Problem Books in Mathematics
Edited by P.R. Halmos
Springer New York
Berlin Heidelberg Barcelona Budapest Hong Kong London
Milan Paris Santa Clara Singapore
Tokyo
Problem Books in Mathematics Series Editor: P.R. Halmos
Polynomials by Edward J. Barbeau
Problems in Geometry by Marcel Berger, Pierre Pansu, JeanPic Berry, and Xavier SaintRaymond
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 19621991
With 39 Illustrations
Springer
Gabor J. Szekely Department of Mathematics Eotvos Lorand and Technical University Mtizeum krt. 68 1088 Budapest, Hungary Series Editor: Paul R. Halmos Department of Mathematics Santa Clara University Santa Clara, CA 95053 USA Mathematics Subject Classification (1991): 1506, 0506, 2606, 5106
Library of Congress CataloginginPublication Data Contests in higher mathematics:Miklos Schweitzer competitions, 19621991 / Gabor J. Szekely (ed.). p. cm.  (Problem books in mathematics) Includes bibliographical references and index. ISBN 0387945881
1. Mathematics  Competitions  Hungary. 2. Mathematics  Problems, I. Schweitzer, Miklbs, 19231945. II. Szekely, Gabor J., 1947 . III. Series. QA99.C62 1995 exercises, etc.
510' .76  dc20
9525361
Printed on acidfree paper. ÂŠ 1996 SpringerVerlag 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 (SpringerVerlag 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. Cameraready 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 0387945881 SpringerVerlag 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 worldfamous scientists. The success of high school competitions led the Mathematical Society to found a collegelevel 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 worldclass 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, 19491961 (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 researchlevel 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 padic 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 pgroup (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<y<
(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,k0
Df(x)=limsup
Df(x)=liminf
h,k>O h+k>O
h+k>O
h,k.0
f (x+h)  f (xk) h+k
h,k>O
Show that V f and D f are Borelmeasurable 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. nco 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) [MiillerPfeiffer] 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 ÂŽR2 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 realvalued 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 realvalued, continuous function on the halfline
0<x<oo,and
= f2(x)dx < oo,
then the function
g(x) = f (x)  2ex 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. Sz6kefalviNagy] 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<408.
J (F.5) [L. FejesT6th] 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 Lebesguemeasurable, 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 ngon 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 ndimensional 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<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. (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,,,,An1
abn_iAn1
b0E
b1A
...
bn_2An2
abn iAn1
b0E
...
ab2A2
ab3A3
B=
abn_2An2
\ab1A
bn_3An3
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 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, P1P2= 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 nspace 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 cointossing 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. FejesToth, 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 noo
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 xaxis in the xyplane is the union of a segment AB in the upper halfplane perpendicular to the xaxis and a segment CD containing B in its interior and parallel to the xaxis. Show that it is impossible to erect a letter T at every point of the xaxis 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 twovariable 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=.
lamn + amn+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 k1K 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 212 > ak > 22 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
no
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. Sz6kefalviNagy]
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 ngon 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 ngon. (G.14) [L. FejesToth] 1968 1. Consider the endomorphism ring of an Abelian torsionfree (resp. torsion) group G. Prove that this ring is Neumannregular 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 Neumannregular 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
a1
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 complexvalued, 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
>
(2e)n.
oÂŤ< 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 rtuples 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 rtuples 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 <k
(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. FejesT6th] 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 + C3 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
1a; 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 halfplane 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 pea9 (A and Âľ 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 ndimensional 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 ndimensional Euclidean space, the square of the twodimensional Lebesgue measure of a bounded, closed, (twodimensional) planar set is equal to the sum of the squares of the measures of the orthogonal projections of the given set on the ncoordinate 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 pgroups (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. Fejesloth] 4. If c is a positive integer and p is an odd prime, what is the smallest residue (in absolute value) of P1
. (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 echain not longer than K. (T.5) [M. Bognar] 6. Let a neighborhood basis of a point x of the real line consist of all Lebesguemeasurable 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 Nmeasure 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 Nmeasure Âľ if any Nmeasure coinciding with p on all elements of 2t is necessarily identical with Âľ. Prove the existence of an Nmeasure Âľ 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 1n >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 FÂ° 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, wellordered subset and that cannot be represented as a union of a countable family of wellordered 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
f1(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 n1
k=1
Prove that there exists a constant K' such that n1
EAk<K'An (n=1,2,...). k=1
(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. X00
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 
X00
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 < 2n1  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 EOÂ° 1 an is convergent. (S.13) [L. Leindler] 10. Let {Â˘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
Âľ <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 realvalued 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 realvalued 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<i<j<k<5 T (AiAjAk) max1<i<j<k<5 T (AiAjAk) '
(T(ABC) denotes the area of the triangle AABC.) (G.18) [J. Surinyi] 9. Let K be a compact convex body in the ndimensional 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. FejesToth, E. Makai] 10. Let T1 and T2 be secondcountable 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 Tiopen set G 3 p (i = 1, 2), there exist a Tiopen set G' and a T3_iclosed 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<i<n
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/n34Othen 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<n f (n). Show that F(n) log n 1oo nloglogn
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. FejesToth] 9. Determine the value of
sup [log E  E log ],
1<g<2
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 2connected 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}째째_o be a uniformly bounded sequence of realvalued 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 rearrangement of the series E째째 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 IanIna
some a > 2. Let us assume that for some 3 > 1/a, we have f (x) _ O(exo) 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 Lebesguemeasurable functions, and only those, extend continuously over T. Show that the real line cannot be an everywheredense 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 twosided 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 rationalvalued, multiplicative arithmetical functions and the set of complex rationalvalued, 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 Lebesgueintegrable functions on [0, 1] such that for any Lebesguemeasurable 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. Sz6kefalviNagy]
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=(i1)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. t100
(F.22) [L. Hatvani] 10. Prove that an idempotent linear operator of a Hilbert space is selfadjoint
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<k<n
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<Ivl<n
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 Lebesguemeasurable subset Hr of measure c of the interval [0, 1]. Prove the existence of a Lebesguemeasurable 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 ndimensional 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. FejesToth, 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<k<n
Prove that lim of (log n) Tn = 0 with probability one. (P.16) [P. Reoo 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 padic 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 halfaxis 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(1x)>xlog2x(1x)log2(1x) 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 complexvalued 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=i11 Xm.=i)>0, i>0. Prove that for arbitrary k > 0 there is an ak > 1 such that
Pn(k)=P( max 0<j<n satisfies the limit relation L
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) = EÂ°Â° _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. FejesTdth, E. Szemeredi] 9. Suppose that all subspaces of cardinality at most R1 of a topological space are secondcountable. 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
nan
(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 ndimensional 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 3space 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 pseudoR.iemannian metric on the set of points of the Euclidean space E3 not lying on the zaxis 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 ndimensional sphere, and to each vertex of T let us assign a nonzero vector of a linear space V. Show that if T has an ndimensional 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, 27rperiodic 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(2IikI+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 1l such that every graph in 7l 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 = G1TG. 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. FejesToth]
1. PROBLEMS OF THE CONTESTS
31
10. Suppose that the T3space 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 nowheredense 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
i0
j=0
xA
V
Y.
o<i<n i#j
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 ndimensional 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 finitedimensional, 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 finitedimensional 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' PIÂ° <n <pÂ°+1
Prove that lim sup f ( n
noo
)
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 = UÂ°_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, 27rperiodic even function. Prove
that if 1
fr(X) + f(x)
f (x + 37r/2)
holds for every x, then f is it/2periodic. (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 threeelement 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 IIATA111 < 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 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 settheoretic intersection and the convex hull of the settheoretic union. We call a lattice ndistributive if n
n
x A (V yi) = V (x A( V yi)) i=O
j=0
O<i<n
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(xh,x+h)nEI>uh and
i(xh,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 ]fink (k < n) as well, where Mk is a hypersurface in some (k + 1)dimensional subspace Ek+1 C Rn+l, Rnk 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<y<1
n
y
(1  u)dQ(u)
(n + 1  i)(Xi:n  Xil: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 Borelmeasurable (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 fourtuples 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 Lebesguemeasurable subsets E of the real line does a positive constant c exist for which sup
00<t<00
<c
L
sup
Leinxf(x)dx
n=O,±1,...
for all integrable functions f on E? (M.17) [G. Halasz] 7. Let V be a finitedimensional 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. FejesToth]
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<k<n, Sk=x}I (x=0,+1,+2,...), So=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(1IZI l \Iwz12/
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 d7r 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 CÂ°Â°). (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 valgebras we have .Fl C_ .F2 C_ C_ A, and for all n, Xn, is an .Fnmeasurable and integrable random variable). Assume that
(n = 2, 3 ... ).
E(Xn+l lFn) = 1 Xn + 1 Xni 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 {<a: a < r.} on A with < = fla<,c <a. Show that if dim(A, <) > 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
n1 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 1periodic 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. FejesToth] 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<a<1,
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 quasiorder if it is reflexive and transitive. The infimum of the quasiorder (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 oneelement 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, selfadjoint 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.10 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 kdimensional 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  x1K(x) for x > 5, where if x E [5, oo) \ UÂ°Â°_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 wellordered. (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 01 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 zaxis, 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(afl1,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 EOÂ° 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 nowheredense 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 infinitedimensional, 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 ÂŽ HTM. Construct a bundle involution P : TTM TTM such that Poh = voP and prove that, for any pseudoRiemannian metric given on the bundle VTM, there exists a unique metric connection V such that
VxPYVyPX =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 mdimensional stationary GaussMarkov 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 >
IAld1
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 bchain 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  xill < 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 6chain of length 4. Also, prove that we cannot always find an interval in which any two elements could be connected by a 6chain 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 Al째), ... , A(째) 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 (Al째), ... , A(째)) E (R2)", there exists a natural number N such that the points AlN)7 ... , (N are consecutive vertices of a convex ngon. (G.46) [B. Csikos] 9. Prove that if all subspaces of a Hausdorff space X are Qcompact, then X is countable. (T.22) [I. Juhasz] 10. Let X and Y be independent identically distributed, realvalued 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 ngon. (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 < Âľ, aÂľn
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=2n1
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
n1
{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/ÂŤ } = 2k,
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/ÂŤ
<x Gn(x)
* 0
as
n4oc.
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 padic numbers.
Solution. The padic number a_,,,,p' + + a_lp1 + ao + alp + + (0 < ai < p) is a padic integer if all the coefficients with a,,.p' + negative index are equal to 0, and it is a unit in the ring of padic integers
if furthermore ao # 0. It is clear that every padic number a can be written in the form a = 3p', where 3 is a padic unit and r an integer. Since the product of padic units is again a padic unit, we deduce that a padic number can be a root of unity only in the case when it is a padic unit. Every root of unity can be written in the form e = a + ap', where r is a positive integer, a is a padic 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 padic 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 padic 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 padic integer, whereas cp, as we can easily see, is a padic unit.
This guarantees that 311 is different from 1. We have proved that in the field of padic 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 padic 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 gP1 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, gP1 = 1 + cep3+1
Let aj+1 be the solution of the congruence gox  cj (mod p). Then
1P1 +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 xp1  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 nth 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 pgroup (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 pgroup, 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 + ...+pk179i+k1)
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 openclosed 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 ÂŽ 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 righthand side, this is clear because h(ri) is an annihilator of R2 and h(ri) E R2. For the lefthand 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 ÂŽRi ÂŽ 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 righthand 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 righthand 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 (twosided) 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
(cyana)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(ees+s)=dedes+ds=dds+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, Rk1 # 0 would be satisfied, but this would imply that Rk1 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 . amlk = ak . amlk = 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 oneelement 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 ÂŽ 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), 630638).
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) xsr times (1), we get
1 + x + ... + xsr1 = (1 + x.f) (g9  grxsr) + xs+l (hs  hr). This means that if there are indices s > r such that
hs  hrEI holds, then 1+x+...+x9r1
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) <cp(b).
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 16step 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),<i} (i=0,1,2,...). Clearly, To = {0}, To C T1 C
C Tk C ...
and
R = U Ti. io
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 leftsided identity. Proceeding in the same way with an arbitrary c: cb = cqb, c = cq, so q is a twosided 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(ab) <, cp(a). On the other hand, W(a) = cp(b(ba)) 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 (twosided) 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 = a1. 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) b1) >, 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(lbq). 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(x1c1) = k. Then, using steps 2 and 12, we have co(xk) = cp(xk1.x) > cp(xk1), so co(rk) 3 k+1. On the other hand, let W(a) = k+ 1 and a = xk1 b+r where cp(r) < cp(xk1) = k.
Then co(xk1b) = W(a  r) = W(a) = k + 1 > k = cp(xk1). Using step 12, we get b Â˘ 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 + xk1 s + r. Here W(a) = k+1; in case W(s) = 1, we have cp(xk1s) = co(xk1) = k; in case W(s) = 0, that is, s = 0, we have cp(xk1 s) = 0, finally, we know W(r) < k. All in all W(a) > max{cp(xk1s), cp(s)}, so using step
8, (p(xk) , co(xkc) = cp(a  xkls  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 = xklak_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(xklak_1 + + xal + ao) < k, ak = 0 follows. By repetition of the argument, we get ak1 = 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 + albi1 + ... + 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°(a1)°, 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 righthand side is divisible by the greatest common divisor of f (x) and g(x), since the lefthand side is divisible by it. But the polynomial on the righthand 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
k1K 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 k1K. 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 21i > ak > 2i 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
<
j1
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: Â°Â°
.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 Neumannregular 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 Neumannregular 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 torsionfree  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 1tol 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 " g1xg 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 < 2n1 holds. The group G is nilpotent. Indeed, for any elements 91, 92.... ,92n1 E G, [... [[91,921,931,
is true, because 91, 92,
,
92n1] = 1
, 92n1 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 wellknown 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(P1)(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 2P11cl
>
f
P1 I L(k>(x) I> 2P1 JJ I Im L(k) (x)
(r
2k)x3  (k )X2 + ((2k  (2k
P1
2P1
k=1
P1
= [f l
Sk
k=1
k=1
P1
fl(r k (k) = k=1
P1 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 + Ck 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 + C6)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
91
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))
s1
= (s + k)mxs+k  EOj (7 + k)mxj+k, j=o
and consequently s1
(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; 81 E7'm,ioj (j + k)
81
n
ri
j:0j E E 1'm,i (j+ k)m [Ji+k
i=0m=0j=0 81
n
j=O
i=om=O
91
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 pgroups (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 pgroup. 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 wellknown 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* = ÂŽCi, 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* = EOÂ° 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 pgroup 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 cp1(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 pgroup. 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 pgroup. 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 pgroup. 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 padic 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 rank2 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 c1 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 pgroups 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 pSylow 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 ÂŽ 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 ÂŽ 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 torsionfree. 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 twosided 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 twosided 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 orderpreserving. But then 0 can only be the identity that we easily verify if we extend in an operationpreserving way to the quotient field of J and observe that in this way 0 becomes an orderpreserving map, fixing all rational numbers.
Solution 2. We prove that the integral domain
JzIx1 '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 righthand side of (2) has a pole of order k1 in 0, thus the lefthand side too), which gives 3xk1 . 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 + (P1 and ( are units in K. So E1 = (c + C1)2 and E2 = ((2 + b2) 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, ... , p1). Repeatedly applying the automorphism (4 r2 to (3), we get (2 r2)a' ((22 + r _22 )b' (r(r22
+rS522)a'
+
=
(S23 +
rSS23)b' (4)
((2 d1
d1
d
)ai
+ S2
= (S 2
+(2
d )b'
= (S
+
(1)bi
(3) and (4) impliy
( + 1)a'db'd = 1 . If a'd ; b'd, then ( + C1 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 lefthand 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 = (p2 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 Âą(OI2 (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
p11
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 p1
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
p1
r U
L: a,j k(ke E2 +
j=n k=1
L:
V p1 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 wellknown (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, Âąj, Âąk} 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<aVm}. In view of the maximality of M, we have (M, a) = L. So for some element m E M we have b <, a V m, therefore bcp < acp V mcp = acp. Similarly, acp <, bcp consequently, acp = bcp, as we stated. (B) DOES NOT IMPLY (A). We construct a counterexample. Let L = {V I V C I[R finite} U{(oo, x] U V I x E R, V C ]ER finite}.
The operations on L should be the settheoretic 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 halflines 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 ndimensional 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 twodimensional 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 onedimensional 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 1to1 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)aVzbVz; (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
zVazAa=((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
bAztooneofa,b,zorbAzaAz. 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 23cycle. 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 2transitive, and as Stab (x, y) contains a 23cycle G is actually 3transitive. We have seen that the 23cycles 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 3transitive. By a wellknown 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 4transitive, 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 3transitive, 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(h1k) = cp(h1g'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 . . . aTlbTlaTbT = 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 righthand side clearly contains the lefthand 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
+)3y01 }) \ 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 ... xnlaxnxn+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 ... xn1 = xm ... xnla
EI.
for
Now let a, b E R and Xm
xn_1 E I be arbitrary elements. Then we
have
(ab)xm ... xn1 = Xm ... xn_l(ab) = (Xm ... xn_la)b = (axm ... xn1)b = (axm)xm+l ... xnlb = b(axm)xm+l ... xn1 = (ba)xm ... xi_1.
Applying once more our previous argument to the resulting identity
(ab  ba)xm .. xn1 = 0
(a, b E R, xm . . xn1 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 twosided 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 settheoretic intersection and the convex hull of the settheoretic union. We call a lattice ndistributive if n
n
x A (V yi) = V (x n( V yi)) j=0
i=0
O<i<n i#i
holds for all elements of the lattice.)
Solution. Let us introduce some notation. Let En1 denote the Euclidean space of n  1 dimensions; for X C_ En1, let C(X) denote the convex hull of X; and let L be the lattice of convex subsets of En1. For an arbitrary H C_ En1, 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 En1, L(H) is ndistributive. Let X, Yo, ... , Yn E L(H) be arbitrary elements of L(H). By Caratheodory's theorem for an arbitrary subset U of En1, 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 ndistributive.
Let p be a latticetheoretic term in k variables, let H C En1, 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 En1 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<n. (Actually, we could obtain u < 1, v < 1, but we don't need this now.) By the induction hypothesis there exist finite sets H,.... , Hu, G1, ... , G C_ Ere1 such that h' E pH; (Xl n H2,.. . , Xk n Hi) and h' E p'' (X1 n .
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 ndistributive 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 Ere1 we have h E pH (X1 nH, ... , Xk n H). Since L(H) is finite and ndistributive we have
h E gH(Xl nH,...,Xk nH)
gEn_1(Xl nE7L1,...,Xk nEn1) = 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 fourtuples 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 wellknown 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+n1
(2)
A2 (X, Y)F(X, Y) + B2 (X, Y)G(X, Y) = R. ym+n1
(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) =
dmn 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 fixedpointfree. By a wellknown result (see, for example, D. Gorenstein, Finite Groups, Harper & Row, New York, 1968, Theorem 10.1.2), for every prime p there exists a pSylow subgroup P of
G with W(P) = P, so G has to be a pgroup. Furthermore, as the center is a characteristic subgroup and pgroups have nontrivial center, G has to be Abelian. Also, in an Abelian pgroup, 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=ÂŽZP.
(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 oneelement 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)=(flfz)(x)+ (flfz)(x+pl+pz)= {11=(flfz)(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 wellknown 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 wellknown 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
q1 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 wellknown 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 wellknown 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)3c1n \ nsC
=
.
Problem A.43. Let A be a finite simple groupoid such that every proper subgroupoid of A has cardinality one, the number of oneelement 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 <k and ai+1,...,ak cA, then B(x1i ... xi, ai+1, ... , ak)
= {(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 oneelement 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 An1, containing Un1. So, by the minimality of n, we have B(xl,... , xn_1, u) = An1. So U is a subset of the set
S= {a E A I Anlx{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), (x7r1, y7r1 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 = Ak1 for every j E k.
(3)
Consider the subgroupoid of A2 of the form
C(u1i... , uk2, x, y) (u1,.. , uk2 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 Uk2 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
Uk2
(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 ... , zk1 E A, such that (z1, ... , zk1, x), (z1, ..., zk1, 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, ... , ak1, u), (a1, ... , ak_1, v) E C .
So for the subgroupoid C* of Ak1 defined by
C = Q X1, ... , xk1, U) n (Xi, ... , xk1, v), we have (a1, ... , ak_1) E C*. On the other hand, because of (4) we have
C* n Uk1 = 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 < k2 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_l1) 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 oneelement subgroupoids {u} (u E U). This contradiction proves D = A. The equality D = A shows that the projection C * prk_{k}C (= Ak1) 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 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).
Solution. Let us denote by 0 the smallest element of P. Let L' = {a E L 0 < a}. Then L' is a (clearly modular) coverpreserving 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 wellknown 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 <Lv q2 also holds. So Q is indeed a coverpreserving sublattice of L'. We are going to embed P in Q. For an element p E P let f (p) = V{a E L' I a < p; a is an atom}. The map f is injective. If namely for two elements P1, P2 E P we have p1 P2, then there is an element r E P such
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 pgroup. 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 ele: al + ... + p e,ie. ai1 + 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 ... , pek1+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 > ... > ek1 + 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 pgroup 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 pgroups 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=z1, and therefore
p2nb2v2
= z2  2z + 1.
Since n >, 1 and z E R, we finally get
1 = p(p2n1b2v2)
 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(an1,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(an1, 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<i<n:
(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 abovementioned 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  an1) 
(an, al,a3+a2,a4+a2ia5a4,...,anan1) ...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  an1) 
 (a2,a1,a3,a4  a3, a5  a4,...,an  an1) r
(a2,al,a3,a4,a5 a4i...,an an1) * ... 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,...,ai1,ai,ai+1, ...,an) 4 (a1, a2 + a1, a3  a2, ... , ai1  ai2, ai1, ai+1,... , an)  (al, a2 + a1i a3 + a1, ... , ai_1  ai2, ai1, ai+1, ... , an) *
(al, a2 + al, a3 + a1i ... , ai_1 + al, ai1, ai+l,... an) a
 (al, a2i a3  a2, ... , ai1  ai2, a1, ai+1, ... , an) p 
*
al, a2, a3i ... , aZ_1  ai2, a1, ai+l, ... , an)
(a1,a2,a3,...,ai1,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 righthand 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 = qlr (a permutation has an inverse), V'(p) = Vi(q1)1li(r), and (b) and (d) show that 1li(q1) 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 psubgroup of T; we are going to show that P
is transitive, thereby proving that T is necessarily a pgroup.
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 wellknown 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 ktuple (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 n1m 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)sts1 = (/3)s1, 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 SchurZassenhaus 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. = (,rci11 .rCi21
m}i
.rcim1 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 SchurZassenhaus theorem, we can simply consider a Sylow 2subgroup.
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<k,1<n).
(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 , fiz (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<k<n we have fk(a) a fk(oo) = oo.
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 Â˘ 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 2.2<a1ifk>1. 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 Â˘ An
then i=ak1. 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 14, 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 rtuples 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 rtuples 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 wellknown theorem of Ramsey. If the rtuples of natural numbers are divided into finitely many classes, then there is an infinite subset N' of N such that every rtuple 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 rtuples 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 relement 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 rtuples, 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 <n
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, iÂ˘j
i=1
xEAj
and so
Â˘x <n+nk=n(k+1). xEAj
Add these up for all j. On the lefthand 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 righthand 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<j<n+l
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<i <n+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 < 2n1  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<i<k},
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,A12Â˘Ail. Obviously,
K=KIUK2UK3UK4.
(1)
We will estimate the cardinality of the set K. The inductional hypothesis implies that (2) jK1j < 2n2  1.
Let I = IA1J. Then IK21 < 2nL1  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 2n11 and 21  2 distinct sets, respectively (they cannot meet Al in Al or 0). Pairing these intersections in all possible ways, we get IK41 <
2n11(21  2) = 2n1 
2n!
Actually, this estimate can be sharpened. If X E K4, then
Y=(XA,)U(AAX)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
XyY=(XA1)U(AAX)
3.2 COMBINATORICS
127
is onetoone 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 <
2n2
 2n11
(5)
Summarizing (1) to (5), we get the desired inequality, IKI < 2n1  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 2n2  1 by the inductional hypothesis. The number of sets X C_ S containing x is 211. At most half of them, that is, at most 2n2, 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 2n1  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 2s1  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 2s1 1 and 2''1  1 ways, respectively, the number of these sets Ai is at most (2s1  1)(2r1  1). Adding up the partial results obtained, we get that the number of all sets Ai is at most 2n1  1. Remark. It can be shown that if k = 2n11, 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 realvalued 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, ... , xn1 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 2connected 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, ... , rn1) 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 Â˘ 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 n1
n
an = t=1
al (7) an1 = E t=0
forn>1. Using the notation bn = an/n!, we have bo = 1 and n1
bn = 1=o
1
(n
l)! bl.
Clearly, bn < 2n, so if Izl < 1/2, then the series EÂ°Â° 0 bnzn is absolutely
3. SOLUTIONS TO THE PROBLEMS
132
convergent. For the sum f (z) of this series, we have 0o n1
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 Ai1)
(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<c=e2 otherwise. This is a contradiction again, which makes the proof complete.
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 n2n
(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 lefthand 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 Âľ(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 alai1, 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 {aiai1 1 i = 2, ... , n}, and so µ(Ga) =
(1)IaI(1)n1= (1)IaI(n1).
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(n1)
r
c. Sgn P2
P3
12=113=1
...
L (P,) PP
1n=1
7
Pi (1)l'1
= fT
9=213=1
(1)131
3)
7
=
3.2 COMBINATORICS
139
where Pk denotes the maximal indegree 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 ndimensional sphere, and to each vertex of T let us assign a nonzero vector of a linear space V. Show that if T has an ndimensional 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(aii))
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 ndimensional 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 ndimensional 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 ndimensional 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 0dimensional simplices (that is, the vertices) of K, then the number of ndimensional 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 Â˘ 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,2Inml
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,
1IAIE>_ 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<n2 2
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
i0
xA V
yi
0<i<n
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 lefthand side of (1) is x and the righthand side of it is not since (vo, vk) is not contracted in the righthand 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 righthand 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 righthand 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 2n2 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) _
n2andd(x,z)=n1. Letx=z0,zl,...,zk=z(k=n2orn1)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 2connected 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=Pk1pi=Pk2pj 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...UAi1)I=IAi((Ain A,)U...U(AinAi))I = IAiI 
n  i  (i  1) = n  2i + 1. Hence,
IA1u...uAn,I =IAiu(A2Al)u(A3AAluA2))U...U(An(Alu...UAn1))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 yxEEifxyÂ˘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 ktuple 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 ErdosKoRado 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={bs:bEB}U{bs2: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)
n2
6
edges. It is known that a planar graph of n vertices has at most 3n6 edges
(if n > 3). Thus, H has at most 6n  12 vertices. From these estimates, we obtain
n(n1) <6n12, 6
that is,
n237n+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
x21
+
1192x2
), 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)
(x21)(11)2x2) J1
dx t(1  192x2)
J0
dt
J0
dt
1+00
=
+Â°Â° 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 zplane (excluding the points z = 1 and z = 1/19 by semicircles open from below) to the wplane 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<v<
(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<v<D,
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 nk1, Mk1, 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
k1 Ismk(1)I > kEmaxIsn,(z)I; 1=0
Izl=k
in addition, we may assume that Mk > nk1 We now define Ck. Since Sink (z) is continuous, for Ck sufficiently close to 1 we also have k1 Sink (Ck) I > k E max I Snt (z) I l=0
jzj=k
We additionally assume that Ck is real, further and
Ck > Ck1
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 EÂ°Â° 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/ _ Snk1 ( 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, k1
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=ni1+1
k1
<2Zmax i_ 1=0
nk
___ I.
i_.\i
IzI=k
C` i_
.
(
kck l
i_n
n>nk
k1
<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 = k1
smk (k)
Snk1
(k) +
{sn,
l (k) sn,1 (k)}
k1
> ISmk(ck)I 2EmaxIsn,(z)I ? (k2)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
k1
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
k2
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+.
k2
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 realvalued 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 realvalued, continuous function on the halfline 0 < x < oc, and 00
f2(x)dx < oo,
fo
then the function g(x) = f (x)

2ex
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 halfline (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
ew
OF f(x)dx+ 100/2 f2(t)dt, o
whence it follows that lim
W00
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<4V8.
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
f1(y) if p < y < f(1),
02(y)
1
if
f(1)<y<q,
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 midparallel 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)m1, we have
t(T) =
2 (2mn 2
\
+ 1) (2  m).
It follows that P1
=4 (2 +m If(x)kIdx< 4m2m2 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(x1) 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, 405407) 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(n1)(x) + ... + fn(x)y(x) = 0
(1)
have the desired properties, and let Â˘(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(n1) (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(n1)(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 ndimensional 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.r20
dyk
Therefore (dkÂ˘r(y)/dyk) is continuous in (oo, r2) and (r2, +oo) and has righthand and lefthand 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 ndimensional 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 Â˘ F, thus the function F(x) has the required properties. Remark. Let F1 and F2 be disjoint closed sets in the ndimensional 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+(t2t1)]=f[(tt1)+t2]=F[f(tt1),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:
(nWl l e J
zQ
> (2  e)n .
o<e<
Solution. Put Cn  kt) e
an(z) = an = E
f
z ,
n = 1,2,....
0<t<
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 + EÂ°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 + ankz
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 ex 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,
(1ct)f'(x+u) < f'(x) <
ctf'(xu)
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,
1ct 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) atJ ext
(3)
We first consider the lefthand side of (3). For fixed t,
f `x+t) et 
lim
ex+t
Hence
lim inf X00
f'(x) x e
x f (x)
= et  1.
> (1  ct) et  1, t
and therefore
limnff (y)>tlimo1(1ct)ett 1I=1. From the righthand side of (3), it similarly follows that lim sup f f (x) ex
X00
1  et 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+klnx6.
(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 halfplane 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
aioo
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
ii'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 <(a2ai)F(/3 )40,
a1
and similarly fcrl i'Yn
J
f (z) dz < (a2  ai)F('y) + 0,
aziry, therefore
J
az+ioo
al+ioo
f (z) dz =
a1ioo
J 12ioo
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 halfplane Re z > 1). Then
B
= pa+boo f(z) dz = i f°° f(a + ij3) d
J
z
Jaioo
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 etz 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;
aioo 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<
1x2
<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 13, 196465 (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 (Âą1) = 0, it follows that 7rn(x) = (1  x2) f (x), where f (x) is a polynomial of degree not exceeding
n2 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) = Âą 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. X00
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) _ Xco 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
Âľ= i1
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 pnumber; let us sum with respect to this. Then
Eµ'_µ'<x E
µ<x
n>1,a; nAi <xµ'
Ai.
With the notation n> 1,aj nai <y
(replacing p' by p) we have
Ep=E£(xµ). µ<x
(1)
µ<x
Apply this to x + 1 and subtract the two relations from each other:
E x<µ<x+l
1t=E(G(x+11)L(xp))+ E G(x+1p). µ<x
x<µ<x+l
3. SOLUTIONS TO THE PROBLEMS
172
The lefthand side is at least x (M(x + 1)  M(x)), and the second sum on the righthand 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<xK
µ<xK
<eEL(xµ)=eEp<exM(x) µ<x
µ<x
and
E (L(x + 1 µ)  L(x  µ)) < L(K + 1)M(x). xK<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 X00
and since e > 0 is arbitrary, lim sup X00
M(x + 1)  M(x) M(x)
= 0.
It remains to prove (2). We may write L(x)
E 1> = ai<x Ai 1<n<Z i
A{<2
Ai
2+
Ai
<ai<x
> Z L(2) + 2 (L(x)  L(2 x)) = 2 xL(x). Now
ai.
L(x + 1)  L(x) n>_1,,\; x<na,<x+l
(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<n<E n<ai<y 1
/
\ (xn
1<n<
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)<L(+1)L(x)o(Ll
=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
i1
0
= f27r
11aEj eio
do
\\\
eino log 1
do.
an
0
Now, if lal > 1, then for n  oo the integral on the righthand 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 aend 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
,
j1
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
27rE
=F1+F2.
E
27r
log M do = 27r log M.
(2)
0
We split F1 again into two parts:
F1=
flogl(z1)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) _
nk
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
nk +b 2nknk1
k+2 k+3 + nk 3_ n+1
1+ak+2 Â°Â°
1
(k+1)6k+2 ( )  (k+1) 16=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 _ nk
< 2M
n+1 k+l
If k > 2, then using the relation t! > ttet 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 <log(2M)+(k+1)logk+l and, consequently,
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 (z1)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, ... , 2ne+1) such that
(z  1)2t =
2n2+1 i2
2nC+1 zk
fj
_ i + 1 2 W2n(Zk)
k=e
(z  zj).
j=Q
Z1,
j34k,2n+1k
0 (t < v < 2ne+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 =
2n1+1 (zi  1)21
j=1
j#i,2ni+l
(zi  zj)
Then for z = zi (i # v) the lefthand side is equal to the righthand side. Indeed, the terms of the sum appearing on the righthand side are zero except for the term with k = i; if k # 2n  i + 1, then this is true because of the factor 2n1+1 (zi  zj) = 0,
11 j=1
j#k,2n+1k
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 2n21+1 and 1/z are 0. This can be done because the two coefficients are the negative of each other since 2n1+1
11
(zj) = 1
j=e
j#k,2n+1k
for all k; together with each zj appearing in the product, its reciprocal z2n+1_j also appears.
Thus, the condition 2ne+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 lefthand and the righthand 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 _
2n1+1
H
(zv  1)21
(zv  zj)
j=1
l
j#v,2nv+l also holds.)
Since lz?  11 = Izi  1/zil = Izi  z2n+1il 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+1i
Relying on this, and setting Â˘ = 7r/n, we obtain the following estimate of Jail (for t < i < n, but also for i > n because of the symmetry):
1
jail =
2n1+1
H
(zi  1)21
(zi  zj)
j=1
j54i,2n+1i 11
2n (zi  zj)
(zi  zj)
fi
(zi  1)
j=2n1+2
j=1
1
(zi  1)21
2n
12
t1
[J[(i A0) 11 [(i+j)0] 2
< j=1
j=0
1
211
[l2
<
1i
 2)
212  1) 2
.2 211
Li  1)211 2
2n
2] 0212 .
. 16211 . 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 2nE+1
2nP+1
fi
fi (z  zj )
(z  zj) <4
9=t
j=2
j54i,2ni+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  z2nj+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(1zj)
2nt+1 4
fi (1zj)
=4
j=1
4(12n+1) 2)21 2
2
zj)
l
< [llj=1(j
n)212
8 <

(coear
{(_2)!()
e1 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
<n 2 2 'M
con 222 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  zÂľ), 1z"1>1
w*(z)=c fi (1z21) 11 (zzÂľ). 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 RIxI
47rR
13
dS
s(0 R)
R2Ix12f(x0) R2Ix0I2 R+IxI (R+IxoI)2 . (RIxl)2f(xo) <f(x0) (R+B)2 R+A 3
jxj
RIxol
RB (RA)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) (1Âą
)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, McGrawHill, 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<t'<x+1,
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 (A4)
(4)
.
From the inequality 0 < (2A1)2 = 4A2 4A+1, it follows that A 1/4 < A2. Substituting this into (4), we finally arrive at
B<3(A41
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 secondorder 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 onesided 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 <MIyxI, 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 Dii 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<c'. Solution 1. First, we verify the following statement.
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) >
qp
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<a'<b<b'ora<a'<b<b'aswell.
3. SOLUTIONS TO THE PROBLEMS
188
Solution 2. Let D = f (b)  f (a)
ba
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<a'<r'<r<b<b'. Because of the Darboux property of derivative functions, f'(x) lies on one side of D for all x E (a, r), and we may suppose that for all such x we have
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'(ba') < f(b)  f(a') <D(ba'), and from here f'(b) < D' < D < f'(a'). Let T E (D', D) such that T # f'(r). Because of the Darboux property, there must be a place t E (a', b) where f'(t) = T. This, however, can be neither in (a', r) (because there
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 PicardLindelof 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, AddisonWesley, 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 > T0 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), 223225).
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}Â°' 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(1x)>xlogex(1x)loge(1x) 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 k0
x)nk < 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)nk) \
From the functional equation on f, we obtain f (xk) =
kxk1 f(x)
and
f ((l
 x)k) = k(1 
x)k1 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 noo 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)nk
(n)
+ k 1092 x + (n  k) 1092 (1 (k =Sn(x)+xlog2x+(1x)log2(x1),
=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 complexvalued 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 rightinvariant mean m on the space of complexvalued, 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 lefthand 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 = y11 ... , x = ym and adding the resulting equalities together, we obtain with
m
Sm(y) = 1: f (xyi)X(xy i) j=1
the identity
E ck
m
k1
n
(Sm(Y)
k=1
+ j=0 E j=mk+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'Â°+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) = limnoo 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)xn1 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 halfline 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 halfline x = x0, y > yo. Write the function in the form
i
a3xy+a4xy a3xy  a2x + al
1+ (a4a2)xy+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
<a2a4.
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 En
<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) _ ÂąTT, (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 f1(t) sin(t f1(t))dt 0
.
o
Solution. Let
T1={(u,v):0<u<x, 0<v<y}, T2 = {(u,v): 0 < u < x,
0 < v < f(u)},
T3={(u,v):0<v<y, 0<u< f1(v)}. Then T1 C T2 U T3, T2 n T3 = 0, and each of these sets is a Borel set, hence for every nonnegative and Lebesgueintegrable function g on [0, 1] x [0, 1], we have /U
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 finitedimensional 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 finitedimensional 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 HahnBanach 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, 27rperiodic
even function. Prove that if 1
f"(x) + f(x) _
f(x + 37r/2)
holds for every x, then f is 7r/2periodic.
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 37rperiodic, and since it is also 27rperiodic, 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
 ftt 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[t1,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 wellknown 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 finitedimensional 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 halfspace 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, Onesided L'approximation, Can. Bull. Math., 25 (1986), 8490.
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(1Izl2
l1wz$
,
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(1s)+1s 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 =
1s 1s
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
1zw 1
 IzI2l
(1
l IZI2
Iwz121 =L\Iwzi2), for all (z, w) E D x B.
1zwJ
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 BanachSteinhaus 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 wellknown differential equation
(1x2)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 HahnBanach 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(tx)dÂľ(x)=J
g(xt)d(x) = 0o
f g(xt)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,
ti
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
ti
a(s)Ix(s)Ids
2 (fÂ° a(t)jx(t) Ids 
f
o
a(t + s)lx(t + s)Ids )
t
 (I cost)  cost) it
2 dtd
fj ro
i
ti
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)
(IcostIcost)f a(s)x(s)ds. t ti (1)
3.3 THEORY OF FUNCTIONS
215
Since the righthand 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 t1
(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 n1/2
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.
t1
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.10 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)
x10 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)
X10 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,Âą1,... 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 righthand 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
1x4'Y =q7,
lim x.10 (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 = k0 ak. By a wellknown 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, m1
52m = 1 (1+c)+c= (1+c)m+c 0
and C)m1 + C2,
82+21 = (1 + C)m + c(1 + from which
f
52m+2m1 c 1+1+c 82
(1 (1 
ql/(21,3))P
ql/(2m1.2))P
(ql/(2_1.3)) (gl/(2m1.2))
.f
2 .2m1 1 P
(3
. 2m1
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
3P2.2P+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,1o 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 HardyLittlewood 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 HardyLittlewood 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 HardyLittlewood 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
1IzI,
and this is what we had to prove.
Remark. The example f  0, g(z) _ (1z)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 Âą zil = 1 = Ily Âą zil, and therefore by (1)(2),
2f(x) = f (x + z) + f (x  z)  2f (z) =c+c2f(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(xy)=a(Ilx+y112)+a(IIxyl12) = 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 twodimensional 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.22an1an
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,
P2SIP, +2S2=0,
........................
PnSIPn_1+...+(1)"1Sn_1P1+(1)nnSn=0.
Hence, it follows that
Slf =Plf = 0, S2f = 1(S1Plf  P2f) = 0,
1 Snf = (Sn1P1f  Sn2P2f + ... + (1)nlpnf) = 0. n
Now, for fixed i E {1, 2, ... , n}, consider the differential operator
P(ai) = (ai a2) ... (a1  an) =a2nS a2n2+S a2n4...+ (  1 ) ' S
.
We see that P(aa)  0 and, therefore, at n f = Slain2 f  S2a2n4 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: 2n3 xl \ /X in1
/Qn(x1, x27 ... , xn) = det
x22n1
x22n3
X2
k x2n1
x2n3
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, n1
0ikQnl(xlIx2,...,xn1) = 0,
k = 1,2,... .
i=1
Expand the determinant Qn according to the first column: n
Qn(xlex 2
o
Xn) n) = 
j=1
()j1x2n1 Qn1(X1, ..., x
1,x{lo
x n)
It follows that n
Ea2kQn(xl, x21... , xn) ii=1 n
n
W 1)
_
2k [2n1 xj Qn1(xl, ... , xj1i xj+1i ... , xn)]
C7i
i=1 j=1 n
_
1)j1
j
d2k
2n1 &x3 kxj
Qn1(X1.... ,xj1ixj+li...,xn)
n
(1) j1xj2n1 j=1
ai2kQn1(x1i...,xj1,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: / x12n2k1
xl2n3 ... xl \
x22n2k1
2n3 x2
\X 2n2k1 n
x2n3
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 11) 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 = fk9k 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 Â˘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(u1)(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) = Pm(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 6chain 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 xz1 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 6chain of length 4. Also, prove that we cannot always find an interval in which any two elements could be connected by a 6chain 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 6chain 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 6chain 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)/2neighborhood 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 6chain of length at most 2, and every interval contain such a pair of points. To this end, let P = UÂ°__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 = UÂ°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 8chain. Suppose that (x, y) E Pi. Then (y  x)/2 = 1/2i. If x, 2z, y form a 6chain, then
8(2+z)>Ix2zl =2 x(2+z)
,
whence by the definition of 6 for j > i we obtain x/2 + z Â˘ 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 8chains 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 firstorder 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
2n+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) = 2n1, 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 n1 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), ... xn1 = xnl(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),...,xn1(xn),xn),
x2 = x2 (xn), ... , xn1 = xn1(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
]fin1 > 0: G(xl, x2, ... , xn) < 0}.
Then
a. H C [0,1]n1, 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, ... , xÂ° ). 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.j1
 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  27ra)  cos ax),
where a is to be defined later. Then f (x) > 0, while f (x) = 0 if x = 2k7r 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
an7r/a
en7r/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}Â°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 = flÂ° oInk. Then Ino
f () = e2nk7r/a I 2  cos l
ak7r
2
 27ra I  cos 2nkir]
a amk =e 2nk7r/a [1cos21r nk where Mk is an integer satisfying the relation
amk a nk
enk7r/a nk
3.3 THEORY OF FUNCTIONS
233
Therefore, using the inequality cos u > 1  u2/2, we obtain
f (?:)
< <
ezÂŤ 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 e7rn/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"nk13 (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 e7rnk/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
e7rnk/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'r7r2E2eank = 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 ThueSiegelRoth 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
,, _
e1rnk/2
(Ne_ir/2)nk
nk
(Ne7r/2)nk
2
2nk
2
Since
=
it follows for any e > 0 that
a <e Mk
nk
k >oo
e7rnk/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 wellknown 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
=Â°Â°,
(3)
3. SOLUTIONS TO THE PROBLEMS
236
1/2
2n1
00
E 2 kl1akl2
< 00,
(4)
n=1 (k=2n1
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=2n1 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
2n1
2n1
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}:
2n1
it
rir
2n1
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
<C
dx
\\\
1/2
2n1
k=n
cos (k + 2) x dx 2 sin 2
2n1
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
j1 > 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
<k+1 (0 <x <ir; k=0,1,...), we obtain that 2'1
I1 >
it/J
91
l(7+1) k=0 = I11  112 .
dx
Dak X 
2'1
1
9=1
ir/7
71
> (k + 1) jakl dx
fit/(j+1) k=0
(9)
One of the most effective ways to stimulate students to enjoy intellectual efforts is the scientific competition. In 1894 the Hungarian Math...
Published on Oct 12, 2016
One of the most effective ways to stimulate students to enjoy intellectual efforts is the scientific competition. In 1894 the Hungarian Math...