So for a periodic function $f$ (of period $1$, say), I know the Riemann-Lebesgue Lemma which states that if $f$ is $L^1$ then the Fourier coefficients $F(n)$ go to zero as $n$ goes to infinity. And as far as I know, the converse of this is not true. My question, then, is this:
Under what conditions on the Fourier coefficients $F(n)$ is the function $f$, defined pointwise as the Fourier series with $F(n)$ as coefficients,
- integrable,
- continuous, and
- differentiable?
On
I'm aware of one necessary and sufficient condition. It comes out of the Spectral Theorem for selfadjoint operators. The domain of $L=\frac{1}{i}\frac{d}{dx}$ on $L^{2}[-\pi,\pi]$ consists of all periodic absolutely continuous functions on $[-\pi,\pi]$ with $f' \in L^{2}[-\pi,\pi]$. $L$ is selfadjoint on this domain. From the Spectral Theorem: $$ \mathcal{D}(L) = \left\{ f \in L^{2}[-\pi,\pi] : \sum_{n}n^{2}|\hat{f}(n)|^{2} < \infty \right\} $$ Therefore, $f \in L^{2}[-\pi,\pi]$ is equal a.e. to an absolutely continuous periodic function $g$ on $[-\pi,\pi]$ with $g' \in L^{2}$ iff $\sum_n n^{2}|\hat{f}(n)|^{2} < \infty$.
It's much harder to say something like this for $L^{p}$ functions where $p \ne 2$.
Here's a partial answer, too long for a comment. I'm not sure about necessary and sufficient conditions for continuity or differentiability, but it's not too hard to find sufficient conditions.
Certainly, if $\sum |F(n)| < \infty$ then the series $\displaystyle \sum F(n) e^{iun}$ converges absolutely and uniformly, by the Weierstrass M-test, so $f$ is continuous in that case.
But this condition does not imply differentiability; indeed, by spacing the nonzero coefficients appropriately, with increasingly long sections of zeros (a so-called lacunary Fourier series), one can make $f$ continuous but nowhere differentiable. Stein and Shakarchi construct such a function in their book Fourier Analysis.
Here is a sufficient condition for $f$ to be continuously differentiable: Decay of Fourier Coefficients and Smoothness