#### 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.