Riemann Removability Theorem

Statement

Let $G\subseteq ℂ$ be a domain, $z_0\in G$, and $f:G\setminus z_0\to ℂ$ be holomorphic. Then $f$ can be holomorphically extended to $z_0$ if and only if there exists a neighborhood $U\subseteq G$ of $z_0$ such that $f$ is bounded on $U\setminus z_0$.

Proof

Let $r>0$ be chosen such that ${\overline{B}}_r(z_0)\subseteq U$, and let $M$ be an upper bound for $f$ on $U$.

We consider the Laurent Series of $f$ around $z_0$. It is

$f(z)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n(z-z_0{)}^n,a_n=\frac{1}{2\pi i}\underset{|w-z_0|=r}{\int }\frac{f(w)}{(w-z_0{)}^{n+1}}dw$

Estimating

$a_n$

gives the so-called Cauchy estimates, namely

$\begin{array}{cc}|a_n|& =\left|\frac{1}{2\pi i}\underset{|w-z_0|=r}{\int }\frac{f(w)}{(w-z_0{)}^{n+1}}dw\right|\\ & \le \frac{1}{2\pi }\underset{|w-z_0|=r}{\int }\frac{|f(w)|}{|w-z_0{|}^{n+1}}|dw|\\ & \le \frac{1}{2\pi }\underset{|w-z_0|=r}{\int }\frac{M}{r^{n+1}}|dw|\\ & =\frac{M}{r^n}\end{array}$

For

$n<0$

, it follows that

$|a_n|\le \frac{M}{r^n}=M{r}^{-n}\to 0,r\to 0$

Thus,

$a_n=0$

for all

$n<0$

, meaning we have

$f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n(z-z_0{)}^n$

, and

$f(z_0):=a_0$

is a holomorphic extension of

$f$

to

$z_0$

.

Translation and Version Control

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• Source: Riemannscher Hebbarkeitssatz - URL:https://de.wikiversity.org/wiki/Riemannscher Hebbarkeitssatz
• Date: 11/26/2024

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