Let $n \in \mathbb{N}$ and $s>0$ non-integer. Consider the Sobolev space given by $$ H^s(\mathbb{R}^n)=\{ f \in \mathcal{S}'(\mathbb{R}^n)\; ; \; \|f\|_{H^s}<\infty\} $$ where $$ \|f\|_{H^s}=\|(1+|x|^2)^{s/2}\mathcal{F}f\|_{L^2(\mathbb{R}^n)}, f \in H^s(\mathbb{R}^n), $$ $\mathcal{F}$ is Fourier transform in $\mathbb{R}^n$ and $\mathcal{S}$ is the Schwartz space. Now, given $U\subset \mathbb{R}^n$ open we define $$ H^s(U)=\{ f \in L^2(U)\; ; \exists \; g \in H^s(\mathbb{R}^n), g(x)=f(x),\; \forall\; x \in U\} $$ with norm $$ \|f\|_{H^s(U)}= \inf_{g_{|U}=f \atop g \in H^s(\mathbb{R}^n)} \|g\|_{H^s(\mathbb{R}^n)}, f \in H^s(U). $$
I seen in a doctorate thesis (without proof) the following result:
Theorem. If $U \subset \mathbb{R}^n$ is a bounded domain and $s>0$ is such that $$ s>n\left( \frac{1}{2}-\frac{1}{q}\right), q\geq 2 $$ then the embedding $H^s(U) \hookrightarrow L^q(U)$ is compact.
Question. If $U=\frac{\mathbb{R}^n}{2\pi \mathbb{Z}}$ is the torus then the theorem above holds? Roughly speaking, the theorem above holds in periodic context?
The best result that I found is the Rellich's Theorem in page $73$ of $[1]$. But this theorem don't answer my question (see the theorem in $[1]$, please). Other references are welcome.
$[1]$ J. Roe. Elliptic Operators, Topology and Asymptotic Methods, Pitman Res.Notes in Math Series 179, Longman Scientific and Technical, Harlow, 1988.
We want to show that if $s>n\left( \frac{1}{2}-\frac{1}{q}\right)$, then $H^s(\mathbb{T}^n)\hookrightarrow L^q(\mathbb{T}^n)$.
The case $q=\infty$ is a consequence of the Sobolev embedding: Indeed in case $s>n/2$, we have $H^s\hookrightarrow C^\alpha$, for an appropriate $\alpha>0$, and this last space embeds compactly into $L^\infty$ by the Arzela-Ascoli theorem.
Therefore we only need to worry about $q<\infty$. By the same argument as above, and the fact that $L^\infty\hookrightarrow L^q$, we may as well assume that $s< n/2$.
Now notice that $$ |1-e^{inh}|\lesssim_{a} |nh|^a, \qquad h\in \mathbb{R}, $$ for all $0<a\leq 1$. Using this we can estimate the Fourier coefficients of $f(x)-f(x+h)=:g(x)$ as $$ |\hat{g}(n)|^2 \lesssim_a |nh|^{2a}|\hat{f}(n)|^2=|h|^{2a} |n|^{2(a-s)}|n|^{2s}|\hat{f}(n)|^2, \qquad n\neq 0. $$ In order to be able to sum this and get the $H^s$ norm of $f$ on the right, we need $a-s\leq 0$ or $a\leq s$ which can always be arranged. With such a choice we obtain $$ \| f-f(\cdot +h)\|_{L^2} \lesssim |h|^a \| f\|_{H^s}, $$ which gives equicontinuity in the $L^2$ norm of the unit ball of $H^s$, and so the embedding is compact in the case $q=2$.
For the general case we interpolate: By the Sobolev embedding, if $s>n(1/2-1/q)$ (and without loss of generality $s<n/2$), then there exists $r>q$ such that $H^s\hookrightarrow L^r$ and $$ \| f\|_{L^r}\lesssim \| f\|_{H^s}. $$ Therefore, for $b$ satisfying $1/q=b/2+(1-b)/r$ we have $$ \| g\|_{L^q}\lesssim \| g\|_{L^2}^b\| g\|_{L^r}^{1-b}\lesssim |h|^{ab}\| f\|_{H^s}, $$ using that $\| g\|_{H^s}\leq 2\| f\|_{H^s}$. This proves the compactness of the embedding.