2
$\begingroup$

We just consider puntured unit disk $\triangle^{*}$ in $\mathbb{C}$. $f$ is a bounded quasiconformal map on $\triangle^{*}$. Why $f$ can extend to the origin,becoming quasiconformal map on the whole disk?

rk:It is easy to see we only have to show that we can define $f(0)$ such that $f$ is continous.Then quasiconformality of new $f$ follows easily!

Improve this question
$\endgroup$
1
  • $\begingroup$ More homework than research, voting to close. $\endgroup$
    – Igor Rivin
    Commented Nov 19, 2011 at 12:04

1 Answer 1

Reset to default
2
$\begingroup$

A simple way is the use the modulus of curve family argument, see Väisälä's book lecture notes on n-dimensional quasiconformal mappings. (Thm. 17.3)

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .