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I am reading a paper 'Periodic nonlinear Schrodinger Equation and Invariant measure' by J.Bourgain.

And I have a questions in the proof of lemma 3.10.

Please click the paper title for the link.

The problem I am attempting to solve is to get (3.12) of the paper.

The proof begin by trying to get (3.12).

Back ground : The goal of the proof is to estimate $$\mathbb{P}\left[ \left\|\sum \frac{g_n(\omega)}{n}e^{2\pi i nx} \right\|_p>\lambda, \left(\sum \frac{|g_n(\omega)|^2}{n^2} \right)^{\frac{1}{2}}<B \right]$$ where $(g_n(\omega))$ is a sequence of normalized Gaussian random variables and $\lambda>0$ and $B>0$ are constants possibly will be changed in the proof. Let's call the event we want to estimate $E$.

Now, we want to find an event, $D$, we can deduce from the event above so that we can use $\mathbb{P}(E)\leq \mathbb{P}(D)$. In order to do that, we split the system into dyadic blocks to use the fact that $$\left\|\sum \frac{g_n(\omega)}{n}e^{2\pi i nx} \right\|_p \leq \sum_{M:dyadic} \left\|\sum_{n\sim M} \frac{g_n(\omega)}{n}e^{2\pi i nx} \right\|_p$$

Now, the problem I am working on is to get (3.12) and (3.13) in the paper by using the two inequalities and the a well known inequality stated below.

$$\left\| \sum_{n\sim M} a_n e^{inx} \right\|_p\leq M^{1/2-1/p} \left\| \sum_{n\sim M} a_ne^{2\pi in x} \right\|_2$$

My attempt to get (3.12) [trying to use (3.13) as well ] up to now is as below.

\begin{align*} \lambda \sigma_M &< \left\| \sum_{n~ M} \frac{g_n(\omega)}{M} e^{inx} \right\|_p \\ &\leq \left\| \sum_{n~ M} \frac{g_n(\omega)}{n} e^{inx} \right\|_p\\ &\leq M^{\frac{1}{2}-\frac{1}{p}} \left\| \sum_{n~ M} \frac{g_n(\omega)}{M} e^{inx} \right\|_2\\ &\leq M^{\frac{1}{2}-\frac{1}{p}} B \end{align*}

Thus, we get $$\sigma_M \frac{\lambda}{B}<M^{\frac{1}{2}-\frac{1}{p}}$$

Here, $(\sigma_M)$ is a sequence given in (3.14) of the paper satisfying $$\sum_{M>M_0} \sigma_{M}<1$$

In order to have (3.12), I would need to remove $\sigma_M$ by using (3.14) but it seems like I went too far to use it. I will be happy to have any idea on this inequality.

Thanks in advance.

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  • $\begingroup$ I would suggest you make your question more self-contained, including defining the symbols that you use. $\endgroup$
    – Leo Moos
    Commented Jun 29, 2021 at 16:45
  • $\begingroup$ @LeoMoos You are right. I should have added more details. I just edited. Thanks! $\endgroup$
    – Lev Bahn
    Commented Jun 30, 2021 at 1:24

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