Let $D$ denote the complex unit disk and $X \subset \mathbb{C}$ some subset. Let us consider a holomorphic motion $i \colon D \times X \rightarrow \mathbb{C}$ (denoted $i_\lambda(z)$) meaning for each fixed $\lambda \in D$ the map $i_\lambda(\cdot)$ is injective, for each fixed $z \in X$ the map $i_{(\cdot)}(z)$ is holomorphic and lastly $i_0 = id_X$. It is known that for such a motion each map $i_\lambda$ is quasi-conformal, i.e. has bounded dilatation: $$\exists K \colon~ \sup_{x \in X} \limsup_{t \rightarrow \infty} \frac{\sup_{y \in \partial B_t(x)} ~d(f(x),f(y))}{\inf_{z \in \partial B_t(x)} ~d(f(x),f(z))} \leq K$$ The proof proceeds as follows:

Fix four distinct points $z_1,\dots,z_4 \in X$ and define $$f(\lambda) = \frac{i_{\lambda}(z_1)-i_{\lambda}(z_3)}{i_{\lambda}(z_1)-i_{\lambda}(z_2)} \cdot \frac{i_{\lambda}(z_2)-i_{\lambda}(z_4)}{i_{\lambda}(z_3)-i_{\lambda}(z_4)}$$ so that $|f(\lambda)|$ is the cross ratio of the points $(i_\lambda(z_1),\dots,i_\lambda(z_4))$. By the properties of $i$, $f$ is holomorphic and omits the values $0,1,\infty$. Thus, the image of $f$ is in $\mathbb{C} \setminus \{0,1\}$, which admits a hyperbolic metric $\rho$. As $f$ is holomorphic, $\rho(f(\lambda),f(0))\leq \rho_D(\lambda,0)$, where $\rho_D$ is the hyperbolic metric on $D$. In particular, the right hand side is a bound independent of the quadruple fixed in the beginning. From this it follows that $i_\lambda$ is quasi-möbius (meaning that the cross ratios of image quarduples are bounded), which implies quasi-conformality (proved by Väisälä).

Quasi-conformality implies Hölder-continuity (sometimes called Mori's theorem, see Lectures on quasi-conformal mappings by Ahlfors). Thus, for any motion, the maps $i_\lambda$ are Hölder continuous.

Now Shishikura claims in his paper "The Hausdorff Dimension of the Mandelbrot set and Julia sets" (https://www.jstor.org/stable/121009?seq=1#page_scan_tab_contents) that the inverse maps $i_\lambda^{-1}$ are also Hölder-continuous (see page 234), but does not provide a proof. Unfortunately, this fact is crucial for the proof of his lemma 3.2 (page 235).

Can someone explain to me how it follows? Cheers!

  • 2
    $\begingroup$ The inverse of a qc map is again qc. $\endgroup$
    – Misha
    Nov 27, 2017 at 13:27
  • $\begingroup$ As far as I'm concerned, this only holds with some additional regularity assumptions on $X$ and $i_\lambda(X)$. In the context, $X$ will be some hyperbolic set for a rational map (and could potentially look horrible). If I'm wrong about this, can you give me a reference for a general statement about qc of the inverse? $\endgroup$
    – Florian R
    Nov 27, 2017 at 14:05

1 Answer 1


The answer is given in @Misha's comment. This is an extended comment. You do not need the theory of quasiconformal (quasisymmetric) maps on arbitrary sets $X$ here, because there is a stronger form of $\lambda$-lemma: every holomorphic motion of any set extends to a holomorphic motion of the whole Riemann sphere, with $K$ estimated in terms of $|\lambda|$. Thus it is quasiconformal on the whole sphere. The inverse of a quasiconformal map is quasiconformal, without any additional restrictions, and all quasiconformal maps are Holder (also without any additional restriction; the Holder exponent is estimated by a function of $K$ alone).


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