# Does the validity of Sobolev inequality on a domain imply it's a Sobolev extension domain?

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?

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.

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