definition of ${\sigma_{II}}$.\
[**ii.**]{} As said previously, when $\sigma(x,\xi)=\sigma(x)$ does not depend on $\xi$, one has ${\mbox{\textnormal{Op}}}({\sigma_{II}})u=T_{(1-\psi(D))u}\sigma$ and thus $\vert {\mbox{\textnormal{Op}}}({\sigma_{II}})u\vert_{H^s}\lesssim 
    \vert u\vert_\infty \vert \sigma\vert_{H^s}$, that is, the endpoint case $t=0$ holds in Prop. \[propII3\] if one weakens the $\vert u\vert_{C^0_*}$-control into a $\vert u\vert_\infty$-control. This is no longer true in general when dealing with general symbols.

Prop. \[propA2\] allows us to reduce the study to the case $m=0$ and to the reduced symbols $p_k(x,\xi)$ given in that proposition.\
By definition of ${\sigma_{II}}$, one has, for all $u\in{{\mathcal S}({{\mathbb R}}^d)}$, ${\sigma_{II}}(x,D)u=\sum_{p\geq-1}v_p$, with $v_p=\sigma_{(p)}(x,D)(1-\psi(D))\psi(2^{-p+N}D)u$ and $\sigma_{(p)}(\cdot,\xi)=\varphi_p(D_x)\sigma(\cdot,\xi)$. Since the spectrum of $v_p$ is supported in $(1-2^{1-N})2^{p-1}\leq \vert\xi\vert\leq (1+2^{-1-N})2^{p+1}$, Lemma \[lemmLP1\] reduces the control of $\vert {\sigma_{II}}(x,D)u\vert_{H^s}$ to finding an estimate on each $\vert v_p\vert_2$, and hence on $$\begin{aligned}
    I&=&\Big\vert \sum_{q\geq -1}\big[\varphi_p(D_x)c_{k,q}\big]
    \lambda_k(2^{-q}D)
    \psi(2^{-p+N}D)(1-\psi(D))u\Big\vert_2\\
    &\leq&\sum_{q\geq -1}\left\vert \varphi_p(D_x)c_{k,q}\right\vert_2
    \left\vert \lambda_k(2^{-q}D)
    \psi(2^{-p+N}D)u\right\vert_\infty.\end{aligned}$$ Remarking that $\lambda_k(2^{-q}\xi)\psi(2^{-p+N}\xi)=0$ when $q\geq p-N+2$, and using Prop. \[propA2\], one deduces $$\label{II1}
    I\leq {\mbox{\textnormal{Cst }}}N^m_{{{2[\frac{d}{2}]+2}},0}(\sigma_{(p)})\sum_{q=-1}^{p-N+1}
    \left\vert \lambda_k(2^{-q}D)
    \psi(2^{-p+N}D)u\right\vert_\infty,$$ where we recall that $\sigma_{(p)}(\cdot,\xi)=\varphi_p(D_x)\sigma(\cdot,\xi)$.\
We now need the following lemma:

\[lemmII2\] Let $A,B>0$ and $\lambda\in C_0^\infty({{\mathbb R}}^d)$ supported in $A\leq\vert\xi\vert\leq B$. Then, for all $t\in{{\mathbb R}}$ and $q\geq -1$, one has, $$\forall u\in C_*^t,\qquad
    \left\vert \lambda(2^{-q}D)u\right\vert_\infty
    \leq C_t \vert {\check}\lambda\vert_{L^1}2^{-qt}\vert u\vert_{C^t_*}.$$

Since $\lambda$ 