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Suppose that you have a quasiconformal mapping $f:D\rightarrow D$ from the open disk to the open disk. Is it true that f always admits an extension $\tilde f:\bar D\rightarrow\bar D$ from the closed disk to the closed disk such that its restriction to the disk boundary is a homeomorphism that is differentiable almost everywhere?

If we assume that the dilatation $\mu = \bar\partial f/\partial f$ is $C^{\infty}$ in $D$, then an old theorem of Gauss implies that $f$ is a disk diffeomorphism. What can we say about the extension of $f$ to the boundary in this case? Does it extend to a circle diffeomorphism? Is it $C^{\infty}$?

Frank
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  • What book are you reading? It will be covered in pretty much any textbook on qc maps, for instance, Lehto-Virtanen (in dimension 2) or Vaisala (in all dimensions). – Moishe Kohan Dec 28 '24 at 00:32
  • Thank's for the reference, it is a very nice one, but it does not seem to contain a discussion of what I would like to know (which probably means that it is not true...). My question is whether there is an extension that is more regular than being simply a homeomorphism, e.g. a circle diffeo, under the assumption that $\mu$ is smooth in the interior. – Frank Dec 28 '24 at 11:10
  • In your comment you are asking a question different from the one in the post (which was about a.e. differentiability, which is true for all self-homeomorphisms of $S^1$). In general, smoothness of $\mu$ in the interior does not imply smoothness on the boundary, moreover, the boundary map need not be even absolutely continuous. – Moishe Kohan Dec 28 '24 at 11:18
  • Thank's a lot for clarifying! – Frank Dec 28 '24 at 15:31

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To get it off the "unanswered list":

  1. Every quasiconformal (qc) map of the open unit disk $D\subset \mathbb C$ to itself extends to a homeomorphism of the closed disk. You can find a proof of this in most textbook on quasiconformal maps. For instance, on page 48 in

Ahlfors, Lars V., Lectures on quasiconformal mappings. With additional chapters by C. J. Earle and I. Kra, M. Shishikura and J. H. Hubbard, University Lecture Series 38. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3644-7/pbk). viii, 162 p. (2006). ZBL1103.30001.

One can also derive it from the harder existence theorem on solvability of the Beltrami equation (extend the Beltrami differential to the open exterior disk $S^2\setminus cl(D)$ by inversion, and solve the corresponding Beltrami equation on $S^2$).

The same extension result holds in higher dimensions (where the Beltrami equation is overdetermined), even for maps of open subsets which are interiors of closed topological balls in $\mathbb R^n$, see for instance in:

Väisälä, Jussi, Lectures on (n)-dimensional quasiconformal mappings, Lecture Notes in Mathematics 229. Berlin-Heidelberg-New York: Springer-Verlag. XIV, 144 p. (1971). ZBL0221.30031.

It is known that the boundary maps of quasiconformal maps $D\to D$ are exactly the quasisymmetric (quasimoebius) maps of the unit circle (this is due to Ahlfors and Beurling):

Beurling, Arne; Ahlfors, Lars V., The boundary correspondence under quasiconformal mappings, Acta Math. 96, 125-142 (1956). ZBL0072.29602.

  1. On the other hand, if $f: D\to D$ is a qc $C^\infty$-diffeomorphism, the boundary map $S^1\to S^1$ need not even be absolutely continuous (it can send zero 1-dimensional measure sets to positive measure sets). This is what happens when you take a (quasiconformal) diffeomorphism between two compact hyperbolic surfaces, lift it to a qc diffeomorphism of $D$ and then extend to the boundary: Unless the original map is isotopic to a conformal map, the boundary extension is not absolutely continuous.

  2. On yet another hand, every monotonic continuous map $\mathbb R\to \mathbb R$ is differentiable almost everywhere (but it can have zero derivative almost everywhere). Hence, boundary maps of qc homeomorphisms of $D$ are differentiable almost everywhere.

Moishe Kohan
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  • Dear Moishe, do you know the reference for your sentence on point two: "Unless the original map is isotopic to a conformal map, the boundary extension is not absolutely continuous". In particular I want to know if the original map is isotopic to identity, then the quasisymmetric extension is absolutely cts. – Ricky Simanjuntak Mar 26 '25 at 16:53
  • @RickySimanjuntak: Then the boundary value map is linear-fractional (belongs to the group of covering transformations). – Moishe Kohan Mar 26 '25 at 18:05
  • My apologies, I just realized when you say "isotopic to conformal map", what you mean is "isotopic to conformal map relative to the boundary". In that case indeed it is trivial to see the boundary extension is absolutely continuous. Do you happen to know other criteria when boundary extension is absolutely continuous? – Ricky Simanjuntak Mar 27 '25 at 07:09
  • I suggest you ask a new question in a post, comments are not meant for this. – Moishe Kohan Mar 27 '25 at 08:42