Verifying the proof of a bilinear estimate in $L^2$

Question

$\newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\ds}{\displaystyle} \newcommand{\Lpn}[2]{\left\lVert#1\right\rVert_{L^{#2}}}$ $\newcommand{\Lptxy}[3]{\left\lVert#1\right\rVert_{L^{#2}_{#3}}}$ $\newcommand{\PQ}[2]{P_{N_{#2}}Q_{L_{#2}}{#1}_{#2}}$ $\newcommand{\PQwV}[2]{P_{N_{#2}}Q_{L_{#2}}{#1}_{#2}(\tau_{#2},\xi_{#2},q_{#2})}$ $\newcommand{\FPQwV}[2]{\Fh{\prc{P_{N_{#2}}Q_{L_{#2}}{#1}_{#2}}}(\tau_{#2},\xi_{#2},q_{#2})}$ $\newcommand{\F}[1]{\mathcal{F} {#1} } \newcommand{\Fh}[1]{\widehat{#1} }$ $\newcommand{\prc}[1]{ {\left( #1 \right)}}$ $\newcommand{\FPQ}[2]{\Fh{P_{N_{#2}}Q_{L_{#2}}{#1}_{#2}}}$ $\newcommand\abs[1]{\left|#1 \right|}$

I am verifying the bilinear estimate (Proposition 3.6) in this paper. 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.

Answer 1

$\newcommand{\R}{\mathbb{R}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\ds}{\displaystyle} \newcommand{\Lpn}[2]{\left\lVert#1\right\rVert_{L^{#2}}}$ $\newcommand{\Lptxy}[3]{\left\lVert#1\right\rVert_{L^{#2}_{#3}}}$ $\newcommand{\PQ}[2]{P_{N_{#2}}Q_{L_{#2}}{#1}_{#2}}$ $\newcommand{\PQwV}[2]{P_{N_{#2}}Q_{L_{#2}}{#1}_{#2}(\tau_{#2},\xi_{#2},q_{#2})}$ $\newcommand{\FPQwV}[2]{\Fh{\prc{P_{N_{#2}}Q_{L_{#2}}{#1}_{#2}}}(\tau_{#2},\xi_{#2},q_{#2})}$ $\newcommand{\F}[1]{\mathcal{F} {#1} } \newcommand{\Fh}[1]{\widehat{#1} }$ $\newcommand{\prc}[1]{ {\left( #1 \right)}}$ $\newcommand{\FPQ}[2]{\Fh{P_{N_{#2}}Q_{L_{#2}}{#1}_{#2}}}$ $\newcommand\abs[1]{\left|#1 \right|}$ $\newcommand{\prdd}{\prod_{j=1}^2}$

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*}