Strong maximum principle - Open and relatively closed and the premise of connectedness

Question

Evans pdes p27 - Strong maximum principle

Let $u\in C^2(U)\cap C(\bar{U})$ be harmonic within $U$.

Suppose $u(x_0)=\text{max}_{\bar{U}}$ for $x_0\in U$.

Then since $u$ is harmonic, it satisfies the mean value property, and hence there is some ball in $U$ where $$M=u(x_0)=-\!\!\!\!\!\!\int_{B(x_0,r)} u dy \leq M$$

Where the equality only holds if $u(y)=M$ for each $y\in B(x_0,r)$. I want to understand if this is the logic that is wanted for the following sentence:

Then the set $\{x\in U: u(x) =M\},$ is both open and relatively closed.

It is open because this is satisfied on the union of balls around each point that is maximal.

It is relatively closed because $u$ is continuous, and we are taking the preimage of $\{M\}$, a closed set in $\Bbb R^1$.

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Secondary question. He says that if $U$ is connected, then the above set is equal to $U$, and hence the result follows. That is fine. But then he says that thus the result follows that $$\text{max}_{\bar{U}} u = \text{max}_{\partial U} u$$

Which if $U$ is connected I agree with. But that equality didn't have $U$ connected as a premise. Did Evans leave this part of the proof open?

Answer 1

For the first question, the set $V=\{x\in U:\, u(x)=M\}$ is open because for every $x_0\in V$ there is a ball $B(x_0,r)\subset V$, which is exactly the definition of open sets. Yes, $V$ is relatively closed because it is the inverse image of a closed set through a continuous function. For your second question. Let $x_0\in \bar U$ be such that $u(x_0)=\max_{\bar U}u$. You always have the inequality $\max_{\bar U}u\ge \max_{\partial \bar U}u$. If $x_0\in \partial U$, then $\max_{\bar U}u=\max_{\partial \bar U}u$. If $x_0\in U$, consider the connected component $W$ that contains $x_0$. By what he proved, $u=M$ on $W$ and so by continuity, $u=M$ on $\partial W\subset \partial U$. So $\max_{\partial \bar U}u\ge \max_{\partial \bar W}u=M$.