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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 $n \geq 3$.

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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.

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  • $\begingroup$ Yes but $u(x,y) = y$ is unbounded in $\mathbb{H}$. I think the boundedness assumption will be necessary somewhere. $\endgroup$ – user596383 Dec 4 '18 at 15:38
  • $\begingroup$ @ImNotThereRightNow : Yes, you'll need further assumptions if you want the conclusion that you hoped for. I believe my answer is correct for the question that you asked. $\endgroup$ – DisintegratingByParts Dec 4 '18 at 16:30
  • $\begingroup$ Thanks but I'm a little confused now. How does this work with regards to my question? I'm assuming that $u$ is bounded. $\endgroup$ – user596383 Dec 4 '18 at 16:34
  • $\begingroup$ @ImNotThereRightNow_ : Ah, I missed your assumption of boundedness in the question. Boundedness will give a Poisson integral representation, which will give you what you want. $\endgroup$ – DisintegratingByParts Dec 4 '18 at 16:38
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    $\begingroup$ @DisintegratingByParts Is there an approach that would also work for more general domains in $\mathbb{R}^n$? $\endgroup$ – Quoka Dec 4 '18 at 17:34

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