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For simplicity let me state the question in this way.

Let $f$ be a holomorphic function on $\{z\in\mathbb{C}:0<|z|<2\}$. Suppose $f$ has an essential singularity at $0$. Do we have $$\int_{0<|z|<1}|f(z)|^p\,\mathrm{d}z=\infty,\qquad\text{for }0<p<2?$$

It can be shown using Hilbert space methods that when $p=2$ this is true (e.g., see here). Then this is true for all $p\geq2$, since the disk has finite measure. However, I wonder whether this is true for $p<2$ also.

Yuxiao Xie
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  • With complex functions we are usually concerned with path integrals, where the measure is given by arc length. Here you seem to want an integral over a disk of the complex plane. Accordingly some definition of what measure $dz$ is representing will be needed. – hardmath May 01 '19 at 14:59
  • @hardmath Presumably it's $2$-dimensional Lebesgue measure. – Robert Israel May 01 '19 at 15:10
  • @hardmath Yes it is the Lebesgue measure. This is just asking whether it lies in the appropriate Bergman spaces. – Yuxiao Xie May 01 '19 at 15:12

1 Answers1

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As clarified in comments, this is using $2$-dimensional Lebesgue measure, which I prefer to denote as $dm(z)$ rather than $dz$.

Let's try $p=1$. If $\Gamma_r$ is the positively-oriented circle of radius $r$ centred at $0$, the coefficient $a_k$ of the Laurent series for $f$ in the punctured disk is

$$ a_k = \frac{1}{2\pi i} \oint_{\Gamma(r)} \frac{f(z)}{z^{k+1}} dz = \frac{1}{2\pi r^k} \int_0^{2\pi} f(r e^{i\theta}) e^{-ik\theta}\; d\theta$$ so that $$ \int_0^{2\pi} |f(r e^{i\theta})|\; d\theta \ge \left| \int_0^{2\pi} f(re^{i\theta}) e^{-ik\theta}\; d\theta\right| = 2 \pi |a_k| r^k $$

and thus if $a_k \ne 0$ for some $k \le -2$ (which is certainly true if the singularity is essential), $$ \int_{0<|z|<1} |f(z)| \; dm(z) = \int_0^1 \int_0^{2\pi} r |f(re^{i\theta})| \; d\theta \; dr \ge 2 \pi |a_k| \int_0^1 r^{k+1}\; dr = \infty$$

Robert Israel
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