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PROBABILISTIC AND CONSTRUCTIVE METHODS IN HARMONIC ANALYSIS AND ADDITIVE NUMBER THEORY

a dissertation submitted to the department of mathematics and the committee on graduate studies of stanford university in partial fulfillment of the requirements for the degree of doctor of philosophy

By Mihail N. Kolountzakis May 1994

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c Copyright 1994

by Mihail N. Kolountzakis

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I certify that I have read this thesis and that in my opinion it is fully adequate, in scope and in quality, as a dissertation for the degree of Doctor of Philosophy. Paul J. Cohen (Principal Adviser)

I certify that I have read this thesis and that in my opinion it is fully adequate, in scope and in quality, as a dissertation for the degree of Doctor of Philosophy. Yitzhak Katznelson

I certify that I have read this thesis and that in my opinion it is fully adequate, in scope and in quality, as a dissertation for the degree of Doctor of Philosophy. Rafe R. Mazzeo

Approved for the University Committee on Graduate Studies:

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Abstract We give several applications of the probabilistic method in harmonic analysis and additive number theory. We also give ecient constructions in place of previous probabilistic (existential) proofs. 1. Using the probabilistic method we prove that there exist integers p1; : : :; pN 0 for which N min X p cos jx = O(s1=3); j x

P

j =1

as s ! 1, where s = Nj=1 pj . This improves a result of Odlyzko who proved a similar inequality with the right hand side replaced by O((s log s)1=3). 2. Similarly we prove that there are frequencies 1 < < N 2 f1; : : :; cN g, for c = 2, for which N X cos x j min = O(N 1=2) x

j =1

and that this is impossible for smaller values of the positive constant c. 3. The previous result is used to prove easily a theorem of Erd}os and Turan about the density of nite integer sequences with the property that any two elements have a dierent sum (B2 sequences). We also generalize this to B2h sequences (of which all sums of 2h elements are distinct). Some dense nite and in nite B2 [2] sequences (only two pairs of elements are allowed to have the same sum) are also exhibited. 4. We prove that for any sequence of integers n1 : : : nN there is a subsequence nm ; : : :; nmr such that 1

X r min cos nmj x C N; x j=1 vii

where C > 0 is an absolute constant. Uchiyama had previously proved this with the right hand side replaced by C N 1=2. Furthermore, our proof is constructive. We give a polynomial time algorithm for the selection of such a subsequence. 5. A set E of positive integers is called a basis if every positive integer can be written in at least one way as a sum of two elements of E . Using the probabilistic method, Erd}os has proved the existence of such a basis E for which every positive integer x can be written as a sum of two elements of E , in at least c1 log x and at most c2 log x ways, where c1; c2 > 0 are absolute constants. We give an algorithm for the construction of such a basis which outputs the elements of E one by one, and which takes polynomial time to decide whether a certain integer is in E or not. 6. We employ the probabilistic method to improve on some recent results of Helm related to a conjecture of Erd}os and Turan on the density of additive bases of the integers. We show that for a class of random sequences of positive integers A (which satisfy jA \ [1; x]j C px), with probability 1, all integers in the interval [1; N ] can be written in at least c1 log x and at most c2 log x ways as a dierence of elements of A \ [1; N 2]. Furthermore, let mk be a sequence of positive integers which satis es the growth condition 1 log m X k < 1: k=1

pm k

We show that, for the same class of random sequences and again almost surely, there p is a subsequence B A, jB \ [1; x]j C x, such that, for k suciently large, each mk can be written in exactly one way as a dierence of two elements of B.

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Acknowledgements I want to express my deep gratitude to my advisor, Professor Paul Cohen, for all that I learned from him during the last four years of my graduate studies. The many discussions that we've had on various subjects, mathematical as well as not, have been invaluable to me. His generous encouragement during periods of frustration helped me greatly through this thesis. I would like to thank Prof. Y. Katznelson for many useful discussions as well as for persistently running the Problem Seminar. I also want to thank Prof. R. Mazzeo for being on the reading committee of my dissertation. Profs. Joel Spencer of the Courant Institute and Andrew Odlyzko of Bell Laboratories have given me valuable advice and comments on my work, for which I am grateful. The support of the Department of Mathematics has been great during my ve years of studies. It is one of the friendliest places I have ever been. My colleagues: Yannis Petridis, Andrew Stone, David Hurtubise, Sergei Makar-Limanov, Bryna Kra, Tatiana Toro, Maia Fraser, all have my thanks for making my life as a student a memorable experience. One has to look back to nd the people who deserve most of the blame that this thesis came to be. My parents Marilena and Nikos Kolountzakis clearly did not insist enough that I do not become a mathematician. It was too little, too late. After my father had already challenged me with problem solving for years, they tried, halfheartedly, to convince me to study a science that promised more nancial security than Mathematics. Their love and support throughout my studies show that they were very pleased with their failure. A failure which was partly due to the faculty of the Department of Mathematics at the University of Crete who managed to distract me from my Computer Science studies ix

and turn me to one of their own. Among them: Profs. Souzana Papadopoulou, Vassilis Nestoridis, Nikos Tzanakis, and the late Stelios Pichorides, who left a clear mark on my mathematical tastes. I owe much of what I am to Lydia Kavraki, my wife. Our parallel studies, for so many years now, would not be possible without her. Making it to this point together means to me that the end of our studies will be the greatest beginning for our life.

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A! os os o :

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Notations The letter C will be used as an absolute constant. It need not represent the same constant in all its occurences, even in the same formula. a = O(b) is equivalent to jaj C jbj, for some constant C . a b means a = O(b). a b means b a. a = o(b) means that a=b ! 0 as a certain parameter, that will be implicit from the context, tends to a limit. bxc is the largest integer not larger than x and dxe is the smallest integer not smaller than x. x? is de ned as min f0; xg (the negative part of x). R (The Lp norms) kf kp = ( jf jp )1=p for 1 p < 1, and kf k1 = esssupjf j, where the integration and the supremum are taken over the domain of de nition of the function f , which will be implicit. If E is a subset of a measure space then jE j denotes its measure. If E is a nite set then jE j denotes its cardinality. We denote sequences of positive integers by capital letters and the individual elements of the sequence by lower case indexed letters:

A = fa1 ; a2; : : :g and we usually assume that a1 < a2 < . We also write

A(x) = jA \ [1; x]j for the counting function of the sequence A. The Fourier coecients are de ned by Z 2 f (x)e?inx dx: fb(n) = 21 0 xiii

R

In particular fb(0) = 21 02 f (x)dx. (f must be in L1 [0; 2 ].) N denotes the set of natural numbers (positive integers), Z the integers, R the real numbers and C the complex numbers. E[X ] denotes the expectation of the random variable X . Pr [A] denotes the probability of the event A, and Pr [A j B ] = Pr [A \ B ]=Pr [B ] denotes the conditional probability of A given B . 1(: : :) is equal to 1 if the condition in the parentheses is true, otherwise it is equal to 0. The acronym SIIRV stands for Sum of Independent Indicator Random Variables. p i = ?1 unless it is used as a running index.

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Contents Abstract

vii

Acknowledgements

ix

Notations

xiii

1 Introduction

1.1 The Prototype Average Value Argument : : : : : : : : : : : : : : : : : : : : 1.1.1 An Edge-Colored Kn with Few Monochromatic Km 's : : : : : : : : : 1.1.2 A Large Sum-Free Subset of a Given Set of Integers : : : : : : : : : 1.2 The Prototype Large Deviation Argument : : : : : : : : : : : : : : : : : : : 1.2.1 An Asymptotic Additive Basis with Small Representation Function : 1.3 The Method of Conditional Probabilities for Derandomization : : : : : : : : 1.4 Introduction to the Problems Studied and Other Related Problems : : : : : 1.4.1 Littlewood's Conjecture and the Cosine Problem : : : : : : : : : : : 1.4.2 The Salem-Zygmund Theorem : : : : : : : : : : : : : : : : : : : : : 1.4.3 The Salem-Zygmund Theorem { Applications : : : : : : : : : : : : : 1.4.4 The Rudin-Shapiro Polynomials. Spencer's Theorem. : : : : : : : : 1.4.5 Uchiyama's Theorem : : : : : : : : : : : : : : : : : : : : : : : : : : : 1.4.6 Bh [g ] Sets of Integers : : : : : : : : : : : : : : : : : : : : : : : : : :

1

1 2 3 4 6 8 10 10 12 15 21 24 26

2 On Nonnegative Cosine Polynomials with Nonnegative Integral Coecients 31 2.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2.2 Proof of the Inequality M (s) s1=3 : : : : : : : : : : : : : : : : : : : : : : 2.3 The construction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : xv

31 33 35

3 The Density of Bh [g] Sequences and the Minimum of Dense Cosine Sums 37 3.1 3.2 3.3 3.4 3.5

Introduction : : : : : : : : : : : : : : : : : : : : : : Dense Cosine Sums : : : : : : : : : : : : : : : : : : An Upper Bound for Fh (n), h Even : : : : : : : : Dense Finite B2 [2] Sequences : : : : : : : : : : : : In nite B2 [2] Sequences with Large Upper Density

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37 39 40 41 42

4 A Construction Related to the Cosine Problem

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5 An Eective Additive Basis for the Integers

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6 On a Problem of Erd}os and Turan and Some Related Results

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Bibliography

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5.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5.2 Probabilistic Proof of Existence : : : : : : : : : : : : : : : : : : : : : : : : : 5.3 Derandomization of the Proof : : : : : : : : : : : : : : : : : : : : : : : : : :

6.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 6.2 Proofs : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :

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50 51 53

57 59

Chapter 1

Introduction 1.1 The Prototype Average Value Argument The probabilistic (counting) method in mathematics in its simplest form is the proof of the existence of a certain object by examining the average behavior of an appropriate collection of candidates. The prototype example of the probabilistic method in this form can be considered to be the following obvious statement.

Proposition 1.1 If x1; : : :; xn 2 R and then for some j

x1 + + xn a n

(1)

xj a:

(2)

The usefulness of the method lies in the fact that the average (1) is often easier to compute than exhibiting a speci c xj for which (2) can be proved to hold. This lack of the power to construct the solution to a speci c problem is an inherent characteristic of the method. Very frequently the probabilistic proof of a theorem is extremely simple compared to a bare hands constructive proof, and that is to be expected since it furnishes less: the mere existence of a solution to a problem rather than the solution itself. Yet, one of the points that this thesis wants to make, is that very often a probabilistic proof can easily be turned into a construction, if one assumes the point of view, that an ecient algorithm is a construction. 1

2

Introduction

Let us rephrase Proposition 1.1 in the following more useful form. The measure space

is equipped with a nonnegative measure dP of total mass 1, and a real random variable X on is just a measurable function X : ! R.

Proposition 1.2 Let X be a real random variable on a probability space ( ; dP ) whose expected value Z E[X ] = X (!)dP (!)

satis es

E[X ] a:

Then there is ! 2 such that

X (!) a:

We remark that because of the obvious linearity property of the expectation of a random variable E[X1 + X2] = E[X1] + E[X2] (whenever the right hand side makes sense), the expected value of quantities of interest in the problems that follow are almost always very easy to compute or at least to estimate very well. Notice that no independence is required of the pair X1 ; X2. (For a de nition of independence of a collection of random variables see for example [41].) We proceed to give some examples.

1.1.1 An Edge-Colored Kn with Few Monochromatic Km's Let n; m, with n m 3; be two positive integers. We denote by Kn the complete graph on n vertices. We want to color the edges of Kn with two colors (say red and blue) so that it contains few monochromatic copies of Km . Of course it is easy to have many monochromatic Km's by coloring every edge with the same color. For each subset A of [n] = f1; : : :; ng with jAj = m we de ne the function (of the coloring)

8 < A = : 1 if A is monochromatic, 0 otherwise:

Then the number X of monochromatic Km 's is

X=

X

A[n]; jAj=m

A :

(3)

The Average Value Argument

3

We color each edge of Kn red or blue with equal probability 1=2 and independently of the m other edges (we toss a fair coin for each edge). The expected value of A is then 2( 12 )( ) and by the linearity of expectation and (3) we get 2

!

E[X ] = mn 21?(m): 2

We have proved:

Theorem 1.1 There is a 2-coloring of the edges of the graph Kn which gives rise to no ? m more than n 21?( ) monochromatic K 's. m

2

m

We shall see later a way of turning this probabilistic proof into an ecient construction of such a coloring.

1.1.2 A Large Sum-Free Subset of a Given Set of Integers A subset E of an additive group is called sum-free if

x + y 6= z; for all x; y; z 2 E:

(4)

The following theorem of Erd}os [16], [2] has a beautiful probabilistic proof, which, to the best of my knowledge, is the only proof known to date. See also [32] for a similar, but computationally more ecient approach.

Theorem 1.2 Let A N be a set of N positive integers. Then there is a sum-free subset

E of A with

jE j > 31 N:

Proof: Let A = fn1 < < nN g and choose any prime p > nN such that p = 3k + 2 for some k 2 N. View the set A as a subset of the multiplicative group of units of the eld Zp

(the integers mod p). Write

S = fk + 1; : : :; 2k + 1g

and notice that jS j > (p ? 1)=3 and S is sum-free as a subset of Zp . Let t be uniformly distributed over Zp = f1; : : :; p ? 1g and write

X = jS \ (t A)j;

4

Introduction

where t A = ft n1 ; : : :; t nN g and the arithmetic is in Zp. Since

X= and

X

j 2S

1(t?1 j 2 A)

E[1(t?1j 2 A)] = p N? 1 ; for all j 2 Zp

(Zp is a multiplicative group), we have

E[X ] = pjS?jN1 > N3 :

This implies that there is t0 2 Zp for which X > N=3. De ne then

E = A \ (t?0 1 S ): It follows that E is sum-free as a set of integers (even more, it is sum-free mod p) and jE j > N=3, as we had to show. 2

1.2 The Prototype Large Deviation Argument Often we associate several quantities X1; : : :; Xn with a random object. Typically their averages E[X1]; : : :; E[Xn] will be easy to compute or estimate and their value will be in the desirable range. Our objective is to have the values of the random variables Xj themselves in that range, simultaneously for all j . Having found a proper distribution of random objects, namely one for which the expected values E[Xj ] are of the desirable magnitude, we still need to bound the probability that some Xj deviates too much from its expected value. That is, we want an upper bound on Pr [jXj ? E[Xj ]j > dj ; for some j ]:

(5)

The maximum allowed deviations dj are problem dependent. It is usually the case that the best upper bound for this probability that we know is n X

j =1

Pr [jXj ? E[Xj ]j > dj ]:

So we aim for this sum (n can be in nite in some cases) to be less than 1. This implies that with positive probability none of the bad events

Bj = fjXj ? E[Xj ]j > dj g; j = 1; : : :; n;

The Large Deviation Argument

5

holds. In particular there is an object for which the quantities Xj satisfy

E[Xj ] ? dj Xj E[Xj ] + dj ; for j = 1; : : :; n: To achieve this we can use several well known Large Deviation Inequalities. The following two are straightforward to prove.

Proposition 1.3 (Markov's Inequality) If X is any nonnegative random variable with nite

expectation then for all > 0

Pr [X > E[X ]] 1 :

(6)

Proposition 1.4 (Chebyshev's Inequality) If X is any real random variable with nite variance 2 = E[(X ? E[X ])2 ] then for all > 0 Pr [jX ? E[X ]j > ] 12 :

(7)

The inequalities of Markov and Chebyshev are rather weak in most cases, but they are applicable to virtually any random variable and this makes them very useful. In the following theorems the random variable X is assumed to be of a special form: a sum of independent random variables.

Theorem 1.3 (Cherno [9], [3, p. 239]) If X = X1 + + Xk , and the Xj are independent indicator random variables (that is Xj 2 f0; 1g), then for all > 0 Pr [jX ? E[X ]j > E[X ]] 2e?c E[X ]; where c > 0 is a function of alone

c = min f? log (e (1 + )?(1+) ); 2=2g: We call a random variable X which, as above, is the sum of indepenent indicator random variables a SIIRV.

Remarks on Theorem 1.3:

1. Observe that if X = X 0 + X 00, where X 0 and X 00 are SIIRV then we have Pr [jX ? E[X ]j > E[X ]] 4e?c min fE[X ];E[X ]g : 0

00

6

Introduction 2. Since there is no dependence of the bound on k (the number of summands in X ), P X , provided that it is easy to prove that the same bound holds for X = 1 j =1 j P1 E[X ] < 1. k j =1

Theorem 1.4 [3, p. 236] Let p1; : : :; pn 2 [0; 1] and the independent zero-mean random

variables X1; : : :; Xn have the distribution

8 < 1 ? pj with probability pj ; Xj = : ?pj with probability 1 ? pj : If X = a1X1 + + an Xn where a1 ; : : :; an 2 [?1; 1] then we have for all a > 0 Pr [jX j > a] 2e?2a =n : 2

Theorems 1.3 and 1.4 are extremely useful. In the next section we show a nice application of Theorem 1.3 to a problem in additive number theory.

1.2.1 An Asymptotic Additive Basis with Small Representation Function A set E of positive integers is called an asymptotic additive basis of order 2 if the representation function

r(x) = rE (x) = jf(a; b) : a; b 2 E & a b & x = a + bgj is strictly positive for all suciently large integers x. In other words all suciently large x can be expressed as a sum of two elements of E . Examples of asymptotic additive bases are the set N itself and the set f1; 2; 4; 6; 8; : : :g. We are interested in bases for which the representation function is small. Notice that in the previous two examples r(x) can be as large as Cx. We present Erd}os' probabilistic proof that there is an asymptotic basis of order 2 such that c1 log x r(x) c2 log x (8) for all suciently large x. The ratio of the two absolute constants c1 and c2 can be made arbitrarily close to 1. De ne the probabilities log x 1=2 px = K x

The Large Deviation Argument

7

for the values of x for which the right hand side is in [0; 1], otherwise let px = 0. The constant K will be determined later in the proof. We de ne a random set E by letting Pr [x 2 E ] = px independently for all x. We show that with high probability the random set E has the claimed property (8). De ne the indicator random variables

j = 1(j 2 E ) with mean values E[j ] = pj . We then have

r(x) =

bX x=2c j =1

j x?j

from which and the independence of j it follows that

E[r(x)] =

bX x=2c j =1

pj px?j :

(9)

Notice also that, for each xed x, r(x) is a SIIRV. Easy calculations on the right hand side of (9) allow the asymptotic estimate

E[r(x)] IK 2 log x;

R where I = 01=2(s(1 ? s))?1=2ds. We now de ne the bad events Ax = fjr(x) ? E[r(x)]j > 21 E[r(x)]g; x = 1; 2; 3; : : :: Using Theorem 1.3 we can bound Pr [Ax ] 2 exp (? 21 c1=2IK 2 log x) = 2x? where = 12 c1=2IK 2. All we have to do now is to choose the constant K large enough to P have > 1. We deduce that x Pr [Ax ] is a convergent series and thus there is n0 2 N for which X Pr [Ax ] < 1; xn0

8

Introduction

which implies that with positive probability none of the events Ax ; x n0 , holds. This in turn implies the existence of a set E N such that 1 IK 2 log x r(x) 3 IK 2 log x 2 2 for all x n0 which concludes the proof. We emphasize the structure of the proof. First we de ned an appropriate class of random objects (random subsets of N). We then showed that the quantities of interest (the numbers r(x); x 2 N) have expected values of the desired size. The last step was to show that, with high probability, none of the quantities of interest deviates much from its expected value.

1.3 The Method of Conditional Probabilities for Derandomization In Section 1.1.1 we saw that there is a 2-coloring of the edges of the complete graph Kn on n vertices such that the number of monochromatic copies of Km is at most

!

n 21?(m) : m 2

Assume that m is xed and our task is to produce such a coloring of Kn . Trying every n possible coloring clearly takes too much time since there are 2( ) possible colorings. Let us describe a very general method of derandomizing the randomized construction that we gave in Section 1.1.1, to get an algorithm for nding such a coloring in time polynomial in n. We keep the same notation. ? Let A1 ; : : :; Ak , where k = mn , be all the copies of Km in Kn (otherwise known as m-cliques), and enumerate all edges of Kn as e1; : : :; e(n). Let the color of edge ej be the random variable cj . We are going to de ne the colors aj 2 fRED; BLUE g one by one, for ? j = 1; : : :; n2 . De ne the events 2

2

Rj = Rj (a1; : : :; aj ) = f(c1; : : :; c(n)) : c1 = a1; : : :; cj = aj g: 2

R0 is the whole probability space. Intuitevely, Rj represents our choices of colors up to the j -th color. As in Section 1.1.1 we de ne the 0-1-valued random variable j to indicate whether Aj P is monochromatic or not. We have X = kj=1 j and we already computed

E[X ] = k21?(m): 2

The Method of Conditional Probabilities

9

We are going to choose the sequence of colors a1; : : :; a(n) so that the function of j

E[X j Rj ]

2

is non-increasing. This is possible for the following general reason (the nature of the variable X is immaterial here):

E[X j Rj?1(a1; : : :; aj?1)] = 1 (E[X j R (a ; : : :; a ; RED)] + E[X j R (a ; : : :; a ; BLUE )]): j 1 j ?1 j 1 j ?1 2 This means that at least one of the choices aj = RED or aj = BLUE will yield E[X j Rj ] E[X j Rj?1]. Which of the two choices works can be decided (here the nature of X plays a

role) since we can explicitly compute E[X j Rj (a1; : : :; aj )] for any colors a1 ; : : :; aj . This computation clearly takes time polynomial in n. We proceed like this until all colors have been xed. Then X is completely determined ! m n X = E[X j R(n) ] E[X j R0] = E[X ] m 21?( ) and our coloring a1 ; : : :; a(n) is thus a solution to our problem. The very general applicability of the previous method, the so called method of conditional probabilities, should be obvious. The most general context is the following. We have n independent random variables 1 ; : : :; n which, without loss of generality, can be assumed to have the distributions 8 < j = : 1 with probability pj ; 0 with probability 1 ? pj : We also have a certain function X = X (1; : : :; n) for which we can eciently compute 2

2

2

E[X j 1 = v1; : : :; m = vm] for any m and v1 ; : : :; vm 2 f0; 1g. In particular we can compute E[X ] = . We can then

eciently compute an assignment

1 = v1 ; : : :; n = vn ; v1; : : :; vn 2 f0; 1g; for which X . See [3, p. 223] for a more detailed description of the method as well as other methods of derandomization.

10

Introduction

1.4 Introduction to the Problems Studied and Other Related Problems We proceed to describe some of the problems that are studied in this thesis.

1.4.1 Littlewood's Conjecture and the Cosine Problem Both Littlewood's Conjecture (now a theorem) and the Cosine Problem concern norms of trigonometric polynomials whose coecients are restricted. Littlewood's Conjecture [26]: For any set of N distinct integers n1 < < nN we have

in x

e + + einN x

1 C log N

(10)

1

for some absolute constant C . R Above we denote by kf k1 the L1 norm 21 02 jf (x)jdx. It is clear that C log N is the most we can expect in the right hand side of (10) since [30, 62]

ix i2x

e + e + + eiNx

1 log N:

The Cosine Problem: For any set of N distinct positive integers n1 < < nN we have p (cos n x + + cos n x ) (11) min 1 N C N x for some absolute constant C . This conjecture was stated by Chowla [11] in 1960. p It is easy to see that C N is the best lower bound that one can expect in (11). Indeed let p

g(x) =

and

f (x) =

X

p

XN

j =1

1j 0. This is super-logarithmic but it is not a power of N yet. It should be pointed out that the two problems are not very similar. Littlewood's Conjecture is a translation invariant problem (kf k1 stays the same if we add a xed integer to all nj ) but the Cosine Problem is not. In fact it is easy to see that if we translate the frequencies nj far away so that (3 ? )n1 nN then jmin f j C ()N . Another case where Chowla's conjecture (11) is known to hold is when the set of frequencies is sum-free [54]. The following variant of the C.P. has been studied by Odlyzko [46]. Let 0 p(x) = p0 + p1 cos x + + pn cos nx

(13)

with p1; : : :; pn nonnegative integers, and de ne

M (s) = p(0) inf p0; s the in mum taken over all choices of pj 2 N. How large must M (s) be? In other words we want to have polynomials as in (13) with as small p0 as possible. Odlyzko proved that M (s) (s log s)1=3 and we shall improve this to M (s) s1=3 in Chapter 2.

1.4.2 The Salem-Zygmund Theorem The following theorem is often used to estimate the size of a random trigonometric polynomial

Theorem 1.5 (Salem and Zygmund [56], [29, p. 69]) Let f1(x); : : :; fn(x), be trigonometric

polynomials of degree at most m, and 1; : : :; n be independent zero-mean random variables

8 < j = : 1 ? pj with probability pj ; ?pj with probability 1 ? pj ;

for some pj 2 [0; 1]. Write

f (x) =

n X j =1

j fj (x):

(14)

Introduction to the Problems Studied

13

Then, for some C > 0,

2 11=23 0n X Pr 64kf k1 C @ kfj k21 log mA 75 ! 1; as m ! 1: j =1

Theorem 1.5 was used in [46] to change the coecients of a polynomial to integers without a big loss:

Corollary 1.1 Let p(x) = p0 + PNj=1 pj cos jx and de ne the random polynomial r(x) so that p(x) + r(x) has always integral coecients (except perhaps the constant coecient):

r(x) =

N X j =1

j cos jx;

with j = 0 if pj is an integer, else

8 < bp c ? pj with probability dpj e ? pj ; j = : j dpj e ? pj with probability pj ? bpj c: Then Pr krk1 (N log N )1=2 ! 1, as N ! 1. For the proof of the Salem-Zygmund theorem we need the following.

Theorem 1.6 Let aij ; i = 1; : : :; n1; j = 1; : : :; n2, be a matrix of complex numbers, such that jaij j 1. Let also p1; : : :; pn 2 [0; 1] and the random variables 1; : : :; n be de ned as in (14). Then with probability tending to 1 as n1 ! 1 n X pn log n ; for all i = 1; : : :; n ; 1 aij j C 2 1 j =1 2

2

2

where C is an absolute constant.

Proof: De ne

Li() =

n X 2

j =1

aij j :

We can clearly work on the real and imaginary parts of the linear forms Li separately, so we assume aij 2 R. De ne the bad events

n

p

o

Ai = jLi()j > C n2 log n1 :

14

Introduction

Using Theorem 1.4 we get Pr [Ai ] 2e?2C n 2

2

log n1 =n2

= 2n1?2C : 2

Now choose the constant C = 1 to get

"[ # X n n Pr Ai Pr [Ai] n2 ; 1

1

1

i=1

i=1

which concludes the proof. 2 To complete the proof of the Salem-Zygmund theorem we note that it is enough to ensure that f (xj ) is small for a suciently dense set of points xj 2 [0; 2 ). Since f is a trigonometric polynomial of degree at most m we can use Bernstein's inequality [30, p. 12]:

0

f 1 mkf k1:

De ne xi = i 102m for i = 1; : : :; 10m and the matrix

aij = fj (xi); i = 1; : : :; 10m; j = 1; : : :; n: Notice that for all i = 1; : : :; 10m

f (xi) =

n X j =1

From this and Theorem 1.6 follows that

h

j fj (xi ) =

n X j =1

j aij :

i

Pr jf (xi )j C (n log m)1=2; for all i ! 1

(15)

as m ! 1. But the event in (15) implies that jf (x)j C (n log m)1=2 for all x 2 [0; 2 ) and for a larger constant C . For assume that jf (x0 )j = kf k1 and that jxk ? x0j 102m : Then, using Bernstein's inequality,

jf (x0) ? f (xk )j 102m f 0 1 210 jf (x0)j and, since 2=10 < 1, we get

kf k1 = jf (x0)j C jf (xk )j C (n log m)1=2:

Remark: Theorem 1.6 has the following constructive equivalent [3, p. 225]. We give it only in the case j = 1; n1 = n2 but it holds in general.


15

Theorem 1.7 Let aij be a real n n matrix, jaij j 1. We can then nd in polynomial time signs 1 ; : : :; n = 1 such that for every i = 1; : : :; n we have n X aij j C (n log n)1=2; j =1 where C is an absolute constant.

This of course means that the Salem-Zygmund theorem is equally eective, so that in most of the applications described below the trigonometric polynomials that are claimed to exist can actually be computed in time polynomial in the parameter of the problem.

1.4.3 The Salem-Zygmund Theorem { Applications The Salem-Zygmund Theorem is used mostly in a form similar to Corollary 1.1 to prove the existence of trigonometric polynomials with small maximum norm and whose coecients are restricted. Typically the situation is as follows. We want to construct a polynomial with certain properties which satis es certain restrictions on its coecients. We are able to construct a polynomial which has the required properties except that its coecients are not exactly what we want. We then add to that polynomial a random polynomial with appropriately chosen coecients. Since our random polynomial will be small in size (by the Salem-Zygmund Theorem) it will not change the nice properties of our original polynomial by much, and at the same time will make its coecients be what we want them to be. We give several examples of this random modi cation.

Bourgain's Sum of Sines with Small L1 Norm Our intention is to nd sums of sines

f (x) = sin n1 x + + sin nN x whose L1 norm is as small a function of N as possible. We present Bourgain's proof [6], [29, p. 79], that there is a set of integers fn1 ; : : :; nN g for which

kf k1 N 2=3: Notice that the best we can expect is kf k1 N 1=2 since kf k1 kf k2 = CN 1=2.

16

Introduction

We use the following variant of the Salem-Zygmund Theorem [29, p. 79] where the variances of the random variables j are taken into account.

Theorem 1.8 If fj ; j are as in Theorem 1.5 and also kfj k1 1 and a2j = E[j2 ] = pj (1 ? pj ) then as m ! 1.

2

n

3 n X X

Pr 4

j fj

C ( a2j log m)1=25 ! 1

j =1

j =1

1

The polynomial that we are going to modify is M sin jx X q(x) = a j ; 2

j =M

where M is a large integer and the parameter a > 0 will be chosen later. It is well known [62, v. 1, p. 182] and easy to prove that there is an absolute constant A, independent of M , such that

kqk1 Aa:

So q (x) strongly satis es the requirement that it is a sine polynomial with small L1 norm. We now modify it to have coecients 0 or 1 by adding to it a suitable random polynomial. De ne the independent, zero-mean random variables (a will be less than M )

8 < j = : 1 ? a=j with probability a=j; ?a=j with probability 1 ? a=j;

for j = M; : : :; M 2, with variance

E[j2] =

a 1 ? a a : j j j

De ne also the random trigonometric polynomial

r(x) =

M X 2

j =M

j sin jx

and notice that the polynomial r(x) + q (x) has coecients 0 or 1:

r(x) + q(x) = sin n1x + + sin nN x;


17

with nj 2 fM; : : :; M 2g. The expected value of N is

E[N ] =

M X 2

j =M

M a X Pr [j > 0] = j a log M: 2

j =M

It is easy to see using either Chebyshev's inequality (E[N 2] can easily be estimated to be a2 log2 M ) or Cherno's inequality, that 1 Pr N 2 a log M ! 1; as N ! 1: Using Theorem 1.8 and the estimate on the variances of the j we get that, with high probability, 0 1 1=2 M a X 2 krk1 C @ j log (M )A Ca1=2 log M: 2

j =M

Thus with probability tending to 1 we have

N Ca log M and

kr + qk1 krk1 + kqk1 Ca1=2 log M + Ca: The best choice for a is then a = log2 M which implies N C log3 M while kr + q k1 log2 M , which concludes the proof.

To the best of my knowledge Bourgain's result has not been improved or proved best possible. I do not know of any other non-trivial (that is o(N )) upper bound for

ksin n1x + + sin nN xk1: It is also not clear whether Bourgain's theorem can be made eective in polynomial time in N . This is so since we applied the Salem-Zygmund theorem on a polynomial of length exponential in N . This sparsity is unavoidable, as it can easily be seen that any sine sum with a small L1 norm has to be sparse (take inner product with a conjugate Dirichlet kernel).

Odlyzko's Nonnegative Cosine Polynomial Here we give Odlyzko's proof [46] that there is a nonnegative cosine polynomial 0 p(x) = p0 + p1 cos x + + pn cos nx

18

Introduction

with p1; : : :; pn nonnegative integers, and such that

p0 (s log s)1=3 where

s = p0 + + pn = p(0):

Several similar results are presented in [46] with dierent restrictions on the coecients pj . Consider the Fejer kernel

KA (x) =

A X

j =?A

A X j j j j j j ijx 1 ? e = 1 + 2 1? A+1 A + 1 cos jx j =1

which is known to be nonnegative. Our initial polynomial will be q (x) = KA (x) where > 1 will be determined later. Write

q(x) = q0 + q1 cos x + + qA cos Ax (notice that q0 = ). We modify q so that it has integer coecients by adding to it the random polynomial r(x) = r1 cos x + + rA cos Ax; where the independent zero-mean random variables rj have the distribution

8 < dq e ? q with probability qj ? bqj c; rj = : j j bqj c ? qj with probability dqj e ? qj :

Then the numbers r1 + q1 ; : : :; rA + qA are always nonnegative integers. It follows by the Salem-Zygmund theorem that

krk1 C (A log A)1=2 with high probability. Write

M = min x ((r1 + q1 ) cos x + + (rA + qA ) cos Ax)

and notice that

M q0 + krk1 + C (A log A)1=2 while if we write s = q (0) + r(0) then (using the fact that KA (0) A) s A ? C (A log A)1=2 A:

(16)


19

The best choice for is the one which equates the two terms in the right hand side of (16), that is = (A log A)1=2. We can now check that

M (s log s)1=3; as we had to show.

Korner's Flat Polynomials with Unimodular Coecients Littlewood [43] had asked whether there exist trigonometric polynomials

f (x) =

N X j =1

aj eijx

with unimodular coecients jaj j = 1; aj 2 C, such that

C1N 1=2 jf (x)j C2N 1=2; for all x 2 [0; 2);

(17)

where C1 and C2 are two absolute constants. If only the upper bound is required in (17) then these were known to exist even with aj = 1 (the Rudin-Shapiro polynomials that we shall see in Section 1.4.4). Korner [37] proved that the answer to this question is indeed arimative. His proof essentially consists of a random modi cation of a polynomial found by Byrnes which almost ts the requirements.

Theorem 1.9 (Byrnes [8], Korner [37, Lemma 4]) If n is the square of an even integer, then there is M 2 N, absolute positive constants C1; C2 and complex numbers c1; : : :; c4n+2M +3 such that (i) M = O(n3=4); (ii) jc1j; : : :; jc2M +3j 1; jc4n?2M +4j; : : :; jc4n+2M +3 j 1; (iii) jc2M +4j; : : :; jc4n?2M +3j = 1; n+2M +3 ijx cj e C2 n1=2: (iv) C1n1=2 Pj4=1

20

Introduction

The only problem with the polynomial

q(x) =

N X j =1

cj eijx ; with N = 4n + 2M + 3;

is that O(M ) = O(N 3=4) of its coecients are of modulus at most 1, while the rest are unimodular as required. We solve this problem by adding to q (x) two random polynomials

r(x) = R(x) =

2M X+3

rj eijx ;

j =1 4n+2 XM +3

j =4n?2M +4

rj eijx ;

where the independent zero-mean random variables rj are such that for each j in the set

f1; : : :; 2M + 3g [ f4n ? 2M + 4; : : :; 4n + 2M + 3g we always have

jrj + cj j = 1:

One way to accomplish this is by letting each rj be equal to

q i 1 ? jcj j2 jccj j j

with equal probability. The Salem-Zygmund theorem then guarantees that

krk1 + kRk1 (M log M )1=2 N 3=8 log1=2 N; with high probability. Consequently the polynomial with unimodular coecients r + q + R, and of degree N , satis es (C1 ? )N 1=2 j(r + q + R)(x)j (C2 + )N 1=2 for any > 0 and suciently large N . This solves the problem for this restricted set of N 's and the general result is easy to obtain by concatenating such a polynomial and a much shorter random one (see [37]). Using a much more elaborate argument, in which the Salem-Zygmund theorem again plays a central role, Kahane improved Korner's result.


21

Theorem 1.10 (Kahane [28], [29, p. 75]) For n suciently large there is a trigonometric polynomial

f (x) =

with jcj j = 1, which satis es

n X j =1

cj eijx

jf (x)j = (1 + o(1))n1=2

for all x 2 [0; 2 ).

1.4.4 The Rudin-Shapiro Polynomials. Spencer's Theorem. The Rudin-Shapiro sequence [55, 57], [30, p. 33], is a sequence of trigonometric polynomials of the form m ?1 2X Pm (x) = j eijx ; j = 1; m = 0; 1; 2; : : :; which have small maximum

j =0

kPmk1 C kPmk2 = C 2m=2; for a constant C . We de ne the double sequence Pm ; Qm inductively as follows. For m = 0 we de ne P0 (x) = Q0 (x) = 1 and

Pm+1 (x) = Pm (x) + ei2mx Qm (x); Qm+1 (x) = Pm (x) ? ei2mx Qm (x): It can be ver ed that

jPm+1 (x)j2 + jQm+1 (x)j2 = 2(jPm (x)j2 + jQm (x)j2): From this it follows immediately that for all m

jPm(x)j2 + jQm(x)j2 = 2m+1 and thus

p kPm k1 2 2m=2:

The Rudin-Shapiro polynomials have degrees which are powers of 2 but for every n we can get (by concatenating them) a polynomial of degree n, whose coecients are 1 and whose p maximum is bounded by a constant multiple of n.

22

Introduction

The Rudin-Shapiro polynomials are very explicitly constructed and their structure is perfectly clear from their simple de nition. Yet, if one changes the problem just a little bit the inductive construction cannot be salvaged at all. Suppose that we want to construct polynomials of degree n, whose coecients are 0 outside a given set E f1; : : :; ng and on E their coecients are restricted to be 1. Clearly the inductive de nition cannot work here. Essentially the best we can do is take a random polynomial and hope for the best. P Let r(x) = j 2E rj eijx be a random polynomial with the independent zero-mean random variables rj being 1 with equal probability. Then the Salem-Zygmund theorem gives that, with high probability, krk1 C (jE j log n)1=2: In the particular case where jE j n the result is clearly suboptimal, at least in the case where E is an arithmetic progression and thus we can have Rudin-Shapiro polynomials on it. It turns out that for any E we can nd signs rj for which krk1 Cn1=2. For this we need the following important theorem of J. Spencer which in certain cases is an improvement over plain randomization.

Theorem 1.11 (Spencer [59], Gluskin [23]) Let aij ; i = 1; : : :; n1; j = 1; : : :; n2 be such that jaij j 1. Then there are signs 1 ; : : :; n 2 f?1; 1g such that for all i n X 1=2 (18) j aij Cn1 : j =1 2

2

Notice that there is no dependence of the bound on n2 . Compare with Theorem 1.6.

Corollary 1.2 Let f1(x); : : :; fn(x); kfj k1 C , be trigonometric polynomials of degree at most m. Then there is a choice of signs 1 ; : : :; n 2 f?1; 1g such that

n

X

j fj

Cm1=2: j =1 1

Proof of Corollary 1.2: For i = 1; : : :; 10m; j = 1; : : :; n de ne aij = fj (xi), where

xi = i 102m . Let 1; : : :; n be the sequence of signs given by Theorem 1.11 for the matrix aij P and write f = nj=1 j fj . There is x0 2 [0; 2 ] such that jf (x0 )j = kf k1 . For some k we have jxk ? x0 j 102m . By Bernstein's inequality, kf 0 k1 mkf k1 , we get

jf (x0) ? f (xk )j 102m f 0 1 210 jf (x0)j


23

which, since 2=10 < 1, implies

n X kf k1 = jf (x0)j C jf (xk )j = C j akj Cm1=2 j =1

and the proof is complete. 2 Using Corollary 1.2 on the functions fj (x) = eijx for all j 2 E we conclude that there exist signs j such that

X

j eijx

C pn:

1

j2E

Thus on any subset E of f1; : : :; ng there is a polynomial with coecients 1 with maximum p less than C n.

Remarks

1. The proof of Spencer's theorem uses in an essential way the fact that the numbers j can take on the values 1. The proof does not work if j can take the values 1 ? pj or pj for some pj 2 [0; 1], while randomization (Theorem 1.6) applies to this case too. This situation can be remedied a little by the following generalization of Spencer's theorem [59, 44].

Theorem 1.12 Let aij ; i = 1; : : :; n1; j = 1; : : :; n2, be such that jaij j 1. Let also p1 ; : : :; pn 2 [0; 1]. Then there is a choice of j 2 f?pj ; 1 ? pj g; j = 1; : : :; n2, such that, for all i, n X a Cn1=2: (19) 1 j ij 2

2

j =1

We essentially prove this theorem in Chapter 2, Section 2.2, in the form of a result on trigonometric polynomials. Despite Theorem 1.12 we do not have a generalization of the form of Theorem 1.8. That is, while Theorem 1.12 is a generalization in the nonsymmetric case, the bound that we get does not involve the \variances" of the j 's. In other words we cannot take advantage of the fact that the pj 's might be small. An example might clarify the situation more. Suppose that we want to nd a polynomial

f (x) =

N X j =1

j eijx

24

Introduction

with small kf k1 and such that j 2 f?pj ; 1 ? pj g for all j = 1; : : :; N . Then the SalemZygmund theorem (in the form of Theorem 1.8) gives

11=2 0N X kf k1 @ pj (1 ? pj ) log N A j =1

while Spencer's theorem (in the form of Theorem 1.12) gives

kf k1 N 1=2: If the numbers pj are bounded away from 0 and 1, then clearly Spencer's theorem does better, but not necessarily if N X j =1

pj (1 ? pj ) = o( logNN ):

It would be very interesting to know what is the best possible result when N X j =1

pj (1 ? pj ) = o(N ):

2. Unlike plain randomization (Theorem 1.6) it is not known whether Spencer's theorem can be eciently derandomized. The reason for this is that the proof is not purely probabilistic. At some point in the proof the existence of two 1 assignments of the j 's is claimed such P that all sums (for all i) nj =1 aij j take approximately the same value for both assignments. No way is known to make this existential claim constructive. 2

1.4.5 Uchiyama's Theorem The following theorem was proved by Uchiyama [61]. It is related to both the Littlewood Conjecture and the Cosine Problem. In Chapter 4 we shall give some constructive analogs and improvements.

Theorem 1.13 Let A = fn1 < < nN g be a set of N positive integers. Then there is a subset E A such that

p

X ijx

(20)

e

C N; j 2E 1 where C is a positive constant.


25

Proof: Let g(x) = Pj2A eijx and f (x) =

X j 2A

j eijx ;

where j = 1 with equal probability and independently. By the triangle inequality it p suces to show that there is an assignment to j that makes kf k1 N . To this end we use Holder's inequality in the form

kf k22 kf k21=3 kf k44=3 : We always have kf k22 = N and

2 X jf j4 = j k ei(j?k)x : j;k

Thus writing (the indices j; k always run through A)

r(x) = we get and

kf k44 =

E[kf k44] =

X x2N

X

x=j ?k

j k

1 Z 2 jf j4 = X r2 (x) 2 0 x2N

E[r2(x)] =

X j ?k=j ?k 0

0

E[j k 0j 0k ]

and the only terms that will survive are those with j = j 0; k = k0 , thus

E[kf k44] = N 2: This implies the existence of an assignment of the j such that kf k44 N 2. Using Holder's inequality for this assignment we get kf k1 N 1=2 which concludes the proof. 2 A very similar proof but applied on cosine sums, instead of sums of exponentials, shows that there is always a subset E A such that

X

p

cos jx

N: j 2E

1

Since the absolute value of the minimum of a cosine sum dominates its L1 norm we also get

X pN: cos jx min x j 2E

26

Introduction

1.4.6 Bh g Sets of Integers [ ]

We introduce some concepts of Additive Number Theory. Let E = fn1 < n2 < g be a nite or in nite set of nonnegative integers. We de ne a corresponding representation function on N

r(x) = rE (x; 2) = jf(a; b) : a; b 2 E & a b & x = a + bgj:

(21)

A set E is in the class B2 if r(x) 1 for all x 2 N. In other words all sums of the form

a + b; a; b 2 E;

(22)

are distinct except for permutation of a and b. It is not hard to see that this condition is equivalent to requiring that all dierences

a ? b; a; b 2 E; a 6= b;

(23)

are distinct. The terminology \Sidon set" is sometimes used to describe B2 sets but we will avoid it since it has a rather dierent meaning in classical harmonic analysis. An immediate generalization of (21) is the following

rh (x) = rE (x; h) = jf(a1; : : :; ah) : aj 2 E & a1 ah & x = a1 + + ah gj: (24) We call a set E a Bh set if rh (x) 1 for all x 2 N. We call it a Bh [g ] set if rh (x) g for all x 2 N. Thus a Bh set is a set of which all sums of the form

a1 + + ah ; aj 2 E; a1 ah; are distinct.

The Density of Finite Bh [g] Sets The main question that we are interested in is \How big can a Bh [g ] subset of f1; : : :; ng be?" Considerably more is known when g = 1 (Bh sets) and among Bh sets much more is known about B2 sets. The reason is roughly that when g > 1 sets of the type Bh [g ] do not admit any characterisation in terms of distinct dierences like (23). Even for B2 [2] sets very little is known. De ne Fh (n) to be the maximum size of a Bh subset of f1; : : :; ng. Also let Fh;g (n) be the maximum size of a Bh [g ] subset of the same set. It is easy to see that

Fh;g (n) Ch;g n1=h :

(25)


27

Indeed by just counting distinct sums we can get

!

Fg;h (n) ghn h which implies

Fh;g (n) (ghh!)1=hn1=h + o(n1=h): The quantity of interest is the constant Ch;g in (25). For h = 2; g = 1 we can do better by counting the distinct positive dierences instead

of the sums. We have which gives

!

F2(n) n 2

p F2(n) 2n1=2 + o(n1=2):

This is far from best possible though. Erd}os and Turan [20] have proved

F2 (n) n1=2 + O(n1=4):

(26)

Not only is the coecient of the major term better but the error term is non-trivial too. On the other hand it has been shown by Chowla [10] and Erd}os [20, Addendum] that

F2(n) n1=2 ? o(n1=2):

(27)

Their method uses a result of Singer [58]. For B4 sets Lindstrom [42] has proved

F4 (n) (8n)1=4 + O(n1=8):

(28)

In Chapter 3 we will prove that for all m

F2m(n) (m(m!))1=2m n1=2m + O(n1=4m):

(29)

The coecient of the major term that can be obtained by just counting distinct sums or dierences is (2m(m!))1=2m. Notice also that we get a non-trivial error term, and that (26) and (28) are subsumed by our result. This result has also been proved independently and by a completely dierent method by Jia [39]. Graham [24] has proved an analogous result for F2m?1 (n): F2m?1 (n) (m!)2=(2m?1)n1=(2m?1) + O(n1=(4m?2)): (30)

28

Introduction

The Density of In nite Bh[g] Sets

p

While it is possible to have a B2 subset of f1; : : :; ng with about n elements, the following theorem of Erd}os shows that the situation is quite dierent if we look at in nite B2 sequences of high lower density.

Theorem 1.14 If the sequence fn1 < n2 < g N is B2 then we have j 0: lim sup j 2 nlog j j

(31)

Thus we cannot have a ( nite or in nite { the in nite sequence can be obtained from nite sequences by a diagonal argument) B2 sequence which satis es for all j

nj j 2 : On the other hand it is easy to see that we can have an in nite B2 sequence with large upper density.

Theorem 1.15 There is a B2 sequence fn1 < n2 < g N for which limjinf nj 2j A; where A is an absolute positive constant.

Erd}os [60], [25, p. 88] proved Theorem 1.15 with A = 4 and Kruckeberg [38], [25, p. 89] gave A = 2. This is still the best known constant, and it is not known whether it can be lowered to A = 1 to match the Erd}os-Turan bound (26). For a long time the B2 sequence with the highest lower density known was the one produced by the so called greedy method. Let n1 = 1 and having found n1 ; : : :; nk choose nk+1 to be the smallest positive integer x that is not in the set

fa + b ? c : a; b; c 2 fn1; : : :; nk gg : It then follows easily that the sequence nj is B2 and that nj j 3. The gap between this sequence and Theorem 1.14 still stands except for the following result of Ajtai, Komlos and Szemeredi [1].

Theorem 1.16 There is a B2 sequence fn1 < n2 < g N such that ! 3 j nj log j :


29

The following theorem of Erd}os and Renyi deals with dense in nite B2 [g ] sequences. The proof is once again probabilistic.

Theorem 1.17 (Erd}os and Renyi [18]) For every > 0 there is an integer g and a B2[g] sequence A = fa1 < a2 < g such that aj j 2+ ; (32) for all j > 0.

Proof: Let 2 (0; 1) be given. Let A be a random set with Pr [x 2 A] = px ; independently for all x 2 N, where

px = x?1=2?=2: Then with high probability A(x) x1=2?=2 for all x, which implies (32). Write, as usual, j = 1(j 2 A). Then we have

r(x) = and we can estimate

bX x=2c j =1

j x?j

E[r(x)] Cx? ;

R where C = 01=2(s(1 ? s))?1=2?=2ds. De ne the bad events

Ax = fr(x) > gg = fr(x) > ( E[rg(x)] )E[r(x)]g:

We now use Theorem 1.3 with observing that We get

= E[rg(x)] ? 1 Cgx; c log ; as ! 1:

Pr [Ax ] 2e?2c E[r(x)] 2e?3 log (E[r(x)]) = 2e?3g log ;

and using the estimate on E[r(x)] we get Pr [Ax ] Ce?Cg log x = Cx?Cg :

30

Introduction

P

Choose now g = C= , for large enough C , to get Pr [Ax ] x?2 and thus x Pr [Ax ] < 1. So there is n0 2 N for which X Pr [Ax ] < 1; xn0

so that with positive probability none of the bad events Ax ; x n0 , holds. Now discard all elements of the random set A up to n0 to get a B2 [g ] set with the desired growth. 2 See also the paper [17] by Erd}os for further results and conjectures on additive number theory.

Chapter 2

On Nonnegative Cosine Polynomials with Nonnegative Integral Coecients 2.1 Introduction We consider nonnegative cosine polynomials of the form 0 p(x) = p0 + p1 cos x + p2 cos 2x + + pN cos Nx ; x 2 [0; 2 ];

P where pj 0. We also write pb(0) = p0 . Notice that p(0) = Nj=0 pj is the maximum of p(x). We are interested in estimating the size of M (s) = p(0) inf pb(0) s

R

for s ! 1. That is, we want to nd polynomials of this form for which p0 = 21 02 p(x)dx is small compared to the maximum of p(x). If no more restrictions are imposed on the cosine polynomial p(x) then M (s) = 0 for all s. This is because the Fejer kernel

KA (x) =

A X 1 ? jj j eijx = 1 + 2 1 ? j cos jx A+1 A+1 j =1 j =?A A X

has constant coecient 1, has KA (0) A and is nonnegative. 31

32

Nonnegative Cosine Polynomials

If we restrict the coecients p1 ; : : :; pN to be either 0 or 1 we have the classical cosine problem (see Section 1.4.1), about which we know that for some > 0 2log s M (s) s1=2:

(33)

The upper bound in (33) is easily proved by considering the polynomial

f (x) = (

A X 1

cos 3j x)2

(34)

A A X X cos (3k + 3l )x + cos (3k ? 3l )x : = A + 21 cos (2 3j x) + j =1 k;l=1

(35)

k>l

All cosines in (35) have distinct frequencies. De ne

f1(x) = f (x) + 21 A ? 12

A X j =1

cos (2 3j x):

Then f1 (x) 0, f1 (0) A2, c f1(0) A and f1 has non-constant coecients which are either 0 or 1. The lower bound in (33) is much harder to prove and is due to Bourgain [7]. Earlier, Roth [54] had obtained M (s) (log s= log log s)1=2. From this point on, we will study the case of p1 ; : : :; pN being arbitrary nonnegative integers. This case was studied by Odlyzko [46] who showed that

M (s) (s log s)1=3:

(36)

See Chapter 1, p. 17, for the proof. Odlyzko studied this problem in connection with a problem posed by Erd}os and Szekeres [19]. The problem is to estimate

Y n a E (n) = inf max (1 ? z k ) ; jzj=1 k=1

where a1 ; : : :; an may be any positive integers. The following inequality holds (see [46]) log E (n) M (n) log(n)

(37)

so that Odlyzko's result implies log E (n) n1=3 log4=3 n. In this chapter we replace the random modi cation in Odlyzko's argument with a more careful modi cation, based, again, partly on randomization. We use Spencer's theorem

Proof of the Inequality M (s) s1=3

33

(Corollary 1.2) which in some cases does better than the Salem-Zygmund theorem. We show in Section (2.2) that, when p1 ; : : :; pN are restricted to be nonnegative integers, we have M (s) s1=3: By (37) this implies log E (n) n1=3 log n. The method employed has appeared in [59, 44] and is also similar to that used by Beck [4] on a dierent problem, posed by Littlewood. In Section 2.3 we give a deterministic procedure which, given a polynomial p(x) with nonnegative, integral Fourier coecients (in other words pj is a nonnegative even integer, for j 1) and with pb(0) (p(0)), for some > 0, produces a sequence of polynomials p = p(0); p(1); p(2); : : : , such that deg p(n) ! 1, p(n) (0) ! 1 and (p(n) )b(0) (p(n) (0)): This shows M (s) Cs1=, with C; dependent on the initial p only. This has appeared in [31].

2.2 Proof of the Inequality M(s) s1=3 Since Corollary 1.2 only allows us to choose random signs, we cannot use it directly (as we used the Salem-Zygmund theorem) to modify the coecients of a polynomial to integers, while controlling the size of the change. In this section we show how to modify the coecients little by little, to achieve the same result. Let > 0 and de ne A X a(x) = KA (x) = aj cos jx: j =0

Suppose > 0 and the nonnegative integer k0 is such that for some nonnegative integers bj

jaj ? bj 2?k j for all j = 1; : : :; A: 0

We shall de ne a nite sequence of polynomials

a(0)(x) = a0 +

A X

j =1

bj 2?k cos jx; a(1)(x); : : : ; a(k ) (x) 0

inductively, so that if

a(k) (x) = a

0+

0

A X j =1

a(jk) cos jx;

34


then for each j = 1 : : :A,

a(jk) = b(jk)2k?k

(38)

0

for some nonnegative integers b(jk). We de ne inductively the coecients of a(k+1) as follows. If b(jk) , j > 0, is even then a(jk+1) = a(jk) . Else de ne

a(jk+1) = a(jk) + (jk) 2k?k ; 0

(39)

where (jk) 2 f?1; 1g are such that

X ( k )

j cos jx

CA1=2:

b k odd j 1

(40)

( )

The existence of the signs (jk) is guaranteed by Corollary 1.2. Notice that (39) implies the preservation of (38) by the inductive de nition. We deduce from (40) that

(k+1) (k)

a ? a 1 C 2k?k A1=2: 0

(41)

The polynomial a(k ) has integral coecients (except perhaps for the constant coecient). Summing (41) we get 0

(k )

(0)

(0) (k )

a ? a 1 a ? a 1 + a ? a 1 A + CA1=2: 0

0

Choose = 1=A to get

(k )

a ? a 1 CA1=2: 0

On the other hand, the coecients of a and a(k ) dier by at most 1 and this implies that for the nonnegative polynomial 0

p(x) = a(k )(x) +

a ? a(k )

1 0

0

we have

p(0) a(0) ? A CA ? A;

pb(0) = +

a ? a(k )

1 + CA1=2: 0

(42) (43)

Select = A1=2 to get pb(0) A1=2 and p(0) A3=2. Since p has integral coecients, we have exhibited a polynomial that achieves M (s) s1=3 , and the proof is complete.

The construction

35

Remark on Cosine Sums

Applying the method of the preceding proof on the coecients of the Fejer kernel KA (x), one ends up with a nonnegative polynomial of degree at most A, which is of the form

p(x) = p0 + 2

k X

j =1

cos j x;

where j 2 f1; : : :; Ag are distinct. We have kKA ? pk1 A1=2 which, since p0 = 1 R 2 2 0 p(x)dx, implies p0 A1=2 and p(0) A: Thus p is a new example of a cosine sum that achieves the upper bound in (33). It is not as simple as the one mentioned in Section 2.1 but the spectrum of it is much denser: 1 1=2 2 A + O(A ) cosines with frequencies from 1 to A. Since the Dirichlet kernel

DA (x) =

A X

j =?A

= 1+2

eijx A X j =1

(44) cos jx

(45)

sin (A + 12 )x (46) sin x2 has a minimum asymptotically equal to ? 34 A, it is conceivable that one may be able to raise this number of cosines from 12 A + O(A1=2) to (1 ? 2 )A + o(A): 3 In other words, since A X min cos jx = ? 32 A + o(A); x =

j =1 2 one must remove at least 3 A cosines from the sum, in order to make its minimum be o(A), in absolute value. However we show in Section 3.2, that 12 A + O(A1=2) is best possible.

2.3 The construction Suppose we are given a polynomial p(x) 0 of degree d, whose non-constant coecients are even nonnegative integers, which satis es

pb(0) (p(0))

36


for some > 0. De ne the sequence of nonnegative polynomials p = p(1); p(2); p(3); : : : , with the recursive formula

p(k+1) (x) = p(k) ((2d + 3)x) p(x):

(47)

Since p has even non-constant coecients, the Fourier coecients of all p(k) are nonnegative integers. The spectrum of the rst factor in (47) is supported by the multiples of 2d + 3, and that of the second factor is supported by the interval [?d; d]. This implies that (p(k+1) )b(0) = (p(k))b(0)pb(0): We obviously have p(k+1) (0) = p(k)(0)p(0). We conclude that for all k 0 (p(k))b(0) = (pb(0))k and p(k) (0) = (p(0))k and consequently

(p(k) )b(0) (p(k) (0)):

So, if s is a power of p(0), we have M (s) s , and for any s we have M (s) Cs, where C = (p(0)). As an example we give

p(x) = 4 + 4 cos x + 2

10 X

j =2

cos jx

which can be checked numerically to be positive and has constant coecient pb(0) = 4 and p(0) = 26. This gives = log 4= log 26 = :42549 . In view of the construction above, nding a single polynomial p with pb(0) (p(0)), with < 1=3, will prove that the result in this Chapter is not the best possible. This example was actually found by a computer but if no more insight is gained into how these good \seed" polynomials look like, the computing time grows dramatically as we increase the degree of the polynomial.

Chapter 3

The Density of Bh g Sequences and the Minimum of Dense Cosine Sums [ ]

3.1 Introduction Let E be a set of integers. For any integer x we denote by rE (x; h) the number of ways x can be written as a sum of h (not necessarily distinct) elements of E . Two sums a1 + + ah and b1 + + bh are considered the same if the aj 's are a permutation of the bj 's. A set E of integers is called a Bh [g ] set if rE (x; h) g for all x. A Bh [1] set is also called a Bh set. We are interested here in the density of Bh [g ] sets. Considerably more is known about B2 sets than general Bh [g] sets ([25, Ch. 2] is the principal reference). The main reason for this is that a set E is B2 if and only if all dierences x ? y , for x; y 2 E; x 6= y; are distinct. Nothing similar is true for B2 [g ] sets, for example. Let Fh (n) be the maximum size of a Bh set contained in f1; : : :; ng. It is obvious that Fh (n) Cn1=h and it is the size of this constant C that we care about in this chapter. For h = 2 Erd}os and Turan [20], using a counting method, have proved p (48) F2(n) n + O(n1=4): p (The constant one obtains by just counting dierences is 2.) For h = 4 Lindstrom [42], using van der Corput's lemma, has proved F4(n) (8n)1=4 + O(n1=8): (49) 37

The Density of Bh [g ] Sequences and Cosine Sums

38

In the other direction it has been shown by Chowla [10] and by Erd}os [20, Addendum] using a theorem of Singer [58] that p p F2 (n) n ? o( n) (50) and more generally it has been proved by Bose and Chowla [5] that

Fh (n) n1=h ? o(n1=h ):

(51)

The Cosine Problem: Chowla [11] has conjectured that for any distinct positive integers n1 < < nN N X p (52) ? min cos nj x C N: x j =1

This conjecture has remained unproved and the best result known to date is due to Bourgain [7]:

? min x

N X

j =1

cos nj x 2log N

for some > 0. It is easy to see that there are sequences fnj g for which the left hand side of (52) is p bounded above by C N . We proved in Chapter 2, p. 35, that there are very dense such sequences: we can have nN 2N . (This can also be proved using (50).) In Section 3.2 we prove that the density above is best possible. In Section 3.3 we use this result to prove the following theorem.

Theorem 3.1 Let h = 2m 2 be an even integer. Then 1=h Fh (n) m(m!)2 n1=h + O(n1=2h):

(53)

Theorem 3.1 contains the results of Erd}os and Turan (48) and Lindstrom (49) as special cases. It was also proved recently by Jia [39] who used an elementary combinatorial argument. In Sections 3.4 and 3.5 we show that allowing g > 1 indeed helps. We exhibit a B2 [2] p p subset of f1; : : :; ng with 2n + o( n) elements and an in nite B2 [2] sequence 1 n1 < < nj < for which limjinf nj 2j = 1:

Dense Cosine Sums

39

3.2 Dense Cosine Sums It was proved in Section 2.2 that for every positive integer N there are positive integers 1 1 < < N 2N such that

? min x

N X 1

p

cos j x C N:

(54)

We now prove that we cannot have more dense cosine sums whose minimum is small in absolute value.

Theorem 3.2 Let 0 f (x) = M + PN1 cos j x, with 1 1 < < N (2 ? )N , for some > 3=N . Then

M > C2 N:

(55)

Proof: We use the following theorem of Fejer [21]: Theorem 3.3 Let p(x) be a nonnegative trigonometric polynomial of degree n and constant term pb(0) = 1. Then p(0) n + 1. The obvious inequality above is p(0) 2n +1. We note that Theorem 3.3 is a corollary of the well known theorem of Fejer and Riesz which states that every nonnegative trigonometric polynomial can be written as the square of the modulus of a polynomial of the same degree. To use Theorem 3.3 we rst need to \smooth" fb. De ne p(x) = f (x)Ka(x) 0, where

Ka(x) =

j j j 1 ? (a + 1) eijx 0

a X

j =?a

is the Fejer kernel of degree a (the parameter a will be determined later). Then deg p (2 ? )N + a; X pb(0) = M + a1 (a ? j ); j a

p(0) = (M + N )(a + 1):

Observe that pb(0) M + a=2 and apply Theorem 3.3 to get

p(0) (1 + deg p)pb(0)


40 or

M + a2 (2 ?(M)N+ +N )aa+ 1 2 ? + (1a + 1)=N a: Let a = N=2 to get

1 1 M 4 ? + 2=N ? 4 N 2=N ) N ( ?16 2

3 16 N;

since > 3=N . 2

3.3 An Upper Bound for Fh(n), h Even Let h = 2m 2 be a xed even integer. We shall give an upper bound for the size of Bh sets contained in f1; : : :; ng. Theorem 3.2 is the main tool. In this section C denotes an arbitrary positive constant which may depend on h only. Let E = fn1; : : :; nk g, 1 n1 < < nk n, be a Bh set. This means that all sums a1 + + ah with aj aj+1 and aj 2 E are distinct. Consequently the sums of the form

a1 + + a m ? b 1 ? ? b m with

aj ; bj 2 E; aj < aj+1 ; bj < bj+1 and ai 6= bj

P P P P are all dierent. Indeed, if m1 aj ? m1 bj = m1 a0j ? m1 b0j we have m X 1

aj +

m X 1

b0j =

m X 1

a0j +

m X 1

(56)

bj

and, since fnj g is a Bh sequence, the collection of terms in the left hand side is the same as that in the right hand side. But the aj 's have been assumed dierent from the bj 's, so we must have aj = a0j and similarly bj = b0j , for all j .

Dense Finite B2 [2] Sequences

41

De ne the nonnegative polynomial

k X inj x h f (x) = e 0j=1k 1m 0 k 1m X X = @ einj x A @ e?inj x A j =1 0j=1 1 X X X = r(x) + 2(m!)2 @ cos ( aj ? bj )xA : aj ; bj satisfy (56)

The polynomial r(x) consists of O(kh?1 ) terms with coecient 1, and thus, for some C > 0,

Ckh?1 +

X

m X

aj ?

m X

bj )x 0: j =1 j =1 aj ; bj satisfy (56) kh =(2(m!)2) ? O(kh?1 ), for the positive cos (

Write < < , N = sums of the form Pm a ?1 Pm b (theyNare all dierent). Using Theorem 3.2 we conclude that 1 j 1 j 1 h h ? 1 ? 1 = 2 (57) mn N 2 ? Ck 2(m!)2 k ? O(k ) : This implies

k (m(m!)2)1=hn1=h + o(n1=h ): The error term can also be bounded as follows. Write k = C1n1=h + R, where R 0 and C1 = (m(m!)2)1=h. Then n = ((k ? R)=C1)h , and substituting this in (57) and matching the second largest terms we get

R = O(k1=2) = O(n1=2h ); which concludes the proof of Theorem 3.1. Theorem 3.1 improves the estimate one gets by just counting the j 's. Indeed, there are N = kh =(2(m!)2) ? O(kh?1 ) dierent j 's in f1; : : :; mng so we only get

k 2m(m!)2

1=h

n1=h + o(n1=h ):

(58)

3.4 Dense Finite B2[2] Sequences As mentioned in Section 3.1, for each n there is a B2 sequence 1 n1 < < nk n with k = pn + o(pn). In this Section we show that if one allows up to 2 sums to coincide we can have denser sequences. We do this by interleaving two dense B2 sequences.


42

Theorem 3.4 For each n there is a B2[2] set B f1; : : :; ng with p jBj = 2n + o(pn):

(59)

p Proof: By (50) there is a B2 set A f1; : : :; bn=2c ? 1g, with jAj = n=2 + o(pn). We

shall show that the subset

B = 2A [ (2A + 1)

of f1; : : :; ng is B2 [2] which proves the theorem. The proof is by contradiction. Assume that we have the non-trivial relations

x1 + y1 = x2 + y2 = x3 + y3 ;

(60)

with xj ; yj 2 B and let z = x1 + y1 . Look at xj + yj mod 2. There are three possible patterns: 0 + 0, 1 + 1 and 0 + 1. If z is even then only 0+0 and 1+1 may appear in (60) and we have either a relation of the pattern 0+0 = 0+0 or a relation of the pattern 1+1 = 1+1. Both cases contradict the fact that A is B2 , the rst after just dividing by 2, the second after canceling the remainders and then dividing by 2. If z is odd then only the pattern 0 + 1 appears in (60) which can be rewritten as 2a1 + (2a01 + 1) = 2a2 + (2a02 + 1) = 2a3 + (2a03 + 1)

(61)

with aj ; a0j 2 A. By canceling 1 and dividing by 2 we have

a1 + a01 = a2 + a02 = a3 + a03 : But A is B2 so for at least one of these relations, say the rst one, we have a1 = a2 and a01 = a02 which contradicts the fact that the rst relation in (61) is non-trivial. 2 Jia [40] recently improved (59). He constructed a B2 [2] set B f1; : : :; ng with

p jBj = 3n + o(pn):

(62)

3.5 In nite B2 [2] Sequences with Large Upper Density The situation is rather dierent for in nite B2 [g ] sequences. Erd}os [60] has proved that there is no in nite B2 sequence fnj g with nj = O(j 2). In nity is not the problem here but

In nite B2 [2] Sequences

43

the fact that we require nj Cj 2 for all j and not just for the last one, as we did with nite sequences. For the upper density of the sequence fnj g Erd}os [60] proved that it is possible to have limjinf nj 2j 4

and Kruckeberg [38] later improved this to

limjinf n2j 2:

j

(63)

It is still unknown whether the number 2 in the right hand side of (63) can be reduced (it cannot be less than 1 by (48)). We now show that for a B2 [2] in nite sequence this is possible.

Theorem 3.5 There is a B2 [2] sequence fnj g with

limjinf nj 2j = 1:

(64)

Proof: The theorem will be proved if we show that any B2 [2] sequence 1 n1 < < nk can be extended to a sequence 1 n1 < < nk < nk+1 < < nl , such that nl = l2 + o(l2).

Write A = fn1 ; : : :; nk g and x = nk . Take B f2x + 1; : : :; x4g to be a B2 set with jBj = x2 + o(x2) (this is possible by (50)). In what follows aj 2 A, bj 2 B and dj 2 D (to be de ned below). Consider the relations of the form

a1 + b 1 = a 2 + b 2 :

(65)

Such a relation may be written as a1 ? a2 = b2 ? b1. But B is a B2 set, so all dierences b2 ? b1 are distinct, which implies that a pair a1 ; a2 2 A may appear in (65) only once. Thus there are O(k2 ) = O(x) of these relations which may involve O(x) elements of B . Let then

D = fb 2 B : b does not appear in any relation of the form (65)g and E = A [ D. Obviously jE j = x2 + o(x2 ). We show that E is a B2 [2] set. First note that the relations of the form

a1 + a2 = a3 + d1 a1 + a2 = d1 + d2

(66)

44


are not possible (the left hand side is too small) and A is itself B2 [2]. This proves rE (a1 + a2 ; 2) 2 for all a1; a2 2 A. It remains to be checked that rE (a1 + d1 ; 2) 2 and rE (d1 + d2; 2) 2. By passing from B to D we eliminated all relations of the form (65) and so the only remaining non-trivial relations that we have to check are of the form

a1 + d1 = d2 + d3:

(67)

These are indeed possible. Assume y = a1 + d1 = d2 + d3 . We have to show that these are the only ways that y can be written as a sum of two elements of E . But this is obvious since y = d02 + d03 is impossible (this would mean d2 + d3 = d02 + d03 which contradicts D in B2 ), y = a01 + a02 is impossible because of size and y = a01 + d01 would mean that a01 + d01 = a1 + d1 which we took care to eliminate in (66). 2 Remark: Because of the result of Section 3.4 the previous theorem is not necessarily best possible.

Chapter 4

A Construction Related to the Cosine Problem In a result related to the Cosine Problem (see Chapter 1, p. 10) Uchiyama [61] proved that for any sequence of N distinct positive integers n1 < < nN there is always a subsequence m1 ; : : :; mr for which r X p ? min cos m x C N: (68) j x j =1

He actually proved the stronger statement

r p 1 Z 2 X cos mj x dx C N: 2 0 j =1

(69)

In this chapter we improve (68). (To appear in [33].)

Theorem 4.1 For any sequence n1; : : :; nN of positive integers there is a subsequence m1 ; : : :; mr such that

? min x

r X j =1

cos mj x CN:

Theorem 4.1 is an obvious corollary of the more general theorem that follows.

(70)

Theorem 4.2 Let wk 0 and w = P11 wk < 1. Then there is a set E of positive integers for which X ? min wk cos kx Cw: (71) x k2E

45

46

A Construction Related to the Cosine Problem

The essential content of this chapter is that the proof of Theorem 4.2 (and consequently of Theorem 4.1) we give is constructive. Indeed there is a simple non-constructive proof of our theorem. Proof of Theorem 4.2 { Non-constructive: (Odlyzko [46]) De ne

f (x) = Then

1 X 1

wk (cos kx)? :

1 1 Z 2 (cos kx)? dx 1 Z 2 f (x)dx = X w k 2 0 k=1 2 0 = ? 1 w:

(72) (73)

Thus there is x0 2 [0; 2 ) with f (x0) ? 1 w. Let E = fk 2 N : cos kx0 0g. Then obviously X cos kx0 ? 1 w: k2E

2

We now give a constructive proof of Theorem 4.2 with a worse constant. (See Remark 1 after the proof for the exact meaning of the word \constructive".) We shall need two lemmas.

Lemma 4.1 Let Ik = (ak ; bk) (0; 1), k = 1; 2; : : :; be intervals of length at least > 0 P w < 1. Then and wk be nonnegative weights associated with them. Let also w = 1 1 k there is an interval J (0; 1), with jJ j = =2, for which X wk 21 w: (74) J Ik

Proof of Lemma 4.1: Let m = b2=c and J = [=2; ( +1)=2), for = 0; 1; : : :; m ? 1. P w . Since w = Pm?2 s there is some m ? 2 for which Write also s =

ak 2J k

0

0

s mw? 1 21 w: Let J = J +1 . Then J satis es (74) since ak 2 J implies J +1 Ik . 2 0

0

0

0

The following lemma is a useful special case of Theorem 4.2.


47

Lemma 4.2 Let a > 0, > 1, 24, Ej0 = [j a; ja) \ N; for j = 0; 1; 2; : : :, and

E0 =

1 [ j =0

Ej0 :

P w < 1 and w = 0 outside E 0. Then there is a set E E 0 Assume also wk 0, w = 1 k 1 k for which X 1 w: (75) ? min w cos kx k x 48 k2E

Proof of Lemma 4.2: First observe that in any interval of length at least 2=k there is a subinterval of length 2=(12k) in which cos kx ?1=2. According to this observation, for all k 2 E00 there is an interval Ik contained in (0; 2=a), of length at least 2=(12a), in which cos kx ?1=2. By Lemma 4.1 ( = 1=(12 )) there is an interval J0 (0; 2=a) of length 2=(24a) for which X X wk 241 wk : (76) J0 Ik

k2E0 0

Let E0 = fk 2 E00 : J0 Ik g. Then

X

k2E0

wk cos kx ? 481

X k2E0

wk ; for all x 2 J0.

(77)

0

Similarly we can nd an interval J1 J0 , with jJ1 j = 2=(12a), and E1 E10 , such that X X wk ; for all x 2 J1 . wk cos kx ? 481 k2E k2E 0

1

1

This is possible since 24 and therefore J0 is big enough to accomodate all frequencies in E10 . In the same fashion we de ne J2 J3 : : :; and E2; E3; : : :: Finally we set E = S1 0 Ej . It follows that (75) is true. 2 We can now complete the proof of the theorem. Proof of Theorem 4.2 { Constructive: Let = 2, = 64 and write for = 0; : : :; 5

A =

1 [

j =0

[j ; j +1 ) \ N:

S Since N = 50 A there is some 0 for which X k2A0

wk 61 w:

(78)

48


An application of Lemma 4.2 with = 2, = 64, a = 1 and the collection of wk for k 2 A furnishes a set E A for which X 1 w: ? min wk cos kx 6 48 x 2 k2E

0

0

2

Remarks 1. The simple proof of Theorem 4.1 mentioned can of course be made constructive by

looking for an x that satis es

N X 1

(cos nk x)? ? 21 N

among the points xj = jh, for j = 0; : : :; b1=hc. But h has to be smaller than Cn?N1 and this leads to an algorithm which in the worst case takes time exponential in the size of the input (which is considered to be the number of binary digits required to write down all n1 ; : : :; nN ). For example if nN = 2N then the algorithm needs time at least C 2N but the size of the input is at most N 2. In contrast, our construction takes time which is polynomial in the size of the input (in other words, polynomial in N log nN ). Assume that we are given N positive integers n1 nN and let L = dlog2 nN e. De ne wj = jfk 2 N : j = nk gj. The algorithm we described consists of the following steps. The notation of Lemma 4.2 is used throughout. 1. Find for which 0 2 f0; : : :; 5g inequality (78) is true. 2. Construct the sequence of intervals J0 J1 and the sequence of sets E0; E1; : : :: This proceeds inductively. Having constructed the interval Jm?1 and the set Em?1 we a. construct the intervals Ink for all nk 2 Em0 , b. nd (as described in Lemma 4.1) a subinterval Jm of Jm?1 which is big and is contained in many of the Ink 's. The set Em consists of those nk 2 Em0 for which Jm Ink . Notice that the sequences Jm and Em have length O(L). After observing that we never need to perform arithmetic with more than O(L) binary digits, it is easy to see that all the above can be carried out in time O(N L2 ), since an algebraic operation on two numbers, with O(L) binary digits each, takes O(L2) time.


49

2. Uchiyama's proof of (68) is probabilistic (see Section 1.4.5). We give an even simpler

constructive proof of (68). (Of course Uchiyama proved the stronger statement (69) about the L1 norm of a subseries.) Assume n1 n2 nN and let be any xed number between 2 and 3, say = 5=2. Observe that if nN n1 then

? min x

N X j =1

cos nj x CN;

P

as can be seen by evaluating the function Nj=1 cos nj x for x = (=2+ )=n1, where = () is a small positive constant. Let 1 = n1 and de ne k 2 fn1 ; : : :; nN g recursively by

k = min fnj : nj > k?1g [ fnN g: Let L be the length of the sequence k . That is let L be the rst equal to nN . Then p either L N or there is some k for which the set

p

A = fnj : k nj < k+1 g

has more than N elements. In the rst case we have

? min x

LX ?1 j =1

p

cos j x CL C N;

since the j 's form a lacunary sequence with ratio > 2. Otherwise, according to the observation above, we have

? min x

X

nj 2A

p

cos nj x C jAj C N;

which completes the proof. 3. It is easy to see that Theorem 4.2 holds also for complex wk , with w = P jwk j < 1 and writing eikx in place of cos kx. Also the minimum in (71) has to be interpreted as the minimum (or maximum) of the real part.

Chapter 5

An Eective Additive Basis for the Integers 5.1 Introduction A set E of nonnegative integers is called a basis if every nonnegative integer can be written as a sum of two elements of E . We write r(x) = rE (x) for the number of representations of x as a + b, with a; b 2 E and a b. Erd}os [14, 15] has proved that there is a basis E such that

C1 log x r(x) C2 log x

(79)

for all positive integers x and for some absolute constants C1; C2. For the proof see Chapter 1, p. 6 (see also [2, p. 106] and [25, Ch. 3]). The most widely known proof (in Chapter 1 and [2, 15, 25]) is probabilistic. It is proved that if we let x 2 E with a certain probability px, independently for all x, then the random set E is such an asymptotic basis (that is (79) is true for suciently large x) with probability 1. Since the probability space used is in nite, the question of whether such a basis exists which is also computable is not addressed by this proof. The original [14] proof though, which has been stated using counting arguments and not probability, uses an existential argument on a nite interval at a time and can thus be readily turned into a construction by examining all possible intersections of E with the interval. But the algorithm which we get this way takes time exponential in n to decide whether n is in E or not. 50

Probabilistic Proof of Existence

51

In this Chapter, we give an algorithm which produces the elements of E one by one and in increasing order, and which takes time polynomial in n in order to produce all the elements of E not greater than n. We use the method of conditional probabilities (Chapter 1, p. 8 and [2, p. 223]) in order to \derandomize" a modi ed proof. The method is not directly applicable to Erd}os's probabilistic proof. We will only show that (79) holds for x large enough, since, then, with the addition of a nite number of elements to the set E we can have it hold true for all positive x. In Section 5.2 we give a probabilistic proof of the existence of a basis with certain properties. In Section 5.3 we apply the method of conditional probabilities to derandomize the proof and arrive to our algorithm. This will appear in [34].

5.2 Probabilistic Proof of Existence We de ne the modi ed representation function r0(x) = rE0 (x) as the number of representations of the nonegative integer x as a sum a + b, with a; b 2 E , g (x) a b, where g(x) = (x log x)1=2. (This is our main dierence from Erd}os's proof. By doing this modi cation we have achieved that the presence or absence of a certain number n in our set E aects r0 (x) for only a nite number of nonnegative integers x.)

Theorem 5.1 There are positive constants c1; c2; c3, with c2 < c3, and a set E of positive integers such that

and

c2 log x r0(x) c3 log x

jE \ [x ? g(x); x]j c1 log x

for all large enough x 2 N.

Proof: In what follows x is assumed to be suciently large. We de ne the random set E by letting 1=2 Pr [x 2 E ] = px = K logx x independently for all x 2 N, where K is a positive constant that will be speci ed later. We are going to show that with positive probability (in fact almost surely but we do not need

52

An Eective Additive Basis for the Integers

this here) the random set E satis es Theorem 5.1. Let x=2 X

= E[r0 (x)] = De ne also

t=g(x)

pt px?t :

s(x) = jE \ [x ? g(x); x]j

and

= E[s(x)] =

First we estimate and for large x. We have

x=2 X t=x= log x

x X t=x?g(x)

pt :

pt px?t x=2 X

x

log x t=x= log x (t(x ? t)) = (1 + o(1))IK 2 log x;

K 2 log

?1=2

R where I = 01=2(s(1 ? s))?1=2ds, and

x=2 X 1

ptpx?t

x=2 X K 2 log x2 (t(x ? t))?1=2 1

= (1 + o(1))IK 2 log x;

which proves = (1 + o(1))IK 2 log x. For we have g(x)) 1=2 Kg(x) log x 1=2 ; Kg(x) log (x ? x x ? g(x) which implies = (1 + o(1))K log x: We de ne the \bad" events

Ax = f r0 (x) ? > g Bx = fs(x) ? > g

Derandomization of the Proof

53

for a positive constant . To bound their probabilities we need the following Theorem 1.3. Since both r0(x) and s(x) are sums of independent indicator random variables we can use the theorem to get Pr [Ax ] 2e?c 2e? c IK log x = 2x? 1 2

and

2

Pr [Bx ] 2e?c 2e? c K log x = 2x? 1 2

where = 12 c IK 2 and = 21 c K . We now let = 1=2 and choose K large enough to make both and greater than 1. Then 1 X Pr [Ax ] + Pr [Bx ] < 1 x=1

which implies the existence of n0 2 N such that, with positive probability, none of the events Ax and Bx , x n0 , holds. In particular there exists a set E for which

=2 r0(x) 3=2 and

s(x) 3=2;

for all x n0 . This implies the conclusion of Theorem 5.1 with c1 = 21 K , c2 = 21 IK 2 and c3 = 32 IK 2. 2 Observe that r0(x) r(x) r0(x) + s(x). We deduce that for the set E of Theorem 5.1 we have c2 log x r(x) (c1 + c3) log x so that (79) is true for E .

5.3 Derandomization of the Proof We keep the notation of the previous section. We showed that for some n0 2 N the complement of the bad event [ B = (Ax [ Bx ) xn0

has positive probability, by establishing the inequality

X

xn0

Pr [Ax ] + Pr [Bx ] < 1:

54

An Eective Additive Basis for the Integers

This implies the existence of a point E in our probability space f0; 1gN which is not in B (there is a natural identi cation between points in the probability space and subsets of N). In this section we are going to show how to construct eciently such a point E . We give an algorithm which at the n-th step outputs 0 or 1 to denote the absence or presence of n in our set E . Denote by 2 f0; 1gN a generic element in our space and by R(a1; : : :; ak ) the event 1 = a1; : : :; k = ak , where a1 ; : : :; ak 2 f0; 1g. It is obvious that for any event D f0; 1gN Pr [D j R(a1; : : :; an?1 )] = pn Pr [D j R(a1 ; : : :; an?1; 1)] + (1 ? pn )Pr [D j R(a1; : : :; an?1 ; 0)]:

(80)

We are going to de ne the sequence an 2 f0; 1g so that the function

bn = bn(a1 ; : : :; an) =

X

xn0

Pr [Ax j R(a1; : : :; an )] + Pr [Bx j R(a1; : : :; an )]

is non-increasing in n. (Notice that the function Pr [Ax j R(a1; : : :; an )] is constant in n when n > x, and is equal to either 0 or 1. The same is true for the events Bx .) Since b0 = Pxn Pr [Ax ] + Pr [Bx ] < 1, the monotonicity of bn implies that 0

X

xn0

Pr [Ax j R(a1; : : :; an; : : :)] + Pr [Bx j R(a1; : : :; an; : : :)] < 1:

The probabilities above are either 0 or 1, so they are all 0, and the point E = (a1 ; : : :; an ; : : :) is not in B . So all that remains to be done is to ensure that bn does not increase. Adding up (80) we get

bn?1 (a1; : : :; an?1 ) = pnbn(a1 ; : : :; an?1 ; 1) + (1 ? pn )bn(a1 ; : : :; an?1; 0); which implies that at least one of bn (a1; : : :; an?1 ; 1), bn (a1 ; : : :; an?1; 0) is not greater than bn?1 (a1 ; : : :; an?1 ). We let an = 1 if the rst number is smaller than the latter, otherwise we let an = 0. Notice that = bn (a1; : : :; an?1 ; 1) ? bn (a1; : : :; an?1 ; 0) =

GX (n) x=n

Pr [Ax j R(a1 ; : : :; an?1; 1)] ? Pr [Ax j R(a1; : : :; an?1 ; 0)] +

+Pr [Bx j R(a1; : : :; an?1; 1)] ? Pr [Bx j R(a1 ; : : :; an?1; 0)];

Derandomization of the Proof

55

where G(n) = (1 + o(1))n2= log n is the greatest integer k such that g (k) n. This is so because the events Ax and Bx , with x > G(n) are independent of 1 ; : : :; n and their probabilities cancel out in the dierence above. We have to decide in time polynomial in n whether 0. This is indeed possible since the expression for has (4 + o(1))n2= log n terms, each of which can be computed in polynomial time as the following Lemma claims.

Lemma 5.1 Let Xk = 1 + + k be a sum of k independent indicator random variables

with Pr [j = 1] = pj , j = 1; : : :; k. Then the distribution of Xk can be computed in time polynomial in k.

Proof: The distribution of Xk is a vector of length k + 1, where the j -th coordinate in the

vector, j = 0; : : :; k, is equal to Pr [Xk = j ]. To compute the distribution of Xk from that of Xk?1 we use the obvious formulas Pr [Xk = j ] = pk Pr [Xk?1 = j ? 1] + (1 ? pk )Pr [Xk?1 = j ]; for j = 1; : : :; k ? 1; Pr [Xk = 0] = (1 ? pk )Pr [Xk?1 = 0] and Pr [Xk = k] = pk Pr [Xk?1 = k ? 1]. It is obvious now that the computation of the distribution of Xk can be carried out in time polynomial in k. (Here we are really assuming that arithmetic operations on the numbers pj can be done in time polynomial in k. See the Remarks at the end of the section for a justi cation of this assumption.) 2 Thus all probabilities of the form Pr [ < Xk < ] can be eciently computed. Observe that having xed 1 = a1; : : :; n = an we have

r0(x) = =

x=2 X

t x?t

t=g(x) x=2 n X X atx?t + t x?t n+1 g(x)

for x ? g (x) > n, otherwise r0 (x) has already been completely determined by the assigned values of 1 ; : : :; n . This means that r0 (x) is a sum of independent indicator random variables and so is s(x). Thus the probabilities of Ax and Bx conditioned on R(a1; : : :; an?1; 1) and R(a1; : : :; an?1 ; 0) can be eciently computed and 0 can be decided in polynomial time, as we had to show.

Remarks:

56

An Eective Additive Basis for the Integers 1. Our de nition of the probabilities px = K (log x=x)1=2 has to be modi ed so that the numbers px can be represented with a number of digits polynomial in x and can also be computed in polynomial time, given x. One such modi cation is to use the p probabilities qx = K 2?bL=2c S , where L = blog2 xc and S = b Lc. The number S can for example be computed in time polynomial in log L (and in particular in x) using a simple binary search of the interval [0; L]. Since px < Cqx < Cpx one can easily prove asymptotic estimates of the form CIK 2 < < CIK 2 and CK log x < < CK log x, which is all our existential proof needs. 2. Ignoring polylogarithmic factors, the time our algorithm needs to decide whether n 2 E , having already found the set E up to n ? 1, is O(n6 ). This is so since the distribution of Xk in Lemma 5.1 can be computed in time O(k2 ). So the computation of any probability of the form Pr [ < Xk < ] can be computed in time O(k2 ). For the computation of we need to evaluate O(n2 ) such probabilities with k = O(n2 ), thus the total time is O(n6).

Chapter 6

On a Problem of Erd}os and Turan and Some Related Results 6.1 Introduction All sequences we consider are sequences of distinct positive integers. We denote by the lower case indexed letter the members of the sequence and by the capital letter the sequence as a set as well as its counting function. For example A = fa1 ; a2; : : :g denotes a sequence of distinct positive integers and A(x) = jA \ [1; x]j denotes its counting function. We de ne

A (x) = jf(a; b) : a; b 2 A; x = a ? bgj;

hA;N (x) = f(a; b) : a; b 2 A \ [1; N 2]; x = a ? bg ; HA (N ) = and

N X x=1

hA;N (x);

rA(x) = jf(a; b) : a; b 2 A; a b; x = a + bgj:

A conjecture of Erd}os and Turan [20] asserts that for any asymptotic basis (of order 2) of the positive integers, that is for any set E N = f1; 2; 3; : : :g for which rE (x) > 0 for all suciently large x, we must have lim sup rE (x) = 1: x!1

57

58

A Problem of Erd}os and Turan

Erd}os [60] has proved that it cannot eventually be true that rE (x) = 1, by showing that for p any sequence E , with E (x) x, we have

HE (N ) N log N

p

(81)

for all N . Indeed, any asymptotic basis E satis es E (x) x and, if rE (x) = 1, all sums we can form with two elements of E (with the exception of a nite number of elements of E ) are distinct. This in turn implies that so are all the dierences, that is E (x) 1 for all x, which makes (81) impossible. Recently Helm [27] proved that (81) is best possible by explicitly constructing a sequence A, with A(x) px, for which HA(N ) N log N (82) for all suciently large N . Helm's proof does not provide any upper or lower bound on the individual hA;N (x) for x 2 [1; N ], but only describes the average behaviour. In addition to this result Helm [27] constructed two sequences B and M , with B (x) px and M (x) log x, for which (m ) = 1, for all k suciently large. B k In this Chapter we shall use the probabilistic method to improve both results of Helm. Throughout this Chapter a random sequence A is de ned by letting x 2 A with probability

8 K < p if x K 2; px = : x 0 otherwise;

for appropriately chosen K , independently for all x. We prove

Theorem 6.1 Let A be a random sequence as de ned above. Then, with probability 1, there

is an integer N0 and positive constants c1; c2; c3; c4 such that

and

p p c1 x A(x) c2 x

(83)

c3 log N hA;N (m) c4 log N

(84)

for all x; N N0 and 1 m N .

This implies the rst result of Helm and with upper and lower estimates on the individual hA;N (m). We also prove

Proofs

59

Theorem 6.2 Let M be any sequence which satis es the growth condition 1 log m X pm k < 1: k

1

(85)

Let also A be a random sequence. Then, with probability 1, there is a subsequence B of A, an integer N0 and positive constants c3; c4 such that

p p c3 x B(x) c4 x

(86)

B (mk ) = 1

(87)

for all x N0 and for all k N0 .

Helm's second result follows from Theorem 6.2, but Theorem 6.2 is much stronger, since we are free to choose the sequence M , subject only to the growth condition (85). This work will appear in [35].

6.2 Proofs Proof of Theorem 6.1: Write j = 1 if j 2 A, j = 0 otherwise, so that E[j ] = pj . Notice that

A(x) = hA;N (m) =

x X

j ; j =1 2 ?m NX j =1

j j+m ;

so that A(x) is a SIIRV and hA;N (m) is the sum of two SIIRV:

hA;N (m) = heA;N (m) + hoA;N (m); where and

heA;N (m) =

m X X j =1 k even

j+km j+(k+1)m

m X X o j+km j+(k+1)m : hA;N (m) = j =1 k odd

60


(We broke up hA;N (m) so that each j appears at most once in each heA;N (m) and hoA;N (m).) Then, as x ! 1,

E[A(x)] =

x X j =1

pj

x K X pj

j =1

x 1 1 X p = K x xp j=x j =1 p 2K x;

since 2 =

R 1 ds=ps. We also have, for m N and N ! 1, 0

E[hA;N (m)] =

2 NX ?m

pj pj+m j =1 2 ?m NX 2 p 1 K j (j + m) j =1 2 NX ?m 1 2 K 2K 2 log N; j j =1

and

E[hA;N (m)] So we have

(1 + o(1))K 2

2 NX ?m

j =1 2 (1 + o(1))K log N:

1 j +m

E[A(x)] 2K px

(88)

(1 + o(1))K 2 log N E[hA;N (m)] (1 + o(1))2K 2 log N

(89)

as x ! 1 and as N ! 1, and for all m N . Notice that E[heA;N (m)] 21 E[hA;N (m)] and E[hoA;N (m)] 1 E[h (m)]. A;N 2 Now x = 1=2 and de ne the bad events

Px = fjA(x) ? E[A(x)]j > E[A(x)]g; QN;m = fjhA;N (m) ? E[hA;N (m)]j > E[hA;N (m)]g;

Proofs

61

for all x; N and m N . Using Theorem 1.3 and the remarks following it we have p Pr [Px ] 2 exp (?c E[A(x)]) 2 exp (? 21 c 2K x) and Pr [QN;m ] 4 exp (? 31 c E[hA;N (m)]) 4 exp (? 61 c K 2 log N ) = 4N ? c K for x and N suciently large. Thus 1 6

1 X x=1

Pr [Px ] +

1 X N X N =1 m=1

Pr [QN;m ]

1 X x=1

p

exp (?c K x) +

1 X N =1

N 1?

1 6

2

c K 2 :

The rst term in the right hand side is nite, and we choose K large enough to make the second term also nite, that is large enough to make 1 ? 16 c K 2 < ?1. Let now 0 2 (0; 1) be arbitrary. Since the right hand side above is nite, we can nd N0 so that

X

xN0

Pr [Px ] +

N X X

Pr [QN;m ] < 0

N N0 m=1 which means that, with probability at least 1 ? 0 , none of the events which appear in holds. We conclude that, with probability at least 1 ? 0 ,

(90) (90)

p

p

(1 + o(1))K x A(x) (1 + o(1))3K x; and

(1 + o(1)) 12 K 2 log N hA;N (m) (1 + o(1))3K 2 log N;

for all x; N N0 and 1 m N . Since 0 was arbitrary this concludes the proof.

Proof of Theorem 6.2: Let > 0. By the proof of Theorem 6.1, with probability at least 1 ? , the random sequence A satis es p p c () x A(x) c () x: 1

We have where

A (m) =

2

1 X j =1

j j+m = Ae (m) + Ao (m);

n X X e r+km r+(k+1)m A (m) = r=1 k even

62


and

n X X o r+km r+(k+1)m : A (m) = r=1 k odd Ao (m) are both SIIRV and that E[Ae (m)]

Notice that Ae (m) and = E[Ao (m)] = 1. An application of the Borel-Cantelli Lemma shows that they are both 1, for all m, almost surely (a.s.), and so is A (m). We shall say that a is used by m if a; a + m 2 A. Let lk be a sequence of integers which tends to 1, and whose rate of growth will be determined later. The sequence lk will depend on the sequence mk only. We de ne f (mk ) to be the least integer a lk which is used by mk (it exists a.s. by the previous argument). For each mk we will remove every a 2 A which is used by mk except if a = f (mk ). We then want to ensure that f (mk ) and f (mk ) + mk will not be removed by any mj . De ne

w(y) =

1 X

j =1

py +1 m : j

P w(y)=y < 1: We rst prove that the growth condition on the sequence mk implies that 1 y=1 1 w(y ) X

y=1

y

=

1 1 X X

1 p y y+m

j =1 y=1 0 mj 1 X X

j

1

1 @ p1 + X p1 A = j =1 y=1 y y + mj y=mj +1 y y + mj

0 mj 1 1 X 1 X X 1 1 @ p + A 3=2 y m y j j =1 y=1 y=mj +1 1 X log mj 1 ! pmj + pmj j =1 < 1: Then de ne the bad event

E1 =

1 [ [ [

k=1 j 6=k ylk

fy 2 A & y + mk 2 A & y + mj 2 Ag:

Clearly no f (mk ) will be removed by any mj , j 6= k, if E1 does not hold. We have to bound the probability of E1 by : 1 X 1 X X 1 q Pr [E1] K 3 k=1 j =1 ylk y (y + mk )(y + mj )

Proofs

63

K3 = K3

1 XX 1 X

1 p y y+m

k=1 ylk j =1 1 X w(y ) X k=1 ylk

j

y :

P w(y)=y < 1 we can choose the numbers l large enough to have But since 1 k y=1 X w(y)

ylk

?3 ?k y K 10

and then Pr [E1] follows. Similarly we can bound by the probability of the event

E2 =

1 [ 1 [ [

k=1 j =1 ylk

fy 2 A & y ? mk 2 A & y + mj 2 Ag:

Avoiding E2 implies that no integer of the form f (mk ) + mk can be removed by any mj , j = 1; 2; : : :. We can now form the sequences

A0 = fa 2 A : a is used by some mk , k N0 and a 6= f (mk )g and

B = AnA0 : If E1; E2 are false then B (mk ) = 1 for all k N0 and all that remains is to ensure that B(x) c3()px. p We shall prove that A0 (x) 21 c1 () x. Note that A0 (x) is not a SIIRV and we cannot

use Theorem 1.3 to bound the probability of large deviations from its mean value (which p is x). Instead, we shall use Markov's inequality: Pr [X E[X ]] 1=, for X 0. It is not really essential in the rst part of the proof, that is in bounding the probabilities of E1; E2.) For n 0 write sn = A0 \ [2n; 2n+1 ) : It is enough to show sn c()2n=2, with c() suciently small. We have

sn

+1 1 2nX ?1 X

j =2n k=N0

j j+mk

64


which implies, for n suciently large,

E[sn ] =

K2

= K2

+1 ?1 1 2nX X

p 1

j =2n k=N0 j (j + mk ) +1 ?1 2nX 1 1 X 1

pj

j =2n

K 22n=2

1 X

k=N0

k=N0

pj + m k

p2n 1+ m : k

We want to bound by, say, the probability of the event

E3 =

1 [

n=1

fsn c()2n=2g:

By Markov's inequality, it suces to show

c?1 ()K 2

1 1 X X

n=1 k=N0

p2n 1+ m : k

But the sum on the left can be written as 1 X 1 X X X 1 1 : ? 1 2 ? 1 2 p p S1 + S2 = c ()K + c ( ) K n n 2 + mk n=1 kN n=1 mk >maxf2n ;N g 2 + mk mk 2n

0

0

If one writes C (K; ) for an arbitrary constant that depends at most on K and , then

S1 C (K; )

and

1 X

2?n=2

k=N0 nlog2 mk 1 2?dlog2 mk e=2 X C (K; ) ?1=2 k=N0 1 ? 2 1 X C (K; ) m?k 1=2; k=N0

S2 C (K; )

X

1 X X

1 pm

k k=N0 2n

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