Question

Hey Folks!

Thanks to Jasun for posting all the notes. They're great. I hope you all enjoyed the workshop too - I was really jealous.

So here's the question. I haven't done much research on this, so I don't know if it's impossible or totally trivial.

A theorem of Ahlfors and Beurling* states that every quasisymmetric map of the circle extends to a quasisymmetric map of the disk. That is, every quasicircle is the boundary of a quasidisk. Now, every bi-Lipschitz map is quasistymmetric, so every bi-Lipschitz image of the circle is the boundary of a quasidisk. But, does every bi-Lipschitz map of the circle extend to a bi-Lipschitz map of the disk?

The obvious thing to try would be to see if the Ahlfors-Beurling** extension for quasisymmetric maps produces a bi-Lipschitz map from bi-Lipschitz boundary data. However, I don't really remember how the Ahlfors-Beurling*** extension works, and since I'm on the road, I can't really look it up (a technical version is in Lehto's book, but Mario explained to me once the "right" way of thinking about it - damn my crappy memory!) Anyway, any insights would be appreciated.

Another way to think about this would be to use some Lipschitz extension theorems. However, the McShane extension wouldn't work because there is no way in that construction to ensure that the extension of a bi-lip homeo is a bi-lip homeo. (one only gets Lipschitz, not bi-Lipschitz, which is a result of the fact that one must apply the extension to the coordinate functions individualy). Maybe Kirzbaum is better? (again, I'm on the road so I can't look up the proof of kirzbaums theorem).

Ok, thanks!

*Thanks to Anonymus for correcting this reference.
**Ditto
***Ditto Ditto