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.


About ioannis parissis

I'm a postdoc researcher at the Center for mathematical analysis, geometry and dynamical systems at IST, Lisbon, Portugal.
This entry was posted in math.CA, math.CO, math.NT, Mathematics, open problem, seminar notes, The Rudin (Hardy-Littlewood) Conjecture and tagged , , , , . Bookmark the permalink.

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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