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.
 A: 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.
A: 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.
