# On the Removability Property ofφ-Natural Domains in Real Banach Spaces

Authors: Yuehui He
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we investigate the removability ofφ-natural domains in Banach spaces. Let E be a real Banach space with dimension at least 2, G a domain in E, and let be a countable subset of G which satisfies a quasi hyperbolic separation condition. Then, G is φ-natural if and only if G is ψ-natural, quantitatively.


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## Body

## 1. Introduction

Throughout the paper, we always assume thatE denotes a real Banach space with dimension of at least 2, and a domain means a nonempty open and connected set in E.The motivation of this study stems from the discussions in [1], where the authors studied the removability of uniform domains and ψ-uniform domains in Banach spaces. The following are the main results in [1].Theorem 1.
([see [1], Theorem 2]). A domain G⊊E is a c1-uniform domain if and only if G0=G\ is a c2-uniform domain, where the uniformity coefficients c1 and c2 depend on each other.Theorem 2.
([see [1], Theorem 1]).A domain G⊊E is a ψ-uniform domain if and only if G0=G\ is a ψ1-uniform domain, where the control functions ψ and ψ1 depend on each other.Here, denotes a sequence of points ofG satisfying a 1/2-quasihyperbolic separation condition in G. Other terminology in the above two theorems and in the rest of this section will be given in Section 2 unless otherwise stated.We recall that uniform domains in Euclidean spacesℝn were introduced by John in connection with his work on elasticity [2]. The term is due to Martio and Sarvas [3]. Roughly speaking, a domain is a uniform domain if it is possible to travel from one point of the domain to another along a curve which is not too close to the boundary and not too long. Up to this time, uniform domains have been studied extensively because of their own characteristics (cf. [4–6]) and their significant applications (see, e.g., [7–13]). To investigate the relationships between conformal invariants and quasi hyperbolic metrics, in [14], Vuorinen introduced the so-called ψ-uniform domains, which are a generalization of uniform domains. See [1,15–19] for the development of ψ-uniform domains in recent years.In [20], to explore the relationships among relative quasimappings, Väisälä introduced a class of more general domains, i.e., the so-called φ-natural domains, and proved that each domain in Banach spaces with finite dimension is φ-natural with φ depending only on dimE, see [[20], Theorem 2.7]. But this property is invalid in Banach spaces with infinite dimension. In fact, Väisälä constructed a domain, a broken tube, in Banach spaces with infinite dimension, which is not natural, see [[21], Example 4.12]. Since all uniform domains and φ-uniform domains are φ-natural, naturally, one will ask whether there is a result similar to Theorems 1 and 2 for the setting of φ-natural domains in E. The purpose of this paper is to investigate this question. Our main result is as follows.Theorem 3.
Suppose thatG⊊E denotes a domain and Pκ is a countable subset of G which satisfies κ-quasi hyperbolic separation condition. Then G is φ-natural if and only if G0=G\κ is ψ-natural, where the control functions φ and ψ depend on each other, together with the constants ν from Theorem 4 and κ.This paper is organized as follows: in Section2, necessary terminology will be introduced and a series of lemmas will be proved. In Section 3, we will prove Theorem 3.


## 2. Preliminaries

In this section, necessary notions and notations in this paper will be introduced, several known results will be recalled, and a series of lemmas will be proved.We say that a domainG⊊E is called c-uniform if there exists a constant c with the property that each pair of points z1, z2 in G can be joined by a rectifiable arc γ in G satisfying (cf. [[18], Section 6]).(1)
minj=1,2ℓγzj,z≤cdGz for all z∈γ,(2)
ℓγ≤cz1−z2,whereγzj,z is the part of γ between zj and z, ℓγzj,z stands for the length of γzj,z, dGz denotes the distance from z to the boundary ∂G of G in E, and ⋅ is the metric in E.The following result is from [18] due to Väisälä.Theorem 4.
([see [18], Theorem 6.5]). For any x0∈E and r>0, all the domains Bx0,r, Bx0,r\x0 and E\x0 in E are ν-uniform with a universal ν, where Bx0,r denotes the ball in E with center x0 and radius r, i.e., Bx0,r=x∈E:x−x0<r.A curve is a continuous mapα:I⟶E from an interval I=a,b⊂ℝ to E. If α is an embedding, then it is said to be an arc. For simplicity, we also use α to denote its image set αI. The length ℓα of α is defined as usual,(1)ℓα=sup∑i=1nαti−αti+1,where the supremum is over all finite sequences a=t1≤t2≤…≤tn≤tn+1=b. We call a curve α rectifiable if its length ℓα<∞.Letα⊂G be a rectifiable curve or arc with endpoints z1 and z2. Then, there are sα:I⟶0,ℓα and αs:0,ℓα⟶α such that αs0=z1, αsℓα=z2, and ℓαz1,αst=t for all t∈0,ℓα, moreover, α=αs°sα. The curve αs is the arc length parameterization of α, see [[22], §2].Ifα is a rectifiable curve or arc in G and if ϱ:γ⟶0,∞ is continuous, the line integral of ϱ along α is defined as follows:(2)∫αϱdz=∫0ℓαϱ°αstdt.The quasihyperbolic length ofα in G is the number.(3)ℓk,Gα=∫αdzdGz,(cf. [23]), where the quasihyperbolic distance kGz1,z2 between z1 and z2 is defined by the following equation:(4)kGz1,z2=infℓk,Gα,where the infimum is taken over all rectifiable arcs α joining z1 to z2 in G.The following result is useful for the discussions in Section3.Theorem 5 ([see [8], Lemma 2.13]).
Suppose thatG is c-uniform domain. Then, for all x, y∈G,(5)kGx,y≤4c2log1+x−ymindGx,dGy.Suppose thatγ is a curve in G⊊E and μ≥1 is a constant. It is called a μ-quasigeodesic (i.e., neargeodesic in [18]) if for any x, y∈γ,(6)ℓk,Gγx,y≤μkGx,y,and μ-quasiconvex for μ≥1 if for any x, y∈γ,(7)ℓγx,y≤μx−y.The following result concerning the existence of quasigeodesics inE is due to Väisälä.Theorem 6.
([see [18], Theorem 3.3]).Suppose that x, y∈G and μ>1. Then there is a μ-quasigeodesic in G joining x and y.Letψ:0,+∞⟶0,+∞ be a homeomorphism. A domain G is called ψ-uniform if for all x, y∈G,(8)kGx,y≤ψx−ymindGx,dGy.(cf. [14]).For a nonempty setA⊂G⊊E, we write(9)rGx,y=x−ymindGx,dGy,rGA=suprGx,y:x,y∈A,and(10)kGA=supkGx,y:x,y∈A.Obviously, for any domainsG and G1 in E, if A⊂G⊂G1 is a nonempty set, then(11)rGA≥rG1A.Moreover, let(12)r¯GA=diamAdGA,where “diam” means “diameter,” and dGA denotes the distance between the set A and the boundary ∂G, i.e., dGA=infdGx:x∈A. Then, it follows from [[20], equation (2.6)] that(13)rGA≤r¯GA≤2rGA.Letφ:0,+∞⟶0,+∞ be an increasing function. A domain G⊊E is called φ-natural if(14)kGA≤φrGA,for every nonempty connected set A⊂G with rGA<+∞.LetP denote a sequence xii=1+∞ of points in G⊊E. If there is a constant κ>0 such that for any pair i≠j∈1,2,3,…,(15)kGxi,xj≥κ,then we say that P satisfies a κ-quasihyperbolic separation condition in G. For such a sequence, in the following of this paper, we always use Pκ to denote it.Lemma 1.
Suppose thatG⊊E denotes a domain and Pκ⊂G. Then for any xi, xj∈Pκ with xi≠xj, we have(16)xi−xj≥κκ+1maxdGxi,dGxj.Proof.
Without loss of generality, we assume that(17)maxdGxi,dGxj=dGxi.Suppose on the contrary that(18)xi−xj<κκ+1dGxi.
Then, it follows from [[21], Lemma 2.2(2)] and the assumption in the lemma that(19)κ≤kGxi,xj≤11−κ/κ+1xi−xjdGxi<κ,which is the desired contradiction. Hence the lemma is true.For simplicity, in the rest of this paper, we apply the following notations. For anyxi∈Pκ, let(20)Bxi,κ4κ+1dGxi=Bxi,λi,the boundary of Bxi,λi is denoted by Sxi,λi, i.e.,(21)∂Bxi,κ4κ+1dGxi=Sxi,λi,and(22)B¯xi,λi=Bxi,λi∪Sxi,λi,where λi=κ/4κ+1dGxi.Now, we are ready to state our next lemma.Lemma 2.
For anyxi, xj∈Pκ with xi≠xj, we have(23)Bxi,2λi∩Bxj,2λj=∅.Proof.
Suppose on the contrary that(24)Bxi,2λi∩Bxj,2λj≠∅.
Let(25)z∈Bxi,2λi∩Bxj,2λj.
Then,(26)xi−xj≤xi−z+z−xj<κκ+1maxdGxi,dGxj,which contradicts Lemma 1. Hence Lemma 2 holds true.Also, we introduce the following notations. For a domainG⊊E and a nonempty set A in G, let(27)D=Bxi,λi:xi∈Pκ,BA=Bxi,λi∈D:Bxi,λi∩A≠∅,XA=xi∈κ:Bxi,λi∈BA,and let(28)SA=z∈E:z∈Sxi,λiforxi∈XA.Obviously, the connectedness ofA implies the one of the set A\D∪SA.We recall the reader the standard convention that(29)diam∅=Card∅=0,for the empty set ∅, where “Card” stands for “cardinality”.Lemma 3.
For a domainG⊊E, if A is a nonempty connected set in G with rGA<+∞, then diamSA<+∞.Proof.
IfCardBA≤1, then the lemma is obvious. In the following, we assume that CardBA≥2.
SinceG⊊E implies dGA<+∞, we deduce from equation (15) that(30)diamAdGA=r¯GA≤2rGA,and so,(31)diamA≤2rGAdGA<+∞.
For eachxi∈XA, it follows from Lemma 2 that(32)diamA≥distSxi,λi,Sxi,2λi≥λi,where distSxi,λi,Sxi,2λi denotes the distance between the sets Sxi,λi and Sxi,2λi (i.e., distSxi,λi,Sxi,2λi=infu−v:u∈Sxi,λiandv∈Sxi,2λi. Thus we get(33)supxi∈XAdiamSxi,λi≤supxi∈XA2λi≤2diamA.
Letz1, z2∈SA. Then there are xi, xj∈XA such that(34)z1∈Sxi,λi\textup andz2∈Sxj,λj.
SinceA is connected, we get(35)A∩Sxi,λi≠∅\textup andA∩Sxj,λj≠∅,and thus,(36)distz1,A≤diamSxi,λi\textup anddistz2,A≤diamSxj,λj,where distz1,A denotes the distance from the point z1 to the set A (i.e., distz1,A=infz1−x:x∈A). This ensures that(37)z1−z2≤distz1,A+diamA+distz2,A≤diamSxi,λi+diamA+diamSxj,λj,which, together with equations (31) and (33), implies(38)diamSA≤5diamA<+∞.
This proves the lemma.Lemma 4.
For a domainG⊊E, let G0=G\κ. Then for any xi∈Pκ and z1, z2∈B¯xi,λi\xi,(39)kG0z1,z2≤4ν2log1+rG0z1,z2,where the constant ν is from Theorem 4.Proof.
Letz1,z2∈B¯xi,λi\xi, and let Di=Bxi,2λi\xi. By Theorem 4, we know that Di is ν-uniform.
Sincez1∈B¯xi,λi, we have(40)distz1,Sxi,2λi≥λi≥z1−xi.
This implies(41)dDiz1=z1−xi.
For anyxj∈Pκwith xj≠xi, since(42)z1−xj≥xi−xj−z1−xi,we infer from Lemma 1 that(43)z1−xj≥3κ4κ+1maxdGxi,dGxj≥z1−xi,and thus,(44)distz1,Pκ=infz−xj:j=1,2,…=z1−xi.
Since(45)dGz1≥dGxi−z1−xi≥z1−xi,we deduce from equation (41) that(46)dDiz1=z1−xi=mindGz1,distz1,κ=dG0z1.
Similarly, we have(47)dDiz2=z2−xi=dG0z2.
Because it follows from Lemma2 that(48)Di⊊G0,we get that (by equations (46)–(48) and Theorem 5)(49)kG0z1,z2≤kDiz1,z2≤4ν2log1+z1−z2mindG0z1,dG0z2=4ν2log1+rG0z1,z2,and hence, the lemma is proved.Lemma 5.
Suppose thatG⊊E is a domain and G0=G\Pκ. Then for any rectifiable curve β⊂G\D with endpoints x and y,(50)kG0x,y≤5+4κℓk,Gβ.Proof.
Since for anyz∈β and xi∈Pκ,(51)z−xi≥λi,we have(52)z−xi≥dGz−dGxi≥dGz−4κ+1κz−xi,and so,(53)z−xiκ5κ+4dGz.
This gives(54)distz,κ=infz−xi:i=1,2,…κ5κ+4dGz,and thus, we get(55)dG0z=mindGz,distz,κ≥κ5κ+4dGz.
This ensures that(56)kG0x,y≤∫βdzdG0z≤5+4κℓk,Gβ,and hence, the lemma is proved.Lemma 6.
Under the assumptions of Lemma5, suppose further that α⊂Sxi,λi is a 2-quasiconvex curve connecting two points u and v in Sxi,λi, where xi∈Pκ. Then,(57)ℓk,G0α≤43κ+2κkGu,v.Proof.
Sinceu and v∈Sxi,λi and ℓα≤2u−v, we see from [[21], Lemma 2.2(3)] that(58)kGu,v≥11+2κ/4κ+1u−vdGxi≥κ+13κ+2ℓαdGxi.
Because for anyz∈α⊂Sxi,λi,(59)dG0z=z−xi=λi,we get(60)ℓk,G0α≤4κ+1κℓαdGxi≤43κ+2κkGu,v.
This completes the proof of the lemma.Lemma 7.
Suppose thatG⊊E is a domain and G0=G\Pκ. Then, for each pair z1, z2∈GD, we have(61)kG0z1,z2≤83κ+2κkGz1,z2.Proof.
Letz1, z2∈G\D. Then it follows from Theorem 6 that there is a 2-quasigeodesic γ in G joining z1 and z2.
Ifγ⊂G\D, then the inequality in equation (61) easily follows from Lemma 5. In the following, we assume that(62)γ⊂G\D.
Then,Bγ≠∅. Furthermore, we prove that Bγ contains a finite number of elements. To this end, let Bxi,λi∈Bγ. Then, we have(63)dG=γ≤distxi,γ+dGxi≤λi+dGxi,and so(64)dGxi≥4κ+15κ+4dGγ,since λi=κ/4κ+1dGxi.
IfBγ contains an element Bxj,λj with xj≠xi, let zi∈γ∩Bxi,λi and zj∈γ∩Bxj,λj. Then, we get (by Lemma 1 and equation (64))(65)zi−zj≥xi−xj−xi−zi−xj−zj≥κ2κ+1maxdGxi,dGxj≥2κ5κ+4dGγ,which, together with Lemma 2, implies that the length of γ would be infinite if Bγ contains infinite number of elements. This contradiction affirms that(66)CardBγ<+∞.
Next, we construct a curveβ in G0 to connect z1 and z2. Since Bγ≠∅, there is an element in Bγ, denoted by B1, such that,(1)
there areξ1, ζ1∈γ∩S1 such that γz1,ξ1 does not intersect any element in Bγ and γz2,ζ1∩B1=∅ (here, recall that S1 denotes the boundary of the ball B1)(2)
there is a 2-quasiconvex curveα1⊂S1 joining ξ1 and ζ1 (cf. [[18], Lemma 2.2])
Ifγz2,ζ1 does not intersect any element in Bγ, then let(67)β=γz1,ξ1∪α1∪γζ1,z2.
Otherwise, there is an element, denoted byB2, in Bγ\B1 such that,(1)
there areξ2, ζ2∈γ∩S2 such that γζ1,ξ2 does not intersect any element in Bγ and γz2,ζ2∩B2=∅(2)
there is a 2-quasiconvex curveα2⊂S2 joining ξ2 and ζ2
Ifγz2,ζ2 does not intersect any element in Bγ, then let(68)β=γz1,ξ1∪α1∪γζ1,ξ2∪α2∪γζ2,z2.
By repeating this procedure, there are elements,B1,…,Bt⊂Bγ (Without loss of generality, assume that t≥3) and a curve(69)β=γz1,ξ1∪α1∪γζ1,ξ2∪α2∪…∪γζt−1,ξt∪αt∪γζt,z2,such that,(1)
all pointsξi and ζi belong to the intersection γ∩Sxi,λi for i∈1,2,…,t(2)
all subcurvesγz1,ξ1, γζi,ξi+1 for i∈1,2,…,t−1 and γζt,z2 of γ do not intersect any elements in Bγ(3)
all curvesαi⊂Sxi,λi for i∈1,2,…,t are 2-quasiconvex curves(4)
β⊂G\D
By Lemma6, we see that for each r∈1,2,…,t,(70)ℓk,G0αr≤43κ+2κkGξr,ζr.
Since bothξr and ζr belong to the intersection γ∩Sr, we deduce that(71)ℓk,G0αr≤43κ+2κℓk,Gγξr,ζr.
Then it follows from Lemma5 that(72)kG0z1,z2≤ℓk,G0β≤43κ+2κℓk,Gγ.
Sinceγ is a 2-quasigeodesic, we obtain that(73)kG0z1,z2≤83κ+2κkGz1,z2,and hence, the lemma is proved.


## 3. Proof of Theorem 3

The purpose of this section is to prove Theorem3. We start the proof of the theorem with two lemmas.Lemma 8.
Under the assumptions of Theorem3, if G is φ-natural, then G0=G\κ is ψ-natural, where ψ=ψφ,ν,κ and the constant ν is from Theorem 4.Here and hereafter, the notationψφ,ν,κ means that the control function ψ depends only on φ, ν and κ.Proof.
LetA be a nonempty connected set in G0 with rG0A<+∞. To prove this lemma, it suffices to show that(74)kG0A≤ψrG0A,where ψt=16ν2log1+t+163κ+2/κφt and the constant ν is from Theorem 4.
For the proof, letx, y∈A be with(75)kG0A≤2kG0x,y.
In the following, we divide the discussions into four cases according to the positions ofx and y in G.Case 1.
Suppose that there isxi∈Pκso that x, y∈Bxi,λi.For this case, sincex, y∈Bxi,λi, by Lemma 4, we have(76)kG0x,y≤4ν2log1+rG0x,y.This, together with equation (75), shows that(77)kG0A≤8ν2log1+rG0A.Case 2.
Suppose thatx, y∈G\D.Under this assumption, it follows from Lemma7 that(78)kG0x,y≤83κ+2κkGx,y,and thus, equation (75) gives(79)kG0A≤163κ+2κkGA.Finally, the assumptionG being φ-natural ensures that(80)kG0A≤163κ+2κφrG0A.Case 3.
Suppose thatx∈Bxi,λi and y∈Bxj,λj with xi≠xj.Sincex, y∈A and A is connected, we see that there are points w1 and w2 such that(81)w1∈A∩Sxi,λiandw2∈A∩Sxj,λj,and because x, w1∈B¯xi,λi, and y, w2∈B¯xj,λj, we infer from Lemma 4 that(82)kG0x,w1≤4ν2log1+rG0AandkG0w2,y≤4ν2log1+rG0A.Moreover, by Lemma7, we have(83)kG0w1,w2≤83κ+2κkGw1,w2≤83κ+2κkGA,and thus, the assumption G being φ-natural guarantees that(84)kG0w1,w2≤83κ+2κφrG0A.Now, we conclude from equations (75), (82), and (84) that(85)kG0A≤2kG0x,w1+kG0w1,w2+kG0w2,y≤16ν2log1+rG0A+163κ+2κφrG0A.Case 4.
Suppose thatx∈G\D and there is xi∈Pκ such that y∈Bxi,λi.Obviously, there is a pointw such that(86)w∈A∩Sxi,λi.Then, equation (75) implies(87)kG0A≤2kG0x,w+kG0w,y,and thus, the assumption G being φ-natural, Lemmas 4 and 7 ensure that(88)kG0A≤8ν2log1+rG0A+163κ+2κφrG0A.By letting(89)ψt=16ν2log1+t+163κ+2κφt,we conclude from the estimates equations (77), (80), (85), and (88) that equation (74) is true, and hence, the proof of this lemma is complete.Lemma 9.
Under the assumptions of Theorem3, if G0=G\κ is ψ-natural, then G is φ-natural, where φ=φψ,ν,κ and the constant ν is from Theorem 4.Proof.
LetA be a nonempty connected set in G with rGA<+∞. In the following, we separate the discussions into three cases based on the number of the elements in the set BA.Case 5.
Suppose thatCardBA=0, i.e., BA=∅.Obviously, in the case,A⊂G\D. As the preparation, we establish a relationship between dG0z and dGz for z∈G\D.Since for anyz∈GD, there must be a point xi∈Pκ such that(90)z−xi≤2distz,κ,we know that(91)dG0z=mindGz,distz,Pκ≥mindGz,12z−xi.If1/2z−xi>dGz, then(92)dG0z≥dGz.Now, we assume that1/2z−xi≤dGz. We consider two possibilities: dGxi≥4κ+1/5κ+4dGz and dGxi<4κ+1/5κ+4dGz. For the first possibility, we have(93)dG0z≥12z−xi≥λi2≥κ25κ+4dGz,and for the remaining possibility, we get(94)dG0z≥12z−xi≥12dGz−dGxi≥κ25κ+4dGz.Hence we conclude that for any z∈G\D,(95)dG0z≥κ25κ+4dGz.Now, it follows that(96)ψrG0A≤ψ25κ+4κrGA,and thus, the assumption G0 being ψ-natural gives(97)kGA≤kG0A≤ψ25κ+4κrGA.Case 6.
Suppose thatCardBA=1.We denote the unique element inBA by Bxi,λi. Since for any z∈Bxi,λi,(98)dBxi,2λiz≥dBxi,2λixi−xi−z≥λi,and(99)dGz≤dGxi+xi−z≤5κ+44κ+1dGxi,we have(100)dBxi,2λiz≥κ5κ+4dGz.Furthermore, Theorems4 and 5 guarantee that(101)kGA∩Bxi,λi≤kBxi,2λiA∩Bxi,λi≤4ν2log1+rBxi,2λiA∩Bxi,λi.Then, we deduce from equation (100) that(102)kGA∩Bxi,λi≤4ν2log1+5κ+4κrGA∩Bxi,λi≤4ν2log1+5κ+4κrGA.IfA⊂Bxi,λi, then equation (102) gives(103)kGA=kGA∩Bxi,λi≤4ν2log1+5κ+4κrGA.In the following, we assume that(104)A⊂Bxi,λi.Then, A∩Sxi,λi≠∅.Next, we are going to establish a relation between the quantitieskGA\Bxi,λi and rGA\Bxi,λi as indicated in equation (133). To this end, we consider two possibilities: diamA\Bxi,λi≥1/8λi and diamA\Bxi,λi<1/8λi.For the former possibility, it follows that(105)diamSxi,λi≤2λi≤16diamA\Bxi,λi,and thus,(106)diamA\Bxi,λi∪Sxi,λi≤diamA\Bxi,λi+diamSxi,λi≤17diamA\Bxi,λi.Furthermore, sinceA\Bxi,λi∪Sxi,λi⊂G\D is connected, the similar reasoning as in the discussions in Case 5 demonstrates that(107)kGA\Bxi,λi∪Sxi,λi≤ψ25κ+4κrGA\Bxi,λi∪Sxi,λi.Still, we need a relationship betweenrGA\Bxi,λi∪Sxi,λi and rGA\Bxi,λi as stated in equation (114). For this, let z∈Sxi,λi. Then it follows that(108)dGz≤dGxi+xi−z≤5κ+44κ+1dGxi,and(109)dGz≥dGxi−xi−z≥3κ+44κ+1dGxi.These imply(110)dGA∩Sxi,λi≤5κ+44κ+1dGxi≤5κ+43κ+4dGSxi,λi,and thus, we infer from A∩Sxi,λi⊂A\Bxi,λi that(111)dGA\Bxi,λi≤dGA∩Sxi,λi≤5κ+43κ+4dGSxi,λi.Therefore,(112)dGA\Bxi,λi∪Sxi,λi=mindGA\Bxi,λi,dGSxi,λi≥3κ+45κ+4dGA\Bxi,λi,which, together with equation (106), implies that(113)rGA\Bxi,λi∪Sxi,λi≤diamA\Bxi,λi∪Sxi,λidGA\Bxi,λi∪Sxi,λi≤175κ+43κ+4r¯GA\Bxi,λi.Hence equation (15) ensures that(114)rGA\Bxi,λi∪Sxi,λi≤345κ+43κ+4rGA\Bxi,λi.By combining equations (107) and (114), we obtain that(115)kGA\Bxi,λi≤kGA\Bxi,λi∪Sxi,λi≤ψ685κ+42κ3κ+4rGA\Bxi,λi.For the latter possibility, that is,diamA\Bxi,λi<1/8λi, to reach the goal, let u∈A∩Sxi,λi, and let(116)F=B¯u,diamA\Bxi,λi,the closure of the ball Bu,diamA\Bxi,λi. Then it follows from the assumption diamA\Bxi,λi<1/8λi that(117)A\Bxi,λi⊂F⊂Bxi,2λi,and so, for any z∈F, we have(118)dGz≥dGxi−z−xi≥z−xi,and(119)z−xi≤z−u+u−xi≤98λi.Then for anyxj≠xi in P, from Lemma 1, we know(120)z−xj≥xj−xi−z−xi≥κκ+1maxdGxi,dGxj−98λi≥z−xi.This and equation (118) imply(121)dG0z=z−xi.In the sequel, we are going to establish a relation between the quantitieskGA\Bxi,λi and rGF as stated in equation (126). For this, let z∈F and u∈A∩Sxi,λi. Then it follows from equation (121) that(122)dG0z≥xi−u−u−z≥λi−diamA\Bxi,λi≥78λi,and(123)dGz≤dGxi+z−u+u−xi≤dGxi+98λi.SincedGxi=4κ+1/κλi, we see that(124)dG0z≥7κ41κ+32dGz,which implies(125)ψrG0F≤ψ41κ+327κrGF.Then, the factA\Bxi,λi⊂F and the assumption G0 being ψ-natural guarantee that(126)kGA\Bxi,λi≤kG0F≤ψ41κ+327κrGF.Next, let us estimate the quantityrGF in terms of rGA\Bxi,λi as indicated in equation (131). Because for any u∈F∩Sxi,λi,(127)dGF≥dGxi−distxi,F≥dGxi−xi−u=3κ+44κ+1dGxi,and(128)dGA\Bxi,λi≤dGxi+distxi,A\Bxi,λi≤5κ+44κ+1dGxi,we obtain that(129)dGF≥3κ+45κ+4dGA\Bxi,λi.This, together with the factdiamF≤2diamA\Bxi,λi, shows that(130)rGF≤diamFdGF≤25κ+43κ+4diamA\Bxi,λidGA\Bxi,λi=25κ+43κ+4r¯GA\Bxi,λi.Then, it follows from equation (15) that(131)rGF≤45κ+43κ+4rGA\Bxi,λi,and thus, equation (126) implies(132)kGA\Bxi,λi≤kG0F≤ψ45κ+441κ+327κ3κ+4rGA\Bxi,λi.Now, we conclude from equations (115) and (132) that(133)kGA\Bxi,λi≤ψ685κ+42κ3κ+4rGA\Bxi,λi.Thus, equation (102) gives(134)kGA≤kGA\Bxi,λi+kGA∩Bxi,λi≤ψ685κ+42κ3κ+4rGA+4ν2log1+5κ+4κrGA.Case 7.
Suppose thatCardBA≥2.This assumption implies thatA\D≠∅. As the preparation, we are going to get an estimate on the quantity kGA with the aid of kGA\D as shown in equation (154).To this end, letx, y∈A be such that(135)kGA≤2kGx,y.Ifx,y∈A\D, then(136)kGA≤2kGx,y≤2kGA\D.In the following, we assume that at least one ofx and y belongs to some element of BA. Obviously, it suffices to consider the two cases: both x and y belong to some element(s) of BA, or y∈A\D and x belongs to an element of BA.For the first case, there must be two elementsBxi,λi, Bxj,λj in BA so that(137)x∈Bxi,λiandy∈Bxj,λj.Ifxi≠xj. Since A is connected, we see that A∩Sxi,λi≠∅≠A∩Sxj,λj. Let(138)w1∈A∩Sxi,λiandw2∈A∩Sxj,λj.Sincex,w1∈B¯xi,λi, by [[21], Lemma 2.2(2)], we get(139)kGx,w1≤4κ+13κ+4x−w1dGxi≤2κ3κ+4.On the other hand, by Lemma1, we have(140)w1−w2≥xi−xj−w1−xi−w2−xj≥κ2κ+1maxdGxi,dGxj,and(141)dGw1≤xi−w1+dGxi≤5κ+44κ+1dGxi.Therefore [[21], Lemma 2.2(1)] leads to(142)kGw1,w2≥log1+w1−w2dGw1≥log7κ+45κ+4,which, together with equation (139), gives(143)kGx,w1≤c1kGw1,w2,where c1=2κ/3κ+4log7κ+4/5κ+4.Similarly, we have(144)kGy,w2≤c1kGw1,w2.These estimates, together with equation (135), ensure that(145)kGA≤2kGx,w1+kGw1,w2+kGy,w2≤22c1+1kGA\D.Ifxi=xj, i.e., x,y∈Bxi,λi, then by [[21], Lemma 2.2(2)], we obtain(146)kGx,y≤11−λix−ydGxi≤2κ3κ+4.SinceCardBA≥2, we see that BA contains two elements of Bxi1,λi1 and Bxj1,λj1 such that xi1≠xj1. Let(147)w3∈A∩Sxi1,λi1andw4∈A∩Sxj1,λj1.Then, the similar reasoning as in the arguments of equation (142) gives(148)kGw3,w4≥log7κ+45κ+4,which, together with equation (146), implies that(149)kGx,y≤c1kGw3,w4,and thus, equation (135) leads to(150)kGA≤2c1kGw3,w4≤2c1kGA\D.For the second case, that is,y∈A\D and x∈Bxi,λi for Bxi,λi∈BA, since CardBA≥2, there is an element Bxj,λj∈BA such that xj≠xi. Because A is connected, we take(151)w5∈A∩Sxi,λiandw6∈A∩Sxj,λj.Then, the similar reasoning as in the arguments of equation (143) gives(152)kGx,w5≤c1kGw5,w6,which implies(153)kGA≤2kGx,y≤2kGx,w5+kGw5,w6+kGy,w6≤2c1kGw5,w6+kGw5,w6+kGy,w6≤2c1+2kGA\D.Now, we deduce from equations (136), (145), (150), and (153) that(154)kGA≤2c1+2kGA\D.Still, we need a relationship between the quantityrGA\D∪SA and the one rGA\D as indicated in equation (176). Recall that the set A\D∪SA is connected.First, we establish a relationship between the quantitiesdiamA\D∪SA and diamA\D as indicated in equation (165).By Lemma3, we see that diamSA<+∞. Let u1, v1∈SA be such that(155)diamSA≤2u1−v1.Then, there arexm and xn∈XA such that(156)u1∈Sxm,λmandv1∈Sxn,λn.We shall show thatxm≠xn. Suppose on the contrary that xm=xn. Then,(157)u1−v1≤diamSxm,λm≤2λm.This, together with equation (155), implies(158)diamSA≤4λm.SinceCardBA≥2, there must an element Sxm1,λm1 in SA such that xm1≠xm. Then,(159)diamSA≥diamSxm,λm∪Sxm1,λm1>xm−xm1≥κκ+1maxdGxm,dGxm1,where the last inequality follows from Lemma 1. This contradicts equation (158). Hence xm≠xn.It follows that (by Lemma1)(160)u1−v1≤u1−xm+xm−xn+xn−v1≤κ2κ+1maxdGxm,dGxn+xm−xn≤32xm−xn.Let(161)u2∈A∩Sxm,λmandv2∈A∩Sxn,λn.Then (by Lemma1),(162)u2−v2≥xm−xn+xm−u2+xn−v2≥xm−xn−κ2κ+1maxdGxm,dGxn≥12xm−xn,and thus,(163)u1−v1≤3u2−v2.This, together with equation (155), gives(164)diamSA≤6u2−v2≤6diamA\D,and so,(165)diamA\D∪SA≤diamA\D+diamSA≤7diamA\D.Also, we need a relationship between the quantitiesdGA\D∪SA and dGA\D as indicated in equation (174). To reach this goal, let u3∈SA be such that(166)dGu3≤2dGSA,and then, there is xt∈XA such that(167)u3∈Sxt,λt,and let(168)v3∈A∩Sxt,λt.Then, we have(169)dGu3≥dGxt−u3−xt≥3κ+44κ+1dGxt,and(170)dGv3≤dGxt+v3−xt≤5κ+44κ+1dGxt.These imply(171)dGSA≥12dGu3≥3κ+48κ+1dGxt,and(172)dGA\D≤dGv3≤5κ+44κ+1dGxt.Therefore,(173)dGSA≥3κ+425κ+4dGA\D,and thus, we obtain(174)dGA\D∪SA=mindGA\D,dGSA≥3κ+425κ+4dGA\D.Now, it follows from equations (165) and (174) that(175)rGA\D∪SA≤diamA\D∪SAdGA\D∪SA≤145κ+43κ+4diamA\DdGA\D=145κ+43κ+4r¯GA\D,and so, equation (15) gives(176)rGA\D∪SA≤285κ+43κ+4rGA\D.Now, we are ready to finish the proof of the lemma. Since the setA\D∪SA⊂G\D is connected, we infer from the similar arguments as in Case 5 that(177)kGA\D≤kGA\D∪SA≤ψ25κ+4κrGA\D∪SA,and thus, equation (176) gives(178)kGA\D≤ψ285κ+43κ+425κ+4κrGA\D.Then, we know from equation (154) that(179)kGA≤2c1+2ψ285κ+43κ+425κ+4κrGA\D≤2c1+2ψ285κ+43κ+425κ+4κrGA.By letting(180)φt=2c1+2ψ685κ+42κ3κ+4t+8ν2log1+5κ+4κt,we conclude from the estimates equations (97), (103), (134), (150), and (179) that the lemma is true.Proof. of Theorem 3.
The necessity part of the theorem follows from Lemma8, and its sufficiency part follows from Lemma 9. Hence the theorem is proved.



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