to do that is contained in the following:

Assume that $0 \leq s \leq 3$. Then the following fixed time estimate holds: $$\sum_{{\lambda}=1}^\infty \lambda^{2(s-1)}\|{{\tilde S}}_{\lambda}({g_{> \sqrt{{\lambda}}}} u)
\|_{L^2}^2 \lesssim (M (\|{\partial}^2 g\|_{L^\infty}))^2 \| u\|_{
H^{s-2}}^2 \label{largeg}$$ where $M$ stands for the maximal function with respect to time. We also have the dual estimate $$\| \sum_{{\lambda}=1}^\infty  {g_{> \sqrt{{\lambda}}}} {{\tilde S}}_{\lambda}f_{\lambda}\|_{H^{2-s}}^2 \lesssim \sum_{{\lambda}=1}^\infty \lambda^{2(1-s)}\| f_{\lambda}\|_{L^2}^2 \label{largegdual}$$

We take a Littlewood-Paley decomposition of both factors, $${{\tilde S}}_{\lambda}(g_{> \sqrt{{\lambda}}} u)  = \sum_{\mu=1}^\infty \sum_{\nu =
\sqrt{{\lambda}}}^\infty {{\tilde S}}_{\lambda}( g_\nu  u_\mu)$$ The $(\mu,\nu)$ term is nonzero only in the following situations:

\(i) $\nu \ll {\lambda}$, $\mu \approx {\lambda}$. Then we estimate $$\| {{\tilde S}}_{\lambda}(g_\nu  u_\mu)\|_{L^2} \lesssim \|g_\nu\|_{L^\infty}
\|u_\mu\|_{ L^2} \lesssim  \nu^{-2}  M (\|{\partial}^2 g\|_{L^\infty})
\|u_\mu\|_{ L^2}$$ and use the square summability with respect to ${\lambda}$ together with the relation $\nu^{-2} \lesssim  {\lambda}^{-1}$.

\(ii) $\nu \approx {\lambda}$, $\mu \ll {\lambda}$. Then $$\| {{\tilde S}}_{\lambda}(g_\nu  u_\mu)\|_{L^2} \lesssim \|g_\nu\|_{L^\infty}
\|u_\mu\|_{ L^2} \lesssim {\lambda}^{-2} M (\|{\partial}^2 g\|_{L^\infty})
\|u_\mu\|_{ L^2}$$ This is tight only when $s=3$ and $\mu=1$, otherwise there is a gain which insures the summability in ${\lambda}$, $\mu$.

\(iii) $\nu \approx \mu \gtrsim {\lambda}$. Then $$\| {{\tilde S}}_{\lambda}(g_\nu  u_\mu)\|_{L^2} \lesssim \|g_\nu\|_{L^\infty}
\|u_\mu\|_{ L^2} \lesssim  \mu^{-2} M (\|{\partial}^2 g\|_{L^\infty}) \|
