The Rudin (Hardy-Littlewood) Conjecture, Notes 2: A closer look at Λ(p)-sets.

This is the second post on Rudin’s conjecture. For the first introductory notes see here. In this post I will try to build some more intuition on {\Lambda(p)}-sets by studying some examples and discussing their properties. We will also discuss some basic question that have been studied in the literature of {\Lambda(p)}-sets.

1. Lacunary sequences and {\Lambda(p)}-sets.

Let us briefly recall the definition of {\Lambda(p)}-sets. For {E\subset \mathbb Z}, let {f:\mathbb T \rightarrow {\mathbb C}} be an {E}-polynomial, that is {\hat f(n)=0} for all {n\notin E} and {f} is a trigonometric polynomial. For {0<p<\infty} we call {E} a {\Lambda(p)}-set if there exists a {q<p} such that

\displaystyle  	\|f\|_p\lesssim_p \|f\|_q, \ \ \ \ \ (1)

for all {E}-polynomials {f}. Remember that if (1) holds for some {0<q<p} then it holds for all such {q} and this is just a consequence of Hölder’s inequality.

It is pretty obvious that not every subset of the integers can be a {\Lambda(p)}-set. In particular the integers themselves are not a {\Lambda(p)}-set and this can be very easily verified by checking against the Dirichlet Kernel for which we have {\|D_N\|_p \simeq N^\frac{1}{p'}}, {p'} being the dual exponent of {1<p\leq \infty} and {\|D\|_1\simeq \log N}.

On the other hand, the easiest example of a {\Lambda(p)}-set is probably a lacunary sequence.

1.1. Hadamard lacunary sequences

Definition 1 Let {A=\{a_1,a_2,a_3\ldots\}} be a sequence of positive integers. The sequence {\{a_k\}_{k=1} ^\infty} is called lacunary in the sense of Hadamard if there exists some constant {\lambda>1} such that

\displaystyle  \frac{a_{k+1}}{a_k}>\lambda,\quad k=1,2,3,\ldots

Now it is a classical result (due to Salem and Zygmund) that a Hadamard lacunary sequence {A} is a {\Lambda(p)}-set. That is, we have

Theorem 2 (Salem, Zygmund) Let {A=\{a_1,a_2,\ldots\}} be a lacunary sequence of positive integers. Then {A} is a {\Lambda(p)}-set for all {0<p<\infty}.


We will present the proof here since it’s illustrative of the combinatorial nature of Rudin’s problem, at least in the case where {p} is an even positive integer. As in the proof of Proposition 10 of the previous post, we define now the numbers {r_k(E,n)} to be the number of the representations of the positive integer {n} in the form

\displaystyle  n=n_{i_1}+n_{i_2}+\cdots+n_{i_k},

where {n_{i_1},\ldots,n_{i_k}\in E}. Here there is a subtle point since different permutations of {n_{i_1},\ldots,n_{i_k}} `count’ as different representations. One can however the number of representations when say {n_{i_1}\leq \cdots \leq n_{i_k}}. All the other representations are just permutations of this one so it is enough to multiply by {k!}.

We have the following lemma:

Lemma 3 Let {E=\{n_1,n_2,n_3\ldots \}} and {f} be an {E}-function. Then for all positive integers {h} we have that

\displaystyle  \begin{array}{rcl}  		\|f\|_{2h} \leq \max_{n}( r_h(E,n))^\frac{1}{2h} \|f\|_2 	\end{array}

Proof: Observe that we have

\displaystyle  \begin{array}{rcl}  		\|f\|_{2h} ^h &=& \big\| \big(\sum_{k\in E} a_k e^{ik\theta}\big)^h \big\|_2 \\ \\ &=&	\bigg( \sum_{n\in\mathbb Z} \big| \sum_{\substack{n=n_1+\cdots+n_h\\n_1,\ldots,n_h\in E}} a_{n_1}\cdots a_{n_h}\big|^2\bigg)^\frac{1}{2}\\ \\ &\leq& 	\bigg( \sum_{n\in\mathbb Z} r_h(E,n) \sum_{\substack{n=n_1+\cdots+n_h\\n_1,\ldots,n_h\in E}} |a_{n_1}|^2\cdots |a_{n_h}|^2\bigg)^\frac{1}{2} \\ \\ &\leq& (\max_{n\in\mathbb Z} r_h(E,n) )^\frac{1}{2}\big(\sum_{n\in E}|a_n|^2\big)^\frac{h}{2}, 	\end{array}

where we have used Cauchy-Schwartz in the first inequality. \Box

Lemma 3 allows us to construct {\Lambda(p)}-sets for {p} an even integer by combinatorial means. In particular it is enough to construct sets {E} such that every equation

\displaystyle n=n_1+\cdots+n_h,

has at most one solution (modulo permutations) with {n_1,\ldots,n_h\in E}. In fact we have the following proposition which is slightly more general:

Proposition 4 (Rudin) Let {E} be a set of non-negative integers and let {h>1} be a positive integer. If {E} is a union of (a bounded number of) sets {E_1,\ldots,E_t} such that for each {i\in\{1,2,\ldots,t\}} the function {r_h(E_i,n)} is a bounded function of {n}. Then {E} is a {\Lambda(2h)}-set.

Proof: Obviously it is enough to prove the Proposition for {t=1} and then use the triangle inequality. However, for {t=1}, the conclusion is just an application of Lemma 3. \Box

We will take up this issue later on, after we complete our discussion on lacunary sequences.

Proposition 4 allows to give an elementary proof of Theorem 2.

Proof of Theorem 2: Let us consider a lacunary set {A} with {n_{k+1}/n_k>h+1}. Then for a positive integer {h>1}, and {n} a positive integer, let us consider the equation

\displaystyle  	a_1n_{k_1}+\cdots+a_m n_{k_m}=n, \ \ \ \ \ (2)

with {n_{k_1}>\cdots > n_{k_m}} in {A} and positive integers {a_j\geq 1} with {a_1+\cdots+a_m=h}. Now assume there are two different representations of {n} as a sum of elements in {A}:

\displaystyle  \begin{array}{rcl}  	a_1 n_{k_1}+\cdots+a_m n_{k_m}=b_1n_{\ell_1}+\cdots+b_\tau n_{\ell_\tau}, \end{array}

with {a_1+\cdots+a_m=h=b_1+\cdots+b_\tau}. Since we have assumed that the two representations are different, there will be a maximum element of {A} which only appears in one of the two representations or which appears more times in one or the other representation). We conclude an equality of the form

\displaystyle  \begin{array}{rcl}  	\gamma_1n_{\nu_1}+\cdots+\gamma_pn_{n_p}=0, \end{array}

where {\gamma_1\geq 1} and {|\gamma_{\nu_j}|\leq h} for {2\leq j \leq p}. However this implies that

\displaystyle  \begin{array}{rcl}  	n_{\nu_1} &	\leq & \gamma_1 n_{\nu_1}=| \gamma_2 n_{\nu_2}\cdots+\gamma_p n_{n_p}| \\ \\ &\leq & h \big(\frac{1}{\lambda}+\frac{1}{\lambda^2}\cdots+ \frac{1}{\lambda^p}\big)n_{\nu_1} \leq h\frac{n_{\nu_1}}{\lambda-1}, \end{array}

which is impossible whenever {\lambda>h+1}. We conclude that for a lacunary sequence {A} with lacunary constant {\lambda>h+1}, the representation of integers in the form (2) is unique up to permuations. we have {r_h(A,n)\leq h!}. However, every lacunary sequence can be written as a union of {s\simeq \log h/\log \lambda} lacunary sequences {A_1,\ldots,A_s} with lacunary constants {\lambda_i>h+1} and {r_h(A_i,n)\leq 1}. Using Proposition 4 we conclude that for every {A}-polynomial {f} we have

\displaystyle  \begin{array}{rcl}  	\|f\|_{2h} \lesssim_{s,h}\|f\|_2, \quad h\in\{1,2,\ldots\}. \end{array}

In fact the implied constant can be calculated to something like {\frac{\log h}{\log \lambda} h^\frac{1}{2}}. \Box

Two more properties of lacunary sets are important in our discussion.

1.2. Sidon sets and {\Lambda(p)}-sets.

Lacunary sets actually satisfy a stronger property than the {\Lambda(p)}-property that is, they are Sidon sets:

Definition 5 Let {E\subset \mathbb Z}. Then {E} is called a Sidon set if

\displaystyle  \begin{array}{rcl}  		\sum_{n\in E}|\hat f(n)|\lesssim \| f\|_\infty, 	\end{array}

for all {E}-polynomials {f}.

The notion of a Sidon set is genuinely stronger than that of a {\Lambda(p)}-set, for all {p<\infty}, that is, there exist sets {E} which are {\Lambda(p)}-sets for {p<\infty} but are not Sidon sets. In [R], Rudin showed the following theorem

Theorem 6 (Rudin) Suppose that {E\subset E} is a Sidon set with constant {B}, that is,

\displaystyle  \begin{array}{rcl}  		\sum_{n\in E}|\hat f(n)| \leq B \|f\|_\infty, 	\end{array}

for all {E}-polynomials {f}. Then {E} is a {\Lambda(p)}-set for all {1<p<\infty}. Furthermore we have that

\displaystyle  \begin{array}{rcl}  		\|f\|_p &\leq& B \sqrt{p} \|f\|_2, \quad 2<p<\infty , \\ 		\|f\|_2 &\leq& 2B \|f\|_1, 	\end{array}

for all {E}-polynomials {f}

Rudin asked whether he opposite of Theorem 6, that i whether a set {E} which satisfies the conclusion of Theorem 6 is necessarily a Sidon set. The answer is YES and this was proved by Pisier in [P]:

Theorem 7 (Pisier) Let {E\subset \mathbb Z}. Then {E} is a Sidon set if and only if

\displaystyle  \begin{array}{rcl}  		\sup_{p>2} \frac{\|f\|_p}{\sqrt{p} \|f\|_2} <+\infty, 	\end{array}

where the supremum is taken over all {E}-polynomials {f}.

It is not very hard to see that lacunary sets are actually Sidon sets. The proof can be found in [R].

2. {\Lambda(p)}-sets of maximal density.

An obvious fact about {\Lambda(p)} set is the inclusion:

\displaystyle  	0<q<p \implies \Lambda(p) \subset \Lambda(q). \ \ \ \ \ (3)

This is because of the definition of {\Lambda(p)}-sets and Lemma 4 of the previous post. A natural question is whether this inclusion is proper:

Question 8 (Rudin) Let {1<p<\infty}. Does there exist a {E\subset \mathbb Z} which is a {\Lambda(p)}-set but not a {\Lambda (q)}-set for any {q>p}?

The question was originally posed by Rudin in [R]. The case {1<p<2} is settled (in the negative) by the following theorem of Bachelis and Ebenstein [BE]:

Theorem 9 (Bachelis, Ebenstein) Let {E\subset \mathbb Z}. Then the set {\{p\in(1,2):E\mbox{ is a } \Lambda(p)\mbox{-set}\}} is an open set.

Thus we are left with the range {[2,\infty)} in Rudin’s question. Rudin himself answers the question in [R], in the special case where {p=2h} is an even integer:

Theorem 10 (Rudin) For any positive integer {h>1} there exist {\Lambda(2h)}-sets which are not {\Lambda(2h+\epsilon)}-sets for any {\epsilon>0}.

We will present a proof of Theorem 10 adopting a slightly different construction that the on in [R] although the ideas are very similar. First observe that if we construct a set {E\subset \mathbb Z} for which {r_h(E,n)} is a bounded function of {n}, then Proposition 4 guarantees that {E} is a {\Lambda(2h)}-set. In Theorem 8 of the the previous post we saw that, for {2<p<\infty}, a {\Lambda(p)} set satisfies {\alpha_E(N)\lesssim_p N^\frac{2}{p}}. This means, that any {N}-term arithmetic progression contains at most {N^\frac{2}{p}} elements of {E}. Based on this fact we give the following definition:

Definition 11 A {\Lambda(p)}-set {E\subset \mathbb Z} has maximal upper density if

\displaystyle  		\limsup_{N\rightarrow +\infty} \frac{\alpha_E(N)}{N^{\frac{2}{p}}}>0. 	\ \ \ \ \ (4)

It is essential to notice here that a {\Lambda(p)}-set {E} of maximal upper density cannot be a {\Lambda(q)}-set for any {q>p} since {\alpha_E(N)\lesssim_q N^\frac{2}{q}} if {E} is a {\Lambda(q)}-set. Thus, in order to answer Rudin’s question 8 in the case where {p=2h} is an even integer it is enough to construct {\Lambda(2h)}-sets of maximal upper density. We will now digress a bit in order to discuss the notion of {B_h}-set.

2.1. {B_h}-sets of maximal size.

For {E\subset E}, let us consider the equation

\displaystyle  	n_1+\cdots+n_h=n,\quad n_1,\ldots,n_h\in E, \ \ \ \ \ (5)

where {n} is an integer and {n_1\leq \cdots \leq n_h}.

Definition 12 A set {E\subset \mathbb Z} is called a {B_h}-sequence if for every {n\in\mathbb Z}, equation (5) has at most one solution such that {n_1\leq n_2\leq \cdots \leq n_h}.

Remark 1 {B_2}-sequences are mentioned in the literature as Sidon-sets. This is not to be confused with the notion of Sidon sets defined earlier in this post. We will stick to the {B_h}-notation to avoid ambiguity.

If {E} is {B_h}-sequences then obviously {r_h(E,n)\leq h!} which is the smallest possible value of {r_h(E,n)}. This interest in {B_h} sequences lies here in the fact that {B_h} sequences are {\Lambda(2h)}-sets. This is an immediate application of by Lemma 3. The following theorem of Bose and Chowla from [BC] constructs finite {B_h}-sets of maximal size:

Theorem 13 (Bose, Chowla) Let {m =p^n} where {p} is prime and {h\geq 2} be a positive integer. There exist {m} non-zero integers

\displaystyle  \begin{array}{rcl}  	d_1=1,d_2,\ldots,d_m, \end{array}

such that the sums

\displaystyle  	d_{i_1}+d_{i_2}+\cdots+d_{i_h},\quad 1\leq i_1\leq i_2 \leq \cdots \leq i_h\leq m, \ \ \ \ \ (6)

are all distinct {\mod (m^h-1)}. By a suitable choice of representatives modulo {m^h-1} the {d}‘s can be chosen to satisfy

\displaystyle  \begin{array}{rcl}  	1=d_1<d_2<\cdots<d_m\leq m^h-1. \end{array}

The proof of this theorem lies beyond the purposes of this post. However the exposition in [BC] is really easy to follow and the article is freely available.

2.2. Constructing {\Lambda(2h)}-sets of maximal upper density

Of course the {B_h}-sequence constructed in Theorem 13 is a finite sequence and as such, it can’t have positive upper density. However, it has maximal size since there is an arithmetic progression of length {N=m^h} which contains {m=N^\frac{1}{h}} terms of the sequence. This will be enough for our purposes here since we can exploit the construction of theorem 13 in order to construct a {\Lambda(2h)} of maximal upper density, although we don’t claim anything about the B_h-properties of this set. The idea is to place such sets in dyadic pieces of the positive integers and glue them together. Since the size of these sets is maximal, {N^\frac{1}{h}}, and the dyadic pieces become arbitrarily large, the upper density of the set will be positive.

The first step is the following proposition:

Proposition 14 For any {h\geq 2} and any positive integer {N} there exists a {\Lambda(2h)}-set {E\subset\{1,2,\ldots,N\}} with {|E|\simeq N^\frac{1}{h}}.

Proof: Fix your favorite prime {p} and define the positive integer {n} by {p^n<N^\frac{1}{h}\leq p^{n+1}}. From Theorem 13 there exists a set {E\subset \{1,2,\ldots,(p^{n})^h-1\}} of size {|E|=p^{n}}. Since {(p^{n})^h-1 \leq N-1} our set is contained in {\{1,2,\ldots,N-1\}}. On the other hand we have that

\displaystyle N^\frac{1}{h} \geq |E|=p^n\geq \frac{1}{p} N^\frac{1}{h},

that is {|E|\simeq N^\frac{1}{h}}. \Box

Corollary 15 (Bourgain) Let {p=2h} where {h> 1}. Then there exists a {\Lambda(2h)}-set of maximal upper density.

Proof: Using Proposition 14 we construct for each {N=2^{k-1}}, {k=1,2,\ldots,} a set {E_k} which is contained in {\{2^{k-1}+1,2^{k-1}+2,\ldots,2^k-1\}}. For this, just translate the sets constructed in the Proposition by {2^{k-1}}. Note that each set {E_k} has size {|E_k|\simeq N^\frac{1}{h}\simeq_h 2^\frac{k}{h}}. Now we define

\displaystyle  	E:=\cup_{k=1} ^\infty E_k. \ \ \ \ \ (7)

Observe that for {\ell=2^k}, the set {E} has at least {\simeq 2^\frac{k}{h}} terms in the arithmetic progression {\{1,2\ldots,\ell\}}. We conclude that {E} has positive upper density. To see that {E} is a {\Lambda(2h)}-set, let {f} be an {E}-polynomial. Invoking the Littlewood-Paley inequalities we have for any

\displaystyle h>1

\displaystyle  \begin{array}{rcl}  \|f	\|_{2h} \lesssim_h \| S(f)\|_{2h}, \end{array}

where {S(f)} is the Littlewood-Paley square function:

\displaystyle  \begin{array}{rcl}  	S(f):=\bigg(\sum_{k=1} ^\infty \bigg| \sum_{2^{k-1}<n\leq 2^k } \hat f(n) e^{inx} \bigg|^2 \bigg)^\frac{1}{2}. \end{array}

Since {f} is an {E}-polynomial and {E\cap \{2^{k-1}+1,\ldots 2^k\}=E_k}, we have for all {k\geq 1}

\displaystyle  \begin{array}{rcl}  	\sum_{2^{k-1}<n\leq 2^k } \hat f(n) e^{inx} =\sum_{k\in E_k} \hat f(n) e^{inx} . \end{array}

Furthermore, since each one of the sets {E_k} is a {\Lambda(2h)}-set with the same {\Lambda(2h)}-constant, we have

\displaystyle  \begin{array}{rcl}  	\bigg\|	\sum_{2^{k-1}<n\leq 2^k } \hat f(n) e^{inx}\bigg\|_{2h} \leq c(h) \big(\sum_{n\in E_i} |\hat f(n)|^2\big)^\frac{1}{2}. \end{array}

We conclude that

\displaystyle  \begin{array}{rcl}  	\|f\|_{2h} &\lesssim_h& \bigg\|\bigg( \sum_{k=1} ^\infty \bigg| \sum_{n\in E_k } \hat f(n) e^{inx} \bigg|^2 \bigg)^\frac{1}{2} \bigg\|_{2h} \\ \\ &\lesssim_h& 	 \bigg( \sum_{k=1} ^\infty \bigg\| \sum_{n\in E_k } \hat f(n) e^{inx} \bigg\|_{2h} ^2 \bigg)^\frac{1}{2} \\ \\ &\lesssim_h & 	\bigg( \sum_{k=1} ^\infty \bigg\| \sum_{n\in E_k } \hat f(n) e^{inx} \bigg\|_{2} ^2 \bigg)^\frac{1}{2}=\|f\|_2. \end{array}

This shows that {E} is a {\Lambda(2h)}-set and completes the proof of Corollary 15 and thus the one of Theorem 10. \Box

Remark 2 In [B], Bourgain extends the result of Rudin, Theorem 10, by proving that for any {p>2} there exist {\Lambda(p)}-sets which are not {\Lambda(p+\epsilon)}-sets for any {\epsilon>0}. For this we need the analogue of Proposition 14 for any {p>2}. Bourgain proves this using the probabilistic arguments. In fact his proof shows that `most’ subsets of {\{1,2,\ldots,N\}} of size {\sim N^\frac{2}{p}} have the {\Lambda(p)}-property.

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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7 Responses to The Rudin (Hardy-Littlewood) Conjecture, Notes 2: A closer look at Λ(p)-sets.

  1. Alex Iosevich says:

    The last reference above should probably read: Rudin, W… instead of Walter, R…

  2. Mark Lewko says:

    Nice post! One very small typo: In remark 2, I think you mean that there exists a \Lambda(p) set that is not a \Lambda(p+\epsilon) set for p>2.

    • Thanks for the correction Mark. Probably there are more. Also, this is a problem that I’m reading and trying to understand right now. As a result there might things in this post that are not very consistent, at least to specialists in the area. Judging from your recent article on the arxiv (with Allison Lewko) , I would guess that you understand B_h-sets a lot better than I do. I would ask you (if you can) to let me know about these inconsistencies as well. For example, the construction I give here is a \Lambda(2h)-set which is (at least in appearances) an infinite union of B_h-sets. Reading your paper I realize that this is probably deceiving, but at the moment I can’t really see what’s going on.

  3. Mark Lewko says:

    Ioannis– Thanks for the comments. The Bose-Chowla sets (glued together with the Littlewood-Paley theory) is an infinite union of B_{h} sets, however it isn’t clear how one might go about showing that it isn’t a finite union of B_{h}[G] sets (for any G). I haven’t thought too much about this, but it seems likely this construction is a finite union of B_{h}[G] sets (you won’t have very many repeated sums between copies of each Bose-Chowla set since each copy is in a distinct dyadic interval). As you remark, in Rudin’s paper he uses the Bose-Chowla construction, but instead of gluing translates together with the Littlewood-Paley inequality, he takes unions of (very) disjoint dilations. He then shows that the resulting set is a B_{h}[1] set (using that the dilations are very far apart) and hence is a \Lambda(2h) set.

    In general, Allison and I found that it is difficult to show a set isn’t a finite union of B_{h}[G] sets (unless, say, the set is trivially too dense to be a finite union of B_{h}[G]). Bourgain’s constructions (from [2]) are a bit more dense than the Bose-Chowla construction. With the Bose-Chowla construction (say with h=2), if I recall correctly, you have that \limsup |S \cap [1,N]| / N^{1/2} > 0 (which is enough to guarantee that S isn’t a \Lambda(4+\epsilon set) but you don’t necessarily have that \liminf S \cap [1,N]| / N^{1/2} > 0. Bourgain’s construction gives you that S \cap [1,N] >> N^{1/2}. This is enough to allow one to conclude that his set isn’t a finite union of B_{2}[1] sets by an old theorem of Erdos. It also seems likely that this is enough to imply that S isn’t a finite union of B_{2}[G] sets (for any G), however I don’t know how to prove this.

  4. Mark. Thanks so much for taking the time and putting the effort to explain the situation! There is a subtle difference that you point out in your comment that I haven’t explained in the post (trying to keep things simple). I use a notion of ‘upper density’ exactly because you can see that the Littlewood-Paley union of the Bose-Chowla sets (say for h=2) satisfy \limsup |S\cap[1,N]|/N^\frac{1}{2}>0 but not necessarily \liminf |S\cap[1,N]|/N^\frac{1}{2}>0. Bourgain in “On Λ(p)-subsets of squares” proves for example that there are maximal density $\Lambda(p)$ sets contained in the squares for any p>2 where he uses the `lower density’, i.e. the \liminf in the previous expressions. I don’t know enough about the proof of these results though to study whether these sets are or aren’t finite unions of B_2[G]-sets.

    PS: When you refer to [2] in your comment, do you mean the same Bourgain article I cite here?

  5. Mark Lewko says:

    Yes, I meant the paper [B] (I’m not sure why I wrote [2]).

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