4
$\begingroup$

I'm wondering if there are known results about the "regularity" (in some sense to be determined) of sub and super levelsets of Sobolev functions $u\in W^{1,p}(\mathbb{R}^d)$. More precisely:

Assume $u\in W^{1,p}(\mathbb{R}^d)$, fix a constant $M\in \mathbb{R}$, and let $$ E_M:=\{x\in \mathbb{R}^d,\hspace{1cm}u(x)>M\}\subset \mathbb{R}^d $$ be the corresponding super-levelset of $u$. What can we say about the regularity of $E_M$? I actually have no precise idea of what "regularity" should mean here, but I am not talking about regularity of the boundary $\partial E_M$ as in the usual context of PDE's...

I think a natural question to ask is for example: how can one guarantee that $E_M$ is open (for a suitable representative of $u$), or at least that $E_M$ has nonempty interior? For example if $p$ is large enough we know by Sobolev inequality that $u$ is continuous so $E_M$ is open. When the gradient regularity deteriorates, how "ugly" could $E_M$ be? While still being measurable $E_M$ could be quite pathological, for example a Smith-Volterra-Cantor set with positive measure but empty interior (well, maybe not, this is actually my question!)

Also, what if we replace $u\in W^{1,p}$ by a weaker difference-quotient-type condition $$ u\in L^p\quad\text{and}\quad \sup\limits_{0\neq h\in R^d}\frac{|\tau_h u-u|_{L^p(\mathbb{R}^d)}}{|h|^{\alpha}}<\infty \text{ for some }\alpha\in(0,1), $$ where $\tau_h u(.)=u(.-h)$ denotes the usual shift???

Thank you in advance for your reference suggestions, input, and comments.

$\endgroup$

1 Answer 1

3
$\begingroup$

One positive answer is that this set is $p$-quasi-open, see some resource about capacity theory, e.g., here: https://math.stackexchange.com/questions/48776/capacity-theory-beginner-resources.

$\endgroup$
1
  • $\begingroup$ thank you Gerw, I will check that out. I was hoping to avoid this kind of things but I convinced myself that there is no easy way out... $\endgroup$ Nov 14, 2013 at 9:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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