1
$\begingroup$

Let $\Omega$ be an open subset of $\mathbb{R}^{n}$ and let $1\leq p<n$. It is well--known that if $\Omega$ is an extension Domain for $W^{1,p}(\Omega)$, then the Sobolev inequality holds in that there exists a constant $C>0$ such that $\|u\|_{L^{\frac{np}{n-p}}(\Omega)}\leq C\|u\|_{W^{1,p}(\Omega)}$ for all $u\in W^{1,p}(\Omega)$. Assuming further $\Omega$ to be bounded, this implies that $u\in L^{q}(\Omega)$ for all $1\leq q \leq np/(n-p)$.

My question is about somewhat the reverse question, namely, given an open and bounded set $\Omega\subset\mathbb{R}^{n}$ and we assume the aforementioned Sobolev inequality to hold for all $1\leq p <n$ and all $u\in W^{1,p}(\Omega)$, is it then necessarily true that $\Omega$ is an extension domain?

$\endgroup$
1
$\begingroup$

No, the converse is not true. Let $\Omega$ be a slit disk in the plane, that is the unit disk minus the radius from $(0,0)$ to $(1,0)$. The Sobolev inequalities hold in this domain, as one can see by applying them in the top and bottom half-disks (which are Lipschitz domains).

However, $\Omega$ is not a Sobolev extension domain. For example, the function written in polar coordinates as $u(r,\theta)=r\theta$, where $0<\theta<2\pi$, is Lipschitz in $\Omega$ but does not admit even a $W^{1,1}$ extension to the plane. Indeed, any extension will be discontinuous on every line segment crossing the removed radius, violating the ACL property of Sobolev functions.

$\endgroup$
  • $\begingroup$ Thanks a lot for the comment - that is exactly what I was looking for! $\endgroup$ – jim1970 Jul 20 '16 at 19:17

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.