# Harmonic functions in the half-plane

Denote by $$\mathbb{H}$$ the upper half-plane $$\mathbb{H} := \left\{ x \in \mathbb{R}^n : x_n > 0\right\}.$$ Suppose that $$u \in C^2(\mathbb{H}) \cap C(\bar{\mathbb{H}})$$ is a bounded harmonic function such that $$u \leq 0$$ on $$\partial\mathbb{H} = \{ x_n = 0\}$$. Is it possible to conclude that $$u \leq 0$$ in all of $$\mathbb{H}$$? I know this is the case for $$n = 2$$ but am unable to establish the general case $$n \geq 3$$.

That's not true in $$n=2$$. For example, $$u(x,y)=y$$ is harmonic in $$\mathbb{R}^2$$, and positive for $$y > 0$$, even though $$u(x,0) \le 0$$. And it's not true in $$\mathbb{R}^n$$ for the same reason.
By assuming that $$u$$ is bounded, then you get the Poisson integral representation of $$u$$ from its boundary function, and that will give you what you want.
• Yes but $u(x,y) = y$ is unbounded in $\mathbb{H}$. I think the boundedness assumption will be necessary somewhere. – user596383 Dec 4 '18 at 15:38
• Thanks but I'm a little confused now. How does this work with regards to my question? I'm assuming that $u$ is bounded. – user596383 Dec 4 '18 at 16:34
• @DisintegratingByParts Is there an approach that would also work for more general domains in $\mathbb{R}^n$? – Quoka Dec 4 '18 at 17:34