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 $\alpha>1/2$.
Best Answer
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 $2^{\nu-1}\le \vert\xi\vert\le 2^{1+\nu}$.