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.
Best Answer
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 $\Omega$.