# Singularities in the punctured unit disc and square integrability

I'm having trouble with the following problem:

Let $f$ be holomorphic on the punctured unit disc, $D$. If $\int_D|f(z)|dA(z)<\infty$, then $z=0$ is either a removable singularity or a simple pole of $f$.

A similar problem is to prove or disprove that if $\int_D|f(z)|^2dA(z)<\infty$, then $f$ has a removable singularity at $0$.

I've tried using the Mean Value Property for $f$ and taking the limit as $z\to0$, but I didn't get anywhere.

-
Have you tried integrating $1/|z|^2$ over the disk? – Matt E Sep 11 '11 at 3:23
Matt E: isn't that infinity? – Steven Sam Sep 11 '11 at 4:27

## 2 Answers

The one with the $|f(z)|^2$ is simpler. Consider the Laurent series $f(z) = \sum_{n=-\infty}^\infty c_n z^n$. By the orthogonality of the functions $e^{in\theta}$ on $[0, 2\pi]$, we have $\int_0^{2\pi} |f(r e^{i\theta})|^2 \ d\theta = 2 \pi \sum_{n=-\infty}^\infty |c_n|^2 r^{2n}$ and $\int_D |f(z)|^2 \ dA = \int_0^1 \int_0^{2\pi} r |f(r e^{i\theta}|^2 \ d\theta \ dr$. Since $\int_0^1 r^{2n+1}\ dr$ is finite for $n \ge 0$ and infinite for $n \le -1$, the only way to have $\int_D |f(z)|^2 \ dA < \infty$ is that all the coefficients for $n \le -1$ are 0.

-

For the first problem, write $c_n$ as an integral around the circle of radius $r$ centred at $0$. If $\int_D |f(z)|\ dA < \infty$, the integral of that times a certain power of $r$ for $r$ from 0 to 1 would be finite ...

-