# A maximum principle for bounded functions in unbounded domain

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.

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.
• 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 – Quoka Dec 5 '18 at 6:47
• 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 $\epsilon>0$. – user254433 Dec 5 '18 at 7:04
• I edited the question. Thanks for the counterexample. Do you see how to generalize the proof for domains in $\mathbb{R}^2$? – Quoka Dec 5 '18 at 18:49
• 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. – Wang Dec 5 '18 at 18:59
• @Wang True, but $ln(x^2+y^2)$ is not bounded – Quoka Dec 5 '18 at 19:22