Let $G \subseteq \mathbb{C}$ be a domain (one may assume that $G$ is bounded and simply connected, if needed). Suppose $(f_n)_{n \in \mathbb{N}}$ is a sequence of bounded (i.e. $\sup_{z \in G} |f_n(z)| < \infty$ for all $n$) quasiconformal mappings $f_n: G \rightarrow \mathbb{C}$ that converges uniformly to a bounded continuous (not necessarily quasiconformal!) mapping $f \in C(G)$, i.e. $$d_{\sup}(f_n, f) \overset{n \to \infty}{\longrightarrow} 0$$ with $d_{\sup}(f_n, f) := \sup_{z \in G} |f_n(z) - f(z)|$ denoting the supremum metric. By a well-known result on sequences of quasiconformal mappings, the limit mapping $f$ will either be a quasiconformal mapping of $G$ into $\mathbb{C}$ or some degenerated function (e.g. constant), provided that the maximal dilatations $K(f_n)$ of the mappings $f_n$ are uniformly bounded, i.e. $K(f_n) \leq K$ for all $n$ and some fixed $K \in [1, \infty)$ (see for example Lehto/Virtanen, Quasiconformal Mappings in the Plane, Theorem 5.3, p. 74). Utilizing the complex dilatation $\mu_n := \frac{(f_n)_{\overline{z}}}{(f_n)_z}$ of the mappings $f_n$, this may also be written as $$ \| \mu_n \|_{L^\infty} \leq k < 1 $$ for corresponding $k := \frac{K-1}{K+1} \in [0,1)$.

My question is: Can it happen in this situation that the maximal dilatations $K(f_n)$ are unbounded (or equivalently, $\| \mu_n \|_{L^\infty} \overset{n \to \infty}{\longrightarrow} 1$)? Or does the uniform convergence of the $f_n$ somehow force the corresponding real sequence $(K(f_n))_{n \in \mathbb{N}}$ to be necessarily bounded?

Any hints, (counter)-examples, reference requests etc. on this (and related) question(s) are highly appreciated!


1 Answer 1


No, the limit $f$ need not be quasiconformal. You can easily construct a sequence of invertible linear maps which converges to a map whose image is (real) 1-dimensional. Among other pathologies: Every homeomorphism $S^2\to S^2$ is the limit of diffeomorphisms (in the uniform topology).

  • $\begingroup$ Thanks for your reply! In a special situation I'm considering, the $f_n$ are supposed to be quasiconformal automorphisms of the domain $G$, i.e. $f_n(G) = G$ for all $n$...does this additional requirement change anything about your answer? Moreover, can you give me a hint on how to construct the linear maps you mentioned explicitely? One more question: How does the "pathology" with the diffeomorphisms fit in the situation of my question? Don't get it....(sorry if this is a silly question of mine...) $\endgroup$
    – ComplexF
    Jan 19, 2018 at 18:21
  • $\begingroup$ @ComplexFlo: No, having an invariant domain does not help at all. As for the linear maps, consider $f_n(x,y)= (x, \frac{1}{n}y)$. As for diffeomorphisms, they are always qc, while homeomorphisms can be nowhere differentiable. $\endgroup$ Jan 19, 2018 at 18:34
  • $\begingroup$ Great, thank your very much for your answers and explanations! :) $\endgroup$
    – ComplexF
    Jan 20, 2018 at 10:20

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