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.
Best Answer
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.