A Proof of Liouville’s Theorem in complex analysis August 31, 2009Posted by Phi. Isett in Uncategorized.
Tags: complex analysis, Liouville's theorem
Liouville’s theorem in complex analysis says that the only bounded holomorphic functions are constant. This is my favorite
Pretend you had a non-constant, bounded holomorphic function on all of . Since is bounded at , Riemann’s theorem on removable singularities implies that extends to a holomorphic (and hence continuous) function on the Riemann sphere , which is compact. If 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 . But this cannot be the case, because has no nonempty subsets which are both open and compact.
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.