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Hello ,we know that for given $h:S^1\to S^1$, we can solve the Dirichlet problem on $\bar{D} $ with the boundary value $h$ and in fact this extension, which is the complex harmonic extension $H=E(h) $ of $h$, is a diffeomorphism of $D$ and homeomorphism of $\bar{D}$.

My question is : suppose $h:S^1\to S^1$ is a k-quasisymmetric homeomorphism, then is the complex harmonic extension $H=E(h)\hspace{2mm} K$-quasiconformal with $K$ depending only on $k$ ?

Answers/resources/papers related to this would be greatly appreciated !

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6 Answers 6

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For the unit disk see: http://poincare.matf.bg.ac.rs/~pavlovic/QC_DIF.PDF For the halp-plane: http://poincare.matf.bg.ac.rs/~pavlovic/QC_DIF.PDF For a generalization to Jordan domains: http://arxiv.org/abs/0910.4950 (see also my paper in MATH z)

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  • $\begingroup$ @ David : Could you send me the link to the paper on quasiconformal harmonic extension for the upper half plane again ? I think you attached the one for the unit disk in place of the upper half plane. Thanks a lot ! $\endgroup$
    – Analysis Now
    Commented Dec 12, 2011 at 15:05
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The answer to the question depends on the Riemannian metric you choose in the target space (but only on the conformal class of the domain). One possibility is to choose the Euclidean metric, as in David's answer, but a more natural one is to consider the hyperbolic plane, as mentioned by Deane Yang. In this second case the question is still open. It was conjectured by Schoen that any quasisymmetric homeomorphism of the circle has a unique quasiconformal harmonic extension to the hyperbolic plane, see Richard M. Schoen, The role of harmonic mappings in rigidity and deformation problems, Complex geometry (Osaka, 1990), Lecture Notes in Pure and Appl. Math., vol. 143, Dekker, New York, 1993, pp. 179–200. MR MR1201611. The uniqueness is known, as for the existence there are newer results than those mentioned in other answers, see Vladimir Markovic, Harmonic diffeomorphisms of noncompact surfaces and Teichm¨uller spaces, J. London Math. Soc. (2) 65 (2002), no. 1, 103–114. MR MR1875138, but no complete answer.

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This is true, if you use the hyperbolic metric on $D$ and the map $h$ is near the identity. See Hardt, Robert(1-RICE); Wolf, Michael(1-RICE), "Harmonic extensions of quasiconformal maps to hyperbolic space.", Indiana Univ. Math. J. 46 (1997), no. 1, 155–163. But I'm not sure whether more is known for dimension 2.

ADDED: The Douady-Earle extension is an elegant way to extend a map of the boundary to a map of the interior, and I believe that it does extend a quasisymmetric map to a quasiconformal one.

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  • $\begingroup$ @ Prof. Yang : Thanks, I have taken a look at the main theorem of Wolf's paper. But I guess by "complex harmonic extension", I meant the solution to the Dirichlet prolem on $\bar{D}$ with $h$ as a boundary value on $S^1$, which is a particular harmonic map, and which is also a diffeomorphism in $D$ if $h$ is a homeomorphism on $S^1$.I was wondering whether this is quasi-conformal if the boundary data $h$ is quasi-symmetric. For the Douady-Earle ( or Barycentric ) extension, which is conformally natural,it is $K$ quasiconformal,$K$ depending only on $k$, where $h$ is $k$-q.s. $\endgroup$
    – Analysis Now
    Commented Apr 28, 2011 at 11:19
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An earlier reference than the ones given by @Deane is:

MR0086869 (19,258c) Beurling, A.; Ahlfors, L. The boundary correspondence under quasiconformal mappings. Acta Math. 96 (1956), 125–142.

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A function f of the unit circle onto itself can be extended to a harmonic quasiconformal mapping if and only if f is bi-lipschitz, see Pavlovic and Kalaj's results... Here we have in mind the euclidean harmonic mappings...

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  • $\begingroup$ @ david : could you please post a link to the paper ? Is it in English ? I went to your homepage but I am unable to download. $\endgroup$
    – Analysis Now
    Commented May 11, 2011 at 21:01
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Correction: A function f of the unit circle onto itself can be extended to a harmonic quasiconformal mapping if and only if f is bi-lipschitz and Hilbert transform H(f') is in L^\infty(S^1).

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