# Removable Singularity

• Jan 1st 2011, 07:42 PM
roninpro
Removable Singularity
Hello. I am looking at the following problem:

Let $f(z)$ be holomorphic in $D(0,1)/\{0\}$ so that

$\displaystyle \int_0^1 |f(re^{i\theta})|^2\text{ d}\theta \leq 1, \text{ for all } 0

Prove that $z=0$ is a removable singularity.

I'd like to use the Gauss Mean Value Property, where

$\displaystyle f(0)=\frac{1}{2\pi} \int_0^{2\pi} f(re^{i\theta})\text{ d}z$

Then I could try to put a bound on $f$ and conclude that the singularity is removable. However, the Gauss Mean Value property only works if the function is holomorphic on a simply connected domain.

I was wondering if there are any suggestions on how to approach this problem. Any ideas would be much appreciated.
• Jan 2nd 2011, 12:14 AM
chisigma
As conuterexample let's consider $f(z)= \sqrt{z}$ . For all $0 is ...

$\displaystyle \int_{0}^{1} |f(r\ e^{i \theta})|^{2} d \theta = r <1$

... but the singularity of $f(*)$ in $z=0$ is not 'removable'...

Kind regards

$\chi$ $\sigma$
• Jan 2nd 2011, 02:16 AM
FernandoRevilla
Quote:

Originally Posted by roninpro
Prove that $z=0$ is a removable singularity.

Perhaps you meant:

$\displaystyle \int_{0}^{2\pi} |f(r\ e^{i \theta})| d \theta = r <1$

In that case, the $A_{-n}\;(n>0)$ coefficient of the Laurent expansion satisfies:

$\left |{A_{-n}}\right |=\dfrac{1}{2\pi}\left |\displaystyle \int_{|z|=r}f(z)z^{n-1}dz\right |\leq \ldots \leq \dfrac{r^n}{2\pi}\rightarrow{0}\;(\textrm{if}\;r\r ightarrow{0})$

So, $z=0$ would be a removable singularity.

Fernando Revilla