# 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

1. if $g$ is the Poisson kernel, is $f\ast g$ a Schwartz function?
2. are there any known sufficient conditions on $g$ to guarantee that $f \ast g$ a Schwartz function?
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1. Which Poisson kernel do you mean? There is a Poisson kernel for many domains.

2. 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, $f*g\in\mathscr{S}$.

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1.By poisson kernel, I mean$P(t,x) = {c_n}\frac{t}{{{{({t^2} + {{\left| x \right|}^2})}^{\frac{{n + 1}}{2}}}}}$ 2.actually, I know that if $g \in$ ${{\rm{\mathscr{S}'}}}$ with compact support, then $f*g$ is Schwartz. But I only want to know the conditions like Poisson kernel. – Hezudao Dec 20 '11 at 14:14
As I mentioned in another comment, $g$ must decay at least as fast as a Schwartz function. $g=P(t,x)$ does not decay fast enough to insure that $f*g$ is a Schwartz function; $f*g$ is smooth enough, but it and its derivatives do not decay fast enough. – robjohn Dec 21 '11 at 2:08
Can you give a proof of the necessity of the fast decay? – Hezudao Dec 21 '11 at 5:16
@Adterram: Suppose that $g(x)=\left(1+|x|^2\right)^{-(n+1)/2}$. If $f(x)\ge0$ has compact support in $|x|<1$ then $|f*g(x)|>\left(1+(|x|+1)^2\right)^{-(n+1)/2}\|f\|_{L^1}$. This does not decay fast enough to be a Schwartz function. – robjohn Dec 22 '11 at 1:06

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 $g\in L^1$.

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The last inequality seems to be unjustified. – Hezudao Dec 20 '11 at 8:49
$g$ must decay at least as fast as a Schwartz function. – robjohn Dec 20 '11 at 9:33
Oh right! What I wrote is just not enough... Sorry. – Dirk Dec 21 '11 at 8:59