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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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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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  • $\begingroup$ just a small question: You said,"Now 2. says..". But what if $U$ is not connected then I can not apply 2 here. $\endgroup$ – user20869 Aug 10 '12 at 7:21
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    $\begingroup$ @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 $\max_{\overline U} u = \max_{\partial U} u$ $\endgroup$ – Sam Aug 10 '12 at 15:03
  • $\begingroup$ what exactly do you mean by "to each component"? Thank you for your patience $\endgroup$ – user20869 Aug 10 '12 at 16:07
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    $\begingroup$ $U$ 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 $\partial U_i \subset \partial U$). $\endgroup$ – Sam Aug 11 '12 at 9:53
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    $\begingroup$ @hulik: Remember that $U_i$ is a component of $U$, and not just some arbitrary open subset. We certainly have $\partial U_i\subset \overline U_i \subset \overline U$. And we also must have $\partial U_i \cap U = \emptyset$, because $U_i$ is a component of $U$, i.e. it is a maximal connected subset of $U$ (if this intersection were not empty, we could extend $U_i$ by adding some open ball around a point in $\partial U_i \cap U$)! So $\partial U_i \subset \overline U \cap U^c = \overline U \setminus U = \partial U$. $\endgroup$ – Sam Aug 11 '12 at 11:58
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I tried to prove the Part 1 as follows. Please check if it is valid.

We prove Part 1 by contradiction. Suppose the conclusion is not true, then $\max_{\overline{U}} u\neq \max_{\partial U} u.$ But since $\partial U\subset \overline{U},$ we have $\max_{\overline{U}} u>\max_{\partial U} u.$ Because $u\in C(\overline{U}),$ with $U$ bounded, it follows that $\overline{U}$ is compact in $\mathbb{R}^n,$ and so there exists $x_0\in \overline{U}$ such that $u(x_0)=\max_{\overline{U}}u.$ Since $\max_{\overline{U}} u>\max_{\partial U} u,$ we infer that $x_0\in U.$ In the case that $U$ is connected, by Part 2, we have $u\Big|_{\overline{U}}=u(x_0),$ which implies that $u\big|_{\partial{U}} =u(x_0),$ and so $\max_{\overline{U}} u=u(x_0)>\max_{\partial U} u=u(x_0),$ which is absurd. In the case that $U$ is disconnected, by the openness of $U,$ there is a non-empty connected component $W,$ such that $x_0$ is in the interior $\text{int }(W)$ of $W.$ Application of Part 2 to $u$ on $W,$ we get $u\Big|_{\overline{W}}=u(x_0),$ which leads to $u\Big|_{\partial W}=u(x_0),$ and so $\max_{\partial W} u=u(x_0).$ Since $\mathbb{R}^n$ is locally connected, we get that $\partial W\subset\partial U,$ and hence $\max_{\partial W}\leq \max_{\partial U}u,$ which implies that $u(x_0)=\max_{\overline{U}} u>\max_{\partial U}u\geq \max_{\partial W}u=u(x_0),$ also a contradiction. Therefore, we have proved that $\max_{\overline{U}} u=\max_{\partial U} u.$

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