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.

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Having had a quick look, does the following work? Put $x= \sum\_i x\_i/3$ and put

$$y(t) = \sum\nolimits\_{i \in {R^1}_t} x_i = \sum\_i \eta\_i(t)x\_i$$

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.

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Thank you! I am thankful (and a bit embarrassed) at how simple this is. –  Mark Lewko Nov 3 '09 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. –  Yemon Choi Nov 3 '09 at 7:37