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?