7
$\begingroup$

I'm studying for an exam and am stuck on the following. If $f$ is holomorphic on the punctured unit disk $D- \{0\}$, and $0$ is an essential singularity does it follow that

$\displaystyle\int_{D -\{0\}} |f(z)|^{2} dA = \infty$

  • 0
    Agree with @joriki above. not really sure what to do with your question but may I point you to the Casorati - Weierstrass theorem on the behaviour of meromorphic functions near essential singularities - http://en.wikipedia.org/wiki/Casorati%E2%80%93Weierstrass_theorem2012-04-07

1 Answers 1

5

I believe I found a solution to my own question. I'll write it out to see if anyone agrees, Basically we can write the integral as

$\displaystyle\int_{D -\{0\}} |f(z)|^{2} dA = 2\pi \int_{0}^{1} \sum_{n=-\infty}^{\infty} |a_{n}|^{2}r^{2n+1} dr$

where $a_{n}$ are the Laurent coeficients Here I used Tonelli Thereom along with the fact that $\{e^{i \theta n} | n\in \mathbb{Z} \}$ are orthogonal.

Since we have an essential singularity $a_{k} \ne 0$ for some $k<0$. Thus we have

$\displaystyle\int_{D -\{0\}} |f(z)|^{2} dA = \infty$

  • 4
    Is it such a formidable effort to write out "coefficients" and "theorem"? I'd estimate that it would ta$k$e you at least an order of magnitude less time than the sum of all the times that your readers spent decyphering the abbreviations.2012-04-07