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The following is taken from an old complex analysis qualifying exam.

Let $\Delta$ denote the open unit disc.

Suppose $f:\Delta\setminus\{0\}\rightarrow \mathbb{C}$ is holomorphic and assume that $$\int_{0<|x+iy|<1}|f(x+iy)|^2dxdy<\infty.$$ Prove that $f$ can be extended uniquely to a holomorphic function on $\Delta$.

I would like to show that $|f(z)|$ is bounded in a neighborhood of 0, and then use Riemann's removable singularity theorem... but this is giving me trouble.

I can use Cauchy's integral formula on $f^2$ to obtain $$|f(z)|^2\leq\frac{1}{\pi R^2}\int_0^{2\pi}\int_0^R|f(z+re^{i\theta})|^2r\,drd\theta,$$ where $R<|z|$. And this double integral is no greater than the given integral, which is finite. However, the $1/R^2$ is preventing me from concluding anything about boundedness near 0.

Any help would be greatly appreciated.

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Note that the hypothesis you have is not that "the image has finite area", but "the squared modulus of the function value has a finite average". It is (at least a priori but perhaps only for non-analytic functions) possible so satisfy the finite-average condition but not the finite-area one from your question title. –  Henning Makholm Apr 24 '12 at 20:53
    
I would like to expand it in laurent series and then calculate that quantity explicitly in terms of the coefficients. It simple, a la the norm of a fourier series. –  mike Apr 24 '12 at 22:02
    
Thanks. Plugging in the Laurent series was definitely the right way to begin. –  John Adamski Apr 26 '12 at 19:39
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Let us represent $f$ as a Laurent series inside the punctured unit disc using polar coordinates. $$f(re^{i\theta})=\sum_{n=-\infty}^\infty a_n(re^{i\theta})^n,\quad 0<r<1$$ Applying the given conditions, we have $$\infty>\int_0^1\int_0^{2\pi}\left(\sum_{n=-\infty}^\infty a_n(re^{i\theta})^n\right)\overline{\left(\sum_{m=-\infty}^\infty a_m(re^{i\theta})^m\right)}r\,d\theta dr$$ $$=\sum_{n=-\infty}^\infty\sum_{m=-\infty}^\infty a_n\overline{a_m}\int_0^1 r^{n+m+1}\left(\int_0^{2\pi} e^{i(n-m)\theta}\,d\theta\right) dr.$$ Note that the integral over $\theta$ is $2\pi$ when $n=m$ and 0 otherwise. Thus, we have $$\infty>2\pi\sum_{n=-\infty}^\infty|a_n|^2\int_0^1 r^{2n+1}\,dr.$$ Now, since the integral over $r$ is infinite for $n\leq-1$, we must conclude that $a_n=0$ for $n\leq-1$. In other words, $f$ does not have a pole or an essential singularity at 0. Thus, $f$ must have a removable singularity at 0, i.e. $f$ can be extended to a holomorphic function on $\Delta$.

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