Q:

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!

A:

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.  

