The problem I am stuck with is as follows:
Let $\Omega\subset\mathbb{R}^n$ be a domain and let $u\in C^1(\Omega)\cap C^2(\Omega-\Sigma)$ be a non-negative function such that $\Delta u=0$ in $\Omega-\Sigma$, where $\Delta$ is the Laplacian and $\Sigma=\{x\in\Omega:u(x)=0\}$. Show that either $u=0$ in $\Omega$ or $u>0$ in $\Omega$
My attempt is as follows:
$\Sigma$ is clearly closed as the pre-image of a closed set via a continuous function, so we wish to show that $\Sigma$ is open so that by connectedness of $\Omega$ it follows that either $\Sigma=\Omega $ or $\Sigma=\emptyset$. Pick any $x\in\Sigma$ and choose a ball $B_R(x)\subset\subset\Omega$, then if $B_R(x)\subset\Sigma$ we are done, so suppose that $\exists x_0\in B_R(x)$ such that $u(x_0)>0$. Now $x_0\in\Omega-\Sigma$ which is open so there is a ball $B_\rho(x_0)$ such that $\Delta u=0$ in $B_\rho(x_0)$.
And I'm stuck here. I have no clue how to get a contradiction and show that $u(x0)=0$, I tried using Harnack's inequality, but I can't have the closure of the ball around $x_0$ touching $\Sigma$ since then Harnack wouldn't hold anymore and applying Harnack otherwise doesn't seem to get me anywhere.
Best Answer
This is a consequence of the strong maximum principle (or Hopf lemma). Note that $\Omega \setminus \Sigma$ is a open set, choose a ball $B_r \subset \Omega \setminus \Sigma$ s.t., $y_0 \in \overline {B_r} \, \cap \Sigma \neq \phi$ (this is possible by choosing a point $x_0$ in $\Omega \setminus \Sigma$ which is closer to $\Sigma$ than $\partial \Omega$ and choose the ball with center $x_0$ and radius $r = \operatorname{dist}(x_0, \Sigma)$). Then by Hopf lemma the directional derivative of $u$ at $y_0$ along the external normal to $\Sigma$ is strictly negative. But $y_0 \in \Sigma$ means it is a point of global minima therefore $\nabla u(y_0) = 0$ which is a contradiction. Therefore, either $\Omega \setminus \Sigma = \phi$ or $\Sigma = \phi$ leading to the two alternatives proposed.