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.

  • $\begingroup$ 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 $s=1$. $\endgroup$
    – Hannes
    Jan 14, 2017 at 14:15
  • $\begingroup$ 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. $\endgroup$
    – Hannes
    Jan 14, 2017 at 14:19
  • $\begingroup$ @Hannes You're right. Thanks for being thorough! $\endgroup$
    – anonymous
    Jan 14, 2017 at 14:26
  • $\begingroup$ On the Cyrillic spelling that had been in the title, see the meta discussion here: meta.mathoverflow.net/questions/3104/… $\endgroup$
    – Todd Trimble
    Jan 15, 2017 at 12:01

2 Answers 2


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

  • $\begingroup$ 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! $\endgroup$
    – anonymous
    Jan 14, 2017 at 15:54
  • $\begingroup$ (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) $\endgroup$
    – anonymous
    Jan 14, 2017 at 15:55
  • $\begingroup$ Whoops, should be Ch. 2.4.1. I'll correct it. $\endgroup$
    – Hannes
    Jan 14, 2017 at 16:02
  • $\begingroup$ @Hannes Are you aware of any other methods, not using interpolation? I am looking for a more direct proof that does not use the compactness of the classical Sobolev $W^1$ into $L^p$. $\endgroup$
    – Guy Fsone
    Jun 19 at 7:36
  • $\begingroup$ @GuyFsone there is the Frêchet-Kolmogorov(-Riesz) theorem which characterizes compact sets in $L^p$ and whose conditions have a direct connection to certain degrees of smoothness, so this could be a more direct way to go. (A generalization of this theorem is also what is behind the Amann paper the other answer by anonymous and possibly there are some helpful calculations in there.) $\endgroup$
    – Hannes
    Jun 21 at 13:29

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.


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