\langle j\rangle^{-N_{0}}2^{-j+k},$$ while using (\[stkg\]) and Hölder we have $$\big\|e^{{i}t\Lambda}\Gamma^{\mu}f_{jk}^{*}\big\|_{L^{\infty}}\lesssim2^{-3m/2}\|\Gamma^{\mu}f_{jk}\|_{L^{1}}\lesssim2^{-3(m-j)/2}\|\Gamma^{\mu}f_{jk}\|_{L^{2}}\lesssim \langle j\rangle^{-N_{0}}2^{-(3m-j+k)/2}.$$ Similarly we have (\[disper2\]).

Now we prove (\[disper3\]) in the case $j\leq m-A\log m$. The case $j\geq m+A\log m$ only needs minor changes. Let $\Gamma^{\mu}f_{jk}=F$ and $\Gamma^{\mu}f_{jk}^{*}=F^{*}$; if $|x|\leq 2^{m}\langle m\rangle^{-A}$, recall that $$\label{express}(e^{{i}t\Lambda}F^{*})(x)=\int_{\mathbb{R}^{3}}e^{{i}t\Lambda(\xi)+{i}x\cdot \xi}\psi(\xi)\widehat{F}(\xi)\,\mathrm{d}\xi,$$ where $\psi(\xi)=\varphi_{[k-2,k+2]}(\xi)$ is a cutoff because $|k|\leq K_{0}$. Since $F(z)$ is supported in $|z|\sim 2^{j}$ (unless $j=\max(-k,0)$, in which case $j=O(1)$ and the estimate will be trivial), we may rewrite the above expression as $$\big(e^{{i}t\Lambda}F^{*}\big)(x)=\int_{\mathbb{R}^{3}}\psi_{0}(2^{-j}z)F(z)\,\mathrm{d}z\int_{\mathbb{R}^{3}}e^{{i}(t\Lambda(\xi)+(x-z)\cdot \xi)}\psi(\xi)\,\mathrm{d}\xi,$$ with another cutoff $\psi_{0}(z)$ supported in the region $|z|\sim 1$. Now fix any $z$ such that $|z|\sim 2^{j}$, we can use Proposition \[ips0\] with $$K=2^{\max(m,j)},\quad n=1,\quad \epsilon,\lambda\sim 1$$ to bound the $\xi$-integral by $2^{-10m}$, which is clearly enough for (\[disper3\]). The same argument also applies for $|x|\geq 2^{m}\langle m\rangle^{A}$.

Therefore, to prove (\[disper3\]), we may assume $|x|\sim 2^{r}$ where $|r-m|\leq A\log m$; without loss of generality we may also assume $x=(\alpha,0,0)$, where $\alpha=|x|$. Decomposing $F$ into $S_{l}F$ as in (\[spherical\]), we may also assume $l\leq 7\delta m$, since otherwise we have$$\|S_{l}F\|_{L^{2}}\lesssim 2^{-Nl/2}\sup_{|\nu|\leq N/2}\|\Omega^{\nu}S_{l}F\|_{L^{2}}\lesssim 2^{-7m/2},$$ from which (\[disper2\]) follows easily. Since the $\psi$ in (\[express\]) is radial, we may write $\psi(|\xi|)\widehat{S_{l}F}(\xi)=g(|\xi|,\xi/|\xi|)$, then we have $$\label{polar}\big(e^{{i}t\Lambda}S_{l}F^{*}\big)(x)=\int_{\mathbb{R}}\int_{\mathbb{S}^{2}}e^{{i}(t\Lambda(\rho)+\alpha\rho\theta^{1})}\rho^{2}\psi_{1}(\rho)g(\rho,\theta)\,\mathrm{d}\rho\mathrm{d}\omega(\theta),$$ where $\mathrm{d}\omega$ is 