# Fractional-order Rellich–Kondrashov Theorem

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.

• I suggest to add that $p^* = p^*(n,s) = \frac{np}{n-sp}$. I am aware that the paper also just writes $p^*$ but at least in my world $p^*$ is usually the Sobolev conjugate corresponding to $s=1$. Jan 14 '17 at 14:15
• Also, adding to my previous comment: I think one should expect $W^{s,p}(\Omega) \hookrightarrow W^{s-\epsilon,q}(\Omega)$ compactly for $q \in [1,p^*(n,\epsilon))$ since you spend only $\epsilon$ of differentiability - assuming that was not what you asked for. Jan 14 '17 at 14:19
• @Hannes You're right. Thanks for being thorough! Jan 14 '17 at 14:26
• On the Cyrillic spelling that had been in the title, see the meta discussion here: meta.mathoverflow.net/questions/3104/… Jan 15 '17 at 12:01

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, 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 $q < p^*(n,\delta)$.

• Beautiful. I especially like how this can be decomposed into four parts, each with an obvious purpose and comprehensible on its own. To me, the key insight is Theorem 2 from section 1.16.4 that enables the third step. The second step from section 1.2.4 is one that is equally important, but so abstractly stated that I would have overlooked its reach. Thanks! Jan 14 '17 at 15:54
• (btw, did you really mean to refer to Ch. 2.8.1 in the beginning? I find a series of inclusions there but it seems the interpolation formula one needs is (8) in 2.4.2 on p.185) Jan 14 '17 at 15:55
• Whoops, should be Ch. 2.4.1. I'll correct it. Jan 14 '17 at 16:02

I've since come across the article

Amann, Herbert. Compact embeddings of vector-valued Sobolev and Besov spaces. Dedicated to the memory of Branko Najman. Glas. Mat. Ser. III 35(55) (2000), no. 1, 161--177. MR1783238

which contains the result

Theorem 5.1: Suppose $E_1 \subset\subset E_0$ If $s_1 > s_0$ and $s_1 - n/p_1 > s_0 - n/p_0$ then $$W^{s_1,p}(X,E_1) \subset\subset W^{s_0,p_0}(X,E_0)\text.$$

under the assumption that $X$ is a smoothly bounded open subset of $\mathbb R^n$ and $E_0$, $E_1$ are Banach spaces.