# Removable Singularity

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

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

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

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

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

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

Then I could try to put a bound on $\displaystyle f$ 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 $\displaystyle f(z)= \sqrt{z}$ . For all $\displaystyle 0<r<1$ is ...

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

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

Kind regards

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

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

Perhaps you meant:

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

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

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

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

Fernando Revilla