# proof of Liouville’s theorem

Let $f:\u2102\to \u2102$ be a bounded, entire function^{}. Then by Taylor’s theorem,

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

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

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

where ${M}_{r}=\mathrm{sup}\{|f(w)|:w\in {\mathrm{\Gamma}}_{r}\}$.

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 |