Let $U \subsetneq \mathbb{R}^2$ be a domain.
Suppose that $u \in C^2(U) \cap C(\bar{U})$ is a bounded harmonic function such that $u \leq 0$ on $\partial U$.
If $U$ is bounded, then the maximum principle yields that $u\leq 0$ in all of $U$.
Is it possible to conclude that $u \leq 0$ in all of $U$ without the assumption that $U$ is bounded? Does anyone have an idea of how to proceed with this?
Thanks!
Update: If $U$ is such that $U^\complement$ contains an open ball, then using the fundamental solution in $\mathbb{R}^2$ and following strategy outlined by @user254433 in the comments, I was able to prove the statement.
Any ideas on how to proceed if $U^\complement$ does not contain an open ball? In case it's helpful: I know that if $U = \mathbb{R}^2\setminus\{p\}$ for some point $p$ then any bounded harmonic function on $U$ is constant.
Best Answer
One problem is that $\partial U$ is "smaller" in the unbounded case, so $u\le 0$ on $\partial U$ becomes less restrictive. For example, if $U=\mathbb R^n$, then $\partial U=\varnothing$, so the condition $u|_{\partial U}\le 0$ implies no restrictions on $u$. So although Liouville's theorem for bounded harmonic functions on $\mathbb R^n$ implies $u\equiv a$ is constant, this constant could be positive.
To make sense of $u|_{|x|=\infty}\le 0$, we could replace it with a limiting condition, like $\limsup_{|x|\to\infty} u(x)\le 0$. Once we do this, it becomes natural to apply the maximum principle on bounded approximations of $U$:
For each $R>0$, let $U_R=U\cap B_R$ be a large subdomain of $U$. Then for each small $\epsilon>0$, we can find a large radius $R=R(\epsilon)$ so that $u|_{\partial B_R}\le \epsilon$, which, since $\epsilon>0$, implies $u|_{\partial U_R}\le \epsilon$. By the maximum principle, $u|_{U_R}\le \epsilon$. Sending $\epsilon\to 0$ gives the desired conclusion.