Show that $\max_{\bar \Omega} |u|=\max_{\partial \Omega} |u|$.

Question

Let $\Omega$ denote the unbounded set $|x|>1$. Let $u \in C^2(\bar \Omega)$, $\Delta u=0$ in $\Omega$ and $\lim_{x\rightarrow \infty} u(x)=0$. Show that $\max_{\bar \Omega} |u|=\max_{\partial \Omega} |u|$.

So, applying maximum principle is an obvious choice, but it seems obvious with the principle, even without the limit condition. So I think I must have missed something. Any help is appreciated.

Answer 1

Any unbounded harmonic function on $\mathbb{R}^n$ shows why the limiting condition is necessary; for example, $u(x_1, ..., x_n) = x_1$. The issue here is that there is another part of the boundary, the point at infinity. This is easier to visualize if you think of $\mathbb{R}^2$ and the Riemann sphere, for example.

This should indicate how to proceed with the proof. Spoilers below.

So the maximum is attained on the boundary, and the value at infinity is zero after we extend the function to make sense there. If $f$ is identically zero, we're done. Else, the maximum doesn't occur at infinity.