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Suppose $f$ is holomorphic on the punctured unit disc with

$$\int_{D^*} |f(z)|^2 dA(z)<\infty $$

where the integration is with respect to the two dimensional Lebesgue measure.

Show that $f$ has a holomorphic extension to the unit disc.

FYI, This is a problem from the August 2022 UMD analysis qualifying exam and can be found here https://www-math.umd.edu/images/pdfs/quals/Analysis/Analysis-August-2022.pdf I am working on it to review for my own qualifying exam and have not been able to solve it, nor found any help online or from others. Any help would be greatly appreciated.

Connor Lockhart
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1 Answers1

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If $f(z)=\sum_{n \in \mathbb Z} a_nz^n$ is the Laurent series of $f$, $M=\int_{D^*} |f(z)|^2 dA(z)<\infty$ and $\rho<1/2$ then $$\int_{\rho \le |z| \le 1/2}|f(z)|^2dA(z)=\int_{\rho}^{1/2}\int_0^{2\pi}\sum_{n,m \in \mathbb Z}a_n\bar a_m r^{n+m}e^{i(n-m)\theta}rdrd\theta$$

But by absolute convergence, we can switch sum and integral so only the $n=m$ terms remain after we integrate in $\theta$ so we get:

$$\int_{\rho \le |z| \le 1/2}|f(z)|^2dA(z)=2\pi\sum_{n\in \mathbb Z}\int_{\rho}^{1/2}|a_n|^2 r^{2n+1}dr$$

In particular we get for every $n <-1$ that $$2\pi |a_n|^2\frac{\rho^{2n+2}-(1/2)^{2n+2}}{-2n-2} =2\pi\int_{\rho}^{1/2}|a_n|^2 r^{2n+1}dr\le$$ $$ \le \int_{\rho \le |z| \le 1/2}|f(z)|^2dA(z) \le M$$

so letting $\rho \to 0$ we must have $a_n=0$ as otherwise $|a_n|^2\frac{\rho^{2n+2}-(1/2)^{2n+2}}{-2n-2} \to \infty$

Similarly for $n=-1$ we get $2\pi |a_{-1}|^2(\log 1/2-\log \rho)$ is uniformly bounded so again letting $\rho \to 0$ forces $a_{-1}=0$

Hence $f(z)=\sum_{n \ge 0}a_nz^n$ is analytic on the unit disc

Conrad
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