the surface measure on $\mathbb{S}^{2}$ and $\psi_{1}$ is another cutoff.

Since $g$ is (qualitatively) a Schwartz function of $\rho$ and $\theta$, we may decompose $g=g_{1}+g_{2}$, where for each $\theta$, $\mathcal{F}_{\rho}g_{1}(\rho,\theta)(\tau)$ is supported in the region $|\tau|\lesssim M_{0}$, and $\mathcal{F}_{\rho}g_{2}(\rho,\theta)(\tau)$ is supported in the region $|\tau|\gg M_{0}$, where $M_{0}=2^{j}\langle m\rangle^{A/8}$. To estimate $g_{2}$, we only need to estimate $$\int_{|\tau|\gg M_{0}}\big|(\mathcal{F}_{\rho}g(\rho,\theta))(\tau)\big|\,\mathrm{d}\tau$$ uniformly in $\theta$. But for each fixed $\tau$ with $|\tau|\gg M_{0}$ we have$$(\mathcal{F}_{\rho}g(\rho,\theta))(\tau)=\int_{\mathbb{R}}e^{-{i}\rho\tau}g(\rho,\theta)\,\mathrm{d}\rho=\int_{\mathbb{R}}e^{-{i}\rho\tau}\psi(\rho)\,\mathrm{d}\rho\int_{\mathbb{R}^{3}}\psi_{0}(2^{-j}z)S_{l}F(z)e^{-{i}\rho(\theta\cdot z)}\,\mathrm{d}z.\nonumber$$ Now if we fix $z$, then the integral in $\rho$ can be bounded by $(|\tau|+2^m)^{-10}$, which will be acceptable. Therefore in (\[polar\]) we may replace the function $g$ by $g_{1}$. Using (\[sobolev\]) and also noticing the bounded Fourier support of $g_{1}(\cdot,\theta)$ for each $\theta$, we have $$\label{sobolev2}\|g_{1}\|_{L_{\theta}^{\infty}L_{\rho}^{\infty}}\lesssim M_{0}^{1/2}\|g_{1}\|_{L_{\theta}^{\infty}L_{\rho}^{2}}\lesssim M_{0}^{1/2}\|g_{1}\|_{L_{\rho}^{2}L_{\theta}^{\infty}}\lesssim2^{l}M_{0}^{1/2}\|g_{1}\|_{L_{\rho}^{2}L_{\theta}^{2}}.$$ If, in the integral (\[polar\]), we restrict to the region $|(\theta^{1})^{2}-1|\geq 1/100$, since the function $\theta\mapsto\theta^{1}$ has no critical point in this region, we can integrate by parts in $\theta$ many times to bound the left hand side of (\[polar\]) by $2^{-10m}$, which implies (\[disper2\]). Therefore, in (\[polar\]), after replacing $g$ by $g_{1}$, we may also cutoff in the region $|\theta^{1}-1|\leq 1/50$ (the case $|\theta^{1}+1|\leq 1/50$ is treated in the same way), on which we can use $(\theta^{2},\theta^{3})$ 