Morrey's inequality for Sobolev spaces of fractional order Let $H^s(\mathbb T)$, where $s\in\mathbb R$,  be the space of $2\pi$-periodic functions, $u(x)=\sum_{k\in\mathbb Z}\hat u_k\,\mathrm{e}^{ikx}$, such that 
$$
\|u\|_{H^s}^2=\sum_{k\in\mathbb Z}(1+k^2)^{s}\lvert \hat u_k\rvert^2<\infty.
$$
Assume now that $s\in \big(\frac{1}{2},\frac{3}{2}\big)$. Is it true that there exist a constant $c=c_s$, such that
$$
\lvert u(x)-u(y)\rvert\le c \|u\|_{H^s}\lvert x-y\rvert^{s-\frac{1}{2}}?
$$
If $s=1$, then the above is a special case of the the celebrated Morrey's inequality.
Reference would be welcome.
 A: The same approach as in Decay of Fourier Coefficients implies Holder Continuity? works.  The starting point is the Cauchy-Schwarz inequality,
$$
|u(x)-u(y)|^2
=\left(\sum_{k\in\mathbb{Z}} (1+k^2)^{s/2}\hat u_k \ \frac{|e^{ikx}-e^{iky}|}{(1+k^2)^{s/2}}\right)^2 \\ \le 
 \sum_{k\in\mathbb{Z}} (1+k^2)^s|\hat u_k|^2 \ \sum_{k\in\mathbb{Z}} \frac{|e^{ikx}-e^{iky}|^2 }{(1+k^2)^s} \\= \|u\|_{H^s}^2 \sum_{k\in\mathbb{Z}} \frac{|e^{ikx}-e^{iky}|^2 }{(1+k^2)^s}
$$
It remains to estimate the last sum by a multiple of $|x-y|^{2s-1}$. To do this, split it into ranges $|k|\le M$ and $|k|>M$ where 
  $M=\pi/|x-y|$.
Low frequencies
Since $|e^{ikx}-e^{iky}|^2\le k^2|x-y|^2$, we get at most 
$$
|x-y|^2 \sum_{|k|\le M} \frac{k^2}{(1+k^2)^s}
\le C|x-y|^2 M^{3-2s} = C'|x-y|^{2s-1}
$$
High frequencies
Since $|e^{ikx}-e^{iky}|^2\le 4$, we get at most 
$$
\sum_{|k|> M} \frac{4}{(1+k^2)^s}
\le CM^{1-2s} = C'|x-y|^{2s-1}
$$
A: This should hold.
Lemma
Define
$$
P_j f (x) = \sum_{\sigma = \pm 1, k=2^j}^{2^{j+1}} \hat{f}(\sigma k) e^{2\pi i \sigma k x}.
$$
If $\lvert f \rvert \leq 1$, then $f \in C^\alpha(\mathbb{T})$ for $0 < \alpha < 1$ iff
$$
\sup_{j \in \mathbb{Z}} 2^{j\alpha} \| P_j f \|_\infty \leq A
$$
for some $A$, and the smallest such $A$ is comparable to the $\alpha$-Holder norm of $f$.
Let $\alpha = s - 1/2$. By Cauchy-Schwarz inequality, we have
$$
\| P_j u \|_\infty \lesssim \| u \|_{H^s} \left( \sum_{k=2^j}^{2^{j+1}} (1 +k^2)^{-s} \right)^{1/2} \lesssim  \| u \|_{H^s}2^{-j \alpha}.
$$
Thus if the lemma holds, the desired statement holds.
There is a proof of the lemma in the continuum case in Schlag and Muscalu's Classical and Multilinear Harmonic Analysis, Vol 1, section 8.3.1, page 205. Looking over the proof, it seems the circle case should also hold, and the proof may even simplify a bit. Admittedly, I have not delved into the details.
