The Rudin (Hardy-Littlewood) Conjecture, Notes 1: Introduction and basic facts.

1. Introduction.

This is the first of a series of posts concerning the Rudin-Hardy-Littlewood Conjecture. To give a taste of the problem right away let us consider ${f}$ to be a trigonometric polynomial of the form

$\displaystyle \begin{array}{rcl} f(\theta)=\sum_{n=1} ^N a_n e^{i n^2\theta}, \quad \theta\in \mathbb T, \end{array}$

where ${a_n\in\mathbb C}$ for ${1\leq n \leq N}$ . The main question we are interested in is whether one has an inequality of the form:

Conjecture 1 (Rudin’s Conjecture) For all ${2<p<4}$ we have that

$\displaystyle \|f\|_p\lesssim _p \|f\| _2, \ \ \ \ \ (1)$

where the implied constant depends only on ${p}$ .

Conjecture 1 was stated in this form by Walter Rudin himself for example in [R] but the the first (and essentially only) results on this question go back to Hardy and Littlewood (see for example [R]).

Inequalities of the form (1) have deep number theoretic implications. For example, let ${\alpha_S(N)}$ (`S’ for squares) denote the maximum number of squares in the arithmetic progression ${a+b,a+2b,\ldots,a+Nb,}$ as we vary over positive integers ${a,b}$ . Then, inequality (1) for a specific ${p\in(2,4)}$ , implies that ${\alpha_S(N)=O_p(N^\frac{2}{p})}$ . Assuming inequality (1) for values of ${p}$ arbitrarily close to ${4}$ we would then conclude that for all ${\epsilon>0}$ we have the bound ${\alpha_S(N)=O_\epsilon(N^{\frac{1}{2}+\epsilon})}$ . Rudin has actually conjectured that ${\alpha_S(k)=O(\sqrt{k})}$ while, at the moment, the best known bound (due to Bombieri and Zannier) is ${\alpha_S(N)=O(N^{\frac{3}{5}+o(1)})}$ . Thus, there are two parallel conjectures, that always go hand in hand:

Conjecture 2 (Squares in Arithmetic Projections) For any ${\epsilon>0}$ we have that ${\alpha_S(N)=O_\epsilon(N^{\frac{1}{2}+\epsilon})}$ .

As we have already observed, conjecture 1 implies conjecture 2. There are also several other number-theoretic and combinatorial implications and connections that we’ll only superficially discuss here.

2. ${\Lambda(p)}$ -sets and Rudin’s conjecture.

We work on the unit circle ${\mathbb T=\mathbb R/ 2\pi \mathbb Z}$ and for an integrable function on ${\mathbb T}$ , the Fourier coefficients of ${f}$ are defined as

$\displaystyle \begin{array}{rcl} \hat f(n)=\frac{1}{2\pi}\int_0 ^{2\pi} f(x)e^{-in\theta} dx,\quad n\in\mathbb Z. \end{array}$

We begin by discussing Rudin’s approach from [R].

Definition 3 Let ${E\subset \mathbb Z}$ . A function ${f:\mathbb T\rightarrow \mathbb C}$ is called an ${E}$ -function if ${f\in L^1(\mathbb T)}$ and ${\hat f(n)=0}$ whenever ${n\notin E}$ . A trigonometric polynomial which is an ${E}$ -function is called an ${E}$ -polynomial. We will denote by ${L^p _E}$ the space of all ${E}$ -functions that belong to ${L^p}$ .

In order to define the notion of ${\Lambda(p)}$ -sets we need the following simple observation:

Lemma 4 Let ${0<q_1<q_2<p<\infty}$ . Then the following are equivalent:

(i) ${\|f\|_p\lesssim_p \|f\|_{q_1}}$ .

(ii) ${\|f\|_p\lesssim_p \|f\|_{q_2}}$ .

Proof: It is obvious that (i) implies (ii). To see that (ii) implies (i) we can interpolate by writing ${\frac{1}{q_2}=\frac{\theta}{ q_1}+\frac {1-\theta}{p}}$ so that

$\displaystyle \begin{array}{rcl} \|f\|_{q_2} \leq \|f\|_{q_1} ^\theta\|f\|_p ^{1-\theta}. \end{array}$

Then by (ii) we have that

$\displaystyle \begin{array}{rcl} \|f\|_p\lesssim_p \|f\|_{q_2} \leq \|f\|_{q_1} ^\theta \|f\|_p ^{1-\theta}, \end{array}$

which in turns implies (i). $\Box$

In other words the property ${\|f\|_p \lesssim_p \|f\|_q}$ for ${q<p}$ only depends on the larger index. This allows us to define ${\Lambda(p)}$ -sets as follows:

Definition 5 Let ${0<p<\infty}$ . A set ${E\subset \mathbb Z}$ is called a ${\Lambda(p)}$ -set if there exists a ${q<p}$ such that

$\displaystyle \begin{array}{rcl} \|f\|_p \lesssim_p \|f\|_q, \end{array}$

for all ${E}$ -polynomials ${f}$ .

Because of Lemma 4, if inequality (1) is true for some ${0<q<p}$ , then it is true for all such ${q}$ . We therefore agree in the following form of the definition.

(i) If ${2<p<\infty}$ then ${E}$ will be called a ${\Lambda(p)}$ -set if $\displaystyle \|f\|_p\lesssim_p \|f\|_2,$ for all ${E}$ -polynomials ${f}$ .
(ii) If ${1<p\leq 2}$ then ${E}$ will be called a ${\Lambda(p)}$ -set if $\displaystyle \|f\|_p \lesssim_p \|f\|_1 ,$ for all ${E}$ -polynomials ${f}$ .

Of course the ${\Lambda(p)}$ -property makes sense for ${p\leq 1}$ but we won’t discuss this here.

2.1. Equivalent formulations

There are several equivalent ways to define ${\Lambda(p)}$ -sets. We list some of them here focusing on the range ${1<p<\infty}$ .

Proposition 6 Let ${1<p<\infty}$ . Then ${E}$ is a ${\Lambda(p)}$ -set if and only if ${L^p _E=L^q _E}$ for all ${q<p}$ .

Proof: Assume first that ${E}$ is a ${\Lambda(p)}$ -set. Obviously it is enough to show that ${L^q _E\subset L^p _E}$ . Assuming that ${f\in L^q _E}$ we see that the Cesáro means of ${f}$ are ${E}$ -polynomials in ${L^q}$ with norms uniformly bounded by the ${L^q}$ norm of ${f}$ . Now by the ${\Lambda(p)}$ property of the set ${E}$ the Cesáro means are ${E}$ -polynomials which are uniformly in ${L^p}$ . We conclude that ${f\in L^p}$ and of course ${f}$ is an ${E}$ -function so we are done.

To prove the other direction just observe that if ${L^p _E = L^q _E=X}$ , then ${\|\cdot \|_p}$ and ${\|\cdot \|_q}$ are two norms in the same Banach space ${X}$ and must therefore be equivalent. $\Box$

The ${\Lambda(p)}$ -property is essentially a restriction phenomenon and that is better illustrated by the following reformulation of the problem. For a set ${E\subset \mathbb Z}$ let us consider the restriction operator ${T_E}$ acting initially on trigonometric polynomials ${f}$ in the following way:

$\displaystyle \begin{array}{rcl} T_E(f)=T_E\big(\sum_{|n|\leq N } a_n e^{in\theta}\big) = \sum_{\substack{n\in E\\ |n|\leq N}} a_n e^{in\theta}, \end{array}$

or ${ \widehat {T_E(f)} = \mathbf 1_E \hat f }$ where ${\mathbf 1_E}$ is the indicator function of the set ${E}$ .

The definition of the ${\Lambda(p)}$ property together with the fact that ${T_E}$ is a self-adjoint operator gives the following equivalent characterization:

Proposition 7 Let ${E\subset \mathbb Z}$ . For any ${1<p<\infty}$ we write ${p'}$ for the dual exponent ${p'=\frac{p}{p-1}}$ .

Let ${2<p<\infty}$ . Then ${E}$ is a ${\Lambda(p)}$ -set if and only if ${T_E}$ extends to a bounded operator from ${L^2}$ to ${L^p}$ . By duality this is equivalent to ${T_E}$ being bounded from ${L^{p'}}$ to ${L^2}$ .

2.2. Arithmetic Progressions and ${\Lambda(p)}$ -sets

As we have mentioned in the Introduction, the ${\Lambda(p)}$ -property of a subset ${E}$ of the integers was considered in connection to the problem of studying how many elements of $E$ we can find in arithmetic progressions of length $N$ . For ${N}$ a positive integer let us define ${\alpha_E(N,a,b)}$ to be the number of terms which ${E}$ has in the arithmetic progression

$\displaystyle a+b,a+2b,\ldots,a+Nb,$

for ~~positive~~ integers ${a,b}$ and ${\alpha_E(N):= \sup_{a,b} \alpha_E(N,a,b)}$ .

Theorem 8 Suppose that ${E\subset \mathbb Z}$ is a ${\Lambda(p)}$ -set for some ${p>2}$ , that is if for all ${E}$ -polynomials ${f}$ we have

$\displaystyle \|f\|_p\lesssim_p \|f\|_2,.$

Then

$\displaystyle \alpha_E(N) \lesssim_p N^\frac{2}{p}.$

Proof: We have two proofs of the theorem. We begin with the one due to Rudin in [R]. Let ${A}$ be the arithmetic progression ${A={a+b,a+2b,\ldots,a+Nb}}$ and suppose that ${E\cap A=\{n_1,\ldots,n_\alpha\}}$ . Observe that by definition ${\alpha=\alpha_E(N)}$ . The `natural choice’ of the ${E}$ -polynomial to use with Rudin’s conjecture is

$\displaystyle \begin{array}{rcl} f(\theta)=\sum_{k=1} ^\alpha e^{in_k\theta}, \end{array}$

whose ${L^p}$ norm we can easily control by the ${\Lambda(p)}$ -property of ${E}$ . Indeed, for any function ${g\in L^{p'}}$ , where ${p'<2}$ is the dual exponent of ${p}$ , we have

$\displaystyle \begin{array}{rcl} |\langle g, f\rangle | =\bigg| \frac{1}{2\pi}\int_0 ^{2\pi} g(\theta)\overline {f(\theta)} d\theta\bigg| \leq \|f\|_p \|g\|_{p'}\lesssim_p \|f\|_2 \|g\|_{p'} =\sqrt{\alpha}\|g\|_{p'}, \end{array}$

where in the before-last inequality we have used the ${\Lambda(p)}$ -property of ${E}$ and the fact that ${f}$ is an ${E}$ -polynomial. On the other hand we have that

$\displaystyle \begin{array}{rcl} \langle g, f\rangle =\sum_{k=1} ^\alpha \frac{1}{2\pi}\int_0 ^{2\pi} g(\theta) e^{-in_k\theta} d\theta = \sum_{k=1} ^\alpha \hat g (n_k). \end{array}$

In order to make these estimates useful we need to find a test function ${g\in L^{p'}}$ whose Fourier coefficients we can easily control. There are several choices here that are possible but let us work with the Fejér kernel ${k_N}$ in the place of ${g}$ . For the Fejér kernel, ${k_N(\theta)=\sum_{|n|\leq N} \big(1-\frac{|n|}{N}\big) e^{in\theta}}$ , we have that ${\|k_N\|_1=1}$ and ${\|k_N\|_2 ^2=\sum_{|n|\leq N}\big(1-\frac{|n|}{N}\big)^2\leq N}$ . Interpolating ${L^{p'}}$ between ${L^1}$ and ${L^2}$ , ${\frac{1}{p'}=\frac{\theta}{1}+\frac{1-\theta}{2}}$ , ${\theta=\frac{2-p'}{p'} }$ , we get

$\displaystyle \begin{array}{rcl} \|k_N\|_{p'} \leq \|k_N\|_1 ^\theta \|k_N\|_2 ^{1-\theta}\leq N^\frac{1}{p}. \end{array}$

This is the desired control of the ${L^{p'}}$ -norm of our test function. How about its Fourier coefficients? Well, since ${\hat k_N(n)\geq\frac{1}{2}}$ for ${|n|\leq \frac{N}{2}}$ , if all the coefficients we were considering were in the range ${|n|\leq \frac{N}{2}}$ then we would be done. This however is not necessarily the case since we are calculating Fourier coefficients corresponding to some frequencies in the set ${\{a+b,a+2b,\cdots,a+Nb\}}$ . We can mend this situation by dilating the Fejér kernel and suitably translating its frequencies. Indeed, observe that the function ${h(\theta):=k_N(b\theta)}$ satisfies ${\hat h (bn)\geq \frac{1}{2}}$ whenever ${|n|\leq \frac{N}{2}}$ . Now, defining ${g(\theta):= e^{i(a+bm)\theta}k_N(b\theta)}$ where ${m=[(N+1)/2]}$ , we have that

$\displaystyle \begin{array}{rcl} \hat g(n_k) &=&\sum_{|n|\leq N } \frac{1}{2\pi } \int_0 ^{2\pi}\big(1-\frac{|n|}{N} \big) e^{i(a+bm+bn-n_k)\theta}d\theta = \\ \\&=& \hat k_N (b(m-l))=\hat h (m-l), \end{array}$

for some ${1\leq l \leq N}$ . For any such ${l}$ we have that ${|m-l|\leq \frac{N}{2}}$ so we conclude that ${\hat g (n_k) \geq \frac{1}{2}}$ for all ${1\leq k\leq \alpha}$ . Now we have fixed the Fourier coefficients of the function but what about its ${L^{p'}}$ -norm? It is easy to see that this hasn’t changed due to the fact that ${b}$ is an integer. Putting all the estimates together we conclude

$\displaystyle \begin{array}{rcl} \frac{\alpha}{2} \leq \langle f,g\rangle \lesssim_p \sqrt{\alpha} N^\frac{1}{p}\implies \alpha\lesssim_p N^\frac{2}{p}. \end{array}$

A similar albeit more elegant way to prove this relies on Proposition 7. Let us define ${P(\theta)=\sum_{k=1} ^N e^{i(a+bk)\theta}}$ . A standard calculation shows that ${\|P\|_p \simeq N^\frac{1}{p'}}$ for all ${1<p<\infty}$ . Assuming that ${E}$ is a ${\Lambda(p)}$ -set for some ${2<p<4}$ we get from Proposition 7 tha ${T_E}$ is a bounded operator from ${L^{p'}}$ to ${L^2}$ . This means that

$\displaystyle \begin{array}{rcl} \|T_E(P)\|_2\lesssim_p \|P\|_{p'}\simeq N^\frac{1}{p}. \end{array}$

However, ${T_E(P)}$ has as many distinct frequencies as the members of ${A\cap E}$ , that is ${\alpha}$ , so that ${\|T_E(P)\|_2 =\sqrt{\alpha}}$ . We conclude that ${\alpha\lesssim_p N^\frac{2}{p}}$ . $\Box$

2.3. Rudin’s conjecure on the set of squares

Let us write down Rudin’s conjecture in the language of ${\Lambda(p)}$ -sets. Let

$\displaystyle S=\{1,2^2.3^2,\ldots\},$

be the set of squares. Then Conjecture 1 reads:

Conjecture 9 (Rudin’s Conjecture) The set of squares ${S}$ is a ${\Lambda(p)}$ -set for all ${2<p<4}$ .

Some remarks are in order. First of all the conjecture is open (to the best of my knowledge) for any ${1<p<4}$ but the interesting number theoretic implications happen only in the range ${p>2}$ . On the other hand, the set of squares ${S}$ is not a ${\Lambda(4)}$ -set so the restriction ${p<4}$ is best possible. This was first observed by Rudin in [R]. We repeat the proof of this fact here using a different argument.

Proposition 10 The set of squares ${S}$ is not a ${\Lambda(4)}$ -set.

Proof: We consider the trigonometric polynomial ${f(\theta)=\sum_{1\leq n \leq N}e^{in^2\theta}}$ which is obviously an ${S}$ -polynomial. Obviously ${\|f\|_2 =\sqrt{N}}$ . On the other hand, we have that

$\displaystyle \begin{array}{rcl} \|f\|_4 ^4=\sum_{\substack{1\leq k,k',\ell,\ell'\leq N\\ k^2+k'^2=\ell^2+\ell'^2}}1=\sum_{n\in\mathbb N}\bigg(\sum_{\substack{1\leq k,k'\leq N\\k^2+k'^2=n}}1\bigg)^2. \end{array}$

Now, for ${k}$ a positive integer, let ${r_2(k)}$ be the number of representations of ${k}$ as a sum of two squares of positive integers

$\displaystyle \begin{array}{rcl} r_2(k):=\sharp \{ (m,n)\in\mathbb N^2:m^2+n^2=k \}. \end{array}$

Then observe that we have

$\displaystyle \begin{array}{rcl} \|f\|_4 ^4= \sum_{n\in\mathbb N} \big(\sharp\{1\leq k,k'\leq N: k^2+k'^2=n\}\big)^2\geq \sum_{1\leq n\leq N^2} r_2(n)^2. \end{array}$

Taking for granted the classical number theoretic asymptotic estimate

$\displaystyle \begin{array}{rcl} \sum_{1\leq n \leq x} r_2(n)^2 \gtrsim x\log x, \end{array}$

we conclude that

$\displaystyle \begin{array}{rcl} \|f\|_4 ^4 \gtrsim N^2 \log N, \end{array}$

which shows in particular that ${S}$ is not a ${\Lambda(4)}$ -set. $\Box$

3. References.

[HL]. Hardy, G. H., Littlewood, J. E., Some problems of diophantine approximation. Acta Math. 37 (1914), no. 1, 193–239.
[R]. W. Rudin, Trigonometric Series with Gaps, Indiana Univ. Math. J. 9 No. 2 (1960), 203–227.

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3 Responses to The Rudin (Hardy-Littlewood) Conjecture, Notes 1: Introduction and basic facts.

Michael Lacey says:

March 16, 2010 at 2:22 am

This is an elegant summary of the problem! I was left wondering what Bourgain proved: MR1029904 (91d:43018)
Bourgain, J.(F-IHES)
“On $\Lambda(p)$ -subsets of squares”

It seems that he proves that the squares have a maximal density allowed by the failure the $\Lambda(p)$ -property for p>4.

- ioannis parissis says:
  
  March 24, 2010 at 11:38 pm
  
  Michael. As mentioned in this post (and is well known), the set of squares $S$ fails to be a $\Lambda(4)$ -set. In the article “On Λ(p)-subsets of squares”, Bourgain shows however that for any $p>4$ , there exist $\Lambda(p)$ -sets of maximal density contained in $S$ . In particular this implies that they are not $\Lambda(q)$ -sets for any $q>p$ . He also proves the corresponding results for any set of the form $\{1,2^j,3^j,\ldots \}$ and also for the set of primes $P$ . Look as well in the subsequent post in relevance to maximal density $\Lambda(p)$ -sets.
  
ioannis parissis says:

March 16, 2010 at 3:37 am

Thanks for the good words! Hopefully I can cover the Bourgain density result in another post. I haven’t quite figured out what is the relation of the density to the $\Lambda(p)$ -property yet. My purpose first is to walk through the only known cases of the conjecture. Technically speaking there is none; however there are some special trigonometric polynomials for which (1) holds. I know that Bourgain has mainly worked on constructing sets $E$ which are $\Lambda(p)$ -sets for some $p>2$ but NOT $\Lambda(q)$ for any $q>p$ .