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I know that a harmonic function $f\in C^2(\Omega)\cap C(\bar\Omega)$ which is constant on boundary is constant, if $\Omega\subset\mathbb{R}^N$ is a bounded and connected domain. Also, if $\Omega$ is not bounded, this property does no longer hold (e.g. $f:\mathbb{\bar H}^2=\{(x,y)\in\mathbb{R}^2\ \vert\ y\geq0\}\rightarrow\mathbb{R}$, $f(x,y)=y$).

I wonder if there are some unbounded domains for which this property remains true? For example, if $\Omega=\{(x,y)\in\mathbb{R}^2\ \vert\ |y|<1\}$, it looks that this may be true, provided we prove $f$ should be bounded.

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The strip and the half plane are conformally equivalent (and the equivalence maps most of the boundary continuously to most of the boundary), so there's no difference between the two for this problem.

Is there an unbounded $\Omega$ such that a harmonic function vanishing on the boundary must vanish? I don't know, but I seriously doubt it.

I'm not sure what you mean by "...provided we prove $f$ should be bounded". In case you didn't know, if $f$ is a bounded harmonic function in a half plane that vanishes on the boundary then $f=0$.

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  • $\begingroup$ Ok, what's the problem with that? $\endgroup$ – David C. Ullrich Nov 27 '17 at 16:55

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