TITLE: Fourier Coefficients and Hölder Continuity
QUESTION [13 upvotes]: 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 $\alpha>1/2$.

REPLY [11 votes]: 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
$2^{\nu-1}\le \vert\xi\vert\le 2^{1+\nu}$.