Research Article | Open Access

# Arcwise Connected Domains, Quasiconformal Mappings, and Quasidisks

**Academic Editor:**Zhi-Gang Wang

#### Abstract

We prove that a homeomorphism is a quasiconformal mapping if and only if is an arcwise connected domain for any arcwise connected domain , and is a quasidisk if and only if both and its exterior are arcwise connected domains.

#### 1. Introduction

We shall assume throughout this paper that is a Jordan proper subdomain of the complex plane with a boundary containing at least three points. For convenience we shall adopt the notation and terminology as in paper [1]. For and , let , be the closure of , , , and . Suppose that is a homeomorphism in , and , let , and .

Suppose that is a constant, we say that is a -arcwise connected domain if each pair of points and in can be joined by an arc with . Here is the Euclidean diameter of . And we call an arcwise connected domain if is a -arcwise connected domain for some .

Arcwise connected domain is an important concept; it has been used extensively in the research fields of superspaces [2], multiobjective programmings [3], color images [4], continuous mappings [5], fixed points theory [6], continuity and differential problems [7], and topology groups theory [8].

is called a quasidisk if there exists a -quasiconformal mapping () such that is the image of the unit disk under .

It is well-known that quasidisks play a very important role in quasiconformal mappings, complex dynamics, Fuchsian groups, and Teichmüller space theory [9–12].

The purpose of this paper is to prove the following Theorems 1 and 2.

Theorem 1. *A homeomorphism is a quasiconformal mapping if and only if is a arcwise connected domain for any arcwise connected domain .*

Theorem 2. * is a quasidisk if and only if both and are arcwise connected domains.*

For convenience we shall introduce the following concepts and they will be used in the next section.

Let be a constant. If for any and , points in can be joined by an arc in ; then we say that is a -inner linearly locally connected domain, denoted by -; If for any and , points in can be joined by an arc in ; then we say that is a -outer linearly locally connected domain, denoted by -.

is said to be a linearly locally connected domain if there exists such that - and -.

Let be a constant; then is said to be a -cigar domain if each pair of points can be joined by an arc for which where is the part of between and , and is the Euclidean distance from to .

We say that is a cigar domain if is a -cigar domain for some .

Gehring and Osgood proved the following result in [13].

Theorem A. * is a quasidisk if and only if is a linearly locally connected domain.*

#### 2. Proof of Theorems 1 and 2

To prove our Theorems 1 and 2, we first establish and introduce several lemmas.

Lemma 3. *If is an arcwise connected domain, then is a cigar domain.*

*Proof. *For any , , let be the hyperbolic geodesic joining and in , for any ; suppose that is a conformal mapping with and
where is the inverse of . Then there exist
by [14, Corollary 10.3] such that is rectifiable with
where is an absolute constant, and is the half open segment joining the origin and ; .

Let , , then and

For above , , there exist a constant and a simple curve joining and such that
by is an arcwise connected domain.

If we denote by the bounded domain with boundary , then one of the points and must be in . Without loss of generality, we may assume that ; then it follows from (5) and (6) that
Inequality (7) leads to

Therefore, is a cigar domain that follows from (8).

Lemma 4. *If is a -cigar domain, then -OLC.*

*Proof. *Let . If -, then there exist , and , , such that and can not be joined by any arc in .

Since is a -cigar domain, there exists an arc such that joining and with
for all .

It is obvious that . Let ; then (9) implies
But
Inequalities (10) and (11) together with the triangular inequality yield
Inequality (12) contradicts with . Hence -.

Lemma 5. *If is a -cigar domain, then -ILC.*

*Proof. *Let . For any , , and , . Denote and . We first prove that must be in the same component of .

If belong to different components of , then must be in the different components of . Let be the line segment which joins and ; then contains a subcurve such that divides into and , and , . This yields

For any , if and , then take

Since is a -cigar domain, there exists an arc joining and with
for all .

Let ; then
This implies
But
Inequalities (17) and (18) together with the triangular inequality yield
so
This is obviously impossible. Hence or , and we obtain
Inequalities (13) and (21) together with imply
This contradicts with . Hence , must be in the same component of , and there exists an arc joining and with

It follows from (23) that

Hence -; this completes the proof of Lemma 5.

Lemma 6. *If is a -cigar domain, then is a -arcwise connected domain.*

*Proof. *For any , , , let , , and . We shall prove that and can be joined by an arc in .

If and can not be joined by any arc in , then and must be in the different components of , and hence and must be in the different components of . There exist points , such that for any arc joining and .

Since is a -cigar domain, hence there exists an arc joining and such that
for all .

Let ; then
It follows from (26) and (27) that
which contradicts with . Hence and can be joined by an arc with
This shows that is a -arcwise connected domain.

Lemma 7. *If is a -quasiconformal mapping, and is a constant, then for any and , we have
**
where , is the inverse of , and is a constant which depends only on and .*

*Proof. *Let be the curve family which joins and in ; denotes the modulus of , . Then from the comparison principle of modulus and the result given in [1, 7.5] we get
According to the properties of -quasiconformal mapping in [1], we have
where is a constant which depends only on .

It follows from (31)–(33) that
The same reason to get (33) implies
It follows from (34) and (35) that

Lemma 8. *Suppose that is a -quasiconformal mapping. If is a -cigar domain for some , then is a -cigar domain. In here is a constant which depends only on and .*

*Proof. *For any points , , let , . Since is a -cigar domain, hence there exists an arc joining and such that for all . Let ; then is an arc joining and in . For any , let ; then with .

Without loss of generality, we assume that ; then and . From Lemma 7 and the fact we know that there exists constant which depends only on and such that . This implies . Therefore,
This shows that is a -cigar domain.

Lemma 9 (see [15]). *Let be a homeomorphism. If there exists a constant such that
**
for all and ; then is a quasiconformal mapping, where denotes the -dimensional measure of .*

Lemma 10. *Let be a homeomorphism. If maps any cigar domain onto a cigar domain , then is a quasiconformal mapping.*

*Proof. *For any and , choose such that

It is easy to see that is a 1-cigar domain if we take for any , where denotes the closed line segment joining and , . By the assumption of Lemma 10 we know that there exists a constant such that is a -cigar domain; hence there exists an arc joining and with
for all . If we choose such that , then (39) and (40) imply
This yields
From the above argument and Lemma 9 we know that is a quasiconformal mapping.

*Proof of Theorem 1. *Consider the following. *Necessity*. For any quasiconformal mapping and any arcwise connected domain , we know that is a cigar domain by Lemma 3. This and Lemma 8 imply that is a cigar domain, and then is an arcwise connected domain by Lemma 6.*Sufficiency*. To prove a homeomorphism is a quasiconformal mapping, making use of Lemma 10, we need only to prove that is a cigar domain for any cigar domain . In fact, for any cigar domain , Lemma 6 implies is an arcwise connected domain. From this and the condition of Theorem 1 we know that is an arcwise connected domain; then from Lemma 3 we know that is a cigar domain.

*Proof of Theorem 2. *Consider the following.*Necessity.* Let be a quasidisk; then there exists a quasiconformal mapping such that .(1)It is obvious that is a 1-arcwise connected domain; then from Theorem 1 we know that is an arcwise connected domain.(2)Since is a 1-cigar domain, hence is an arcwise connected domain; then Theorem 1 implies that is an arcwise connected domain.*Sufficiency*. Suppose that both and are arcwise connected domains.(1)Since is an arcwise connected domain, hence is a cigar domain by Lemma 3; then Lemma 5 implies that .(2)Since is an arcwise connected domain, hence is a cigar domain by Lemma 3; then Lemma 4 implies that .

From the above and together with Theorem A we know that is a quasidisk.

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants 61374086 and 11171307 and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004.

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#### Copyright

Copyright © 2014 Yu-Ming Chu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.