yourMATHsolver: Liouville's theorem (complex analysis)

In complex analysis, Liouville's theorem, named after Joseph Liouville, states that every bounded entire function must be constant. That is, every holomorphic function f for which there exists a positive number M such that |f(z)| ≤ M for all z in C is constant.

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.

Proof

The theorem follows from the fact that holomorphic functions are analytic. Since f is entire, it can be represented by its Taylor series about 0

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

where (by Cauchy's integral formula)

$a_k = \frac{f^{(k)}(0)}{k!} = {1 \over 2 \pi i} \oint_{C_r} \frac{f( \zeta )}{\zeta^{k+1}}\,d\zeta$

and C_r is the circle about 0 of radius r > 0. We can estimate directly

$| a_k | \le \frac{1}{2 \pi} \oint_{C_r} \frac{ | f ( \zeta ) | }{ | \zeta |^{k+1} } \, |d\zeta| \le \frac{1}{2 \pi} \oint_{C_r} \frac{ M }{ r^{k+1} } \, |d\zeta| = \frac{M}{2 \pi r^{k+1}} \oint_{C_r} |d\zeta| = \frac{M}{2 \pi r^{k+1}} 2 \pi r = \frac{M}{r^k},$

where in the second inequality we have invoked the assumption that |f(z)| ≤ M for all z and the fact that |z|=r on the circle C_r. But the choice of r in the above is an arbitrary positive number. Therefore, letting r tend to infinity (we let r tend to infinity since f is analytic on the entire plane) gives a_k = 0 for all k ≥ 1. Thus f(z) = a₀ and this proves the theorem.

Pages

Sunday, June 24, 2012

Liouville's theorem (complex analysis)

No comments:

Post a Comment