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Assume that $\Omega$ is a bounded connected domain and $\partial \Omega \in C^{\infty}$. Denote $\Gamma_1,\Gamma_2,\cdots,\Gamma_n$ are $n$ connected components of $\partial\Omega$. This notation leads to $\partial \Omega=\cup^n_{i=1}\Gamma_i$. Consider the following problem. \begin{cases} \Delta \phi&=0\\ \phi|_{\partial \Omega}&=g(x),\quad g\in C^{\infty} \end{cases} Denote $\mathcal{S}$ is subset of $\{1,2,\cdots,n\}$ and $\mathcal{S}^c$ is $\{1,2,\cdots,n\}\setminus \mathcal{S} $. If we know that $$ \min_{i\in \mathcal{S}}\inf_{x\in \Gamma_i}\phi(x)\ge \max_{i\in \mathcal{S}^c} \sup_{x\in \Gamma_i} \phi(x)+\delta, $$ where $\delta$ is a positive constant, how to prove that \begin{align*} \sum_{i\in \mathcal{S}}\int_{\Gamma_i}\nabla \phi\cdot \vec{n}d\sigma>0? \end{align*} where $\vec{n}$ denotes the outer normal of $\partial \Omega$.

My effort: I meet this question when I read a paper. This paper say that it is a standard comparison principle exercise, but I still don't know how to solve this question. When $g(x)$ is a step function, we may consider Hopf Lemma or strong maximum principle. However, $g$ is a function. Any comments will be welcome. the equation (2.51) of page 2992 in this paper

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Let $\psi$ be harmonic in $\Omega$, with $\psi=\phi$ on $\cup _{i \in S} \Gamma_i$, $\psi=m$ on $\cup_{i \not \in S}\Gamma_i$, where $m=\max_{i \not \in S} \max_{\Gamma_i} \phi$. By comparison, $\psi \geq \phi$ and then $\nabla \psi \cdot n \leq \nabla \phi \cdot n$ on $\cup _{i \in S} \Gamma_i$. Then $$ 0=\int_{\partial \Omega}\nabla \psi\cdot n=\sum_{i \in S}\int_{\partial \Gamma_i} \nabla \psi\cdot n+\sum_{i \not \in S}\int_{\partial \Gamma_i} \nabla \psi\cdot n. $$ The function $\psi$ attains its minimum at any point of $\cup_{i \not \in S} \Gamma_i$, hence Hopf's lemma gives $\nabla \psi\cdot n<0$ and each integral on $\partial \Gamma_i, i \not \in S$ is negative. Then $$ \sum_{i \in S}\int_{\partial \Gamma_i} \nabla \phi\cdot n \geq \sum_{i \in S}\int_{\partial \Gamma_i} \nabla \psi\cdot n >0. $$

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