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
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
posted by Kevin at 12:37 AM
3 Comments:
Ahlfors and Beurling (not Bers) treated quasisymmetric maps of circle T onto itself. In this case the bi-Lipschitz version is trivial -- if f:T->T is bi-Lipschitz, then its 1-homogeneous extension is bi-Lipschitz too (i.e. set F(z)=|z|f(z/|z|) for complex z).
However, you probably wanted to extend a general bi-Lipschitz map f:T->C. Let Omega be the domain bounded by f(T). Let Phi be the conformal map of the disk D onto Omega. Since f(T) is a quasicircle (it is even a chord-arc curve), Phi extends to a quasisymmetric map of T onto f(T). Define g:T->T by g(z)=Phi^{-1}(f(z)). Now g is a quasisymmetric map of T onto itself, so it has a quasiconformal extension G:D->D. It remains to prove that Phi(G(z)) is bi-Lipschitz, as it obviously agrees with f on T.
For this last part it is important that G is not just any quasiconformal extension. Roughly speaking, the derivative of G at any point z in D must be comparable to the stretch factor (under g) of the circular arc of length 1-|z| centered at z/|z|. To put it another way, G must be bi-Lipschitz in the hyperbolic metric of the disk. Fortunately, any reasonable extension (Ahlfors-Beurling, Douady-Earle, or piecewise linear) has this property. For details see either "Extension of quasisymmetric and Lipschitz embeddings..." by Tukia or "Hardy spaces, A_infinity, and singular integrals..." by Jerison and Kenig.
Wow! Thanks for your comments, Anonymus. If you are the same anonymus that has posted on my or Janus' blogs before, you know that I am terrible with names (remember how many times it took me to correctly spell Cannon?). I'll correct my question to correctly credit Beurling (I guess I've said "Ahlfors-Bers" so many times that it's become a habit...)
Anyway, it looks like you spent a while writing that, and I really appreciate your help. What you write makes sense to me- though I did have to go back and think about why the uniformizing map Phi from the inside of f(T) to the disk extends to a homeo of the boundaries...
Thanks again!
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