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.

[B] Bourgain, J., Bounded orthogonal systems and the ${\Lambda(p)}$ -set problem, Acta Math. 162 (1989), no. 3-4, 227–245
[BC] Bose, R. C., Chowla, S., Theorems in the additive theory of numbers, Comment. Math. Helv. 37 (1962/1963)
[BE] Bachelis, G., Ebenstein, S., On ${\Lambda (p)}$ sets, Pacific J. Math. Volume 54, Number 1 (1974), 35-38.
[R] Rudin, W., Trigonometric Series with Gaps, Indiana Univ. Math. J. 9 No. 2 (1960), 203–227.

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

Alex Iosevich says:

March 25, 2010 at 12:08 am

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

- ioannis parissis says:
  
  March 25, 2010 at 12:10 am
  
  True:) thanks Alex.
  
Mark Lewko says:

March 26, 2010 at 3:13 am

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 $p>2$ .

- ioannis parissis says:
  
  March 26, 2010 at 11:08 am
  
  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.
  
Mark Lewko says:

March 27, 2010 at 6:58 am

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.

ioannis parissis says:

March 28, 2010 at 5:07 pm

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?

Mark Lewko says:

March 28, 2010 at 8:47 pm

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