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?


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.


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.

  • $\begingroup$ Sorry I should have been clearer in my question. So this is inspired from this question. I was wondering if it would be sufficient to only have the condition on the boundary for $U\subsetneq \mathbb{R}^n$ (i.e. without assuming $u\mid_\infty \leq 0$). I remember seeing that this holds for $n=2$ but even in that case I'm not sure how to prove it $\endgroup$ – Quoka Dec 5 '18 at 6:47
  • $\begingroup$ I see. I think you need more conditions on $U$: if we take $U=\mathbb R^n\setminus B_1(0)$, then an inverted fundamental solution $u(x)=1−|x|^{2−n}$ would be a counterexample. Maybe assuming $U$ is convex would work, or less generally, maybe assume that $U$ is a cone. In the half-space case, the proof is like this: for each $\epsilon>0$, observe that $u(x)-\epsilon x_n\to-\infty$ as $x_n\to\infty$, so by the maximum principle, $u\le \epsilon x_n$ for all $\epsilon>0$. $\endgroup$ – user254433 Dec 5 '18 at 7:04
  • $\begingroup$ I edited the question. Thanks for the counterexample. Do you see how to generalize the proof for domains in $\mathbb{R}^2$? $\endgroup$ – Quoka Dec 5 '18 at 18:49
  • $\begingroup$ An other counterexample, the function $\ln(x^2 + y^2)$ is harmonic and positive outside the unit disk, and it vanishes on the domain’s boundary. $\endgroup$ – Wang Dec 5 '18 at 18:59
  • $\begingroup$ @Wang True, but $ln(x^2+y^2)$ is not bounded $\endgroup$ – Quoka Dec 5 '18 at 19:22

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