True. The Fourier transform of a <compactly supported distribution> is a smooth function of at most polynomial growth. A sufficiently negative Sobolev weight makes its square integrable, so every $u\in\mathcal E'(\mathbb R^n)$ belongs to $H^t(\mathbb R^n)$ for some $t$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 327 / 2 / b / vii / Solution

 Solution (source code)