When is the convolution with a tempered distribution again a tempered distribution? If $f$ is a Schwartz function on $\mathbb R^n$ and $g \in L^1(\mathbb R^n)$, then


*

*if $g$ is the Poisson kernel, is $f\ast g$ a Schwartz function?

*are there any known sufficient conditions on $g$ to guarantee that $f \ast g$ a Schwartz function?

 A: *

*Which Poisson kernel do you mean?  There is a Poisson kernel for many domains.

*A sufficient condition for $f*g\in\mathscr{S}$ is that $f\in\mathscr{S}$ and $g\in L^1$ with compact support.
Suppose $g(x)=0$ for $|x|\ge R$ and $\|g\|_{L^1}=G$. Then for any multi-indices $\alpha$ and  $\beta$,
$$
\begin{align}
\left|x^\alpha(f*g)^{(\beta)}(x)\right|
&=\left|x^\alpha\left(f^{(\beta)}*g(x)\right)\right|\\
&\le G\max\left(C_{\alpha,\beta}\left|\frac{x^\alpha}{(x-R)^\alpha}\right|,C_{0,\beta}\left|x^\alpha\right|\right)\\
&\le G\max(C_{\alpha,\beta},C_{0,\beta})(R+1)^{|\alpha|}
\end{align}
$$
Thus, $f*g\in\mathscr{S}$.
A: Let's check differentiability: If $f$ is $C^\infty$ and $(\partial^\alpha f)*g$ exists for every $\alpha$, then $f*g$ is also $C^\infty$.
Now let's check the decay:
$$ |x^\beta \partial^\alpha(f*g)(x)| = |x^\beta| |(\partial^\alpha f)*g(x)| \leq |x^\beta| \int |\partial^\alpha f(x-y)||g(y)|dy \leq |x^\beta|\sup_x |\partial^\beta f(x)| \|g\|_{L^1}.$$
This shows that $f*g$ is Schwartz for $f$ Schwartz and $g\in L^1$.
