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