1 Introduction

Our idea of studying Q(G) was initiated by a paper of Gaier (1984, see [4]) about the conformal automorphism group

$$\begin{aligned} \Sigma (G):= \left\{ f \in Q(G) \ \big | \ f \text { is a conformal mapping of } G \text { onto } G \ \right\} , \end{aligned}$$

which was motivated by the question, whether the identity \({\text {id}}_G\) of the domain G can uniformly be approximated by conformal automorphisms of G. Naturally, the same question arises for quasiconformal automorphisms of G using thereby the results about \(\Sigma (G)\) as a subspace of Q(G).

Group–theoretic and topological information can be transferred between quasiconformal automorphism groups by conjugation:

Theorem 1

Let F be a conformal mapping from the unit disk \(\mathbb {D}\) onto G. Then the mapping

$$\begin{aligned} \Phi : Q(\mathbb {D}) \longrightarrow Q(G), \, f \longmapsto \Phi (f):= F \circ f \circ F^{-1} \end{aligned}$$

is a group isomorphism.

Furthermore, the mapping \(\Phi\) is continuous if and only if the domain G has prime ends only of the first kind.

Proof

We only have to show the statement about continuity of \(\Phi\). For this we use the known fact that G has prime ends only of the first kind if and only if the Riemann mapping F is uniformly continuous. First, assume G has prime ends only of the first kind. Then, analogously to the proof of Satz 2b) in Gaier (1984, [4]) we get also for \((f_n)_n\) in \(Q(\mathbb {D})\) converging uniformly on \(\mathbb {D}\) to \(f \in Q(\mathbb {D})\), \(h_n:= \Phi (f_n)\) and \(h:= \Phi (f)\)

$$\begin{aligned} \begin{aligned} d_{\sup }(h_n, h)&= \sup \limits _{z \in \mathbb {D}} \left| (F \circ f_n)(z) - (F \circ f)(z) \right| \\&\le \omega _F \left( \sup \limits _{z \in \mathbb {D}}\vert f_n(z) - f(z)\vert \right) = \omega _F \left( d_{\sup }(f_n, f) \right) \end{aligned} \end{aligned}$$
(1)

where \(\omega _F\) denotes the modulus of continuity of F. As F is uniformly continuous on \(\mathbb {D}\) this yields the continuity of \(\Phi\). Secondly, assume \(\Phi\) is continuous. Then, the assertion directly follows by Satz 2b) in Gaier (1984, [4]) since also the restriction of \(\Phi\) to \(\Sigma (\mathbb {D})\) with \(\Phi \left( \Sigma (\mathbb {D})\right) = \Sigma (G)\) is continuous. \(\square\)

Information on the theory of prime ends, originally due to Carathéodory, can be found e.g. in Pommerenke (1992, [9]).

As in the case of \(\Sigma (G)\) with Gaier (1984, cf. [4, Hilfssatz 4, p. 234]) also the space Q(G) becomes somehow topologically homogenous by the right multiplication, i.e. the operation \(f \longmapsto f \circ g\) for fixed \(g \in Q(G)\):

Proposition 2

The right multiplication in Q(G) is an isometry.

Gaier (1984) concluded that \(\Sigma (G)\) is a topological group if G has prime ends only of the first kind (cf. [4, p. 235, Satz 5a)]). In the quasiconformal setting, it turns out that a related statement remains valid as well:

Theorem 3

If G is a Jordan domain, then Q(G) is a topological group.

Proof

By the same arguments as Gaier (1984) used for \(\Sigma (G)\) (cf. [4, Satz 4]) we get the characterization that Q(G) is a topological group if and only if

$$\begin{aligned} \forall \, (\varphi _n)_{n \in \mathbb {N}} \subseteq Q(G) \, \forall \, f \in Q(G): \left( \varphi _n \overset{n \rightarrow \infty }{\longrightarrow }\ {\text {id}}_G \Longrightarrow f \circ \varphi _n \overset{n \rightarrow \infty }{\longrightarrow }\ f \right) . \end{aligned}$$
(2)

Now let \(\epsilon > 0, f \in Q(G)\) and \((\varphi _n)_{n \in \mathbb {N}} \subseteq Q(G)\) converge to \({\text {id}}_G\). The domain G has prime ends only of the first kind such that due to Theorem 1 there exists such a sequence.

Since G is a Jordan domain, f extends homeomorphically to \(\overline{G}\), and, by the compactness of \(\overline{G}\), f is uniformly continuous on \(\overline{G}\), i.e. there is a \(\delta > 0\) such that \(\left| f(z_1) - f(z_2) \right| < \epsilon\) for \(\vert z_1 - z_2\vert < \delta\) in G. For sufficiently large n we get \(\vert \varphi _n(z) - z\vert < \delta\) for all \(z \in G\) and therefore

$$\begin{aligned} d_{\sup }(f \circ \varphi _n, f) = \sup \limits _{z \in G} \vert f(\varphi _n(z)) - f(z)\vert \le \sup \limits _{z_{1,2}\in G, \vert z_1 - z_2 \vert < \delta } \vert f(z_1) - f(z_2) \vert \le \epsilon . \end{aligned}$$

\(\square\)

2 Incompleteness of Q(G)

Gaier (1984) showed that \(\Sigma (G)\) is always complete, regardless of the boundary structure of G (cf. [4, Satz 1b), p. 229]). However, in this section we will show that Q(G) is always incomplete even for domains with prime ends only of the first kind. For the proof we use a special family of quasiconformal mappings:

Definition 1

Let \(\rho \in C([0,1])\) be a strictly increasing function of the interval [0, 1] onto itself. Then the mapping

$$\begin{aligned} f_\rho : \mathbb {D} \longrightarrow \mathbb {D}, \; z \longmapsto f_\rho (z) := \frac{z}{\vert z \vert }\rho ( \vert z \vert ),\ f(0) := 0, \end{aligned}$$
(3)

is called a (general) radial stretching of \(\mathbb {D}\).

Obviously, the mapping \(\rho\) provides the deviation of \(f_\rho\) from being conformal as shown by the following result (to be found in Astala et. al. 2008, cf. [1, Section 2.6, pp. 28–29]):

Lemma 4

For each piecewise \(C^1\)–mapping \(\rho \in C([0,1])\) as in Definition 1 the corresponding radial stretching \(f_\rho\) belongs to \(Q(\mathbb {D})\) with complex dilatation

$$\begin{aligned} \mu _{f_\rho }(z) = \frac{z}{\overline{z}} \cdot \frac{\vert z \vert \rho '(\vert z \vert ) - \rho (\vert z \vert )}{ \vert z \vert \rho '( \vert z \vert ) + \rho ( \vert z \vert )} \end{aligned}$$

for almost every \(z \in \mathbb {D}\), where \(\rho '\) denotes the (a.e. existing) derivative of \(\rho\).

The radial stretching visualized in Fig. 1 is build with

$$\begin{aligned} \rho (x) = {\left\{ \begin{array}{ll} \sqrt{\frac{x}{2}}, &{} x \in [0, \frac{1}{2}) \\ 2(x - \frac{1}{2})^2 + \frac{1}{2}, &{} x \in [\frac{1}{2}, 1] \end{array}\right. }. \end{aligned}$$
Fig. 1
figure 1

Radial stretching on a cartesian grid in \(\mathbb {D}\)

Full size image

Theorem 5

The space Q(G) is always incomplete.

Proof

First, the case \(G = \mathbb {D}\) will be treated. Consider the sequence of strictly increasing and continuous piecewise linear self mappings of [0, 1] defined by

$$\begin{aligned} \rho _n(x) := {\left\{ \begin{array}{ll} 2x, &{} x \in [0, \frac{1}{4}], \\ \frac{1}{2} \left( \frac{x}{n} + 1 - \frac{1}{4n} \right) , &{} x \in (\frac{1}{4}, \frac{3}{4}], \\ \left( 2 - \frac{1}{n} \right) x - 1 + \frac{1}{n}, &{} x \in (\frac{3}{4}, 1], \end{array}\right. } n \in \mathbb {N}, \end{aligned}$$
(4)

which converges uniformly on [0, 1] to the non–injective limit function

$$\begin{aligned} \widetilde{\rho }(x):= {\left\{ \begin{array}{ll} 2x, &{} x \in [0, \frac{1}{4}], \\ \frac{1}{2}, &{} x \in (\frac{1}{4}, \frac{3}{4}], \\ 2x - 1, &{} x \in (\frac{3}{4}, 1]. \end{array}\right. } \end{aligned}$$

The corresponding sequence of radial stretchings \(\left( f_{\rho _n}\right) _{n \in \mathbb {N}}\) belongs to \(Q (\mathbb {D})\) because of Lemma 4 and due to the uniform convergence of \((\rho _n)_{n \in \mathbb {N}}\) we get

$$\begin{aligned} d_{\sup }(f_{\rho _n}, f_{\widetilde{\rho }}) = \sup \limits _{z \in \mathbb {D}} \left| f_{\rho _n}(z) - f_{\widetilde{\rho }}(z) \right| = \sup \limits _{x \in [0,1)} \left| \rho _n(x) - \widetilde{\rho }(x) \right| \overset{n \rightarrow \infty }{\longrightarrow }\ 0 \end{aligned}$$

where analogously \(f_{\widetilde{\rho }}(z):= \widetilde{\rho }(\vert z \vert ) \frac{z}{\vert z \vert }\) with \(f_{\widetilde{\rho }}(0):= 0\). However, the limit function \(f_{\widetilde{\rho }}\) is not injective and therefore does not belong to \(Q(\mathbb {D})\). This shows the incompleteness of the metric space \(Q(\mathbb {D})\). For a general domain G, let \(z_0 \in G\) be a fixed inner point and let \(B \subsetneqq G\) be an open ball centered at \(z_0\) in G. Clearly the sequence \((f_{\rho _n})_n\) of the proof’s first part together with its limit function \(f_{\widetilde{\rho }}\) can be transferred to B via conformal equivalence, denoted by \(g_n\) and \(g_{\rho }\), respectively. Then, define quasiconformal automorphisms \(h_n\) of G by

$$\begin{aligned} h_n(z) = {\left\{ \begin{array}{ll} g_n(z), &{} z \in B \\ {\text {id}}_G(z), &{} z \in G \backslash B \end{array}\right. } \end{aligned}$$

and likewise for \(h_{\rho }\) and \(g_{\rho }\). Now, the sequence \(\left( h_n \right) _{n\in \mathbb {N}}\) converges uniformly to the non-injective limit function \(h_{\rho }\) on G. \(\square\)

3 Separability of Q(G)

In [11] Volynec (1992) shows that in case of the space of conformal automorphisms the topological property of separability geometrically means that the boundary cannot be complicated in the sense of prime end theory, i.e. G has prime ends only of the first kind. This equivalence can be extended to the space of quasiconformal automorphisms:

Theorem 6

The space Q(G) is separable if and only if G has prime ends only of the first kind.

Proof

If the metric space Q(G) is separable, then, in particular, the subspace \(\Sigma (G)\) is separable, which is equivalent to G having prime ends only of the first kind by Volynec (1992) [11, Theorem 3, p. 201]. For the other direction, we first consider the case \(G = \mathbb {D}\). Each mapping \(f \in Q(\mathbb {D})\) extends homeomorphically to \(\overline{\mathbb {D}}\), see e.g. Lehto (1987, cf. [6, p. 13]), such that \(Q(\mathbb {D})\) is a subspace of the metric space \(C_b(\overline{\mathbb {D}})\) of bounded continuous complex–valued functions on \(\overline{\mathbb {D}}\). It is known from topology (see e.g. Conway (1990), cf. [3, Theorem 6.6, p. 140]) that \(C_b(\overline{\mathbb {D}})\) is separable. This implies the separability of \(Q(\mathbb {D})\). Finally, in the general case for a domain G with prime ends only of the first kind, the separability of Q(G) follows from Theorem 1, since Q(G) is the continuous image of the separable space \(Q(\mathbb {D})\) by \(\Phi\). \(\square\)

Naturally, this result raises the question for concrete countable and dense subsets of Q(G) in the case that G has prime ends only of the first kind, and how such automorphisms may be constructed, especially in the case of \(Q(\mathbb {D})\).

4 Path–connectedness of Q(G)

Dealing with the topological question about the decomposition of Q(G) into path–components we first encounter the result of Schmieder (1986, cf. [10]) that the subspace \(\Sigma (G)\) is path–connected if and only if G has prime ends only of the first kind. For the whole space Q(G) we cannot show the same equivalence, but at least that also the whole space Q(G) consists of only one path–component in case of G having a simple boundary. In order to prove the corresponding Theorem 10 below we need some preparation.

At first, for the construction of paths in Q(G) it is natural to use liftings of paths from \(\Sigma (G)\):

Lemma 7

Let G be a domain having prime ends only of the first kind. Then for every \(f \in Q(G)\) and every \(\sigma \in \Sigma (G)\), the mappings f and \(\sigma \circ f\) can be joined by a path in Q(G).

Proof

By Schmieder (1986, cf. [10, Corollary, p. 199]), \(\Sigma (G)\) is path–connected, such that the identity \({\text {id}}_G\) and \(\sigma\) can be joined by a path in \(\Sigma (G)\), say \(\gamma : [0,1] \longrightarrow \Sigma (G)\). Thus the mapping

$$\begin{aligned} \widetilde{\gamma }: [0,1] \longrightarrow Q(G), \, t \longmapsto \widetilde{\gamma }(t):= \gamma (t) \circ f \end{aligned}$$

is continuous due to the isometric right multiplication in Q(G) (cf. Proposition 2) and clearly joins f and \(\sigma \circ f\). \(\square\)

Furthermore, the composition with conformal automorphisms will also be used to construct somehow normalized paths in \(Q(\mathbb {D})\), see Branner and Fagella (2014, cf. [2, pp. 42–43]).

Definition 2

Let \(G \subsetneq \mathbb {C}\) be a simply connected domain and \(z_1, z_2 \in G\) distinct points. A quasiconformal mapping \(f: G \longrightarrow \mathbb {D}\) is called normalized w.r.t. \(z_1, z_2\) if \(f(z_1) = 0\) and \(f(z_2) > 0\).

Obviously, by the Measurable Riemann Mapping Theorem for each Beltrami coefficient \(\mu \in \mathbb {B}_{L^\infty }(G)\), i.e. open unit ball of \(L^\infty (G)\), there exists a unique normalized quasiconformal mapping \(f: G \longrightarrow \mathbb {D}\) with \(\mu _f = \mu\) almost everywhere in G. The main idea for proving the Theorem 10 below is to build intermediate paths in \(\mathbb {B}_{L^\infty }(G)\) and then to conclude from pointwise on uniform convergence for the corresponding normalized quasiconformal mappings. Therefore, we finally need

Proposition 8

(Branner and Fagella (2014) [2, Theorem 1.30(b), p. 43]) Let \(\Lambda\) be an open subset of \(\mathbb {R}\), \(G \subsetneq \mathbb {C}\) be a Jordan domain, \(z_1, z_2 \in G\) distinct points and \((\mu _t)_{t \in \Lambda }\) be a family in \(\mathbb {B}_{L^\infty }(G)\). Suppose \(t \longmapsto \mu _t(z)\) is continuous for every fixed \(z \in G\) (whenever defined). Moreover, assume there exists \(k < 1\) such that \(\Vert \mu _t \Vert _{L^\infty (G)} \le k\) for all \(t \in \Lambda\), and denote by \(f_t: G \longrightarrow \mathbb {D}\) the quasiconformal mapping normalized w.r.t \(z_1, z_2\) and satisfying \(\mu _{f_t} = \mu _t\) almost everywhere in G. Then \(t \longmapsto f_t(z)\) is continuous for every fixed \(z \in G\).

and

Proposition 9

(Näkki and Palka (1973) [8, Corollary 4.4, p. 432]) Let \(G' \subseteq \mathbb {C}\) be a domain with finitely many boundary components which is finitely connected on the boundary. Furthermore, let \((f_n)_{n \in \mathbb {N}}\) be a sequence of K–quasiconformal mappings of a domain G onto \(G'\) converging pointwise in G to a homeomorphism f. Then the sequence \((f_n)_n\) converges uniformly on G to f.

By means of the results above we now can prove our theorem about path–connectedness in Q(G).

Theorem 10

The space Q(G) is path–connected if the domain G has prime ends only of the first kind.

Proof

First, the case \(G = \mathbb {D}\) will be considered.

Let \(f, g \in Q(\mathbb {D})\) with complex dilatations \(\mu _f, \mu _g \in \mathbb {B}_{L^\infty }(\mathbb {D})\). W.l.o.g, assume that \(\mu _f \not = \mu _g\); otherwise, \(g = \sigma \circ f\) for some \(\sigma \in \Sigma (\mathbb {D})\) by the Measurable Riemann Mapping Theorem and f and g can be joined by a path in \(Q(\mathbb {D})\) by Lemma 7. For an arbitrary, but fixed \(z^* \in \mathbb {D}\), we find \(\sigma _f, \sigma _g \in \Sigma (\mathbb {D})\) such that \((\sigma _f \circ f)(0) = 0 = (\sigma _g \circ g)(0)\) and \((\sigma _f \circ f)(z^*) > 0\) and \((\sigma _g \circ g)(z^*) > 0\) with \(\mu _{\sigma _f \circ f} = \mu _f\) resp. \(\mu _{\sigma _g \circ g} = \mu _g\) almost everywhere in \(\mathbb {D}\). Now Lemma 7 provides paths \(\gamma _f, \gamma _g: [0,1] \longrightarrow Q(\mathbb {D})\) joining f with \(\sigma _f \circ f\) and g with \(\sigma _g \circ g\).

Next, let \(\Lambda := (-a, 1+a)\) for some fixed \(a > 0\) and using the straight line

$$\begin{aligned} \gamma : [0,1] \longrightarrow \mathbb {B}_{L^\infty }(\mathbb {D}), \; t \longmapsto \gamma (t):= t \mu _g + (1-t) \mu _f, \end{aligned}$$

we construct a new path \(\Gamma\) in \(\mathbb {B}_{L^\infty }(\mathbb {D})\) by

$$\begin{aligned} \Gamma : \Lambda \longrightarrow \mathbb {B}_{L^\infty }(\mathbb {D}), \; t \longmapsto \mu _t:= {\left\{ \begin{array}{ll} \mu _f, &{} t \in (-a, 0) \\ \gamma (t), &{} t \in [0,1] \\ \mu _g, &{} t \in (1, 1 + a) \end{array}\right. } \end{aligned}$$

which satifies \(\sup _{t \in \Lambda } \Vert \mu _t \Vert _{L^\infty (\mathbb {D})} < 1\). For each \(t \in \Lambda\) there exists a unique normalized (w.r.t. \(0, z^\star\)) \(\phi _t \in Q(\mathbb {D})\) with \(\mu _{\phi _t} = \mu _t\) a. e. in \(\mathbb {D}\) and \(\phi _t = \sigma _f \circ f\) for \(t \in (-a, 0)\) and \(\phi _t = \sigma _g \circ g\) for \(t \in (1, 1+a)\). Obviously, by the definition of \(\Gamma\) the mapping \(t\longmapsto \mu _t(z)\) is continuous for each \(z\in \mathbb {D}\) such that the same is true for \(t\longmapsto \phi _t(z)\) by Proposition 8.

Due to the uniform boundedness of \(\Vert \mu _t \Vert _{L^\infty (\mathbb {D})}\) for all \(t \in \Lambda\) the whole family \((\phi _t)_{t \in \Lambda } \subseteq Q(\mathbb {D})\) is K–quasiconformal for a suitable K. Then, by Proposition 9, pointwise convergent sequences \(\left( \phi _{t_n}\right) _n\) are even uniformly convergent on \(\mathbb {D}\) such that the induced mapping \(H: \Lambda \longrightarrow Q(\mathbb {D}), \; t \longmapsto \phi _t\) is continuous with respect to the topology of uniform convergence on \(Q(\mathbb {D})\).

Eventually, appropriately combining the paths \(\gamma _f\), \(\gamma _g\) and H yields a path form f to g. Hence \(Q(\mathbb {D})\) is path–connected.

Finally, for an arbitrary domain G having prime ends only of the first kind, the bijective mapping \(\Phi : Q(\mathbb {D}) \longrightarrow Q(G)\) is continuous by Theorem 1. Thus Q(G) is path–connected as well. \(\square\)

Naturally, the question for the corresponding counterpart rises: Does G necessarily have prime ends only of the first kind if the space Q(G) is path–connected? If this question should happen to be answered negatively, which domain with at least one prime end not of the first kind could have a path–connected space of quasiconformal automorphisms?

5 Discreteness of Q(G)

For the space of conformal automorphisms \(\Sigma (G)\) in [4] Gaier (1984) gives examples for both discrete and non-discrete spaces which intimately depend on the structure of the boundary of the underlying domain G.

For the much larger space Q(G), however, we get that the subgroup

$$\begin{aligned} Q^\infty (G) := \left\{ f \in Q(G) \left| \; f \text { is a } C^\infty \text {--diffeomorphism} \right. \right\} \end{aligned}$$
(5)

is dense in Q(G) from the following result of Kiikka (1983, cf. [5, Theorem 1, p. 252]).

Proposition 11

Let \(G, G' \subseteq \mathbb {C}\) be domains, let \(f: G \longrightarrow G'\) be a K–quasiconformal mapping and \(\epsilon > 0\). Then there exists \(\widetilde{K} \ge 1\) and a \(\widetilde{K}\)–quasiconformal \(C^\infty\)–diffeomorphism \(\widetilde{f}: G \longrightarrow G'\) such that \(\left| f(z) - \widetilde{f}(z) \right| < \epsilon\) for every \(z \in G\).

This implies now

Theorem 12

The space Q(G) does not have any isolated element.

Proof

Choose \(f \in Q(G)\setminus Q^\infty (G)\), e.g. an appropriate radial stretching.

Using Proposition 11 we can find a sequence \((f_n)_{n \in \mathbb {N}} \subseteq Q^\infty (G)\) with pairwise distinct elements \(\ne f\) converging to f uniformly on G. Then, for each \(g\in Q(G)\) by the isometry of right multiplication in Q(G) (cf. Proposition 2) the sequence \((f_n \circ f^{-1} \circ g)_n \subseteq Q(G)\) converges to g without meeting g. \(\square\)

Naturally, the question raises how to construct explicit quasiconformal automorphisms uniformly approximating \({\text {id}}_G\), especially in the case of G having at least one prime end not of the first kind, since then \(\Sigma (G)\) might be discrete following Gaier (1984, cf. [4, Satz 9, p. 254]).

6 Compactness in Q(G)

For a subset \(\emptyset \ne M \subseteq Q(G)\) let \(K(M):= \sup \limits _{f \in M} K(f) \in [1, + \infty ]\) denote the maximal dilatation of M.

Example 1

In case of \(Q(\mathbb {D})\) the special family of radial stretchings

$$\begin{aligned} M = \left\{ f \in Q(\mathbb {D})\ \left| \ f(z) = z \vert z \vert ^{K-1}\ \text { with }\ K \in [1, + \infty ] \right. \right\} \end{aligned}$$

can homeomorphically be mapped onto \([1, + \infty )\) in the evident way. Therefore, if K is restricted to a compact set, also the corresponding subset of M will be compact in \(Q(\mathbb {D})\).

This example suggests the conjecture that the maximal dilatation of a compact subset of Q(G) is necessarily uniformly bounded from above. However, Lehto and Virtanen (1973, cf. [7]) construct an example for non–good approximation of quasiconformal mappings, which can be used as a counterexample for this conjecture:

Example 2

Following Lehto and Virtanen (1973, cf. [7, p. 186]), in the open unit square \(R:= (0,1) \times (0,1)\) for each \(m\in \mathbb {N}\) there exists a sequence \(\left( h_n^m\right) _{n\in \mathbb {N}}\) in Q(R), which uniformly converges to \({\text {id}}_R\), whereas the corresponding complex dilatations satisfy \(\left| \mu _n^m\right| \equiv 1-\frac{1}{m}\) a.e. for all \(n\in \mathbb {N}\) resp. \(K(h_n^m)\ge 2m-1\).

Then, by an usual diagonal argument we can find a sequence \(\left( h^m_{n_m}\right) _{m\in \mathbb {N}}\) also converging to \({\text {id}}_R\) but with maximal dilatation \(K\left( h^m_{n_m}\right)\) tending to \(+\infty\).

Therefore, the sequence \(\left( h^m_{n_m}\right) _{m\in \mathbb {D}}\) together with \({\text {id}}_R\) forms a compact set with unbounded maximal dilatation.

For compact subgroups, e.g. the group of inner rotations, we can affirm this conjecture:

Theorem 13

Let \(G \subsetneq \mathbb {C}\) be a bounded Jordan domain and \(H \subseteq Q(G)\) be a compact subgroup.

Then \(K(H) < + \infty\).

Proof

The set H can be written as countable union of closed sets, i.e.

$$\begin{aligned} H = \bigcup \limits _{n = 1}^\infty H_n\ \text { with }\ H_n = \{ f \in H \; \big | \; K(f) \le n \} \ \text { for }\ n \in \mathbb {N}, \end{aligned}$$

because for a sequence of functions from \(H_n\) converging in H the limit function must also belong to \(H_n\) by the Hurwitz–type theorem for K–quasiconformal mappings, to be found e.g. in Lehto (1987, cf. [6, Theorem 2.2, p. 15]). Furthermore, the compact set H is a complete metric space, so that by the Baire Category Theorem there exists an index \(N \in \mathbb {N}\) such that the subset \(H_N\) has non–empty interior in the subspace topology of H. This means that there exists \(\epsilon > 0\) and \(\widehat{f} \in H_N\) such that for the open ball \(U:= B_\epsilon (\widehat{f}) \subset Q(G)\) we have

$$\begin{aligned} \widehat{f} \in U \cap H \subset H_N. \end{aligned}$$

Now, for \(g \in H\), consider the left multiplication in Q(G) with \(g \circ \widehat{f}^{-1}\), i.e.

$$\begin{aligned} L_g: Q(G) \longrightarrow Q(G), \; h \longmapsto L_g(h):= g \circ \widehat{f}^{-1} \circ h \end{aligned}$$

Furthermore, \(L_g\) is a homeomorphism of Q(G) onto itself, since Q(G) is a topological group by Theorem 3. This implies that the image \(L_g(U)\) with \(L_g(\widehat{f}) = g \in L_g(U)\) is open in Q(G).

Then, the family of open sets \(\left( L_g(U) \cap H \right) _{g \in H}\) is an open cover of the compact space H, which yields a finite subcover

$$\begin{aligned} H \subset \bigcup \limits _{j = 1}^m \left( L_{g_j}(U) \cap H \right) = \bigcup \limits _{j = 1}^m L_{g_j}(U \cap H). \end{aligned}$$

Hence, every \(f \in H\) can be written as \(f = g_j \circ \widehat{f}^{-1} \circ h\) for some \(j\in \{1,\ldots ,m\}\) and \(h \in U \cap H \subseteq H_N\) implying

$$\begin{aligned} K(f) \le K(g_j) \cdot K(\widehat{f}^{-1}) \cdot K(h)&\le \left( \max \limits _{j = 1, \dots , m} K(g_j) \right) \cdot K(\widehat{f}) \cdot N \\&\le \left( \max \limits _{j = 1, \dots , m} K(g_j) \right) \cdot N^2 < + \infty . \end{aligned}$$

This gives an upper bound for the maximal dilatation of the whole subgroup H. \(\square\)

As with the previously established results in this paper, the question raises whether the necessary compactness criterion of Theorem 13 is also valid for domains with a more complicated boundary.