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

 $f(z)=\sum_{k=-\infty}^{\infty}c_{k}(z-a)^{k}$

be the Laurent series  of $f$ centered at $a$. We will show that $c_{k}=0$ for $k<0$, 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

 $\operatorname{Res}((z-a)^{n}f(z),a)=\frac{1}{2\pi i}\lim_{\delta\to 0^{+}}% \oint_{|z-a|=\delta}(z-a)^{n}f(z)\mathrm{d}z.$

This is equal to zero, because

 $\displaystyle\left|\oint_{|z-a|=\delta}(z-a)^{n}f(z)\mathrm{d}z\right|$ $\displaystyle\leq$ $\displaystyle 2\pi\delta\max_{|z-a|=\delta}|(z-a)^{n}f(z)|$ $\displaystyle=$ $\displaystyle 2\pi\delta^{n}\max_{|z-a|=\delta}|(z-a)f(z)|$

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

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

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

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

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 ProofOfRiemannsRemovableSingularityTheorem 2013-03-22 13:33:03 2013-03-22 13:33:03 pbruin (1001) pbruin (1001) 5 pbruin (1001) Proof msc 30D30