Let $\Omega\subseteq \mathbb{R}^n$ be a bounded domain with smooth boundary. Fix $1< p < n$ and let $p^\ast$ denote the sobolev conjugate of $p$.
Now, fix a small number $r>0$ and suppose $u\in W_0^{1,p}(\Omega)$. By the Sobolev inequality, we also know that $u\in L^{p^\ast}(\Omega)$. Furthermore, extending $u$ to be $0$ outside of $\Omega$ I may suppose that $u\in W_0^{1,p}(\mathbb{R}^n)\cap L^{p^\ast}(\mathbb{R}^n)$.
Consider now $$ \sup_{y\in \mathbb{R}^n}\int_{B(y,r)}\lvert u\rvert^{p^\ast} $$ which we know to be finite since $u\in L^{p^\ast}(\mathbb{R}^n)$. First, I was able to show that the supremum is achieved at some fixed point $y\in\mathbb{R}^n$. I can show this by taking a sequence $(y_m)$, considering a converging sub-sequence and then applying the dominated convergence theorem.
Is it possible to show that the supremum is acheived at some point $y\in \Omega$?
It seems intuitive that this should be the case but I am unsure how to prove it.
For $r$ fixed, here is a counterexample. Consider any $u$ with support $\Omega = \{ 1\le |x|\le 2\}$ and set $r=2$. Then the maximiser is at $x=0\notin\Omega$, as any deviation means you miss part of the support. It would seem then that you need to tune $r$ based on $\Omega$ for a positive result.