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

Posted by Phi. Isett in Uncategorized.
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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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Comments»

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).

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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