Chemistry - Liouville's theorem (complex analysis)

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

$f(z) = \sum_{k=0}^\infty a_k z^k$

by Taylor series about 0, which implies

$|a_k| = \left|{1 \over 2 \pi i} \oint_{C_r} {f(z)\over (z-0)^{k+1}}\,dz\right| = {1 \over 2 \pi} \left|\oint_{C_r} {f(z)\over z^{k+1}}\,dz\right|$

where C_r is the circle about 0 of radius r. By moving the absolute value inside of the integral, we find

$\left| \oint_{C_r} {f(z)\over z^{k+1}}\,dz\right| \le \oint_{C_r} {|f(z)|\over |z|^{k+1}}\,dz.$

Now we can use the assumption that |f(z)| ≤ M for all z (since f is given to be bounded), and the fact that |z|=r on the circle C_r. We get

$\oint_{C_r} {|f(z)|\over |z|^{k+1}}\,dz \le \oint_{C_r} {M\over r^{k+1}}\,dz.$

Then,

$|a_k| \le {1 \over 2\pi} {M \over r^{k+1}} {2\pi r} = {r M \over r^{k+1}} = {M \over r^k}.$

Let r now tend to infinity so the circle C_r gets ever larger. If k is greater than 0, M/r^k tends to zero and so a_k must be zero.

However, if k=0, r⁰ = 1 (r ≠ 0 as r tends to infinity), so a₀ is the only term left in the Taylor series, and we have our result.

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Categories: Complex analysis | Theorems

Liouville's theorem (complex analysis)

Proof

See also

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