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!


1 Answer 1


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 $f_n: D\to D$.

For a proof see for instance Chapter 6, Theorem 3, in

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.

  • $\begingroup$ 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! $\endgroup$
    – ComplexF
    Commented Oct 20, 2020 at 14:48
  • $\begingroup$ @ComplexFlo: What is your definition of a quasiconformal map? $\endgroup$ Commented Oct 20, 2020 at 15:13
  • $\begingroup$ Usually I use the geometric definition as it is given e.g. in Lehto/Virtanen $\endgroup$
    – ComplexF
    Commented Oct 20, 2020 at 15:40
  • $\begingroup$ 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? $\endgroup$
    – ComplexF
    Commented Dec 16, 2020 at 17:21
  • 1
    $\begingroup$ @ComplexFlo: (1) Yes. (2) That's one way to prove it. But this works even if the domain has nontrivial topology and its group of self-homeomorphisms has more than two components. There is a homological argument (which works in all dimensions and for oriented manifolds) that if a uniform (on compacts) limit of orientation-preserving homeomorphisms is a homeomorphism, then it necessarily preserves orientation. $\endgroup$ Commented Dec 16, 2020 at 19:01

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