Fourier Coefficients and Hölder Continuity Suppose we are given the Fourier coefficients of an $L^2$ function on the circle.  Are there necessary and sufficient conditions on the coefficients that allow us to determine that $f$ is Hölder continuous of order $\alpha$? 
Note that the necessary condition $|\hat{f}(n)| \leq C_f|n|^{-\alpha}$ is not sufficient. For example if $\hat{f}(n)=|n|^{-2/3}$ for all $n$ then $f$ is an $L^2$ function whose Fourier series does not converge absolutely.  Therefore $f$ cannot be Holder continuous of order $\alpha>1/2$. 
 A: I think that is a hard question, even in the case of just showing $f$ is continuous - there is a recent book that mentions this in detail by Stein, E.M. and Shakarchi, R. (Fourier Analysis: An Introduction) which is one place to understand the subtleties of this issue.  
One thing I will mention is that the Sobolev embedding theorem implies sufficient conditions for Holder continuity.  If, for example, $n^2 |\hat{f}(n)|^2$ is summable ($f \in H^1$), then $f$ is $C^{0,\alpha}$ for $\alpha<\frac{1}{2}$.  More generally, you can find conditions based on the following idea:
$|f|_\alpha \leq \sum_n |\hat{f}(n)| |n|^\alpha = \sum_n |n|^{\alpha+\frac{1}{2}+\epsilon}|\hat{f}(n)| \frac{1}{|n|^{\frac{1}{2}+\epsilon}} \leq  \sum_n |n|^{2\alpha+1+2\epsilon}|\hat{f}(n)|^2 \sum_n \frac{1}{|n|^{1+2\epsilon}}$
Therefore if $|n|^{2\alpha+1}|\hat{f}(n)|^2$ is summable then $f$ is Holder continuous of any order strictly less than $\alpha$.
A: I suspect, as Daniel Spector said, that to find a characterization of Holder continuity in terms of Fourier coefficients is very hard. A nice necessary condition is a theorem by Bernstein asserting that if $f$ is $\alpha$-Holder for $\alpha>\frac{1}{2}$ then its Fourier coefficients are in $\ell^1(\mathbb{Z})$ (see Katznelson "An introduction to harmonic analysis" for a proof).
A: There is an excellent characterization of Hölder spaces via the Fourier transform, using Besov spaces. Let $\alpha\in (0,1)$: a function $u$
defined on $\mathbb R^n$ belongs to $L^\infty\cap C^\alpha$ if and only if it belongs to $B^\alpha_{\infty,\infty}$, i.e.
$$
\sup_{\nu\in \mathbb N}2^{\nu\alpha}\Vert\phi_\nu(D_x) u\Vert_{L^\infty}<+\infty,\quad\text{i.e.
the sequence} (2^{\nu\alpha}\Vert\phi_\nu(D_x) u\Vert_{L^\infty})_{\nu\in \mathbb N}
\in \ell^\infty.
$$
Here $\phi_\nu$ stands for a Littlewood-Paley decomposition:
$$
1=\sum_{\nu\in \mathbb N}\phi_\nu(\xi),
$$
$\phi_0$ is compactly supported and for $\nu\ge 1$,
$\phi_\nu(\xi)=\phi(2^{-\nu}\xi)$ where $\phi$ is supported in the ring 
$1/2\le \vert\eta\vert\le 2$ so that $\phi_\nu$ is supported in the ring
$2^{\nu-1}\le \vert\xi\vert\le 2^{1+\nu}$.
