Let $H^s(\mathbb T)$, where $s\in\mathbb R$, be the space of $2\pi$-periodic functions (or distributions), $u(x)=\sum\_{k\in\mathbb Z}\hat u\_k\,\mathrm{e}^{ikx}$, 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 $s\in \big(\frac{1}{2},\frac{3}{2}\big)$. I wish to find out whether there exists a constant $c=c\_s$, such that
$$
\lvert u(x)-u(y)\rvert\le c \lvert u\rvert\_{H^s}\lvert x-y\rvert^{s-\frac{1}{2}},
$$
where 
$$
\lvert u \rvert\_{H^s}^2=\sum\_{k\in\mathbb Z}\lvert k\rvert^{2s}\lvert \hat u\_k\rvert^2.
$$


If $s=1$, then this is Morrey's inequality.


Any reference?


Note. I have asked this question in [math.stackexchange](https://math.stackexchange.com/questions/1821243/morreys-inequality-for-sobolev-spaces-of-fractional-order) without any luck.



Morrey's inequality for Sobolev spaces of fractional order
I have produced an answer, not the most elegant one. 


Let $x\ne y$, then we have that
$$
\lvert u(x)-u(y)\rvert = \left|\sum\_{k\in\mathbb Z}
\hat u\_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right|\le
\left|\sum\_{\lvert k\rvert \le |x-y|^{-1}}
\hat u\_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right|+\left|
\sum\_{\lvert k\rvert \ge |x-y|^{-1}}
\hat u\_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right|.
$$
We shall exploit the fact that 
$$
\big|\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big|\le
\min\big\{\lvert k\rvert\lvert x-y\rvert,2\big\}.
$$
For the first term we have two cases:


Case I. $s \le 1$,
$$
\left|\sum\_{\lvert k\rvert \le |x-y|^{-1}}
\hat u\_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right| \le
\sum\_{\lvert k\rvert \le |x-y|^{-1}}
\lvert \hat u\_k\rvert \lvert k\rvert \lvert x-y\rvert \\ =
\lvert x-y\rvert \sum\_{\lvert k\rvert \le |x-y|^{-1}}
\lvert \hat u\_k\rvert \lvert k\rvert^s \lvert k\rvert^{1-s} \\ \le 
\lvert x-y\rvert \,
\left(\sum\_{\lvert k\rvert \le |x-y|^{-1}}\lvert k\rvert^{2-2s}\right)^{1/2}
\left(\sum\_{\lvert k\rvert \le |x-y|^{-1}}
\lvert k\rvert^{2s}\lvert \hat u\_k\rvert^2 \right)^{1/2}\\
\lesssim \lvert x-y\rvert \, \|u\|\_{H^s} 
\left(\frac{2}{\lvert x-y\rvert^{3-2s}}\right)^{1/2} \\
=2^{1/2}\|u\|\_{H^s}\lvert x-y\rvert^{s-1/2}
$$
Case II. $1<s<3/2$. We have
$$
\left|\sum\_{\lvert k\rvert \le |x-y|^{-1}}
\hat u\_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right| \le
\sum\_{\lvert k\rvert \le |x-y|^{-1}}
\lvert \hat u\_k\rvert \lvert k\rvert \lvert x-y\rvert\\ =
\lvert x-y\rvert \sum\_{\lvert k\rvert \le |x-y|^{-1}}
\lvert \hat u\_k\rvert \lvert k\rvert^s \lvert k\rvert^{1-s} \\ \le 
\lvert x-y\rvert \,
\left(\sum\_{\lvert k\rvert \le |x-y|^{-1}}\lvert k\rvert^{2-2s}\right)^{1/2}
\left(\sum\_{\lvert k\rvert \le |x-y|^{-1}}
\lvert k\rvert^{2s}\lvert \hat u\_k\rvert^2 \right)^{1/2} \\
=\lvert x-y\rvert \, \|u\|\_{H^s} 
\left(\frac{4s}{(2s-1)\lvert x-y\rvert^{3-2s}}\right)^{1/2} \\
=\left(\frac{4s}{2s-1}\right)^{1/2}\|u\|\_{H^s}\lvert x-y\rvert^{s-1/2}
$$
For the second term we have
$$
\left|
\sum\_{\lvert k\rvert \ge |x-y|^{-1}}
\hat u\_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right|\le 
2\sum\_{\lvert k\rvert \ge |x-y|^{-1}}
\lvert \hat u\_k\rvert=2\sum\_{\lvert k\rvert \ge |x-y|^{-1}}
\lvert \hat u\_k\rvert \lvert k\rvert^s \lvert k\rvert^{-s}\\ \le 2\,
\left(\sum\_{\lvert k\rvert \ge |x-y|^{-1}}\frac{1}{\lvert k\rvert^{2s}}\right)^{1/2}
\left(\sum\_{\lvert k\rvert \ge |x-y|^{-1}}
\lvert \hat u\_k\rvert^2 \lvert k\rvert^{2s}
\right)^{1/2}\\ \le 2\cdot\left(\frac{2\lvert x-y\rvert^{2s-1}}{2s-1}\right)^{1/2}\|u\|\_{H^s} \\
=\frac{2^{3/2}}{(2s-1)^{1/2}}\cdot\lvert x-y\rvert^{s-1/2}\|u\|\_{H^s}
$$
Altogether, for every $s\in(1/2,3/2)$, there exists a $c\_s>0$, such that
$$
\lvert u(x)-u(y)\rvert\le c\_s\lvert x-y\rvert^{s-1/2}\|u\|\_{H^s},
$$
for all $u\in H^s(\mathbb T)$.


Note. We have used the following rather crude inequalities


a. For $s>0$,
$$
\sum\_{k=1}^n k^s\le n^{s+1}.
$$


b. For $s>1$, 
$$
\sum\_{k=n}^\infty \frac{1}{k^s}\le \frac{s}{(s-1)n^{s-1}}.
$$


c. For $0<s<1$
$$
\sum\_{k=1}^n \frac{1}{k^s}\le \frac{(2-s)n^{1-s}}{s-1}.
$$


