Given a square-integrable function $f$, let $\hat{f}(n)$ denote its Fourier coefficients. It is well-known that differentiability of $f$ implies that the $\hat{f}(n)$ satisfy certain bounds, i.e. they decay rapidly. I am interested in a clean implication going in the other direction: Is there a bound on the $\hat{f}(n)$ that guarantees $f$ must be smooth of some order?
For example, to keep things concrete, what is a sufficient condition on $\hat{f}(n)$ that guarantees $f$ is Lipschitz continuous? Analogous assertions for higher-order smoothness (e.g. H\”older, Sobolev, etc.) are welcome as well.
I found this related question, but the accepted answer is not precise. I am looking for a precise implication.