Let $S'(\mathbb{R}^n)$ be the set of all tempered distributions than the map $$ F: S'(\mathbb{R}^n)\rightarrow S'(\mathbb{R}^n)\\ FT(\varphi)\rightarrow T(\hat \varphi)\ $$ where $\hat \cdot$ denotes the fourier transform. Showthat $F$ is a continous, bijective operator.
It's clear that this holds for regular distributions, but how do I show this for distrubtions that are not regular ?
Would appreciate any help/hints