0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Is $a$ equal to $p$? $\endgroup$
    – Deane Yang
    Dec 31, 2012 at 1:36
  • $\begingroup$ I encounter some problem in editting, e.g. \{\}, \frac{1}{2} $\endgroup$
    – woodbass
    Dec 31, 2012 at 1:40
  • $\begingroup$ Your $a$ depends on the metric. $\endgroup$ Dec 31, 2012 at 1:48
  • $\begingroup$ Yes, $a$ dependant on the metric, but for all complete metric, I belive $a \leq 1$. $\endgroup$
    – woodbass
    Dec 31, 2012 at 1:53

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Oh, the solution is so simple. I can never bilieve it. $\endgroup$
    – woodbass
    Dec 31, 2012 at 9:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.