# 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.
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.