proof of Liouville’s theorem

Let $f:ℂ\to ℂ$ be a bounded, entire function. Then by Taylor’s theorem,

$$f(z)=\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n\text{ where }{c}_n=\frac{1}{2\pi i}{\int }_{{\mathrm{\Gamma }}_r}\frac{f(w)}{w^{n+1}}𝑑w$$

where ${\mathrm{\Gamma }}_r$ is the circle of radius $r$ about $0$, for $r>0$. Then $c_n$ can be estimated as

$$|c_n|\le \frac{1}{2\pi }\mathrm{length}({\mathrm{\Gamma }}_r)\sup \left\{\left|\frac{f(w)}{w^{n+1}}\right|:w\in {\mathrm{\Gamma }}_r\right\}=\frac{1}{2\pi }\mathrm{\hspace{0.17em}2}\pi r\frac{M_r}{r^{n+1}}=\frac{M_r}{r^n}$$

where $M_r=\sup \{|f(w)|:w\in {\mathrm{\Gamma }}_r\}$.

But $f$ is bounded, so there is $M$ such that $M_r\le M$ for all $r$. Then $|c_n|\le \frac{M}{r^n}$ for all $n$ and all $r>0$. But since $r$ is arbitrary, this gives $c_n=0$ whenever $n>0$. So $f(z)=c_0$ for all $z$, so $f$ is constant.$\mathrm{\square }$

Title	proof of Liouville’s theorem
Canonical name	ProofOfLiouvillesTheorem
Date of creation	2013-03-22 12:54:15
Last modified on	2013-03-22 12:54:15
Owner	Evandar (27)
Last modified by	Evandar (27)
Numerical id	5
Author	Evandar (27)
Entry type	Proof
Classification	msc 30D20