A question regarding Bourgain's paper on $\Lambda(p)$-subsets

Question

I'm trying to understand Bourgain's proof of Proposition 1.10 on page 304-307 in On $\Lambda(p)$-subsets of squares which states

Given $p>4$, we have the estimate \begin{align} \left\|\sum_{n=1}^N a_ne^{in^2t} \right\|_{L^p(\mathbb{T})} \lesssim N^{\frac{1}{2}-\frac{2}{p}}\left( \sum^N_{n=1}|a_n|^2\right)^{\frac{1}{2}}. \end{align}

More precisely, I am trying to understand the following on page 305

Let $t_1, \ldots, t_R$ be $1/N^2$-separated points in $[0, 1]$ satisfying \begin{align} \left|\sum^N_1 a_n e^{2\pi i n^2t_r} \right|>\delta N^{\frac{1}{2}} \ \ (1\leq r \leq R). \ \ \ \ \ \ (4.11) \end{align}

Here we already assumed $\sum |a_n|^2 \leq 1$ and $0<\delta<1$. Then the paper continues

Our purpose is to estimate $R$. By linearization, (4.11) yields \begin{align} \sum_{1\leq r, r'\leq R}\left|\sum^N_1 \exp[2\pi i n^2(t_r-t_{r'})] \right|>\delta^2NR^2. \ \ \ \ \ (4.12) \end{align}

Here I have a potentially very elementary question.

• What does he mean by linearization in this context, that is, how did he deduced (4.12) from (4.11)?

Answer 1

The method given by Mayank Pandey is pretty much it, but here's an alternative presentation:

There exists some choice of signs $c_r\in \mathbb{C}$ with $\lvert c_r\rvert =1$ such that

$$ \delta N^{1/2}R < \sum_{1\leq r\leq R}\lvert \sum_{1\leq n\leq N} a_n e(n^2t_r) \rvert = \sum_{1\leq n\leq N}a_n \sum_{1\leq r\leq R}c_r e(n^2t_r). $$

By the Cauchy-Schwarz inequality

$$ \delta^2NR^2 < \left(\sum \lvert a_n\rvert^2\right)\left(\sum_{1\leq r,r'\leq R}c_rc_{r'}\sum_{1\leq n\leq N}e(n^2(t_r-t_{r'}))\right). $$

The result follows by the triangle inequality and fact that $\sum \lvert a_n\rvert^2\leq 1$. Note that it was not needed in this argument that the points $t_r$ are $1/N^2$-separated.

The 'trick' of changing the order of summation 'through an absolute value sign' (by introducing the sign factors $c_r$ which are then discarded in the final conclusion), following by Cauchy-Schwarz and/or Holder's inequality, is extremely powerful and appears often throughout Bourgain's work (and elsewhere - it also appears under a slightly different guise in the large sieve-type techniques).

It is usually what Bourgain refers to when he invokes linearisation. I'm not sure if the use of this term for this technique appears outside of Bourgain's work -- presumably it refers to being able to create information about linear relations between the $t_r$ from just a pointwise bound.

Answer 2

I think it can be shown as follows. Write $$c_r = \overline{\left(\sum_{n\le N} e(n^2t_r)\right)}$$ (where we write $e(t) = e^{2\pi i t}$ for convenience. Then, we obtain that $$\delta^2 NR < \sum_r c_r\sum_n a_n e(n^2t_r) = \sum_{n}a_n\sum_r c_re(n^2t_r)\le\left(\sum_n \bigg|\sum_r c_re(nt_r)\bigg|^2\right)^{1/2} $$ by Cauchy-Schwarz. Then, since we have that $|c_r| > \delta N^{1/2}$, expanding out the square we obtain that $$\delta^4N^2R^2\le \sum_{r, r'} c_r\overline{c_{r'}}\sum_{n} e(n^2(t_r - t_{r'})) \le \delta^2 N\sum_{r, r'}\bigg|\sum_{n} e(n^2(t_r - t_{r'})\bigg| $$ so (4.12) follows by rearranging. I'm not sure if this is the linearisation in the sense Bourgain means.