We prove that a homeomorphism f:R2→R2 is a quasiconformal mapping if and only if f(D) is an arcwise connected domain for any arcwise connected domain D⊆R2, and D is a quasidisk if and only if both D and its exterior D*=R2∖D¯ are arcwise connected domains.
1. Introduction
We shall assume throughout this paper that D is a Jordan proper subdomain of the complex plane R2 with a boundary containing at least three points. For convenience we shall adopt the notation and terminology as in paper [1]. For x∈R2 and 0<r<∞, let B2(x,r)={z∈R2:|z-x|<r}, B¯2(x,r) be the closure of B2(x,r), S1(x,r)=∂B2(x,r), B2(r)=B2(0,r), and B2=B2(1). Suppose that f is a homeomorphism in R2, x∈R2 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 that b≥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:R2→R2 such that D is the image of the unit disk B2 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 Theorems 1 and 2.
Theorem 1.
A homeomorphism f:R2→R2 is a quasiconformal mapping if and only if f(D) is a arcwise connected domain for any arcwise connected domain D⊆R2.
Theorem 2.
D is a quasidisk if and only if both D and D*=R2∖D¯ are arcwise connected domains.
For convenience we shall introduce the following concepts and they will be used in the next section.
Let c≥1 be a constant. (1) If for any x0∈R2 and 0<r<+∞, points in D∩B¯(x0,r) can be joined by an arc in D∩B¯(x0,cr); then we say that D is a c-inner linearly locally connected domain, denoted by D∈c-ILC; (2) If for any x0∈R2 and 0<r<+∞, points in D∖B2(x0,r) can be joined by an arc in D∖B2(x0,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.
Let b>0 be a constant; then D is said to be a b-cigar domain if each pair of points x1,x2∈D can be joined by an arc γ⊆D for which
(1)minj=1,2dia(γ(xj,x))≤bd(x,∂D)∀x∈γ,
where γ(xj,x) is the part of γ between xj and x, and d(x,∂D) is the Euclidean distance from x to ∂D.
We say that D 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 Theorems 1 and 2
To prove our Theorems 1 and 2, we first establish and introduce several lemmas.
Lemma 3.
If D is an arcwise connected domain, then D*=R2∖D¯ is a cigar domain.
Proof.
For any z1, z2∈D*, let γ be the hyperbolic geodesic joining z1 and z2 in D*, for any z∈γ∖{z1,z2}; suppose that f:B2→D* is a conformal mapping with f(0)=z and
(2)f-1(γ)⊆R={z:z∈R2,Imz=0},
where f-1 is the inverse of f. Then there exist
(3)x1∈{z:z∈S1,Imz>0},x2∈{z:z∈S1,Imz<0}
by [14, Corollary 10.3] such that αj=f([0,xj)) is rectifiable with
(4)l(αj)<a0d(z,∂D*),j=1,2,
where a0 is an absolute constant, and [0,xj) is the half open segment joining the origin O and xj; j=1,2.
Let yj=f(xj), α=α1∪α2, then yj∈∂D and
(5)l(α)≤l(α1)+l(α2)<2a0d(z,∂D*).
For above y1, y2∈∂D*=∂D, there exist a constant b≥1 and a simple curve β⊆D joining y1 and y2 such that
(6)dia(β)≤b|y1-y2|≤bl(α)
by D is an arcwise connected domain.
If we denote by D0 the bounded domain with boundary α∪β, then one of the points z1 and z2 must be in D0. Without loss of generality, we may assume that z1∈D0; then it follows from (5) and (6) that
(7)dia(γ(z1,z))≤dia(D0)=dia(∂D0)iiiiiiiiiiiiiiiiiii≤dia(α)+dia(β)iiiiiiiiiiiiiiiiiii≤l(α)+dia(β)iiiiiiiiiiiiiiiiiii<2a0(b+1)d(z,∂D*).
Inequality (7) leads to
(8)minj=1,2dia(γ(zj,z))<2a0(b+1)d(z,∂D*).
Therefore, D* is a cigar domain that follows from (8).
Lemma 4.
If D is a c-cigar domain, then D∈(2c+2)-OLC.
Proof.
Let b=2c+2. If D∉b-OLC, then there exist y0∈R2, 0<r<∞ and x1, x2∈D∖B2(y0,r), such that x1 and x2 can not be joined by any arc in D∖B2(y0,r/b).
Since D is a c-cigar domain, there exists an arc γ⊂D such that γ joining x1 and x2 with
(9)minj=1,2dia(γ(xj,x))≤cd(x,∂D)
for all x∈γ.
It is obvious that γ∩S1(y0,r/b)≠∅. Let y∈γ∩S1(y0,r/b); then (9) implies
(10)d(y,∂D)≥1cminj=1,2dia(γ(xj,y))≥1cminj=1,2|xj-y|≥1c(1-1b)r.
But
(11)d(y0,∂D)≤rb.
Inequalities (10) and (11) together with the triangular inequality yield
(12)1c(1-1b)r≤d(y,∂D)≤|y-y0|+d(y0,∂D)≤2br,b≤2c+1.
Inequality (12) contradicts with b=2c+2. Hence D∈(2c+2)-OLC.
Lemma 5.
If D* is a c0-cigar domain, then D∈(16c0+21)-ILC.
Proof.
Let δ=8c0+10. For any u∈R2, s>0, and z1,z2∈D∩B¯2(u,s), z1≠z2. Denote z=(1/2)(z1+z2) and r=|z1-z2|. We first prove that z1,z2 must be in the same component of B¯2(z,(1/2)δr)∖D*.
If z1,z2 belong to different components of B¯2(z,(1/2)δr)∖D*, then z1,z2 must be in the different components of B¯2(z,(1/2)r)∖D*. Let β be the line segment which joins z1 and z2; then β contains a subcurve α⊆D* such that α divides D* into D1 and D2, and dia(Dj)≥(1/2)r(δ-1), j=1,2. This yields
(13)minj=1,2dia(Dj)≥12r(δ-1).
For any x∈α, if D1⊈B2(x,(2c0+2)dia(α)) and D2⊈B2(x,(2c0+2)dia(α)), then take
(14)xj∈Dj∖B¯2(x,(2c0+2)dia(α)),j=1,2.
Since D* is a c0-cigar domain, there exists an arc γ⊆D* joining x1 and x2 with
(15)minj=1,2dia(γ(xj,w))≤c0d(w,∂D*)
for all w∈γ.
Let y∈γ∩S1(x,dia(α)); then
(16)|y-xj|≥(2c0+1)dia(α),minj=1,2dia(γ(xj,y))≤c0d(y,∂D*),mmmmmmmmmmmj=1,2.
This implies
(17)d(y,∂D*)≥2c0+1c0dia(α).
But
(18)d(x,∂D*)≤dia(α).
Inequalities (17) and (18) together with the triangular inequality yield
(19)2c0+1c0dia(α)≤d(y,∂D*)≤|y-x|+d(x,∂D*)iiiiiiiiiiiiiiiiiiiiii≤2dia(α),
so
(20)dia(α)≤0.
This is obviously impossible. Hence D1⊆B¯2(x,(2c0+2)dia(α)) or D2⊆B¯2(x,(2c0+2)dia(α)), and we obtain
(21)minj=1,2dia(Dj)≤2(2c0+2)dia(α).
Inequalities (13) and (21) together with dia(α)≤r imply
(22)δ≤8c0+9.
This contradicts with δ=8c0+10. Hence z1, z2 must be in the same component of B¯2(z,(1/2)δr)∖D*, and there exists an arc γ⊆D joining z1 and z2 with
(23)dia(γ)≤δr=δ|z1-z2|≤2δs.
It follows from (23) that
(24)γ⊆D∩B¯2(u,s+dia(γ))⊆D∩B¯2(u,(2δ+1)s)=D∩B¯2(u,(16c0+21)s).
Hence D∈(16c0+21)-ILC; this completes the proof of Lemma 5.
Lemma 6.
If D is a b-cigar domain, then D* is a (4b+3)-arcwise connected domain.
Proof.
For any z1, z2∈D*, z1≠z2, let z0=(z1+z2)/2, r=|z1-z2|/2, and d=4b+3. We shall prove that z1 and z2 can be joined by an arc in D*∩B¯2(z0,dr).
If z1 and z2 can not be joined by any arc in D*∩B¯2(z0,dr), then z1 and z2 must be in the different components of D*∩B¯2(z0,dr), and hence z1 and z2 must be in the different components of D*∩B¯2(z0,(d/2)r). There exist points y1, y2∈∂D*∖B¯2(z0,(d/2)r) such that β∩B2(z0,r)≠∅ for any arc β⊆D joining y1 and y2.
Since D is a b-cigar domain, hence there exists an arc γ⊆D joining y1 and y2 such that
(25)minj=1,2dia(γ(yj,y))≤bd(y,∂D)
for all y∈γ.
Let y0∈γ∩B2(z0,r); then
(26)minj=1,2dia(γ(yj,y0))≤bd(y0,∂D),d(y0,∂D)≤|y0-zj|≤2r,(27)minj=1,2dia(γ(yj,y0))≥minj=1,2|yj-y0|≥minj=1,2(|yj-z0|-|y0-z0|)≥(d2-1)r.
It follows from (26) and (27) that
(28)d≤4b+2,
which contradicts with d=4b+3. Hence z1 and z2 can be joined by an arc α⊆D*∩B¯2(z0,dr) with
(29)dia(α)≤2dr=(4b+3)|z1-z2|.
This shows that D* is a (4b+3)-arcwise connected domain.
Lemma 7.
If f:R2→R2 is a K-quasiconformal mapping, and c≥1 is a constant, then for any x∈R2 and 0<r<+∞, we have
(30)L(x,f,cL(x′,f-1,r))≤ar,
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.
Let Γ be the curve family which joins S1(x,cL(x′,f-1,r)) and ∂(f-1(B2(x′,r))) in B2(x,cL(x′,f-1,r))∖f-1(B2(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)≥2πlog(cL(x′,f-1,r)/l(x′,f-1,r)),M(Γ′)≤2πlog(l(x,f,cL(x′,f-1,r))/r).
According to the properties of K-quasiconformal mapping in [1], we have
(32)KM(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,cL(x′,f-1,r))r≤(cc′)K.
The same reason to get (33) implies
(35)L(x,f,cL(x′,f-1,r))l(x,f,cL(x′,f-1,r))≤c′.
It follows from (34) and (35) that
(36)L(x,f,cL(x′,f-1,r))r≤c′(cc′)K=a.
Lemma 8.
Suppose that f:R2→R2 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 points y1, y2∈D′, let x1=f-1(y1), x2=f-1(y2). Since D is a b-cigar domain, hence there exists an arc γ⊆D joining x1 and x2 such that minj=1,2dia(γ(xj,x))≤bd(x,∂D) for all x∈γ. Let γ′=f(γ); then γ′ is an arc joining y1 and y2 in D′. For any y∈γ′, let x=f-1(y); then x∈γ with minj=1,2dia(γ(xj,x))≤bd(x,∂D).
Without loss of generality, we assume that dia(γ(x1,x))≤bd(x,∂D); then γ(x1,x)⊆B¯2(x,bd(x,∂D)) and γ′(y1,y)=f(γ(x1,x))⊆B¯2(y,L(x,f,bd(x,∂D))). From Lemma 7 and the fact L(x,f,bd(x,∂D))≤L(x,f,bL(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,bd(x,∂D))≤(b′/2)d(y,∂D′). This implies γ′(y1,y)⊆B¯2(y,(b′/2)d(y,∂D′)). Therefore,
(37)dia(γ′(y1,y))≤b′d(y,∂D′),minj=1,2dia(γ(yj,y))≤b′d(y,∂D′).
This shows that D′ is a b′-cigar domain.
Lemma 9 (see [15]).
Let f:R2→R2 be a homeomorphism. If there exists a constant c>0 such that
(38)[
dia
(f(B2(x,r)))]2≤c·m[f(B2(x,r))]
for all x∈R2 and 0<r<∞; then f is a quasiconformal mapping, where m[f(B2(x,r))] denotes the 2-dimensional Lebesgue measure of f(B2(x,r)).
Lemma 10.
Let f;R2→R2 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 any x∈R2 and 0<r<∞, choose y∈B2(x,r) such that
(39)dia(f(B2(x,r)))≤3|f(y)-f(x)|.
It is easy to see that B2(x,r) is a 1-cigar domain if we take γ=[x1,x]∪[x2,x] for any x1,x2∈B2(x,r), where [xj,x] denotes the closed line segment joining xj and x, j=1,2. By the assumption of Lemma 10 we know that there exists a constant b>0 such that f(B2(x,r)) is a b-cigar domain; hence there exists an arc γ⊆f(B2(x,r)) joining f(x) and f(y) with
(40)min{dia(γ(f(x),z)),dia(γ(f(y),z))}≤bd(z,∂f(B2(x,r)))
for all z∈γ. If we choose z0∈γ such that dia(γ(f(x),z0))=dia(γ(f(y),z0)), then (39) and (40) imply
(41)dia(z0,∂f(B2(x,r)))≥dia(γ(f(x),z0))biiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii≥dia(γ(f(x),f(y)))2biiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii≥|f(x)-f(y)|2biiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii≥dia(f(B2(x,r)))6b.
This yields
(42)B2(z0,dia(f(B2(x,r)))6b)⊆f(B2(x,r)),π(dia[f(B2(x,r))]6b)2≤m(f(B2(x,r))),[dia(f(B2(x,r)))]2≤36b2πm(f(B2(x,r))).
From the above argument and Lemma 9 we know that f is a quasiconformal mapping.
Proof of Theorem 1.
Consider the following.
Necessity. For any quasiconformal mapping f:R2→R2 and any arcwise connected domain D⊆R2, 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:R2→R2 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 Theorem 2.
Consider the following.
Necessity. Let D be a quasidisk; then there exists a quasiconformal mapping f:R2→R2 such that D=f(B2).
It is obvious that B2 is a 1-arcwise connected domain; then from Theorem 1 we know that D=f(B2) is an arcwise connected domain.
Since B2 is a 1-cigar domain, hence (B2)* is an arcwise connected domain; then Theorem 1 implies that f((B2)*)=D* is an arcwise connected domain.
Sufficiency. Suppose that both D and D* are arcwise connected domains.
Since D is an arcwise connected domain, hence D* is a cigar domain by Lemma 3; then Lemma 5 implies that D∈ILC.
Since D* 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.
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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