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Hello, I asked this question before, but didn't get any response, so I took the liberty of asking once again , with slightly modified version of the question:

Consider an orientation-preserving quasiconformal homeomorphism $f$ of the open unit disk $\mathbb{D}\subset \mathbb{C}$ with the complex dilatation/Beltrami cofficient $\mu, ||\mu||_{L^\infty(\mathbb{D})}<1, \mu \in C^0(\mathbb{\bar{D}})$ , i.e. $\mu$ is continuous on the closed unit disk $\mathbb{\bar{D}}$. Is it true that the restriction of $f$ on the boundary $S^1$ has continuous (ordinary) derivative on $S^1$, i.e., is $f\in C^1(S^1)$ ?

I guess one might start with continuously extending $\mu$ to all of $\mathbb{C}$, then solve the Beltrami equation on $\mathbb{C} $ with that extended $\mu$, but then I am getting stuck:because I guess the solution to this new equation might not be $C^1(\mathbb{C})$ ?? ( Look at Examples 15.1 in the book "Elliptic PDE and Quasiconformal Mappings" by K. Asltala, T. Iwaniec and G. Martin.

May be, to start with, we can ALSO assume that $\mu=0$on $S^1$. Then is the answer to my question yes ?

Could you please cite a reference of your proof to my question or give a counterexample ? Thanks a lot !!

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IMHO, either this or… should be closed, and answers/comments left on the remaining one – Yemon Choi Jun 21 '12 at 2:21
@ Yemon Choi : these two questions are different, in the other question which I asked a while back, I was looking for $C^1(\bar{D})$ regularity, which I already got the counter examples of. Here I am asking a weaker question, namely, whether the restriction of $f$ on $S^1$ is $C^1$, so I am not looking at the $C^1 $ regularity upto the boundary, I am only looking whether the restriction of the solution to the boundary is $C^1.$ P.S. I already deleted the question ( with the same content ) which I asked two days back. – Analysis Now Jun 21 '12 at 2:29
Sorry; my mistake. – Yemon Choi Jun 21 '12 at 4:35
I have two suggestions: 1. Take a look at the linearization of the Beltrami equation, $v_{\bar{z}}=\nu$, where $v$ is a vector field. The point is that linearly interpolating from $\mu$ to $0$ you get a 1-parameter family of qc maps, differentiating this family you get time-dependent vector fields. Thus, positive answer in linear case could lead to positive answer to your question. 2. MO is clearly a wrong place for this question since nobody here is doing hard-core qc analysis. Write email to few experts, I think, you know who they are, and see what they say. – Misha Jun 22 '12 at 0:43
Dear Prof. Kapovich, thanks for your comment. 1) What did you mean by linearization of the Beltrami equation, because Beltrami equation $f_{\bar{z}}= \mu. f_z $ is a linear equation in $f$, if $\mu$ is taken to be the same in each case. 2) I has e-mailed to people whom I knew are pioneers in this area, but unfortunately that was not of much help to me. – Analysis Now Jul 10 '12 at 4:04
up vote 2 down vote accepted

You may be interested in looking at the article "On boundary correspondence under quasiconformal mappings" by V.Ya. Gutlyanskii and V.I. Ryazanov. In particular, their Corollary 1 concerns what can be said when the Beltrami differential extends uniformly continuous to some part of the boundary.

They also mention the example $$ f(z) = z\cdot (1-\log|z|),$$ which defines a quasiconformal function near zero whose complex dilatation is continuous near zero, but which is not differentiable at zero.

Note that this example is real on the real axis, so it easily yields a counterexample to your question (unless I have misunderstood something!).

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Thanks you Lasse Rempe-Gillen. I will take a look at it. – Analysis Now Nov 6 '12 at 3:51

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