The following is known:



> 
> Let $s \in (0,1)$ and $p \in [1,\infty)$ be such that $sp < n$. Let $q \in [1, p^\*\_{n,s})$ with $p^\*\_{n,s} = np/(n-sp)$, $\Omega \subset \mathbb R^n$ be a bounded extension domain for $W^{s,p}$ and $\mathscr F$ be a bounded subset of $L^p(\Omega)$. Suppose that
>  $$
> \sup\_{f \in \mathscr F} \int\_\Omega \int\_\Omega
> \frac {|f(x) - f(y)|^p} {|x-y|^{n+sp}}\,dx\,dy < \infty
> $$
>  Then $\mathscr F$ is pre-compact in $L^q(\Omega)$.
> 
> 
> 


Consequently, the embedding $W^{s,p}(\Omega) \to L^q(\Omega)$ is compact for $s \in (0,1]$ (where the case $s = 1$ is the classical theorem and proved differently). The theorem above can be found, e.g., as Corollary 7.2. in the summary article



> 
> Di Nezza, Eleonora; Palatucci, Giampiero; Valdinoci, Enrico. *Hitchhiker's guide to the fractional Sobolev spaces*. Bull. Sci. Math. 136 (2012), no. 5, 521–573. [MR2944369](http://www.ams.org/mathscinet-getitem?mr=2944369), doi: [10.1016/j.bulsci.2011.12.004](http://dx.doi.org/10.1016/j.bulsci.2011.12.004), [arxiv](https://arxiv.org/abs/1104.4345)
> 
> 
> 


**My question is now**: Is the following generalisation true?



> 
> Let $s \in (0,1]$, $\varepsilon > 0$ and $p \in [1, \infty)$ be such that $\varepsilon p < n$. Let $q \in [1,p^\*\_{\varepsilon,n})$, $\Omega \subset \mathbb R^n$ be a bounded extension domain for $W^{s,p}$. Then the embedding $W^{s,p}(\Omega) \to W^{s-\varepsilon,q}(\Omega)$ is compact.
> 
> 
> 


**Some thoughts**: For $p = q = 2$ and $s < 1$, and limited to the closure of $C^\infty\_0(\Omega)$, the argument should go something like this: Let $M$ be a bounded subset in $H^s$. We want to show that $M$ is pre-compact in $H^{s-\varepsilon}$. Write
$$
\|u\|\_{H^s\_0} \doteq \|(-\Delta)^{s/2}u\|\_{L^2}
= \|(-\Delta)^{\varepsilon/2}(-\Delta)^{(s-\varepsilon)/2}u\|\_{L^2}
\quad \text{and} \quad
\|u\|\_{H^{s-\varepsilon}\_0} \doteq \|(-\Delta)^{(s-\varepsilon)/2}u\|\_{L^2}\text.
$$
The theorem mentioned earlier showed that if $\sup\_{w \in K} \|w\|\_{H^{s-\varepsilon}}$ is finite, then $K$ is pre-compact in $H^0 = L^2$. But this is precisely our setting with $K = (-\Delta)^{(s-\varepsilon)/2}M$.


Hence I would expect the more general claim I made above to be true as well. I would have just expected this to be stated as a corollary somewhere and have not been able to find it. So this is either too obvious to even warrant a comment or not true.



Fractional-order Rellich–Kondrashov Theorem
The answer is **yes** if $\Omega$ admits a uniform extension operator for $W^{1,p}$ and $L^p$. I suspect that it is not written down in the Hitchhiker's guide because they seem to avoid interpolation. 


Suppose $0 \leq s \leq 1$ and $1 < p < \infty$. We know that $$W^{s,p}(R^d) = \bigl(L^p(R^d),W^{1,p}(R^d)\bigr)\_{s,p}$$ (see the [tome of Triebel](https://books.google.de/books/about/Interpolation_theory_function_spaces_dif.html?id=ZzvvAAAAMAAJ&redir_esc=y), Ch. 2.4.1, with $W^{s,p} \dot{=} B^s\_{p,p}$ there, cf. Remark 4 in Ch. 2.5.1). 


Now let $\Omega$ be an extension domain, simultaneously for $W^{1,p}$ and $L^p$ with the same extension operator. Then the above interpolation formula transfers to the function spaces on $\Omega$ via retraction/coretraction (Ch. 1.2.4 in Triebel), that is, $$W^{s,p}(\Omega) = \bigl(L^p(\Omega),W^{1,p}(\Omega)\bigr)\_{s,p}$$ for $0 \leq s \leq 1$ (in fact, the extension operator reveals itself to also work for the $W^{s,p}$ scale this way). 


Here comes the kicker: Thanks to Rellich-Kondrachov, $W^{1,p}(\Omega)$ embeds compactly into $L^p(\Omega)$ and this is very nicely compatible with interpolation, because it tells us that
$$W^{s,p}(\Omega) = \bigl(L^p(\Omega),W^{1,p}(\Omega)\bigr)\_{s,p} \hookrightarrow\hookrightarrow W^{s-\epsilon,p}(\Omega) = \bigl(L^p(\Omega),W^{1,p}(\Omega)\bigr)\_{s-\epsilon,p}$$ for every $0 < \epsilon \leq s$! (Ch. 1.16.4 in everyone's favorite book). Now you just have to apply the usual embeddings (Ch. 2.8.1) for $W^{s-\epsilon,p}(\Omega)$, so $$W^{s-\epsilon,p}(\Omega) \hookrightarrow W^{s-\delta,q}(\Omega)$$ for $\delta > \epsilon$ and $s - \epsilon - \frac{n}p \geq s-\delta - \frac{n}{q}$, so $q \leq p^\*(n,\delta-\epsilon)$. Since $\epsilon$ was arbitrary, you obtain $q < p^\*(n,\delta)$.


