1 - C) < 2(x2 - C) for a loss. Thus if x1 < x2, then, for the new amounts xi and x2, we have xi < x2. Define x = (a/2n) 2C (1 < a < 2' -1 , a is odd, n = 1, 2,... ). We will
prove by induction that for such an x, f (x) = a/2n = x/2C. For n = 1, f((1/2) 2C) = f (C) = 1/2, since starting from C the player will have 2C with probability 1/2. Suppose the formula holds for n = k (k > 1), that is, f (x) = f ((a/2 k) 2C) = a/2k. If x = (a/2 k+1) 2C and a < 2k (that is, x < C), then the player will win 2x forints or lose everything with probabilities 1/2, 1/2, respectively. So f (x) = 2 f (2k 2C) + 2 f(O) = 2k+1
2C
If x = (a/2 k+1) 2C and 2k < a < 2k+1 (a $ 2k since it is odd), then the inequality x > C implies
if (a-2 2k
A( xX )) = 2 f (2C) + 2 f 1
I
\
- 2C I =
1a-2k
1
2
+
2
2k
=
a_ x 2k+1
2C
Suppose now xo 0 (a/2n) 2C (1 < a < 2' - 1, a is odd), and suppose f (xo) # xo/2C, but, for example, f (xo) > xo/2C Then there should exist a number having the form a/2n such that 2C < 2n < f (xo),
(1)
and it contradicts the fact that f (x) is nondecreasing. Namely, because of (1), xo < (a/2n) 2C, and, on the other hand, 2n
= f (2n 2C) < f (xo)
We can obtain a similar contradiction for f (xo) < xo/2C, therefore f (x) = x/2C (0 < x < 2C).
3.8 PROBABILITY THEORY
409
Solution 2. Introduce the notation g(x) = f (2Cx) (0 < x < 1). Since f (0) = 0 and f (2C) = 1, we get g(0) = 0 and g(1) = 1. It is easy to verify
that g(x)
2
+
2g(2x - 1)
if x E (2',1].
(2)
To see this, observe that if x E (0, 1/2] , the gambler wins or loses 2Cx forints with probability 1/2, so g(x) = 29(2x) + 2g(0) = 2g(2x).
If x E (1/2, 1], the gambler will have 2C or (2x - 1)2C forints with equal probabilities 1/2 , so
g(x) = 29(1) + 2g(2x - 1) = 2 + 2g(2x - 1). We will next show that the only bounded solution of the functional equation (2) is the identity function g(x) = x (from which we get the same solution as before).
Put h(x) = g(x) - x. Then h(x) is bounded and 2h(2x)
h(x)
if x E [0, 2],
h(2x - 1) if x E (2,1].
(3)
We are going to show that h(x) = 0. Since h(x) is bounded, both M = supXE[0,1] h(x) and m = infxE[o,1] h(x) are finite, and
h(2x) < M
h(x)
2
if x E [0, 1], 2
2h(2x - 1) < M if x E c1,1].
The definition of M implies that M < M/2, that is, M < 0. A similar argument shows that m > 0, that is, h(x) - 0. Problem P.S. For a real number x in the interval (0, 1) with decimal representation
0.a1(x)a2(x)...an(x).... denote by n(x) the smallest nonnegative integer such that an(x)+lan(x)+2an(x)+3an(x)+4 = 1966.
Determine fo n(x)dx. (abcd denotes the decimal number with digits a, b, c, d.)
Solution 1. The integral is understood as a Lebesgue integral. Introduce the following notation:
Ln = {x : x E (0, 1), n(x) = n}, An = {x: x E (0, 1),
an+1(x)an+2(x)an+3(x)an+4(x) = 1966}.
3. SOLUTIONS TO THE PROBLEMS
410 )\
is the Lebesgue measure, An = \(L,,), an = \(An), where (n = 0,
1, 2, ... ). Observe that Ln and An are the unions of finitely many intervals, so they are measurable sets. First, we show that Y-°O o )1n = 1, that is, the integrand is defined almost everywhere. If
H={x:xE(0,1), ak+1(x)ak+2(x)ak+3(x)ak+4(x) # 1966, k = 1,2,...}
and
Cn={x:xE(0,1), a4k+1(x)a4k+2(x)a4k+3(x)a4k+4(x) ¢ 1966, k = 1,2,...},
then H C Cn (n=0,1,2.... ) and A(Cn)
- (104 104-1)n+1
Therefore, limn . A(Cn) = 0, \(H) = 0. The lengths of the intervals in An are equal to 10-(n+4), the number of these intervals is 10" (we can choose the first n decimal digits in ion different ways), and therefore an = 10-4.
Obviously, An f1 Ln = Ln (n = 0,1.... ), and if n - 3 < k < n - 1, then An fl Lk = 0 since two sequences 1966 cannot overlap. If 0 < k < n - 4, then ,\(An n Lk) = \(An-k-4),\(Lk) = 10-4\k, since the assumption that x is in Lk restricts only the first (k + 4) decimal digits of x. So n-4
n
10-4 = \(An) = E A(An n Lk) = : 10-4\k + \n, k=0
k=0
that is, n-4 104I\n = 1
-
ak k=0
By the equality Ek o Ak = 1, we get
= \k = 104I\n+4
(1)
k=n+1
For the integral of n(x), 00
00
n'\n =
n(x)dx = n=0
1 00
n=Ok=n+1
\k.0
3.8 PROBABILITY THEORY
411
Hence, by (1), o0
f n(x)dx = E0"104An+4 = 1O4 0
n=0
An n=4
=104(1-A0-A1-A2-A3), and since )o = Al = A2 = A3 = 10-4, we clearly get 1
L
n(x)dx = 104 - 4 = 9996.
Solution 2. We solve the problem in a more general form. Let a = s1iS2,...,Sk be an integer given in the decimal system, and for x E (0, 1) define the function n(x) as follows: let n(x) be the smallest nonnegative integer such that an(x)+lan(x)+2 ... an(x)+k = S1S2 ... Sk.
We are going to calculate the value of ff n(x)dx. Let (Il, A, P) be the probability space of the Lebesgue-measurable subsets of the interval (0, 1) endowed with the Lebesgue measure. Then on the set S1 = (0, 1), the digits al, a2i ... , ak, ... are independent and
P(ak-a)
10
(k=1,2,...; i=0,1,...,9).
On this space, n(x) is a random variable, and its expectation is
I
1
n(x) dx.
Put Pn = P(n(x) = n) and Qn = P(an+lan+2 ... an+k = a) = 1110k
(n=0,1,...).
Let l be the smallest natural number such that shifting a by 1 positions,
the digits on the same positions coincide. (That is si = si+1 for all i = 1, 2, ... , k - 1). Obviously, 1 < 1 < k . Rewrite kin the form k = pl + r, where 0 < p, 1 < r < 1. The a consists of p blocks r = T1-s2 .... sl of length 1, and the first r digits of this block r. Notice that Qn = (POQn-k + PiQn-k-1 + ... + Pn-kQO)
+
(Pn-k+r10-(k-r)
+
Pn-k+r+110-(k-r-1)
(n = k,k + 1,...) since the event an+lan+2 ... an+k = a
can occur in two different cases, as follows:
+... + Pn_110 ' + Pn)
(2)
3. SOLUTIONS TO THE PROBLEMS
412
(a) Block a appears only from the (i + 1)th index (0 < i < n - k), and after n - k - i digits it occurs again. (The probability of this case is (b) Block a appears from the (n - k + r + il)th index (i = 0,1,. .. , p), thus an-k+r+il+1an-k+r+il+2 ...an+r+il = a,
and it appears again starting from the (n + 1)th index. Note that besides the previous assumption, it is also necessary that the values of the undefined (k - r - il) digits are given. The probability of this case is Pn_k+r+il10k-l-il Assume that Pn = Qn = 0 if n = -1, -2, .... Then (2) holds also for 0 < n < k. Multiply (2) by zn, add up the resulting equations for all values of n, and introduce the notation 00
Q(Z) = E
P(z) _ E Pnzn,
Qnzn
n=0
n=0 to get P
Q(Z) = zkp(z)Q(z) +
zk-r-i1P(z)10-(k-r-ti).
i=o
Using the obvious formula 00
- = lok zn = 10k(1 - z)'
Q(z) _
1
1
we have
P(z) _
1
zk + iok (1 - Z)
o
zk-r-il10-(k-r-il)
From P(1) = 1 we can see that n(x) is defined almost everywhere, and P
n(x)dx = P'(1) _ -k + E 10r+il 10
i=0
This last equality shows that the integral is always an integer. If a = 1966,
then k=l=r=4,p=0, and thus 1
n(x)dx = 104 - 4 = 9996. Jo
Ifa=1961,then k=4,l=3,p=r= 1, and thus n(x)dx = 10 + 104 - 4 = 10,006. Ji
3.8 PROBABILITY THEORY
413
Ifa=1919,thenk=4,l=r=2,p=1,thus
r
1
= 102 + 104 - 4 = 10,096,
and, finally, if a = 1111, then k = 4, 1=r= 1, p = 3, thus 1
L
n(x)dx = 10 + 102 + 103 + 104 - 4 = 11, 106.
Remarks.
1. The problem can obviously be generalized for any number system not just the decimal one. 2. One can reduce the problem to compute the expectation of the first return time of a recurrent sequence of events. In this case, the value of the integral can be determined using the theorem by Erdos, Feller, and Pollard. Problem P.6. Let f be a continuous function on the unit interval [0, 1]. Show that 1
moo
n
1
...
fr
I dx1 ... dxn = f 2 I (1
n
10
10
and li
JJ0' 0
...J
1
1
f (e
e
0
Solution 1. Let k be an arbitrary positive integer and n > k. Consider the multinomial expansion of (E 1 xi)'. The number of terms that contain all variables with powers not exceeding one is
n(n- 1)... (n-k+1)
=nk+O(nk-1)
The integrals of these terms on the unit cube
Rn={(xl,...,xn): 0<xi<1} (i=1,...,n) are equal to 1/2k. The number of terms containing at least one xi with a power higher than 1 is not greater than n nk-2 = nk-1. The integrals of these terms on Rn are not greater than 1. From this observation, we get
x1+...+xn n
asn -+oo.
11
k
l
dxl...dxn -*
\2/
k
3. SOLUTIONS TO THE PROBLEMS
414
An easy calculation shows that 1
0
...
f
'a
1
1
(n xl ... xn)kdxl ... dx,, _
x
0
i=1
k/indxi
1
-'
= (1 + n)
0
( 1\k e
asn ->oo. Hence, we have shown that the statements are valid for f (x) = xk. It is also obvious that the statements hold for constant functions, and if they are valid for two continuous functions, they are valid for their linear combinations as well. So we have proved the theorem for polynomials.
If the function f is continuous on [0, 1], then by the Weierstrass approximation theorem, for any positive e there is a polynomial p(x) such that I f (x) - p(x) I < E/3 (0 < x < 1). Since the statements are valid for polynomials, there is a K such that f 1...
J0
1p
Cxl+'n +xnl
0
2J
dx1...dxn,-7 (1) <
1
<
p(xl...x)dxl...dx-pJ1...
e
0
3,
3,
for any n > K. Therefore, for any n > K,
f1 f(x1+ n+xn)dxl...dxn-.1 x
f1 f1
...I p\ 1
n
0
J0
P
dx1...dxn
2(1)
f (21)
(1 dxl...dxn-p\2)
< E,
and similarly,
j...jf(xixn)dxidxn_f() 1
1
< E.
Now the proof is completed.
Solution 2. Let Sl, ... , fin, ... be mutually independent, uniformly distributed random variables on the interval (0, 1). Then the variables rin = log en (n = 1, 2, ...) are also mutually independent, and they have the same distribution. Since E(en) = 1/2 and E(rjn) _ -1 (n = 1, 2, ... ), a theorem by Kolmogorov implies that
cn
__
51+...+en
n
-+
1
2
3.8 PROBABILITY THEORY and
Vn
771+...+17n
_
415
-1
n
with probability 1. By the assumptions the function f is bounded on the interval [0, 1] and continuous at x = 1/2 , and the function g(x) = f (ex) is also bounded on (- 00, 0] and continuous at x = -1. Using the Lebesgue theorem, we get
E(f(cn)) -* E(f(1)) = .f(2),
E(g(V')) -> E(g(-1)) = E(f(e)) _ f(e), and, finally, the definitions of cn, cp;,, and g imply that \ f(1)=E(f(Vn))=J ... 2 f1
0
f(X1+...+xn\dxl...dxn,
1
n
0
J
and
f (e J =
11. ..
1 f( n xl ... xn)dx1 ... 0
Remark. The statements also hold in the more general case when f is bounded, integrable, and continuous at x = 1/2 and x = 1/e, respectively. In the first statement we can also drop the condition of boundedness (as was pointed out by Laszlo Lovasz).
Problem P.7. Let A1, . . . , An be arbitrary events in a probability field. Denote by Ck the event that at least k of A1i . . . , An occur. Prove that n
n
11 P(Ck)
fj P(Ak)-
k=1
k=1
Solution. We start with three remarks: Comment A. The statement is true for n = 2, that is,
P(A + B)P(AB) < P(A)P(B). To see this, use the notation AB = x, AB = y, and AB = z. Then our inequality becomes
(P(x) + P(y) + P(z))P(x) < (P(x) + P(y))(P(x) + P(z)), which obviously holds.
Comment B. If Al D ... D An, then Ck = Ak.
3. SOLUTIONS TO THE PROBLEMS
416
Comment C. Ck is identical to the event that at least k occur from the events A, + A2, A, A2, A3, ... , An, that is, from the events A, + A2, A, A2, A3, ... , An, exactly as many occur as from the events A1, A2, ... , An. This statement is trivial. Our proof is based on the above comments. Put A° = Ai (i = 1, . . . , n), and assume that the events Ai , ... AA are already defined. Next, define the events ,A`+1 as follows: choose a pair Aq , A of events, i < j, Ai+',
for which Al' +1
A' , A AZ (that is, Aµ and A' are incomparable) and = Aµ + A , A +1 = A A and Ak+1 = Ak for k # i, j.
define Ai Continue this procedure until there exists at least one pair of incomparable events. This procedure will terminate in finitely many steps, since in each step the number of incomparable pairs decreases. This observation follows from the facts that Ai +1 and A +1 become comparable, and any event is comparable to at least as many of the events Aµ + A, Aa AJ , as of Ai and By comment A, n
n
H P(Ak+1)
11 P(Ak).
k=1
k=1
(1)
Observe that comment C implies that the events Ck remain unchanged, and because of comment B they are identical to the final events Aq . Therefore, (1) implies the original statement.
From the proof of comment A, we see that if 11k=1 P(Ak) # 0 and P(Aq A7) # 0, then strict inequality holds in (1). Hence, equality holds if either 11k=1 P(Ak) = 0 or the original system of events {Ak} are already "ordered." Remarks.
1. If multiplication is replaced by addition, then in the statement we always have equality. That is, n
n
P(Ai).
P(Ci) i=1
i=1
2. Let Sk(x1, ... , xn) be the kth elementary symmetric polynomial of the variables x1, ... , xn. Then Sk(P(Cl), ... , P(Cn))
Sk(P(A,),... , P(An)).
Slight modifications of the above proof can be applied to show the validity of this more general statement.
Problem P.8. Let A and B be nonsingular matrices of order p, and and ri be independent random vectors of dimension p. Show that if , ri and A + 913 have the same distribution, if their first and second let
3.8 PROBABILITY THEORY
417
moments exist, and if their covariance matrix is the identity matrix, then these random vectors are normally distributed.
Solution. Put C = A + r7B. Without loss of generality, we can suppose that E(i7) = E(() = 0. Since and 77 are independent, Var
77B)] = A* Var
B* Var (77)B,
where Var (c), Var (77), and Var (() denote the covariance matrices of , 77, and C, respectively. They are equal to the identity matrix, so
A*A+B*B = I. Thus, for arbitrary vector t, (tA*, tA5) + (tB*, tB*) = (t, t),
and by the regularity of A and B, ItA*I< Its
ItB*I < Its
(t # 0),
(1)
where 1.1 denotes the Euclidean length of vectors. Since and 77 are independent, E(ei(t,tA+vB))
= E(ei(tA'O)E(ei(tB',l1)).
If ¢(t) denotes the common characteristic function of , 77, and c, then W(t) = cv(tA*)cv(tB*)
(2)
We are going to show that W(t) # 0. Assume that cp(t) has real roots. Then its continuity implies that there is a root to with smallest absolute value. Since cp(0) = 1, to $ 0. Thus, (1) implies that co(toA*)
0,
co(toB*)
0,
and this contradicts equation (2). Consider the function log cp(t) + 2 (t, t),
and choose the branch of the logarithmic function for which 0(0) = 0. The assumptions imply the existence of vectors dd
VI(t) =
`P(t),
(0) =
0
3. SOLUTIONS TO THE PROBLEMS
418
and matrices fz
cP (t)
dtz
co"(0) = -Var (
co(t),
Therefore,
) _ -E.
[(t)l dz
dt
J t=o
t-o
= 0,
and hence li(t) = o((t,t))
if
Itl , 0.
(3)
We are going to show that O(t) - 0. Suppose O(t) 0 0. Then there exists a positive E and a t # 0 such that IV,(t) I > E(t, t). Because of the continuity of co(t), there exists such a t1 with minimal absolute value , and (3) implies that t1 # 0. From inequality (1), we conclude that I,0(t1A*)I < E(t1A*)t1A*)
and
I',(t1B*)l < E(t1B*,t1B*). From (2),
,O(t) =,O(tA*) + ,(tB*), so
I'(tl)I <- ,(t1A*)I+I ,(t1B*)I <E[(t1A*,t1A*)+(t1B*,t1B*)] =E(tl,t1). I
This inequality contradicts the definition of t1. Consequently, fi(t) - 0, and so co(t) = e- 2L (t,t).
Problem P.9.. Let 1, z, ... be independent random variables such that Eln = m > 0 and vz < oo (n = 1, 2, ... ). Let {an} be a sequence of positive numbers such that an -+ 0 and >°O_1 an = oo. Prove
that
Jim
P
n
1: akSk = 00 n- oo
= 1.
k=1
Solution. We can solve the problem using the Kolmogorov inequality, but in the solution given below we only use the Chebyshev inequality. Instead of an -+ 0 and Var (£n) = v.2, it is sufficient to assume that {an}
is bounded and Var ( n) = a2. For instance, suppose 0 < an < 1. Put Sn = Ek=1 ak, and define the sequence NJ by S,.,, < n2 < Sr..+1. Then
nz-1<Sr,.
(1)
3.8 PROBABILITY THEORY
419
If limnk -1 akG = oo does not hold, then there is a K such that n CC
akSk < K
(2)
k=1
occurs infinitely many times. It is sufficient to show that for any fixed K, this event has probability 0 since the union of these events for n = 1, 2.... includes all event sequences for which limn E akG = oo does not hold.
Let Ak be the event that there is an n between rk and rk+1 such that (2) holds. Note that (2) holds infinitely many times if infinitely many Ak occur. Denote the probability of Ak by Pk. We have to show that with probability one only finitely many Ak occur. We prove this by verifying that E' 1 Pk is convergent. In this case, >k>m Pk --> 0, and >k>m Pk is greater than the probability that infinitely many Ak occur. Therefore, this latest probability is equal to 0. (Actually, here we have proved and applied the Borel-Cantelli lemma.) Put 7)k = Eik aiSi and Ok = Ei k rk+1 ai j. If (2) holds for some rk < z < rk+1 with z replacing n, then either rk
aie
2
kzm
i=1
or
z
E aiSiC-k23, i=rk+1
supposing k2m > 6K. In the last case, O k = Ei k rk+1 ai I fi I > k2m/3, so at least one of the inequalities Ok > k2m/3 and 11k < k2m/2 is valid. Thus, Pk <
P(qk <
3)
l k2mI+P
k2m
2
(3)
We have to show that on the right side both terms are small. It is obvious that E(77k) = mSrk and Var (7)k) < cr2Srk since ai < 1 implies that Var (7Ik)
rk
rk
Q2 E ai
Q2 E ai
i=1
i=1
= o2Srk.
Furthermore,
E(Ieil) <E(1+E1) =
1+m2+Q2 =A
and
Var(Ie
<E(ei)
3. SOLUTIONS TO THE PROBLEMS
420
Since the absolute values of independent random variables are independent, E(V)k) <-
(Srk+l - Srk)
max
rk
(Srk+l -
Srk)A = 0(k)
and rk+1 q/'
Var (00 <
a2
i=rk+l
rk
+l Var(Iei1) C
(Srk+l
-Srk)A=0(k).
Now we can use the Chebyshev inequality and (1) to see that E(rlk) k2z )
P(77k
Var (ilk) - k2Q,
k2m,
k3 2
P(177k
- E(ilk)I ?
k2
P(ibk > 3) < P(I'k - E('k)I ?
) = 0(
z1
),
z k3 ) = O(T2
if k is large enough. From these relations, the convergence of E Pk follows.
Remarks.
1. One can sharpen the proof to see that >kk
an = 0
i-(1+E1'-4l og (S.
/
(which is still weaker than boundedness) does.
P.10. Let
be independent, uniformly distributed, random
variables in the unit interval [0, 11. Define
h(x) = n #{k: t9k < x}. Prove that the probability that there is an x0 E (0, 1) such that h(xo) = xo,
is equal to 1 - n.
Solution 1. Let us denote by Ik the interval ((k - 1)/n, k/n). Consider the elementary events that for every i, i9i belongs to Ik,, where ki is given for every i. So we decomposed the original sample space to the union of
n' disjoint events of probability 1/nn (events with probability zero are omitted). These events will be called atoms.
3.8 PROBABILITY THEORY
421
1 = h(x) is an integer multiple of 1/n, Since the value (1/n) in order to check whether h(xo) = xo holds for some x0, it is sufficient to know which intervals Ik, contain t9i, that is, which atom will be the result of the experiment series. They have the same probability; therefore, it is sufficient to determine the number of atoms for which there is no k such that exactly k i9ti occur until k (1 < k < n). Denote by An (1 < k < n) the number of atoms for which there is no k such that the first k intervals contain exactly k t9i values. Let Bk denote the number of atoms for which there are exactly k i9t values in U3
nn - An = y k=1
n k
Ak(n - k)n-k,
Al=1.
(1)
We are going to verify that An = nn-1 by mathematical induction. It is easy to check that Al = 10 and A2 = 21 or A3 = 32. Suppose that we have already verified this relation up to n - 1. In order to show this equation for n, we have to show that n-1
nn -
nn-1
k=1
nkk-1(n - k)n-k (k)
(2)
It is easy to see by interchanging k and n - k that this equation is equivalent to n-1
()k2(n
nn-1(n - 1) _
-
k)kk-2(n
- k)n-k-2
(3)
k= 1
We will prove this equality by a well-known theorem of Cayley. Consider the trees having the numbers 1, 2, ... , n as vertices. (A tree is a connected
graph not containing a circle.) Two graphs are different if they do not have the same (i, j) pairs as edges. By the Cayley theorem, there are nn-2 different such trees. Now choose in every tree in all possible different ways one vertex and one edge. We will call these graphs vertex-edge-signed trees, and consider these graphs different if they are different before our vertexedge selection, or the selected vertex-edge pair is different. Since a tree has n - 1 edges, we have nn-1(n - 1) different vertex-edge-signed trees, and this number is exactly the left-hand side of (3). Let us count the number of these vertex-edge-signed trees in another way. By omitting the signed edge, we get a tree with k vertices with one signed vertex and a normal tree with (n - k) vertices (k = 1, 2, ... , n - 1). On the other hand, if we choose
3. SOLUTIONS TO THE PROBLEMS
422
k vertices out of n, define a tree on them, select a vertex and define a tree from the remaining n - k vertices, and connect the two trees by an edge, which will be called the signed edge, then we get a vertex-edge-signed tree. Thus, the two procedures are the reverse of each other. In this way, we get n-1
11 1 k=1 (k)
k kk-2 (n - k)n-k-2 k(n - k)
different graphs, which is the right-hand side of (3).
Solution 2. The first part of the solution is the same, but here we give an algebraic proof of (2).
Lemma 1. If 0 < j < m, then
E ()-k 1)kk' = 0. k=1
Proof. Denote by D the operation of differentiation followed by a mul-
tiplication by x. Thus Dm(xk) = kmxk. It is easy to see by induction that if x0 is a root of the polynomial p(x) with multiplicity mo, then x0 will also be a root of Dm(p(x)) with multiplicity at least mo - in. Apply this observation to the polynomial p(x) = (1 - x)m and the operator Dj, j < m. Then we get that x = 1 is a zero of the polynomial D1.((1 - x)m) _ 1: k=1
(k)(-1)kkjxk. /J
This completes the proof of Lemma 1. Lemma 2. n-1
(n)
k1
k
kk-1(z
- k)n-k =
nzn-1 - nn-1
Proof.
n-1
n)
1: I k kk-1(z - k) k=1\ /
n-k
--
n-1
n-k
( n - k 1 zj(-k)n-k-.7
j1=0
{\n
(n)kk-1
k=1
_
1"
n-ln-k
E k!j!(n k=1 j=o
- k - j/J
n
z".(-1n-k-jkn-l-j
k - j)!
3.8 PROBABILITY THEORY
n-i n-k
1: 1: k=1 j=O
(n) (m;3)Zi(_1)n-k-ikn-1-i
j
n n-k (n)
= 1:1: k=1 i=0
n-1 n-j
423
j
k')Z,(-1)n-k-jkn-1-j _ nn-1
(n
(n3)(n-j)zj(_,)n-k-jkn-1-i _ nn-1
= j=0 k=1
n-2n-j
_
(;)
\I
n - 3) i (+ nzn-1
n -2
_
if (n) -1)n j=0
=
3
nzn-1
k=1
(n -71)kkn-j-1) -
nn-1 +
nzn-1
k
- nn-1
since, by Lemma 1, the inner sum is equal to zero.
Substitute z = n to get (2).
Problem P.11. We throw N balls into n urns, one by one, independently and uniformly. Let Xi = Xi(N,n) be the total number of balls in the ith urn. Consider the random variable
y(N,n)= min Xi - NI. 1
n
Verify the following three statements:
(a) If n -+ oo and N/n3 -> oo, then
P y(N' n) < 2 n
1 - e-x
2/'r
for all x > 0.
n
(b) If n -> oo and N/n3 < K (K constant), then for any e > 0 there is an A > 0 such that P(y(N, n) < A) > 1 - E.
(c) If n-.ooand N/n3->0then P(y(N, n) < 1) -> 1.
Solution. The basic idea of the proof is the following: Xi is a B(p, N) (binomial distribution with parameters p, N), where p = 1/n. If the Xi's were independent, the statements could be proved easily. But of course
3. SOLUTIONS TO THE PROBLEMS
424
+ X,,, = N. However, we will show that k of they are not, since X1 + them can be considered independent for sufficiently small values of k. Suppose that the Xis are independent and their distribution is B(p, N). Let al, ... ak be natural numbers in the interval [0, NJ,
a=P(Xi=ai,i=1,...k), a'=P(X'=ai,i=1,...,k). Introduce some additional notations: k
k
a=Eai, ai=pN+bi, b=>bi i=1
i=1
We will show that a - a' under the conditions k = 0(1), bi = o(-,I-N), and
N > n. Obviously,
N!
k
pa (1
(fl=1 ai!)(N - a)!
- kp) N-a
and k
a'_
a4))
i=1
so
ftINpa(1_p)kN-a
a' _ N!k-1(N - a)!(1 7k a (l 1i-1(N - ai)!)(1 - kp)N-a p)kN-a
The value of ,3 will be estimated by the Stirling formula:
t! =
27rt tte-t+0(1/t)
The error of log t, that is, the difference of the logarithms of the true and estimated ,Q values can be determined from the O(1/t) terms as O
k
1
N - ai
+k-1 N
1
N-a
0( 1 N
01
()'
since
ai = pN + o(v), a = kpN + o(VN--), p = 1 -> 0 n
and N > n implies N --+ oc . The terms a-t and the value of the f terms can be given as
Nk-1(N-a) _ r1k 1(N-ai)
V1- r,
1(1- f)
27r are cancelled, and
3.8 PROBABILITY THEORY
425
So, finally,
N N(k-1) (N - a) N-a (1 f 1(N - ai)"`-ai (1 - kp)N-a k N-ai . lli=1(l - Ni - Ni (1 p)kN-a (1 b N-a k b N-ai =(1-N(1-kp)) .11(1-N(li p)) p)kN-a
0-
(la N-a 7 kp)N-a
Since each term converges to 1, we can apply the formula log(1 + e)
ce + O(e2). The value of the second-order term is k
2
2
b=
b + Ei=1
O
= o(1).
N
The value of the first-order term is
-b(N-kpN-b) N(1-kp)
k
bi(N - pN - bi)
-
N(1-p)
+ b2
k
N(1-kp)
b2
N(1-p) -o(1),
which implies that log/3 = o(1).
Solution to part (a). Put
ak=P(IXi - pNI < x
c= P
N3
k),
(Nn) < x VnNs
Recall that if A1, ... , An are arbitrary events and
Si =
E
P(Ai1,...,AZJ),
1
2s+1
2s
(-1)S3 < P(A1 ... An) < 1:(-1)iSj j=0
j=0
(see, for example, K. Bogndrne, J. Mogyorodi, A. Prekopa, A. Renyi, D. Szdsz, Exercises in Probability Theory (in Hungarian), Tankonyvkiado, Budapest, 1971, Problem 1.2.11.c). So 2s+1
(-W (n),j
2s
<1-c<
()aj
3. SOLUTIONS TO THE PROBLEMS
426
Note that
ak-a'k=P lXi'-pNl<x 3;i=l,...,k
a
since we can apply the previous asymptotic relation for any system Xi pN = bi, when IbiI < x N/n3. Since a' = a'k, with
a' =P IX' - pNj < x
n3
,
((k
)k+1)
+(nj) 7
The first sum is the first k terms of (1 - a')', and substituting them by
(1 - a')'" itself, the error will be only O((k+l)a'k+1) From the local De Moivre-Laplace theorem we conclude that
(here we use the fact that N/n3 -+ oo). Select an arbitrary, small e > 0. If k is sufficiently large, then the first error term is less than e, the second term is o((l+a')") = o(1), and finally,
log(1 - a') = -a' + O(a'2) , n log(1 - a') = -1/ x + o(1), which proves the assertion.
Solution to part (c). We are going to prove for an arbitrary, small e > 0
that there exist b and no such that for n > no and N/n3 < b, we have P(y(N, n) < 1) > 1 - E. If we wish to apply the previous reasoning, we face the difficulty that is not small enough, and (1 + a')" is not bounded if a' is too large. We can, however, overcome this difficulty since these quantities can be replaced by (k+i)a'k+l
k+1
[A
and
k+1 ]) (c N
(i+C)
[A/]
VVVVVV N
respectively. Consider the probability that for any of the first [A.fN/n] urns, I Xi - N/nI < 1. (If b is small enough, then A N/n < n). Here
P(,Xi -
nI
<1) =c N,
3.8 PROBABILITY THEORY
427
where c remains between positive bounds for N > n. (We can assume this since otherwise the assertion is obvious.) Then the proof can be completed as it was demonstrated in the proof of (a).) Solution to part (b). We can assume that with the 6 given in the previous part N/n3 > 6, otherwise, the previous result implies the desired inequality. Now a' = cA/n, where c is between positive bounds (they depend on 6 and K). The proof then can be completed with this a' in the same way as was shown at the end of the solution to part (a).
Problem P.12. Determine the value of
sup [log E - E log
1<£<2
where l= is a random variable and E denotes expectation.
Solution. Since log a is concave in
and 1 < < 2,
logy = log[(2 - l;) + ( - 1) 2] > ( - 1) log2, which implies that E(logl;) > [E(l;) - 1] log 2. Note that 1 <
2,
and therefore log
E(log ) < log E(l;) -
1] log 2
< max [log t - (t - 1) log 2] 1
=logto-(to-1)log2=K where to = 1/ log 2 E [1, 2]. We will next show that K is the least upper bound. Consider the random variable l=o defined by P(eo = 1) = 2 - to and P(lo = 2) = to - 1. Then log E(lo) = log to and E(log o) = (to - 1) log 2, therefore, K = - log log 2 - 1 + log 2. Remark. A similar proof shows that if W(t) is a concave and continuous function on the closed interval [a, b], then
sup {co(E(e)) - E(co(e))} = max
-b-
.a co(a)
b-
A further generalization is also possible for continuous but not necessarily concave functions cp.
Problem P.13. Find the limit distribution of the sequence q,, of random variables with distribution P (?ln = arccos(cos2
(2j - 1)7r)) = 2n
(arccos(.) denotes the main value.)
J
1
n
(i = 1...... n).
3. SOLUTIONS TO THE PROBLEMS
428
Solution. Let
Sn
be defined as 272n1 =
Then 7 1 n
=
n
(j=1,2,...,n).
Since n E (0, 1) (n = 1, 2, ... ), and the
arccos(cos2
function arccos(cos2 7rx) maps the interval (0,1) into (0, 7r/2), P(7?n < x) = 0 for x < 0 and P(7?n < x) = 1 for x > 7r/2. Suppose x E [0, 7r/2]. Since cos x is strictly decreasing in the interval [0,7r], arccos x is strictly decreasing in the interval [-1, 1]. The distribution of Sn is symmetric with respect to 1/2, therefore, P(7jn < x) = P(arccos(cos2 7rG) < x) = P(COS2(7rG) > COS x)
= P(cos(7ren) >
cOS x) + P(cos(7rG) < - COS X)
= P(7rG > arccos(- cos x)) + P(7rG < arccos( COs x)) arcco cos x )
= 2P(7rn < arccos cos x) = 2P(n <
For xe(0, 7r/2), arccos cos x E (0, 7r/2); thus only the value of P(n < y)
for ye(0,1/2) is to be determined. Let k be the largest integer such that k/n < y. Then n
2j-1 < k
n
P(n
n-
n
n
Therefore, limn-o. P(en < y) = y, and consequently,
ifx<0,
0
lim P(qn < x) _ n-oo
arccos cos x
1 1
if x E (0, 2 ),
ifx> z.
El
Remarks.
1. The random variable qn may take the same value twice.
2. The numbers \j = cos((2j - 1)/2n)7r (j = 1, 2, ... , n) are the zeros of the Chebyshev polynomial Tn = cos(n arccos x), and the corresponding Cotes numbers are An,j = 7r/n. Using the characteristic function of 7jn, one can easily prove the following.
Theorem. Let f be a continuous function in [-1, 1], and let n be a random variable defined by n
(j=1,...,n), (n=1,2,...).
Assume that is a random variable with density function 11(7r VI - x2) in the interval (-1, 1) and zero otherwise. Then the sequence converges weakly to the random variable f ().
3.8 PROBABILITY THEORY
429
In the problem, select f (x) = arccos x2 so that the limit distribution of the sequence {rij is the distribution of ri = arccosl;2. Simple calculations show that the density function of 77 is 1
1-t.22
if
xe(0, 2) and 0 otherwise.
Problem P.14. Let p and v be two probability measures on the Borel sets of the plane. Prove that there are random variables bl, S2, 711, 772 such
that (a) the distribution of ( i, 2) is p and the distribution of (771, r/2) is v, (b) i < 111, C2 < n2 almost everywhere, if and only if µ(G) > v(G) for all sets of the form G = U%` 1(-oo, xi) x (-oo, yi).
Solution. The necessity part of the assertion is obvious, since if _ (Sl S2) and ,q = (Mi rj2), then qiG implies eG, so
v(G) = P(rifG) <
µ(G).
First, we prove the sufficiency part of the assertion for the special case and yi, ... ,yn, where it and v are concentrated to the finite set x1i ... yl,... , yn and respectively. Define the graph G with vertices x1,.. . , where the vertices xi and yj are connected by an edge if and only if xi _< yj ((a, b) < (c, d) means a < c and b < d). Define a random variable ( , ri) that
takes the values (xi, yj) with probability ai,j. These numbers aij satisfy the following relations:
(a) aij > 0; (b) Ean--1 aij = v(y,); (c) Ej1 aij = p(xi); (d) If aij > 0, then there is an edge between xi and yj. By the KSnig-Egervary theorem, such numbers aij exist if and only if f o r any set Y C {y1, ... , y, } the p measure of the set X consisting of all points xi connected with the elements of Y is at least v(Y). Assume that Y = {yi, ... , yk}
with yi = (yi, yi )
and
G = Uk 1(-0o, y) x (-00,y ) Then a point xi is connected to Y if and only if xi E G. Therefore,
E p(xi) = µ(X n G) = p(G) > v(G) > v(Y). xiEX
Consequently, there are numbers aij having properties (a)-(d), and any vector variable (, that has the values (xi, y3) with probabilities aij satisfies the requirements of the theorem.
3. SOLUTIONS TO THE PROBLEMS
430
Now consider the general case. Denote by F(x, y) the distribution function of the underlying variable and put F1,
(X/' x") = !-W-oc, x') x (-oo, x"))
F,.(y', y") =
v((-oo, y') x (-oo, y")).
The existence of a distribution function F satisfying
(ii)
F(x, oc) = F. (x) F(oc, y) = F, (y)
(iii)
F(y, y) = F, (y)
(i)
solves the problem, since for a random variable (C, r7) having distribution F, £ < q holds with probability one because F(y, y) = F(oo, y).
Divide the square [-n, n] x [-n, n] into n4 small squares of sides 1/n. Define An as follows: concentrate the /t measure of every point into the closest node. (More precisely, consider the set of all points of the plane that are closest to a given vertex, and concentrate the Fc measure of this set to the given vertex. If a point is in the same distance from two or more vertices, then select the one that has the smallest abscissa among the vertices with smallest ordinate.) Let vn be defined analogously. Then (µn, vn) satisfies the requirements of the theorem, since they are measures concentrated in finitely many points. i]n) such that n < qn , the distribution Therefore, there is a variable
of G is An, and the distribution of 77n is vn. Let Fn be the distribution Then function of Fn (Y, Y) = Fn (00, Y),
furthermore, Fn(x', x", 00, oc) = An
00, X') X (-00, x")),
Fn (00, 00, Y', y") = vn((-oo, y') x (- 00, y"))
Let S be a countable and everywhere-dense set of the continuity points of
F and F. Select a subsequence nti such that F(a, b) = limi .
Fn; (a, b)
exists for all a, b E S. Extend the definition of F by the relation
F(x, y) = sup{ lim Fn; (a, b) : a < x, b < y, a, b e S}. 8-00
It is easy to verify that F is a distribution function and that it satisfies (i), (ii), (iii). Thus, the proof is completed.
3.8 PROBABILITY THEORY
431
Problem P.15. Let X1, X2, ... , Xn be (not necessarily independent) discrete random variables. Prove that there exist at least n2/2 pairs (i, j) such that H(Xi+X1)> 1 min {H(Xk)}, 3 1
where H(X) denotes the Shannon entropy of X.
Solution. Consider the graph G with vertices {1, 2, ... , n}, and assume that i and j are connected with an edge if and only if the inequality H(Xi + X3) > 1 min H(Xk) 3 1
(1)
does not hold. There is no loop in G, since for all i
H(Xi + Xi) = H(Xi) > 1 min H(Xk). - 3 1
We are going to show that there is no triangle in G. Let i, j, and 1 be different elements of {1, 2, ... , n}. Obviously, Xi = 2 ([Xi + Xj] + A + X1 - [Xi + X1]), that is,
H(Xi) = H([X2 + Xj] + [Xi + X1] - [Xj + Xi])
H(Xi+X3)+H(Xi+Xi)+H(XX+X1) < 3 max{H(XX + X3), H(XZ + X1), H(XJ + Xi)}. So
1 min H(Xk) < 1 H(Xi) 3 31
< max{H(Xi + Xj), H(X2 + X1), H(XJ + X1)}.
Therefore, there are two vertices among i, j, and l that are not connected. By a well-known theorem of Pal Turin (see, Mat. Fiz. Lapok 48, (1941), pp. 436-452), the number of edges of such a graph is not larger than n2/4.
Consequently, the number of pairs (i, j) such that (1) holds is at least n2 - 2. (n2/4) = n2/2. Remark. The bound n2/2 is the best possible. Let X be a random variable such that H(X) 0, and define X1=X2=...=X[2]
= X,
X[1+1 = ... = Xn = -X2 If either 1 < i < n/2 < j< nor 1 < j < n/2 < i < n, then H(Xi+X3 The number of such pairs (i, j) is equal to [n2/2].
= 0.
3. SOLUTIONS TO THE PROBLEMS
432
Problem P.16. Let SI, b.... be independent, identically distributed random variables with distribution P(61 = -1) = P(61 = 1) = 2
WriteSn =6+6 +
+ G n (n= 1,2,...), So =0, and max Sk .
Tn = V
O
Prove that lim inf (log n) Tn = 0 with probability one. n-oo
Solution. Denote maxa
E P{(f(n))-1,2'Sf(n) > g(f(n))} < oo n=1
and 00
P (Sf(n+1) - Sf(n))* +
f (n)g(f(n)) <
f(n+1)
n=1
c
logf(n+1)
_ 00
(b)
for all E > 0. If such sequences g(n), f (n) are found, then by the BorelCantelli lemma with probability one, only finitely many of the events Sf(n)
> g(f(n))
f(n)
and infinitely many of the events (Sf(n+1) - Sf(n))* +
f(n)g(f(n))
E
< log f(n + 1)
f(n + 1)
occur. Consequently, infinitely many of the events Sf(n+1)
< (Sf(n+1) - Sf(n))* + Sf(n)
f(n+1)
f(n+1)
E
<
logf(n+1)
will occur with probability one. Hence, it is sufficient to find sequences g(n), f (n) satisfying (a) and (b). We are going to apply the approximation
\\
P I Sn <
x\ 1
/
2
1
70
edu
3.8 PROBABILITY THEORY
433
(see, for example, A. Renyi, Foundation of Probability, Holden Day, San Francisco, 1970). This approximation holds if x is replaced by a sequence xn = o(n 3 ). So
P
(Sf(n+i) - Sf(n))* +
f(n)9(f(n))
f(n + 1)
> P (Sf(n+1) - Sf(n))* <
f(n+1)- f(n)
j
E
< log f (n + 1)
_
E
f (n)9(f (n))
logf(n+1)
f(n+1)
e_ 22/2du > 2 A,
0
where A
_
E
.f (n)9(f (n))
logf(n+1)
f(n+1)
is small enough. Therefore, we have to find sequences f (n) and g(n) such
that °O
1
1: logf(n+1)
(i )
oo
if (n)92(f (n)) < o0
(ii)
f(n + 1)
n=1
C
00
n=1
f (f (n))
(iii)
00
e-"2/2du <
oo.
It is easy to see that, for example, g(f (n)) = n3 and f (n + 1) = (n!) 14 satisfy these relations.
Problem P.17. Let the sequence of random variables {Xm, m > 01, X0 = 0, be an infinite random walk on the set of nonnegative integers with transition probabilities
i>0' 4i=P(Xm+1=i-1IXm=i)>0, i>0.
Tai = P(Xm+1 = i + 1 I Xm. = i) > 0,
Prove that for arbitrary k > 0 there is an ak > 1 such that
Pn(k)=PI\o<j max Xf=k
lim
Pn(k)ak < oo
1
L-+oo L
n=1
.
3. SOLUTIONS TO THE PROBLEMS
434
Solution. Introduce the notation
Pn (k) = P( max X. < k), 0<j
P*n(k)=P(maxXj
oP*n(k). It is easy to see that the
following inequality holds: k
P*n(k)pipi+l ... Pi+k+1 + PP+k+l (k) < Pn (k). i=O
Define Ek = omin jPiPi+1 ... Pi+k+l }.
Then 0 < Ek < 1 and P`+k+l(k) < (1 - Ek)Pn(k)
Using mathematical induction, one can easily show that for all positive integers n,
,n(k) < (1 -
Ek)11(k+1)-2
If 1
< crk < [
1
]1/(k+1)-2
,
1-Ek
then
P.(k)ak < (1 -
Ek)2n-2
and this inequality implies the statement.
Problem P.18. Let Yn be a binomial random variable with parameters n and p. Assume that a certain set H of positive integers has a density and that this density is equal to d. Prove the following statements: (a) limn-,,, P(Yn E H) = d if H is an arithmetic progression. (b) The previous limit relation is not valid for arbitrary H. (c) If H is such that P(Yn E H) is convergent, then the limit must be equal to d.
Solution. (a) By the assumptions of the problem,
P(Yn=k)= (fl)pkqn_k
(q = 1 -p).
For any given k, limn- 0 P(Yn = k) = 0; consequently, limn-.". P(Yn E H) does not change if finitely many elements of H are changed. We
3.8 PROBABILITY THEORY
435
can therefore assume that H is a complete residue class with respect to a module D:
H={k: k=-amodD}. Then
(fl)pkqn_k.
P(Yn E H) _
(1)
k=-a( D)
Denote by e a primitive Dth unit root. It is known that D-1 (k-a)v
E
=
V=0
D if k { 0
if k
a(D), a(D).
We can write (1) in the following form: 00 D-1
P(Y E H) _ E
1:
k=0 v=0
(fl)pkqn_kE(k_a)v k
D-1
00
L_00
kk=O
_ D E -av 1:
(n) (W)kgn-k
D-1
=D
E-av(q
'=0
+ EVp)n -' D = d,
since in the last sum, for v > 0, the absolute values of the corresponding terms are less than one, so they converge to zero as n -* oo. b) We construct a sequence H with zero density such that
I'm supn-00 P(Yn E H) = 1. The expectation and variance of Yn are np and npq, respectively, so the Chebyshev inequality implies that
P(I Yn - np > A npq) 5 a2 If An is a sequence such that limn-0 An = oo, then
lim P(I Yn - npJ
n-oo
Let nk be an arbitrary increasing sequence of natural numbers, and let
c/ H = U (nk - [nk/4], nk + [nk/4]) k=1
By the previous observation, k
E H) = 1,
and so lim sup P(Yn E H) = 1. n--+oo
3. SOLUTIONS TO THE PROBLEMS
436
Finally, if nk increases sufficiently fast (for example, nk = 10k), then H has zero density. (c) The majority of the contestants verified that if H has density, then lim P(Yn E H) = d in the sense of Cesaro, that is,
_
Sn
En P(Y H) v=1
n
(n--goo).
n
From this property, the original statement follows, since for arbitrary convergent sequence, the Cesaro limit is the same as the usual limit. One can also prove that we do not need to assume the existence of the density of H, since if P(Yn E H) is convergent, then H automatically has a density that equals limn-,,. P(Yn E H). A more general result can also be verified: for an arbitrary set H of natural numbers, define h(x) = I In E H : n < x}1. Then Sn
lim
n-oo n
- h(np) = 0. np
Thus H has a density if and only if P(Yn E H) is convergent in the sense of Cesaro.
After these remarks, we can start to prove part (c). We are going to prove that if the density of H is d, then pn = P(Yn E H) --+ d in the sense of Abel, that is, E°° o(pn -Pn-1) = d in the sense of Abel, = d. If we prove this result, then we are ready, since limn-00 P(Yn E H) = c implies P(Yn E H) --+ c in the sense of Abel, that is, c = d. For JxJ <1, )xn
namely, lima-.l-o EnO°_o(Pn -Pn-1
00
00
G(x) = E(Pn - Pn-1)xn = (1 - x) E pnxn n=0
n=0 00
x)n =0 ExEEH x)
(fl)pkqnkxn k
E
E
kEH
n=0
00 n(n - 1)...(n - k + 1)(gx)n-k
(Px)
x)kE
k! 1
k
(1-qx)k+1
1-x
_
1-qxk
px
k
-qx) .
Substitute z = px/(1 - qx). If x --+ 1 - 0, then z ---f 1 - 0, and further-
more (1 - x)/(1 - qx) - 1 - z. So X
lim G(x) = z--.1-0 lim (1 - z) 1-0
zk
(2)
kEH
if this last limit exists. Let an be the indicator sequence of H, that is, an equals 1 if n E H,
3.8 PROBABILITY THEORY
437
and equals 0 if n H. The limit of an in the sense of Cesaro is the density of H, which is d. Then, by a theorem of Frobenius, the limit of the sequence an in the sense of Abel also exists and equals d. This Abel
limit is the right-hand side of (2). Consequently, lim.,_l_o G(x) = d, which was to be proved.
Problem P.19. Let {G1}°1=1 be a double sequence of random variables such that (i, j, k, l = 1, 2, ... ).
Eeijek1 = O((log(21i - ki + 2) log(2Ij - lI + 2)) -2)
Prove that with probability one,
m n kl ---> 0
as max(m, n) -+ oo.
k=1 1=1
Solution. Let b+m c+n S(b, c; m, n) =
where
E G1
and m, n > 1.
b, c > 0,
k=b+11=c+1
Using assumption (1) and the identity thickmuskip=1.4mu b+m c+n
(
E(S2(b,c,m,n)) _ E E
b+m c+n-lc+n-1
E E E E(G14,1+j)
k=b+ll=c+1 j=1
k=b+ll=c+1
c+n b+m-lb+m-k
+
EEE
E(Glek+i,1) I
i=1
1=c+1k=b+1
b+m-1c+n-1b+m-kc+n-1
+2k=b+ll=c+1 EEE i=1
E(GiG+i,l+j) j=1
+E(G,l+jG+i,l)) , it is easy to see that
m-1 n-1 1 E(S2 (b, c, m, n)) = O{mn F F i=0 j-0 (log(1 + 2i) log(1 + 2j))2 }
= O{mn (log
mn 2m)2 (log 2n)2
}
(m, n = 1, 2, ...) .
Put M(b, c; m, n) = max max IS(b, c, k, l)1, 1
(2)
3. SOLUTIONS TO THE PROBLEMS
438
and use the following result (Theorem 4 in F. Mdricz, Momemt inequalities
for the maximum of partial sums of random fields, Acta. Sci. Math. 39 (1977), pp. 353-366). Assume that the nonnegative function f (b, c; m, n) (b, c, m, n E N) satisfies the following inequalities:
f(b,c,m,n), f(b,c;m,n),
f(b,c;h,n)+ f (b + h, c; m - h, n) f(b,c,m,n)+ f (b, c + i; m, n - i)
forb>0,c>_O,and1
K(1) = X(1),
A(1) = A(1).
Form> 2 andn>2, K(m) = X(h) + K(h - 1), A(n) = A(i) + A(i - 1),
i = [2 (n + 2)]
([. ] denotes the integer part of real numbers). Assume that for some y > 1, and for all b, c > 0 and m, n > 1, E(IS(b,c;m,n)I") <_ K^1 (m) XT (n) f(b,c;m,n).
Then E(M"(b, c; m, n)) < K7(m)A"(n) f (b, c; in, n)
for all b, c > 0 and in, n > 1. In our case, choose f (b, c; in, n) = mn and X(m) = V/m--/ log 2m, .fi(n) _ V,'n-/ log 2n, ry = 2. Note that f (b, c; m, n) is an additive set function on the rectangle [b + 1, b + m] x [c + 1, c + n]. Furthermore, for 2P < m < 2P+i P
K( m ) < K(2P+' - 1)
P
= E X (2P) _ k=0
2P" '° = O ( log2m =(p+1) O
k=0
2k/2
+1
Consequently, E(M2 (b, c; m, n ) = O{ mn
(log 2m)2 (log 2n)2
}
In the case b = c = 0, define S(m, n) = S(0, 0; in, n) = Em i E
(3) Sk1
Since (2) holds, °°
°°
1
22k221
k=01=0
E(S2 (2 k 21)) = O(1)
°)
°°
1
22k221
E 22k221 (k + 1)2(1 + 1)2 k=01=0
< 00
3.8 PROBABILITY THEORY
439
A theorem of B. Levi implies that with probability one, 1
2k21
S(2k21) --+ 0
max(k, l) --+ oo.
as
(4)
To complete the proof, we have to show that 1
max
max
2k21 2k <m<2k+1 21
IS(m, n)
- S(2k 21)1 -- 0 ,
(5)
with probability one, as max(k, 1) -+ oo. Consider the following decomposition: S(m, n) = S(2k, 21) + S(2k, 0; m -2 k , 2') + S(0, 21; 2k, n - 21) + S(2k, 21; m - 2k, n - 21),
from which we see that the left-hand side of (5) is not greater than 221 1 {M( 2k 0.2 k21)+M(021.2 k 21)+M(2 k 21.2 k 21)}.
By (3), 00
00
E(M2 (2k,
1
21)) < 21; 2k,
22k221
oo,
k=01=0
and the theorem of B. Levi again implies that 2k21
M(2k21, 2k21) -4 0
(6)
with probability one, as max(k, 1) -+ oc. Notice that 1-1
M(2 k' 0; 2k, 21)
M(2k, 2q,
2k,
2q),
4=-1
where, by definition, M(2k, 2-1; 2k, 2-1) = M(2k, 0; 2k,1). A Toeplitz lemma and (6) imply that with probability one l-1
2k21
M(2k, 0; 2k, 21) <
q=-1
211q 2k2q M(2k, 24; 2k 2q) -, 0
as max(k, 1) --+ oc. A similar argument shows that 2k21M(0,21;2k21) , 0, as max(k, 1) -+ oo, with probability one. Combining (4) and (5), we complete the proof.
3. SOLUTIONS TO THE PROBLEMS
440
Problem P.20. Let P be a probability distribution defined on the Borel sets of the real line. Suppose that P is symmetric with respect to the origin, absolutely continous with respect to the Lebesgue measure, and its density function p is zero outside the interval [-1, 1] and inside this interval it is between the positive numbers c and d (c < d). Prove that there is no distribution whose convolution square equals P. Solution. Assume indirectly that there exists such a distribution Q. Then Q([-1/2,1/2]) = 1, and its moments satisfy the relations l/2
Al =
J
10 xkdQ(x)
xkdQ(x) <1. £112
Thus,
limsup'` I kkl M1-
0
Therefore, the characteristic function coQ of Q is analytic on the real line. Since P is symmetric, cpp is real. The equality cpp(0) = 1 and the continuity of cop imply that, for any x with sufficiently small absolute value, cpp(x) > 0. For such x values cpp(x) = (cpQ(x))2, so cpQ(x) is also real. Since coQ is analytic, it is real everywhere, which implies that c 2 p (x) > 0 for all x. By a well-known theorem (see for example E. Hewitt, and K. Stromberg, Real and Abstract Analysis, Springer, 1965, p. 409), we know that if a density function is bounded and its characteristic function is nonnegative, then its characteristic function is integrable. Therefore, cpp is integrable, so the density function p should be equal to a continuous function almost everywhere, but because of its "jumps" at -1 and 1, this is impossible.
Remark. For many related negative and positive decomposition results see I. Z. Ruzsa and G. J. Szekely, Algebraic Probability Theory, Wiley, New York, 1988 and G. J. Szekely, Paradoxes in Probability Theory and Mathematical Statistics, Reidel (Kluwer), Dordrecht, 1986.
Problem P.21. Let po, p,.... be a probability distribution on the set of nonnegative integers. Select a number according to this distribution
and repeat the selection independently until either a zero or an already selected number is obtained. Write the selected numbers in a row in order of selection without the last one. Below this line, write the numbers again in increasing order. Let Ai denote the event that the number i has been selected and that it is in the same place in both lines. Prove that the events
Ai (i = 1, 2,...) are mutually independent, and P(Ai) = pi. Solution. We will show that for any k (k = 1, 2, ... ), and any sequence P(Ai1Ai2
... Aik) = pilpi2 ...pik.
3.8 PROBABILITY THEORY
441
Modify the experiment in such a w a y that in addition to zero and a repetition of a number, we also stop a t selecting a n y of the numbers il, i2 ,- .. , ik. Consider now a (finite) outcome of the original experiment in which Ail Ail ... Aik occurs. (The probability of infinite outcome is equal to zero.) From the original sequence, omit the values i 1, i2, ... , ik. In this way, an arbitrary outcome of the modified experiment can be uniquely obtained. On the other hand, it is easy to see that there is only one w a y to insert the numbers il, i 2 ,- .. , ik in such way that event Ail Ail ... Aik occurs, that is, the numbers i1, i 2 , ... , ik are placed in their increasing locations. This one-toone correspondence implies that the probability of the event AilAi2 ... Aik is equal to pilpi2 ... pik 1.
Problem P.22. Let X1,. .. , Xn be independent, identically distributed, nonnegative random variables with a common continuous distribution func-
tion F. Suppose in addition that the inverse of F, the quantile function < Xn:n be Q, is also continuous and Q(O) = 0. Let 0 = Xo:n < Xl:n < the ordered sample from the above random variables. Prove that if EX1 is finite, then the random variable y
1 [ny]+1
O = sup o
(n + 1 - i)(Xi:n -- Xi-1:n) -
n i=1
fo(1 - u)dQ(u)
tends to zero with probability one as n -+ oo.
Solution. Introduce the notation [ny]+1
HH(y) =
(n + 1 - i)(Xi:n - Xi-l:n) 7E
i=1
and
HF(y) =
y - u)dQ(u). f (1 0
Thus, HF(1) = E(X1) and Hn(1) = limHn(y) yti
1
n
= - EXi,
(1)
n i=1
that is, the problem generalizes the strong law of large numbers for nonnegative random variables. Because of the continuity of F, the independent random variables Yi = F(Xi), 1 < i < n, are uniformly distributed on the interval (0, 1). Denote by En(y) = n-l x {k : 1 < k < n , Yk < y} the empirical distribution function of Y1, Y2,. .., Yn, and let Un(y) = { l
Yk:n
if kn l < y!5 n k = 1, ... , n,
Yn:n
ify=1
3. SOLUTIONS TO THE PROBLEMS
442
be their empirical quantile function. Further, let
1
Fn(x)=n-1x{k and
Qn(TJ) _
{
if k-'
Xk:n Xn:n
ry
if y = 1
be the empirical distribution and quantile function of the original sample, respectively.
Consider the random function U.. (y)
Gn(x) = f
(1- En(u))dQ(u)
0
By the continuity of F and Q, JQ(un(y))
Q,.(y) (1 - Fn(x)dx
(1 - En(F'(x)))dx =
Gn(y) =
0
with probability one, where the exceptional set (where the original sample
elements coincide) is independent of y. If (k - 1)/n < y < k/n for some integer 1 < k < n, then this last integral is the following: X1-
k
(1 - Fn(x))dx =
Xs n
(1 - Fn(x))dx,
k
where k
(1 i=1
i
n
1)(Xi:n - Xi-l:n) = Hn(y)
Since Gn(1) = Hn(1) with the common value given in (1), we conclude
that for n=1,2,..., P{ sup IHn(y) - Gn(y)I = 01 = 1. 0
It is sufficient to show that for n -+ oo,
An = 0
An* < sup IGn(y) - HF(U(y))I + sup IHF(Un(y)) - HF(y)I 0
= Q(1) +,N(2) ,
O
(2)
3.8 PROBABILITY THEORY
443
Notice that for 0 < y < 1, Un(y) has only n different values from the interval [0, 1]; therefore, Only < On 3)
= sup f y(1- En(u))dQ(u) - f y(1- u)dQ(u)I. 0
0
0
Let 0 < e < 1 be an arbitrary value. Then
i$)
< J (1 - En(u))dQ(u) + J 1
E
(1 - u)dQ(u)
1 1
E
+ Q(1 - e) sup Iy - E. (y)
Since Q(1 - e) < oo, the third term tends to zero with probability one (Glivenko-Cantelli theorem, see, for example, A. Renyi, Probability Theory, Akademiai Kiadd, Budapest, 1970, VII. §8). The first term (I(A) denotes the indicator function of the event A) can be written as
J
(1 - En(u))dQ(u) =
1 E
E
1ni=1 Jl
(1 - I({Ui < u}))dQ(u),
1 E
and because of the strong law of large numbers, this quantity tends to 1 fi E(1 - u)dQ(u) with probability one.
In summary,
f
P{limsupLn3) <2 (1 - u)dQ(u)} = 1, 1_E n-oo where the upper bound can be made arbitrarily small by choosing a sufficiently small E since E(X1) < oo. Thus, we have proved that the first term of (2) tends to zero (.41) -> 0) with probability one as n - oo. It is known (see, for example, M. Csorgo and P. Revesz, Strong Approximations in Probability and Statistics, Akademiai Kiado, Budapest, 1981, p. 162) that the Glivenko-Cantelli theorem is also valid for the quantile function of a uniformly distributed sample. Therefore, sup lUU(y) - yI --+ 0
0
with probability one, as n -> oo. Using the fact that the integral fo (1 u)dQ(u) is a continuous function of y, for the second term in (2) we get P{ lim
n-.oo
One)
= 0} = 1,
which completes the proof.
Remark. Suppose n machines start working at time t = 0, and let Xl:n < X2:n < ... be the failure times for these machines. Then nHn(y) is the total time until the ([ny] + 1)th failure. The best published result (N. A. Langberg, R. V. Leon, and F. Proschan, Characterization of nonparametric classes of life distribution, Annals of Probability 8(1980), pp. 1163-1170, Theorem 3.2) shows only pointwise convergence, that is, for all fixed 0 < y:5 1, Hn(y) -+ HF(y) with probability one, as n -+ oo.
3. SOLUTIONS TO THE PROBLEMS
444
Problem P.23. Let X0, X1, ... be independent, identically distributed,
nondegenerate random variables, and let 0 < a < 1 be a real number. Assume that the series k=0
is convergent with probability one. Prove that the distribution function of the sum is continuous.
Solution. Define 00
00
Z = > ak-1Xk
and
Y = E akXk. k=0
k=1
Then Y = Xo + aZ, where Y and Z have the same distribution, and X0 and Z are independent. Assume that the distribution function of Yis not continuous. Define
p=maxP(Y=a), a and denote by a1, ... , an the points for which
P(Y=aj)=p. Then
p=P(Y=aj)=P(Xo+aZ=aj)
1:
P(Xo = x)P(Z = a
P(Xo=x)>O
x
a
P(Xo = x) max P(Z = at - x) < p.
<
i
P(Xo=x)>O
a
Therefore, we have equality everywhere. This implies >xER P(X0 = x) 1, that is, X0 has a discrete distribution, and
P(Z= a,-x)=p
if
a
P(Xo=x)>0.
Put P(Xo = x) > 0},
K = {x :
Pa={x: P(Z=aax)=p}. Then
C n
K
nP,
j=1
3.8 PROBABILITY THEORY
445
Since IPal=nand P,,,=P0+a, K C Po +a.,,
that is, K - aj C Po
,
j = 1,...,n.
This relation can hold only if IKI = 1, that is, if X0 is degenerate.
Remark. For another proof and some related results see I. Z. Ruzsa and G. J. Szekely, Algebraic Probability Theory, Wiley, New York, 1988, Section 5.5.
Problem P.24. Let X1, X2, ... be independent random variables with the same distribution:
P(Xi = 1) = P(Xti = -1) = 2
(i = 1, 2, ... ).
Define
So =0,
Sn
e(x,n)_ Ilk: 0
(n=1,2 .... ), (x=0,±1,±2,...),
and
a(n) _ I {x : l; (x, n) =1} 1
(n = 0,1, ... ).
Prove that P(lim inf a(n) = 0) = 1
and that there is a number 0 < c < oo such that P(limsup a(n)/logn =
c)=1. Solution. Represent the values of So, S1i ... in a two-dimensional coordinate system as follows: starting from the origin, make a step to the right, and move up or down if Xn = +1 or Xn = -1. Note that a(n) counts the points whose second coordinates are visited exactly once by this random walk.
First, we show that
P(lim inf a(n) = 0) = 1, which means that the events An = {a(n) = 0} occur infinitely many times with probability one. In the moments when the random walk returns to zero, there are infinitely many such points with probability one, the value
of a(n) is at most one, since only the extreme values can occur once. However, the probability that between two returns the random walk visits the extreme values at least twice is positive. Therefore, among the infinitely many instants, there will be one, and thus there will be infinitely many ones, when the value of a(n) is zero.
3. SOLUTIONS TO THE PROBLEMS
446
Let us consider the second statement. We will proceed via several lemmas. Denote by An the set of second coordinates visited only once, that
is, An = {x : there is a k, 1 < k < n, such that Sk = x, and for j $ k, Sj # x}. Then a(n) = IA 11.
Lemma 1. limn-. P(Sj > 0, 0 < i < n, Sn - Sj > 0, 0 < j < n) _ c*>0. Proof. Define the following conditional distribution: µn (dx, dy) = P
1
Sn = dx,
1
sup Sk = dy
V/n o
jSj >0
1 <_ j< n
If n = 2m + 1, then
P(Sj>0,0<j0,0<j0,0<j<m,Sm> SUP (Sn-Sk)-(Sn-Sm), m0,m<j sup (Sk-Sm)) o
= P(Sj > 0, 0 < j < m) P(Sn - Sj > 0, m < j < n),um * ,um(A), where
A = {(x1, yi, x2, y2) : xi > y2 - x2, x2 > yi - xl}. We know that the sequence µn of measures tends weakly to a measure p* without atoms and has a positive value on the open subsets of the set {(x, y) : x > 0, y > 0}. On the other hand, 9 P(Sj>0,0<j<m)P(Sn-Sj>0,m<j
so the statement of the lemma holds for integers of the form 2m + 1. For integers of the form 2m, the proof is similar.
Lemma 2. Suppose k - alogn, ak(n) = (°kn)). Then for arbitrary y > 0, there is an no = no(k,77) such that for any n > no, [(c* - 77) logn]k < Eak(n) < [(c* + ii) logn]k.
Proof. Put
C(r,t)={S,.<S3<St,r<j
0 < j < t},
D2(t)={Sj>St,
t<j
Then
iK
P(C(r, t)) = P(C(0, t - r)) = t c* r (1 + o(1)), P(Di(t)) =
-(1 +o(t))
P(D2(t)) =
VIn - -t ('+o(')).
(K = 2/3
447
3.8 PROBABILITY THEORY
Put
A+ =Anf1{z:z>0}, a (n) _ (a+k(n))
a+(n) = I A+ I ,
< 3k < n, use the notation
and f o r 0 < jl < j2 <
Bjl,...,jk = Dl(jl)C(jl,j2) ... C(jk-l,jk)D2(jk) Then
>
Eat (n) =
P(Bj1,...,jk),
O<jl<...<jk
since the event Bj1 ,..,jk means that
(il, ... ,jk) C A,+, and the event a(n) = l contains exactly (k) such events. Put
P(C(j1,j2)) ... P(C(jk-l,jk)) = U(0,l - j).
U(j,l) _ j=j1 <...<3k=t So
n-r
00
U(0, r) E P(D1(j))P(D2(r + j)) j=0
r=0
n-r 1 const E U(0, r) E n
j=1
r=0
n
1
7n=---3 ---r
< const E U(0, r) r=0
k-1 n
const
< [(c* + 17) log n] k.
P(C(01 A) j=0
I
On the other hand,
Ea+ (n) >
P(Dl(jl)C(jl, j2) ... C(jk-1,jk)D2(jk)) 0<j1
n/3k>j=-j=-1>0
conSt
n/3
E P(DI (j)) j=1
> const [(c* - 2) log
k-1
n3
P(C(O1 i)) j=1
nk-1 3k 1
> [(c* - rl) log n] k
since k - a log n, so [(c* - 77/2)/(c* - q)] a 10g n > log n if n is large enough.
3. SOLUTIONS TO THE PROBLEMS
448
We can get the same result for An in a similar way. So
Al plus maybe one more point, if Sn > 0,
I So the statement of the lemma is valid for a(n).
Lemma 3. For all K > 0 and e > 0, there is an no = no(K,e) such that for all n> no, n-(K/c')-e < P(a(n)
< K logz n) < n-(K/c*)+e.
Proof. Put k = (K log n) /c*. Then
P (c(n) > K logz n) = P(ak(n) > (Klo2n))
< Eal kg(n))
If n is large enough, then we can use Lemma 2 to get tog((c'e)/(c'+n)) < n-(K/c*)+e
P(a(n) > K logz n) <
For the proof of the lower bound, put q,,,, = P(a(n) = m). Then m
Eak (n) _ E q. m
(k)'
Put k' = (K/c*) (1 + E2) log n. Then for any sufficiently small e, 1
(M)
Eak, (n)
m>(k+e) loga n and
m
1
qm k, < 3 Eak' (n) m
because if k" = ((K + e) /c*) log n, then k" > k' and 4m
m>(K
loga n
qm (kmil(mkl)
Ck'/
(k")
<
((K+e) log 2 n)
((K+) toga n)
l
k"
((K+ek) 'logs n)
4m m
< l((K+e) logs n) Eak (n) k"
Ck J
3.8 PROBABILITY THEORY
449
Again using Lemma 2 with 77 = e3, we get qm
2 25Kc*
(M) < (c` log n)k' exp{(-
+ 0(e:2)) log n}
m>(K+c) log2 n 3 Eakw (n)
if n is sufficiently large. On the other hand, K loge n
(m qm1 k/) 1: m< K log2 n \
k'
)
L
qn+1mk
).
((K logs n) ( k / m
Using Lemma 2 again, we get
qm(M) ki < 3 Eak, (n). m
These inequalities imply
q , . k J > 3Eak,(n). K log2 n<m<(K+E) loge n So
. T,
P(a(n) > K log 2 n) >
qm (M)
1
log n k'
K loge n<m<(K+E) loge n
>
(K+E)
)
E(ak'(n)) > in i(1+E3)log[er T(1+E3)-2 IKC}}ee
- 3 ((K+E) loge n) - 3 k'
This last expression tends to 1 for E > 0 and 77 > 0, so we have checked the lower bound. Using the above lemmas, we are able to prove that
P Clim sup a(n) = c = 1. loge n
Put
Mn=inf{j:Sj >n}. Then lim Mn = 0
and
M(n+l)2 - Mn2 > 2n + 1.
Consider the instant when the random walk reaches the height k2 for the first time. Start a k-step walk from this point, and denote it by Uk , that is,
Uk = {0, SMk2+1 - SMk2 , ... , SMk2+k - SMk2 }.
3. SOLUTIONS TO THE PROBLEMS
450
These walks are independent from each other. Put Bk
_
the walk Uk visits exactly once at least (c* - E) log2 k points with positive coordinates
Then P(Bk) > k-1+E/2 if k is large enough. So > P(Bk) = oo, hence P(lim sup Ak) = 1. But on the event Ak a(Mk2 + k) > (c* - E) log2 k, and since Mk2 + k < ks + k for sufficiently large values of k,
a(n)
c* - e
loge n >
64
for infinitely many n with probability one. The upper bound of Lemma 3 with K = c* + e shows
P(a(n) > (c* + e) loge n) < 00,
that is, lim sup a(2) < c* + e,
log n
for large values of n with probability one. Recall that the Kolmogorov 0-1 law implies that lim sup a (n) /log 2 n is constant with probability one; therefore,
P (limsup
a(2) =cl=1 log n
with a suitable c*/64 < c < c*.
Remark. Unfortunately, the original formulation of the second statement of the problem was incorrect. It contained a log divisor instead of loge. The proof of the correct statement appeared in P. Major, On the set visited once by a random walk, Probab. Th. Rel. Fields 77, (1988), pp. 117-128. The present proof follows this work. Problem P.25. Let (S2, A, P) be a probability space, and let (Xn, .'n) be an adapted sequence in (Q, A, P) (that is, for the a-algebras.Fn, we have ,r1 C_ .F2 c ... C_ A, and for all n, Xn is an Fn-measurable and integrable random variable). Assume that E(Xn+1I.Fn) =
2Xn +
2Xn-1
(n = 2, 3 ... ).
Prove that supnEIXnI < oc implies that Xn converges with probability one as n -4 oc.
Solution. Put Yn = Xn + (1/2) Xn_1 (n = 2, 3, ... ). Then Yn is an Fn-measurable and integrable random variable. Furthermore, E(Yn+1I.Fn) = E
(x+1 + 2XnIFn) = 2Xn + 2Xn-1 + 2Xn = Yn
for n > 2. That is, (Yn,.Fn,
n = 2, 3 ...) is a martingale.
3.8 PROBABILITY THEORY
451
Since supra E(IYnI) < (3/2) sup, E(I XnI) < oo, the martingale convergence theorem implies that Yn is convergent with probability one, and lim Yn(w) = Y(w) n-oo if w E S2', where P(S2') = 1. Here Y is a random variable that is finite with probability one, and therefore we may assume that it is finite for w E W. We will show that for w E S2', the sequence Xn(w) is also convergent. Let w E 1' be given, and an = X, (w). We will prove that if the sequence bra = an+(1/2) an_1 converges to c, then an converges to (2/3) - c. Assume
first that c = 0. Then for any E > 0, there is an N such that Ibnl < e/2 as
n > N. So 1
Ian+1I = Ibn+i - 1anl < Ibn+1I + 2IanI <
E+2anl
for all n > N. Using this inequality for n = N,. .. , N + k - 1, we get IaN+k I
<-
2
+ - + ... + 2k + 12 I < E + 12 I 4
< 2E
if k is large enough. However, for all n > N + k, Ianl < 2E. That is, the sequence converges to zero.
If c # 0, then the same proof applies for the sequence an = an - (2/3) C.
Problem P.26. Let X1, X2,... be independent, identically distributed random variables such that Xi > 0 for all i. Let EXi = m, Var(Xi) = a2 < oo. Show that, for all 0 < a < 1, a2a2 / X1 + ... + Xn li nVarQ n mz( 1-a )
Solution. Put Xn = (X1 +
+ Xn)/n and Sn = X1 +
+ Xn. Notice
first that
[(Xi +..+Xn/l) n
=nE[(X-m+ma-E(X))2] J
= nE(Xn - ma)2 - n(E(Xn - ma))2. First we show that the second term tends to zero as n - oo. The quadratic Taylor polynomial of (1+z)a with remainder term implies that for z > -1, 1- (1 +z)a < (1- a)z2 - az. From the Jensen inequality and this relation, we get
0 < ma - E(Xn) = maE (1
- (;y) M 2
3. SOLUTIONS TO THE PROBLEMS
452
which implies the assertion. Consider next the first term. Introduce the
notation Zz = (Xi/m)-1 and Zn = (Z1+Z2+ +Zn)/n. Then Zi > -1, E(Zj) = 0, Var (Zi) = Q2/m2, furthermore for arbitrary 0 < e < 1, nE(X" ma)2 = m2anE((1 + Zn)a - 1)2
-
= m2a[nE(((1 + Zn)a - 1)2llznl<e)
+ nE(((1 + Zn)a In the next step, we will use the following inequalities. In the first term,
alzl/(1 + zj) < 1(1 + z)a - 11 < ajzj/(1 - jzj) for jzj < 1. In the second term, I (1 + z) a -11 < z a for z > -1. Observe, furthermore, that if Iz j < e,
then jzj/(1 + Izi) > zI/(1 + e) and Izj/(1 - jzj) < zj/(1 - e), therefore, m 2a [n(1
z
_
+ e)2 E(Z n 1
'<E )]
< n E(Xnm a)2
_
az
< M 2a [n(1
E)2E(7'nhIZ,.I<e)
Therefore, it is sufficient to show that z
nE(ZnIlZnl<e) -+
and 2
nE(ZnallZnl>e) -' 0.
Let 0 be the standard normal distribution function. Then for large n,
r ( nZn 1
az E
- ) = m2 n > nE(Z2I n I-9nj!5e m2 = nE(Z2) a2
> mz E since
(
MZn
(o
a
2
y2dO(y)
m2
m
I
11\ Q/M J
2
2
z
a
nZn/(Q/m) -+ 0 in probability. In addition, e2anE
(
< ne2aE
Zn Zn
2
III>1 2
= ne2a-2E(Z2I n Since
z
mz =nE(Zn) we get 2
nE(ZnII Z..I <_e) asn -goo.
-'m 2;
Il
l>
) -+ 0.
... )<_emn
3.8 PROBABILITY THEORY
453
Problem P.27. Let F be a probability distribution function symmetric with respect to the origin such that F(x) = 1 - x-1K(x) for x > 5, where
K(x) -
1
if x E [5, oo) \ UO0_5(n!, 4n!),
n,
if x E (n!, 2n!], n > 5, if x E (2n!, 4n!), n > 5.
3- Zn,
Construct a subsequence ink} of natural numbers such that if X1, X2, ... are independent, identically distributed random variables with distribution function F, then for all real numbers x lim P cc
x
1
=
X j < 7rx
l nk j=1
1
2
+ 1 arctan x.
r
Solution. Observe first that the independence of X1, X2, ... implies that for any nk > 1 and t # 0 the characteristic function of the random variable nk Ynk
_ ink EXj j=1
can be given as
E(ett ^k) = (1 -
(1/7) ' hnk (tllr)ItIJJ\nk. nk
The assumed symmetry implies that, for any s # 0,
(1-E(etnkx')=
link(s)= ISI
I3I J-5
nk I
cc
1-cosnkx dF(x)
I
1-cosnkx I dF(x)-2
I
(1-cosy)d(yKl Iylnk" )5e
nk
\s
The above form of the characteristic function is useful since by a suitable choice of subsequence ink} we can guarantee that these functions converge to a-ItI. Note that this is the characteristic function of the Cauchy distribution, which is the limit distribution given in the problem. (See A. Renyi, Probability Theory, Akademiai Kiado, Budapest, 1970, IV. §10, VI. §2.) The first term of the above representation of hnk tends to zero if for k --+ oo, nk ` oc. It is easy to see that if K - 1, then the second term (with the choice nk = k) tends to the limit
°° 1 - cosy
2 o
y2
dy = 7r
454
3. SOLUTIONS TO THE PROBLEMS
Thus, in this case nk = k would be a suitable choice. Therefore, it is sufficient to choose the sequence {nk} so that for any y > 0 and s 0,
K((y/lsl)-nk)- 1ask-goo. Let {ak} be an arbitrary sequence of positive integers such that ak -f 00 and ak/k --+ 0 as k -+ oo. Then for arbitrary x > 0 and sufficiently large
k, 4k! < akk!x < (k + 1)! and for any such k, K(akk!x) = 1. That is, if we choose nk = akk!, k = 1, 2, ... , then the continuity theorem of P. Levy, implies that hflk (s) -a 7r for k -+ oo. Therefore, E(eit" k) -+ a-Itl for
arbitrary t # 0 as k -+ oo, and the repeated application of the continuity theorem completes the proof.
Problem P.28. Let a E C, IaI < 1. Find all values of b E C for which there exist probability measures with characteristic function 0 satisfying 0(2) = a and 0(1) = b.
Solution. If 0 is a characteristic function with the required properties, then 0(0) = 1, 0(-1) = b, and 0(-2) = a, and the self-adjoint matrix 0(1)
o(O)
qS(2)
0-1)
0(1) 0(0)
00) 0(-2) 0(-1)
I=
1
b
a
b
1
b
a
b
i
is positive semidefinite. So its determinant is real and nonnegative:
1+ab2+ab2-2bb-aa> 0.
(1)
If a = u + vi and b = x + yi, then (1) can be rewritten as (2 - 2u)x2 - 4vxy + (2 + 2u)y2 < 1 - u2 - v2.
(2)
Let a' be a complex number such that its square equals a. It is easy to see that if IaI = 1, then (1) holds for the points of the interval between a' and -a' (since IbI < 1, it does not hold for the other points of the connecting line). If IaI < 1, then (1) holds for the points in the interior and on the curve of the ellipse with major axis 2 + 2Ia1 and foci a' and -a'. These statements are shown next. Assume first that lal = 1. Let the random variable X be defined as arg a'. Then Ox (1) = a' and OX (2) = a'2 = a, so b = a' is suitable. A similar proof shows that b = -a' also satisfies the required properties. Next assume that lal < 1, and let b be a boundary point of the ellipse. In this case, we have equality in (1). If b2 - a = re2ai, then (1) implies
that r2 = (b2 - a)(b2 - a) = lb14 - ab2
= Ib14 - 21b
that is,r=1-IbI2.
12
1
+ = (Ib12 -1)2,
-
ab2 + Ia12
3.8 PROBABILITY THEORY
455
Put c =Re a-aib, w1 =a+ axecosc, and w2 =a -axccosc. Furthermore, let
1
pi =
P2 =
Im a-"b
2+21
-c2'
1
Im a-"b
2
2 1-c2
Since lbl < 1, p1 and P2 are positive. Define X as follows: P(X = w1) = p1
and
P(= w2) = p2
Then the characteristic function 0 of X satisfies
(1) = pie"' +p2ew2i = eai(piearccosc +p2earccosc) = e' ((pi + P2) cos(arccos c) + i(pi - P2) sin(arccos c)) = eai((p1 + p2)c + 2i (p, - p2)V'1 - ) = e«i((Re a-aib) + i(Im e-ccib)) = b and
0(2) = ple2wli +p2e2w2i = e2ai(Pie 2arccosc
+p2e-2arccosc)
= e2aa ((pl + p2) cos(2 arccos c) + i (p1 - p2) sin(2 arccos c) ) = e2ai((p1 + p2)(2c2 - 1) + i(pi - p2)(2c 1 - c2)) = e2«i(2(Re a-"b)2 - 1 + 2i(Re e-aib)(Im e-aib))
= e2ai((e-aib)2 + le-aibl2 - 1) = b - e2ai(1 - lbl2) =b2-re2ai=b2-(b2-a)=a. Hence 0(1) = b and 0(2) = a. Finally, we show that the set of the suitable b values is convex, and consequently, the interior points are also good. Let b1 and b2 be two suitable points, and let Q1 and Q2 be two probability distributions such that OQl (2) = OQ2 (2) = a, OQl (1) = b1i and r¢Q2 (1) = b2.
Define Q3 = )tQ1 + (1 - \)Q2 with 0 < A < 1; then Q3 is also a
probability measure, OQ,(2) = a, and 0Q,(1) = )b1 + (1 - ))b2.
Problem P.29. Let Y(k), k = 1, 2,... be an m-dimensional stationary Gauss-Markov process with zero expectation, that is, suppose that Y(k + 1) = A Y(k) + e(k + 1),
k = 1,2,...
Let Hi denote the hypothesis A = Ai, and let Pi(0) be the a priori probability of Hi, i = 0, 1, 2. The a posteriori probability P1(k) = P(H1IY(1), ... , Y(k)) of hypothesis H1 is calculated using the assumptions P1(0) > 0, P2(0) > 0, P1(0) + P2(0) = 1 .
3. SOLUTIONS TO THE PROBLEMS
456
Characterize all matrices A0 such that P{limk-,,. P1(k) = 1} = 1 if Ho holds.
Solution. Let Y(1) be an m-dimensional, normally distributed random vector with zero expectation and covariance matrix D. The so-called "white noise" e(k), k = 1, 2, ... , is independent from the past of the process Y(k), that is the s(k)'s are m-dimensional, normally distributed random vectors independent from Y(1) and from each other, and they have zero expectation and covariance matrix Q. The background of the problem is as follows: from hypothesis H1 and H2, we accept the one for which the posterior probability converges to 1. We examine the robustness of our procedure, that is, if Ho holds, then our decision must result in the hypothesis that is "closer" to H0.
Suppose that Q is regular. Since the process is stationary, each Y(k) has the same covariance matrix D, k = 1, 2, .... From (1) we obtain the following matrix equation: D = E(Y(k + 1)Y* (k + 1)) = E([A Y(k) + s(k + 1)] [A Y(k) + e(k + 1)]*) = E(A Y(k)Y*(k)A*) + E(E(k + 1)e*(k + 1)) = ADA* + Q.
Hence Q = D - ADA*; therefore, D is also regular. Since Q is positive definite and D is at least semidefinite, for any x E R" such that Dx = 0, 0 < x*Qx = x*Dx - x*ADA*x = -(A*x)*D(A*x) < 0,
that is, x = 0. Since P1(0) + P2 (0) = 1, P1(k) + P2 (k) = 1 for all k = 1, 2, ... , and therefore limk..,, P1(k) = 1 if and only if limk.,,. P1(k)/P2(k) = oo. The Bayes theorem implies that, for i = 1, 2,
P(HijY(1) =yi,...,Y(k) = Yk) =
PY(1),...,Y(k)IH;(y1,...,Yk)Pi(0), PY(1),...,Y(k) (Y1, ... , yk)
where py(1),...,Y(k) (or py(1)....,y(k)IH;) denotes the probability density func-
tion of variables Y(1), . . . , Y(k) (or the corresponding conditional density function under Hi). This relation implies that
P(HIIY(1) =y1,...,Y(k) =yk) _ PY(1),...,Y(k)IH,(Y1,...,yk)P1(0) P(H2I Y(1) = y1, ... , Y(k) = yk)
PY(1),...,Y(k)jH2 (y1, ... , Yk)P2(0) .
/ (2)
Under the hypothesis Hi, the random variables
Y(1) : N(0, D),
Y(k + 1) - A2Y(k)
N(0, D)
are independent. (Here N(M, S) denotes the normal distribution with expectation M and covariance matrix S. In the following, we will denote the density function of N(M, S) by pN(M,s))
3.8 PROBABILITY THEORY
457
In this case, PY(i),...,Y(k)I Hi (y1, ... , Yk)
= PN(o,D) (Y1)PN(o,Q) (y2 - Aiy1) ... PN(o,Q) (Yk - A:yk-1) 1
e-zyiD-1y1
.
1
(27r)mIQIe
-2(Y2-Aiyl)*Q-1(Y2-Aiy1)
(2-7r --I DI
Xx
1
(yk-Aiyk-1)*Q-1(yk-Aiyk-1)
2
(21r)jQI e 1
(21r)kmIDIIQIk-1
C-2 [yiD-1y1+E' 2(yi-Aiyi-1)*Q-1(yi-Aiyi-1)]
Substituting this relation into (2) and using the variables Y(1), . . we get the following ratio of the posterior probabilities: P1(k) =PP1 (O) exp( 1 P2 (k) 2 2 (0)
.
,
.
Y(k),
k
E[Y(7) - A2Y(j - 1)]*Qi-1[Y(j) - A2Y(j - 1)] j=2
k
E[Y(7) - A1Y(7 - 1)]*Q-1[Y(j) - A1Y(7 - 1)]). j=2
That is, if the hypothesis Ho holds, that is, the process is based on the matrix A0. Then with the random variable L(Y(1), ... , Y(k)) k-1
_ E[(Ao - A2)Y(7) + IF(i + 1)]*Q-1[(Ao - A2)Y(7) + IF(i + 1)]f-1 k-1
- E[(Ao - A1)Y(7) +,F(7 + 1)]*Q-1[(Ao - A1)Y(7) + E(7 + 1)] j=1
we get
=exp (L(Y(1) ...,Y(k)) \I
.
The question to be answered is as follows: when does the limit relation
m L(Y(1),... , Y(k)) = oo hold with probability one?
The process (Y(k), e(k+1) k = 1, 2, ...) is an ergodic Gauss-Markov process. By the ergod theorem, lim 1 L(Y(1), ... , Y(k))
k-+oo k
= E([(Ao - A2)Y(1) + e(2)]*Q-1 [(Ao - A2)Y(1) + E(2)])- E([(Ao - A1)Y(1) +s(2)]*Q-1[(Ao - A1)Y(1) +E(2)]) = tr [(A0 - A2)*Q-1(Ao - A2)D] - tr [(A0 - A1)*Q-1(Ao - A1)D] = L
3. SOLUTIONS TO THE PROBLEMS
458
with probability one. If L > 0, then limk_,,,. P1(k)/P2(k) = oo with probability 1, and this is the same as limk_ o. P1(k) = 1.
Remark. For L < 0, we get limk ,,,. P2(k) = 1 with probability 1. One
can also prove that for L = 0, neither Pl(k) nor P2(k) tends to 1 with probability one, that is, no appropriate decision between Hl and H2 can be made.
Problem P.30. Let X and Y be independent identically distributed, real-valued random variables with finite expectation. Prove that
EJX +YJ > EJX -Y1. Solution 1. For any real numbers a and ,Q, ja + /jI - la-01 = 2 sign(a/3) min{jcaj, I/3 }. Therefore,
E(JX +Yj) - E(JX -Yj) =E(IX +YJ - IX -Yj) =2E(min{IXj, Yj}sign(XY))
=2J0 'IP(IXI > t, JYJ > t,XY > 0) - P(JXI > t, JYJ > t,XY < 0)}dt.
Since X and Y are independent, identically distributed variables, and P(Z > t) = P(Z > t) almost everywhere,
E(JX +YJ - IX -Yj) =2
=21
j{P(X > t)]2 + [P(X < -t)]2 - 2 P(X > t)P(X < -t)}dt
> t) - P(X
-t)2dt > 0.
E(JX +Yj) > E(JX -Yj).
Solution 2. This problem has a simple solution, if we use the Fourier transform of generalized functions. We may assume that X has a nice density function f (x), say an infinitely many times differentiable function with finite support. Then the statement of Problem P.30 can be rewritten as
f IxI(f *f)(x)dx> J IxI(f *f )(x)dx,
3.8 PROBABILITY THEORY
459
where * denotes convolution, and f - (x) = f(-x). This formula can be rewritten by means of the Plancherel formula if JxJ is considered as a generalized function. The Fourier transform of Jxi, when it is considered as a generalized function, equals -2Q-2. (See, for example, I. M. Gelfand, G. E. Shilow, Verallgemeinerte Funtionen (Distributionen) I. VEB Deutscher Verlag der Wissenschaften, Berlin (1960) Vol. 1). The Plancherel formula, which is actually the definition of the Fourier transform of generalized functions, and the definition of Q-z as a generalized function (see formula (5) on page 60 of Gelfand and Shilow's book) state that the last formula can be rewritten as
r
2
f2(-a) - 2f2(0)] da
0
-2
f
1
-
{f(a)f-() +
(-o) - 2f(O)f- (0)] da,
where-denotes the Fourier transform. A simple calculation shows that the last formula is equivalent to the relation
f
' [f(ar) - f(-U)]
z
dar < 0
This relation clearly holds, since f(Q) - f (-v) is a purely imaginary number.
Remark. The second solution was suggested by Peter Major.
Problem P.31. Let X1, X2, ... be independent, identically distributed random variables such that, for some constant 0 < a < 1,
P {Xi = 2k/-} = 2-k,
k=
1, 2, .. .
Determine, by giving their characteristic functions or any other way, a sequence of infinitely divisible, nondegenerate distribution functions G,,, such that sup
-00<X<00
P
nl/a
Solution 1. Put S,,, = X1 +
<x -Gn(x)
->0
as
n ->oo.
+ X,, n = 1, 2. .... The characteristic
function
p(t) = E (eitxl)
°°
kk=1
ck/n>
eit2
2k
t E R.
Introduce the sequence ryn = n/2 flog n1 , n = 1, 2, ... , where lul = min{n E N : u < n}. Obviously, 1/2 < ryn < 1 for all n. The independence
3. SOLUTIONS TO THE PROBLEMS
460
assumption implies that for all t the value of the characteristic function of the random variable Sn/nl/a is 00
1 + Deit2(1/')k/nl/' -1) 2k
) = cpn (1/n 1/a ) =
cpn(t) = E(e
k=1 00
1+
=
it2a/a)(k-rlognj)/7n/°
(
1
2 Flog nj
e
1+n
1
(eit2' /°/ryl,/°
n
_)
1
2k- Flog ni
2
- 1) n
r=- Flog nl +1
For a given 1/2 < 'y < 1, define 00
1)
h--1 (t) _
r=-oo
tER
2
and put
7(T) = eh1(t)
tER
Since for all t E IR, 00
Ih7(t)I
27 E
2-r+71-1/alt
-0 <2y+ r=0
r=1
-t
1_1/a
1-21-1/'
ItI,
the above definition is correct, and .y(.) is the characteristic function of the random variable 00 2r/a
Z.y = E r=-oo
,yl/a Yr('Y),
where the Yr(y)'s are independent Poisson-distributed random variables with expectation E(Yr(ry)) = -y/2r, r = 0,±l,±2,... . For the sake of simplicity, Z.y can be defined as the limit in distribution of the partial sums
n
2r/a
E ,y1/aYr('Y) r=-n By the continuity theorem of P. Levy it is easy to see that this is possible. Obviously, Zr is an infinitely divisible random variable. Observe that, for any fixed t E R, .y (t) is continuous as the function of as a bivariable function is continuous on ry on the interval [1/2, 1]. the domain [1/2] x R). We are going to prove that the distribution function H.y of Z.y, which is uniquely defined by the relation
7(t) =
eitxdH (x),
t E R,
3.8 PROBABILITY THEORY
461
is continuous on the whole real line for all -y E [1/2, 1]. Simple calculations show that 7
-1 a 2
f
°°
1
dt = 00
J 2
°°
00
exp
r=-oo
0o
f f
1
ds
J
- cos(s2))
ds
r=-oo
oo
exp -ry E(1 - cos(s2-k/,))2" ds
0
<
\
(1
f exp
/1
(eis2 °_ 1)
ry
k=0 oo
-ry E (1 - cos(s2-k/,))2k
exp
°
ds,
(1)
k=,c(s)
where for arbitrary s > 0, ac(s) denotes the smallest integer k > 1 such that log(s/7r) < k - 1. Since ac(s) < 2 + log(s/-7r) and s2 -k < 7r/2, for all k > ic(s), s2-k/a < 7r/2. Thus, using the inequality
(0<x<2),
1-cosx> 2x2 we get 00
(1
-
Sz
cos(s2-k/,,)) 2k >
E
T2
k=rc(s)
k=rc(s)
4 6221-,c(s) >
T2
Hence
ra
s2
2
k
0C
k=ic(s)
2
(2)
ir
f00 I
1
2(1 2/a)k >
.y(t)Idt < 2'y1/a
e-try/7rsds
< oo.
0
The continuity of H., follows from the inversion formula of characteristic functions. (See, for example, A. Renyi, Probability Theory, Akademiai Kiadd, Budapest, 1970.) Consider the random variables Zry with distribution functions G,,(x) = H7 (x), x E R , and characteristic functions XFn (t) _ 7n (t), t E IR, n
1,2,... . Since for n -* oo
t-
00
(eit2'/°/7,1,x°
h 7n()
- 1) ^In2r , 0
r=- [log n) +1
for arbitrary t E R, it is easy to see that
On (t)-an(t)-*0
as
n ->oo.
(3)
3. SOLUTIONS TO THE PROBLEMS
462
Finally, introduce
A(F., Gn) = where
sup
-oo<X<00
I F'n(x) - Gn(x)I
r
Fn(x)=P j Sn <x},
xER,
and consider a sequence {n'} of positive integers tending to 00. Since 1/2 < ryn, < 1, the Bolzano-Weierstrass theorem implies that there is a subsequence {n"} C {n'} such that for some ry E [1/2, 1], ryn --> ry as n" -4 00. In this case, A(Fn,,,
A(Fn,,, Hy) + A(Gn,,, H-r)
and Tn (t) = yn (t) -+ .y (t) for all t E R as n" -> 00.
From the continuity theorem of Levy, we know that Gn (x) _ H.yn (x) --> H.7(x) at any continuity point of the limit function. Since the limit function is continuous everywhere, it converges uniformly by a well-known theorem of Polya. Hence
A(Gn", Hy) -> 0
as
n" -> oo.
The relation (3) and a similar argument imply that
A(Fn,,,H.y)->0
as
n"->oo.
Since the sequence {n'} was arbitrary, A(I''n, Gn) --> 0
as
n -> oo.
Solution 2. Define the sequence Z.y of random variables in the same way as above: 'Yn =
n
2
flog nl '
L,, = [logn],
Z.y =
n-1/«
00
E 2(L^+r)/«yr(ryn),
r=-oo
where the Yr(ryn)'s are independent Poisson-distributed random variables The "three-series theorem" implies with expectation ryn/2r = that the sum defining Z.y is convergent. We will use estimations (1) and (2), which guarantee that the absolute value of the characteristic function
Wt) of the distribution function Gn (x) of the random variable Z.y is integrable, and this integral remains bounded independently of n. The distribution functions Gn(x) are obviously infinitely divisible, and we will show that they satisfy the requirements of the problem. This assertion will follow from the following two statements.
3.8 PROBABILITY THEORY
Put Sn = n-1/a En=1 Xk
463
and let Fn(x) denote its distribution func-
,
tion.
(a) There exists a sequence En -> 0 of positive numbers such that for all x E (-oo, oo), Gn(x - En) - En < Fn(x) < Gn(x + En) + En.
(b) For all n, the density function gn(x) of Gn(x) exists, and there is a real M that is independent of n, and for all x E (-oc, oo), gn(x) < M. First, we show that the requirements of the problem follow from (a) and (b). Indeed,
Fn(x)-Gn(x) = Fn(x)-Gn(x+En)+Gn(x+En)-Gn(x)
En+M En, (4)
Gn(x)-Fn(x) = Gn(x-En)-Fn(x)+Gn(x)-Gn(x-En)
En+M En. (5)
Since En -> 0, from (4) and (5) we have the limit sup., I Gn (x) - Fn (x)0,I - . which was to be proved. Statement (b) follows from estimations (1) and (2).
Proof of statement (a). Let dn be a sequence of real numbers that tends to infinity slowly enough. Put Ln -dn
Znl) = n-1/a > 2(Ln+r)/ayr(_Yn), r=-oo 00
Zn2) = n-1/a Zn3) = Z-rn
E
2(Ln+r)/ayr(7n)
- Znl) - Zn2) n
S(1)
= nl
EXkI(Xk<2(Ln-dn)/a)yk(7'n), k=1 n
S(2)
= n-1/a E XkI(Xk>2Ln+dn)/c)yk('Yn), k=1
Sn3) = Sn
- S(1) - S(2)
where IA is the indicator function of the set A.
The relations P(5,2) # 0) -> 0, P(Zn2) # 0) -+ 0, E(Sn1)) -+ 0, E(Znl)) -+ 0, and simple calculations imply that Sn - Sn3)
0
and
Zyn - Z(3)
where = denotes the stochastic convergence. Observe that Sn3) can be written in the form Ln+dn
21/avin-1/a,
Sn3) _
l=Ln-dn
01
(6)
3. SOLUTIONS TO THE PROBLEMS
464
where vl = 01j, 1 < j < n , xj = 21/'J, 1 = 1, 2, ... , and #A denotes the cardinality of the set A. Denote the distribution function of by F(3) (x), and that of Z,3) by Gn) (x). We are going to prove that Var (F(3) (x), G'3) (x)) -' 0,
(7)
where Var ({pi}, {qi}) denotes the distance >i Ipi - qiI of the discrete distributions {pi} and {qi}. From relations (6) and (7), statement (a) follows immediately. We only have to prove the convergence in (7). Denote by distr{Yj(ryn), IiI < dn} and distr{vl+L,,,Ill < dn} the joint distributions of the random variables {YY (yn)} and {vl+L }, respectively, (Ii 1 < dn, 111 < dn). The limit relation (7) follows from the convergence Var (distr {Yj (ryn),
distr {vl+L },
Ii 1 < dn},
(8)
111 < dn).
This last statement can be proved as follows. Since dn tends to infinity slowly enough, by restricting the range {pl , Ill < dn} of the random variables appearing in (8) by log2 n, the resulting error does not affect the validity of the following estimation:
P(vl+L..
n!
pl , I4I
d
d,.
111=-dnPL! (n-El-dnpl)!
d
(l_2_L+l) d
t--drift
)
11 2-P1
l=-d
dII
)n_Edn
l=-d -n2-L"-1
(n2
e
pl!
= (1+0(log3 n
1+0
n log3n)/
do l
f lP(Y(7n)=pl)
n
= (1+0 The proof is complete.
(P(Yn(-Yn)=pl ,
IlI -dn)
3.9 SEQUENCES AND SERIES
465
3.9 SEQUENCES AND SERIES Problem S.I. Let the Fourier series a 2 + E(ak cos kx + bk sin kx) k>1
of a function f (x) be absolutely convergent, and let 2
2
2
ak + bk > ak+1 + bk+1
Show that
1
(k = 1,2,...).
f02, (f(x + h) - f(x - h))2dx
(h > 0)
h
is uniformly bounded in h.
Solution. By the Denjoy-Lusin theorem, >' 1(Iak! + jbki) < oo. Hence 00
10k<00
(Pk=
ak+bk; k=1,2,...).
(1)
k=1
So we obtain El 1,02k < oo. By the Riesz-Fischer theorem and the completeness of trigonometric functions, it follows that (f (x))2 and thereby (f (x + h) - f (x - h))2 is integrable. Easy calculation shows that
f(x+h) - f(x- h) - 2E(bksinkhcoskx-aksinkhsinkx). k=1
The Parseval formula implies that 2ir
00
r
J0
(f (x + h) - f (x - h))2 dx = 41r E Pk sin2 kh.
(2)
k=1
By the condition ok > ek+1 (k = 1, 2, ... ), we obtain kPk < 1 Pa (k = 1, 2, ... ). So by (1), kok = 0(1). Using sin2 x < lxi, by (1) we obtain 00
OK)
kQk = O(h).
ok sin2 kh < IhI k=1
k=1
By this and (2), we conclude the theorem.
Remark. Instead of the absolute convergence of the Fourier series and
Pk > Ok+1 (k = 1, 2,... ), it is enough to suppose that E' 1 kPk < oo. Gabor Halasz and Tibor Nemetz remarked that by the conditions of the theorem (1/h) fo (f (x + h) - f (x - h))2 dx = o(h) holds.
3. SOLUTIONS TO THE PROBLEMS
466
Problem S.2. Let y1(x) be an arbitrary, continuous, positive function on [0, A], where A is an arbitrary positive number. Let
(n=1,2.... ).
yn(t) dt
yn+1(x) = 2 f 0
Prove that the functions yn(x) converge to the function y = x2 uniformly on [0, A].
Solution. We will not only prove the theorem but also approximate the rate of the convergence.
Let yi(x) and yi*(x) be given continuous functions on [0,A] such that 0 < yi(x) < yl*(x),for all 0 < x < A. Denote by yi(x),y2(x),...,yn(x),... and by yi*(x),y2*(x),...,y,**(x), . the functions obtained from yl(x) and y**(x) by the above iteration. Obviously, y,*, (x) < y,** (x) for all 0 < x < A. So it is enough to prove that the claim of the problem holds for a constant function of value C > 0, since if we choose C to be the maximum (minimum)
of y1(x), then the nth element of the function series received from the constant function will be larger (smaller) than yn(x) at each x. So let Y1 (x) - C, where C > 0, and let Y2 (x), Y3 (x), . . . , Yn (x) be the functions obtained by the iteration. We can show by induction that Yn(x) =
Cn
x2-(1/2"-')
where cn is defined by the recursion cn+1
=
Cn 1
(n = 1, 2, ... ; c1 = C).
(1)
2nn
Take the 2n+1th power of each side of (1): 2n+1 Cn+1 = cn2n
1 1
2n11
(1 -
The multiplier of Cn2n is positive and converges to a finite positive value (to e2) as n goes to infinity. So the quotient of the neighboring elements of the , . .. cnn, ... remains between two positive constants. Therefore, series ci, c22
by appropriate constants m and M,
mn < c n < Mn (n = 1, 2, ...)
.
n
Hence, by mn/2n < cn < Mn/2n
(n = 1, 2, ... ),
it follows that cn =
e°("/2n).
(2)
3.9 SEQUENCES AND SERIES
467
Obviously,
Yn(x) - x2 =
X2-(1/2'-') . (en - 1) +
(x2-(1/2n-1)
- x2).
By (2), for all 0 < x < A, x2_(1/2n
')=0.(cn-1
2Icn-11
2n )
It is easy to show by derivation that the function x2-(1/2n-') - x2 achieves
its maximum on the interval (0,1) at xo = (1 - 2 2"-1, and 1)
') - xo2 =
2-(1/2n-1)
2-(1/2n-1)
(1
- x0
/Zn
1))
< I' -
(1
-
211
)J = 2
ForA>l, 1<x
Thus,
1
1
I
n(X) x= 0 y2
1
02 1
n
n 2n
Hence, it follows that, starting from the function y1(x), which satisfies the conditions of the problem, we get yn(x) -
x2 = 0 (
),
where the constant included in the sign 0 depends only on A, and on
the maximum and minimum of y1 (x). So the claim of the problem is proved.
Remarks.
1. Gabor Halasz proved the following generalization of the problem. If the function G(y) is monotone increasing and continuous for all y > 0,
G(0)=0,G(y)>0fory>0, and (
f G(y) dy < +oo, 0
but
00
J
1
G1
dy = +oo,
0
then obtaining the functions yn(x) by the recursion X
yn+1(x) = f G(yn(t))dt, 0
3. SOLUTIONS TO THE PROBLEMS
468
it follows that the functions yn(x) uniformly converge to the unique solution of the differential equation y'(x) = G(y), which goes through (0,0) and does not agree with the constant function (x = 0) in any neighborhood of the origin. 2. If we suppose that yl (x) is nonnegative instead of positive, we can state the following:
Let b be the upper bound of those x values for which y1 (u) = 0 for all
0 < u < x, and let f (x, 6) be the function such that f (x, 6) = 0 for 0 < x < 6 and f (x, 6) = (x - 8)2 for b < x < A; then the functions y.,, (x) uniformly converge on [0,A] to the function f (x, b).
Problem S.3. Let E be the set of all real functions on I = [0, 1]. Prove that one cannot define a topology on E in which fn ---+ f holds if and only if fn converges to f almost everywhere.
Solution. Let I2m+k =
k1 k
[-_,] 2n,
2m
(m = 0,1,2.... ; k = 1, 2, ... 2 ),
and denote by fn the characteristic function of I. Then If"} will nowhere converge to 0; but if If,,,, } is an arbitrary subsequence and xnk E Ink, where {xnk, } converges to a number x E I, then obviously {fnkt } -* 0 everywhere but at x, so in particular, almost everywhere. So if T is a topological space
and xn 74 x, xn E T, x E T, then there exist, a neighborhood of x such that for some subsequence Xnk , xnk ¢ V, and hence for any subsequence xn,,, xnk{ 74 x. Remarks.
1. The solution shows that the same statement is true if in the definition of almost-everywhere convergence we give the role of the zero-measurable sets to a set system R, where for any x E I, {x} E R but I ¢ R. 2. With a slight modification of the solution, one can show that the statement holds when I is the set of real numbers or a closed interval.
Problem S.4. Let the continuous functions fn (x), n = 1,2,3,..., be defined on the interval [a, b] such that every point of [a, b] is a root of fn(x) = fm(x) for some n m. Prove that there exists a subinterval of [a, b] on which two of the functions are equal.
Solution 1. Denote by En,,.,, the zero set of fn(x) - f,,,,(x) for n # m. These sets can be enumerated. Let
M1 ,M2,M3i...,Mk,... be an enumeration of these sets. If Ml is a nowhere-dense set on [a, b], then there exists a subinterval [al, b1] of [a, b] that is disjoint from Ml. If M2 is a
3.9 SEQUENCES AND SERIES
469
nowhere-dense set on [a, b], then there exists a subinterval [a2i b2] of [a1i b1]
that is disjoint from M2. If all of the Mk are nowhere dense on [a, b], then we can continue this method. The intersection of all the intervals [a, b] ID [a1, b1] ID ... ID [ak, bk] I ...
is not empty and does not contain a root of fn (x) = fm (x) for any n # m. This is a contradiction. So there is an MV that is dense on some [c, d] part of [a, b], that is, the roots of the corresponding equality fjy(x) = f,, (x) are dense on [c, d]. By the continuity of the functions, it follows that fw(x) = f,- (x) on [c, d].
Solution 2. The union of the sets Enm of the previous solution is the interval [a, b]. If none of the sets En,,,, is dense on any subinterval of [c, d], then their union cannot be an interval, since an interval cannot be a union
of countably many nowhere dense sets. So there is a set that is dense on some [c, d] subinterval of [a, b]; that is, the set of the roots of the corresponding fw(x) - fj (x) is dense on [c, d]. Since the functions are continuous, each point of [a, b] is a root. Remark. Many participants used the Baire category theorem to generalize the problem.
Problem S.5. If >m=_m lam I < oc, then what can be said about the following expression? 1
lim
+00
E lam-n + am-n+1 + ... + am+n
2n + 1 m=-00
Solution. Since I m°=-m I am I = v < oo, Lm=-m am = S exists. Let 1
Cn =
+00
2n + 1 m=-00
lam-n + am-n+1 + ... + am+n I
We will show that limn-00 Cn = A. Let e be an arbitrary positive number. Then there is a natural number M such that Iaml < e. JmJ>M
Hence, for n > M,
...I+
(2n+1)Cn= Iml>n+M
I...I+ n+M>!mJ>n-M
I...I. lml
3. SOLUTIONS TO THE PROBLEMS
470
So
E lam,-n + ... + am+n l < E am-n + ... + am+n Iml>n+M
Iml>n+M
<(2n+1)
a,,, l < (2n + 1)E, Iml>M
and
+... + am+n l <
lam-n
or < 4MQ.
n+M>ImI>n-M
n+M>ImI>n-M
Since m - n < -M < M < m + n if lml < n - M, it follows that if lm l < n - M, then llam-n+...+am+nl-ISll< E
laml<E,
ImI>M so
lam_n+...+am+nl
- (2n-2M+1)lSll
Iml<
< E Ilam-n+"'+am+nl-lSll<(2n-2M+1)E. Iml
Therefore,
1(2n + 1)Cn - (2n - 2m + 1)ISI I < (2n + 1)E + 4Ma + (2n - 2M + 1)E. Dividing by (2n + 1) and taking the limit as n oo, lim sup lCn - HSl l < E + E = 2E. n-.oo
Since e is an arbitrary positive number, we conclude that
lim Cn = ISl
n--.oo
Remark. Let (M, or, µ) be a commutative measurable group. Namely, M is an additive group, a is an invariant a-algebra of the subsets of M under addition and p is an invariant measure, where p(M) > 0. Suppose that the function S : M x M - M x M defined by S(x, y) = (x, x + y) maps measurable sets of M x M into measurable sets of M x M. Let Al, A2, ... (Ai E a) be a sequence of sets symmetric about 0 (that is, Ai = -Ai) such that UloAi = M; Ai C Ai.+1; 0 < p(Ai) < oo; p(Ai+1 - Ai) = 0(µ(A2)); Ai + A; C Ai+j. Claim. If f (x) E L1(M, p), then
1
p(An)
f ff
(x + y) dFiy)I dµ. =
ffdp
I
This can be proved by similar methods. Mikl6s Simonovits gave this generalization.
3.9 SEQUENCES AND SERIES
471
Problem S.6. Let a1, a2i -,an be nonnegative real numbers. Prove that
Eai
Eai-1
i=1
i=1
i=1
i=1
Solution 1. If one of the ai is zero, then the statement follows immediately from the known inequality between the nth power mean and lower power means. It also follows that equality holds if all the other aj numbers are equal. So we can suppose ai > 0 f o r i = 1, 2, ... , n. Denote by Ek the sum of the kth power of the numbers a1, a2i ... , an, and by Sk their kth elementary symmetrical polynomial. We generalize the inequality:
n (Il1)>1Ek_1_<m5k+(k_1)()Ek, 1 < k < n .
(1)
The problem is the special case of inequality (1) for k = n.
For n = 1, 2, k = 1, equality holds. So we can suppose n > 3, k > 2. We introduce the following notation. If a1,... , an > 0 are real numbers, then denote by [al, ... , a, ] the sum (1/n!) E a l ... a ", where we sum over all permutations i1 , . , in of 1, ... , n. Lemma. If n > 3 and v, 04,
> 0, b > 0, are real numbers, then
[v+26,0,0,a4.... ]-2[v+6,b, 0,a4,...]+[v,6,6,a4,...]>0.
(2)
Equality holds if all of the ai are equal.
Proof. We cut the sum (2) into (3) (n - 3)! terms, so it is enough to prove that for arbitrary real numbers b1, b2, b3 > 0, b1+26 + b2+26 + b3+26 - (b1+6 b2 + b2+6b1 + bi+6 b3 + b3+6b6 1
1
+ b2+6b3 + b3+6b2) + bib2b3 + bsb2b3 + bib2b3 > 0,
where equality holds if b1 = b2 = b3. This is straightforward by the following obvious inequality:
xµ(x - y)(x - z) + yµ(y - x)(y - z) + z''(z - x)(z - y) > 0,
for x, y, z > 0, µ > 0. Equality holds if x = y = z. We conclude the lemma by changing x, y, z, and p into ai, a2, a3, and v/b, respectively. The ideas in the proof of the lemma appears in G. H. Hardy, J. E. Littlewood, Gy. Pdlya, Inequalities, Cambridge University Press, Cambridge, 1959. To prove inequality (1), write it in the following form:
k - 1)
1: aia -1 < nSk +
((k_1)() -
())
Ek .
(3)
3. SOLUTIONS TO THE PROBLEMS
472
The following equalities hold by the definition of [al, .... an]: 1
n(n -- 1)
ajar -1 = [k - 1,1, 0 L.- 0 1#j
n-2
n=[k,0,...0, 1
L.
n-1
(1 Sk = 1,...1,0,...0
(4)
n-k
k
If we substitute the inequalities (4) for (3), and making use of the relation (k) _ n . (k-1)+ we can write (3) in the following form:
(n-1)n[k-1,1,0,...0 < 1,...1,0,...0 +I n-2
k
n-k
k-1-
[k,0,...0. (5)
n
\
n-1
By (n - 1) (k/n) = k - 1 - (k/n) + 1, we can write (5) as
[k-1,1,0,...0 - 1,...1,0,...0 n-2
n-k
k
(k-1-k)[k,0,...0 -[k-1,1,0,...0. n
(6)
n-2
n-1
We introduce the following notation:
of = [t + 1,1,1, ...1, 0.... o ] - [t,1,1, ...1, 0.... 01 k-t-1 n-k+t-1
k-t
n-k+t-1
n>3, k>2, t=0,1,...,k-1. By substituting v = t - 1, 6 = 1 into the lemma, it is straightforward that
0 = 00 < Di < ... < pt -1
(7)
Equality holds if all of the a2s are equal. We can write (6) in the following form: k-2
Ck-1- knll Akk-11
This inequality holds, since by (7), k-2
E Ot < (k - 2)Ok-1 and k - 2 < k - 1 - k. t-o
n
3.9 SEQUENCES AND SERIES
473
It is straightforward that if n > 3 and k > 2, then equality holds if k = n and all of the ais are equal.
Remark. Inequality (1) was first proved by Laszlo Lovasz. He also proved it for the case when some of the ais are zero, but he did not examine the cases of equality. The previous proof is the generalization of Peter Gacs's proof. He used this method for proving the case k = n, that is, for proving Problem S.6. Solution 2. If one of the ais is zero, then the statement follows immediately from the Chebyshev inequality for uniformly ordered sequences. It also follows that equality holds if all the other numbers aj are equal. So we can suppose ai > 0 f o r i = 1, 2, ... , n. Let n
n
n
n
i=1
i_1
i=1
i=1
F(a1,... an)=nfl ai+(n-1)ai-ai>aa-1. By the homogeneity of F, it is enough to prove that
=n.
F(a1i...,an) > 0 if
(1)
F(a1i . . . , an) = 0 holds for n = 1, 2. Let us suppose n > 3. It is enough to prove (1) in those cases when F has a local minimum at a1+ +an = n, that is, when 1
n
ai
dF dai
n
= (n - 1)(a - ai -1) + flai = Aai
(i = 1, ... , n)
(2)
i_1
for some real A.
By the Descartes sign rule and H tai > 0, the polynomial (n - 1)x (n - 1)xn-1 + Ax + fz 1 ai cannot have more than two positive roots. So by (2) it is enough to prove that
F(x,... x, y, ... , y) > 0 for x, y > 0. k n-k By the homogeneity, it is enough to prove that g(x) > 0, where
g(x)=F(x,...x,1,...,1), k
0<x<1. g(1)=1.
n-k
If we show g(x) > 0 for 0 < x < 1, k # 0, k 54 1, then it also follows that in case n _> 3, ai > 0, equality holds if all of the ais are equal. Let 0 < x < 1, 0 < k < n. By rearranging, we get g(x)=k(n-k)(xn-x"i-1-x+1)-kxn+nxk-n+k.
3. SOLUTIONS TO THE PROBLEMS
474
Hence,
9(x) = k(n - k)(1 - xn-1) - n(1 +
+ xk-1) + k(1 + ....+ xn-1)
k(n - k)xn-1 > 0, since
k(n-k) > (n-k)(1+...+xk-1) and k(xk+ +xn-1) >
k(n-k)xn-1.
Equality holds in both inequalities if k = 1, k = n-1, that is, iff n = 2.
Solution 3.
`t
n L ai i=1
i=1
ai -1
ati -
=2 1` L(ai - aj ) (ait-2 + ait-3 aj+... + at-2 ) 2
i=1 j=1
i=1
for arbitrary t, since aj)2(ai-2
(ai -
a -
+ ... +
alai-1
(1)
i, j = 1, 2, ... n.
for
If
n
and
I:ai=1, i=1
then
n
flai = i=1
n
n (an+1 + ai -
an+1) ? an+1 + an+1 E(ai - an+l )
i=1
i=1
n-1 = an+l (1 - n) + an+1 n
(2)
We shall use induction. Equality holds for n = 1, 2. Suppose that n > 2, an+1 < an < < al, and that the inequality of the problem holds for n. We are going to prove it for n + 1. By the homogeneity, we can suppose that 1 ai = 1. By the induction, n
n
n +nHai-Eai-1>0.
an
(n-
i=1
i=1
(3)
i=1
We have to prove n+1
n+1
n+1
n+1
i=1
i=1
i=1
i=1
nEaZ+1+(n+1)flai->ai>0 >0.
(4)
3.9 SEQUENCES AND SERIES
475
The left side of (4) can be written in the following form: n
n
n Eai
+ na; +i + nan+1
ii=1
n
i=1=1 ai + an+1
ai
n
an+1)(E a2 + a;+1)
- (a +
(5)
i=1
By (3), it is enough to prove that (5) is at least as large as an+1 times the left side of (3). So, by rearrangement, we have to prove that n
n
n
(n > ai +1 -
i=1
i=1
i=1
n
- an+1(n > ai) -
an)
ai -1) i=1
n
+ an+1(fl ai - (1 - n)a'n+l - a; +i) ? 0.
(6)
i=1
By (1) and ai > an+1 (i = 1, 2, ... , n), the difference of the first two terms is n
n
1
2
n-2
(ai - aj )2 [a n-1 + ai
aj +
+ aj n-2
i=1 j=1 n-2
- an,+1(ai
n-3
+ ai
aj + ...
n-2)] > 0.
(7)
By (2), the third term is also nonnegative. By (7), if n > 2, then equality holds iff a1 = a2 = = an. If an+1 > 0 and a1 = a2 = = an, then the third term of (6) can be zero only in the case ai - an+1 = 0; otherwise, strict equality would hold in (2).
Problem S.7. Let f (x) _> 0 be a nonzero, bounded, real function on an Abelian group G, 91, ... , gk are given elements of G and Al, ... , Ak are real numbers. Prove that if Aif(gix) > 0
holds for all x E G, then k
kAi>0. i_1
Solution 1. We can suppose that f (gl) > 0. Denote by An the set of those elements that can be written in the form gi' , ... , gk k , where the maximum
3. SOLUTIONS TO THE PROBLEMS
476
absolute value of the numbers al, ... , ak is n, where n > 0 is an integer. Denote by S(H) the sum >xEH AX), where H is a finite set. inf S(An+1) - S(An-1) - 0 S(An)
n>0
holds, since if for some s > 0 and for all n > 0,
- S(An-1)
S(An+1)
S(An)
>E
would hold, then S(An+1) > S(An-1) + ES(An) > (1 + e)S(An-1)
and so S(A2n+1) ? (1 + E)nS(A1)
would hold, which is a contradiction, since S(An) < sup f (x) (2n + 1)k. xEG
By (1), k
S (9i A n)
i=1
1
1: 1: .if(9ix) ? 0,
S(An) x1A i=1
y S(An)
that is, k
k
iL=1
i=11
S(9iAn)
hence
S(An) - S(92An) I- S(An+1) - S(An-1) S(An) S(An)
and therefore k
k
i=1
i=1
A
S(An+1) - S(An-1) S(An)
It follows that k i=1
-/
\i > -
k
i=1
1
,\j
inf
n>0
-
S(An+1) S(An-1) S(An)
- 0.
Solution 2. Let us again suppose f (g1) > 0. It is enough to consider the subgroup G' generated by the elements 91,...,gk. We are going to
3.9 SEQUENCES AND SERIES
477
define a discrete translation-invariant measure on G'. Let E > 0, and let al ... ak be integers. For x = gi' . . . g"'`, denote by µ6(x) the number +e)-(kall+ '+lakl). The measure of G' is finite since (1 00
1: XEG
00
E (1 +
µ6(x) <_ 2k
E)-(al+...+ak)
ak=0
a,=0
cc
00
= 2k E (1 + e) a' ...
(1 + ak=0
a1=0
and so 0 < S = J f(x)dµe(x) < oo. G'
Furthermore, µe(9ix) = (1 + E)tlµe(x),
and so
pe(9ix) - µe(x)1 < spe(9ax). Integrating inequality (1) on G', we get 0 <
f
k
k
Aif(9ix) dye(x) C
G, z=1 k
1+
J
f f(x)dµe(x) G'
k
/I/'
I Ai I
i=1
\i
i=1
i=1
G`
k
[t
.i + SE
Ef (9ix) dµe (9ix) = S
I
Ai
I
i=1
So bye -+ 0, k
SEAi>0 i=1
it follows that
Ai>0. i=1
Remarks.
1. Peter Gacs and Andras Simonovits remarked that the statement of the problem holds for an arbitrary group if k = 2. 2. Peter Gacs and LAszlo Lovasz showed that the statement generally does not hold for a noncommutative group. 3. L6szl6 Lovasz, Endre Makai, and Imre Ruzsa showed that f (x) has to be bounded. /. Laszl6 Lovasz discussed situations where the converse theorem holds.
3. SOLUTIONS TO THE PROBLEMS
478
Problem S.8. Show that the following inequality holds for all k > 1, real numbers a1, a 2 ,--. , ak, and positive numbers x1, x2, ... , xk.
Exi In
i=1
EaixiIn xi < i=1
k
E x1-a.
xi
i
i=1
i=1
Solution 1. By the known inequality relating weighted arithmetic and geometric means,
F,=i XiYi >
,i=1 xi
(ftyi)
kl
(xi > 0, yi > 0, i = 1, 2, ... k; k > 1).
Li=1 xi
i=1
If we substitute yi = xi a` for i = 1, ... , k and take -1 times the logarithm of each side, then we get the claim of the problem.
Solution 2. We are going to prove the following generalization of the problem.
Generalization. Let (X, S, p) be a measure space, where µ(x) > 0. Let x(8)1-a(9), x(s) > 0 and let a(s) be measureable functions on X, and x(s), a(s)x(s) ln(x(s)) are integrable functions on X. In this case,
f x(s)dp In
f
f a(s)x(s) In x(s)dp
(s)1_a(9)d,a < x
x
f d(s)dp
(1)
x
Equality holds if x(s)-a(') is a constant for almost all s E X. Proof. The following inequality holds for all positive d:
d-1>Ind. Put
(2)
f a(s)x(s) In x(s)dp u = exp x f x(s)dp
x
By substituting d = u. x(s)-a('), multiplying the inequality by x(s), and integrating each side on X, we get the following equality:
UJ
x(s)1-a(') dp
- J x(s)dp > 0.
(3)
This is equivalent to the claim of the generalization. Equality holds in (2) if d = 1. So equality holds in (3) (that is, in (1)) if U. x(s)-a(9) is a constant for almost all s E X, that is, if x(s)-a(9) is a constant for almost
allsEX.
3.9 SEQUENCES AND SERIES
479
Problem S.9. Construct a continuous function f (x), periodic with period 21r, such that the Fourier series of f (x) is divergent at x = 0, but the Fourier series off 2(x) is uniformly convergent on [0, 21r]. The background of the problem. The following statement is a special case of the Wiener-Levy theorem. If a positive function f (x) is periodic by 27r and the Fourier series of f2 (x) is absolutely convergent, then the Fourier series of f (x) is also absolutely convergent. The following question arose. Can the absolute convergence be replaced by uniform convergence? The problem above states that it cannot be replaced without assuming the positivity. (R. Salem constructed an f (x) whose Fourier series is uniformly convergent, but the Fourier series of f2(x) is not.) There are more ways to solve the problem. Laszo Lovasz slightly modified Lipdt Fejer's known construction: By adding the Fejer polynomials under a relatively strong gap condition, he received a continuous function f (x), whose Fourier se-
ries is divergent at x = 0, and the Fourier series of f2(x) is uniformly convergent. It is also nontrivial to prove the uniform convergence of the Fourier series of f2 (x). Lajos Posa and Peter Gacs solved the problem in an essentially similar way: They considered the function f (x), instead of its Fourier series and used the observation that if the following inequality holds for a continuous, 27r-periodic function g(x), then the Fourier series of g(x) is uniformly convergent: I
9
1
y (x)-9OI< (log
Ixi-vl)i+E
(See the Dini-Lipschitz condition in I. P. Natanson, Constructive Function Theory 1-3, 1964-65, VII. 3§.) Now we prove the claim.
Solution 1. Denote by sn Y; x) the nth partial sum of the Fourier series of f (x), where f (x) is periodic by 21r and integrable, that is,
f 7r
1
sn(f;x) = it
f(2d)
sin(2n + 1)?9 dV. sinV
(0)
0
If, on [0, 27r], (2n + 1)x
fn(x) = sin
(1)
2
then
sin2 (2n + 1)V
1
sn(fn, 0) = it
,f
sin O
do
1
>
7r
rr/2
1 f sin2(2n + 1)V 7r
19
0
f
sin 79
d19
0
0
>
/Zsin2(2n + 1)t9
n7r
d?9 >
1
ir
f 0
sin2 0
d?9 > Al log n, (2)
480
3. SOLUTIONS TO THE PROBLEMS
where Al and later A2, A3, ... are positive constants. Note that if Ig(x)I 1 is integrable, then for all x,
Isn(g;x)I
(3)
The function we are going to construct will have the form
=
F(x) = E c sin
2n, + 1
x,
(4)
V=1
where c,,, n are still undefined. Outside [0, 27r] we will get the function by periodic continuation. The restriction (v
Ic"I < 4-"
= 1,2,...)
(5)
and F(0) = F(2,7r) = 0 will guarantee the continuity of F(x) and the absolute convergence of (4). Let cl = 1, c2 = 1/4, n1 = 1, and suppose that cl, C2.... , cµ_1, cµ and n1, n2, .... nµ_1 are already defined. Since the functions fn(x) - defined in (1) - are twice differentiable and their Fourier series are uniformly convergent, the following inequality holds at each x, for m > A3
A=1,2.... , v - 1 .
I sm(fn,.; x)I :52,
(6)
Denote by n" the smallest positive integer for which cµ log nµ >
(7)
Al
and
nµ > 2 + max{A3i 3nµ_1} .
(8)
If cµ+1 = min
1LA
1
4 , log
nµ
(9)
then (5) is straightforward. So F(x) is defined. We will use the fact that the following numbers are pairwise different:
ni + nk + l and nj - n,n
(1 < i < k, 1 < m < j).
(10)
By nl = 1 and by (8), the numbers above are different if 1 < i < k < 2, 1 < m < j < 2. If they are different for ni, nk, n nm, where i, k, j, m are less then µ, then by joining nµ to ni, nk, n3, nm, we get the following sums and differences:
>nµ-n2>...>nµ-nµ_1.
3.9 SEQUENCES AND SERIES
481
Even the smallest of them, nµ - nµ-1, is larger than the largest of the previous ones, nµ-1 +nµ_1 + 1. So the numbers at (10) are indeed pairwise different.
By (4) and (5), n + 1)x}
CµCv{cos(nµ -
F2 (x) = 2 µ,v ao
2
E cv + i=1
-2
c
c, c COS(nµ 1
cv cos(2nv + 1)x - 1 cµcv cos(nµ + nv + 1)x. 1
v=1
Since the numbers at (10) are pairwise different, this gives the Fourier series of F2(x). By (5), this is absolutely convergent. We still have to prove
that the Fourier series of F(x) is divergent at x = 0. To see this, divide the series into three parts: µ-1
00
cjsn(fnj; 0) +
Sn(F; 0) = Cµsn(fn; 0) +
j=1
Cjsnµ (fn,,; 0)
j=µ+1
=U1+U2+ U3. By (2) and (7), U1 > µ.
By (6) and (8),
(12)
µ-1
2>cj<3
1U21
(13)
j=1
Finally, by (3) and (5), 00
IU31 < A2 log nµ
c3,
A4cµ+l log nµ < A4.
(14)
j=µ+1
So by (11), (12), (13), and (14), the Fourier series of F(x) is divergent
atx=0.
Solution 2. We will define the function f (x) instead of its Fourier series, such that (0) will hold for f2(x), f (0) = 0, and 7r
lim up n-oo
f
sin nt f (t)dt > 0. t
(15)
0
f (x) will be an even function. Its nth subsum Sn(0) at 0 thus has the following form:
3. SOLUTIONS TO THE PROBLEMS
482
Sn (0) =
Jf(2t)sin2+1tdt. sin t
7r 0
If a continuous function is convergent, then its Fourier series and the Fejer means will converge to the same value, to the value of the function. So to see the divergence of Sn (0), it is enough to prove that Sn (0) 74 0. To prove this, we modify the kernel:
sin(2n+1)t sint
= sin2ntcott+cos2nt =
sin2nt ( 11 +sin2nt cott - J +cos2nt. t t
cott - (1/t) can be modified to be continuous at 0, so by applying the Riemann lemma twice, we get
i
7r
lim sup f f (2t)
sin(2n + 1)t sin t
n--oo
/
dt = lim sup J f (2t) n
0
sin 2nt
t
dt
0
2
= lim scup
n-
f f
f (2t)
sin 2nt
t
dt
0 IT
= lim sup p
f (t)
sir nt
dt > 0.
0
Thus Sn(0) 74 0, that is, Sn(0) is divergent if (15) holds. Since f (n) is even, it is enough to construct it on [0, 7r]. Let an =
Ire-n3,
an = n-2,
Pn =
en2
(n > 0)
and
In = [an, do-1], g(x) = 6n sin pnx . Define gn(x) = gn* (x) between the second and the last roots of g*(x) on In, and gn(x) = 0 elsewhere. We will define f (x) as a sum of a subsequence of the sequence {gn(x)}. Construct a sequence n1 < n2 < ... of natural numbers such that k
fk(x) dx < o
if fk = >gn;,
(16)
i=1
r sin pnj x
J
x
9nk+1(x) dx
< o2 -k
for all j < k,
(17)
0
where p is a sufficiently small constant, say o = 10-5. To see that such a sequence exists, observe that, on the one hand, by Pn --+ oc, by the continuity of gn(x)/x, and by the Riemann lemma, all terms of the integral in (16)
3.9 SEQUENCES AND SERIES
483
converge to zero; on the other hand, by the continuity of (sin 0.,,x)/x, the integral in (17) converges to 0 for any fixed j. Define f (x) _ Ek o g,,,,. (x). Obviously, f (x) is everywhere continuous, even at 0. We will show that the Fourier series of f (x) does not converge to 0 at 0, that is, it is divergent. For the reader's convenience, denote by Sn ~ Tn if Sn/Tn -+ 1. By (16) and (17), the difference of fo (sin pnkt)/t dt and fa nk-1 911k (t) (sin pnkt)/t dt is nk a maximum of 2e. ank-1 gnk (t) sinpnkt
f
dt
bnk sine enkt dt
J
ank
3 Ire-nk 7fenk-(nk-1)3
= bnk
Jf
sine u U
du
7r
e3nk -3nk -1
bnk
2
2 nk_2nk
sine u du
m
nt=1 3-7r
7r
0
31r
=2
So the Fourier series of f (x) is divergent at 0. Now we are going to show that 2
If2(x)-f2(y)I=
1
(log(Ix-YD/
(18)
'
that is, the Fourier series of f2(x) is uniformly convergent. Let x > y and x E In, y E I7n for some n > r. We move y to the sine curve pn sin p.,, t of I7n such that y lies on the same quarter period of the sine curve as x, but f2(x) remains the same. Ix - yI does not increase if n = m or n > M. So we have to prove (18) only in the case when x and y are on the same In and Ix - yj < lr/pn. By the Lagrange mean value theorem, If2(x)
- f2(y)I = bnI sinpnx - sinpn2yl <- 2b2npnl x - yJ = 2b2npn
- yI(log Ix - yi)2 )2
(19)
(log 1=lye
Since h(log h)2 monotonically increases if h < e-2, by (19) we have to prove
(18) only in the case IX - yI = 7r/pn; but this case is trivial. This finishes the solution. Remark. Solution 1 proves more, as it proves that the Fourier series of f 2 (x) is absolutely convergent.
3. SOLUTIONS TO THE PROBLEMS
484
Problem S.10. Prove that for every 19, 0 < 19 < 1, there exist a sequence An of positive integers and a series En 1 an such that (1) An+1 - An > (An)", 00
(ii)
lim r -4 1-0
anra exists, n=1
0c
(iii) E an is divergent. n=1
The background of the problem. Andras Simonovits noted that An+1 > cAn (c > 1) and the (C,1) summability together guarantee the convergence.
Moreover, if we substitute x = e-b in G. H. Hardy, Divergent Series, Clarendon Press, Oxford, 1949, §7.13, then Theorem 114 can be stated as follows. Given c > 1 constant and a sequence of natural numbers, such that An+1 > cAn (c > 1), En_=1 anrAn convergent in the unit circle and its limit exists at r -+ 1 + 0, then E°° 1 an is convergent. So this theorem, conjectured by Littlewood and proved by Hardy and Littlewood, states that the Abel summability and the convergence are equivalent notions in the case of sequences satisfying the Hadamard gap condition An+1 > cAn (c > 1). Our problem states that the Hadamard gap condition cannot be substituted with a much weaker condition. The participants gave two kinds of generalization for the problem. Some of them substitute the Abel summability with the stronger (C,1) summability; others showed that for any sequence An satisfying lim inf (An+1 - An) /An = 0,
there exists a sequence EO° 1 an that satisfies the conditions of the problem. This latter statement is also a generalization of the theorem, since if An = e\/n-, then
An+1-An=e n+1_e`f =(e n+1 " -1)e`r =
1) = (1 + (o(1)) 2--,
and therefore,
An+1 - An
An+1
An An
An
-4 m fnr
,9
-1
n
Solution. We will prove the following statement, which is slightly weaker than the previous two generalizations. Theorem. For any series {Ak} that satisfies lim inf
Ak+1k Ak
= 0,
there exists a divergent series En00=1 an that is (C,1) summable and an = 0 only if n is not in {Ak}.
3.9 SEQUENCES AND SERIES Proof.
485
Let 1 < m1 < m2 < m3 < ... be a subsequence of A1, ... , Ak, .. .
such that
2k-1
k1: mn<m2k
m2k - m2k_1 <
and
k
m2k
n=1
1
Choose a,,,, to be (-1)k+1, the others to be zero. Since E; 1 aj is 0 infinitely many times and 1 infinitely many times, En 1 aj is divergent. Denote En 1 aj by sn. Since
0
< 1,
to is monotone decreasing in [m2k, m2k+1) and monotone increasing in [m2k_1, m2k). Therefore, to prove t,,,. --+ 0, it is enough to show that tm2k -p 0:
tm2k =
(m2 - m1) + (m4 - m3) + ... + (m2k - m2k-1)
< 1 +1 k
m2k
k
- 0.
Hence, tm --+ 0, that is, E, 1 aj is (C,1) summable.
Problem S.11. Let 0 < ak < 1 for k = 1, 2,.... Give a necessary and sufficient condition for the existence, for every 0 < x < 1, of a permutation pry of the positive integers such that 00
air. (k) ,)k
k=1
Solution. Obviously, the following condition is sufficient: inf ak = 0, k
sup ak = 1. k
We are going to prove that this condition is also necessary. Let x E (0, 1) bk and be fixed. For a permutation 7r, let dk = 2k Lx -
(bi
+
b2
+... + bk
(do = x).
If we choose the permutation 7r such that 0 < dk < 1 holds for all k, then obviously
cc
x =
bk
2k.
k=1
Let us suppose that 7r(1), 7r(2), ... , ir(k) are already defined such that 0 < do, d1i ... , dk < 1. We want to define -7r(k + 1) such that 0 < dk+1 < 1. This is equivalent to 2dk - 1 < bk+1 < 2dk,
(1)
486
3. SOLUTIONS TO THE PROBLEMS
and also equivalent to 2(1 - dk) - 1 < 1 - bk+1 < 2(1 - dk).
(2)
We have to take care that ak will be used exactly once in the construction. Let a, be the element with the smallest index such that at ¢ {b1, b2i ... ,
bk}. If 2dk - 1 < at < 2dk, then we can choose bk+1 = a,. If at > 2dk, then let n be the smallest natural number such that 2n+ldk > a,; then 2n+ldk < 2a1 < 1 + a,, so 2n+1dk
- 1 < at < 2n+ldk.
Since infk ak = 0, bk+l, bk+2, ... , bk+n can be chosen sufficiently small, such that by d3+l = 2dk - bj+1i dj+n will be close enough to 2ndk. So we can achieve 2dk+n - 1 < at < 2dk+n.
Thus, we can choose bk+n+l = at.
If a, < 2dk - 1, or equivalently 1 - a, > 2(1 - dk), then we repeat the above process by substituting 1 - a,,1 - aj,1 - d3 in the role of a,, aj, dj, respectively, and by using (2) instead of (1); and in the meantime, we use the equality 1 - dj+l = 2(1 - d3) - (1 - b3+1). By supk ak = 1, we can choose the numbers 1 - bj to be sufficiently small, that is, b3 is sufficiently close to 1. Repeating the process above, we obtain the required permutation.
Problem S.12. Let Al < A2 < ... be a positive sequence and let K be a constant such that n-1
ak < Kay
(n = 1, 2, ... ).
(1)
k=1
Prove that there exists a constant K' such that n-1
E Ak < K'An
(n = 1,2,...).
(2)
k=1
Solution 1. We can suppose that K is an integer. We are going to prove by induction that (2) holds if K' = 8K. The theorem is straightforward if n > 8K, since K > 1. Let n > 8K. We observe that by the monotonicity of {An} and by (1), n-1 4K A2 2
A2
Ak < KAz
4K C k=1
3.9 SEQUENCES AND SERIES
487
SO )'n-4K > )n/2. By this, by the induction, and by the monotonicity of {An},
n-1
E
n-4K-1
Ale
=
n-1 Ak +
k=1
Ak
k=n-4K
k=1
n-1
< 8KAn-4K +
Ak < 4KAn + 4KAn = 8KAn. k=n-4K
Thus, the solution is complete.
Solution 2. The monotonicity of {an} is superfluous. We are going to prove the following. If µi > 0, then the necessary and sufficient condition of
n-1
Epj
(*)
t=1
is that there exist c > 0 real and r natural numbers such that, for each n, µn+l > cµn µn+r > 2µn.
and
(a) (b)
This implies the statement of the theorem, since if An+1 > cAn and 2 2 An+r > 2an, then An+1 > y C)tn and''n+2r > 4an, SO )n+2r > 2An. > An+r (a) is necessary: by (*) and by µi > 0, trivially µn-1 < Kµn, so (a) holds if c = 1/K.
(b) is necessary: by (*) again, 1
/in+1 >
1
Kµn, µn+2 >
Kµn,
... , µn+r-1 >
Adding them together, we get
/in+1 + ... + /in+r-1 >
K
1
µn+r-1) >
Thus, (b) holds if r > 2K2 + 1. (a) and (b) are sufficient: define µo = µ-1 = n-1
r
r
00
i=1 j=0 r
i=1 j=0 r
i=1
K2
= 0.
µ2-i 7
<21:µn-i<21: ca =Kµn i=1
r-1 µn
o0
Pi = E E lin-i-jr < 1 1 i=1
Kµn
r-1 Ani
so by (*),
/in+r > K (/Un+l + ... +
1
r K=21:c-t
i=1
In the last step, we used that, by (a), A,1,+, > c1µn,, holds. This finishes the solution. Moreover, a similar argument shows that
the theorem remains true if the exponent 2 is replaced by any exponent
a>0.
3. SOLUTIONS TO THE PROBLEMS
488
Problem S.13. Given a positive, monotone function F(x) on (0, oo) such that F(x)/x is monotone nondecreasing and F(x)/xl+d is monotone nonincreasing for some positive d, let An > 0 and an > 0, n > 1. Prove
that if 00
n
/
i AnF I an L J k 1< 00,
`
n=1 or
k=1
An
(1)
I
In
00
00AnF >ak k A < 00, n=1
(2)
An
k=1
then E°° 1 an is convergent.
Solution 1. If x > 1, then F(x)lx > F(1). If x < 1, then F(x)lxl+d > F(1). We can suppose that F(1) = 1.
F(x) > x if x > 1,
(3)
F(x) > xl+d if x < 1.
(4)
and
First, we prove the statement of the problem when (1) holds. Denote by >n the sum over those numbers n that satisfy n k
an
>1.
k=1 An
Denote by Ell the sum over the other numbers n. Then, by (1) and (3), n
n
00 > > 'AnF ant Ak/An n
E'an E Ak > >'anAi.
n
k=1
n
k=1
Since Al > 0,
Ean <00.
(5)
n
By (1) and (4), n
n
00 > E"AnF an E Ak/An > n
"an+dAnd n
k=1
Divide the sum En into two parts:
E
Ell =
(1) +
E"(2) ,
Ak k=1
(6)
3.9 SEQUENCES AND SERIES
489
where the first sum runs over those numbers n that satisfy (1+d)/d
n
an < An/ E Ak k=1
and the second runs over the other numbers n. Then by one of the theorems of Abel and Dini, and by (1 + d)/d > 1, n
00
(1)an < n=1
n
t Ak
An/
< 00.
(7)
k=1
By (6), 00 >
1+d
n
(2)ananAnd
E Ak n
k=1
In
> (2) an
n
(I+d)/d
nd(E Ak)I+d =
Ak
An/
n
d
(2)an.
n
k=1
k=1
Comparing this with (5) and (7), we conclude that En=1 an is convergent.
Now, we prove the statement of the problem when (2) holds. If an = 0 for all but finitely many n, then the theorem is trivial; otherwise, we can delete those An, an pairs for which an = 0. So we can suppose that an > 0-
Let X = In: an > 1}- If n.X, then n
akAk
> anAn > 1,
k=1 An
An
and so by (2) and (3), 00 >
An
nEX
n akAk k=1
n = > >akAk >
An
nEX k=1
an,, An,,,
nEX
where no is the smallest element of X. So X is a finite set and the sequence
an is bounded. Let An = anan. Then (2) can be written in the following way: 00
An F n=1 an
ran
\
1
n
Ak) < oo. An
By the boundedness of the sequence an, (1) is satisfied with the substitution
An = An and, as we proved before, from this it follows that En=1 an is convergent.
3. SOLUTIONS TO THE PROBLEMS
490
Solution 2. We prove the statement of the problem when (2) holds, but we will not use (1). Solution 2 is the same as Solution 1 through formula (4). Denote by >' the sum over those numbers n that satisfy
E
akAk X
k=1
> 1.
(5)
n
By (2) and (3), we obtain n
n
00 > E 'AnF n
?
akAk/An
/
n
k=1
> E /anoAno = anoAno
> akAk k=1
L: '1.
n<no
n<no
By choosing no such that ano # 0, it follows that (5) holds only for finitely many n. So, by (2) and (4), 00
1+d
n
00 > E n=1
oo
An = k=1
n=1
n
n
(n: d
1+d
akAk
(6)
k=1
Hence, assuming that not all of the an's are zero, it follows that 00
An d<00, n=1 and therefore
.1n->00.
(7)
Let nq be the largest natural number n such that 2q < An < 2q+1; if such an n does not exist, then nq can be chosen arbitrarily. We will see that in the latter case the multiplier of Anq is zero. Continuing (6), A-d
00' n=1
1+d
n
00
>
E akAk
1+d
00 A
q=1
k=1
00
d
ak,\k k:2q
f 00
> E 2-d(q+1)2q(1+d)S1+d =
2-d [ 2gS.1+d9
q=1
4
q=1
where
E
Sq =
ak.
k:2q <.\k <2q+1
So there is a positive K for which
I
00
q=1
2
< K.
3.9 SEQUENCES AND SERIES
491
Hence, for any natural number q, 2gS9+d
< K.
Therefore
Sq < K2-g/(1+d)
which trivially implies 00
E Sq <
00-
q=1
By the definition of Sq and by (7), there is an no such that >9 1 Sq > Therefore, E°° 1 an < 00, which finishes the solution.
E°° no an.
Remarks.
1. Originally, Laszlo Leindler placed part (1) of this problem at the competition committee's disposal. Atilla Mate noticed that the statement also follows from (2).
2. Laszlo Babai states a similar theorem for integrals: Let F(x) be a positive monotone function on (0, oo) such that F(x)/x is monotone nondecreasing and there is a positive d for which F(x)/x1+d is monotone nonincreasing. Furthermore, let A(t) and A(t) be absolutely continuous functions on (0, oo), A(t) monotone nondecreasing, A'(t) > 0 almost everywhere, and limt_,o A(t) = limt-.o A(t) = 0; then any of 00
J
A'(t)F(A (t)A(t))dt
< 00
0
and 00
J0
t
A'(t)F(A t)
J
A'(x) A'(x) dx) dt < 00
0
implies limt+00 A(t) < oc. (By the monotonicity of F(x)/x, the functions after the f signs are measurable.)
Problem S.14. Let f (x) = E00 1 a, /(x + n2), (x > 0), where 1 < oc for some a > 2. Let us assume that for some,3 > 1/a, we have f (x) = 0(e-gy as x - oc. Prove that an is identically 0.
E°°
lanln-a
Solution. Obviously, the series defining f (x) is convergent at each x 54 -n' (n = 1, 2, ... ), so f (x) is meromorphic on the whole plane. Let 00
F(x) = rl (1 + x n=1
,
h(x) = F(x)f (x).
3. SOLUTIONS TO THE PROBLEMS
492
Since a > 2, this product is convergent for all x. Easy calculation shows that F(x) is an entire function, so for any e > 0, 1(x) = O(eIxI1
}e)
The function h(x) is also entire, and there is a sufficiently small a such that, for x > 0, I h(x) I = If (x) I
-
I F(x) I = O(e-x 3)O(e-xlla+E) _ o(1).
(1)
Furthermore, for any x, 00
Jh(x)
E Ianna
E InamnIIn (i+L) n=1
O(elxll/a}e).
F(IxI) =
So if M(r) = max IF(x)I, then log M(r)
r)
_
r1/«+E
O(
Nrr
)
= O(1)
(2)
So by a Phragmen-Lendelof type of theorem (Gy. Polya and G. Szeg6,
Problems and Theorems in Analysis, Springer, Berlin, 1976, vol. I, 111.6.332), (1) and (2) can hold for an entire function h(x) only if h(x) is constant. So by (1), h(x) = 0. Hence f (x) = 0. Therefore, if for some n, an # 0, then f (z) has a pole at z = -ne. But this is a contradiction, so a1=a2=...=0.
Problem S.15. Given a positive integer m and 0 < 6 < 7r, construct a trigonometric polynomial f (x) = ao+>m 1(an cos nx+bn sin nx) of degree m such that f (0) = 1, f6
Solution 1. Let 19 = max{6/2,1/m}. Extend the following function periodically to all real numbers: O(x)
={ 1-11, 0,
if IxI <19,
if 19<1xI<'Ir.
This nearly satisfies our purpose, it just is not a trigonometric polynomial. Let us examine the Fejer kernel of its Fourier series. Let
r f (x) = co
/
f
cp(x - t)Km(t) dt,
3.9 SEQUENCES AND SERIES
493
where sin2 m'2 1 t
Km (t) _
2(m + 1) sin2
and co is chosen such that f (0) = 1. Obviously,
f
1/2m
,v/2
1= f(0) > cot f Km(t) dt >
-co
Km(t) dt
2 -1/2m
-,9/2
trivially if tI < 1/2m. Then
IKm(t)I>lOm. Hence,
1=f(0)>
20
co<20. ,
Since cp is absolutely continuous,
f
ir
f'(x) = co So
V(x - t)Km(t) dt.
f Km(t) < f a
x+19
If'(x)I < co
W
19
Km(t) dt <
80 19
-7r
X-19
Estimate f6<,x,
f
7r x+6
7r
If(x)Idx2
f If(x)Idx<2 f fKm(t)dtdx 6 x-6
6
f
ar
Km(t) dt < 419
f Km(t) dt = 4m a
If 19 = 6/2, then
f
or++9
If (x) I dx < 419
f
Km (t) dt = 26
6-19
7r
f+
2 Km (t) dt.
6/2
Using elementary estimates on the interval 0 < t < it + 6/2 (< (3/2)ir), Km (t) < r30
3. SOLUTIONS TO THE PROBLEMS
494
and so
7r+s/2 26
00
Mt2
J
s/2
dt=240.
s/2
This finishes the solution.
Solution 2. Start from the function h(x)
x(02 - x2)2, if x < 0, if 0 < I
{ 0,
< 7r,
where the parameter 0 will be defined later. Let Cneinx
h(x)
n=-oo and
s(x)
Cneinx
n=-l We will show that i can be chosen in such way that f(x) = s2(x)/s2(0) satisfies the conditions of the problem. Obviously, f (x) is a trigonometric polynomial of degree not more than m, and f (0) = 1. We will estimate the difference of s(x) and h(x). The Cauchy-Schwartz inequality yields
s(x)-h(x)I
E Icnnl <
+oo
1
In <
InI>1
(cnn)2
EI cnnl2
n=-oo
N n1>l
nI<1
2
By Parseval's formula, + oo
(cnn)2
27r
J
I h'(x) I2 dx < 1
Is(x) - h(x) I <
Is'(x) - h'(x) I <
l0 103.
(1)
Fa2 ls(
x)l
2
3.9 SEQUENCES AND SERIES
495
Moreover,
Vj29j-3
zi + 2K = -A,
so
60
-o
If'(x)Is2(o)Is(x)Ils'(x)I < suffices if
0 > b.
(2)
This inequality and (1) also hold if we choose i = max{b, 9/l}. The integral of If (x) I can be estimated in two ways. If 0 = b, then
f If(x)I dx= s2(o) J
Is2(x)Idx=
6
6
6
f (s(x)-h(x))2dx
82(o)
r
s2o) f(s(x) - h(x))2 dx =
s210)
i ICnI2
-r 21r
l2
Inl>l
22
°O
21r
1
Ls n cn
If i = 9/l ± b, then
f
I f (x) I dx <
6
f
-r
l S2
(o) dx = 2(
Ian l2
n=-l
00
<21r EIcnI2=4 f h2(x)dx<8L = l2. r -O° This finishes the solution.
Problem S.16. Prove that if M
1: an
holds for a sequence {an} of nonnegative real numbers with some positive integer N, then ai+r > pai f o r i, p = 1, 2, ... , where iN
ai =
E n=(i-1)N+1
an
(i = 1, 2, ... ).
3. SOLUTIONS TO THE PROBLEMS
496
Solution 1. Since ai+p = a(i+p-1)N+1 + ... + a(i+p)N 1
(i+p-1)N
N
n=1
n=1
> NN E an=
1
(i+p-1)N+1
N
n=1
(i+p)N
a,+...+
an n=1
2N
an+ E an+...+ n=N+1
(i+p-1)N
E
+
an=al+a2+
+ai+p-1,
n=(i+p-2)N+1
it follows that ai+p
al + a2 +
+ ai+p-1.
Hence ai+l > ai, and therefore ai+p > ai + ai+1 + ... + ai+p-1 > pai
Solution 2. Similar to the previous solution, i-1
ai
ak
(1)
2p-1E ak
(2)
k=1=1
We are going to prove that ai+p
k=1
for any i and p. Indeed, (2) is the same as (1) for p = 1, and if (2) holds for p, then by i+1
2p-1 I` ak = 2p-1 (c.i + ai+p+l >_
i k=1
k=1
i ak
> 2p-12
ak k=1
it also holds for p+1. Since ai > 0, from (2) it follows that ai+p > 2p-tai, which finishes the solution.
Remark. We can similarly prove the following, more general statement. Let {an} be a sequence of nonnegative real numbers such that for some c, n-1
an>c>ai. i=1
Define ai as we did in the problem. What is the largest constant dp such that for all such sequences the following holds? ai+p ? dpai
3.9 SEQUENCES AND SERIES
497
In the problem, c = 1/(N - 1). First, we will find a cp such that n
ai
an+p ? ep i=1
e1 = c suffices for p = 1. Since n-1
n
ai > ep(1 + c)
an+p+l ? ep i=1
ai, i=1
it is not a bad idea to choose ep+l = ep(1 + c), that is, ep = c(1 +
c)P-1
Then (i+p)N
iN
N
ai+P =
N
e(P-1)N+j
an > E e(P-1)N+j > an >j=1
n=(i+p-1)N+1
j=1
n=1
ai.
So we can choose dp such that N
dp =
C)(P-1)N+j-1 = (1
C(1 +
+ C)(P-1)N[(l + C)N
j=1
It is easy to see that if we define al = 1, an = c(l+c)n-2 (n > 2), then for any N and p, ap+l = dpal holds, so the inequality cannot be sharpened. As stated, in the problem c = 1/(N - 1), so
N
(P-1)N
dp- (N-1)
N
N
(N-1) -1 >eP-1(e-1),
but even for c = 1/N it is true that dp >
2P-1.
Problem S.17. Let S. = E; 1 bjz,' (v = 0, ±1, ±2.... ), where the bj are arbitrary and the zj are nonzero complex numbers. Prove that
ISoI < n max ISv I. o
Solution. Let 11j 1(z - zj) = Ek=oakzk and maxo>k>n Iakl = Iaml. Obviously, la7z.I > 1. Then n
n
akSk-m
n
= j akbj zj -m =
k=0
n
bjzj
m
akz3 = 0. k=0
k=0
Hence n
ak
n
(-a )Sk-m < E I Sk-m I< n max I Sv I. 0
Remark. One can show that equality holds if bl = b2 = the set of numbers zk is the same as {a e(2vri)/((n+l)j) (j where a is an arbitrary constant of absolute value 1.
= bn and n)},
3. SOLUTIONS TO THE PROBLEMS
498
Problem S.18. Let p > 1 be a real number and R+ = (0, oc). For which continuous functions g : ][8+ -+ R+ are the following functions all convex?
Mn(x)
_ E 19(
)x+1
P
E 19( i
X = (xl, ... , xn+l) E
1[8++1,
n - 1, 2, ...
Solution. We prove that Mn (n = 2, 3, ...) are all convex if g is a constant function. If g is a constant function, then by Minkovsky's inequality, Mme,(x+y) <- Mme,(x)+M. (y)
(x,y E R'+1),
(1)
and by Mn(ax) = aMM,,(x) (a E R+,x E R++1), it follows that Mn is convex. Conversely, if all of the MM,, are convex, then we will show the following.
If M2 (x, u, 1) is a convex function of x for all u E (1- 6,1 + 6), b > 0, then the function g is a constant. Taking each side of the inequality
M2(axl + (1 - a)x2i u,1) < aM2(xl, u,1) + (1 - a)M2(x2i u,1) (X1, x2 E R+, U E (1 - 6,1 + 6), a E (0,1)) to the power p and using the inequality
at + (1 - a)s < (atP + (1 - a)sP)1IP (t,s > 0, a E (0,1)) between the arithmetic mean and the n-th power mean, we obtain
M2(axl+(1-a)x2,u,1)
< a 9(u)(1 - up)
9(!) + 9(u)
+ 1 - a g(u)(1 - up) )
9(u) + 9(u)
,
for x1i x2 E R+, u E (1 - b, 1 + b), a E (0, 1). If u < 1, then divide by g(u) (1- uP) and take the limit u -* 1- 0. Since g is continuous, we get 1
9(ax1 + (1 - a)x2) + g(1) < a9(x1) + g(l) + (1 -
a)
9(x2) + 9(1)
Using the same method for u < 1, we get the opposite inequality. There-
fore, equality holds. So the function f = 1/(g(x) + g(1)) satisfies the following Jensen's equality:
f (axl + (1 - a)x2) = of (xl) + (1 - a)f (x2)
(X1, X2 E R+, a e (0, 1)).
3.9 SEQUENCES AND SERIES
499
Since f is continuous and positive, f (x) = Cx + D, where C and D are nonnegative constants. So Cx + D = g(x) + g(1) < g(1)
2 f (1) = 2(C + D).
This does not hold for large values of x when C > 0. Therefore, C = 0 and g is constant.
Remark. Some of the participants remarked that we do not have to suppose the continuity of g, since it follows from the convexity of Mn. Moreover, g is continuous if M2 (x, u, 1) is a convex function of x for u E
Problem S.19. Suppose that R(z) = -.an Zn converges in a neighborhood of the unit circle {z : IzI = 1} in the complex plane, and R(z) = P(z)/Q(z) is a rational function in this neighborhood, where P and Q are polynomials of degree at most k. Prove that there is a constant c independent of k such that 00
lanl < ck2 max IR(z)j. n=-oo
Solution. Denote by D the unit disc of the complex plane D = {z
:
IzI < 1}, and by 8D its boundary. We can suppose that IR(z)I < 1 on the perimeter of the unit circle. So we have to prove that EO° _. IanI < ck2. If we prove
lanl < ck2, n=- inf
then we can apply it to R(1/z)/z, so that we obtain E°° o Ianl < ck2, which finishes the solution. We know that an
21
y r
-(Z) dz,
(1)
where we can integrate along 8D, or along such a curve t, such that R(z) does not have a pole in the region bounded by 01) and r. Now we will try to choose skillfully r c D. By the previous results,
1:
n=-oo
-L lanI -<27r =
1
27r
i r
IR(z)I(1 + zI + ... + IzI2 + ...) Idzl
J IR(z)I 1 IdzlI r
3. SOLUTIONS TO THE PROBLEMS
500
We are looking for a P for which the integrand is not too big in the rightmost integral. Because the multiplier is 1/(1 - jzl), we will have to bring P as far from the unit circle as possible. The only problem is that we do not know anything about the behavior of R(z) far away from 8D. Denote the poles of R in the unit disc by q1, q2, ... , qh, where the points are included in the sequence with appropriate multiplicity. Obviously, h < k. The Blaschke product h
B(z)
z - qv
-Vil-qvz
is holomorphic in D, vanishes at the points qv, and has absolute value not more than 1 in the points of the unit circle. So S(z) = R(z)Q(z) is holomorphic in D and has absolute value not more than 1 in the boundary of the unit circle. Thus, by the maximum modulus principle, IS(z)l < 1, for z E D, and so IR(z)I <
(IzI < 1).
7h lz-g
(2)
l 1v=1
This is the approximation we were seeking. Let
t=a{zED
:
z
qv
>1
-_L
,
1
IfzEr,then by(2), JR(z)I
(i+)k<e. (1- k+1 )k =
(3)
The curve defined above is not necessarily a Jordan curve, but it consists of finitely many closed Jordan curves. Nonetheless, R(z) will be holomorphic in the region bordered by r and 8D. So all of the consequences of (1) will hold. Obviously, t consists of some subcurves of the curves
z - qv
1 -qvz
If qv = 0, then r is a circular arc. But it is also a circular curve when qv # 0, since IF, is obtained from the circular arc jwI = k/(k + 1) by the transformation z - qv
1-qvz
,
(4)
and such a transformation maps a circle onto a circle. (Of course, qv is not the centerpoint of r.. Moreover, if we are considering Q as the
3.9 SEQUENCES AND SERIES
501
Figure S.1.
Poincare model of the hyperbolic geometry, then the transformation (4) is a conformal transformation of the hyperbolic space, so q will be the nonEuclidean center of I',,. The non-Euclidean radius of r,, is approximately log k, see Figure S.1.) Since F C U"r,,, - by (3), 1
e
lanl
j
IdzI
<1 - IzI e
h
27r L'-lrV
IdzI
1 - Izl
.
IF
The latter sum can be approximated by substituting (4) and some calculations as
r 1-Il
I
j *1 `2J 1
Id
=2 k
1z 12
1
2
12 < 1-(kk
)227r<87rk.
cr4
(Here we had to do "some calculations" to prove the equality above. The equality also shows that curve IdzI/(1 - Iz12) is invariant under the conformal transformation of the hyperbolic plane.) So IanI < 4ek2,
n=-co
and as we noted above it follows that 00
IanI < 8ek2.
n=-oc
Remark. The inequality above is not sharp. Gabor Somorjai showed with a better curve IF that 00
l an I < ck log k.
n=-oc
3. SOLUTIONS TO THE PROBLEMS
502
Problem S.20. Let Kn (n = 1, 2, ...) be periodical continuous functions of period 27r, and write
fa 2
kn(f; x) =
f (t)Kn(x - t)dt.
0
Prove that the following statements are equivalent: (i) fo I kn (f ; x) - f (x) I dx -+ 0 (n -+ oo) for all f E L1 [0, 27r]. (ii) kn (f; 0) - f (0) for all continuous, 27r-periodic functions f .
Solution 1. (i)- (ii): Denote by C[0; 2ir] the space of the continuous functions under
the supremum norm. Let us examine the sequence of operators L1 ---4 L1:
Ilkn(';')II =
sup
Ilkn(f;')IILI
(1)
IIfIILI<<1 27r
<
<
sup
fIL,<0 11
f
2n
27r
0
IKn(t)I dt.
0
On the other hand, the functions if x E [0; 6],
6,
f 6 (x) - { 0,
if x E (S; 27r),
by the continuity of Kn(t), satisfy 27r
IIkn(f5;')IIL1 - f IK.(t)I dt
(a
0),
0
and so 27r
Ilkn('i')II =
fIKntIdt. 0
By applying the Banach-Steinhaus theorem to the sequence 2ir
we obtain f IKn(t)I dt < C, where C is the same constant for all n. 0
If f is continuous, then 27r
kn(f; x) - kn(f; x + h)I = J(f(t)_f(t_h))K(x_t)dt 0
< Csup t I f(t) - f(t - h)I, and so the functions kn (f; x) are equicontinuous.
3.9 SEQUENCES AND SERIES
503
Let us suppose that (ii) does not hold. Then there is a continuous f, an e > 0, and a sequence {vn} such that k,,,, (f; 0) > f (0) + e for n = 1, 2,... . By the uniform continuity of {k n Y; x) II there is a 6 such that, for all n,
k,,, (f; x) > P X) + 2 if O <x<6 (n= 1, 2, ... ). Hence,
f b
21r
JIkvn (f; x) - f (x) I dx > 0
(k-n (f; x) - f (x)) dx > 6
(n = 1, 2.... ),
0
but this contradicts (i). (ii)-*(i): It is well known that the norm of the linear functional kn(.; 0) C[0; 27r] --+ R is fo " I Kn (t) I dt. So by the Banach-Steinhaus theorem, ff " I Kn (t) I dt < C (n = 1, 2, ... ), where C is a constant. Since 2
r f(t)kn(x0
kn(f, xo) = f 0 = f27r f(xo
- t) dt
+ t)Kn (-t) dt = kn (f (x0 + t); 0) 0
is continuous for all f , kn (f; x) -* f (x) holds everywhere. But kn Y; x) C sup(f ), so by Lebesgue's theorem, 2,r
I:;
Ikn(f, x) - f (x) I dx - 0 (n - > oo, f E C[0, 27r]).
Let f E Ll [0, 27r] and let e > 0 be arbitrary. Choose a continuous function g in such a way that 119 - f IILI < e holds. Then, by (1),
f
27r
I kn (f, x) - f (x) I dx < I2 I kn (f, x) - kn (9; x) I dx + f27r I kn (9; x) - 9(x) I dx,
f
27r
I9(x) - f (x)I dx
+
Ikn(9;x)-9(x)Idx+II9-f1ILI.
By the previous argument, the middle term above goes to 0 if n -* oo, so
f
27r
I kn(f, x) - f (x) I dx
becomes arbitrarily small for sufficiently large n. This finishes the solution.
Solution 2. We are going to use the following version of the BanachSteinhaus theorem. Let X and Y be Banach spaces, and let A, An : X -* Y
3. SOLUTIONS TO THE PROBLEMS
504
(n = 1, 2, ...) be linear bounded operators. The sequence {An} pointwise converges to A if there is a constant C and a closed system S C X such
that IIAn,II<_C
(n=1,2,...),
(1)
Anx-4Ax (n--*oo)
(2)
hold for all x E S. (S C X is called a closed system if the linear hull of S is dense in X.)
Let An : L, --4L, be the operator k.,,(.; ), A' be an identity on L1, R be the functional 0), and A2 : C[0, 27r] -+ R be
An : C[0, 27r]
the mapping f -+ f (0). As we showed in the previous solution, IIA 11 = lIAnII = fo" IKn(t)I dt, so (1) holds for {An} and {An} at the same time. Let S = {em = etmx},°,°=1. Since S is closed in L, and in C[0, 27r], and 2a
(A'em)(x) _
L
= eimx
etmtKn(x
2a
- t) dt = 1 eim(x+t)Kn(-t) dt 0
2"
J
etmtKn(-t) dt = em(x)A2em,
0
it follows that f2n
`41emIILi =
em(x)I dx
J0
2a
= IA22em -A 2 eml j
=
27rIIA2em
lem(x)I dx
0
- A2emIIR,
so (2) holds for {An} and A'. Furthermore, (2) also holds for {An} and A2 simultaneously. So we proved the equivalence of the pointwise convergences
An , Al and An , A2. 00
E]
Problem S.21. Let us assume that the series of holomorphic functions
E fk(z) is absolutely convergent for all z E C. Let H C C be the set of k=1
those points where the above sum function is not regular. Prove that H is nowhere dense but not necessarily countable.
Solution. Let gn(z) = En 1 fk(z) and g(z) = i' 1 fk(z). Then gn(z) is holomorphic and gn(z) - g(z) for all z E C. To prove the first part of the statement of the problem, we have to show that inside each circle K of C, there is a circle k in which g(z) is holomorphic. Let cp(z) = supra 9n(z)1This exists by the convergence and is finite for all z E C. Let Ak = {z E KI k - 1 < W(z) < k} (k = 1, 2.... ). By Baire's theorem, there exists an N such that AN is dense inside some circle k C K. If cp(z0) > N for
3.9 SEQUENCES AND SERIES
505
some z0 E k, then Ign(zo) I > N for some n, and by the continuity of gn(z) there is a neighborhood of z0 where also ep(z) > N, which contradicts the
fact that AN is dense in k. So cp(z) < N if z E k, and so Ign(z)I < N if z E k. By the Vitali-Montel theorem, there is a subsequence of {gn(z)} that is equiconvergent inside k. Since gn(z) is convergent, this subsequence converges to g(z). Hence, by Weierstrass's theorem, g(z) is holomorphic inside k. So we proved that H is nowhere dense. Remarks.
1. In the solution, we did not use the fact that the series of functions
Ek
fk(z) is absolutely convergent. To prove the second part of the 1 statement of the problem, we will need the following result. Mergelyan's theorem. Let K be a compact set of C, such that its complement is connected, and let f be is a continuous complex function
on K, such that it is holomorphic inside K. Then for any e > 0 there exists a polynomial P, for which If (z) - P(z) I < E for all z E K. (See W. Rudin, Real and Complex Analysis, McCraw-Hill, London, 1970.)
Let A = {zjIm z < 0}, B = {zIIm z > 0}, C = {z: z is real}. There are series of compact sets {An}, {Bn}, and {Cn} (n = 1, 2, ... ), such that An C An+1, Bn 9 Bn+1, Cn C Cn+1, C\ (An U Bn U Cn) is connected for n = 1, 2, .. . and cc
UAn=A, UBn=B, 1
UCn=C.
1
1
So An U Bn U Cn is compact, its complement is connected, and (1, if z c: An U Bn,
f (z) -
0,
ifzECn.
Hence, by Mergelyan's theorem, there is a polynomial PE,n for any e > 0 and n, such that 11 - PE,n I < e, if z E An U Bn, and I PE,n (z) I < e, if z E C n (n = 1, 2, ... ). Let Qk(z)-Qk-1(z),..., f1(z) = Q1(z), f2(z) = Q2(z)-Q1(z),...,fk(z) = where Qn(z) = P(1/2n).n(z) (n = 1, 2, ...) is a series of polynomials. Then :%=1 fk(Z) = Qn(z), so 00 1 1fz is not real, f k (z) 0, if z is real. k=1
Here H is the set of real numbers, and is hence uncountable. On the other hand, for any z e C there exists an N such that z e An U Bn U Cn if n > N, so in this point z, cc
00
k=1
00
1
I QN+k(z) - QN+k-1(z)I C E 22N+k1 <00;
I fN+k(z)I = k=1
hence, E' 1 fk (z) is convergent at z.
k=1
3. SOLUTIONS TO THE PROBLEMS
506
2. We can prove a stronger statement: the measure of H can be positive. Let R be the real axis, I the imaginary axis. It is enough to show that if S is a nowhere-dense subset of I, then there is a sequence if (n)}n l of holomorphic functions that satisfies the conditions of the problem and R x S contained in the H set obtained from this sequence If (n)}°°_1. So let S be a nowhere-dense, open set of I. Then I\S- is a dense set of I, so it is a union of countably many disjoint intervals; let these be ik = (ak, bk) (k = 1, 2, ... ). (We can suppose that ak and bk are finite.) k)
be monotone decreasing, {rl(k) } ° 1 monotone increasing, 1 such that rikl < I(k) and rikl -+ ak, sikl -+ bk if l -+ oc. Let {an}n 1 be a sequence of natural numbers, such that it contains each natural number infinitely many times. Let n be an arbitrary number. Let an = k, denote by l the cardinality of {m : m < n, a,,,, = k}, and let Let {r( I
Kn = {zlr2,- 1 < Imz < s2,- 1, IRezJ < n}, An = {zllmz = r2l), IRezI < n}, Bn = {zlImz = 821 , IRezI < n},
Ln = {zlr2i _ 1 > Imz > min(ak - n, -n), IRezI < n},
Mn={zl821-1
is connected in C-. So by Runge's theorem, there is a polynomial fn such that Ifn(z)I <
Zn
if
ZEKnULnUMn
and
I fn(z) - nI < 2
if
z E An U Bn.
So
(i) The functions fn are polynomials, and hence holomorphic on C.
(ii) For any z E C, there exists an No such that z E Kn U Ln U Mn for any n > No. So I fn(z)I < 1/2n for any n > no, and En'=1 Ifn(z)I < 00.
(iii) Let zo E R x S arbitrary. Let e > O and U E = { z l I z - zo I < e}. There is an index ko such that UE/2 fl (R x iko) # 0; hence, UEl{z: Imz=ako}'A 0 or U5 fl{z: Imz=bko}#0. Let us suppose that UE fl {z : Imz = ako } # 0 and z1 E UE fl {z : Imz = ako}. (The other case can be discussed similarly.) Let no > max(IRezoI +
e, IIm zo I + e) such that if n > no and an = ko, then Kn fl UE # 0. Let Eno 9no = n= 1 9n Since 9no is continuous at z1, there is a 0 < S < e such that for all z E V6 = {z: Iz-z1I < S}, Igno(z)-gn1I < 1. Let ni > no such that and = ko, n1 > 5 + 2I9no (z1) I and Kn, fl V6 54 0. Let Al E and A2 E aKnl fl U6, where aKnl denotes the boundary of Kn1.
An,
fl U6
3.9 SEQUENCES AND SERIES
If n > no and n
507
n1, then Ifn(Ai)I < 1/2n and I fnl(Al) -nil < 1/anl.
So
If(al) - 9n(zl) -nil < I9no(A1) - 9no(zl)I + Ifni(A1) - n1I 00
+
fn(A1) < 1 +
n>no,n#nl
n=no+1
7
< 2.
On the other hand, I fn (\2) I < 1/2n if n > no, so 00 Ifn(A2)I < 2.
If (A2) - 9no(z1)I < I9no(t2) - 9n(zl)I +
n=no+1
Hence,
If(A,)I >- I9no(zl) +n1I - If(A,) - 9no(zl) - n1I >- ni - I9no(zl)I - 2
and If(A2)1 < If(A2) - 9no(zl)I + I9no(zl)I < 2 + I9no(zl)I
So
If(al)I - If(A2)I >- nl - 4 - 2gno(zl) > 1. Therefore, the total variation of If I is more than 1 in U. Since b > 0 is arbitrary, If I is not continuous at zo. So If I is not continuous in the points of R x S. That is, { fn}0 1 satisfies the conditions of the problem, and the set H, obtained from If (n)}°n°__1, contains R x S. By this, the statement of the remark is proved.
Problem S.22. Let f (x) be a nonnegative, integrable function on (0, 27r) whose Fourier series is f (x) = ao + E' 1 ak cos(nkx), where none of the positive integers nk divides another. Prove that IakI ao.
Solution. Let Kn (x) = 1/2 + (1 - +1) cos x +... be the nth Fejer kernel. It is known that Kn (x)
0. By integrating, we get 0 < fo " f (x)Kn (nkx) =
7r(ao + (1 - 1/(n+ 1))an,,), so if n - oo, then ank < ao. Kn(x - 7r) =
1/2-(1-1/(n+1))cosx+..., so 2,t
0J
f(x)K(nkx-7r)=7r
/ao-(1-_--4-1)ank)
0
Therefore, if n -* oo, then an,< < -ao. 0
3. SOLUTIONS TO THE PROBLEMS
508
Problem S.23. We are given an infinite sequence of 1's and 2's with the following properties: (1) The first element of the sequence is 1. (2) There are no two consecutive 2's or three consecutive 1's. (3) If we replace consecutive 1's by a single 2, leave the single 1's alone, and delete the original 2's, then we recover the original sequence. How many 2's are there among the first n elements of the sequence?
Solution. Since, by the above conditions, the first n - 1 elements of the sequence determine the nth one, there can be only one sequence. It is easy to see that the following sequence of numbers 1 and 2 satisfies the conditions of the problem: in this sequence, the index of the nth 2 is larger
by 2 or 3 than the index of the (n - 1)th 2 if the nth element of the sequence is 1 or 2, respectively. The following condition is equivalent to (3):
(3') If we change all 1's into 12 and all 2's into 112, then we obtain the original sequence.
Denote by f (n) the number of 2's among the first n elements. We are going to prove that, for all n, f(f(n) + 2n) = n,
(a)
f(f(n) + 2n - 1) = n - 1,
(b)
f (f (n) + 2n + 1) = n.
(c)
To prove this, change the first n elements of the sequence as described
in (3'). Hence, we obtain 3 f (n) + 2 (n - f (n)) = f (n) - 2n elements, such that the number of 2's is n, so we proved (a). By (3'), this f (n) - 2n ends with 12, and by (2), the (f (n) - 2n + 1)th element is 1, so we obtain (b) and (c). Obviously,
f(1)=0 and f(2)=1.
(d)
Now we are going to prove that if function g : N -> N satisfies the conditions (a)-(d), then g = f . f (n) = g(n) follows from (d) for n = 1, 2. Let us suppose that we proved f (k) = g(k) for all k < n. Since h(k) = f (k) + 2k, h(k + 1) - h(k) < 3, and so there is a k < n natural number for which n = f (k) + 2k + En, where en = 0, ±1. Then, by (a), (b), or (c) and the induction, f (n) = f (f (k) + 2k + En)
k,
if En = 0, 1,
k-1, if En=-1.
Moreover,
g(n) = g(.f (k) + 2k + En) = g(g(k) + 2k + En) _
Hence, f (n) = g(n).
k,
if En = 0, 1,
k-1, if En=-1.
3.9 SEQUENCES AND SERIES
509
Finally, we are going to show that g(n) = [(/ - 1)n + 1 - 1/vF2] satisfies the conditions (a)-(d), so f (n) = [(v'2- - 1)n + 1 - 1/'] holds. Obviously, g(1) = 0 and g(2) = 1. To verify (a), write g(n) in the form
g(n) = (v-1)n+1-1/f]-e(n), whereE(n) E [0,1]. (However, e(n) = 0 cannot hold, but we do not need this fact.) Hence,
g(g(n) + 2n) = [(/ - 1) (v - 1)n + 1 - 1 - E(n) + 2n) + 1 -
]
= [n+(1-E(n))(v12_ - 1)] =n. We can similarly verify (b) and (c).
Remark. After writing down the first few elements of f (n), it becomes plausible that f (n) is the integer part of some linear function. If we substitute the function f (n) = [an + b] into (a) and (b), we obtain that (a) - 1, b = 1 - 1/v obtained in the and (b) hold only for the pair a = solution.
Problem S.24. Let ao, al, ... be nonnegative real numbers such that
a,=00. n=0 For arbitrary c > 0, let
k
nj (c) =min k : c j <
E ai
,
j = 1, 2, ...
.
i=0
Prove that if E 0 a? < oo, then there exists a c > 0 for which E °1 a,, l,i < oo, and if ,°° 0 ai = oo, then there exists a c > 0 for which E ° 1 a,, W = 00.
Solution. Let
k
Sk =
ai,
i=0
and denote by Xk,j the characteristic function of the interval Ik,j Let [(1/j) - Sk-1, (1/j) . Sk].
00
f : (0, 00) ---* (0, +00] .
f (c) = E ani (C) ; j=1
Since Xk,j(c) = 1 iff nj(c) = k, the function f = >k j akXk,j is Lebesguemeasurable. So 00
]f(c)dc= a
akA(Ik,j U [a, b]) k=oo,j=p
3. SOLUTIONS TO THE PROBLEMS
510
We are going to give a lower and upper estimate for the value of the integral.
Lower estimate. We sum over only those pairs k, j for which Ik,j C [a, b],
so if a < 1 Sk-1 < 1 Sk < b, 3
3
or, equivalently, Sk
<
Sk-1
<
a
b
Note that
ak
A(Ik,i) = .
So b
00
sk-r1/a
00
ak > > ak
J f (c) do > k=O j
J
_
k=0
a
lx
dx
(sk/b)+1 00
00
a2 In Sk-1/a = 0 a2kIn k=0
Sk/b + 1
k
k=0
bSk-1 ask + ab
Upper estimate. We sum over only those pairs k, j for which Ik,i n [a, b]
0,
s
1 Sk-1 < b and a< 3
3
Sk-1 < j < Sk a
b
b
Ja
f (c) dc < 0" ak2
1
<
k=0
'ak /I 1 + 12 + In ask-1bsk+ ab
k=0
\\\
If lim sup an = a > 0, then an > a/2 for infinitely many n. So a2 = oo. Since infinitely many terms of the sum EO_ ani(,/2) are larger than an, f (a/2) = oo. So let us suppose liman = 0. Then sn/sn_1 -> 1. Therefore, lim In asks
lim In
+ ab
bs
as k 1 + ab
In a > 0, b
In a
3.9 SEQUENCES AND SERIES
511
So there exist real numbers A > 0, B > 0, C > 0, and D such that 00
E Cln askbsk-1 ) ak > (a2A+B, + ab k=0
k=0
k=0 \
2
+ln asbsab) ak < k 1+
(a) C+D. (k"=00
k
So if a2 < oo, then fe f (c) dc < oo. Therefore, f (c) < oo almost everywhere, and if ak = oo, then fe f (c) dc = oo. So for any a < b, the set 00
{ j=1 E ani (C) > m} is a dense set for an any m. However, the function an,,(,) is a left-upper-semicontinuous function of c, so the following sum is also left-upper-semicontinuous: 00
1: and (c) j=1
Therefore, the set {any (,) > m} is dense and open. So the set {f = oo} is of second category and therefore is not empty.
Problem 5.25. Let 2/(/+ 1) < p < 1, and let the real sequence {an} have the following property: for every sequence {en} of 0's and ±1's for which En 1 enp' = 0, we also have E°° 1 enan = 0. Prove that there is a number c such that an = epn for all n.
Solution. We start with a definition. Definition. Let L denote the set of all positive and strictly decreasing sequences {ln} for which L = En'- 1 In < oo. We call such a sequence interval filling if for every x E [0, L] there is a sequence En E {0, 1} with
x = E', Enln. We need three lemmas.
Lemma 1. If a sequence {ln} E L satisfies 00
An < E Ai
(n =
2,...
i=n+1
then it is interval filling. Proof. For x E [0, L] we define the numbers En inductively as follows: 1, En
=
1
I 0,
if Ei=1 EiAi + An < x; if En 1 X. i i=1 Ei i >
3. SOLUTIONS TO THE PROBLEMS
512
For every n for which En = 0, n-1
0c
0 <x - > eiAi <x i=1
EiAi < An, i=1
hence if there are infinitely many such n, then x = Ei00 Ei limn_.o li. If, however, there are only finitely many such n's, then for the largest one n-1
x-
00
L:
EiAi < An <
00
Ai =
i=n+1
i=1
from which
EiAi' i=n+1
00
x <
Enln,
n=1
so our claim holds even in this case. Lemma 2. For 2/(,Vr5- + 1) < p < 1, the sequence
p, p2, p3, ... is interval filling, and for any natural number N, the sequence 2
PAP,
P N-2 ,PN-1 ,P N+1 ,PN+2
is interval filling too.
Proof. By Lemma 1, it is enough to verify that 00
pn < E pz
for all
nEN
i=n+1
and
00
Pn-1 <
E pi
for all
ncN
i=n+1
are satisfied. The first of these is equivalent to 1 < p/(1-p) and the second
one to 1 < p2/(1 - p), hence each of these is true if 2/(s+ 1) < p < 1. Lemma 3. Let A and B be nonempty disjoint sets such that their union
is the set of the natural numbers, and let 2/(f + 1) < p < 1. Then there exist nonempty sets A' C A, B' C B with the property
[: pi = [: pi tEA'
(1)
iEB'
Proof. Let L = p/(1 - p) and x = >iEApi Let us choose an N E A for which x < L - p.NIf A is finite, then N can be the largest element in
3.9 SEQUENCES AND SERIES
513
A; otherwise, we can choose any element in A with sufficiently large index. Since Lemma 2 says that the sequence A e
p2
,e
p N-2 p N-1 '
p N+1 , p N+2
is interval filling, there is a set C C N \ {N} for which
Y:pZ = x = > pi. iEC
iEA
Let A' = A \ C, B' = C \ A. Then (1) clearly holds, and because N E A', the set A' is not empty. From this and (1), it follows that B' is not empty, either.
After these preparations, we turn to the proof of the statement in the problem. For every c E R, the sequence a' = an - epn also satisfies the assumptions of the theorem. Suppose, on the contrary, that the claim is not true. Then we can choose a real c for which there is an n with a' > 0 and also another one with a' < 0. Let
A={nEN:a;,>0}, B={nEN:a'n<0}. On applying Lemma 3, we get two sets A' and B' with the properties stated there, and with the aid of these we define if n E A';
11, en = -1,
if n E B';
otherwise.
1 0,
Then
cc
Eeip' = Epl- Epi=0, i=1
but
iEB'
iEA'
00
Eeiai = i=1
ai iEA'
ai > 0, iEB'
and this contradicts the hypothesis of the problem. The contradiction obtained proves the claim.
Remark. The result is true for any 0 < p < 1 (see Z. Daroczi, I. Kdtai, T. Szabd, On Completely Additive Functions Related to Interval-filling Sequences, Arch. Math. 54 (1990), 173-179).
Problem S.26. Let S be the set of real numbers q such that there is exactly one 0-1 sequence {an} satisfying cc
E anq-n = 1. n=1
Prove that the cardinality of S is 2"0.
3. SOLUTIONS TO THE PROBLEMS
514
Solution. Let {bn} be a sequence such that b1 = b2 = 1, b3n = 1, b3n+1 = 0, and b3n+2 = 0 or 1 (n = 1, 2, ... ). There is exactly one number q > 1 such that
E
n=1
= 1.
qn
Such a q exists, since the series EO_1 bnxn is convergent if xj < 1, continuous, it takes the value 0 at x = 0, and goes to infinity if x --+ 1-. We will prove that if there is a sequence {an} with elements 0 or 1 and 00
a"F
Fn n=1 g then an = bn for all n.
Let us suppose that this is false. Let k be the smallest number such that ak # bk. Then
an1: n 1: 00
00
qn
n=k
n=k
qn
Since 00
g5_g2 qn - q+q2+q3n = q4+q3+q2-q bn
1
1
1
n=1
n=1
that is, q5 - q4 - q3 - 2q2 + q + 1 > 0 holds,
1- q3-q2-q-1 q3-1 q3-1 q2
+q
-1-
(q3-q2-q-1)(q3-1)
(q3-1)2
=1-q(q5-q4-q3-2q2+q+1)+1 <1. (q3 - 1)2
So
Suppose that bk = 0 and ak = 1. If n > 1, then at least one of bn, bn+1, and bn+2 is zero, so bn+1
bn
bn+2
1
+
qn + gn+1 + qn+2 - qn
1
qn+1'
Hence CK)
n=k
bn_ [qn
L qn Lm=01\ n=k+1
1
q< qk m=1
and this is a contradiction.
(q3r2+q31)
1
1
qk+3+n+2 J 00
qn
n=k
3.9 SEQUENCES AND SERIES
515
Otherwise, suppose that bk = 1 and ak = 0. Since at least one of bn, bn+1i and bn+2 is 1, bn+1 bn+2 qn + qn+1 + qn+2 bn
>
1
qn+2'
so
00
n=k
bn
qn
1
m=1
1
O°
> qk
_
1
> qk +
1
qk
qn+3m
°°
(i+
m=1
1
1
q3m °°
1
Y- g3-2 + q3-1 + M=1
00
n=k+1
an
J.
M=1 cc
n=k
1
q3m
1
°°
- qk
1
qn
n=1
an an
and this is also a contradiction. So only the sequence {bn} satisfies the conditions. The cardinality of different sequences {bn} is 2x0. If two {bn} sequences
are different, then the corresponding numbers q are also different. So the cardinality of S is maximum 2k°. But it cannot be more than this, since the cardinality of the set of all real numbers is 2°.
Problem 5.27. Given an > an+1 > 0 and a natural number p, such that a lim sup an < µ, n
aµn
prove that for all E > 0 there exist natural numbers N and no such that, for all n > no the following inequality holds: Nn
n
Eak <EEak. k=1
k=1=1
Solution. We have to find a natural number no such that for any E > 0 there exists a natural number N such that < E
nN
k=1 ak EN
=1 ak
holds for all n > no, where condition (1) holds. Choose a real number v in the interval (limsup, (an/anµ,), t) . Define a natural number M such that an/anµ < v, for all n > M. Therefore, for all natural numbers l > 0 and n > M, the inequality an/v' < anµt holds.
3. SOLUTIONS TO THE PROBLEMS
516
First, we show that the series Ek=1 ak is divergent. Obviously, Mµ l
Mµl
Mµ1
am > Mp E aMµt >u1 v 1: ak > E k=1
k=1
k=1
am
= MaM µ v
v
Here M and am are fixed, so when 1 -> oo, then from v < p it follows that the right side of the inequality also goes to infinity, that is, we estimated the series Ek=1 ak from below by a divergent series. This divergence provides us with a number K, for which 1 ak < M+1 ak; so for any n, > K, En k=1 ak :5 2 k=M+1 ak. For given e, choose l such that 1/e < (p/v)1/2. Therefore, 1
2 (µ/v)l
=
>k=M+1 ak/v1 _
En 2
k=M+1 ak
r
< rkµMµl a(Lklv`J+1)pl < n
2 Ek=M+1 ak
En
Ek=M$L µ
2 Ek=M+1 ak
< Ek= ak n 2 Ek=M+1 ak Ek=1 ak n
ak
a(Lk/vlj+1)/v'
3.10 TOPOLOGY
517
3.10 TOPOLOGY
Problem T.1. Prove that any uncountable subset of the Euclidean nspace contains an uncountable subset with the property that the distances between different pairs of points are different (that is, for any points P1 # P2 and Q1 # Q2 of this subset, F1-P2 = Q1Q2 implies either P1 = Q1 and P2 = Q2, or P1 = Q2 and P2 = Q1). Show that a similar statement is not valid if the Euclidean n-space is replaced with a (separable) Hilbert space.
Solution. For the proof of the first statement of the problem, we say that a subset of the Euclidean n-space En has property T if the distances between all pairs of points of this subset are different. By induction on
n, we prove that if all subsets of a set H(C En) that have property T are countable, then H is countable. The case n = 0 is trivial. Suppose that the statement is true for n - 1(> 0). Let H(C_ En) be a set such that all subsets with property T are countable. By Tukey's lemma, there exists a maximal subset M of H with property T. By maximality of M, any point of H \ M either has equal distance from two distinct points of M or has a distance from a point of M that equals the distance between some pair of points in M. Therefore, H is covered by the perpendicular bisector hyperplanes of pairs of points of M, together with the spheres centered at and passing through points of M. The set of these hyperplanes F i and spheres S i is countable (i = 1, 2, ... ), since M is countable. Each H n S; is a countable set. Indeed, the preceding argument can be applied to H n S, instead of H (since H n Sj satisfies all assumptions made on H); therefore, the set of Si's and F's that replace the preceding Si's and Fi's and cover H n SS is countable; by the induction hypothesis H n F' and (since S3 n Sz is contained in a hyperplane of En) H n S; n Sz are countable
sets. Thus, H n S; is countable since it can be covered by a countable family of countable sets. Further, by the induction hypothesis, each H n F3 is a countable set. Therefore, H is a countable set since it is covered by a countable family of countable sets.
In order to prove the second statement of the problem, it suffices to construct an uncountable subset of the Hilbert space spanned by the orthonormal basis ei (i = 1, 2.... ), such that the set of distances between all pairs of points in this subset is countable. To this end, consider an uncountable set 2t of infinite sets of natural numbers with pairwise finite intersection. It is well known that such a set f2t exists. (For instance, the set of bounded, monotone sequences of rational numbers has this property.) Each set A(E 2t) determines a vector E°O1(1/2i) eq; in the Hilbert space, where ai runs through A in increasing order. The set K of these vectors is uncountable. On the other hand, the distance between any pair of elements of K is the square root of a rational number, for if the vectors 1(1/2i) ear are determined by the sets a = E', (1/2') ea; and a' A and A' (E 2t), then
518
3. SOLUTIONS TO THE PROBLEMS
Ila-all2=11a112=I1a112-2(a,a')=3-2
12ij
a;=ai
where in the last term we have a finite sum since A n A' is finite.
Remark. Istvan Juhasz and Bela Bollobas pointed out that a similar argument proves the following generalization of the first statement of the problem: If m is a regular cardinal number, then any subset of cardinality m of the Euclidean n-space has a subset of cardinality m in which all distances between pairs of points are different; further, they considered generalizations of the statement to other metric spaces. Problem T.2. A sentence of the following type is often heard in Hungarian weather reports: "Last night's minimum temperatures took all values between -3 degrees and +5 degrees." Show that it would suffice to say, "Both -3 degrees and +5 degrees occurred among last night's minimum temperatures." (Assume that temperature as a two-variable function of place and time is continuous.) Remark. The formulation of the problem allows for various models. The proof is simplest when the country is assumed to be compact; not assuming compactness requires a different proof and yields a more general theorem. The argument needs some modification if the time interval is replaced with a (not necessarily metrizable) compact space. In the following solutions, the space is connected and the time is compact. The proof is carried out for metric spaces in the first one and for arbitrary topological spaces in the second one. Solution 1. Since any continuous image of a connected space is connected, and in the real line only the intervals are connected sets, it is sufficient to prove that the function that assigns to each point the minimum temperature attained there during the night is continuous.
Let E be a connected metric space, I be a compact metric space, R be the real line, and f : I --* R be a continuous function. For any fixed x E E, the function f (x, t) is continuous on the compact space I; therefore,
g(x) = mintEj f (x, t) exists. We show that g(x) is continuous. If this were not the case, then there would exist an e > 0 and a sequence {xk} (xk E E, k = 1, 2, ...) such that x, --* x, but I9(xk) - 9(x)I ? e. We consider two cases: (a) There is a subsequence of {x,,,} such that
9(xfk)
9(x) - E.
If f(xlk,tk) = g(xIIk), then the sequence {tk} has an accumulation point t E I, and we can assume that tk -> t. Then f(xnk,tk) = 9(xfk) C 9(x) - e C f(x,t) - E,
and f cannot be continuous.
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519
(b) There is a subsequence of {x,,,} such that g(xnk) ? g(x) + E. If f(x,t) = g(x),then f(xflk,t) >_ g(xnk) > g(x) + E > f(x,t) + E,
and f cannot be continuous. Since at least one of the cases must hold, we have a contradiction, which proves the statement.
Solution 2. Now let E be a connected topological space and I be a compact topological space. We keep other notations of Solution 1.
We prove again that g(x) is continuous on E. For e > 0, u c E, and v c I, let U(u, v) and V (u, v) be respective neighborhoods of u and v such that for all (x, t) E U(u, v) x V (u, v), we have
If(u,v)-f(x,t)I <E. For u fixed, the neighborhoods V (u, v) (v E I) cover I. Then, by its compactness, I is covered by a finite number of them:
V(u,vi),V(u,v2),...,V(u,vk). Then u = flk 1U(u, vi) is a neighborhood of u, and if x E U, then Ig(x) g(u) I < E, for if t c I, then t c V (u, vj) for some j, and then, since (x, t), (u, t) E U(u, vj) x V (u, vj), we have
if(x,t) - f(n,t)I < E. Remarks.
1. Assuming space and time to be compact topological spaces, Attila Mate considered temperature to take values in a metric space and proved, under these conditions, that the function g(x) is continuous. Compactness of space is not essential in this model. 2. Gyorgy Vesztergombi proved that the statement of the problem remains true if the definition of night depends on place (astronomical night), provided that the beginning and the end of the night are continuous functions of place.
3. Miklos Simonovits gave an example to show that it is essential to assume compactness of night. (If the night is not compact, then, of course, we have to speak of local infimum instead of local minimum.)
Let E = [-1. + 1], I = (-oo, +oo), f (x, t) = g(x)
and g(x) is not continuous.
' t) = E if (x { t
e-t2X2. Then
1
for x = 0,
0
forx # 0,
3. SOLUTIONS TO THE PROBLEMS
520
4. Juhasz and Posa gave examples that show that it is not sufficient to assume partial continuity of f (x, t). Juhasz's example is the following:
put E = I = [-1, +1], and
f(x,t)=
11
ifxOandt>0,
1+1 ifx>0and-x0and-1
In this case, 9(x)
1
for x < 0,
0
for x > O
is not continuous, and the statement of the problem is not true.
Problem T.3. Let A be a family of proper closed subspaces of the Hilbert space H = 12 totally ordered with respect to inclusion (that is, if L1, L2 E A, then either L1 C L2 or L2 C L1). Prove that there exists a vector x E H not contained in any of the subspaces L belonging to A. Solution 1. More generally, we prove the statement for separable Banach spaces. Let B be a separable Banach space, and let R be a system of subspaces L of B that satisfies the requirements of the problem. Suppose that ULERL = B. Consider a countable, everywhere-dense subset {XI, X2.... } of B. By recursion, define a countable increasing sequence of elements of R as follows. Let L1 be an element of B that contains x1. Suppose that the subspaces L1, ... , Ln (E R) have already been defined. Let L('+') be an element of R that contains xn+1. (Such an element exists by the assumptions.) One of the subspaces L 1, ... , L,a, L(n+1) contains all others; let this one be denoted by Ln+1. Then L. C _ L,,,+1 and x,a E Ln (n = 1, 2, ... ). Now, let L be an arbitrary element of R. Since L is a proper closed subspace of B, it cannot contain all elements of the dense set {x1, X2.... }. Suppose, say, that Xk V L. By definition, xk E Lk, so L C_ Lk since the system B is ordered with respect to inclusion. This means that U°__1Li = B. Now, for all natural numbers n, let f,a be a continuous linear functional on B for which IIfnII = n and fn (x) = 0 if x E L. (Such functionals exist. For example, take an arbitrary element yn in the complement of L. Since Ln is closed, the distance d of yn from the subspace Ln is positive. Consider now the linear subspace [yn] + Ln, where [yn] denotes the one-dimensional subspace generated by yn, and let .fn°) ()tyn - x) = nAd,
where x E Ln and A is a complex number. A ) is obviously linear on [yn] + Ln and fn°) (x) = 0 for x c Ln, and fn') II =sup
nd
IAdl
IAynxll
sup
Ilyn
xIl
infxCL.Ilyn - xII = n.
3.10 TOPOLOGY
521
By the Hahn-Banach theorem, f (0) is extendable to a functional f,,, defined
on the whole B that has the required properties.) Now, if x E B, then f,,,(x) = 0, since, by U°_1Li = B, x is contained in some Li, and the sequence {L,,,} of subspaces is increasing with respect to inclusion. So, the sequence {f ,,, } of functionals is pointwise convergent; therefore, by the Banach-Steinhaus theorem, sup, 11f,,,11 < oo. But this contradicts the
choice off,,. Solution 2. We prove the following generalization of the problem. Generalization. Let M be a separable topological space of second Baire
category, and let 2t be a system of nowhere-dense closed subsets of M ordered by inclusion. Then f1LE%L # M. (It is obvious that the space 12 and the family of subspaces in the problem and, more generally, separable Hilbert and Banach spaces with a similar
family of subspaces all satisfy the hypotheses of this statement; indeed, these spaces equipped with the norm topology are separable topological spaces of second Baire category and, in any of these spaces, a proper closed subspace is nowhere dense.)
Proof. Suppose that, to the contrary, ULE21L = M, and consider a countable, everywhere-dense set R = {x1, X2.... } in M. Put
Mk=U{LE2t:xkEL} (k=1,2,...). Then Mk is a nonempty, nowhere-dense subset of M for all k. Therefore, U' 1 Mk is of first category, and 00
U Mk
M.
k=1
On the other hand, if L E 2t, then L j4 M. L cannot contain R, that is, there is a natural number k such that xk V L. Since 2t is ordered with respect to inclusion, L C Mk follows from the definition of Mk. Thus, ULE21L C U
1Mk, a contradiction.
Remark. Several contestants remarked that the statement of the problem is not true for nonseparable Hilbert spaces. Attila Mate gave the following example. Let H be a nonseparable Hilbert space. Then H is isomorphic to a Hilbert space K ® L2 (W1 ), where K is a suitably chosen Hilbert space, and W1 is the set of countable ordinals considered as the discrete measure space in which the measure of all singletons is 1. Then
U (K ®L2( ) = K (D L2(Wl). Ewi
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3. SOLUTIONS TO THE PROBLEMS
Problem T.4. Let K be a compact topological group, and let F be a set of continuous functions defined on K that has cardinality greater than continuum. Prove that there exist x0 E K and f g E F such that
f (xo) = 9(xo) = max f (x) = maxg(x). xEK
xEK
Solution. In the proof, we use the following theorem by Paul Erdos.
Theorem. If the edges of a complete graph of cardinality greater than continuum are labeled with natural numbers, then there exists an uncountable complete subgraph with all edges labeled with the same number. See,
for example, P. Erdds, A. Hajnal, and R. Radio, Partition Relations for Cardinal Numbers, Acta Math. Acad. Sci. Hung. 16 (1965), Theorem 1. If Fx denotes the system of functions in F whose maximum is x, then it is clear that the cardinality of Fx is greater than continuum for some real number x. Suppose that f-x fl g=x = 0 for all pairs of functions f g E Fx. Consider the complete graph with the functions in Fx as vertices. If f # g E Fx, then there exists a natural number n such that f>x-(l/n) fl 9>x_(1/n) = 0; otherwise, the closed sets { f>x-(1/n), g>x-(1/n) }n=1,2,... would form a family with the finite intersection property whose intersection,
by compactness, would be nonempty, but this intersection is contained in the set f=x fl g=x. By the theorem quoted above, there is an n and
an uncountable F' C Fx such that all edges of F' are labeled with n. Thus, the sets f.,>,/n, f E F' are nonempty, open, and pairwise disjoint. This contradicts the well-known fact that K admits a finite Haar measure, and therefore any family of pairwise disjoint, nonempty, open sets in K is countable. So, the statement of the problem is true for some pair of functions f 54 g E F. In fact, we proved the following theorem.
Theorem. The statement of the problem holds for a compact space K if K satisfies the following condition: any family of pairwise disjoint, nonempty, open sets in K is countable.
Problem T.5. Prove that two points in a compact metric space can be joined with a rectifiable arc if and only if there exists a positive number K such that, for any E > 0, these points can be connected with an E-chain not longer than K.
Solution. If A and B are two points of the metric space, then let t(A, B) denote their distance. By a rectifiable arc joining A and B we mean a homeomorphic image of the real interval [a, b], where a and b are mapped to A and B, and for any subdivision a = to < t1 < t2 < ... < t,,,=1 < t,,, = b, denoting the image of ti by Ti, we have n-1
1: t(Ti,Ti+1) < K i=o
3.10 TOPOLOGY
523
for some fixed K. The infimum of such numbers K is called the length of the arc. Therefore, the half of the problem stating that if two points can be joined with a rectifiable arc, then for any e > 0 they can be joined with an a-chain of length at most K, is obvious. Now suppose that, for any e > 0, A and B are joinable with an a-chain not longer than K. Let L = {H0,. .. , H,,, } be a sequence of points in the metric space. We use the following notation:
q(L) = max t(Hi-1, Hi) 1
and
n
K(L) _
>t(Hi-1,Hi).
i=1
For 0 < h < 1, let L(h) denote the Hk for which k+1
k
t(H1-1, H1) < h K(L) <: t(Hz-1, Hi), i=1
i=1
and let L(1) = Hn. Then, obviously,
t(L(H1)L(H2)) < Ihl - h2I . K(L) + 4(L) Now choose an indefinitely refining sequence So of chains connecting A and B, where the lengths of these chains do not exceed a fixed constant K. The
hypotheses in the problem guarantee the existence of such a sequence of chains. Arrange the rational points of the interval [0, 1] into a sequence. Then let Sk be an indefinitely refining sequence So of chains connecting A and B such that for k = 1, 2, ... , (a) Sk is a subsequence of Sk_1i and (b) L(h) is convergent in the first k rational points when the chains are taken from Sk and k is kept fixed. (More precisely, instead of L(h), we should write Ln,k(h), where this
denotes L(h) for the nth chain of Sk. In (b), k and h are fixed while n -> oo.) The sequence of chains Sk can be defined for all k by compactness of the space. Now define the function f (h) on the set of rational numbers in [0, 1] by the formula
f (h) = lim Lk,n(h) if h is one of the first k rational numbers. This definition is correct by (a). Let h1 and h2 be rational numbers in [0, 1]. Choose a sufficiently large k so that h1 and h2 occur among the first k rational numbers. Then
t(f(h1),f(h2))
K. Ihl -h21
Since a uniformly continuous function defined on a dense subset of the interval [0, 1] can always be continuously extended over the whole interval
524
3. SOLUTIONS TO THE PROBLEMS
[0, 1], provided the target space is compact, the function f (h) is extendable,
and the last estimate remains valid. The image of the interval [0, 1] is a continuous curve connecting the points A and B, and from the estimate it immediately follows that all approximating chains have length at most K. Therefore, the curve is rectifiable of length < K. In general, it is not true that we obtain an arc. But if we choose the above K to be the smallest constant such that, for any e > 0, A and B are joinable with an e-chain not longer than K, then we get an arc. Indeed, otherwise there would exist u and v in [0, 1] (u # v), with f (u) = f (v). Then we could delete the image of the open interval (u, v) from the curve constructed above, and we would get a rectifiable curve from A to B whose length K+ is less than K. Then, by our initial observations, for any e > 0, A and B would be joinable by an e-chain shorter than K+. This contradicts the minimality of K, proving the statement of the problem. Remarks. 1. Laszlo Lovasz proved that, instead of compactness, assuming only local compactness and completeness of the space, the statement remains valid. By deleting a chord from a disk in the plane, we obtain a locally compact space in which two points on different sides of the chord cannot be joined with an arc although they are joinable with an arbitrarily fine e-chain. Therefore, it does not suffice to assume only local compactness of the space. Completeness alone is not sufficient either. To show this, Lovasz gave the following example.
Example. Let eo, e1, .
.
.
, en, ... be an orthonormal basis in the Hilbert
space 12. Consider the segments connecting 0 with the endpoints of the
vectors el, e2.... , and similarly the segments connecting eo with the points el+eo,e2+eo, .... Divide the segment between en and en+eo into n equal parts, that is, consider the points en + (i/n) eo when i = 1, 2, ... , n - 1.
The segments and points just defined form a closed set in the Hilbert space; therefore, they define a complete metric space. In this space the points 0 and eo for any e > 0 are joinable with an a-chain not longer than
3, but they cannot be joined with an arc. This space can be made into a connected counterexample by connecting the points en + ((i - 1)/n) eo and en + (i/n) eo with suitably chosen, nonrectifiable arcs. 2. A sketch of the proof of Lovasz's generalization is the following.
Proof. We call a point accessible if, for some t, B and the point are joinable with an arc of length t and, for any e > 0, the point and A are joinable with an a-chain not longer than K - t, where K is the minimal constant used above. Of the accessible points x and y we say that x is finer than y if, in the definition of accessibility of x, the arc from B can be chosen through y. Then, using Zorn's lemma, we define a "finest" point x0, which we show to be necessarily A. Assuming the contrary, take a compact neighborhood U of x0 that does not contain A. Then define a point z on the boundary of U that is a limit point of certain points of chains from x0 to A in the definition of accessibility of xo. We show that z is accessible
3.10 TOPOLOGY
525
and finer than x0. To this end, it suffices to see that x0 and z are joinable with a sufficiently short arc. But we can apply the result already obtained for compact spaces to the points x0 and z, and this proves the statement.
Problem T.6. Let a neighborhood basis of a point x of the real line consist of all Lebesgue-measurable sets containing x whose density at x equals 1. Show that this requirement defines a topology that is regular but not normal.
Solution. (1) Let m(A) denote the measure of the Lebesgue-measurable set A. Let A and B be basis neighborhoods of the point x, we show that A fl B is as well. Since A fl B is measurable and x E A fl B, we only have to show that its density at the point x is 1, that is, m((A fl B) fl I)
m(I)
-4
1
when the interval I shrinks to x:
m(AflBfll) m(I)
-1
m(I) - m(I - (A f1 B)) m(I)
_ m((I-A)U(I-B)) m(I) m(I - A) + m(I - B) m(I) m(I) The last expression converges to 0 when I shrinks to x. So A fl B is indeed a neighborhood of x. In order to obtain a topology, we show that any neighborhood A of x contains a neighborhood B of x that is a neighborhood of all of its points. Let B be the set of points in A where the density of A is 1. Obviously, x E B, and by the Lebesgue density theorem, A - B has measure 0. Therefore, B is also measurable, and at points of B, B has the same density as A, that is, 1. So, we have a topology indeed. Let E denote this new topology, and call its open sets E-open while keeping the adjective "open" for open sets in the usual topology. Similarly, we distinguish closed and E-closed sets. It is not obvious but it is true that the E-open sets are measurable. We can prove this as follows.
Let H be an E-open set. We can assume that H is bounded; then the outer measure m(H) of H is finite. Now put a = sup{m(C) : C C H, C is measurable},
and for all n choose Cn so that C" C H and m(CC) > a - (1/n). Then S = UnCC is a subset of H with the property that every measurable
3. SOLUTIONS TO THE PROBLEMS
526
subset of H-S has measure 0. (S is usually called the measurable core of H.) If V C H is measurable and Q = H - S, then, by (V fl Q) U (V fl s) _ V, V fl Q is measurable, and therefore it has measure 0. So,
m(V) = m(V fl s). Q decomposes as the union of two sets Z and X, where the former has measure 0, and the density of Q at all points of the latter is 1. In order to prove that H is measurable, we show that Q has measure 0, and to this end, we need X to be empty. This will imply measurability of H.
Arguing by contradiction, we suppose that p E X. Let I denote an interval shrinking to p. By definition of X, m(Q fl I) = (1 + o(1))m(I). Now let U be an E-neighborhood of p contained in H. Then
m(U fl I n s) = m(U fl I) = (1 + o(1))m(I), so u fl S has density 1 at p. Therefore, by I - (U fl s) D Q fl I, we have
m(Q fl I) < m(I - (U n s)) = o(m(I)), which contradicts our assumption. (2) We show that the topology E is regular. Suppose that K is an E-closed set, and x V K. Then x has a neighborhood A that does not meet K. So, m(AfI) = (1+o(1))m(I), where the interval I shrinks to x. Thus, m(A fl I) = o(m(I)). (Here A denotes the complement of the set A.) K C A, so m(K fl I) < m(A fl I), and therefore
m(K fl I) m(I)
0
Put x,,, = x - (1/n) and yn = x + (1/n), n = 1, 2,... . It is well known that, for any measurable set X and e > 0, there exists an open set G such that X C G and m(G) < m(X) + e. Choose the open set Gn in the interval [xn_1i Xn+21 so that K fl [xn, xn+1] C Gn and
m(Gn) < m(K n [xn, xn+1]) +
1
2
hold. For n = 1, we require Gl C (-oo, x2]. Choose the sets Hn in a similar way for the intervals [yn+l, yn]. Put
B=(-oo,xl)UGIUG2U...UH1UH2U...U(yl,oo); then B is open and contains K. Since open sets are E-open (because they have density 1 at every point), it will suffice to prove that the density of B at x is 0. Then the complement of the E-closure of B and B will be disjoint E-neighborhoods of x and K, respectively. It is enough to show
that m(I fl B)/m(I) -> 0 for I = [x, x + k] (k -> 0), because a similar argument shows the corresponding estimate for the left-hand intervals,
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527
and these two together imply the same for arbitrary intervals I shrinking
to x. Assume y,,,+1 < x + k < Then m(I) = k > 1/(m + 1). We have m(B n I) < Em-2 m(H2), since H2 n [x, ym] = 0 for i < m - 2. Therefore cc
m(B n i) <
00
m(K n [yi + 1, y2]) +
i=m-2
i=m-2 1
i
= m(K n [x, yn - 2]) + and so
m((II) )
1
2i
7
2,,,_s
< m(K n [X, ym-2]) + 2 1
m+1
m+1
If I shrinks to x, then m -+ oo, so both terms in the right-hand side of the last inequality converge to 0. This is obvious for the second term; for the
(3)
first one, use the already established convergence m(K n I)/m(I) -p 0 and the fact that (m + 1)/(m - 2) < 4 if m > 3. So, B indeed has density 0 at x, and the space is regular. We show that the space is not normal; namely, that an E-closed set K1 of second Baire category and an E-closed everywhere-dense set K2 cannot be separated by E-open sets. Since a countable set has measure 0, and thus is E-closed, the existence of such a set K2 is clear. The existence of such a K1 is also well known: for every n, there exists a nowhere-dense set of measure greater than 1 - 1/n in [0, 1]; the union of these is a first category set of measure 1; and the complement of this in [0, 1] is of second category and has measure 0 and is therefore E-closed. Since K1 is chosen to have measure 0, K2 can be chosen in the complement of K1, then K1 and K2 are disjoint. Suppose that there exist two disjoint E-open sets C1 and C2 with K1 C C1 and K2 C C2. Then, for any x E C1, there is a neighborhood A C C1 of x for which
M(An I) m(I)
--+ 1 ,
an d thus
m(C1 n I)
m(I)
-1
when I shrinks to x. Therefore, by K1 C C1, for any x E K1, there exists an no such that for n > no, m (C1 n [x - (1/n), x + (1/n)]) >
that is,
/
M /C1
1
- 2'
2/n 1 n X_ n,1 x + nJ I> 1
1
n
(n > no).
If we divide the elements of K1 into a countable number of subsets depending on whether the last inequality holds from a certain index,
3. SOLUTIONS TO THE PROBLEMS
528
then at least one of these subsets is dense in an interval [a, b]. Thus, there exists an no such that the last inequality holds from this no on an everywhere-dense subset of the interval (a, b). Since K2 is everywhere dense, we can choose an element y of K2 in (a, b). Since y E C2, there is an n > no with m (C2 fl [y - (1/n), y + (1/n)]) > 3
2/n
4'
that is, 1
m C2 fl y - y n'
+ n) ?
3
4n K1 is dense in (a, b), so there exists an x in the interval (y, y + (1/8n)) for which
mICln[x n,x+nJ)>n holds. The length of the interval I = [y - (1/n), x + (1/n)] is at most (2/n) + (1/8n). C1 and C2 are disjoint and m(C1 f1 I) > n
and m(C2 fl I) > 2n
Therefore,
C2+8 n >m(I)>m(C1f1I)+m(C2nI)> (1+2)
n,
a contradiction, which proves that the topology E is not normal. Remarks.
1. In order to solve the problem, it is not necessary to prove that E-open sets are measurable, but this simplifies the proof at various points. 2. Lajos Posa proved the nonnormality statement using a cardinality argument. The sketch of his proof is as follows. Proof. There exists a set of cardinality c = and of measure 0. All subsets of this set are E-closed; therefore it can be partitioned into two disjoint E-closed sets in 2` different ways. If the space were normal, then for each such partition we could find two disjoint E-open sets that separate the two parts. It can be proved that the pairs consisting of the E-closures of the separating E-open sets corresponding to different partitions are different.
It is also easy to see that the E-closure of every E-open set equals the E-closure of an F,-subset, which only differs by a set of measure 0. Since the cardinality of the set of F,-sets is only c, the cardinality of the set of all pairs of closures of F,-sets is also only c. Therefore, the cardinality of the set of pairs of separating E-open sets is only c, a contradiction. 3. Laszlo Lovasz proved complete regularity (which is stronger than reg-
ularity). He noticed that the above technique yields that if A and B are disjoint E-closed sets, then they can be separated by an open set containing B and an E-open set containing A. Then the proof of the Urysohn lemma applies.
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529
Problem T.7. Suppose that V is a locally compact topological space that admits no countable covering with compact sets. Let C denote the set of all compact subsets of the space V and U the set of open subsets that are not contained in any compact set. Let f be a function from U to C such that f (U) C U for all U EU. Prove that either (i) there exists a nonempty compact set C such that f (U) is not a proper subset of C whenever C C_ U EU, (ii) or for some compact set C, the set
f-1(C) = U{U E U: f(U) C C} is an element of U, that is, f -1(C) is not contained in any compact set.
Solution. The statement is trivial. Indeed, if f -1(0) = V, then (ii) holds.
If not, then for an arbitrary x E V - f-1(0), (i) holds with C = {x}. (We do not use that V is not or-compact, only noncompactness of V is necessary.)
Remark. LAszlo Babai proves that if, under the hypotheses of the problem, (ii) is not true, then (i) holds with a set C of two elements. The proof of this is fairly difficult.
Problem T.8. Let T1 and T2 be second-countable topologies on the set E. We would like to find a real function or defined on E x E such that 0 < a(x, y) < +oo, Q(x, z) < o (x, y) + a (y, z)
Q(x, x) = 0, (x, y, z E E),
and, for any p E E, the sets Vi (p, e) = {x : Q(x, p) < E}
(E > 0)
form a neighborhood base of p with respect to TI, and the sets
V2(p,E)={x:0'(p,x)<E} (E>0) form a neighborhood base of p with respect to T2. Prove that such a function or exists if and only if, for any p E E and Ti-open set G 3 p (i = 1, 2), there exist a Ti-open set G' and a T3_i-closed set F with p E
G'CFCG.
Solution. Suppose that there exists a function Q(x, y) with the required properties. Let p E E, G be open in T, p E G. Then there exists E > 0 with Vi(p, E) C G. Let 0 < 6 < E, G' = U (p, 6), and let F be the T3_iclosure of G'. Using the triangle inequality, it easily follows that G', F, G
3. SOLUTIONS TO THE PROBLEMS
530
indeed satisfy the hypotheses. (The assumption on second countability is not needed in this direction.) In what follows, G always denotes an open set and F a closed set. Superz scripts indicate which topology this means. For example, G(1) denotes the Tz-closure of the Tl-open set G(1). We prove that the hypothesis given in the problem is sufficient for the existence of Q. The proof is a modification of the well known-proof of the Urysohn metrization theorem. Tikhonov lemma. If F(1) nF(z) = 0, then there exist G(1) and G(z) such that G(1) i F(z), G(z) D F(1), and G(1) n G(z) = 0.
Proof. Let p E F(i). Then p E E - F(3-i), this set is T3_i-open, so by the assumption there exists a G' = (p) such that p E G' (3-i) (p) and G'(3-Z)
G'(3-i)(p)(Z)
C E - F(3-i), that is, G'(3-a)(p)(i) n F(3-i) = 0
To each such G'(j) (p), assign an element UW of the countable basis in T : p E UU) C G'(j) (p). We have countably many sets UU); these can be arranged as U(j), n = 1, 2, ... , (j = 1, 2). We also have oo
U(i) (3-i) nF(i)=0.
F (3-i) c u Un(z)
(1)
n=1
Put U() = U(x)
- J I U(31)(y) k I
k=1
These Un('*) are T-open, they also satisfy (1), and obviously U. n U(m3*i) 0. Therefore, the sets
_
00
G(i) = U Un
(i = 1, 2)
n=1
clearly satisfy the requirements of the lemma. Urysohn lemma. If F(1) C G(z), then there exists a real function p(x, y)
on E x E such that 0 < p(x,y) < 1,
p(x,x) = 0,
P(x, z) < P(x, y) + P(y, z)
(x, y, z E E);
(2)
the sets 01) (p, E) = {x : p(x, p) < E}
(E > 0)
(3)
(E > 0)
(4)
G(z).
(5)
are Tl-open; the sets Vn(z) (p, E) = {x : p(p, x) < E}
are Tz-open; an d
p(x, y) = 1
if x E F(1), Y
3.10 TOPOLOGY
531
FMMM, G12) = G(2), Proof. Introduce the notations Got) = 0, Fol) Fil) = E. Let D be a countable dense subset of the-closed interval [0, 1]: D = fro, rl, r2i ... }, where ro = 0, rl = 1.
By recursion on n, we define the sets
and F(1) so that Gn(2) C F(1)
and, for rn, < r,,,,, hold. These inclusions are true in the Th C GM) cases already defined (n, m = 0, 1). If the sets with subscripts less than n (n > 2) have already been defined, then find the neighbors of rn among ro, r1, ... , r,i_1. Let them be rk and r1: rk < rn, < rl, 0 < k, 1 < n - 1. Apply the Tychonoff lemma to the sets Fkl) and E - G12): by rk < rj and the induction hypothesis, Fk(1) n (E - G12)) = 0, so by the the Tychonoff lemma, there exist
and Gnl) with
G(2) j Fkl),
n Gn,:) = 0 and
G(,:) D E
-
that is, using the notation F(1) _ E - Gn :) Fn(l) C G(2)
and Gn2) n G,:)
0 means that G(2) C F(1).
It is clear that the validity of the induction hypothesis is inherited to these sets, that is, the recursive definition is correct. For x E E, put g(x) = inf({rn, : x E
U {1}),
f (x) = sup({rn : x V Fn(l) } U {0}),
(0 < f (x), g(x) < 1). It is clear that f I FM = gIF(1) = 0,
f J E - G(2) = gI E - G(2) - 1.
Further, if n > 2, then g(x) < rn,
x E G(2)
= x E F(1) = f (X) < rte,
so f (x) < g(x) for all x E E. But if f (x) < g(x) for some x E E, then there would exist rte,, rm E D such that
f(x)
a contradiction. Therefore,
3. SOLUTIONS TO THE PROBLEMS
532
For x, y E E, put if f (x) >- f (y),
0,
P(x, y) =
1 f (y) - f (x), if f (x) < f (y)
Obviously, 0 < p(x, y) < 1, p(x, x) = 0, and (2) and (5) hold. To prove (3) and (4), by (2) it suffices to show that p E int(i) V(a) (p, E) (p E E, e > 0) (i = 1, 2). For i = 1: Let e > 0, p E E. Then V(1) (p,
E) = {x : p(x, p) < E} = {x : f (p) - f (x) < E} D E - F(1) E) p
if n is chosen so that f (p) - E < r,, < f (p). In case f (p) = 0, V(1) (p, e) _ EE) p. E - F n(1) and E are Ti-open. For i = 2: VP(2) (p,
E) = {x : p(x, p) < E} = {x : f (X) - f (p) < E}
= {x : g(x) - g(p) < E} D
if 9(p) < rn < 9(p) + E. In case 9(p) = 1, This proves the Urysohn lemma.
V(2) (p, E)
p
= E.
We turn now to the proof of the theorem. Let P be the set of all pairs (U(2), V(ii), where UW and VW are elements of the countable basis for the topology T and U(-,
(3 -z)
C V(i)
(for i = 1, 2).
Then P is countable: P = {(U1,V1), (U2,V2),... I-
Now for k = 1, 2, ... we define a function pk(x, y). Put (Uk, Vk) = V(')) If i = 2, then apply the Urysohn lemma to
the pair of sets F'1) = U7k(2iGig' = V(2), and call pk the function p given
by the lemma. If i = 1, then put F(2) = GM = Vile, and apply the version of the Urysohn lemma where the superscripts (1) and (2) are interchanged and p(a, b) is replaced with p(b, a) everywhere. (Then (3) and (4) simply interchange and (2) remains the same.) Call Pk the function p thus obtained. Put 00
2k Pk(x, y) =E k=1 1
o (x, y)
3.10 TOPOLOGY
533
It is clear that 0 < o (x, y) < 1, o(x, x) = 0 and, by (2), u(x, z) < a (x, y) + o (y, z) holds for all x, y E E.
We prove that fore>0andpEE
(i=1,2).
p E int(z) U (p, E)
Let N be sufficiently large so that >k N
1
< 2. Then
N
i(P,E)
nv )(p,2) k=1
But, by (3) and (4), the right-hand side is a Ti-open set containing p. It only remains to show that if p E int(i) U, then p E V (p, e) C U for some e > 0. Since p E int(i)U, there is a G(i), then, by the assumptions, there is a G'(z) in the countable basis for Ti, such that p E Gt(z)
Z)
C G(i) C U.
Therefore, (G'('), G()) = (0), V(z)) E P for some k. So, using p E G'(') and (5), Vi
(p,
1
2k
C p(k)(P,1) C G(i).
Problem T.9. Prove that there exists a topological space T containing the real line as a subset, such that the Lebesgue-measurable functions, and only those, extend continuously over T. Show that the real line cannot be an everywhere-dense subset of such a space T.
Solution. 1. Let { fi : i E I} be the set of all Lebesgue-measurable functions, Xi be a copy of the real line with the usual topology (i E I), and
T = fXi. iEI
If h : R -> T is the map defined by
pi(h(x)) = fi(x), where pi is the projection onto Xi, then h is injective and h(R) can be considered as a copy of the real line. We prove that T is as required. Let f be a measurable function, say, f = fi. Then f = piI h(R) and a continuous extension of this over T is pi. Conversely, if k : T -> R is continuous, then kIh(lR) is measurable. Indeed, by a well-known theorem
(see, for example, R. Engelking, Outline of General Topology, PWNPolish Sci. Publ, Warsaw, 1968, p. 98, Problem R), k only depends on
3. SOLUTIONS TO THE PROBLEMS
534
countably many coordinates, that is, k = k' o p, where p : T --> T' is the projection onto the product T' = 11iE1, Xi, where I' C_ I is countable and k' : T' -> R is continuous. Now,
{x : k(h(x)) > c} = {x : k'(p(h(x))) > c} = {x : p(h(x)) E U} where UU is open in V. The sets of the form H Yi,
iEl' where Yi = Xi with a finite number of exceptions when Yi is an interval
with rational endpoints, form a countable basis in V. Therefore, the set {x : k(h(x)) > c} can be written as a countable union of sets of the form {x : p(h(x)) E 11iEl' Yi}. But
x : p(h(x)) E fl Yi
= IX: fi(x) E Yi (i E I)},
iEl'
being a union of countably many measurable sets, is measurable. Thus, {x : k(h(x)) > c} is measurable. 2. Suppose that T has the required properties and R C T is dense. Since all measurable functions are continuous, it is obvious that the topology of T induces the discrete topology on R. Let p c T. Let f * denote the extension of the function f (x) = x over T (this is unique since R is dense in T). Further, consider the function
g(x) =
if x = f * (p), if x f*(p)
10, i
and let g* be its continuous extension over T. Put UP =
{q: Ig*(p) - g*(q)I < 1, If *(p) - f*(q)I <
g*(q) + 1
}
.
Let x E Up n R (since R is dense, such an x exists); we show that x = f * (p). Indeed, if x
f * (p), then
Ig(x)I = x _ f*(p) Ig*(p)I + 1, > I
which contradicts that I g* (p) - g* (x) I < 1. So Up n R = { f * (p)}. From this it also follows that f * (q) = f * (p) for every q E Up; indeed, UgnUpnR
is nonempty (since R is everywhere dense), but its only element must equal both f * (q) and f * (p). So f * (p) = f * (q). Now for an arbitrary, nonmeasurable, real function cp, the function 0* = f* (
0
is a continuous extension of cp, because this function, being constant on the neighborhood Up of any point p E T, is continuous. Thus, nonmeasurable functions also have continuous extensions over T, and this is a contradiction.
3.10 TOPOLOGY
535
Problem T.10. Let A be a closed and bounded set in the plane, and let C denote the set of points at a unit distance from A. Let p E C, and assume that the intersection of A with the unit circle K centered at p can be covered by an arc shorter than a semicircle of K. Prove that the intersection of C with a suitable neighborhood of p is a simple arc of which p is not an endpoint.
Solution. Let ab be the minimal arc of K containing K fl A (possibly a = b). Introduce Cartesian coordinates in the plane so that p is the origin, a and b lie in the left half-plane symmetrically with respect to the horizontal axis, and in case a 54 b, b lies in the upper half-plane. We claim that with a suitable 6 > 0, the intersection C1 of C with the upper half-plane and with the disc of radius 6 centered at p is a simple arc starting from p that only meets the horizontal axis at p. Then a similar statement is true for the lower half-plane, and this proves the theorem. In order to prove our claim, it suffices to show that, for any r < 6, the set C meets the upper semicircle Kr of radius r centered at p in a single interior point. In this case, assigning to points of C1 their distance from p is a topological map from the compact set C1 onto the interval [0, b], since it is continuous and bijective. Let T be a convex plane sector, containing a and b in its interior, with vertex at p, and with the horizontal coordinate axis as axis of symmetry. Then, obviously, p(p, A - T) = 1 + el > 1. Further,
there is an e2 > 0 such that if x E T, say, in the upper half-plane, and o(x, p) < E2, then the triangle pxb has an obtuse angle at p; Therefore, o(x, b) < p(p, b) = 1, and so o(x, A) < 1. Now let b < min(el, e2). If r < b, then K, fl Cl is nonempty. Indeed, let c and d be the endpoints of Kr in the left and right half-planes, respectively. Then o(d, A) < 1 by the above. On the other hand, if x E A - T, then o(x,c)
e(p,x) - o(p,c) > 1+6-6 > 1;
and if x E A fl T, then the triangle xpc has an obtuse angle at p, so o(c, x) > o(p, x) > 1.
Thus, o(c, A) > 1. By continuity of o, there is a point y on the arc Kr for which o(y, A) = 1, that is, y E C1. It also follows that c, d V C1.
On the other hand, suppose that y1, Y2 e C1 fl Kr. By the above, y1, y2 V T. Let yl be the one of y1i y2 that is closer to d, and let t E A be a point such that o(y2i t) = 1. Then t E T, because o(p, t) <- o(P, y2) + o(y2, t) < 1 + 6 < 1 + el.
Since o(yl, t) > 1 = o(y2i t), the point t is on the same side of the perpendicular bisector of the segment yl y2 (which goes through p) as y2. Therefore, the angle y2ptL, measured in positive orientation, is less than 180 degrees
and obviously is greater than 90 degrees. Then the triangle y2pt has an obtuse angle at p, and so o(y2i t) > p(p, t) > 1, a contradiction.
536
3. SOLUTIONS TO THE PROBLEMS
Problem T.11. Suppose that 'r is a metrizable topology on a set X of cardinality less than or equal to continuum. Prove that there exists a separable and metrizable topology on X that is coarser than T. Solution 1. We say that a set A is separated by a family of subsets if for any a, b E A, a # b, there exists a set in the family that contains a but does not contain b. If the cardinality of A is at most continuum, then A is always separated by a countable family of subsets. Indeed, embed A into R and take the subsets corresponding to intervals with rational endpoints. First, we show that there exists a sequence of closed subsets in X that separates X. It is well known that every metrizable space admits a adiscrete basis. Let B = Un__1Bn be a basis in X such that B,, consists of pairwise disjoint, nonempty sets. Then the cardinality of Bn is less than or equal to continuum, so we can take a family {Bn : i = 1, 2, ... } (a family of families of sets) that separates Bn. Let U, be the union of the elements of then U, is an open set with respect to the topology T. The family {Un} separates X. Indeed, let x, y E X, x # y. Since B is a basis, there is an n and V E Bn such that x E V and y V V. There may or may not exist a V' E Bn with y E V', but there is at most one such V', since Bn is a disjoint family. Choose a B; such that V E Bn and V' V Bn (or, an arbitrary B; containing V if there is no V'). Then, obviously, x E U, and y V U, . As the family {Un} separates X, so does the family consisting of the complements of the sets U, . Arrange these complements into a sequence F1, F2,... . Then {Fn} is a family that separates X and consists of closed sets. Let d be a bounded metric on X that induces the topology T. Consider the following map f : X -+ 11 :
f (x) =
Cd(Fi, x) d(F2, x) 21
22
,.
Then f is injective, because if x # y there would exist an Fn with x E Fn and y V Fn, and then d(Fn, x) = 0 # d(Fn, y). The map f is continuous, for if D denotes the distance in 11, then 00
D(f (x), f (y)) = > 2-n I d(Fn, x) - d(Fn, y) I n=1
E 2-nd(x, y) = d(x, y) n=1
Consider now the topology on X determined by f as the inverse image topology. Since 11 is a separable metric space (and separability is a hereditary property of metric spaces) and f is injective, this topology is separable and metrizable and, since f is continuous, it is coarser than T.
Solution 2. For any cardinal number m > 0, we denote by J(m) the following metric space: the points of J(m) are the pairs (i, x), where i E I
3.10 TOPOLOGY
537
(I is an arbitrary set of cardinality m) and x E [0, 1], with the points of the form (i, 0) identified. The metric d of J(m) is defined as y))
_
x+y,
ifi
j,
if i = j. In the proof, we use the theorem that asserts that every metric space of ix - YI,
weight m is topologically embeddable into the topological product of countably many copies of the space J(m). (See, for example, R. Engelking, Outline of General Topology, PWN-Polish Sci. Publ, Warsaw, 1968, p. 197, Theorem 7.) Let P denote the following property of topological spaces (X, T): there
exists on X a separable metrizable topology coarser than r. It is obvious that all subspaces of a space with property P and topological products of countable families of spaces with property P also have property P. It is also clear that metric spaces satisfying the hypotheses of the problem are of weight less than or equal to c (continuum). Since, by the theorem cited above, all metric spaces of weight < c are embeddable into the topological product of countably many copies of J(c), it suffices to show that J(c) has property P. It is obvious that J(c) is homeomorphic to the metric space whose underlying set is the unit disc and in which the distance between two points on the same radius is their usual distance, and the distance between two points on different radii is the sum of the norms of the points. The usual topology of the unit disc is coarser than this topology and is separable and metrizable.
Problem T.12. Suppose that all subspaces of cardinality at most R1 of a topological space are second-countable. Prove that the whole space is second-countable.
Solution. We argue by contradiction. Suppose that the topological space
X is not second-countable but that all subspaces of cardinality < R1 are. We construct a certain subspace Y of cardinality < R1, and secondcountability of Y will lead to a contradiction. The subspace Y is the union of subspaces Y« (a < w1) to be defined by transfinite recursion. Together with the Y«, we simultaneously define a family 9« of open sets in X so that 9« IY« is a basis for Y. (If f is a family of subsets in X, and Z C X, then 1-1I Z = {H f1 Z : H E f} is the trace of f in Z.) The recursion is as follows: Yo = go = 0. If the countable YY and 9g have already been defined for < a < w1, then define Y« and 9« the following way. The family UC<«9g, being countable, is not a basis for X, therefore there exists a point p E X and a neighborhood V of p such that, for any element G of Ug<«9g containing p, we have G 0 V. Choose a point PG E G \ V for each such G, and put
<«
U9e
<«
538
3. SOLUTIONS TO THE PROBLEMS
It is clear that Y,, is countable; therefore, by the assumptions, it is secondcountable. Let g, be a countable family of open sets in X such that 9aIY,, is a basis in Y. Consider the subspace Y = U,, <,,1 Ye, . It obviously has cardinality < 1Z 1, so it is second-countable. The crucial observation now is that Ua<,,,1 Ga IY is a basis for Y. Let us postpone the proof of this statement and show first how to complete the solution, assuming that this observation is valid. Since Y is second-countable, the basis U,,<,1 ga IY contains a countable is a basis basis. Therefore, for a sufficiently large a, the family
is a basis for Y,,. But this is a contradiction since u< ,,G I Y,,, does not contain a basis of neighborhoods for the point p used in the definition of Y. It only remains to show that U,<,1 Ga IY is a basis for Y. This statement is immediate from the following lemma. Let Y be a second-countable topological space, and let Y = U,<",1Ye,, where YQ C Y,3 for a < /3. For a < wl, let G,, be a family of open sets in Y such that GO IY,, is a basis for YQ. Then U,,<",1 g,, is a basis for Y.
for Y. Then, with this a,
The proof is again by contradiction. Suppose that U,,<1G is not a basis for Y. We construct a sequence {V,, : a < wl } of open sets with the property Vp \ UQ> VQ # 0 (/3 < wl ), but such a sequence cannot exist in a second-countable space. The sets Vim, together with a sequence of points rQ E Vim, are defined by transfinite recursion as follows. Since U,<,1g,, is not a basis for Y,
there exists a point q E Y and a neighborhood U of q such that G 0 U whenever q E G E U,,,
U. Thus, rp E V,3 \ U,,,>,3V,, (3 < wl). But this contradicts that Y admits a countable basis {B1, B2, ... }. Indeed, for each a choose a Bna with r,,, E B,,,a C V. Since there are only countably many B,,,, there exists a > /3 with n, = n,3 = n. But then rp E Bn and r,3 # Vc D Bn, which is impossible.
Remark. Unfortunately, the phrase "at most" was missing from the text of the problem when it was posed for the competition. The statement is not true in that version; a counterexample is provided by the set of natural numbers endowed with the topology in which the nonempty, open sets are the sequences of density 1. This is not a second-countable space, but the hypothesis is vacuously true. The contestants usually noticed this defect and considered the correctly modified version of the problem.
Seven solutions were submitted. Best results were obtained by Emil Kiss. He proved the statement for T3-spaces among other arguments that contained all the ideas necessary to prove the statement in full. Nandor Simanyi proved the statement assuming the continuum hypothesis and reg-
3.10 TOPOLOGY
539
ularity of the space. Vilmos Totik used the hypothesis that every point has a basis of neighborhoods of cardinality N1, and Zoltan Szabo assumed firstcountability of the space.
Problem T.13. Suppose that the T3-space X has no isolated points and that in X any family of pairwise disjoint, nonempty, open sets is countable. Prove that X can be covered by at most continuum many nowhere-dense sets.
Solution 1. By transfinite recursion on a < wl, we define a family of open sets in X indexed by sequences of length a consisting of natural numbers,
that is, by functions f : a --+ w. For a = 0, the only such function is the empty set; put Go = X. If a is a limit ordinal and f : a -4 w, then put Gf = int (lgGg), where g ranges over restrictions of f to all smaller ordinals. If a has the form a = Q + 1 and G f = 0 for some f : ,3 -* w, then we define Gg = 0 for all extensions g : 03 - w of f. If Gf # 0, then we can define a sequence {Hl, H2, ... } of open sets such that H,,CGf,H,,,f1H 01,H,,0 0 foralln54 m, and Hl, H2,... is a maximal system of nonempty, open sets in Gf. (Here we used regularity of X and the property formulated in the problem. It is easy to avoid that this
system is finite.) Let the sets Hl, H2i ... be indexed with the extensions 91,92.... of f over 0 + 1; the cardinality of these is also count ably infinite: {Hl, H2,. .. } = {Ggl, Gg2, ... }. So the sets Gf are defined for all functions
f :a-+ w (a<wl). It is obvious that GfflGg=0if f,g:a-*wand
f
g, and that Gf 2 Gg if f C g. First, we prove that there is no p E X that is contained in Gf with some f : a --> w for all a < wl. Indeed, let us suppose the contrary. Then, by our preceding remarks, for all a < wl there exists precisely one function fa : a -4 w with p E C1, and these functions are restrictions of a single function F : wl -+ w. Then none of the sets G f. is empty, and
if we define the functions ga : a + 1 :-+ w by ga(d) = fa(d) ( < a) and g. (a) = fa+l (a) + 1, then the sets Gga are pairwise disjoint, nonempty, open sets, which contradicts the assumptions. Thus, each point p "dies out" at a certain level below w1. Therefore, the space X can be covered as follows:
U Gg Ua<wl U f:a-w U (flCCf). X=a<wl U f:a-.w U Gf-g:a+1-.w gj f
limit
9vf
It is immediate that this is a covering by 2° sets. It remains to show that the summands are nowhere-dense sets. The set
G f - U Gg g:a+1-.w
gjf
540
3. SOLUTIONS TO THE PROBLEMS
is nowhere dense because otherwise it would have a nonempty interior, and
the family of the sets G9 would not be maximal. For the other type of summands, denoting the restriction of f to Q by g,3 for Q < a, we have
n Gsv g n
,3
G97o,
'3+1
C I I Gsv R
that is, the summand is a closed set minus its interior, which, being the boundary of a closed set, is nowhere dense.
Solution 2. Let {Ga : a < Al be a well-ordering of the set of nonempty, open sets of X. Since there are no isolated points, for each a < A there exists a pair GT), G,(1) of nonempty, open sets with GT) n G,(1) = 0 and Gq°) U GT C_ G. Let p c X be arbitrary, and define the set' H(a, p) by transfinite recursion on a as follows. Suppose that the sets H(0, p) are already defined for Q < a. If (U0
the sets H(a, p) are pairwise disjoint. Let the set of these be jag(p) : < }, where cp = cp(p) < w1. We claim that if p and q are points with ap(p) = cp(q) such that, for every < cp(p), H(ag(p), p) = G(1e(p) implies H(ag(q),q) = GaaE(gI (that is, for each < cp(p), the points p and q "ramify in the same direction"), then Fp = F9. Taking this statement for granted, we can complete the proof by the observation that the cardinality of 0-1 sequences indexed by countable ordinals is continuum; therefore the number of different Fp's is at most continuum. It remains to prove our claim. It will suffice to show, by transfinite induction, that ag (p) = aC (q). If a((p) = a((q) holds for all ( < C, then obviously H(a((p),p) = H(a((q),q), and ac(p) is the first ordinal
a > sup{a( : ( < } with Ga n p)) = 0; similarly, aC(q) is the first a with Ga n (U(
Problem T.14. Construct an uncountable Hausdorff space in which the complement of the closure of any nonempty, open set is countable.
3.10 TOPOLOGY
541
Solution 1. Let X be an arbitrary uncountable set. It is easy to see that there exists a topology on X with the required properties if and only if there exists a family A of subsets of X that satisfies the following conditions:
(1) JAS=worA=OforallAEA; (2) A is closed under finite intersections;
(3) for every pair a, b c A, a # b, there exist A, B E A such that a E A, (4) if A' C A and A E A\ {O}, then (UA') fl A = O implies I U A'I < w. Indeed, if A has properties (1)-(4), then X endowed with the topology generated by the basis A satisfies the requirements of the problem. Conversely, if r is a topology required by the problem, then the family A of all countable T-open sets satisfies the conditions (1)-(4). Now choose X to be the set wl of countable ordinals. Let {(b0,, c,,)
a E wl } be a sequence of all pairs of elements of wl with b0 # c,,,. By transfinite recursion, we construct a sequence of families A,,, : a E wl of subsets of wl so that for all a E wl, we have (10,) JAI=w orA=O for allAeA0,, (20,) UQ.<0,Ap C A.,
(30,) A, is closed under finite intersections, (40,) IA0I < w,
(50,) there exist B, C E A, with bc. E B, c0, c C, and B fl C = 0, (60,) if A E A, \ Up<0,Ap and A' E Up<0,A0, then JA fl A'I = w. Assuming that we have constructed such a sequence, we show that A = U0,<,IA0, satisfies conditions (1)-(4). (1), (2), and (3) are obvious. In order to check (4), let A' C A and A E A \ {0}, with (UA') fl A = 0. Let
a E wl be the smallest ordinal such that A E Ate. We shall prove that UA' c UA0,. To this end, it is sufficient to see that for each A' E A', there
exists B E A,, with A' C B. Let A' E A' be arbitrary, and let Q E wl denote the smallest ordinal B E Ap such that B D A' and B fl A = 0. (The definition of /3 is correct since there exists B E A with B D A' and B n A = 0; for example, B = A' is such.) Then Q > a would contradict
(60,), so3
Lemma. If JEl = w and E is a nonempty, countable family of subsets
of E, then there exist subsets B and C of E such that B fl C = 0 and
IBfAI=ICflAl=w for allAEe. Proof. Indeed, put e = {An : n E w} (possibly with multiple appearances), and for n c w, let An = UkEWAnk be a partition of An, into pairwise disjoint infinite sets. Arrange the sets Ask (n, k c w) into a sequence {A' : n c w}. Since each A' is infinite, by recursion we can define the elements bn,, cam, E w such that bn,, cn, E A' \ ({bk : k < n} U {ck : k < n}) for
all n c w. Then B = {bn : n E w} and C = {cn, : n E w} are as required.
Turning now to the construction of the sequence {A,, : a < wl }, let B0, Co C wl be two arbitrary countably infinite sets such that bo E Bo,
3. SOLUTIONS TO THE PROBLEMS
542
co E Co and Bo n Co = 0. Then Ao = {Bo, Co, 0} satisfies conditions (l0)-(6o). Suppose that for some ordinal a E w1, a > 0, we already have defined the sequence {A0 :,3 < a} so that conditions (1Q)-(60) are fulfilled
for all 0 < a. Then, by applying the lemma to E = U0
Solution 2. Let D be a countable, dense subspace of the product space 2"1, which exists by the Marczewski-Pondiczery theorem. Let F = { fa
aEw1}C2"1 \D such that f, I (wl \ 'Y)
fR I (wl \ 7)
and f I (wl \ 'Y)
fa I (wl \ 7)
for all a, Q, y E wl, a 5 4,3, and f E D. Choose D U F to be the underlying set of the space X to be defined; let D be open in X, and let its subspace topology coincide with the topology inherited from 2"1. It remains to define basis neighborhoods for fa (a E w1). To this end, let B = {B,,, : a E wl } be the family of elementary open sets in the product
space 2", let the finite sets Ta (a E w1) be the supports of the Ba, and put A. = sup(Uj< T0) + 1. Then let the neighborhoods of fc. be the sets of the form If, I U (B n D), where B ranges over the elements of B that contain f0, and are supported in w1 \ A. It is easy to check that we have defined a Hausdorff topology. By the construction, for all B,,, E B, we have
clx(BanD)D{f,3 :i3Ew1\(a+1)}, and thus the closure of any nonempty, open set in X has a countable complement.
Remark. It can be shown that the maximum cardinality of spaces satisfying the hypotheses of the problem is 210.
Problem T.15. Let W be a dense, open subset of the real line R. Show that the following two statements are equivalent:
(1) Every function f : R --+ R continuous at all points of R \ W and nondecreasing on every open interval contained in W is nondecreasing on the whole (2) R \ W is countable.
Solution. Suppose first that F = R\W is uncountable. Then we construct a continuous function f : ][8 -* R that is constant on every subinterval of W and that is nonconstant and decreasing on the whole We may assume that F has no isolated points (otherwise we pass to the set of its points of condensation). Let W = UV, where V is a countable
3.10 TOPOLOGY
543
family of pairwise disjoint, open intervals. Consider the natural ordering of V inherited from R. Since F is nowhere dense and perfect, this ordering of V is dense (that is, between any two elements of V, there exists a further element of V); therefore, there exists an order-preserving map q --+ VQ from
the set Q of rational points of the interval [0, 1] into V. For x E R, put
f(x)=1-sup{qEQ: (-oc,x]f1V954 0}. The function f is obviously constant on elements of V, f (x) = 1 for x E Vo, f (x) = 0 for x E Vi, and f is monotone nonincreasing. Moreover, f takes all values t E [0,1], since f (x) = t for x = sup U{V. : q E Q fl [0,1 - t]}. Therefore, f has no jumps and f is continuous.
Suppose now that JR \ W = {an : n E N} is countable. Let e > 0 be arbitrary; we show that for any pair x, y E R, x 54 y, we have f (x) < f(y) + E.
Since f is continuous at the points an (n E N), for each n E N there exists an open neighborhood Un of an such that for all u, v E Un the inequality f (u) - f (v) < e/2n+1 holds. Let W = UV, where V is a countable family of open intervals such that the restriction of f to any element of V is nondecreasing. The family U = V U {Un : n E N} is an open cover of the real line R; consider a finite subfamily U' = V U {Un : n < no} of U such that V C V and U' covers the interval [x, y]. Let L be the set of points y' in [x, y] for which there exists a finite sequence x = zo < zi < . . < zk = y' such that [zi_1, zi] C U(i) with some U(i) E U' for all 1 < i < k. It is easy
to see that L is an open-and-closed subset of [x, y] containing x, and so L = [x, y]. Therefore, there exists such a sequence z 0 , . . . , zk with zk = y. Then no
k
(f ('zi-1) - f (zi)) < E
P X) - f (y) = i=1
9n+1
< e.
n=O
Remark. Statement (2)x(1) of the problem can also be proved by transfinite induction on isolated points of R \ W. In connection with a few incorrect solutions, it seems worth noting that such an induction does not necessarily terminate at the first infinite ordinal.
Problem T.16. Let n > 2 be an integer, and let X be a connected Hausdorff space such that every point of X has a neighborhood homeomorphic to the Euclidean space Rn. Suppose that any discrete (not necessarily closed) subspace D of X can be covered by a family of pairwise disjoint, open sets of X so that each of these open sets contains precisely one element of D. Prove that X is a union of at most R1 compact subspaces.
Solution. Let X be a space that satisfies the hypotheses of the problem. First, we prove the following.
544
3. SOLUTIONS TO THE PROBLEMS
Statement (*). If F is an arbitrary subspace of X, then there exists a family G of open subsets of X homeomorphic to Rn such that (ug) n F is dense in the subspace F, and the family G is o-disjoint, that is, g is a union of countably many families consisting of pairwise disjoint sets.
If F = 0, then this statement is obvious. If F 54 0, then consider a maximal family U of pairwise disjoint, nonempty, separable, relative open
subsets of the subspace F. (Such a family exists by Zorn's lemma.) For each U E U, let S(U) = {xn,(U) : n = 0, 1.... } be a countable, dense subset of U. For each n, the set S,,, _ {xn(U) : U E U} is a discrete subspace of F and, thus, of X. By the hypothesis of the problem, there exists a family Gn = {Gn(U) : U E U} of pairwise disjoint, open subsets of X, such that xn(U) E Gn(U) for all U E U. Since X is locally Euclidean, we may assume that Gn consists of sets homeomorphic to Rn. Then the family G = U°°_oGn satisfies the requirements in (*), since the closure of
(ug) n F i U°°_0Sn in the subspace F contains the closure of UU in F, and, by maximality of U, the set UU is dense in F. We define a sequence { (Fe, G,,) : a E w1 } by transfinite recursion as follows. (As usual, w1 denotes the set of all countable ordinals.) Let Fo =
X and go be a maximal family of pairwise disjoint, open subsets of X homeomorphic to Rn. If a E wl, a > 0 and {(FQ,90) : Q < a} has already been defined, then put Fa = X \ Ups,,(UGp), and let G,, be a v-disjoint family of open sets of X homeomorphic to Rn such that (UGa) nF,, is dense in Fa. (The existence of such a family G,,, is guaranteed by (*).)
We show that X = U,1 (UG,,). Arguing by contradiction, assume that there exists a point x E X \ U-E- (UGa), and consider a neighborhood V of x that is homeomorphic to Rn. Since (UGa) n F,, is dense in Fa _ X \ Up
homeomorphic to Rn, and G* is a union of at most l1 disjoint families. Since a subspace homeomorphic to Rn can only meet countably many disjoint open sets, each G E G* meets at most l1 members of G*. Finally, consider the equivalence relation on g* in which G, G' E G* are equivalent if and only if there exists a finite sequence Go, ... , Gk E G* such
that Go = G, Gk = G', and Gi n Gi+l # 0 (i = 0,1, ... , k - 1). Using the result of the last paragraph, induction on k shows that every equivalence class contains at most R1 sets from g*. On the other hand, if G C G* is an equivalence class, then UG is open and closed in X, and then g = g* by connectedness of X. Therefore, the cardinality of g* is at most 1Z1.
Thus, X is a union of at most R1 subsets homeomorphic to Rn, and therefore it is a union of at most l1 compact sets.
3.10 TOPOLOGY
545
Remarks. 1. The most well-known nonmetrizable topological manifold, the long line, is an example of a space that satisfies the hypotheses of the problem but is not a union of count ably many compact subspaces. 2. Several contestants proved the statement of the problem using the assumption that all separable subspaces of X are second-countable. Gabor Moussong showed that there exists a model of ZFC where this assump-
tion holds. Note, however, that by a construction of M. E. Rudin and P. H. Zenor in some other models of ZFC there exist separable and not second-countable topological manifolds that satisfy the hypotheses of the problem. 3. Gabor Moussong noticed that the statement of the problem follows from the continuum hypothesis, since the cardinality of a connected topological manifold of dimension > 1 is continuum.
Problem T.17. A map F : P(X) -> P(X), where P(X) denotes the set of all subsets of X, is called a closure operation on X if for arbitrary A, B C X, the following conditions hold: (i) A C F(A);
(ii) A C B = F(A) C F(B); (iii) F(F(A)) = F(A). The cardinal number min{ CAI : A C X, F(A) = X} is called the density ofF and is denoted by d(F). A set H C X is called discrete with respect
to F if u
F(H - {u}) holds for all u E H. Prove that if the density of the closure operation F is a singular cardinal number, then for any nonnegative integer n, there exists a set of size n that is discrete with respect to F. Show that the statement is not true when the existence of an infinite discrete subset is required, even if F is the closure operation of a topological space satisfying the T1 separation axiom.
Solution. (a) We prove the statement by induction on n. For n = 0, the statement is obvious; assume that it is true for n. Let the density of the closure
operation F on X be A with cf (A) = n < A. Let JAI = A, with F(A) = X, and well-order A in order type A. By recursion, define a sequence B = {x : < Al by x£ = min (A - F ({x,, : q < l;})). Then F(B) D F(A) = X. Let C C B such that ICI = , and C is cofinal in B, and for Y C B put F'(Y) = F(Y U C) fl B. It is easy to see that F' is a closure operation on B and that d(F') _ A. By the induction hypothesis, there exists a set {x£; : i < n} that is discrete with respect to F'. Let Cn > i (i < n) be such that xC E C. Then the set {xC, i < n} is discrete since xC V F ({x£; : i < n}) C F ({xn : q < En}), and x£i V F ({x£, : i # j, j < n}) C :
F'({xC3 :i# j, j
546
3. SOLUTIONS TO THE PROBLEMS
exists by Zorn's lemma.) Call F C A closed if, for each H E N, IF n HI = w implies F D H. It is easy to see that by this a T1 topological space is defined in which there are no infinite discrete subsets. It remains to prove that the density is A. Let A0 C A, with IAoI < A. For < wl, we define a sequence A by the following recursion: if EN:IHnA,.I =w}, and if is a limit =71+1, then ordinal, then put A£ = U{Aq :77 < }. Since IA,7+11 _< IAqIw, it is easy to see that IACI < IAoIw < A for < wl. On the other hand, A,,,, is obviously closed, and therefore A0 is not dense.
Problem T.18. Suppose that K is a compact Hausdorff space and K = U'--OA,,, where An is metrizable and An C A,,,, for n < m. Prove that K is metrizable. Solution. First, we prove the following lemma. Lemma. If a subspace H of a metric space A is not separable, then there exists an uncountable, discrete, closed subspace Z C H.
Proof. Let Zn be a maximal subset of A such that d(x, y) > 1/n for all x, y E Zn, x # y. If a E A, then d(a, x) < 1/2n can hold for at most one point of Zn, so Zn is discrete. On the other hand, by maximality of Zn, for each h E H there exists x E Zn such that d(h, x) < 1/n, so U°_IZ, is dense. If H is not separable, then this implies that Zn is uncountable for some n. Now we prove that An is separable for all n. Indeed, assuming that it is not separable, there is an uncountable discrete closed subset Zn in An; this set is dicrete in An+1 and is not separable, so there exists an uncountable,
discrete, closed subset Zn+l in Zn. Thus, we obtain a sequence Z,n such that Z,,,, 3 Z,,,, for ml < m2, and Z,n is an uncountable, discrete, closed subset of A,,,,. Since K is compact, the set Z;,, of accumulation points of Z,n is nonempty; therefore n°O=nZm 0. On the other hand, Z;,, nA,n = 0 since Z.. is discrete and closed in An, and therefore 0, and which is a contradiction. Fix k and consider the set A n n Ak (k < n), which is a separable metric space and therefore has a countable basis {Bn,,,, : m E w}. Let Gn,m be an open set in Ak such that Bn,m = Gn,m n An. The family {Gn,m : n > k, m E w} is a subbasis for a Hausdorff topology on Ak, which is coarser than the compact topology of Ak; therefore, these two topologies are equal. This means that Ak is a second-countable, compact topological space. Let {fn,k : n E w} be a family of continuous functions on Ak that separates the points. Let Fn,k be a continuous function on K for which Fn,klAk = fn,k
(Such functions Fn,k exist by the Tietze theorem.) The functions Fn,k separate the points of K; therefore the metric defined by these functions induces the topology of K.
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Problem T.19. Let U denote the set If E C[0,1] : If (x) I < 1 for all x E [0, 1]}. Prove that there is no topology on C[0,1] that, together with the linear structure of C[0, 1], makes C[0,1] into a topological vector space in which the set U is compact. Solution. We assume that topological vector spaces are Hausdorff spaces. Arguing by contradiction, assume that C[0,1] is a topological vector space in which U is compact. Then U is closed since the topology is Hausdorff. For each f E C[0, 1] and 6 > 0, the set Uf,6 = {g E C[0,1] : lg(x) - f (x) 1 < b for all x E [0, 1]}
is also compact and closed. Define the functions fn, gn E C[0,1] for each natural number n > 2 as
if0<x<2-n,
10 f,, (x) =
nx + 1 - 2
if 2 - n < x <
1
if1<x<1,
10 gn(x) = nx - 2 1
2,
if0<x< 2, if 2 < x < 2 + n, if
2+n
<x<1.
Put Kn = {g E C[0,1] : gn(x) < g(x) < fn(x) for all x E C[0,1]} (n > 2). fl Then K2 D K3 3 , and Kn is compact and closed since Kn = Ugn_1,1. So nn'=2Kn # 0. Let f E nn'=_2Kn. Then f (x) = 0 for x < 1/2, and f (x) = 1 for x > 1/2, which is impossible, since f is continuous. This contradiction shows that U cannot be compact. Problem T.20. Let 4 be a family of real functions defined on a set X such that k o h E 4 whenever fi c 4 (i E I) and h : X - 1[81 is defined by the formula h(x)i = fi(x), and (1) k: h(X) ---> 1[8 is continuous with respect to the topology inherited from the product topology of R1. Show that f = sup{gj : j E J, gj E implies f E 4. Does this statement m E M, hn,, E remain true if (1) is replaced with the following condition? (2) k: h(X) -> 1[8 is continuous on the closure of h(X) in the product topology.
Solution. We may obviously assume that J and M are disjoint. Define p : X -+ JJUM by p(x)j = gj (x), for j E J, and p(x),,+. = h,,, (x), form E M. Then, for ally E p(X), we have sup{yj : j E J} = inf{y,n : m E M}. Define k : p(X) -> JR by k(y) = sup{yj : j E J} = inf{ym : m E M}. Then f = k o p, so, in view of (1), it suffices to show that k is continuous on p(X). Let y E p(X) and E > 0. We shall define a neighborhood S of y in
p(X) such that Ik(z) - k(y)l < E for all z E S. Choose jo c J with
548
3. SOLUTIONS TO THE PROBLEMS
gjo(y) > k(y) - (e/2) = sup{gj(y) : j E J} - (E/2), and choose mo E M with h,,,.o(y) < k(y) + (E/2) = inf{hm(y) : m c M} + (e/2). Such jo and mo exist by the definition of supremum and infimum. Put
S = {z ep(X) : xjo -Yjo1 < 2 and Izmo - ymoI < 2 Then, for z E S, we have
k(z)=sup{zj:jEJ}>zjo>yjo-2>k(y)-e and
that is, k(z) - k(y)I < E.
Therefore, for every y E p(X) and e > 0, there exists a neighborhood S
of y such that Ik(z) - k(y)I < e for all z E S, that is, k is continuous on p(X). This solves the first part of the problem. If condition (1) is replaced with (2), then the statement is false. Indeed, let X = N be the set of natural numbers, and let 4) consist of all functions f : N -p ]R such that the sequence { f (n)}n 1 is convergent. We claim that this is a counterexample. Suppose that fi E 4) (i E I), that h : N R is defined by h(n)i = fi(n)
(i E I), and that k : h(X) -> R is continuous. Let b be the element of R1 such that bi = lim fi. Then, by the definition of h, h(n)i -+ bi for all i E I; therefore, h(n) -4 b in the product topology. This implies that b is contained in the closure of h(X) since it is the limit of a sequence in h(X). By continuity of k on h(X), we have that k(h(n)) -+ k(limh(n)) = k(b), that is, the sequence k(h(n)) is convergent, and therefore k o h E This shows that the hypotheses of the problem are satisfied. Let J = M = N; let gj (n) = 1, for n < j and n even, gj (n) = 0 otherwise; and let hm(n) = 0, for n < m and n odd, hm(n) = 1 otherwise. These functions are indeed in (P, since gj(n) --+ 0 and hm(n) -+ 1 for all j E J and
m E M. On the other hand, sup{gj(n) : j E J} = inf{h,.. (n) : m E M} equals 1 if n is even and equals 0 if n is odd. This sequence is not convergent;
therefore f = sup{gj(n) : j E J} = inf{h,,,(n) : m E M} does not belong to
Problem T.21. Characterize the sets A C R for which
A+B={a+b:aeA, bEB} is nowhere-dense whenever B C R is a nowhere dense set.
Solution. We prove that the sets with this property are the bounded sets with countable closure. More precisely, we show that for a set A C R the following are equivalent:
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549
(i) If B C R is nowhere dense, then so is A + B; (ii) For any sequence an (n = 1, 2, ...) of positive numbers, there exists a finite family of intervals II (j = 1, ... , k) that covers A and such that the length of Ij is II, I = aj (j = 1, 2, ... , k); (iii) A is bounded and its closure is countable. (i) = (ii): Suppose that (ii) does not hold. Then there exists a sequence consisting of positive numbers an (n = 1, 2, ...) such that for any finite family of intervals Ij, jIj I = aj (j = 1,2,... , k) we have A ¢ Uj=1Ij. Using this, we shall construct a nowhere-dense set B such that A + B contains all rational numbers, which contradicts (i). Arrange the rational numbers into a sequence rn (n = 1, 2.... ). Now we define the numbers xn and yn by induction on n. Let x1 E A be arbitrary, and y1 = r1 - x1. Suppose that k > 1 and the numbers y1, y2, , yk-1 have already been defined. Then, by our assumption, the intervals
h- (rk - yJ
2
,
rk - yJ +
2I' j
=1,2,...,k-1
do not cover A. Therefore, we can choose an element xk E A \ Finally, put Yk = rk - xk. Then, with B = {yk : k E N}, the set A + B contains all rational numbers since rk = xk + yk E A + B. We show that
every point of B is isolated. Indeed, for j < k, we have xj (rk - yj aj /2, rk - yj + aj /2). Therefore, yk = rk - xk V (yj - a3/2, yj + a3/2). This means that the set B contains at most j points in the neighborhood of radius aj/2 of the point yj, and thus yj is an isolated point of B. This implies that B is nowhere dense. (ii) (iii): If (ii) is true; then A is obviously bounded. Arguing by contradiction, suppose that (iii) does not hold. Then the closure of A is
uncountable and, by the Cantor-Bendixson theorem, contains a nonempty,
bounded, and perfect set P. Let us call a sequence (a1i a2, ...) of positive numbers insufficient if for any family of closed intervals Ij, I3 I = a3 (j = 1, 2, ... , k), we have P ¢ In order to arrive at a contradiction, by induction we shall define an infinite sequence of positive numbers (a1i a2, ...) whose all finite initial segments are insufficient. Let a1 be a
positive number that is smaller than the diameter of P; then (a1) is obviously insufficient. Suppose that n > 1 and we have already defined an insufficient sequence (a1i a2.... , an). We show that there exists a number an+1 such that (a1i a2, ... , an+1) is insufficient. Suppose this is not true; then for every k E N and an+1 = 1/k there exists a sequence of intervals Ij,k, JIj,k = aj (j = 1, 2, ... , n+ 1) such that P C U +1 Ij,k holds. We may assume that the intervals Ii,k all meet the set P, and thus they are all contained in a bounded set. By passing to a suitable subsequence, we may also assume that for each fixed j = 1, 2, ... , n + 1, the sequence of intervals Ij,k converges to an interval I j when k --f oo. Then II; = aj, for j = 1, 2, ... , n, and In+1 consists of a single point. Since (a1, a2, ... , an) is insufficient, the set P is not contained in U 1I3 . Moreover, being perfect, P has infinitely many points in the complement of U 1II. So we can choose a point
550
3. SOLUTIONS TO THE PROBLEMS
x E P \ U±i
But then, for sufficiently large k, x 0 U +1 Ij,k, which . contradicts the choice of the intervals Ii,k. This proves that the sequence (ai, a2i ... ,1/k) is insufficient, completing the induction. (iii) = (i): Suppose that, to the contrary of (i), with a nowhere-dense set B, the set A + B is dense in some interval I. Let A0 denote the closure of A; then A0 + B is also dense in I. Define the sets A,,, for all countable ordinals as follows: If /3 > 0 is an ordinal such that A,, has already been defined for all a < 3, then let Ap = n,<,3A0, if 0 is a limit ordinal, and
let A0 be the set of accumulation points of A,, if 0 = a + 1. Since A0 is countable, there exists a countable ordinal ry such that A., = 0. Then obviously A.1 + B = 0 and A., + B is not dense in I. Let ,3 denote the smallest ordinal for which Ap + B is not dense in I; then ,3 > 0. Let J be a subinterval of I with (A0 + B) n J = 0, and let 0 < E < IJ1/3. Let U(H, S) denote the neighborhood of radius S of the set H. It is easy to see that U(H, S) + B C U(H + B, S) holds for every set H and b > 0. Thus, U(A0, e) + B C U(A0 + B, E), and therefore U(A0, E) + B does not meet the middle third K of J. Suppose that 0 is a limit ordinal. Since {A,,, : a < 3} is a nested sequence
of compact sets whose intersection is A0, there exists a < 0 with A, C U(A0, E). Then A,, + B does not meet K. But this is impossible, since a < ,3 and A,,, + B is dense in I. Finally, suppose that 3 = a + 1. Then V = A, \ U(Ao, E) is a finite set, since all accumulation points of A,, are in A0. Then the set V + B is nowhere dense. Therefore, from (A,, + B) fl K = (V + B) fl K, it follows that (A, + B) fl K is nowhere dense. But again this contradicts that a < /3 and that A,, + B is dense in I. This completes the proof. Problem T.22. Prove that if all subspaces of a Hausdorff space X are or-compact, then X is countable.
Solution. Since X itself is a union of countably many compact sets, it is sufficient to prove the statement for compact spaces X. Suppose that X has no subset that is dense in itself. Then there is a well-ordering on X in which all initial segments are open. Indeed, suppose that the points pp E X have already been defined for all ,3 < a and that
Xa = {pp : /3 < a} # X. Then there exists a point pa E X \ X0. that is not an accumulation point of X \ X0,, and so p0. E G. C_ X. U {p.} with a suitable open neighborhood G0, of p,,. This recursion defines a wellordering of X, and UQ<0,GQ C X. = U0<0,{p0} C_ Up<0,Gp shows that X4 = Up<0.Gp is an open set for each a. An initial segment of order type
wi in X cannot be or-compact since a compact subset in X must have a greatest element. Therefore, X is countable. Thus, it suffices to prove that X does not contain a subset dense in itself. Arguing by contradiction, assume that there exists such a subset Y. Then Y is dense in itself and closed. We define a continuous function f : Y -+ [0, 1] as follows. Since Y is a compact Hausdorff space, for any two distinct points p and q there exist open sets Gp, Gq and closed sets FP,
3.10 TOPOLOGY
551
Fgsuch that pEGp9Fp,gEGg9Fq,andFpnFq=0. So, for all finite 0-1 sequences lei}, we can define closed sets AE1i... En C Y so that AEI,...,e.,o n AEI,...,En,l =0
and
AE1....,e, ,O U AEl ...,e, ,l C A,, ---,En
hold for all natural numbers n and el, ... , en E 10, 11. For any infinite 0-1 sequence e1,E2,...,the set 00
A(El, E2....) = n A,
62,--En
n=1
is nonempty and closed. If x E A(E1, E2.... ),then put °O
f (x)
1
Ei 2i
i=1
The function f maps A(El, E2, ... ), which is a closed set since it is an intersection of closed sets, continuously onto the interval [0, 1]. Then f extends continuously over V with Im f = [0, 1]. The interval [0, 1] contains 2x° or-compact sets (that is, Fe-sets), since the number of closed sets in [0, 1] is 2x°. Therefore, there exists a set Z C [0, 1] that is not or-compact; then its inverse image f -1(Z) C V cannot be a-compact, which contradicts the hypothesis of the problem. This proves that X contains no subset that is dense in itself.
3. SOLUTIONS TO THE PROBLEMS
552
3.11 SET THEORY Problem K.1. Does there exist a function f (x, y) of two real variables that takes natural numbers as its values and for which f (x, y) = f (y, z) implies x = y = z ?
Solution 1. First, we show that it is sufficient to construct for each real number c, two sets Ac and Bc of natural numbers that satisfy the following conditions: (1) 1 V Ac, 1 V Bc for all c,
(2) A,nB,=O forallc, (3) A.nBd54 0forc#d. Indeed, put f (c, c) = 1, and for c # d, let f (c, d) be an arbitrary element of the nonempty set Ac n Bd. Now, if
f(x,y)=f(y,z)=m01, then
mEAxnBy
and
mEAynB,,
that is,
mEA,nB.nAynBZ cAynBy=O, which is a contradiction. But m = 1 is only possible when x = y = z.
In order to construct Ac and B,, map the set R of rational numbers bijectively onto the set N = 12,3.... }; let n(r) denote the image of r E R. Assign to each real number c a sequence {rk(c)}- 1 that converges to c and consists of rational numbers different from c. Let A, = {n(rk(c)) : k = 1, 2, ... } c N,
B,=N-Ac. It is easy to see that these sets Ac and Bc satisfy conditions (1), (2), and (3). Thus, we have constructed a function with the required property.
Solution 2. Arrange the set of rational numbers into a sequence r1, r2, . Define the function f (x, y) the following way: If x = y, put f (x, y) = 1; If x < y, put f (x, y) = 2i, where i is smallest possible with x < ri < y; If x > y, put f (x, y) = 2i+1, where i is smallest possible with x > ri > y. We show that f (x, y) is as required. It is obviously defined on the reals, and its range is the set of natural numbers. Suppose that
f(x,y)=f(y,z)=m. If m = 1, then x = y = z is the only possibility. If m > 1 and m = 2i, then
x
and
y
3.11 SET THEORY
553
which is a contradiction. Similarly, if m > 1 and m = 2i + 1, then
x>ri>y
and
y>ri>z,
which is again a contradiction. Therefore, x = y = z follows.
Solution 3. We will decompose the plane into a countable family of pairwise disjoint subsets with the following property: if one of the lines parallel to the coordinate axes through the point (y, y) intersects one of the subsets, then the other does not, except when the subset is the line x = y, which will be one of the subsets. It suffices to construct such a partition of the open half-plane below the line x = y, because by reflecting in the line x = y and adding the line x = y, we obtain the required partition of the whole plane.
Put
Ao={(x,y):x>0, y<0},
Ak={(x,y):x>k, k-1
An = { (x, y) : An k =
{
2k-1 2n
2k
< x < 2n ,
2k-2 2n
-2k+1 <X< -2k+2 )n , 2n
2k-1 2n
,
-2k -2k + 11 2n < y C 2n j
where k, n = 1, 2, .... It is obvious that the union of these sets is the open half-plane below the line x = y. This defines a partition of the whole plane, as we said above. Label the sets of this partition by the natural numbers, and assign to the point (x, y) the label of the set containing (x, y) as the value of the function f (x, y). If x, y, z are real numbers with f (x, y) = f (y, z), then the points (x, y) and (y, z) are in the same subsets, but then (x, y) (and (y, z)) is on the line through the point (y, y) parallel to the x-axis (y-axis, respectively), and any of the subsets can only meet one of these, except for the subset
{(x,y):x=y}, but then x=y=z. Remarks.
1. Obviously, the existence of such a function is a question of cardinality. In general, it makes sense to ask whether there exists a function f (x, y)
defined on a set A of cardinality m that takes its values in a set B of cardinality n (both m and n are infinite) and for which f (x, y) = f (y, z) implies x = y = z. J. Gerlits, L. Lovasz, L. Posa, and M. Simonovits proved that such a function exists if and only if m < 2°. Here we present the proof by Posa.
3. SOLUTIONS TO THE PROBLEMS
554
Let m < 2". It suffices to give the construction for a set A of cardinality 2". Let a be the smallest ordinal of cardinality n. We may assume that A is the set of sequences of the numbers 0 and 1 of order type a. Define the function f (x, y) as follows:
If x = y, put f (x, y) =0; If x # y, then let 13 be the smallest ordinal where the elements in x and y are different, let j be the 13th element of x (j = 0 or 1), and put f (x, y) = (13, j). The range off is the set of the symbol 0 and the ordered pairs (13, j), where 13 < a and j = 0 or 1. Therefore, it has cardinality n and can be mapped bijectively onto B. f (x, y) = f (y, z) = (13, j) means that all elements before 13 agree both
in x and y, and in y and z, and the 13th element is j in x and y. But then the 13th elements of x and y are equal, which is a contradiction. Therefore, f (x, y) = f (y, z) = 0, which implies that x = y = z. (This part of the statement is also proved by Bollobas and Juhasz.)
Suppose that m > 2", and there exists a function required in the problem. Consider the range of f (x, y) when x is kept fixed. This is a subset of B, and since the cardinality of all possible x's is greater than the cardinality of the set of all subsets of B, for some x1 $ x2 the range of f (xii y) equals that of f (x2, y). So, the value f (x1, x2) is contained in the range of f (x2i y), that is, for some yo, f (x1, x2) = f (x2i yo), which is a contradiction. 2. Bela Bollobas and Miklos Simonovits notice that there exists a function F(xl,... , xn) defined on the real numbers that takes its values in the natural numbers, and for which F(a, x2i ... , xn) = F(yi, a, y3, ... , yn) = ... = F(u1,... , un-1, a)
implies a = xi = yi =
= ui (i = 1, 2, ... , n). For example, n
F(x1i... , xn) _ 11
pj(ijxi,xi)
i,j=1
is such a function, where the pij (i, j = 1, ... , n) are different primes, and f (x, y) is the function defined in either solution of the problem.
Problem 1t.2. Prove that there exists an ordered set in which every uncountable subset contains an uncountable, well-ordered subset and that cannot be represented as a union of a countable family of well-ordered subsets. Solution. Let R be the set of all limit ordinals less than w1i and for a E R, let fa(n) be a monotone increasing sequence (of order type w) of ordinals converging to a. Order R as follows: for a,13 E R, a -< 13 if f, (n) < fp(n) holds for the first n with fa(n) 54 f1i(n).
3.11 SET THEORY
555
Any uncountable subset X of R contains a well-ordered subset of order type w1 with respect to the ordering -. Indeed, let
XIn = { (ao, al, ... , an) : there exists a E X, such that f, ,(m) = a(m) for all m < n}. For n sufficiently large, X In is uncountable, since for large enough n the ordinal
ryn=sup{fa(n): aEX} equals w1, since, by f,, (n) - a,
sup{ryn:n<w}=sup{a: aEX}=w1. Order XIn lexicographically, that is,
(ao,...,an) -<' (00,
On)
if am <,3m for the smallest m with am # /3m. Then X In is well-ordered and, if n is large enough for X In to be uncountable, it contains a wellordered subset of order type w1. Let X' be a subset of X such that for any
(ao,...,an) E XIn, there exists a unique a E X' with f.(0) = ao,...,fa(n) = an.
Then the ordered set (X', -<) is isorr. )rphic to (X In, -<'). Since we have seen that the latter contains a well-ordered subset of order type w1i the same is true of the former. This is what we wanted to prove. Now we show that R is not a union of a countable family of well-ordered subsets. First we make a digression. A function that maps a set of ordinals into the ordinals is called regressive if f holds for all ordinals # 0 in the domain of f . If the domain of f is a subset of w1 and the set
{ : f() < Al is not cofinal with Wl for any p < w1, then f is called divergent. X C wl is called thin if there exists a regressive divergent function defined on X; otherwise, it is called stationary. The set wl is stationary. Indeed, assuming that f is a regressive function on w1, the sequence
S'f(01 f(f(0),... , being a descending sequence of ordinals, can contain only a finite number of elements different from zero. Therefore,
W1 = U Xn n<w
556
3. SOLUTIONS TO THE PROBLEMS
where
Xo = {0}, and
X.+1
= {e : f (e) E Xn}. Assuming further that f is divergent, induction on n shows that sup Xn < w1 for all n, which obviously is a contradiction. From this it easily follows that the set R of limit ordinals less than w1 is also stationary. It quickly follows from the definition that the union of a countable family
of thin sets is also thin. Further, if X is stationary and f is a regressive function on X, then for some p < w1 the set
is stationary (this is a special case of Fodor's theorem). Indeed, assume that all Xµ are thin, and let fµ be a regressive divergent function defined on X. Then the function g defined on X by max(p, f,, (C))
(C E Xµ, p < wi)
is regressive. It is divergent, too, because for all v < w1 the set
{C:g(c)
is not cofinal with wl, since neither one of the summands in the righthand side is, because the functions fµ are divergent. This contradicts the assumption that X is stationary, which proves our statement. Let X C R be stationary. We show that X is not well ordered with respect to the ordering -<. Put Y_1 = X and
Yn={aEYn_1 : fa(n)=6n}, where 6n is the smallest ordinal for which Yn is stationary (such 6n exists by the Fodor theorem proved above).
Put
X'=X- UYn, n<W
where
Y,'={aEYn-1:fa(n)<6n}. By the choice of 6n, Y,' is not stationary, and neither is X - X'. Put 6 = sup{6n : n < w}.
It is obvious that 6 < w1. Let a E X' be greater than 6. (X' is stationary and so it is cofinal with w1.) Since fa(n) -p a, for some n < w we have f, (n) > 6 > 6n,
3.11 SET THEORY
557
so f, (m) = 6,,,, cannot hold for all m. It follows from a E X' that fa (m) >
8,,, for the smallest m with f, (m) # 6m. Thus, for this m, a is greater than all elements of Y,,, with respect to the ordering -<. Put X0 = X, ao = a, and Xl = Y,,,,. Since Y,,,, is stationary, we can repeat the above argument with Xl instead of X0. We obtain a1 and X2,
then a2 and X3, and so on. Since ao >- a1 >- a2 >- ..., X is not wellordered with respect to the ordering -<, and this is what we wanted to prove.
Now, if R = Un< R,,, then some R is stationary, and this R, is not well-ordered with respect to -<. This proves the second statement of the problem.
Remark. Laszlo Babai showed that for each cardinal number is > w, there exists an ordered set B of cardinality is such that all subsets of cardinality r. contain a well-ordered subset of cardinality K and B is not a union of a family of cardinality less than K of well-ordered subsets.
Problem R3. Let < be a reflexive, antisymmetric relation on a finite set A. Show that this relation can be extended to an appropriate finite superset B of A such that < on B remains reflexive, antisymmetric, and any two elements of B have a least upper bound as well as a greatest lower bound. (The relation < is extended to B if for x, y E A, x < y holds in A if and only if it holds in B.)
Solution. B will consist of all the elements and subsets of A without identifying the elements with the corresponding singletons. In what follows, lowercase letters denote elements, whine capitals mean subsets. Define < as follows:
a < b,
as given,
a
Problem 1t.4. Let .F be a family of subsets of a ground set X such that UFE.FF = X, and (a) if A, B E.F, then A U B C C for some C E .F;
(b) if A E F (n = 0,1, ... ), B E F, and Ao C Al C .... then, for some
k>0,A,,,nB=AknBforalln>k.
3. SOLUTIONS TO THE PROBLEMS
558
Show that there exist pairwise disjoint sets X., (7 E F), with X = U{X.y : 7 E F}, such that every X y is contained in some member of .F, and every element of.F is contained in the union of finitely many X.y's.
Solution. Let < be a well-ordering of F. Let F be the set of all finite subsets of F. Well order F as follows:
71,72EF,7154 72,71={A1i...,A,}, 72={B1,...IBm}1
An <... 72 if Ai = Bi for 1 < i < m. If, on the other hand, i < m is the least number with Ai 0 Bi then put 7i < 72 or 71 > 72 if Ai < Bi or Ai > Bi. A straightforward checking shows that this well orders F. We notice that 71 < 72 implies 71 < 72. For 7 E F, we define Y.y E F as follows. If y = {A}, put Y.y = A. Assume that Y6 is defined for b < 7. Then let Yy be a member of F, covering every Yp (,Q < 7). This is possible by (a). Now put X.y = Yy \ U6<.yY6. The only nontrivial thing to prove is that
every A E F is covered by finitely many Xp. We prove by transfinite induction on -y that Yy is covered by finitely many X6's. If this first fails
for 7, there exists /31 < 02 < ... < 7 such that Y.y n Xp # 0. Let Y y = {Al,... , Ak}, A ,>. . .> Ak. For every n there is a 1 < j < k such that
A1i...,Aj_1E/n,
A.7Oi3
.
We can assume, by shrinking, that j is the same for every n. Put ijn = max(01, ... , Qn) < 7. As 7 is minimal, Yp meets finitely many X. As Yn, C Yn2 C
..., by (b) there is an m such that Y,,m nYy = Ynm+, nYy = ....
Xnm+: n Yy # 0 (i = 0, 1, ...) that is, Y1m meets X'7_' X,7-+,, ... , a contradiction. so Y,,m
Ynm+: n Yy
Problem R.5. Show that there exists a tournament (T, -4) of cardinality N1 containing no transitive subtournament of size N1. (A structure
(T, -) is a tournament if - is a binary, irreflexive, asymmetric, and trichotomic relation. The tournament (T, -4) is transitive if - is transitive, that is, if it orders T.) Solution. We use the existence of,a Specker type, that is, an ordered set (A, <) of size ail, not containing subsets similar to w1, wi, any uncountable subset of the reals. See (P. Erdos, A. Hajnal, A. Mate, R. Rado, Combinatorial set theory: Partition Relations for Cardinals, Akademiai Kiado, Budapest, 1984, p. 326.) Enumerate A as {a(a) : a < w1 }, and let {x(a) : a < w1} be different real numbers in [0, 1]. If a < /3, put a +- ,Q iff either x(a) < x(/3) and a(a) < a(/3) or x(a) > x(/3) and a(a) > a(,3).
3.11 SET THEORY
559
Let B C A be uncountable. Let X = {a < w1: a(a) E B}. Claim 1. There is an a E X such that {/3 E X : a(a) < a(0)
and
x(a) < x(,3)}
is uncountable.
Proof. For a E B, let f (a) be the least t E [0,11 such that x(3) < t for all but countably many 3 E X, a(,3) > a. Since f is a non-increasing real-valued function on (A, <), it can only have countably many different values; otherwise, there would be an uncountable subset of A, similar to a set of reals. Hence f is constant on an uncountable B0 C B. Define X0 analogously to X. We can choose ao E X0 such that {,Q E Xo: x(/3) > x(ao)} is uncountable; otherwise (A, <) would contain a subset of type wi.
Now we can choose a E Xo such that a(a) > a(ao) and x(a) < x(ao). Since g(a(a)) = g(a(ao)) > x(a), there are uncountably many /3 E X such
that a(3) > a(a) and x(Q) > x(a).
Claim 2. There are uncountable X0, X1 C X such that if a E X0, Q E X1, then a(a) < a(,3) and x(a) < x(/3).
Proof. Let U be the set of those a E X such that {/3 E X : a(a) < a(0)
and x(a) < x(,0)}
is countable, and let L be the set of those a E X such that {,Q E X : a(,3) < a(a)
and x(,3) < x(a)}
is countable. If U or L is uncountable, we get a contradiction by Claim 1, so we can select a E X \ U \ L, and put Xo = {,Q E X : a(,3) < a(a)
X1 = {/3 E X : a(a) < a(/3)
and x(/3) < x(a)}, and x(a) < x(,3)} .
By an application of Claim 2 to X0 and the Specker type (A, >), one can find Yo, Y1 C_ Xo such that a E Yo, ,0 E Y1 imply a(a) < a(,3), and x(a) > x(Q). Now select a E Yo, 3 E X1, -y E Y1 with a < Q < y. Then a <- /3 +- y and -y +- a, so our tournament is not transitive on B. Remark. The result in the problem was first proved by R. Laver.
Problem R.6. Assume that R, a recursive, binary relation on N (the set of natural numbers), orders N into type w. Show that if f (n) is the nth element of this order, then f is not necessarily recursive.
Solution. We use the following well-known facts. There is a set A C N that is recursively enumerable, that is, the range of a recursive function, but
560
3. SOLUTIONS TO THE PROBLEMS
is not recursive, namely, its characteristic function is not recursive. For every infinite r.e. set A there is a recursive function enumerating A's elements in a one-to-one manner. An infinite set possesses an increasing such enumeration if and only if it is recursive. Take a (necessarily infinite) r.e. but not recursive set A, and a function g enumerating A without repetition. Put aRb if g(a) < g(b). Clearly, R is recursive and orders N into type w. If f (n)=the nth element by this order, and f is recursive, then the recursive h(n) = g(f (n)) would enumerate A's elements in increasing order, which is impossible.
Problem R.7. Let H be the class of all graphs with at most 2K0 vertices not containing a complete subgraph of size 1Z1. Show that there is no graph
H E H such that every graph in H is a subgraph of H.
Solution. We have to show that if (V, G), a graph with IVI = 2s0, does not contain a complete graph on R1 vertices, then there is a graph (W, H) with these properties such that (W, H) cannot be embedded into (V, G). Let W be the set of those functions injecting a countable ordinal (possibly 0 # 0) into a complete subgraph of G. To define H, join two such functions if one extends the other. Clearly, Ja->VI
JWJ= a<wl
First, we show that (W, H) does not contain a complete R1-gon. Assume
that If,,: a < wl } are pairwise joined, and that the ordinals Dom (fe) are in increasing order. By the construction of H, for,3 < a, f,, extends fQ, so the union of the fa's gives a function f : w1 --+ V onto a complete R1-gon in (V, G), a contradiction. Next we prove that (W, H) may not be embedded into (V, G). Assume that h: W -4V is such an embedding. By transfinite recursion on a < w1, we define fa : a -4V, fc. E W, and x,, = h(fa) E V in such a way that fa extends fp for i < a. Put fo = 0, fo E W. If fp (i3 < a) are defined, then
{xo:,3
the requirements. But then {x,,: a < W11 is a complete subgraph in G, a contradiction. Remark. This is an unpublished result of R. Laver. Problem N.8. For which cardinalities ,c do antimetric spaces of cardinality is exist? (X, o) is called an antimetric space if X is a nonempty set, o : X2 -+ [0, oo) is a symmetric map, o(x, y) = 0 holds iff x = y, and for any three-element subset {a1, a2, a3} of X
o(alf,a2f) + o(a2f,a3f) < o(aif,a3f) holds for some permutation f of {1, 2, 3}.
3.11 SET THEORY
561
Solution. For 0 < ,c < 2H0. First, if X C_ R is nonempty, then (X, o) is antimetric, where o(x, y) = (x - y)2 and clearly I X I can take any value in the given interval. For the other direction, let (X, o) be antimetric and IXI > 2K0. Color the complete graph on X by the integers as follows. Let {x, y} get color k if 21c < o(x, y) < 2k+1 By the Erdos-Rado theorem (see P. Erdds, A. Hajnal, A. Mate, R. Rado, Combinatorial Set Theory: Partition Relations for
Cardinals, Akademiai Kiado, Budapest, 1984, p. 98), there exist x, y, z such that {x, y}, {x, z}, {y, z} get the same color. But then, x, y, z do not meet the requirement on antimetricity.
Problem 3t.9. If (A, <) is a partially ordered set, its dimension, dim (A, <), is the least cardinal K such that there exist K total orderings { Rio, then there exist disjoint Ao, Al C A with dim(Ao, <), dim(Al, <) > No. Solution. Assume indirectly that (A, <) is a counterexample of minimal cardinality A. Without loss of generality, A = A. Put
I={BC A:dim(B,<)
Claim 1. If B,,, E I (n = 0,1, ...) and B is such that every pair of B is covered by some Bn, then B E I. Proof. Straightforward from the definition.
Claim 2. There are Bo, B1 E I such that Bo n B1 = 0 and Bo U B1 ¢ I.
Proof. If not, then I is a a-complete primideal containing every subset of cardinality < A. For a < A, let
Put
I1={CC B1:BoUCEI}. Then B1 ¢ Il and every subset of B1 of size < is is in I,. Repeating the argument of Claim 2, we get that there exist Co, C1 E Il such that Co n C1 = 0 and Bo U Co, Bo U Cl ¢ I. Repeating the above argument for Co, Bo this time, one gets Do, D1 C Bo, Do n Dl = 0 such that Do U Co, D1 U Co E I, Do U D1 U Co ¢ I. As every pair of Do U Dl U C1 is contained either in Bo or in Do U C1 or in D1 U C1, by Claim 1, either Do U C1 or D1 U C1 is not in I. In the former case, Do U C1 and D1 U Co are two disjoint sets not in I, and we conclude similarly in the other case, too.
3. SOLUTIONS TO THE PROBLEMS
562
Problem 12.10. A binary relation -< is called a quasi-order if it is reflexive and transitive. The infimum of the quasi-order (Q, -<) is the greatest subset J C_ Q such that (i) for every B E Q there is an A E J with A - B, and (ii) A -< B, A, B E J imply B -< A. Let X be a finite, nonempty alphabet, let X* be the set of all finite
words from X, and let P be the set of infinite subsets of X*. For A, B E P, let A - B if every element of A is a (connected) subword of some element of B. Show that (P, -<) has an infimum, and characterize its elements.
Solution. As follows, subword will always mean connected subword. Let a < b denote that a is a subword of b. Call A E P minimal, if for all w E A, all but finitely many subwords of the elements of A contain w, as subword. Claim 1. If A is minimal, B
A, then A
B.
Proof. Assume that w E A and k is so large that every subword in A longer than k, as B -< A, is a subword, so w < b. Claim 2. If A E P is not minimal, then there exists a B E P such that B
Abut A.AB. Proof. Let w be a subword of A such that
H={h: there is an aEA,h
Proof. Such a B is given by the following algorithm. Step 1: Let X = {al, ... , an}. If, among the subwords of A, infinitely many do not contain a1 i let Hl be their set. H1 E P, H1 -< A, and it does not contain al. Then get successively H2, H3, ... using a2, a3, .... If Hn_1 is defined, it may only contain an, so it is minimal. We can therefore assume the existence of Hi_1 such that all but finitely many subwords of Hi_1 contain ai. Put b1 = ai, B1 = Hi_1 -< A, and there follows Step 2. Here Step k: If bk_1i Bk_1 are given, and b1 < ... < bk_1i Bk_1 -< B1 - A are given, and all but fi-itely many subwords of Bt contain bt, consider the word bk_1a1. If infinitely many subwords of Bk_1 omit it, let Hi -1 be this set. bk_1a1 therefore is not a subword of Hi-1. Repeating this to bk_1a2i we get H2 -1, etc. If we can reach Hn=i, then there exists a natural number in, such that in every subword of length 2m a bk_1 can be found in the first m positions, but it must be followed by an. So we can select Hk-1 as Bk and bk_1ai as bk. By this process, we get b1 < b2 < .... B1 >- B2 ... . Put L = {b1i b2, ... }. Then L A and L is minimal. In fact, for bi c L , E Bi, so only finitely many subwords of them omit bi. bi+i, bi+2, We get that the minimal members constitute the infimum. Claim 3 gives (i), Claim 1 (ii), and the maximality holds as if A is nonminimal, B -< A
3.11 SET THEORY
is such that A 7k B, C is minimal, C - B, then C -< A, A
563
C, so the
addition of C would violate (ii).
Problem N.11. Define a partial order on all functions f : R -* R by the relation f -< g if f (x) < g(x) for all x E R. Show that this partially ordered set contains a totally ordered subset of size greater than 2s0 but that the latter subset cannot be well-ordered.
Solution. For the first part of the problem, it suffices to find a family of more than 20 subsets of R totally ordered by c, as then the characteristic function will do the job. In order to show that if is is an infinite cardinal, then there are #c+ subsets of a set of size K, ordered by C, let A be the least
cardinal such that 2A > ic. By Cantor's theorem, .\ exists and is < K. If we find a totally ordered set of size f+, with a dense subset of cardinality < #c, we are done by the Dedekind cuts. For such a set, take the .\ -> {0, 1}
functions, ordered lexicographically. A dense subset is formed by those functions that take 0 from a place onward. Their number is
1: 21`'
For the second part, assume that f0, < fp for a < 0 < (2"0)+. Let x0, E R satisfy f,, (x,,) < f,,+1(xa), for (2k0)+ many a, xa = x, and so the f,, (x) are (2'0) + different elements of IR, which is impossible.
Index of Names A
C
Abel, N. H., 489 Aczel, J., 192 Adam, A., 16 Adian, S. I., 82 Ajtai, M., 30
Cayley, A., 421 Clifford, A. H., 85 Corradi, K., 10 Coxeter, H. S. M., 91 Csakany, B., 8, 22, 28, 46 Csaszar, A., 2, 4, 7, 12, 13, 18, 21, 30, 42 Csikos, B., vii, 46, 52 Csorgo, M., 443 Csorgo, S., 35, 42, 48 Czach, L., 12 Czedli, G., 46, 51 Czipszer, J., 2
B Babai, L., 25, 32, 50, 51, 282, 491, 529, 557 Baer, R., 63, 81 Bajmoczi, E., 51 Balogh, A., 39 Balogh, Z., 32, 33, 51 Bara, T., 51 Beck, J., 51 Beleznay, F., 52 Benczur, A., 53 Berlekamp, E. R., 374 Bernstein, I. N., 332 Bernstein, S., 158 Birkas, Gy., 53 Biro, A., 53 Bod6, Z., 52 Bognar, J., vii Bognar, M., 13, 20, 21, 27 Bognarne, K., 425 Bohus, G., 52 Bolla, M., vi Bollobas, B., 50, 263, 518, 554 Boltjainskii, V. I., 184 Borges, R., 3 Boros, E., 51 Bosznay, A., 32 Brindza, B., 298 Buczolich, Z., 52 Burnside, W., 96, 118
D
Daroczy, Z., 12, 20, 25, 38, 192, 513
Deak, J., 51 Denes, J., 33 Deny, J., 207 Dini, U., 489 Domokos, M., 53 Domosi, P., 44 Drasny, G., 53 Druszt, E., 34
E Elbert, A., 50 Elek, G., 52 Engelking, R., 533, 537 Eotvos L., v Erdelyi, A., 183 Erdos, J., 11, 14, 19 Erdos, L., 52, 53 ror
INDEX OF NAMES
566
Erdos, P., 7, 11, 12, 15, 16, 19, 23, 26, 33, 36, 38, 43, 44, 145, 522, 558, 561
F Fazekas, I., 39, 45 Fejer, L., v, 151, 479 Fejes-Toth, G., 30, 40 Fejes-Toth, L., 4, 7, 9, 11, 13, 18, 20, 24, 27, 30, 36, 40 Feller, W., 406 Fischer, E., 471 Fleiner, T., 53 Foia§, C., 396 Frankl, P., 51 Freud, R., 50 Fried, E., 2, 7, 9, 17, 23, 27, 41 Fritz, J., 50 Fuchs, L., 2 Fi redi, Z., 51
Hajos, Gy., 3 Halasz, G., 18, 21, 24, 30, 33, 36, 40, 46, 50, 465, 467 Halmos, P. R., 348 Harcos, G., 53 Hardy, G. H., 217, 470, 484 Hatvani, L., 23, 29, 41, 48, 190 Hausel, T., 53 Hayman, W. K., 182 Heppes, A., 7 Hetyei, G., 52 Hewitt, E., 440 Horvath, L., 35 Hosszu, M., 3 Huhn, A., 31, 35 Huppert, B., 107
I Ivanyos, G., 52
J G
Gacs, P., 50, 51, 473, 477, 479 Galvin, F., 17, 30 Garay, B., 44 Geher, L., 9 Gelfand, I. M., 459 Gereb, M., 51, 52 Gerencser, L., 50 Gerlits, J., 553 Gesztelyi, E., 12, 26, 32, 37 Gondocs, F., 51 Gorenstein, D., 104 Greenleaf, F. R., 194 Gyires, B., 12, 20 Gyori, E., vii, 51 Gyori, I., 15 Gydry, K., 11, 25, 38
H Haar, A., v Hajdu, G., 53 Hajnal, A., 7, 10, 14, 20, 21, 23, 27, 32, 35, 39, 522, 558, 561
Hajnal, P., 52
Jaglom, I. M., 184 Joo, I., 36, 43 Juhasz, I., 9, 13, 24, 27, 31, 46, 50, 518, 520, 554
K Kalina, J., 27 Karman, T., v Karolyi, Gy., 52 Karteszi, F., 25 Katai, I., 10, 38, 513 Katona, Gy., 23, 39, 50 Keleti, T., 53 Kelly, D., 39
Kemperman, J. B. H., 336 Kennedy, P. B., 182 Kerchy, L., 29, 41, 48 Kertesz, A., 19, 84 Kery, G., 50 Kiss, E., 51, 52, 538 Knuth, E., 50 Koljada, K. I., 35 Kollar, J., 51, 56, 540 Komjath, P., vii, 42, 51, 52
INDEX OF NAMES
Komlos, J., 21, 50 Komornik, V., 51, 297 Kos, G., 53 Kovics, B., 38 Kovics, I., 9 Kovics, S., 52, 53 Krisztin, T., 34, 47, 48, 51, 52, 242
L
Laczkovich, M., 17, 24, 27, 33, 36, 39, 43, 44, 45, 50, 51 Langberg, N. A., 443 Laver, R., 559, 560 Leindler, L., 8, 15, 16, 22, 47, 491 Lempert, L., 24, 27, 43, 51 Leon, R. V., 443 Levi, B., 439 Linnik, Yu. V., 107 Littlewood, J. E., 470 Losonczi, L., 12, 26 Lovasz, L., 20, 22, 23, 28, 29, 36, 50, 51, 263, 415, 473, 477, 479, 524, 553
Lukics, E., 42
567
Montagh, B., 113 Moor, A., 9 Mori, T. F., 32, 33, 37, 43, 46, 51 Moricz, F., 16, 29, 47, 438 Moussong, G., vii, 52, 545
N Nagy, P., 29 Nagy, Zs., 51 Neumann, J., v Novikov, S. P., 82 Natanson, I. P., 168, 479 Nemetz, T., 465
0 Odor, T., 52
P Pach, J., 43 Pales, Zs., 31, 38, 52 Palfy, P. P., 31, 36, 39, 51, 52, 297 Pap, Gy., 37 Pelikan, J., vii, 17, 20, 30, 50, 51, 81
M Magyar, A., 52, 53 Magyar, Z., 51, 52 Major, P., 21, 450, 459 Majoros, L., 53 Makai, E., 18, 24, 50, 51, 477 Makay, G., 53 Makkai, M., 10, 13 Maksa, Gy., 25, 38 Marki, L., 28 Mate, A., 11, 15, 50, 491, 519, 521, 558, 561
Mate, E., 50 McLean, R. P., 85 Medgyessy, P., 19 Megyesi, L., 22 Michaletzki, Gy., 40 Miklos, D., vi, 52 Mocsy, M., 52, 53 Mogyorodi, J., 425
Petruska, Gy., 17, 50 Pinkus, A., 36, 209 Pinter, L., 29 Pintz, J., 51 Pollak, Gy., 3, 41 Polya, Gy., 180, 462, 470, 492 Pontriagin, L. S., 189 P6sa, L., 23, 28, 50, 51, 479, 520, 528, 553
Prachar, K., 107 Prekopa, A., 425 Preston, G. B., 85 Proschan, F., 443 Pyber, L., 42, 45
R Rado, R., 145, 522, 558, 561 Rado, T., v Ramsey, F. P., 122, 134
Ritz, L., v Redei, L., 3, 8, 13, 25, 103
INDEX OF NAMES
568
Renyi, A., 3, 4, 8, 11, 425, 433, 443, 453, 461 Revesz, P., 14, 19, 24, 37, 443 Revesz, Sz. Gy., 41 Reviczky, J., 51 Riesz, F., 394 Riesz, M., v
Riman, J., 38 Rudin, M. E., 545 Rudin, W., 505 Ruzsa, I. Z., vii, 27, 39, 45, 50, 51, 440, 445, 477 S
Saks, S., 330, 336 Salem, R., 479 Schrijver, A. A., 28 Schottky, W., 228 Schur, I., 382 Schweitzer, M., v, vi Seb6, A., 51 Seress, A., 52 Shapiro, H. S., 27 Shilow, G. E., 459 Shockey, H., 207 Sidon, S., 135 Sigray, I., 52, 53 Simanyi, N., 51, 52, 290, 538 Simonovits, A., 477, 484 Simonovits, M., 50, 470, 519, 553, 554
Somorjai, G., 27, 51, 501 Sos, V. T., 30, 33, 34 Stromberg, K., 440 Suranyi, J., 2, 9, 13, 18 Sylvester, J. J., 382 Szabados, J., 22, 24, 30, 40, 45 Szabo, E., 52 Szabo, Gy., 38, 44 Szab6, L. I., 46, 52 Szabo, T., 53, 513 Szab6, Z., 28, 34, 35, 52, 53, 539 Szasz, D., 50, 425 Szegedy, M., 52 Szeg6, G., v, 180, 492 Szekely, G. J., vi, 32, 440, 445
Szekely, L., 34, 36 Szekelyhidi, L., 25, 44 Szekeres, Gy., 7, 30, 33, 40 Szemeredi, E., 27 Szendrei, A., 41, 47 Szendrei, M., 51 Szenes, A., 52 Szenthe, J., 16 Szep, G., vii Szigeti, F., 50 Szilard, L., v Szilasi, J., 39, 45 Szokefalvi-Nagy, B., 4, 9, 22, 394, 396
Szucs, J., 22, 23, 31, 35, 50
T Tamassy, L., 12, 25 Tandori, K., 4 Tardos, G., 52 Teller, E., v Terjeki, J., 34 Tomko, J., 26 Torocsik, J., vii, 52, 53 Totik, V., vii, 29, 34, 36, 39, 41, 42, 46, 47, 48, 51, 209, 297, 539
Turin, P., 2, 3, 8, 10, 14, 18, 158, 431
Tusnady, G., 14 Tuza, Zs., 51, 297
V Vamos, P., 50 Varga, J., 52 Varlet, J., 27 Veres, S., 51 Vesztergombi, Gy., 50, 519 Vitali, G., 336, 510 Vu, H. V., 53
W Weiss, B., 39 Wiegandt, R., 34, 43 Wielandt, H., 96, 97
Wigner, E., v
Zempleni, A., 52 Zenor, P. H., 545 Zorn, M., 86, 345
Problem Books in Mathematics
(continued)
Demography Through Problems by Nathan Keyfitz and John A. Beekman
Theorems and Problems in Functional Analysis by A.A. Kirillov and A.D. Gvishiani
Exercises in Classical Ring Theory by T. Y. Lam
Problem-Solving Through Problems by Loren C. Larson
A Problem Seminar by Donald J. Newman
Exercises in Number Theory by D.P. Parent
Contests in Higher Mathematics: Miklos Schweitzer Competitions 1962-1991 by Gabor J. Szekely (editor)
