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, doi: 10.1016/j.bulsci.2011.12.004, arxiv

**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.