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I have a question on the Schwartz-Christoffel formula.

The map is a conformal map from the unit disk to polygons. Let me give you a specific example. In fact, \begin{align*} \phi(z)=\int_{0}^z (1-u^n)^{-2/n}\,du \end{align*} maps $\mathbb{D}$ onto the interior of a regular polygon with $n$ sides.

We can know the modulus of continuity of general conformal maps. The following result is well known.

Let $\mathbb{D} \subset \mathbb{C}$ be the unit disk centered at the origin. A conformal map $f$ defined on $\mathbb{D}$ is $\alpha$-Hölder continuous ($\alpha \in (0,1]$) if and only if there exists $L \in (0,\infty)$ such that \begin{align*} |f'(z)| \le L(1-\|z\|^2)^{\alpha-1},\quad z \in \mathbb{D}, \end{align*} where we denote by $\|\cdot\|$ the Euclidean metric on $\mathbb{C}$.

According to this result, $\phi$ defined above is $(1-2/n)$-Hölder continuous. That is, there exists $C>0$ such that \begin{equation} \| \phi(x)-\phi(y)\| \le C\|x-y\|^{1-2/n} \end{equation} for any $x,y \in \mathbb{D}$. However, I do not know the Hölder continuity of $\phi^{-1}$. I think that the index should be bigger than $1-2/n$. Can we show this? What is the specific and optimal index?

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The inverse map satisfies $|\phi^{-1}(z)-\phi^{-1}(a)|\leq c|z-a|^{n/(n-2)}$ at any vertex $a$ (reciprocal to the exponent of the direct map). Since this exponent is $>1$ the inverse is just Lipschitz everywhere. You do not need any general theorems for this. At the vertex the inverse map behaves like a power. At all other points it is smooth (analytic).

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  • $\begingroup$ Thank you for your comment. If the inverse map $\phi^{-1}$ is $n/(n-2)$--Hölder continuous, it should be a constant function. Am I misunderstanding something? $\endgroup$
    – sharpe
    Feb 11, 2020 at 14:19
  • $\begingroup$ Do you mean that $\phi^{-1}$ is a Lipschitz continuous function with respect to $\|\cdot\|$? $\endgroup$
    – sharpe
    Feb 11, 2020 at 16:51
  • $\begingroup$ @Sharpe: yes, I edited. $\endgroup$ Feb 11, 2020 at 20:01
  • $\begingroup$ Thank you for the clarification. I understood. If we consider a conformal map of the unit disk to a convex domain, the inverse map is likely to be Lipschitz continuous. $\endgroup$
    – sharpe
    Feb 11, 2020 at 21:37
  • $\begingroup$ @sharpe: Yes, this generalization seems to be correct. $\endgroup$ Feb 12, 2020 at 12:47

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