A Proof of Liouville's Theorem in complex analysis

A Proof of Liouville’s Theorem in complex analysis August 31, 2009

Posted by Phi. Isett in Uncategorized.
Tags: complex analysis, Liouville's theorem
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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.

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1. Akhil Mathew - November 16, 2009: But if I’m not mistaken the whole fact that $f$ extends to the point at infinity, which implicitly uses the Riemann removable singularity theorem, relies on the representation of $f$ as a power series (which uses the Cauchy formula).

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2. Phi. Isett - November 16, 2009: One can prove the removable singularity theorem directly from Cauchy’s integral formula, by showing the formula

$g(z) = \frac{1}{2 \pi i} \int \frac{f(\zeta)}{(\zeta - z)} d\zeta$

– where the above contour integral winds once around a neighborhood of $z$ in which $f$ remains holomorphic — defines a continuous (and holomorphic) function of z even into the possible singularity. This is the only proof I know, actually, and it doesn’t discuss analyticity, but just like analyticity of holomorphic functions, it follows almost immediately from the representation formula. What would be interesting would be to see a proof of these facts about holomorphic functions without relying on the explicit representation formula, maybe something to do with the notion of capacity in the case of removable singularities.

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