Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

1 Answer

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.

share|improve this answer
Oh, the solution is so simple. I can never bilieve it. –  woodbass Dec 31 '12 at 9:24
add comment

Your Answer


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.