# Arcwise Connected Domains, Quasiconformal Mappings, and Quasidisks

Authors: Yu-Ming Chu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We prove that a homeomorphismf
:
R
2
→
R
2 is a quasiconformal mapping if and only if f
(
D
) is an arcwise connected domain for any arcwise connected domain D
⊆
R
2, and D is a quasidisk if and only if both D and its exterior D
*
=
R
2
∖
D
¯ are arcwise connected domains.


---

## Body

## 1. Introduction

We shall assume throughout this paper thatD is a Jordan proper subdomain of the complex plane R
2 with a boundary containing at least three points. For convenience we shall adopt the notation and terminology as in paper [1]. For x
∈
R
2 and 0
<
r
<
∞, let B
2
(
x
,
r
)
=
{
z
∈
R
2
:
|
z
-
x
|
<
r
}, B
¯
2
(
x
,
r
) be the closure of B
2
(
x
,
r
), S
1
(
x
,
r
)
=
∂
B
2
(
x
,
r
), B
2
(
r
)
=
B
2
(
0
,
r
), and B
2
=
B
2
(
1
). Suppose that f is a homeomorphism in R
2, x
∈
R
2 and 0
<
r
<
∞, let  L
(
x
,
f
,
r
)
=
max
⁡
|
y
-
x
|
=
r
|
f
(
y
)
-
f
(
x
)
|, and l
(
x
,
f
,
r
)
=
min
⁡
|
y
-
x
|
=
r
|
f
(
y
)
-
f
(
x
)
|.Suppose thatb
≥
1 is a constant, we say that D is a b-arcwise connected domain if each pair of points x and y in D can be joined by an arc γ
⊆
D with dia
(
γ
)
≤
b
|
x
-
y
|. Here dia
(
γ
) is the Euclidean diameter of γ. And we call D an arcwise connected domain if D is a b-arcwise connected domain for some b
≥
1.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].D is called a quasidisk if there exists a K-quasiconformal mapping (K
≥
1) f
:
R
2
→
R
2 such that D is the image of the unit disk B
2 under f.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 Theorems1 and 2.Theorem 1.
A homeomorphismf
:
R
2
→
R
2 is a quasiconformal mapping if and only if f
(
D
) is a arcwise connected domain for any arcwise connected domain D
⊆
R
2.Theorem 2.
D is a quasidisk if and only if both D and D
*
=
R
2
∖
D
¯ are arcwise connected domains.For convenience we shall introduce the following concepts and they will be used in the next section.Letc
≥
1 be a constant. (
1
) If for any x
0
∈
R
2 and 0
<
r
<
+
∞, points in D
∩
B
¯
(
x
0
,
r
) can be joined by an arc in D
∩
B
¯
(
x
0
,
c
r
); then we say that D is a c-inner linearly locally connected domain, denoted by D
∈
c-ILC; (
2
) If for any x
0
∈
R
2 and 0
<
r
<
+
∞, points in D
∖
B
2
(
x
0
,
r
) can be joined by an arc in D
∖
B
2
(
x
0
,
r
/
c
); then we say that D is a c-outer linearly locally connected domain, denoted by D
∈
c-OLC.D is said to be a linearly locally connected domain if there exists c
≥
1 such that D
∈
c-ILC and D
∈
c-OLC.Letb
>
0 be a constant; then D is said to be a b-cigar domain if each pair of points x
1
,
x
2
∈
D can be joined by an arc γ
⊆
D for which
(1)
min
⁡
j
=
1,2
dia
(
γ
(
x
j
,
x
)
)
≤
b
d
(
x
,
∂
D
)
∀
x
∈
γ
,
where γ
(
x
j
,
x
) is the part of γ between x
j and x, and d
(
x
,
∂
D
) is the Euclidean distance from x to ∂
D.We say thatD is a cigar domain if D is a b-cigar domain for some b
>
0.Gehring and Osgood proved the following result in [13].Theorem A.
D is a quasidisk if and only if D is a linearly locally connected domain.


## 2. Proof of Theorems1 and 2

To prove our Theorems1 and 2, we first establish and introduce several lemmas.Lemma 3.
IfD is an arcwise connected domain, then D
*
=
R
2
∖
D
¯ is a cigar domain.Proof.
For anyz
1, z
2
∈
D
*, let γ be the hyperbolic geodesic joining z
1 and z
2 in D
*, for any z
∈
γ
∖
{
z
1
,
z
2
}; suppose that f
:
B
2
→
D
* is a conformal mapping with f
(
0
)
=
z and
(2)
f
-
1
(
γ
)
⊆
R
=
{
z
:
z
∈
R
2
,
Im
⁡
z
=
0
}
,
where f
-
1 is the inverse of f. Then there exist
(3)
x
1
∈
{
z
:
z
∈
S
1
,
Im
⁡
z
>
0
}
,
x
2
∈
{
z
:
z
∈
S
1
,
Im
⁡
z
<
0
}
by [14, Corollary 10.3] such that α
j
=
f
(
[
0
,
x
j
)
) is rectifiable with
(4)
l
(
α
j
)
<
a
0
d
(
z
,
∂
D
*
)
,
j
=
1,2
,
where a
0 is an absolute constant, and [
0
,
x
j
) is the half open segment joining the origin O and x
j; j
=
1,2.
Lety
j
=
f
(
x
j
), α
=
α
1
∪
α
2, then y
j
∈
∂
D and
(5)
l
(
α
)
≤
l
(
α
1
)
+
l
(
α
2
)
<
2
a
0
d
(
z
,
∂
D
*
)
.
For abovey
1, y
2
∈
∂
D
*
=
∂
D, there exist a constant b
≥
1 and a simple curve β
⊆
D joining y
1 and y
2 such that
(6)
dia
(
β
)
≤
b
|
y
1
-
y
2
|
≤
b
l
(
α
)
by D is an arcwise connected domain.
If we denote byD
0 the bounded domain with boundary α
∪
β, then one of the points z
1 and z
2 must be in D
0. Without loss of generality, we may assume that z
1
∈
D
0; then it follows from (5) and (6) that
(7)
dia
(
γ
(
z
1
,
z
)
)
≤
dia
(
D
0
)
=
dia
(
∂
D
0
)
iiiiiiiiiiiiiiiiiii
≤
dia
(
α
)
+
dia
(
β
)
iiiiiiiiiiiiiiiiiii
≤
l
(
α
)
+
dia
(
β
)
iiiiiiiiiiiiiiiiiii
<
2
a
0
(
b
+
1
)
d
(
z
,
∂
D
*
)
.
Inequality (7) leads to
(8)
min
⁡
j
=
1,2
dia
(
γ
(
z
j
,
z
)
)
<
2
a
0
(
b
+
1
)
d
(
z
,
∂
D
*
)
.
Therefore,D
* is a cigar domain that follows from (8).Lemma 4.
IfD is a c-cigar domain, then D
∈
(
2
c
+
2
)-OLC.Proof.
Letb
=
2
c
+
2. If D
∉
b-OLC, then there exist y
0
∈
R
2, 0
<
r
<
∞ and x
1, x
2
∈
D
∖
B
2
(
y
0
,
r
), such that x
1 and x
2 can not be joined by any arc in D
∖
B
2
(
y
0
,
r
/
b
).
SinceD is a c-cigar domain, there exists an arc γ
⊂
D such that γ joining x
1 and x
2 with
(9)
min
⁡
j
=
1,2
dia
(
γ
(
x
j
,
x
)
)
≤
c
d
(
x
,
∂
D
)
for all x
∈
γ.
It is obvious thatγ
∩
S
1
(
y
0
,
r
/
b
)
≠
∅. Let y
∈
γ
∩
S
1
(
y
0
,
r
/
b
); then (9) implies
(10)
d
(
y
,
∂
D
)
≥
1
c
min
⁡
j
=
1,2
dia
(
γ
(
x
j
,
y
)
)
≥
1
c
min
⁡
j
=
1,2
|
x
j
-
y
|
≥
1
c
(
1
-
1
b
)
r
.
But
(11)
d
(
y
0
,
∂
D
)
≤
r
b
.
Inequalities (10) and (11) together with the triangular inequality yield
(12)
1
c
(
1
-
1
b
)
r
≤
d
(
y
,
∂
D
)
≤
|
y
-
y
0
|
+
d
(
y
0
,
∂
D
)
≤
2
b
r
,
b
≤
2
c
+
1
.
Inequality (12) contradicts with b
=
2
c
+
2. Hence D
∈
(
2
c
+
2
)-OLC.Lemma 5.
IfD
* is a c
0-cigar domain, then D
∈
(
16
c
0
+
21
)-ILC.Proof.
Letδ
=
8
c
0
+
10. For any u
∈
R
2, s
>
0, and z
1
,
z
2
∈
D
∩
B
¯
2
(
u
,
s
), z
1
≠
z
2. Denote z
=
(
1
/
2
)
(
z
1
+
z
2
) and r
=
|
z
1
-
z
2
|. We first prove that z
1
,
z
2 must be in the same component of B
¯
2
(
z
,
(
1
/
2
)
δ
r
)
∖
D
*.
Ifz
1
,
z
2 belong to different components of B
¯
2
(
z
,
(
1
/
2
)
δ
r
)
∖
D
*, then z
1
,
z
2 must be in the different components of B
¯
2
(
z
,
(
1
/
2
)
r
)
∖
D
*. Let β be the line segment which joins z
1 and z
2; then β contains a subcurve α
⊆
D
* such that α divides D
* into D
1 and D
2, and dia
(
D
j
)
≥
(
1
/
2
)
r
(
δ
-
1
), j
=
1,2. This yields
(13)
min
⁡
j
=
1,2
dia
(
D
j
)
≥
1
2
r
(
δ
-
1
)
.
For anyx
∈
α, if D
1
⊈
B
2
(
x
,
(
2
c
0
+
2
)
dia
(
α
)
) and D
2
⊈
B
2
(
x
,
(
2
c
0
+
2
)
dia
(
α
)
), then take
(14)
x
j
∈
D
j
∖
B
¯
2
(
x
,
(
2
c
0
+
2
)
dia
(
α
)
)
,
j
=
1,2
.
SinceD
* is a c
0-cigar domain, there exists an arc γ
⊆
D
* joining x
1 and x
2 with
(15)
min
⁡
j
=
1,2
dia
(
γ
(
x
j
,
w
)
)
≤
c
0
d
(
w
,
∂
D
*
)
for all w
∈
γ.
Lety
∈
γ
∩
S
1
(
x
,
dia
(
α
)
); then
(16)
|
y
-
x
j
|
≥
(
2
c
0
+
1
)
dia
(
α
)
,
min
⁡
j
=
1,2
dia
(
γ
(
x
j
,
y
)
)
≤
c
0
d
(
y
,
∂
D
*
)
,
mmmmmmmmmmm
j
=
1,2
.
This implies
(17)
d
(
y
,
∂
D
*
)
≥
2
c
0
+
1
c
0
dia
(
α
)
.
But
(18)
d
(
x
,
∂
D
*
)
≤
dia
(
α
)
.
Inequalities (17) and (18) together with the triangular inequality yield
(19)
2
c
0
+
1
c
0
dia
(
α
)
≤
d
(
y
,
∂
D
*
)
≤
|
y
-
x
|
+
d
(
x
,
∂
D
*
)
iiiiiiiiiiiiiiiiiiiiii
≤
2
dia
(
α
)
,
so
(20)
dia
(
α
)
≤
0
.
This is obviously impossible. Hence D
1
⊆
B
¯
2
(
x
,
(
2
c
0
+
2
)
dia
(
α
)
) or D
2
⊆
B
¯
2
(
x
,
(
2
c
0
+
2
)
dia
(
α
)
), and we obtain
(21)
min
⁡
j
=
1,2
dia
(
D
j
)
≤
2
(
2
c
0
+
2
)
dia
(
α
)
.
Inequalities (13) and (21) together with dia
(
α
)
≤
r imply
(22)
δ
≤
8
c
0
+
9
.
This contradicts with δ
=
8
c
0
+
10. Hence z
1, z
2 must be in the same component of B
¯
2
(
z
,
(
1
/
2
)
δ
r
)
∖
D
*, and there exists an arc γ
⊆
D joining z
1 and z
2 with
(23)
dia
(
γ
)
≤
δ
r
=
δ
|
z
1
-
z
2
|
≤
2
δ
s
.
It follows from (23) that
(24)
γ
⊆
D
∩
B
¯
2
(
u
,
s
+
dia
(
γ
)
)
⊆
D
∩
B
¯
2
(
u
,
(
2
δ
+
1
)
s
)
=
D
∩
B
¯
2
(
u
,
(
16
c
0
+
21
)
s
)
.
HenceD
∈
(
16
c
0
+
21
)-ILC; this completes the proof of Lemma 5.Lemma 6.
IfD is a b-cigar domain, then D
* is a (
4
b
+
3
)-arcwise connected domain.Proof.
For anyz
1, z
2
∈
D
*, z
1
≠
z
2, let z
0
=
(
z
1
+
z
2
)
/
2, r
=
|
z
1
-
z
2
|
/
2, and d
=
4
b
+
3. We shall prove that z
1 and z
2 can be joined by an arc in D
*
∩
B
¯
2
(
z
0
,
d
r
).
Ifz
1 and z
2 can not be joined by any arc in D
*
∩
B
¯
2
(
z
0
,
d
r
), then z
1 and z
2 must be in the different components of D
*
∩
B
¯
2
(
z
0
,
d
r
), and hence z
1 and z
2 must be in the different components of D
*
∩
B
¯
2
(
z
0
,
(
d
/
2
)
r
). There exist points y
1, y
2
∈
∂
D
*
∖
B
¯
2
(
z
0
,
(
d
/
2
)
r
) such that β
∩
B
2
(
z
0
,
r
)
≠
∅ for any arc β
⊆
D joining y
1 and y
2.
SinceD is a b-cigar domain, hence there exists an arc γ
⊆
D joining y
1 and y
2 such that
(25)
min
⁡
j
=
1,2
dia
(
γ
(
y
j
,
y
)
)
≤
b
d
(
y
,
∂
D
)
for all y
∈
γ.
Lety
0
∈
γ
∩
B
2
(
z
0
,
r
); then
(26)
min
⁡
j
=
1,2
dia
(
γ
(
y
j
,
y
0
)
)
≤
b
d
(
y
0
,
∂
D
)
,
d
(
y
0
,
∂
D
)
≤
|
y
0
-
z
j
|
≤
2
r
,
(27)
min
⁡
j
=
1,2
dia
(
γ
(
y
j
,
y
0
)
)
≥
min
⁡
j
=
1,2
|
y
j
-
y
0
|
≥
min
⁡
j
=
1,2
(
|
y
j
-
z
0
|
-
|
y
0
-
z
0
|
)
≥
(
d
2
-
1
)
r
.
It follows from (26) and (27) that
(28)
d
≤
4
b
+
2
,
which contradicts with d
=
4
b
+
3. Hence z
1 and z
2 can be joined by an arc α
⊆
D
*
∩
B
¯
2
(
z
0
,
d
r
) with
(29)
dia
(
α
)
≤
2
d
r
=
(
4
b
+
3
)
|
z
1
-
z
2
|
.
This shows that D
* is a (
4
b
+
3
)-arcwise connected domain.Lemma 7.
Iff
:
R
2
→
R
2 is a K-quasiconformal mapping, and c
≥
1 is a constant, then for any x
∈
R
2 and 0
<
r
<
+
∞, we have
(30)
L
(
x
,
f
,
c
L
(
x
′
,
f
-
1
,
r
)
)
≤
a
r
,
where x
′
=
f
(
x
), f
-
1 is the inverse of f, and a
=
a
(
K
,
c
) is a constant which depends only on K and c.Proof.
LetΓ be the curve family which joins S
1
(
x
,
c
L
(
x
′
,
f
-
1
,
r
)
) and ∂
(
f
-
1
(
B
2
(
x
′
,
r
)
)
) in B
2
(
x
,
c
L
(
x
′
,
f
-
1
,
r
)
)
∖
f
-
1
(
B
2
(
x
′
,
r
)
)
¯; M
(
Γ
) denotes the modulus of Γ, Γ
′
=
f
(
Γ
). Then from the comparison principle of modulus and the result given in [1, 7.5] we get
(31)
M
(
Γ
)
≥
2
π
log
⁡
(
c
L
(
x
′
,
f
-
1
,
r
)
/
l
(
x
′
,
f
-
1
,
r
)
)
,
M
(
Γ
′
)
≤
2
π
log
⁡
(
l
(
x
,
f
,
c
L
(
x
′
,
f
-
1
,
r
)
)
/
r
)
.
According to the properties of K-quasiconformal mapping in [1], we have
(32)
K
M
(
Γ
′
)
≥
M
(
Γ
)
≥
M
(
Γ
′
)
K
(33)
L
(
x
′
,
f
-
1
,
r
)
l
(
x
′
,
f
-
1
,
r
)
≤
c
′
=
c
′
(
K
)
,
where c
′
(
K
) is a constant which depends only on K.
It follows from (31)–(33) that
(34)
l
(
x
,
f
,
c
L
(
x
′
,
f
-
1
,
r
)
)
r
≤
(
c
c
′
)
K
.
The same reason to get (33) implies
(35)
L
(
x
,
f
,
c
L
(
x
′
,
f
-
1
,
r
)
)
l
(
x
,
f
,
c
L
(
x
′
,
f
-
1
,
r
)
)
≤
c
′
.
It follows from (34) and (35) that
(36)
L
(
x
,
f
,
c
L
(
x
′
,
f
-
1
,
r
)
)
r
≤
c
′
(
c
c
′
)
K
=
a
.Lemma 8.
Suppose thatf
:
R
2
→
R
2 is a K-quasiconformal mapping. If D is a b-cigar domain for some b
>
0, then D
′
=
f
(
D
) is a b
′-cigar domain. In here b
′
=
b
′
(
b
,
K
) is a constant which depends only on b and K.Proof.
For any pointsy
1, y
2
∈
D
′, let x
1
=
f
-
1
(
y
1
), x
2
=
f
-
1
(
y
2
). Since D is a b-cigar domain, hence there exists an arc γ
⊆
D joining x
1 and x
2 such that min
⁡
j
=
1,2
dia
(
γ
(
x
j
,
x
)
)
≤
b
d
(
x
,
∂
D
) for all x
∈
γ. Let γ
′
=
f
(
γ
); then γ
′ is an arc joining y
1 and y
2 in D
′. For any y
∈
γ
′, let x
=
f
-
1
(
y
); then x
∈
γ with min
⁡
j
=
1,2
dia
(
γ
(
x
j
,
x
)
)
≤
b
d
(
x
,
∂
D
).
Without loss of generality, we assume thatdia
(
γ
(
x
1
,
x
)
)
≤
b
d
(
x
,
∂
D
); then γ
(
x
1
,
x
)
⊆
B
¯
2
(
x
,
b
d
(
x
,
∂
D
)
) and γ
′
(
y
1
,
y
)
=
f
(
γ
(
x
1
,
x
)
)
⊆
B
¯
2
(
y
,
L
(
x
,
f
,
b
d
(
x
,
∂
D
)
)
). From Lemma 7 and the fact L
(
x
,
f
,
b
d
(
x
,
∂
D
)
)
≤
L
(
x
,
f
,
b
L
(
y
,
f
-
1
,
d
(
y
,
∂
D
′
)
)
) we know that there exists constant b
′
=
b
′
(
b
,
K
) which depends only on b and K such that L
(
x
,
f
,
b
d
(
x
,
∂
D
)
)
≤
(
b
′
/
2
)
d
(
y
,
∂
D
′
). This implies γ
′
(
y
1
,
y
)
⊆
B
¯
2
(
y
,
(
b
′
/
2
)
d
(
y
,
∂
D
′
)
). Therefore,
(37)
dia
(
γ
′
(
y
1
,
y
)
)
≤
b
′
d
(
y
,
∂
D
′
)
,
min
⁡
j
=
1,2
dia
(
γ
(
y
j
,
y
)
)
≤
b
′
d
(
y
,
∂
D
′
)
.
This shows that D
′ is a b
′-cigar domain.Lemma 9 (see [15]).
Letf
:
R
2
→
R
2 be a homeomorphism. If there exists a constant c
>
0 such that
(38)
[
dia
(
f
(
B
2
(
x
,
r
)
)
)
]
2
≤
c
·
m
[
f
(
B
2
(
x
,
r
)
)
]
for all x
∈
R
2 and 0
<
r
<
∞; then f is a quasiconformal mapping, where m
[
f
(
B
2
(
x
,
r
)
)
] denotes the 2-dimensional L
e
b
e
s
g
u
e measure of f
(
B
2
(
x
,
r
)
).Lemma 10.
Letf
;
R
2
→
R
2 be a homeomorphism. If f maps any cigar domain D onto a cigar domain D
′
=
f
(
D
), then f is a quasiconformal mapping.Proof.
For anyx
∈
R
2 and 0
<
r
<
∞, choose y
∈
B
2
(
x
,
r
) such that
(39)
dia
(
f
(
B
2
(
x
,
r
)
)
)
≤
3
|
f
(
y
)
-
f
(
x
)
|
.
It is easy to see thatB
2
(
x
,
r
) is a 1-cigar domain if we take γ
=
[
x
1
,
x
]
∪
[
x
2
,
x
] for any x
1
,
x
2
∈
B
2
(
x
,
r
), where [
x
j
,
x
] denotes the closed line segment joining x
j and x, j
=
1,2. By the assumption of Lemma 10 we know that there exists a constant b
>
0 such that f
(
B
2
(
x
,
r
)
) is a b-cigar domain; hence there exists an arc γ
⊆
f
(
B
2
(
x
,
r
)
) joining f
(
x
) and f
(
y
) with
(40)
min
⁡
{
dia
(
γ
(
f
(
x
)
,
z
)
)
,
dia
(
γ
(
f
(
y
)
,
z
)
)
}
≤
b
d
(
z
,
∂
f
(
B
2
(
x
,
r
)
)
)
for all z
∈
γ. If we choose z
0
∈
γ such that dia
(
γ
(
f
(
x
)
,
z
0
)
)
=
dia
(
γ
(
f
(
y
)
,
z
0
)
), then (39) and (40) imply
(41)
dia
(
z
0
,
∂
f
(
B
2
(
x
,
r
)
)
)
≥
dia
(
γ
(
f
(
x
)
,
z
0
)
)
b
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
≥
dia
(
γ
(
f
(
x
)
,
f
(
y
)
)
)
2
b
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
≥
|
f
(
x
)
-
f
(
y
)
|
2
b
iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
≥
dia
(
f
(
B
2
(
x
,
r
)
)
)
6
b
.
This yields
(42)
B
2
(
z
0
,
dia
(
f
(
B
2
(
x
,
r
)
)
)
6
b
)
⊆
f
(
B
2
(
x
,
r
)
)
,
π
(
dia
[
f
(
B
2
(
x
,
r
)
)
]
6
b
)
2
≤
m
(
f
(
B
2
(
x
,
r
)
)
)
,
[
dia
(
f
(
B
2
(
x
,
r
)
)
)
]
2
≤
36
b
2
π
m
(
f
(
B
2
(
x
,
r
)
)
)
.
From the above argument and Lemma 9 we know that f is a quasiconformal mapping.Proof of Theorem1.
Consider the following.
Necessity. For any quasiconformal mapping f
:
R
2
→
R
2 and any arcwise connected domain D
⊆
R
2, we know that D
* is a cigar domain by Lemma 3. This and Lemma 8 imply that f
(
D
*
) is a cigar domain, and then D
′
=
f
(
D
)
=
(
f
(
D
*
)
)
* is an arcwise connected domain by Lemma 6.
Sufficiency. To prove a homeomorphism f
:
R
2
→
R
2 is a quasiconformal mapping, making use of Lemma 10, we need only to prove that D
′
=
f
(
D
) is a cigar domain for any cigar domain D. In fact, for any cigar domain D, Lemma 6 implies D
* is an arcwise connected domain. From this and the condition of Theorem 1 we know that f
(
D
*
) is an arcwise connected domain; then from Lemma 3 we know that (
f
(
D
*
)
)
*
=
f
(
D
)
=
D
′ is a cigar domain.Proof of Theorem2.
Consider the following.
Necessity. Let D be a quasidisk; then there exists a quasiconformal mapping f
:
R
2
→
R
2 such that D
=
f
(
B
2
).(1)
It is obvious thatB
2 is a 1-arcwise connected domain; then from Theorem 1 we know that D
=
f
(
B
2
) is an arcwise connected domain.
(2)
SinceB
2 is a 1-cigar domain, hence (
B
2
)
* is an arcwise connected domain; then Theorem 1 implies that f
(
(
B
2
)
*
)
=
D
* is an arcwise connected domain.
Sufficiency. Suppose that both D and D
* are arcwise connected domains.(1)
SinceD is an arcwise connected domain, hence D
* is a cigar domain by Lemma 3; then Lemma 5 implies that D
∈
ILC.
(2)
SinceD
* is an arcwise connected domain, hence D is a cigar domain by Lemma 3; then Lemma 4 implies that D
∈
OLC.
From the above(
1
) and (
2
) together with Theorem A we know that D is a quasidisk.



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*Source: 419850-2014-08-28.xml*