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Math Help - Removable Singularity

  1. #1
    Senior Member roninpro's Avatar
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    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<r<1

    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.
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  2. #2
    MHF Contributor chisigma's Avatar
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    As conuterexample let's consider f(z)= \sqrt{z} . For all 0<r<1 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
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by roninpro View Post
    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
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