$C^{1,\alpha}$-regularity of certain function related to harmonic extension appearing as inverse function - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T20:46:06Z http://mathoverflow.net/feeds/question/104506 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104506/c1-alpha-regularity-of-certain-function-related-to-harmonic-extension-appea $C^{1,\alpha}$-regularity of certain function related to harmonic extension appearing as inverse function Analysis Now 2012-08-11T18:09:08Z 2012-08-11T19:02:20Z <p>The question I am going to ask refers to the following paper :</p> <p><a href="http://www.springerlink.com/content/hw88761257310165/" rel="nofollow">http://www.springerlink.com/content/hw88761257310165/</a>.</p> <p>For a fixed orientation-preserving homeomorphism $f$ of the unit circle $S^1$,define the function $G: \mathbb{D}\times \mathbb{D} \to \mathbb{D}$ by: $G(z,w):= \int_{S^1}\frac{f(t)-w}{1-\bar{w}f(t)}. p(z,t)|dt|$, where $p(z,t)= \frac{1}{2\pi}.\frac{1-|z|^2}{|z-t|^2}$ is the Poisson kernel . It is shown in the above paper that, for a fixed $z\in S^1$, $G(z,w)$ has a unique $w$ -zero in the open unit disk $\mathbb{D}$, which we denote by $\Phi(f)(z)$.</p> <p>My question is : if I assume that $f\in C^{1,\alpha}(S^1)$, then is $\Phi(f)\in C^{1,\alpha}(\mathbb{D})$ ? The reason I am even suspecting this to be true is the following : Note that for fixed $w\in \mathbb{D}, h:t\mapsto\frac{f(t)-w}{1-\bar{w}f(t)}$ is the left-composition of $f$ with a conformal automorphism $c: p\mapsto \frac{p-w}{1-\bar{w}.p}$of $\mathbb{D}$, hence $h$ is $C^{1,\alpha}(S^1)$, because $f\in C^{1,\alpha}(S^1)$ and $c\in C^{\infty}(\mathbb{\bar{D}})$ [ c has pole outside the unit disk]. Also, note that for fixed $w$, $G(z,w)$ is the complex harmonic extension of the circle homeomorphism $h: t\mapsto\frac{f(t)-w}{1-\bar{w}f(t)}$. So, by Kellog's theorem (cited in any standard PDE book, see for example Gilberg-Trudinger), we have the complex harmonic extension $G$ of $h$ is $C^{1,\alpha}(\mathbb{D})$.</p> <p>But the above automatically does $\textbf{not}$ guarantee that if $f\in C^{1,\alpha}(S^1)$, then $\Phi(f)\in C^{1,\alpha}(\mathbb{D})$. Note that $\Phi(f)$ comes an implicit function of something we know $C^{1,\alpha}$-regularity about. Can we say anything about $\Phi(f)$ ? I understand the above comment is slightly vague, but I would appreciate if you have seen some similar situations like this before and can cite it or help me with this. Thank you !</p>