# Can quasiconformal mappings converge uniformly to a homeomorphism that is NOT quasiconformal?

My question concerns the following situation: Let $$G$$ be a domain in $$\mathbb{C}$$ and $$f_n: G \rightarrow \mathbb{C}$$ be a sequence of quasiconformal mappings. Suppose that $$f_n$$ converges uniformly on $$G$$, i.e. with respect to the supremum metric on $$G$$, to a homeomorphism $$f: G \rightarrow \mathbb{C}$$. My question is:

Is it possible that the limit mapping $$f$$ is NOT quasiconformal, even though being a homeomorphism?

Note that the answer is surely "NO" if the $$f_n$$ are all $$K$$-quasiconformal for some fixed $$K < \infty$$, by well-known convergence results on quasiconformal mappings; in this case, the limit mapping $$f$$ will be $$K$$-quasiconformal again. Hence the interesting part of my question is when the condition "the $$f_n$$ are $$K$$-quasiconformal mappings" is dropped, i.e. the maximal dilatations of the $$f_n$$ are not uniformly bounded by some constant $$K$$...I do not know what could possibly happen in this situation (maybe the situation at hand actually forces the $$f_n$$ to be $$K$$-quasiconformal?), unfortunately all convergence results on quasiconformal mappings I am aware of deal with the situation that the $$f_n$$ are all $$K$$-quasiconformal. Any kind of help is highly appreciated - thanks in advance!

Of course it is possible even when $$G$$ is the open unit disk (more precisely, a bounded convex domain with polygonal boundary). Let $$D$$ denote the closure of $$G$$ and $$f: D\to D$$ be any homeomorphism. Then
$$f$$ is the uniform limit of PL homeomorphisms $$f_n: D\to D$$.
Moise, Edwin E., Geometric topology in dimensions 2 and 3, Graduate Texts in Mathematics. 47. New York - Heidelberg - Berlin: Springer-Verlag. X, 262 p. DM 45.00; $19.80 (1977). ZBL0349.57001.. The homeomorphisms $$f_n$$ obviously restrict to quasiconformal maps of $$G$$. There are many ways to find a homeomorphism $$f: D\to D$$ which is not quasiconformal on $$G$$, you can even find one which is nowhere differentiable. Edit. Since you asked: Suppose that $$G$$ is a triangulated polygonal domain in $${\mathbb C}$$ and $$f: G\to G'\subset {\mathbb C}$$ is an orientation-preserving homeomorphism which is affine on each triangle in the triangulation. Then $$f$$ is quasiconformal. The simplest way to prove this that I know is to use the analytical definition of quasiconformality. There are two things that you need to verify (and I leave the verification for you to work out): a. Each orientation-preserving affine map $$h$$ is quasiconformal: You prove this by verifying that its complex dilatation $$\mu_h= \frac{h_{\bar{z}}}{h_z}$$ is constant and has sup-norm $$<1$$. From this you conclude that a piecewise-affine map $$f$$ as above is differentiable almost everywhere in $$G$$ and satisfies the property that $$||\mu_f||_{L_\infty(G)}<1.$$ b. $$f$$ is absolutely continuous on every coordinate line in the plane. To prove this, you verify that each piecewise-continuous function of one real variable is absolutely continuous. An alternative solution is to verify that $$f$$ is bi-Lipschitz and then conclude that $$f$$ is quasi-symmetric which then implies quasiconformality. • Sorry for bothering you once again, but I don't really see why it is "obvious" that the PL homeomorphisms$f_n$are quasiconformal maps when restricting them to the interior of the unit disk. Is there some easy proof for this that I'm just not able to see? Thanks in advance! Commented Oct 20, 2020 at 14:48 • @ComplexFlo: What is your definition of a quasiconformal map? Commented Oct 20, 2020 at 15:13 • Usually I use the geometric definition as it is given e.g. in Lehto/Virtanen Commented Oct 20, 2020 at 15:40 • Thanks for the explanation! Two more, hopefully last questions: 1.) QC maps are orientation-preserving (op) by definition, so if a sequence of qc maps converges uniformly to a homeomorphism, then this limit map is necessarily op as well, right? 2.) The homeomorphism group$H(\overline{G})$of the closure of a polygonal domain$\overline{G}$in$\mathbb{C}$is a topological group (endowed with the uniform topology induced by sup metric); thus the identity connected component is closed in$H(\overline{G})\$. This is why (in view of my question 1) you can restrict to o.p. PL homeomorphisms, right? Commented Dec 16, 2020 at 17:21