TITLE: Removable singularity at 0 if the image of the punctured unit disc has finite area
QUESTION [5 upvotes]: 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
  $$\int_{0<|x+iy|&lt1}|f(x+iy)|^2dxdy<\infty.$$
  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.

REPLY [2 votes]: Let us represent $f$ as a Laurent series inside the punctured unit disc using polar coordinates.
$$f(re^{i\theta})=\sum_{n=-\infty}^\infty a_n(re^{i\theta})^n,\quad 0&ltr&lt1$$
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 $\Delta$.