of Prop. \[propA2\], one obtains that $\vert c_k(\cdot)\vert_{H^s}\leq n_{{{2[\frac{d}{2}]+2}},s}(\sigma)$. Since $${\mbox{\textnormal{Op}}}(c_k(x)e^{i\xi\cdot k}{\mathbf 1}_{\{\vert\xi\vert\leq 1\}}(\xi))u
=c_k(\cdot)({\mathbf 1}_{\{\vert\xi\vert\leq 1\}}(D)u)(\cdot+k),$$ the result follows from the next lemma:

\[lemmII0\] Let $u,v\in{{\mathcal S}({{\mathbb R}}^d)}$ and assume that $\widehat{v}$ is supported in the ball $\{\vert\xi\vert\leq A\}$, for some $A>0$. Then for all $s\in {{\mathbb R}}$, one has $$\vert u v\vert_{H^s}\leq {\mbox{\textnormal{Cst }}}\vert u\vert_{H^s}\vert v\vert_\infty.$$

Write $uv=\sum_{q\geq -1}v\varphi_q(D)u$; except the first ones, each term of this sum has its spectrum included in an annulus of size $\sim 2^q$. Thanks to Lemma \[lemmLP1\], the $H^s$-norm of the product $u v$ can therefore be controlled in terms of $\vert v\varphi_q(D)u \vert_{L^2}$, $q\geq -1$. Since these quantities are easily bounded from above by $\vert v\vert_\infty\vert \varphi_q(D)u\vert_{L^2}$, the lemma follows from another application of Lemma \[lemmLP1\].

We now turn to study ${\mbox{\textnormal{Op}}}({\sigma_I})$. As already said, ${\sigma_I}$ is the paradifferential symbol associated to $\sigma$ so that it is well-known that ${\mbox{\textnormal{Op}}}({\sigma_I})$ maps $H^{s+m}$ into $H^s({{\mathbb R}}^d)$ for all $s\in{{\mathbb R}}$ (see [@Bony; @Meyer] and Prop. B.9 of [@Metivier-Zumbrun]). However, since we need a precise estimate on the operator norm of ${\mbox{\textnormal{Op}}}({\sigma_I})$, we cannot omit the proof.

\[propII2\] Let $m\in{{\mathbb R}}$ and $s_0>d/2$, and let $\sigma\in \Gamma_{s_0}^m$.\
If ${\sigma_I}$ is as defined in 