Recall, $a\in C^\infty(\mathbb R^n\times \mathbb R^n)$ is a symbol of order $m\in\mathbb R$ if for every $\alpha, \beta\in\mathbb N_0^n$ there is $C_{\alpha, \beta}>0$ such that $$|\partial^\beta_\xi\partial^\alpha_x a(x, \xi)|\leq C_{\alpha, \beta} (1+|\xi|)^{m-|\beta|},$$ for all $(x, \xi)\in \mathbb R^n\times \mathbb R^n$. Given such a symbol, one defines: $$k(x, z):=\mathscr{F}^{-1}_{\xi\mapsto z} [a(x, \xi)]=\int_{\mathbb R^n} e^{2\pi i\xi\cdot z} a(x, \xi)\ d\xi,$$ where $\mathscr{F}^{-1}$ stands for the inverse Fourier transform.
Notice, if $m<-n$ then $(1+|\cdot|)^m$ is integrable so that $k(x, z)$ is well defined.
For $m\geq -n$ I read we must understand the inverse Fourier transform in the sense of distributions. I guess that means $$<k(x, \cdot), \phi\rangle=\int_{\mathbb R^n} a(x, \xi) \check{\phi}(\xi)\ d\xi,$$ right? In this case, for every $x$ we get a distribution $k(x, \cdot)\in\mathscr{D}^\prime(\mathbb R^n)$ because of the rapid decrease of $\check{\phi}$. Is the map $$\mathbb R^n\longrightarrow \mathscr{D}^\prime(\mathbb R^n), x\longmapsto k(x, \cdot),$$ continuous?
Thanks.