and ${\mbox{\textnormal{Op}}}({\sigma_R})$, where ${\sigma_{lf}}$, ${\sigma_I}$, ${\sigma_{II}}$ and ${\sigma_R}$ are the four components of the decomposition (\[A6\]).\
The operator ${\mbox{\textnormal{Op}}}({\sigma_{lf}})$ is handled as follows:

\[propII1\] Let $m\in {{\mathbb R}}$ and $s_0>d/2$, and let $\sigma\in \Gamma_{s_0}^m$.\
[**i.**]{} The operator ${\mbox{\textnormal{Op}}}({\sigma_{lf}})$ extends as an operator mapping any Sobolev space into $H^s({{\mathbb R}}^d)$, for all $s\leq s_0$. Moreover, $$\forall t\in {{\mathbb R}},\quad\forall s\leq s_0,\quad \forall u\in H^t({{\mathbb R}}^d),
    \qquad
    \left\vert {\mbox{\textnormal{Op}}}({\sigma_{lf}})u\right\vert_{H^s}\lesssim n_{0,s}(\sigma)
    \vert u\vert_{H^t},$$ where $n_{0,s}(\sigma)$ is defined in (\[A2\]).\
[**ii.**]{} If $\sigma$ is ${{2[\frac{d}{2}]+2}}$-regular at the origin, the following estimates also hold, for all $s\leq s_0$, $$\vert {\mbox{\textnormal{Op}}}({\sigma_{lf}})u\vert_{H^s}
    \lesssim n_{{{2[\frac{d}{2}]+2}},s}(\sigma)\vert u\vert_{C_*^0}
    \lesssim n_{{{2[\frac{d}{2}]+2}},s}(\sigma)\vert u\vert_\infty.$$

By definition, one has $$\begin{aligned}
    {\mbox{\textnormal{Op}}}({\sigma_{lf}})u(x)&=&(2\pi)^{-d}\int_{{{\mathbb R}}^d}e^{ix\cdot\xi}
    \psi(\xi)\sigma(x,\xi)
    \widehat{u}(\xi)d\xi\\
    &=&(2\pi)^{-d}\int_{\vert\xi\vert\leq 1}e^{ix\cdot\xi}
    \psi(\xi)\sigma(x,\xi)
    \widehat{u}(\xi)d\xi.\end{aligned}$$ Since ${\displaystyle}\vert x\mapsto e^{ix\cdot\xi}\sigma(x,\xi)\vert_{H^s}\leq 
    {\mbox{\textnormal{Cst }}}{\langle\xi\rangle}^{\vert s\vert}\vert \sigma(\cdot,\xi)\vert_{H^s}$, one has $$\vert {\mbox{\textnormal{Op}}}({\sigma_{lf}})u\vert_{H^s}\leq {\mbox{\textnormal{Cst }}}n_{0,s}(\sigma)
    \int_{\vert\xi\vert\leq 1}{\langle\xi\rangle}^{\vert s\vert}
    \vert \widehat{u}(\xi)\vert d\xi.$$ One can then obtain the first estimate by a simple Cauchy-Schwarz argument.\
In order to prove the second estimate, remark that a simple expansion in Fourier series shows that $$\psi(\xi)\sigma(x,\xi)=
    {\mathbf 1}_{\{\vert\xi\vert\leq 1\}}(\xi)\sum_{k\in{{\mathbb Z}}^d}
    \frac{1}{(1+\vert k\vert^2)^{1+[\frac{d}{2}]}}
    c_k(x)e^{i\xi\cdot k},$$ where $ {\mathbf 1}_{\{\vert\xi\vert\leq 1\}}$ is the characteristic function of the ball $\{\vert\xi\vert\leq 1\}$ and $$c_k(x)=(1+\vert k\vert^2)^{1+[d/2]}(2\pi)^{-d}
    \int_{[-\pi,\pi]^d}e^{-i\xi\cdot k}\psi(\xi)\sigma(x,\xi)d\xi.$$ Using methods similar to those used in the proof 