# Real Analysis – How to show that a temperate distribution is a function of schwarz?

Given a temperate distribution $$T$$, to show that it is a function of Schwartz, is it enough to prove for all $$f$$ Schwartz, $$T (f) = int g f$$ for some people $$g$$ Schwartz? Now if $$T$$ is a temperate distribution such as $$(1+ | x | ^ 2) ^ pT (f) in H ^ {q} ( mathbb R ^ n)$$ for all $$p, q$$. Is it true that $$T$$ is Schwartz?

To directly produce a Schwartz function corresponding to this one seems to be difficult because there is no obvious candidate in my opinion. Rather, I propose to show that it is easier to show that the usual definition of a Schwartz function applies to derivatives in the distribution (we do not know if this is enough). Is it a better way?