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## A Proof of Liouville’s Theorem in complex analysisAugust 31, 2009

Posted by Phi. Isett in Uncategorized.
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2 comments

Liouville’s theorem in complex analysis says that the only bounded holomorphic functions are constant.  This is my favorite

Proof:

Pretend you had a non-constant, bounded holomorphic function $f$ on all of $\mathbb{C}$.  Since $f$ is bounded at $\infty$, Riemann’s theorem on removable singularities implies that $f$ extends to a holomorphic (and hence continuous) function on the Riemann sphere $\mathbb{C} \cup \{ \infty \}$, which is compact.  If $f$ were not constant, the open mapping theorem would apply to this extension, and the image of the Riemann sphere would be an open subset of $\mathbb{C}$.  But this cannot be the case, because $\mathbb{C}$ has no nonempty subsets which are both open and compact. $\Box$

There are some downsides to this proof.  It does not rely on the theory of Riemann surfaces (not a downside).  It does, however, rely on some relatively (though not truly) heavy machinery, and in order to be a correct proof, one isn’t allowed to use Liouville’s theorem to develop this machinery (but it is possible).  Liouville’s theorem follows rather immediately from Cauchy’s integral formulae, and I don’t personally know how to establish things like analyticity, Riemann’s theorem and the open mapping theorem without this tool (although I would be interested if anyone else does!).  I also don’t think it really generalizes very well to other PDE for which a Liouville theorem holds.