$\newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\ds}{\displaystyle} \newcommand{\Lpn}[2]{\left\lVert#1\right\rVert\_{L^{#2}}}$
$\newcommand{\Lptxy}[3]{\left\lVert#1\right\rVert\_{L^{#2}\_{#3}}}$
$\newcommand{\PQ}[2]{P\_{N\_{#2}}Q\_{L\_{#2}}{#1}\_{#2}}$
$\newcommand{\PQwV}[2]{P\_{N\_{#2}}Q\_{L\_{#2}}{#1}\_{#2}(\tau\_{#2},\xi\_{#2},q\_{#2})}$
$\newcommand{\FPQwV}[2]{\Fh{\prc{P\_{N\_{#2}}Q\_{L\_{#2}}{#1}\_{#2}}}(\tau\_{#2},\xi\_{#2},q\_{#2})}$
$\newcommand{\F}[1]{\mathcal{F} {#1} }
\newcommand{\Fh}[1]{\widehat{#1} }$
$\newcommand{\prc}[1]{ {\left( #1 \right)}}$
$\newcommand{\FPQ}[2]{\Fh{P\_{N\_{#2}}Q\_{L\_{#2}}{#1}\_{#2}}}$ $\newcommand\abs[1]{\left|#1 \right|}$


I am verifying the bilinear estimate (Proposition 3.6) in [this paper](https://arxiv.org/pdf/1302.2933.pdf). The first steps were clear except the last one.


$$\Lptxy{\prod\_{j=1}^2 \PQ{u}{j}}{2}{t x y}
= \Lptxy{\F(\prod\_{j=1}^2 \PQ{u}{j})(\tau,\xi,q)}{2}{\tau \xi q}$$
$$=\Lptxy{\ds\circledast\_{j=1}^3 \PQ{u}{j}(\tau,\xi,q)}{2}{\tau \xi q}$$
$$ = \Lptxy{\ds\int\_{\R^4} \ds\sum\_{q\_1,q\_2 \in \Z^2}{\prod\_{j=1}^2 \FPQwV{u}{j}} d\nu}{2}{\tau \xi q}$$
$$ \leq \prc{ \ds\int\_\R \ds\sum\_{q} \abs{\int\_{\R^4} \sum\_{q\_1,q\_2 \in \Z^2}d\nu }^2 d\tau d\xi}^\frac{1}{2} \prc{ \ds\int\_\R \ds\sum\_q \abs{{\int\_{\R^4} \sum\_{(q\_1,q\_2) \in \Z^2} \prod\_{j=1}^2 \FPQwV{u}{j} d\nu }}^2 d\tau d\xi }^\frac{1}{2}$$
$$ \leq \sup\_{\tau,\xi,q}\prc{ \int\_{\R^4} \sum\_{q\_1,q\_2 \in \Z^2}d\nu}^\frac{1}{2} \prc{ \ds\int\_\R \ds\sum\_q \abs{{\int\_{\R^4} \sum\_{(q\_1,q\_2) \in \Z^2} \prod\_{j=1}^2 \FPQwV{u}{j} d\nu }}^2 d\tau d\xi }^\frac{1}{2}$$


I could not justify how the last step can be achieved! Namely,


$$\prc{ \ds\int\_\R \ds\sum\_q \abs{{\int\_{\R^4} \sum\_{(q\_1,q\_2) \in \Z^2} \prod\_{j=1}^2 \FPQwV{u}{j} d\nu }}^2 d\tau d\xi }^\frac{1}{2} \lesssim \prod\_{j=1}^2 \Lptxy{\PQ{u}{j}}{2}{t x y}.$$


Any help is appreciated.



Verifying the proof of a bilinear estimate in $L^2$
$\newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\ds}{\displaystyle} \newcommand{\Lpn}[2]{\left\lVert#1\right\rVert\_{L^{#2}}}$
$\newcommand{\Lptxy}[3]{\left\lVert#1\right\rVert\_{L^{#2}\_{#3}}}$
$\newcommand{\PQ}[2]{P\_{N\_{#2}}Q\_{L\_{#2}}{#1}\_{#2}}$
$\newcommand{\PQwV}[2]{P\_{N\_{#2}}Q\_{L\_{#2}}{#1}\_{#2}(\tau\_{#2},\xi\_{#2},q\_{#2})}$
$\newcommand{\FPQwV}[2]{\Fh{\prc{P\_{N\_{#2}}Q\_{L\_{#2}}{#1}\_{#2}}}(\tau\_{#2},\xi\_{#2},q\_{#2})}$
$\newcommand{\F}[1]{\mathcal{F} {#1} }
\newcommand{\Fh}[1]{\widehat{#1} }$
$\newcommand{\prc}[1]{ {\left( #1 \right)}}$
$\newcommand{\FPQ}[2]{\Fh{P\_{N\_{#2}}Q\_{L\_{#2}}{#1}\_{#2}}}$ $\newcommand\abs[1]{\left|#1 \right|}$
$\newcommand{\prdd}{\prod\_{j=1}^2}$


There is a mistake in to which integration applying the Cauchy-Schwarz inequality.


\begin{align\*}
 &\Lptxy{\ds\int\_{\R^4} \ds\sum\_{q\_1,q\_2 \in \Z^2}{\prod\_{j=1}^2 \FPQwV{u}{j}} d\nu}{2}{\tau \xi q}\\
& \leq \Lptxy{\prc{\int\_{\R^4} \ds\sum\_{q\_1,q\_2 \in \Z} d\nu}^\frac{1}{2} \prc{\int\_{\R^4} \ds\sum\_{q\_1,q\_2 \in \Z^2} \abs{\prdd \FPQwV{u}{j}}^2 d\nu}^\frac{1}{2} }{2}{\tau \xi q}\\
& \leq \sup\_{\tau,\xi,q}\prc{ \int\_{\R^4} \sum\_{q\_1,q\_2 \in \Z^2}d\nu}^\frac{1}{2} \Lptxy{ \prc{\int\_{\R^4} \ds\sum\_{q\_1,q\_2 \in \Z^2} \abs{\prdd \FPQwV{u}{j}}^2 d \nu}^\frac{1}{2} }{2}{\tau \xi q}\\
& \leq \sup\_{(\tau,\xi,q)}\abs{A\_{\tau,\xi,q}}^\frac{1}{2} \prc{\int\_{\R^2}\sum\_{q} \abs{\prc{\int\_{\R^4} \sum\_{q\_1,q\_2 \in \Z} \abs{\prdd \FPQwV{u}{j}}^2 d \nu}}d\tau d\xi}^\frac{1}{2} \\
& \leq \sup\_{(\tau,\xi,q)}\abs{A\_{\tau,\xi,q}}^\frac{1}{2} \prc{\int\_{\R^4}\sum\_{q\_1,q\_2} \abs{\prod\_{j=1}^2\FPQwV{u}{j}}^2\prc{\int\_{\R^2} \sum\_q \abs{\FPQ{u}{3}}^2 d\tau d\xi} d\nu}^\frac{1}{2}\\
&= \sup\_{(\tau,\xi,q)}\abs{A\_{\tau,\xi,q}}^\frac{1}{2} \Lptxy{\FPQ{u}{3}}{2}{\tau \xi q} \prc{\int\_{\R^4}\sum\_{q\_1,q\_2} \abs{\prod\_{j=1}^2\FPQwV{u}{j}}^2d\nu}^2\\
&= \sup\_{(\tau,\xi,q)}\abs{A\_{(\tau,\xi,q)}}^\frac{1}{2} \prod\_{j=1}^3 \Lptxy{\PQ{u}{j}}{2}{t x y}.
\end{align\*}


