# Bounded harmonic function is constant

Thanks a lot!

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-1 No effort shown in this question at all. – user38268 Jan 28 '12 at 12:40

If $u$ is a harmonic function then there exists a conjugate function $v$ and an analytic function $f=u+iv$. Thus $\exp(f)$ is bounded, applying the Liouville's Theorem shows that $\exp(f)$ is constant.It's easy to prove that $f$ is constant, as well as $u$.

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E. Nelson, "A Proof of Liouville's Theorem", Proc. Amer. Math. Soc. 12 (1961) 995

9 lines long. Not the shortest paper ever, but maximizes importance/length ...

Edward Nelson's paper is freely and legally available here:

$\bullet\$ pdf file,

$\bullet\$ html page.

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Dear GEdgar: I hope you don't mind my edit. (+1) – Pierre-Yves Gaillard Jan 28 '12 at 14:24
Very nice and short proof. – Beni Bogosel Jan 28 '12 at 14:54