Liouville's theorem (complex analysis)
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Liouville's theorem in complex analysis states that every bounded (i.e., there exists a real number M such that |f(z)| ≤ M for all z in C) entire function (a holomorphic function f(z) defined on the whole complex plane C) must be constant.
Liouville's theorem can be used to give an elegant short proof for the fundamental theorem of algebra.
The theorem is considerably improved by Picard's little theorem, which says that every entire function whose image omits at least two complex numbers must be constant.
In the language of Riemann surfaces, the theorem can be generalized as follows: if M is a parabolic Riemann surface (such as the complex plane C) and N is a hyperbolic one (such as an open disk), then every holomorphic function f : M → N must be constant.
Proof
Given f, we have
- <math>f(z) = \sum_{k=0}^\infty a_k z^k<math>
by Taylor series about 0, which we also have
- <math>|a_k| = \left|{1 \over 2 \pi i} \oint_{C_r} {f(z)\over (z-0)^{k+1}}\,dz\right| = {1 \over 2 \pi} \left|\oint_{C_r} {f(z)\over z^{k+1}}\,dz\right|<math>
where Cr is the circle about 0 of radius r. By the estimation lemma,
- <math>\left| \oint_{C_r} {f(z)\over z^{k+1}}\right| \le M_rL <math>
where Mr is the maximum value of f over Cr. If f is bounded over all C, with maximum X, we have for all r, Mr ≤ X from the boundedness condition, and so
- <math> |a_k| \le {1 \over 2\pi} {X \over r^{k+1}} {2\pi r} = {r X \over r^{k+1}} = {X \over r^k}<math>
Let r tend to infinity so Cr encloses all of C. So, for some fixed k, |ak| tends to 0 as r tends to infinity only if k is greater than 0, for if k were 0, |a_0| = X, it is the only term in the Taylor series then, and we have our result.
See also Joseph Liouville.
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