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.