that $$\label{A3}
        \forall \xi\in{{\mathbb R}}^d,\qquad
        {\mbox{\textnormal{Sp }}}\sigma(\cdot,\xi)\subset 
        \{\eta\in{{\mathbb R}}^d,\vert\eta\vert\leq\delta\vert\xi\vert\}.$$

If $\psi\in C_0^\infty({{\mathbb R}}^d)$ is as in (\[LP1\]), then the spectral condition (\[A3\]) implies that for all $\sigma\in\Sigma_s^m$, one has $$\label{A4}
    \forall \xi\in{{\mathbb R}}^d,\qquad
    \sigma(\cdot,\xi)=
    \sigma(\cdot,\xi)*\left[(2\delta{\langle\xi\rangle})^d
    {\check}{\psi}(2\delta{\langle\xi\rangle}\cdot)\right].$$ It is classical (Bernstein’s lemma) to deduce that for all $\alpha,\beta\in{{\mathbb N}}^d$, one has $$\label{A5}
    \forall \xi\in{{\mathbb R}}^d,\vert\xi\vert\geq 1/4,\quad
    \left\vert \partial_x^\alpha\partial_\xi^\beta
    \sigma(\cdot,\xi)\right\vert_{L^\infty}
    \leq {\mbox{\textnormal{Cst }}}M^m_{\vert\beta\vert}(\sigma)
    {\langle\xi\rangle}^{m-\vert\beta\vert+\vert\alpha\vert},$$ where $M^m_{\vert\beta\vert}(\cdot)$ is defined in (\[A2bis\]).

It is well known that symbols of $\Gamma_s^m$ can be smoothed into paradifferential symbols of $\Sigma_s^m$. In order to give a precise description of the difference between these two symbols (and of the operator associated to it), we split every $\sigma\in \Gamma_s^m$ into four components: $$\label{A6}
    \sigma(x,\xi)={\sigma_{lf}}(x,\xi)+{\sigma_I}(x,\xi)+{\sigma_{II}}(x,\xi)+{\sigma_R}(x,\xi),$$ with, for some $N\in{{\mathbb N}}$, $N\geq 4$, $$\begin{aligned}
    \label{A7}
    {\sigma_{lf}}(\cdot,\xi)&=&\psi(\xi)\sigma(\cdot,\xi),\\
    \label{A8}
    {\sigma_I}(\cdot,\xi)&=&\sum_{p\geq -1}\psi(2^{-p+N}D_x)\sigma(\cdot,\xi)
    (1-\psi(\xi))\varphi_p(\xi),\\
    \label{A9}
    {\sigma_{II}}(\cdot,\xi)&=&\sum_{p\geq -1}\varphi_p(D_x)\sigma(\cdot,\xi)
    (1-\psi(\xi))\psi(2^{-p+N}\xi),\\
    \label{A10}
    \sigma_R(\cdot,\xi)&=&\sum_{p\geq-1}\sum_{\vert p-q\vert\leq N}
    \varphi_q(D_x)\sigma(\cdot,\xi) (1-\psi(\xi))\varphi_p(\xi),\end{aligned}$$ where $\psi\in C_0^\infty({{\mathbb R}}^d)$ is a bump function satisfying (\[LP1\]).\
We also need sometimes a further decomposition of ${\sigma_R}$ as ${\sigma_R}=\sigma_{R,1}+\sigma_{R,2}$, with $$\label{decompreste}
    \sigma_{R,1}(\cdot,\xi):=(1-\psi(D_x)){\sigma_R}(\cdot,\xi)
    \quad\mbox{and}\quad
    \sigma_{R,2}(\cdot,\xi):=\psi(D_x){\sigma_R}(\cdot,\xi);$$ note that $\sigma_{R,2}$ is given by the *finite* sum $$\sigma_{R,2}(\cdot,\xi)=\sum_{p\leq N+1}\sum_{\vert p-q\vert\leq N}
    \psi(D_x)\varphi_q(D_x)\sigma(\cdot,\xi)(1-\psi(\xi))\varphi_p(\xi).$$

\[rempara\] [**i.**]{} The fact that the sum of the four terms given in (\[A7\])-(\[A10\]) equals $\sigma(x,\xi)$ follows directly from (\[LP3\]).\
[**ii.**]{} 