# Has the composition of a sequence of quasiconformal mappings with unbounded dilatation and another q.c. mapping still unbounded dilatation?

Let $$G \subseteq \mathbb{C}$$ be a bounded domain in $$\mathbb{C}$$. Consider for $$m \in \mathbb{N}$$ a sequence of quasiconformal mappings $$f_m: G \longrightarrow \mathbb{C}$$ with unbounded maximal dilatation, i.e. $$K(f_m) \longrightarrow + \infty$$ for $$m \to + \infty$$; thereby, it is not of interest to me whether the sequence $$(f_m)_m$$ actually converges to a limit function (and in which sense) or not, but if things are easier to state, convergence can be assumed. Furthermore, let $$g: G \longrightarrow G$$ be a $$K$$-quasiconformal mapping for some (finite) $$K \in [1, +\infty)$$ which is non-trivial, i.e. $$g \not = \operatorname{id}_G$$. Now the composition $$f_m \circ g: G \longrightarrow \mathbb{C}$$ is defined, giving rise to consider the sequence $$(f_m \circ g)_m$$ of quasiconformal mappings.

My question is: Will this new sequence of quasiconformal mappings have unbounded maximal dilatation as well, i.e. is it always true that $$K(f_m \circ g) \longrightarrow + \infty$$ for $$m \to + \infty$$? As far as I know, one only has the inequality $$K(f_m \circ g) \leq K(f_m) K(g)$$ at hand.

Thanks in advance for any help!

## 1 Answer

Yes, this follows from the inequality you cited, the observation that $$f_m = (f_m \circ g) \circ g^{-1}$$, and the fact that $$K(g) = K(g^{-1})$$.