MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f:\mathbb{S}^1 \to \mathbb{S}^1$ be an orientation-preserving homeomorphism. Denote by $H(f)$ the complex harmonic extension/solution in $\mathbb{D}$ to the Dirichlet problem with boundary data $f$, which can also be shown to be a homeomorphism of $\bar{\mathbb{D}}$ [Choquet's theorem].

Now,it follows from a well-known theorem of Kellog-Warschowski that if $f\in C^{k,\alpha}(\mathbb{S^1})$, then $H(f) \in C^{k,\alpha}(\mathbb{D})$.

My question is : assuming $k=1$, what is the limiting matrix of the (total) derivative matrix $DH(f)_p$ as $p\to \zeta \in \mathbb{S^1},p \in \mathbb{D}$ ? From some other kind of extensions I have seen before, I would take a guess that $DH(f)_p \to f'(\zeta).Id$ as $p\to \zeta$, where $Id$ denotes the $2$ by $2$ identity matrix.

A reference to your answer will be highly appreciated!

share|cite|improve this question

Your guess is wrong. Take real $f$, for example, then harmonic extension is also real, and its derivative is singular. In general, I do not expect a simple answer. And certainly the answer cannot depend on the local properties of $f$ near $\zeta$ only. Indeed, leave $f$ unchanged on one half of the circle, but change it on another half. It is clear that the derivative at the points of the first half will change. So the limit of the derivative at a point cannot be computed from the knowledge of f only near this point.

share|cite|improve this answer
Thanks for your answer, but I needed to change the question to suit the particular cases that is required for me. I know other type of extensions (not harmonic extension) where the limit depends on $\zeta$ only.Do you think in my modified question, there is a chance that the limit should be as I mentioned ? – Analysis Now Sep 24 '12 at 4:55
Where is your modified question? I don't see it. What kind of extension do you want to consider now? – Alexandre Eremenko Sep 24 '12 at 13:11
The modified question is : $f:S^1\to S^1$ is now a $C^{1,\alpha}$ diffemorphism. I am still considering its complex harmonic extension $H(f)$ and am asking whether $DH(f)_p \to f'(\zeta).Id$ as $p \to \zeta, p \in \mathbb{D}, \zeta \in S^1$ ? – Analysis Now Sep 24 '12 at 13:54
The other extension I am talking about comes from theory of Riemann surfaces, where there is a particular kind of extension called Douady-Earle extension $F(f)$ where $DF(f)_p \to f'(\zeta)$ as $p \to \zeta$, where $f$ is a $C^{1,\alpha}$ circle diffeomorphims (actually $C^1$ suffices. – Analysis Now Sep 24 '12 at 13:57
What is the difference between the "modified question" and the original question? I don't see any. – Alexandre Eremenko Sep 25 '12 at 11:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.