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Evans stated the strong maximum principle as follows: $U\subset\mathbb{R}^n$ a bounded and open set. If $u\in C^2(U)\cap C(\overline{U})$ is harmonic within $U$. Then,

  1. $\max_{\overline{U}}u=\max_{\partial U}u$
  2. if $U$ is in addition connected and there exists a point $x_0\in U$ such that $u(x_0)=\max_{\overline{U}}u$ then $u$ is constant within $U$.

I understand the proof of $2$. But why does this already imply 1?

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up vote 1 down vote accepted

The maximum is attained, because $\overline U$ is compact. And there are only two possibiliies: Either the maximum is attained at some interior point or it is not attained for any interior point (in which case it has to be on the boundary).

Now 2. says that $u = \mathrm{const}$ in the first case. So in particular, we must have $\max_{\overline U} u = \max_{\partial U} u$.

In the second case this equality is also true, trivially.

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just a small question: You said,"Now 2. says..". But what if $U$ is not connected then I can not apply 2 here. – user20869 Aug 10 '12 at 7:21
@hulik: You are right! But it's not a serious problem: If $U$ is not connected, we can apply 2. to each component of $U$, separately, and still get the result $\max_{\overline U} u = \max_{\partial U} u$ – Sam Aug 10 '12 at 15:03
what exactly do you mean by "to each component"? Thank you for your patience – user20869 Aug 10 '12 at 16:07
$U$ splits into a number of connected components, each of which is open. So $U = \bigcup_i U_i$, where $U_i$ are the components. And $\max_{\overline U_i} u = \max_{\partial U_i} u \le \max_{\partial U} u$ for each $i$ (the last inequality follows because $\partial U_i \subset \partial U$). – Sam Aug 11 '12 at 9:53
@hulik: (forgot to ping) – Sam Aug 11 '12 at 10:30

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