# proof of Liouville’s theorem

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

 $f(z)=\sum_{n=0}^{\infty}c_{n}x^{n}\mbox{ where }c_{n}=\frac{1}{2\pi i}\int_{% \Gamma_{r}}\frac{f(w)}{w^{n+1}}\,dw$

where $\Gamma_{r}$ is the circle of radius $r$ about $0$, for $r>0$. Then $c_{n}$ can be estimated as

 $|c_{n}|\leq\frac{1}{2\pi}\operatorname{length}(\Gamma_{r})\operatorname{sup}% \left\{\left|\frac{f(w)}{w^{n+1}}\right|\,\colon\,w\in\Gamma_{r}\right\}=\frac% {1}{2\pi}\,2\pi r\frac{M_{r}}{r^{n+1}}=\frac{M_{r}}{r^{n}}$

where $M_{r}=\operatorname{sup}\{|f(w)|\colon w\in\Gamma_{r}\}$.

But $f$ is bounded, so there is $M$ such that $M_{r}\leq M$ for all $r$. Then $|c_{n}|\leq\frac{M}{r^{n}}$ 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.$\square$

Title proof of Liouville’s theorem ProofOfLiouvillesTheorem 2013-03-22 12:54:15 2013-03-22 12:54:15 Evandar (27) Evandar (27) 5 Evandar (27) Proof msc 30D20