# 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 (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 without any luck.

The Sobolev space $H^s(\mathbb T)$ is $W^{s,2}(\mathbb T)$ and is also the Besov space $B^{s}_{2,2}(\mathbb T)$. On the other hand the set of Hölder continuous functions with index $\alpha\in (0,1)$ is the Besov space $B^{\alpha}_{\infty,\infty}(\mathbb T)$. For $s\in(1/2,3/2),$ you have the following continuous inclusion, $$B^{s}_{2,2}\subset B^{s-\frac12}_{\infty,\infty},$$ which is a classical result, whose proof is immediate using the Littlewood-Paley decomposition: if $(\phi_\nu)_{\nu\in \mathbb N}$ is such a decomposition (here $\phi_\nu(\xi)=\phi(\xi 2^{-\nu})$ so that $1=\sum_\nu \phi_\nu(\xi)$, with $\phi\in C^\infty_c)$, the norm of a function $u\in B^{s}_{2,2}(\mathbb T)$ is $$\bigl(\sum_{\nu\in \mathbb N}2^{2\nu s}\Vert{\phi_\nu(D)u}\Vert_{L^2}^2\bigr)^{1/2}$$ whereas the norm in $B^{s-\frac12}_{\infty,\infty}$ is $\sup_{\nu\in \mathbb N}2^{\nu (s-\frac12)}\Vert{\phi_\nu(D)u}\Vert_{L^\infty}$. We have with $\psi\in C^\infty_c$ equal to 1 on the support of $\phi$, $$2^{\nu (s-\frac12)}\Vert{\phi_\nu(D)u}\Vert_{L^\infty} =2^{\nu (s-\frac12)}\Vert{2^\nu\hat\psi(2^\nu \cdot)\ast \phi_\nu(D)u}\Vert_{L^\infty}\\ \lesssim 2^{\nu (s-\frac12)}\Vert2^\nu\hat\psi(2^\nu \cdot)\Vert_{L^2} \Vert{\phi_\nu(D)u}\Vert_{L^2}\lesssim 2^{\nu s}\Vert{\phi_\nu(D)u}\Vert_{L^2}$$ so that the following continuous injections hold $$B^{s}_{2,2}\subset B^{s}_{2,\infty}\subset B^{s-\frac12}_{\infty,\infty}.$$

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. $s \le 1$, $$\left|\sum_{\lvert k\rvert \le |x-y|^{-1}} \hat u_k\big(\mathrm{e}^{ikx}-\mathrm{e}^{iky}\big)\right| \le \sum_{\lvert k\rvert \le |x-y|^{-1}} \lvert \hat u_k\rvert \lvert k\rvert \lvert x-y\rvert = \lvert x-y\rvert \sum_{\lvert k\rvert \le |x-y|^{-1}} \lvert \hat u_k\rvert \lvert k\rvert^s \lvert k\rvert^{1-s} \\ \le \lvert x-y\rvert \, \left(\sum_{\lvert k\rvert \le |x-y|^{-1}}\lvert k\rvert^{2-2s}\right)^{1/2} \left(\sum_{\lvert k\rvert \le |x-y|^{-1}} \lvert k\rvert^{2s}\lvert \hat u_k\rvert^2 \right)^{1/2} =\lvert x-y\rvert \, \|u\|_{H^s} \left(\frac{2}{\lvert x-y\rvert^{3-2s}}\right)^{1/2} \\ =2^{1/2}\|u\|_{H^s}\lvert x-y\rvert^{s-1/2}$$ 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 $0<s<1$ $$\sum_{k=1}^n \frac{1}{k^s}\le \frac{(2-s)n^{1-s}}{s-1}.$$

This works for $s\ge 1$, by Sobolev embedding for fractional orders: if $u\in H^s=W^{s,2}$, then $u'\in L^p$ with $1/p=3/2-s$. Now your inequality follows by applying Hölder's inequality to $u(y)-u(x)=\int_x^y u'$.