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# $C^{1,\alpha}$-regularity of certain function relatedtoharmonicextension appearing as inverse function

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The question I am going to ask refers to the following paper :

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 $\Phi(f)(z)$.

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 $C^{1,\alpha}(\mathbb{D})$.

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 !

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# $C^{1,\alpha}$-regularity of certain function appearing as inverse function

The question I am going to ask refers to the following paper :

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 $\Phi(f)(z)$.

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 $G(z,w)$ 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 $C^{1,\alpha}(\mathbb{D})$.

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 !