Riemann Removability Theorem

Statement

Let $G\subseteq \mathbb {C}$ be a domain, $z_{0}\in G$, and $f\colon G\setminus {z_{0}}\to \mathbb {C}$ 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 ${\bar {B}}_{r}(z_{0})\subseteq U$, and let $M$ be an upper bound for $f$ on $U$.

We consider the Laurent Series of $f$ around $z_{0}$. It is

$f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-z_{0})^{n},\qquad a_{n}={\frac {1}{2\pi i}}\int _{|w-z_{0}|=r}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw$

Estimating $a_{n}$ gives the so-called Cauchy estimates, namely

${\begin{array}{rl}|a_{n}|&=\displaystyle \left|{\frac {1}{2\pi i}}\int _{|w-z_{0}|=r}{\frac {f(w)}{(w-z_{0})^{n+1}}}\,dw\right|\\&\leq \displaystyle {\frac {1}{2\pi }}\int _{|w-z_{0}|=r}{\frac {|f(w)|}{|w-z_{0}|^{n+1}}}\,|dw|\\&\leq \displaystyle {\frac {1}{2\pi }}\int _{|w-z_{0}|=r}{\frac {M}{r^{n+1}}}\,|dw|\\&=\displaystyle {\frac {M}{r^{n}}}\\\end{array}}$

For $n<0$, it follows that

$|a_{n}|\leq {\frac {M}{r^{n}}}=Mr^{-n}\to 0,\quad r\to 0$

Thus, $a_{n}=0$ for all $n<0$, meaning we have $f(z)=\sum _{n=0}^{\infty }a_{n}(z-z_{0})^{n}$, 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