remains to show that $$X^{s,\theta} \cdot X^{1-s,1-\theta} \subset
X^{1-s,1-\theta} + L^2 H^{2-s-\theta}$$ We consider a product $S_{\lambda}u_{\lambda}S_\mu v_\mu$ which we decompose as in the previous proof. Because of the lack of symmetry we now need to consider three cases.

[**Case 1**]{}. Here we estimate $ S_\mu (S_{\lambda}u_{\lambda}S_{\lambda}v_{\lambda})$ where $\mu
\lesssim {\lambda}$. By Corollary \[lessd\] we obtain $$\|S_{\lambda}u_{\lambda}S_{\lambda}v_{\lambda}\|_{L^2 L^\frac{3}{2}} \lesssim \|S_{\lambda}u_{\lambda}\|_{L^4 L^3} \|S_{\lambda}v_{\lambda}\|_{L^4 L^3} \lesssim
{\lambda}^{\theta-\frac{2}{3}} \|u_{\lambda}\|_{\tilde X^{s,\theta}_{{\lambda}}}
\|v_{\lambda}\|_{\tilde
 X^{1-s,1-\theta}_{{\lambda}}}$$

Using then Sobolev embeddings we obtain

$$\|S_\mu(S_{\lambda}u_{\lambda}S_{\lambda}v_{\lambda})\|_{L^2 H^{2-s-\theta}} \lesssim
\mu^{\frac 83 -s-\theta}{\lambda}^{\theta-\frac{2}{3}} \|u_{\lambda}\|_{\tilde
X^{s,\theta}_{{\lambda}}} \|v_{\lambda}\|_{\tilde
 X^{1-s,1-\theta}_{{\lambda}}}$$

[**Case 2**]{}. Here we bound $S_\mu u_\mu S_{\lambda}v_{\lambda}$, $\mu \ll {\lambda}$. The product is localized at frequency ${\lambda}$, and the analysis is almost identical to Case 2 in Proposition \[xxx\].

[**Case 3**]{}. Here we bound $S_{\lambda}u_{\lambda}S_\mu v_\mu$, $\mu \ll {\lambda}$. The same argument applies, the only difference is that we gain some extra $\mu/{\lambda}$ factors.

We continue with the Moser estimates in , which follow from

Assume that $s > 2$ and $\frac12 < \theta < s-\frac32$ . Let $\Gamma$ be a smooth function. Then $$\|\Gamma(u)\|_{X^{s,\theta}} \lesssim C(\|u\|_{L^\infty}) (1 +
\|u\|_{X^{s,\theta}}^5)$$

We write $$\Gamma(u)-\Gamma(v) = (u-v) f(u,v)$$ and $$f(u,v) - f(x,y) = (u-x) g_1(u,v,x,y) + (v-y)g_2(u,v,x,y)$$ where $f$, $g_1$ and $g_2$ are smooth functions. Then 