Doubt about the strong maximum principle on unbounded sets

Question

I have a purely theoretical doubt, because the references I read didn't make it so clear to me. When we are analyzing the maximum principle, two ways are suggested:

$\bullet$ (Weak maximum principle): Let $\Omega \subset \mathbb{R}^n$ be open and bounded. If $u \in C^2(\Omega) \cap C^0(\overline{\Omega})$, with $-\Delta u = 0$ in $\Omega$, then $$\max_{\overline {\Omega}} u = \max_{\partial \Omega} u.$$

$\bullet$ (Principle of Strong Maximum): Let $\Omega \subset \mathbb{R}^n$ be open and connected. If exists $x_0 \in \Omega$ such that $$u(x_0) = \displaystyle\max_{\overline{\Omega}} u,$$ then $u$ is contant in $\Omega$.

Doubt: Is it also necessary for the strong maximum principle that $\Omega$ be limited? If yes, why?

Answer 1

The proof I am aware of (Ian's comment) of Strong Maximum Principle does not use boundedness.

But anyway, suppose the Strong Maximum Principle only for bounded, open, connected sets. Let $\Omega$ be unbounded, open, and connected, with $u(x_0)=\max_{\overline \Omega}u$ for some $x_0\in\Omega$. Open connected in $\mathbb R^n$ is path connected. Let $\Omega_n$ be the open, connected set of all points in $\Omega$ which can be joined to $x_0$ by a path of length $< n$ . Now we can apply the "SMP for bounded sets" on each $\Omega_n$ to deduce $u|_{\Omega_n} \equiv u(x_0)$. As $\bigcup_{n=0}^\infty \Omega_n = \Omega$, this proves SMP for unbounded sets.

For your reference I copy the standard proof of SMP from my notes:

Theorem (Strong maximum principle). Let $\Omega$ be open and connected (possibly unbounded), and let $u$ be subharmonic in $\Omega$. If u attains a global maximum value in $\Omega$, then $u$ is constant in $\Omega .$

Proof. Define $M:=\max _{\Omega} u$, and $A:=u^{-1}\{M\}=\{x \in \Omega: u(x)=M\}$. By the assumption of the theorem, $A$ is non-empty. Since $u$ is continuous, it follows that $A$ is relatively closed in $\Omega$ (i.e., there exists a closed $F \subset \mathbb{R}^{n}$ such that $\left.A=F \cap \Omega\right)$. We now show that $A$ is also open. Indeed, let $x \in \Omega$ and let $r$ be sufficiently small that $\overline{B_{r}(x)} \subset \Omega$. By the mean value property for subharmonic functions, we have $$ 0=u(x)-M \leq \frac{1}{\left|B_{r}(x)\right|} \int_{B_{r}(x)}(u(y)-M) d y $$ Since $M=\max _{\Omega} u$, we have $u(y)-M \leq 0$ for all $y \in B_{r}(x)$. This implies that $$ 0=u(x)-M \leq \frac{1}{\left|B_{r}(x)\right|} \int_{B_{r}(x)}(u(y)-M) d y \leq 0 $$ Since $u$ is continuous, this can only happen if $u=M$ in $B_{r}(x)$. So that $B_{r}(x) \subset A$, which implies that $A$ is open. Since $\Omega$ is connected and $A \neq \emptyset$ is both open and closed, it follows that $A=\Omega$, and hence $u \equiv M$ in $\Omega$. This finishes the proof.