\in W^{k,\infty}(\{\vert\xi\vert\leq 1\}; H^{s_0}({{\mathbb R}}^d))$.

It is quite natural to introduce the seminorms $N_{k,s}^m(\cdot)$ and $M^m_{k,l}(\cdot)$ defined for all $k,l\in{{\mathbb N}}$, $s\in{{\mathbb R}}$ and $m\in {{\mathbb R}}$ as $$\label{A1}
    N_{k,s}^m(\sigma):=\sup_{\vert\beta\vert\leq k}
    \sup_{\vert\xi\vert\geq 1/4}
    \langle\xi\rangle^{\vert\beta\vert-m}
    \left\vert \partial_\xi^\beta\sigma(\cdot,\xi)\right\vert_{H^s}$$ and $$\label{A2bis}
    M^m_{k,l}(\sigma):=\sup_{\vert \beta \vert\leq k}
    \sup_{\vert\xi\vert\geq 1/4} {\langle\xi\rangle}^{\vert\beta\vert-m}
    \left\vert\partial_\xi^\beta 
    \sigma(\cdot,\xi)\right\vert_{W^{l,\infty}}.$$ To get information on the low frequencies, we also define $$\label{A2}
    n_{k,s}(\sigma):=
    \sup_{\vert\beta\vert\leq k,\vert\xi\vert\leq 1}
    \left\vert \partial_\xi^\beta \sigma(\cdot,\xi)\right\vert_{H^s}
    \mbox{ and }
    m_k(\sigma):=\sup_{\vert\beta\vert\leq k,\vert\xi\vert\leq 1}
    \vert\partial_\xi^\beta \sigma(\cdot,\xi)\vert_{\infty}.$$ Note that $M^m_{k,l}(\sigma)$ and $m_k(\sigma)$ still make sense when $\sigma$ is the symbol of a Fourier multiplier (i.e. if $\sigma(x,\xi)=\sigma(\xi)$). When $l=0$, we simply write $M^m_k(\sigma)$ instead of $M^m_{k,0}(\sigma)$.\
Finally, the notation $N_{k,s}^m(\nabla_x^l\sigma)$, $l\in{{\mathbb N}}$, stands for $\sup_{\vert\alpha\vert=l}N^m_{k,s}(\partial_x^\alpha\sigma)$; we use the same convention for the other seminorms defined above.

Associated to the class $\Gamma_s^m$ is the subclass of paradifferential symbols $\Sigma_s^m$ (in the sense of [@Bony; @Meyer], see also [@Meyer-Coifman3] and Appendix B of [@Metivier-Zumbrun]). In the definition below, the notation ${\mbox{\textnormal{Sp }}}$ is used to denote the spectrum of a function, that is, the support of its Fourier transform.

\[defiA2\] Let $m\in{{\mathbb R}}$ and $s_0>d/2$. A symbol $\sigma(x,\xi)$ belongs to the class $\Sigma_{s_0}^m$ if and only if

-   One has $\sigma\in\Gamma_{s_0}^m$,

-   There exists $\delta\in(0,1)$ such 