insured by our hypothesis.

[**Case 2**]{}. Here we consider products of the form $S_\mu v_\mu S_{\lambda}u_{\lambda}$ where $\mu \ll {\lambda}$. Then the product is localized at frequency ${\lambda}$. The summation with respect to ${\lambda}$ is trivial, but not the one with respect to $\mu$. We write $$S_\mu v_\mu S_{\lambda}u_{\lambda}=  S_\mu v_{\mu} S_{\lambda}u_{{\lambda},<\mu} +
\sum_{d = \mu}^{\lambda}S_\mu v_{\mu} S_{{\lambda}} u_{{\lambda},d}$$ Using Lemma \[md\] and Lemma \[xmd\] we obtain $$\begin{split}
\|S_{\lambda}u_{\lambda}S_\mu v_\mu&\|_{\tilde X^{s,\theta}_{{\lambda}}}^2 \lesssim
\|S_\mu v_{\mu} S_{\lambda}u_{{\lambda},<\mu}\|_{\tilde X^{s,\theta}_{{\lambda},\mu}}^2
+  \sum_{d = \mu}^{\lambda}\|S_\mu v_{\mu} S_{{\lambda}} u_{{\lambda},d}\|_{\tilde X^{s,\theta}_{{\lambda},d}}^2
\\ &\lesssim \mu^{2\theta-1}
\|S_\mu v_\mu\|^2_{M_{<\mu}}   \|u_{{\lambda},<\mu}\|_{\tilde X^{s,\theta}_{{\lambda},<\mu}}^2
+ \|S_\mu v_\mu\|^2_{M_{\mu}} \sum_{d=\mu}^{\lambda}\|u_{{\lambda},d}\|_{\tilde
  X^{s,\theta}_{{\lambda},d}}^2
\\ &\lesssim  \mu^{2\theta-1}\|S_\mu
v_\mu\|^2_{M_{<\mu}}\sum_{d=1}^{\lambda}\|u_{{\lambda},d}\|_{\tilde  X^{s,\theta}_{{\lambda},d}}^2
\\ &\lesssim \mu^{3+2\theta-2s} \|v_\mu\|^2_{ {{\tilde{X}}}^{s,\theta}_{\mu}}\sum_{d=1}^{\lambda}\|u_{{\lambda},d}\|_{\tilde  X^{s,\theta}_{{\lambda},d}}^2
\end{split}$$ The summation with respect to $\mu$ is trivial since $\theta < s-\frac32$.

We next prove .

Assume that $s > 2$ and $\frac12 < \theta < s-\frac32$ . Then we have the multiplicative estimate $$X^{s,\theta} \cdot X^{s-1,\theta-1} \subset X^{s-1,\theta-1}$$

By duality this reduces to the multiplicative estimate $$X^{s,\theta} \cdot (X^{1-s,1-\theta}+L^2 H^{2-s-\theta})  \subset
X^{1-s,1-\theta} + L^2 H^{2-s-\theta}$$ Since $s > 2$ we have the fixed time multiplication $$H^s  \cdot  H^{2-s-\theta} \subset H^{2-s-\theta}$$ which implies the space-time bound $$L^\infty H^s \cdot L^2  H^{2-s-\theta} \subset L^2  H^{2-s-\theta}$$ Due to the energy estimate for $X^{s,\theta}$ it 