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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.

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  • $\begingroup$ 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? $\endgroup$
    – SC2020
    Oct 21 '20 at 8:25
  • $\begingroup$ 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. $\endgroup$
    – hexaquark
    Oct 21 '20 at 13:44
  • $\begingroup$ I think $C^2(\Omega)\cap C(\bar{\Omega})$ and $C^2(\Omega)\cap C(\partial\Omega)$ are the same. $\endgroup$
    – SC2020
    Oct 21 '20 at 14:47
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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 $\Omega$.

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