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?

  • $\begingroup$ Are you assuming $u$ is continuous on $\bar \Omega$ or merely on $\Omega$? $\endgroup$ – Anthony Carapetis Nov 2 '17 at 6:35
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    $\begingroup$ You also need some assumption to rule out $n=1,\;u(x) = -|x|.$ I think just requiring $n \ge 2$ is enough, without requiring $u$ to have continuous extension to $\bar \Omega.$ $\endgroup$ – Anthony Carapetis Nov 2 '17 at 7:26
  • $\begingroup$ @AnthonyCarapetis I’m assuming continuity just on $\Omega$. $\endgroup$ – Flowsnake Nov 2 '17 at 14:08

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).


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