# Proof related to maximum value property of harmonic function (PDEs)

### Problem statement

Suppose $$\Omega \subseteq \mathbb{R}^{n}$$ is bounded, and path-connected , and $$u \in C^{2} (\Omega)\cap C(\partial \Omega)$$ satisfies $$\begin{cases} -\Delta u = 0 \quad &\text{in } \ \Omega,\\ u = g \quad &\text{on } \ \partial \Omega. \end{cases}$$ Prove that if $$g\in C(\partial \Omega)$$ with $$g(x) = \begin{cases} \ge 0 \quad &\text{for all } x \in \partial\Omega,\\ >0 \quad &\text{for some} \ x \in \partial \Omega. \end{cases},$$ then $$u(x) > 0 \quad \text{ for all } \ x\in \Omega.$$

### Attempt at solution

By definition the closure is $$\overline{\Omega} = \Omega \cup \partial\Omega$$, the domain is then bounded by $$\partial \Omega$$. The function $$u$$ is harmonic so $$u$$ satisfies the Mean-Value-Property. It follows that we can apply the weak/maximum principle.

By the weak maximum principle, $$\min\limits_{\overline{\Omega}} u = \min\limits_{\partial \Omega} u,$$ $$u$$ on the boundary is $$g$$, which is bounded below by $$0$$, therefore \begin{align} u(x) \ge \min\limits_{\partial \Omega} u = 0 &\implies u(x) \ge 0 \ \ \forall x \in \overline{\Omega} \\ &\implies u(x) > 0 \ \ \forall x \in \overline{\Omega} \backslash{\partial \Omega} \tag{1}\\ &\implies u(x) >0 \ \ \forall x \in \Omega \end{align}

I feel like I'm missing something in this proof, I particularly am not sure how to properly justify $$(1)$$ or if even the justification holds at all.

• In your problem, do you mean $u\in C^2(\Omega)\cap C(\bar{\Omega})$ instead of $u\in C^2(\Omega)\cap (\partial \Omega)$? And furthermore, in your problem you want to show that $u>0$ on $\partial\Omega$ but you show that $u>0$ in $\Omega$. The first one doesn't really make sense since $g\geq0$ is given, so I'll guess you mean the second one? Oct 21 '20 at 8:25
• Thank you I edited the second point since it was a typo. The first point no $u \in C^{2} (\Omega) \cap C(\partial \Omega)$ indeed. Oct 21 '20 at 13:44
• I think $C^2(\Omega)\cap C(\bar{\Omega})$ and $C^2(\Omega)\cap C(\partial\Omega)$ are the same. Oct 21 '20 at 14:47

Instead of the weak maximum principle you can apply the strong maximum principle. Assume that there exists a point $$x_0\in\Omega$$ such that $$u(x_0)\leq 0$$, then there exists a point $$\tilde{x}_0 \in \Omega$$ such that $$u(\tilde{x}_0) = \min_{\bar{\Omega}} u$$ since $$g\geq 0$$. The strong maximum priciple now states that $$u$$ is constant within $$\Omega$$, hence $$u(x)=u(\tilde{x}_0) \leq 0$$ for all $$x\in \Omega$$. However, this contradicts the existence of points $$y$$ in $$\partial\Omega$$ with $$g(y)>0$$. Thus, $$u>0$$ in $$\Omega$$.