# Decoupling lemma for the Lambda(p) problem

I'm attempting to work through Bourgain's paper "Bounded orthogonal systems and the $\Lambda(p)$-set problem". There is a step in the proof of the decoupling lemma that I am stuck on, and thought someone might be able to quickly clarify it. If someone is aware of an alternate exposition of this lemma, please let me know. Since I can't use latex in the post I have temporarily put a copy of the paper up at: http://lewko.wordpress.com/files/2009/11/bounded-orthogonal-systems-and-the-ap-set.pdf .

My question is: How do you derive the first inequality in the proof of Lemma 4, from 3.2?

I understand that $$| \sum x_{i} -\sum\nolimits_{i \in R^{1}} x_{i} |= |\sum (1- n_{i}) x_{i} | = |\sum ( n_{i}-1) x_{i} |,$$ but it seems you need something more like $$| \sum x_{i} -\sum\nolimits_{i \in R^{1}} x_{i} |= |\sum ( n_{i}-1/3) x_{i} |$$ to derive the inequality. I'm sure I'm missing something simple.

In addition, once I have this first inequality in the proof of Lemma 4 I'm not entirely sure how the next inequality follows from this one. I am assuming, once I figure out one of these, I'll be able to figure out the other as well. But any comments would be helpful.

I am aware of the exposition of Quéffelec in "Analyse harmonique: groupe de travail sur les espaces de Banach invariants par translation". However, this doesn't seem to illuminate the point. The proof of Lemma 4 is self-contained, so you shouldn't need to understand the rest of the paper to understand the question. If it's any encouragement I point out that my question is about the second sentence of the proof. The first sentence is "The argument is straightforward."

Update: Yemon gave a very nice proof of the first inequality. Unfortunately, I still don't see how to use this to bound the left-hand-side of 3.4 by the expression below the line "Hence, by 3.1...". Any hints or suggestions are appreciated.

Having had a quick look, does the following work? Put $$x= \sum_i x_i/3$$ and put $$y(t) = \sum_{i \in R^1_t} x_i = \sum_i \eta_i(t)x_i$$ and try to substitute these into (3.2).

Observe that \begin{aligned} |x| + |y(t)| = | \frac13 \sum_i x_i | + | \sum_i \eta_i x_i | & \leq | \frac13 \sum_i x_i | + | \sum_i x_i / 3 | + | \sum_i (\eta_i - 1/3)x_i | \\ &\leq | \sum_i x_i | + | \sum_i (\eta_i - 1/3)x_i | \end{aligned} and this should give what we want on the RHS of the formula you're asking about.

• Thank you! I am thankful (and a bit embarrassed) at how simple this is. Nov 3, 2009 at 7:28
• Oh, I remember once trying to read one of Bourgain's papers, and his style is such that I rapidly lost confidence in my ability to work out which steps were hard and which just had missing parts. If it makes you feel better, I had to think for quite a while before arriving at the above. Nov 3, 2009 at 7:37

As for the second part, the following works. The goal, inequality (3.4), can be written as $$\newcommand{\E}{\mathbb E}$$ $$|\E[\phi_1(X)\phi_2(Y)\phi_3(Z)]-\phi_1(\E X)\phi_2(\E Y)\phi_3(\E Z)| \le C(p)\big(1+|\sum x_i|+|\sum y_i| + |\sum z_i|\big)^{p-\delta}$$ where $$X = \sum_{i\in R_t^1}x_i$$, etc. The idea is to use the identity $$ABC - abc = (A-a)BC + a(B-b)C + ab(C - c)$$ with $$A=\phi_1(X),B=\phi_2(Y),C=\phi_3(Z)$$, and $$a = \phi_1(\E X),b=\phi_2(\E Y),c=\phi_3(\E Z)$$, and use the triangle inequality, estimating the three addends on the right-hand side using the assumptions on the $$\phi_\alpha$$. For that, it is helpful to notice, e.g., $$|\phi_1(\E X)-\phi_1(X)| \le C(1+|\sum x_i|)^{p_1-\delta}(1+|\sum (\eta_i-\frac13)x_i|)^p,$$ which follows immediately from the first line of Bourgain's proof. Ultimately then, the rest of the argument is using Khinchine's inequality and the conditions on the $$\phi_\alpha$$ to match the inequality (3.4).

For example, \begin{align*} \E(1+|X-\E X|)^p \le (1+C(p)|x|)^p < c, \end{align*} and so on.