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,.


\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.

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About ioannis parissis

I'm a postdoc researcher at the Center for mathematical analysis, geometry and dynamical systems at IST, Lisbon, Portugal.
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3 Responses to The Rudin (Hardy-Littlewood) Conjecture, Notes 1: Introduction and basic facts.

  1. Michael Lacey says:

    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.

    • 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.

  2. 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.

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