# Cauchy sequence of quasi-conformal automorphisms --> Conclusion on corresponding sequence of maximal dilatation?

Let $G \subsetneq \mathbb{C}$ be a simply connected, bounded domain in $\mathbb{C}$. Denote by $Q(G)$ the set of all quasi-conformal automorphisms of $G$, i.e.

$$Q(G) := \left\{ f: G \rightarrow G \, | \, f \text{ is } K\text{-quasiconformal for some } K \in [1, \infty) \right\}$$

(Note that, in particular, $Q(G)$ contains the set of all conformal automorphisms of $G$). This set can be given some structure: On the one hand, if equipped with the composition of functions, denoted by $\circ$, $(Q(G), \circ)$ becomes a group. On the other hand, when introducing the supremum metric

$$d_{\sup}(f,g) := d(f,g) := \sup \limits_{z \in G} |f(z) - g(z)|$$

for $f, g \in Q(G)$ on the domain $G$, the tuple $(Q(G), d)$ becomes a metric space.

Now suppose you have a Cauchy sequence $(f_n)_{n \in \mathbb{N}} \subseteq Q(G)$ in $Q(G)$ with respect to $d$. Let $K_n := K(f_n)$ denote the maximal dilatation of $f_n$ in $G$, i.e.

$$K_n := K(f_n) := \sup \limits_{\overline{Q} \subseteq G} \frac{M(f_n(Q))}{M(Q)}$$

where $Q$ is some quadrilateral in $G$ and $M$ denotes the conformal modulus of such a quadrilateral $Q$ (see Lehto/Virtanen, "Quasiconformal Mappings in the Plane", p. 15 ff., for example; Of course, there are also several other ways to express the maximal dilatation!).

My question is: Can one draw any conclusion on the corresponding real sequence $(K_n)_{n \in \mathbb{N}} \subseteq [1, \infty)$ and some of its properties, in particular (un)-boundedness and convergence/divergence?

Any help or hint on this topic (or also related topic!) would be highly appreciated! Thanks in advance!

Not much can be inferred about the behavior of derivatives on the basis of uniform convergence. For example, let $G$ be the square $\{x+iy:|x|,|y|<1\}$ and consider maps of the form $f_n(x+iy)=h_n(x)+iy$ with $h$ increasing. Then uniform convergence of $f_n$ is equivalent to uniform convergence of $h_n$. The maximal dilatation of $f_n$ is $$K_n=\max(\sup h_n', 1/\inf h_n')$$ It's easy to construct piecewise affine $h_n$ such that $h_n(x)\to x$ uniformly but $K_n\to\infty$. (Or $K_n$ alternates between being $n$ and $1$, so there is no limit at all.) Simply put, uniform convergence to a nice map does not prevent $f_n$ from having small wrinkles that make dilatation large.
It is true that if $K_n$ are uniformly bounded, then the limit is quasiconformal, but you probably already knew that: I'm sure Lehto & Virtanen have a proof.
• Hey Gerry, thanks for your answer, and sorry for my late reply! The functions $h_n$ then have to be bijective mappings from the interval $[-1, 1]$ to itself, right? Can you name an example for a concrete sequence of the $(h_n)_n$, or a reference to a paper with examples? Indeed, i know the result that if the $K_n$ are bounded, the limit mapping will be quasi-conformal as well. My problem is that I don't know if the sequence of the $K_n$ is bounded or not ;-) I'd like to show that this is actually the case, but so far I didn't succeed on this task.. Commented Apr 11, 2017 at 18:52
• If I may ask an additional question concerning your answer: How did you derive the formula for the maximal dilatation of the $f_n$, namely $$K_n = \max(\sup h'_n, \ 1/ \inf h'_n)$$ in your answer? Is there some "easy" formula? Commented Jan 26, 2018 at 11:02