have$$\sup_{|\mu|\leq N/2+2}\big\|e^{it\Lambda}\Gamma^{\mu}S_{l}f_{jk}^{*}\big\|_{L^{\infty}}\lesssim2^{-(N/2-2)l},$$ (\[disper3\]) follows by interpolation.

Finally, (\[disper5\]) is proved in the same way as (\[disper3\]), with changes made in a few places: since $k\geq K_{0}$ and $\rho=|\xi|\sim 2^{k}$, in (\[polar\]) the factor $\rho^{2}$ will count as $2^{2k}$, and the measure of the set where $|\Xi(\rho)|\lesssim \epsilon''$ is now $2^{3k}\epsilon''$ instead of $\epsilon''$ since $|\Lambda''(\rho)|\sim 2^{-3k}$. Moreover in (\[seteps\]) we can set $\epsilon'=2^{-(r+k)/2}\langle m\rangle^{2A}$ instead of $2^{-r}\langle m\rangle^{2A}$, thus in total we have a loss of $2^{4k}$ compared with the proof of (\[disper3\]), thus we have $$\sup_{|\mu|\leq N/2+2}\big\|e^{{i}t\Lambda}\Gamma^{\mu}f_{jk}^{*}\big\|_{L^{\infty}}\lesssim 2^{-(3m+j)/2}2^{4k+l}\langle m\rangle^{-N_{0}+A}.$$ Since we trivially have $$\sup_{|\mu|\leq N/2+2}\big\|e^{{i}t\Lambda}\Gamma^{\mu}f_{jk}^{*}\big\|_{L^{\infty}}\lesssim \min(2^{-(N/2-2)k},2^{-(N/2-2)l}),$$ we easily get (\[disper5\]) by interpolation.

\[cor\] Suppose $f=f(x)$ is a function with $\|f\|_{X}\lesssim \varepsilon$, and let $v=e^{{i}t\Lambda}f$ with $\Lambda$ equaling one of the $\Lambda_{\alpha}$, then we have$$\sum_{|\mu|\leq N/2}\|\Gamma^{\mu}v\|_{L^{\infty}}\lesssim\frac{\varepsilon}{(1+|t|)\log^{(N_{0}-2)}(2+|t|)}.$$

Let $1+|t|\sim 2^{m}$, and we decompose$$v=\sum_{(j,k)\in\mathcal{J}}e^{{i}t\Lambda}f_{jk}^{*}.$$ Now (\[disper1\])-(\[disper5\]) in particular implies$$\big\|e^{{i}t\Lambda}\Gamma^{\mu}f_{jk}^{*}\big\|_{L^{\infty}}\lesssim\varepsilon(1+2^{k/2})^{-1}(\max\langle m\rangle,\langle j\rangle)^{-N_{0}}2^{-\max(m,j)}$$for each $|\mu|\leq N/2$. Therefore $$\begin{aligned}
\|\Omega^{\mu}v\|_{L^{\infty}}&\lesssim &\sum_{(j,k)\in\mathcal{J}}\varepsilon(1+2^{k/2})^{-1}\max(\langle m\rangle,\langle j\rangle)^{-N_{0}}\cdot2^{-\max(m,j)}\nonumber\\
&\lesssim&\varepsilon\sum_{j\geq 0}2^{-\max(m,j)}\max(\langle m\rangle,\langle j\rangle)^{-N_{0}}\cdot\sum_{k=-j}^{\infty}(1+2^{k/2})^{-1}\nonumber\\
&\lesssim&\varepsilon\sum_{j\geq 0}2^{-\max(m,j)}(\max\langle m\rangle,\langle j\rangle)^{-(N_{0}-1)}\nonumber\\
&\lesssim &\varepsilon2^{-m}\langle m\rangle^{-(N_{0}-2)},\end{aligned}$$ which is what we need.

\[bilinear\] For the bilinear operator $T$ defined by $$\label{def}\mathcal{F}T(f,g)(\xi)=\int_{\mathbb{R}^{3}}K(\xi,\eta)\widehat{f}(\xi-\eta)\widehat{g}(\eta)\,\mathrm{d}\eta,$$ we have the followings:

1.  If $\|\mathcal{F}_{\xi,\eta}^{-1}K\|_{L^{1}}\leq 1$, and $$p,q,r\in[1,\infty],\qquad \frac{1}{r}=\frac{1}{p}+\frac{1}{q},$$ then we have $$\label{biliest}\|T(f,g)\|_{L^{r}}\lesssim\|f\|_{L^{p}}\|g\|_{L^{q}}.$$

2.  If $K$ satisfies $$\sup_{\xi}\int_{\mathbb{R}^{3}}|K(\xi,\eta)|^{2}\,\mathrm{d}\eta+\sup_{\eta}\int_{\mathbb{R}^{3}}|K(\xi,\eta)|^{2}\,\mathrm{d}\xi\lesssim 1,$$ then we have $$\label{biliest2}\|T(f,g)\|_{L^{2}}\lesssim\|f\|_{L^{2}}\|g\|_{L^{2}}.$$

3.  Suppose 