Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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 2

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.

share|improve this answer

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 ...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.