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Question: Let $\Delta$ be the unit disc in $\mathbb{C}$ and $\rho(z)|dz|^2$ be a complete conformal metric on $\Delta$ where $\rho(z)$ is continuous on $\Delta$. Let $a$ be the infimum of $p (p>0)$ such that $\iint_\Delta |\rho(z)|^pdxdy=+\infty.$

I guess that $a\leq 1$. Of course, generally $a$ depends on the complete metric $\rho(z)$. For example, w.r.t. the Poincare metric, $a=\frac{1}{2}.$ Also, one may consider the infimum of $a$.

Note. We only assume that $\rho(z)$ is continuous and complete on the unit disc.

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Is $a$ equal to $p$? –  Deane Yang Dec 31 '12 at 1:36
    
I encounter some problem in editting, e.g. \{\}, \frac{1}{2} –  woodbass Dec 31 '12 at 1:40
    
Your $a$ depends on the metric. –  Alexandre Eremenko Dec 31 '12 at 1:48
    
Yes, $a$ dependant on the metric, but for all complete metric, I belive $a \leq 1$. –  woodbass Dec 31 '12 at 1:53

1 Answer 1

up vote 5 down vote accepted

Completeness implies that $$\int_{1/2}^1\sqrt{\rho(r,\theta)}dr=\infty$$ for all $\theta$. So, for a complete metric, $$\int_\Delta\sqrt{\rho}=\int_0^{2\pi}\int_0^1\sqrt{\rho(r,\theta)}rdrd\theta=\infty.$$ Thus $a\leq 1/2$.

For Poincare metric $\rho=1/(1-r^2)^2$, so $\alpha=1/2$, and this is best possible.

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Oh, the solution is so simple. I can never bilieve it. –  woodbass Dec 31 '12 at 9:24

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