benke g, chang d c - on the sum of sine products - j. math. anal. appl. 284 (2003), 647-655

8/9/2019 Benke G, Chang D C - On the Sum of Sine Products - J. Math. Anal. Appl. 284 (2003), 647-655

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J. Math. Anal. Appl. 284 (2003) 647655

www.elsevier.com/locate/jmaa

On the sum of sine products

George Benke and Der-Chen Chang ,1

Department of Mathematics, Georgetown University, Washington, DC 20057-1233, USA

Received 23 September 2002

Submitted by K. Jarosz

Abstract

Given N 1 and d 1, we show the existence of positive integers kj for 1 k N and 1

j d such that

N

k=1

d

j=1

sin kjxj CN(d+1)/(2d+1)

for all real x1, . . . , xN, where C is a constant that depends only on d. This extends a result of Bourgain

for the case d= 1. 2003 Elsevier Inc. All rights reserved.

Keywords: Trigonometric polynomial; Sums of sines; Random variable; Probability measure

1. Introduction

Bourgain [1] has shown the existence of positive integers k

such that

N

k=1sin kx

CN2/3

* Corresponding author.E-mail addresses: [email protected] (G. Benke), [email protected] (D.-C. Chang).

1 Partially supported by a William Fulbright Research Grant and a NSF Grant DMS9622249.

0022-247X/$ see front matter 2003 Elsevier Inc. All rights reserved.

doi:10.1016/S0022-247X(03)00384-6


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648 G. Benke, D.-C. Chang / J. Math. Anal. Appl. 284 (2003) 647655

for all real x, where C is an absolute constant. On the other hand, the trivial lower bound

CN1/2 cannot be achieved since Konyagin [2] has shown that for N sufficiently large, and

any set of distinct positive integers 1, . . . , N there exits an x such that

0.15

Nlog N

loglog N

Nk=1

sin kx.

In this paper we extend Bourgains result to higher dimensions. We will show that there

exist positive integers kj for 1 k N and 1 j d such that

N

k=1

d

j=1

sin kj xj CN(d+1)/(2d+1),where C depends only on d. Questions concerning bounds on sums of sines are delicate.

For example it can be shown that if k grows polynomially or exponentially in k then

Nk=1 sin k x = O(N). The (random) construction in this paper produces k whichgrow like exp(k1/3) to give Nk=1 sin kx = O(N2/3). On the other hand, mere gapconditions on the k are not enough. It can be shown that if

Nk=1 sin kx = O(N)

with < 1 then there exists a set of k = 1 such that N

k=1 sin(k + k)x = O(N).These questions regarding bounds on the supremum norm of sine sums are related to

and to some extent motivated by questions regarding bounds on the norm of the Hilbert

transform on finite-dimensional translation invariant spaces of continuous functions. More

specifically, the question is: how large can the Hilbert transform be on a 2N-dimensional

translation invariant space of continuous functions? If the domain of the functions is the

real line, then Bourgains result shows that it can be O(N1/3). If the domain of the func-

tions is d-dimensional Euclidean space, then our result shows that the norm of the Hilbert

transform can be O(Nd/(2d+1)). The authors are grateful to the referee for helpful com-ments and a careful reading of the manuscript.

2. Proof of the main theorem

In order to make this paper as self-contained as possible, we have chosen to give (the

short) proofs of several well-known lemmas. The following three lemmas are well known.

Lemma 2.1. LetX be a random variable with mean satisfying

|x

| 1. For any real

E

e(X) e

2/2.

Proof. Let

() = log E

e(X)= log E[eXe] = log E(eX) .

It follows that

(z) = E(XezX)

E(ezX)


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G. Benke, D.-C. Chang / J. Math. Anal. Appl. 284 (2003) 647655 649

and

(z) = E(ezX)E(X2ezX) [E(XezX)]2

[E(ezX)]2 E(X2ezX)

E(ezX) 1

since |X| 1. Also (0) = 0 and (0) = 0, and we have

(z) =z

0

(y)dy

z0

dy = z,

()=

0

(z)dz

0

z dz=

2

2.

It follows that E(e(X)) = e() e2/2 and the proof of the lemma is therefore com-plete.

Lemma 2.2. Leta 0, R, X : R be a random variable. Then

(a) P(X ) ea E(eaX );(b) P(|X| ) ea{E(eaX ) + E(eaX )}.

Proof.

E(eaX ) = X

eaX dP+ XeaX dPP(X )e(a)().

Hence

P(X ) eaE(eaX ). (2)Combining (1) and (2), we have

P|X| = P(X) + P(X ) eaE(eaX ) + eaE(eaX ).

Lemma 2.3. Let N > 0 and let Y1, . . . , Y q be random variables such that E(ebYk )

eNb2/2 for all real b andk = 1, . . . , q . Then

P|Yk|

1

2.

Proof. By Lemma 2.2, given 0,


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P|Yk| , for some k = 1, . . . , q

qk=1

P|Yk|

ea

qk=1

E(eaYk ) + E(eaYk )

for any a 0. By hypothesis, this last expression is less than

ea2qeNa2/2. (3)

Setting a = /N and then setting = 2Nlog4q , (3) becomes 1/2. Thus,

P|Yk| 0 (depending only on

and ) such that forkA

(a) uk+1 > uk;(b) C1k

1ek uk uk1 C2k1ek ;

(c) C3k1ek

(kA)(kA1)ukk1

(kA+1)(kA)uk+1uk C4k

1ek

.

Lemma 2.5.For

1

3

1 and1

0 there exists a constant C depending only on and such that

Nk=1

k ek

CN1+ eN

.

Proof. Since x ex

is monotonic for x > 27, we can estimate the sum by an integral of the

formN

0 x ex

dx . Change of variables, integration by parts and elementary estimations

give the result.

Lemma 2.6. Let mk, k = 0, . . . , N , denote integers such that0 < m0 < m1 < < mN,and let rk denote real numbers such that r1 > > rN > 0. Let an = rk for mk1 n < mk for 1 n N , and let an

=0 otherwise. Define intervals I0

=(/m1,

],

Ij = (/mj+1,/mj ] forj = 1, . . . , N 1, andIN = [0,/mN1). For x IjnZ

an sin nx

24 1mjj

k=1(rk rk+1)m2k + rj+1mj+1

with the following conventions:

(a) If a lower index exceeds an upper index in a summation, the sum is void;

(b) rN+1mN+1 is defined to be 0.


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Proof. Let

S(x) =nZ

an sin nx =N

k=1rk

mk1j=mk1

sin j x.

Defining rN+1 to be 0, this can be rewritten as

S(x) =N

k=1(rk rk+1)

mk1j=m0

sin j x

= 1sin x/2

N

k=1

(rk rk+1) sin(mk + m0 1)x

2sin

(mk m0)x2

.

Now suppose (/mj+1) < x (/mj ) where 1 j N 1. We write S(x) = S1(x) +S2(x) where S1(x) is the sum over 1 k j and S2(x) is the sum over j + 1 k N.For such x

1

sin x/2

sin (mk + m0 1)x2 sin (mk m0)x2

(mk + m0 1)(mk m0)x24sin(x/2)

mk(mk 1) m0(m0 1) 2

4mj.

Hence

S1(x) 2

4mj

j

k=1

(rk rk+1)mk(mk 1) 2

4mj(r1 rj+1)m0(m0 1).

Also

S2(x) 1sin(x/2)

Nk=j+1

(rk rk+1) =1

sin(x/2)rj+1

xrj+1 rj+1mj+1.

This gives for (/mj+1) < x (/mj ) and 0 j N 1

S(x) 24

1

mj

jk=1

(rk rk+1)mk(mk 1) m0(m0 1)

mj(r1 rj+1)

+ rj+1mj+1.If 0 x (/mN) then S(x) = S1(x) and the above estimates for S1(x) still apply. There-fore, the last estimate for S(x) is valid for j = N where rN+1 = 0 so that rN+1mN+1 = 0even though mN+1 has not been defined. If (/m1) < x < then S(x) = S2(x) and weget |S(x)| r1m1. Hence we complete the proof of the lemma.

Lemma 2.7. Let 13 < 1. Let d 1 and N > 1 be integers. Let A 0 and define

mk = [e(A+k) ] fork = 0, 1, . . . , N , where [] denotes the greatest whole integer function.Suppose A is large enough to give m0 < m1 < < mN. Define

rk =k1/d (k 1)1/d

mk mk1


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and

an =

rk, mk1 n < mk, k = 1, . . . , N ,0, n < m0 ornmN.

Then there exists constants A andC such that for all xnZ

an sin nx

CN(1/d).Proof. Let ck = rk rk+1. We will apply Lemmas 2.42.6. As in Lemma 2.4, let uk de-note [ek ]. First consider

1

mj

jk=1

ckm2k =

1

uA+j

A+jk=A+1

k1/d (k 1)1/d

uk uk1 (k + 1)1/d k1/d

uk+1 uk

u2k.

Setting = 1/d in Lemma 2.4 and = (1/d) 1 in Lemma 2.5 gives

1

mj

jk=1

ckm2k C

e(A+j)

A+jk=A+1

k(1/d)1ek C(A + j )(1/d) CN(1/d),

where C depends only on and d. Next, if 1 q N, Lemma 2.4 gives

rq mq =

q1/d (q 1)1/dmq mq1

mq D

q(1/d)1

(A + q)1e(A+q)

e(A+q)

D(A + q)(1/d)

1

DN(1/d)

1

,where D depends only on and d. Inserting these estimates into Lemma 2.6 gives theresult.

The following lemma is well known.

Lemma 2.8. Let f(x) =k akeikx be a real trigonometric polynomial on [0, 2 )d. Sup-pose

sup|k1|, . . . , |kd|

M.

Then there exist points x1, . . . , xq where q < C M d andC depends only on d, such that

f

2 max

j=1,...,qf (xj ).Proof. Choose z such that |f(z)| = f. Let K be an integer not less than 4 dM, andlet the xj be the lattice points in [0, 2 )d with coordinate spacing equal to 2/K . There areq = Kd such points. Consider the xj that is closest to z. Then by the mean value theoremthere is a cj on the line segment from z to xj such that

f(z) f (xj ) =d

q=1

f

tq(cj )(zq xj q ).


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Let f be the random trigonometric polynomial in [0, 2 )d defined by

f(x) =N

k=1S(x,k).

Note that the degree M of f is less than dmN. Let xj for j = 1, . . . , q be the latticepoints in [0, 2 )d as defined in Lemma 2.8. We will estimate the expectation of the randomvariables Zj = f (xj ):

E(Zj ) =N

k=1E

S(xj , k)=

Nk=1

kLk

S(xj , k)k(k).

Since k(k) =dj=1 (kj),E(Zj ) =

Nk=1

kLk

d

j=1sin kjxj

d

j=1(kj )

=

BN

dj=1

sin(j xj )

(j )=

JN() sin x

d.

Recalling that = 1/(1 +2d) and using Lemma 2.7 gives E(Zj )CN(d+1)/(2d+1). Next,by Lemma 2.1 we have

EexpZj E(Zj )= EexpN

k=1

S(xj , k)

ES(xj , k)

=N

j=1E

exp

S(xj , k) E

S(xj , k)

Nk=1

exp(2/2) = exp(N2/2).

Therefore, by Lemma 2.3,

PZj E(Zj )2Nlog4q, for j = 1, . . . , q 1

2

and

P

maxj=1,...,q

f (xj ) E(Zj )+2Nlog4q 12

.

By Lemma 2.8, f C maxj=1,...,q |f (xj )|. Also q CeN = eN1/(2d+1) and|E(Zj )| CN(d+1)/(2d+1). Therefore there exists a constant C such that

Pf CN(d+1)/(2d+1)

1

2.

We remark that the probability 1/2 can be replaced by probability 1 for any > 0, bychoosing a larger (-dependent) constant C.


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References

[1] J. Bourgain, Sur les sommes de sinus, Sem. Anal. Harm. Publ. Math. dOrsay 84-01 (1984), exp. no. 3.

[2] S.V. Konyagin, Estimates of maxima of sine sums, East. J. Approx. 3 (1997) 6370.

benke g, chang d c - on the sum of sine products - j. math. anal. appl. 284 (2003), 647-655

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