Consider the punctured ball, $\Omega = \{\textbf{x} \in \mathbb{R}^n : 0 < ||\textbf{x}|| \leq 1\}$. Suppose that $u : \Omega \rightarrow \mathbb{R}$ is continuous, bounded from above and harmonic in $\Omega$. I want to show that the maximum principle holds,
$$ \max\limits_{\bar{\Omega}}u = \max\limits_{||\textbf{x}||=1}u $$
Since $u$ is bounded and harmonic, would Liouville's Theorem apply here? Because then $u \equiv M$ in $\Omega$ and by continuity, $u \equiv M$ on $\partial \Omega$.
What is different about this problem compared to the one without the punctured domain?
Liouville's theorem is a statement about harmonic functions on all of $\mathbb{R}^n$. Your function is harmonic on a strict subset. Thus you cannot apply the theorem. Thank god, or else a lot of things would be very boring.
Without the puncture, your problem would be to show that $$ \max_{\overline{\Omega}} u=\max_{\partial D}u=\max_{S^{n-1}}u $$ since your boundary is now the outer circle bounding your disc, rather $S^{n-1}$ and the origin in your case (the boundary is the closure of a set minus its interior).