0
$\begingroup$

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 $\Bbb R^1$.


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?

$\endgroup$
2
$\begingroup$

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 $\max_{\partial \bar U}u\ge \max_{\partial \bar W}u=M$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.