proof of Riemann’s removable singularity theorem

Suppose that $f$ is holomorphic on $U\setminus \{a\}$ and ${lim}_{z\to a}(z-a)f(z)=0$. Let

$$f(z)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_k{(z-a)}^k$$

be the Laurent series of $f$ centered at $a$. We will show that $c_k=0$ for , so that $f$ can be holomorphically extended to all of $U$ by defining $f(a)=c_0$.

For any non-negative integer $n$, the residue of ${(z-a)}^{n}f(z)$ at $a$ is

$$\mathrm{Res}({(z-a)}^{n}f(z),a)=\frac{1}{2\pi i}\underset{\delta \to 0^{+}}{lim}{\oint }_{|z-a|=\delta }{(z-a)}^{n}f(z)dz.$$

This is equal to zero, because

$\left|{\displaystyle {\oint }_{|z-a|=\delta }}{(z-a)}^{n}f(z)dz\right|$ $\le$ $2\pi \delta \underset{|z-a|=\delta }{\max }|{(z-a)}^{n}f(z)|$

$=$ $2\pi {\delta }^n\underset{|z-a|=\delta }{\max }|(z-a)f(z)|$

which, by our assumption, goes to zero as $\delta \to 0$. Since the residue of ${(z-a)}^{n}f(z)$ at $a$ is also equal to $c_{-n-1}$, the coefficients of all negative powers of $z$ in the Laurent series vanish.

Conversely, if $a$ is a removable singularity of $f$, then $f$ can be expanded in a power series centered at $a$, so that

$$\underset{z\to a}{lim}(z-a)f(z)=0$$

because the constant term in the power series of $(z-a)f(z)$ is zero.

A corollary of this theorem is the following: if $f$ is bounded near $a$, then

$$|(z-a)f(z)|\le |z-a|M$$

for some $M>0$. This implies that $(z-a)f(z)\to 0$ as $z\to a$, so $a$ is a removable singularity of $f$.

Title	proof of Riemann’s removable singularity theorem
Canonical name	ProofOfRiemannsRemovableSingularityTheorem
Date of creation	2013-03-22 13:33:03
Last modified on	2013-03-22 13:33:03
Owner	pbruin (1001)
Last modified by	pbruin (1001)
Numerical id	5
Author	pbruin (1001)
Entry type	Proof
Classification	msc 30D30