Let $\Omega\subset\mathbb{R}^2$ be open and of class $C^1$. The Sobolev embedding theorem implies that if $u\in W^{k,2}(\Omega)$ and if $k\in\mathbb{N}: k\geq 2$, then $u$ is continuous.

Question. Does there exist a similar result for fractional Sobolev Spaces? For example, if $u\in W^{1+\theta,2}(\Omega)$ for some $\theta\in (0,1)$, then can we say that $u$ is continuous?

  • $\begingroup$ That looks like (a variant of) the Hardy–Littlewood–Sobolev inequality. Have you searched for this result in Stein's book? Another obvious place to look into is Hitchhiker's guide to the fractional Sobolev spaces. Or Samko's book on hypersingular integrals. $\endgroup$ Jan 13 at 12:00
  • $\begingroup$ @MateuszKwaśnicki Thank you for the references. I'll have a look. I found something close in Hitchhiker's guide to the fractional Sobolev Spaces (Thm. 8.2), but it's not exactly what I'm after. $\endgroup$
    – Nirav
    Jan 13 at 12:18
  • $\begingroup$ @Francesco, 14 edits of (mostly) old questions in a few hours, pushing newer questions off the front page. Please don't do that. $\endgroup$ Feb 19 at 11:59

If $\Omega$ is a "nice" domain in $\mathbb R^n$ and $u \in W^{1+\theta,p}(\Omega)$ with $\theta \in (0, 1)$, then both $u$ and the weak gradient $\nabla u$ are in $W^{\theta,p}(\Omega)$, and hence, by the Hardy–Littlewood-Sobolev inequality (Theorem 6.7 in the Hitchhiker's guide to the fractional Sobolev spaces), $u$ and $\nabla u$ are in $L^{p^\star}(\Omega)$, with $p^\star = n p / (n - \theta p)$. It follows that $u$ is in $W^{1,p^\star}$, and consequently, by the usual Sobolev inequality, $u$ is continuous whenever $p^\star > n$. This boils down to $p > n - \theta p$, or $\theta > \tfrac np - 1$.

In your case, $n = p = 2$, so one only needs $\theta > 0$.

  • $\begingroup$ Can when say something similar for when $u\in W^{\theta,2}(\Omega)$ with $\theta\in (0,1)$ and $n\geq 2$? $\endgroup$ Sep 27 at 11:49
  • $\begingroup$ @SahibaArora: I am afraid I do not know what do you mean "something similar". Hardy–Littlewood–Sobolev only gives $u \in L^{p^\star}$ with $p^\star=np/(n-\theta p)$, and this is sharp, I suppose. (Here $n \geqslant 2 > \theta p$ when $p = 2$. If, however, we allow $p$ to be greater than $n$, then, indeed, we get continuity of $u$ for $\theta$ large enough, namely, $\theta > n/p$.) $\endgroup$ Sep 27 at 13:32
  • $\begingroup$ Your answer considers $W^{\theta,2}(\Omega)$ for $\theta\in (1,2)$, I am interested in the case $\theta\in (0,1)$ with $n\geq 2$. $\endgroup$ Sep 27 at 19:22
  • $\begingroup$ @SahibaArora: Sure, but what would you like to know about $u$? There's no hope for continuity (as indicated in my previous comment), so what could possibly serve as "something similar"? $\endgroup$ Sep 27 at 19:28
  • $\begingroup$ Sorry, I missed the part that it is sharp. Thank you! $\endgroup$ Sep 27 at 19:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.