Question about compactly supported distribuions

Question

Let u be a distribution with compact support and let f be a Schwartz function: Is it true that the convolution of f with u is a Schwartz function?

Answer 1

Yes. A distribution with compact support has finite order: it's just a finite collection of compactly supported Radon measures that get integrated against various derivatives of test functions. So, if you fix a Schwartz function $f$ and feed the translate $\tau_x f$ into the distribution, the value you get decays faster than any power of $|x|$.