Common Fixed Point Results for Quasi-contractions Involving Altering Distance Fu

(School of Mathematics and Information Science,Leshan Normal University,Leshan 614000,China;School of Mathematics and Statistics,Wuhan University,Wuhan 430072,China)

§1. Introduction

As a fundamental result in fixed point theory,Banach contraction mapping principle is extended and generalized by many authors in various directions.Alber Y and Guerre-Delabriere S[1]introduced the concept of weakly contractive mappings and proved some fixed point theorems in Hilbert spaces.Rhoades B E[2]studied the fixed point problems of weakly contractive mappings and obtained some fixed point theorems in complete metric spaces.After Khan M S et al[3]introduced the notion of altering distance functions in metric spaces,some more general contractive conditions were studied.At the same time,many authors also studied common fixed point problems for two or more mappings satisfying certain contractive conditions,and some fixed point and common fixed point results for mappings satisfying weakly contractive conditions were obtained(see[4-7]).The initial research of common fixed point problems depended on the assumption that all of maps are commutative,Jungck G[8]de fined the concept of weakly compatible pairs of mappings,which is weaker than commutativity.Recently,several authors studied generalized weak contractive mappings in an ordered metric space which is a metric space endowed with a partially ordering,and obtained some coincidence point results.For details,see[9-12].The notion of cone metric spaces was introduced by Huang L G et al[13]and some fixed point theorems were obtained in cone metric spaces.Subsequently,some fixed point and common fixed point results were generalized from metric spaces to cone metric spaces.Han Y et al[14]obtained some new theorems for Lipschitz type mappings in cone metric spaces.Choudhury B S et al[15-16]generalized the notion of altering distance functions from metric spaces to cone metric spaces and obtained some fixed point and common fixed point results for generalized weak contractive mappings in cone metric spaces or in partially ordered cone metric spaces.In this paper,we shall show that some common fixed point results for four mappings satisfying quasi-contractive conditions under an altering distance function in partially ordered cone metric spaces.

§2. Preliminaries

Let E be a real Banach space,P a subset of E,and θ the zero element in E.If

(i)P is closed,nonempty and

(ii)a,b∈R,a,b≥0 and x,y∈P imply that ax+by∈P;

(iii)x∈P and−x∈P imply that x=θ;then P is called a cone.We denote the interior of P by intP.If intP is called a solid.

接受新教学法实习学生的理论知识及临床技能分数，均优于传统教学法组学生，差异具有统计学意义（P＜0.05），见表1。

Given a cone P⊂E,we can de fine a partial ordering“≤”with respect to P on E as follows:x≤y if and only if y−x∈P,x

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The cone P is said to be normal if there is a number K>0 such that for all x,y∈E,θ≤x≤y impliesThe least number satisfying this inequality is called the normal constant of P.The cone P is said to be regular if every increasing sequence which is bounded from the above is convergent,that is,if{xn}is a sequence such that x1≤ x2≤ ···≤ xn≤ ···≤p for some p∈E,then there exists x∈E such that kxn−xk→0 as n→∞.Equivalently,the cone P is regular if and only if every decreasing sequence which is bounded from the below is convergent.Every regular cone is a normal cone.

De finition 2.1[13]Let X be a nonempty set,E a real Banach space and P a cone in E.Suppose the mapping d:X×X→E satis fies following conditions.

(i)θ≤d(x,y)for all x,y∈X and d(x,y)=θ if and only if x=y;

(ii)d(x,y)=d(y,x)for all x,y∈X;

(iii)d(x,y)≤d(x,z)+d(z,y)for all x,y,z∈X;then d is called a cone metric on X and(X,d)is called a cone metric space.

根据本组研究结果,我们认为在采用注射器抽吸负压状态下同轴取材活检更容易成功,并且不增加穿刺活检相关并发症,而有助于明确椎体压缩骨折病因,减少椎体压缩骨折病因的误诊和漏诊,使患者获得更长期的良好疗效。因此,经皮椎体成形术中在负压状态下同轴取材活检是一种安全易行、提高诊断准确性的活检方法,应该在椎体压缩骨折行椎体成形术时常规进行。

ψ :P → P is an altering distance function and ϕ :intP ∪{θ} → intP ∪ {θ}is a continuous function with the properties as follows.

By a same arguments as in Caes 2,we get r=θ.Therefore

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(ii){xn}is a Convergent sequence if for every c in E with θ¿ c,there is a positive integer number N such that d(xn,x)¿c for all n>N and some point x in X.We denote this byor xn→x as n→∞.

A cone metric space X is said to be complete if every Cauchy sequence in X is convergent.

Proposition 2.1[13]Let(X,d)be a cone metric space and P a normal cone with normal constant number K.Let{xn}be a sequence in X.Then

(i){xn}converges to x for some x∈X if and only if d(xn,x)→θ(n→∞).

(ii){xn}is a Cauchy sequence if and only if d(xn,xm)→θ(n,m→∞).

Remark 2.1Let E be a real Banach space with cone P.The following properties are often used.

Similarly,we can obtain that

(ii)Let a,b,c∈E.If a≤b and b¿c,then a¿c.

(iii)Let a∈E.If θ≤ a¿c for each c∈ intP,then a= θ.

(iv)Let a∈ P and λ be a real number with 0< λ <1.If a≤ λa,then a= θ.

(v)Let θ≤ x ≤ y and a be a real number with a≥ 0,then θ≤ ax ≤ ay.

(vi)Let θ ≤ xn≤ ynfor n=1,2,···,and limn→∞xn=x,limn→∞yn=y,then θ ≤ x ≤ y.

(vii)P is a normal cone if and only if xn≤ yn≤ znand limn→∞xn=limn→∞zn=w imply limn→∞yn=w.

De finition 2.3[16]Let ψ :P →P be a function.

(i)ψ is called strongly monotone increasing if x ≤ y if and only if ψ(x)≤ ψ(y)for all x,y∈P.

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(ii)ψ is said to be continuous at x0∈P if for any sequence{xn}in P,xn→x0implies ψ(xn)→ψ(x0)as n→∞.

De finition 2.4[16]The function ψ :P →P is called an altering distance function if

(i)ψ is continuous and strongly monotone increasing.

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(ii)ψ(t)= θ if and only if t= θ.

大数据时代下的开放性创新实验室教学管理，其不仅仅只是展现在实验设备上面的革新，它往往还会出现在教学模式上，具体来讲也就是，相关教师可以根据大数据的信息，科学有效的提前做好有关实验器材设备的准备工作，并且在具体的实验开展过程中，明确的了解学生的实验数据，这是因为学生所研究的实验，最终都会以数据的形式保存在相应的数据信息库当中，这种信息的保存与每位学生都是相互联系的，这样大大地方便了教师的研究观察。

Proposition 2.2[15]Let(X,d)be a cone metric space and P a regular cone such that d(x,y)∈ intP for x,y ∈ X withSuppose that function ϕ :intP ∪ {θ} → intP ∪{θ}satis fies the following properties.

(i)ϕ(t)= θ if and only if t= θ;

(ii)ϕ(t)¿t for t∈ intP;

家长进行家长助教时要做好充分的物质准备，教师应该指导家长准备好充足的物质材料，并且明确各个材料投放的具体环节，对教材的应用做到心中有数。如，对于幼儿的操作材料，教师应该先告知家长班级人数，请家长按照多于班级人数的份额准备材料，并按小组分门别类地放好，以方便操作。让家长做好充足的物质准备，便于活动开展。

(iii)Either ϕ(t)≤ d(x,y)or d(x,y)≤ ϕ(t)for t∈ intP ∪{θ}and x,y∈ X.

If{xn}is a sequence such that{d(xn,xn+1)}is monotonic decreasing,thenis convergent to either r=θ or r∈intP.

De finition 2.2[13]Let(X,d)be a cone metric space,{xn}is a sequence in X.We say that

Proposition 2.3[16]Let(X,d)be a cone metric space and ϕ :intP ∪{θ}→ intP ∪{θ}be a function with following properties.

(i)ϕ(t)= θ if and only if t= θ;

(ii)ϕ(t)¿t for t∈ intP.

Then a sequence{xn}in X is a Cauchy sequence if and only if for each c∈ E with θ¿ c,there exists a positive integer number n0such that d(xn,xm)¿ ϕ(c)for all n>m>n0.

De finition 2.5[17]Let(X,„)be a partially ordered set and F,G,H:X → X be mappings such that F(X)⊆H(X)and G(X)⊆H(X).We say that F and G are weakly increasing with respect to H,if for all x∈X,we havefor each y∈H−1(Fx)andfor each y∈H−1(Gx).If F=G,we say that F is weakly increasing with respect to H.

De finition 2.6[10]Letbe a partially ordered set and F,G,H:X → X be mappings such that F(X)⊆H(X),(F,G)is said to be partially weakly increasing with respect to H,if for allfor any y∈H−1(Fx).

Let X be a nonempty set and F,G:X→X be two mappings.If w=Fx=Gx for some x in X,then x is called a coincidence point of F and G,and w is called a point of coincidence of F and G.

De finition 2.7[18]The mappings F,G:X→X are weakly compatible if,for every x∈X,FGx=GFx holds whenever Fx=Gx,that is,F and G commute at every their coincidence point.We say that{F,G}is a weakly compatible pair.

A partially ordered cone metric space is a cone metric space(X,d)endowed with a partial ordering,we denoted it by(X,„,d).

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De finition 2.8Let(X,„,d)be a partially ordered cone metric space and F:X → X be a mapping.We say that F has the property(A)if{xn}is a sequence in X such that{Fxn}is a nondecreasing sequence,converging to Fx0for some x0∈ X,thenfor n=1,2,···,and

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§3. Main Results

Let X be a set and F,G,T,S:X→X be four mappings such that F(X)⊆T(X),and G(X)⊆S(x).Let x0be a arbitrary point in X,then there exist x1,x2∈X such that Fx0=Tx1,Gx1=Sx2.By the induction,we can construct two sequences{xn},{yn}in X such that y2n=Fx2n=Tx2n+1,y2n+1=Gx2n+1=Sx2n+2,n=0,1,2,···.{yn}is called a(F,G)−(T,S)-sequence with initial point x0.We always suppose that P is a solid in E.

Theorem 3.1Letbe a partially ordered complete cone metric space and P a regular cone such that d(x,y)∈intP for x,y∈X withSuppose that F,G,S,T:X→X are four mappings such that F(X)⊆T(X),G(X)⊆S(X)and,both T(X)and S(X)are closed subsets in X.If the following conditions are satis fied.

(1)There exists u(x,y)∈MF,G,S,Tsuch that

for x,y∈X such that Sx and Ty are comparable,where

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(i)ϕ(t)= θ if and only if t= θ;

(ii)ϕ(t)¿t for t∈ intP;

(iii)Either ϕ(t)≤ d(x,y)or d(x,y)¿ ϕ(t)for t∈ intP ∪{θ}and x,y∈ X;

(2)(F,G)is partially weakly increasing with respect to T and(G,F)is partially weakly increasing with respect to S;

(3)S and T have the property(A);

(4){F,S}and{G,T}are weakly compatible pairs.

Then F,G,S and T have a common fixed point.

ProofLet x0be a arbitrary point in X,and{yn}be a(F,G)−(T,S)-sequence with initial point x0.Sincefrom the condition(2),we can obtain that.Hence,Sx2nand Tx2n+1are comparable.From(3.1),we have

where

and n=1,2,···.For every n,there are three possible cases.

Case 1Then

By using the strongly monotone property of ψ,we have

Case 2Then

From the strongly monotone property of ψ,it follows thatwhich implies that

Case 3By using triangular inequality,d(y2n−1,y2n+1)≤then

Also we can obtain that

(i)Let a,b,c∈E.If a¿b and b¿c,then a¿c.

From(3.3)and(3.4),{d(yn+1,yn)}is a monotonic decreasing sequence.By Proposition 2.2,there exists r∈P with either r=θ or r∈intP such that d(yn+1,yn)→r as n→∞.

Next,we show that r=θ.Since

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for n=1,2,···.(3.2)implies that there is a sequence{nk}of positive integer numbers with nk

点击菜单[选择][按位置选择]，打开选择对话框，在[目标图层]勾选建筑物层，[源图层]选择阴影栅格转换而成的矢量图层，[空间选择方法]应选 “目标图层要素的质心在源图层要素内”；根据需要，设置其余参数，点击 [确认]完成查询。建筑物已选中。右击查询后的建筑物图层，选择[数据]→[导出数据]，打开工具对话框，完成设置，输出不符合规范的建筑物。如图12中，深色表示不符合规定的建筑。

Case 1for k=1,2,···.By taking k → ∞ in the inequality above,we have ψ(r)≤ ψ(r)− ϕ(r)from the continuity of ψ and ϕ,and Remark 2.1(vi).Hence ϕ(r)= θ,that is,r= θ.

Case 2for k=1,2,···.By taking k → ∞ in the inequality above and using the continuity of ψ and ϕ,and Remark 2.1(vi),also we have ψ(r)≤ ψ(r)− ϕ(r).So r= θ.

Case 3for k=1,2,···.By using the triangular inequality,.From the strongly monotonic property,we obtain that

(i){xn}is a Cauchy sequence if for every c in E with θ¿ c,there is a positive integer number N such that d(xn,xm)¿c for all n,m>N.

as n→∞.Now,we show that{yn}is a Cauchy sequence.From(3.5),it is sufficient to prove that{y2n}is a Cauchy sequence.If not,by proposition2.3,there exists c ∈ E with θ¿ c such that,for each positive integer number k,there exist positive integer numbers nkand mkwith nk>mk>k such that d(y2mk,y2nk)¿ ϕ(c)is not true.From a property of ϕ,we have ϕ(c)≤ d(y2mk,y2nk).For further,we can choose nksuch that nkis the smallest in such positive integer numbers,then

On the other hand,by using the triangular inequality,we obtain

By taking k→∞in(3.7)and using Remark 2.1(vii),we get

as k→∞.Similarly,by using the triangular inequality,we can also obtain

and

By taking k→∞in the six inequalities above and using Remark 2.1(vii),we have

and

as k→∞.Since Sx2nkand Tx2mk+1are comparable,from(3.1),we have

where

and k=1,2,···.There is a sequence{ki}of positive integer numbers with ki

Case 1for i=1,2,···.ThenBy taking i→ ∞ and using(3.9),(3.10)and the continuity of ψ and ϕ,we get ψ(ϕ(c))≤ ψ(ϕ(c))− ϕ(ϕ(c)),hence ϕ(ϕ(c))= θ.It yields that c=θ.

Case 2for i=1,2,···.Then

By taking i→ ∞ and using(3.5),(3.9),(3.10)and the continuity of ψ and ϕ,we have ψ(ϕ(c))≤ψ(θ)− ϕ(ϕ(c)).Since θ≤ ψ(ϕ(c)),we get ϕ(ϕ(c))= θ,that is,c= θ.

Case 3for i=1,2,···.Then

By taking i→ ∞ in the inequality above and using(3.8)∼(3.11)and the continuity of ψ and ϕ,we obtain ψ(ϕ(c))≤ ψ(ϕ(c))− ϕ(ϕ(c)),which is true unless ϕ(ϕ(c))= θ.Hence c= θ.

Now we come to a contradiction,therefore{yn}is a Cauchy sequence.

From the completeness of X,there exists a point w∈X such that yn→w as n→∞.Since y2n∈T(X),y2n+1∈S(X)and,T(X)and S(X)are closed in X,there exist z1,z2∈X such that w=Tz1=Sz2.From the condition(3)of the theorem,Tx2n+1and Sz2(=Tz1)are comparable.By(3.1),we have

where

and n=1,2,···.There is a sequence{nk}of positive integer numbers with nk+1>nk(k=1,2,···)such that the one of following three cases is true:

Case 1fork=1,2,···.ThenBy taking k → ∞ and using the continuity of ψ and ϕ,we obtain ψ(d(Fz2,w)≤ ψ(θ)−ϕ(θ)= θ.So d(Fz2,w)= θ,that is,Fz2=w.

Case 2for k=1,2,···.Then

By taking k → ∞ in the inequality above,we getFrom the strongly monotone property of ψ,d(w.Fz2)= θ.So Fz2=w.

Case 3for k=1,2,···.Then

By taking k → ∞ in the inequality above,also we obtainHence Fz2=w.Therefore,Fz2=Sz2=w,z2is a coincidence point of F and S and w is a point of coincidence of F and S.

Similarly,since Sx2nand Tz1(=Sz2)are comparable from the condition(3)of the theorem,as we do it in the above,we can obtain that z1is a coincidence point of G and T and w is a point of coincidence of G and T.

If F and S are weakly compatible,then

Sincefrom(3.1)and(3.12),we have

where

If u(w,z1)=d(Fw,w),then ψ(d(Fw,w))≤ ψ(d(Fw,w))−ϕ(d(Fw,w)).From the strongly monotone property of ψ,we have ϕ(d(Fw,w))= θ,which implies that d(Fw,w)= θ.Hence Fw=w.

If u(w,z1)= θ,then ψ(d(Fw,w)) ≤ ψ(θ)− ϕ(d(Fw,w))= −ϕ(d(Fw,w)).Since θ ≤ψ(d(Fw,w)),which means that ϕ(d(Fw,w))= θ.Also we have Fw=w.

From(3.12),we can obtain that Sw=Fw=w.Similarly,we have Tw=Gw=w from that G and T are weakly compatible,that is,w is a common fixed point of F,G,S and T.

This completes the proof of the theorem.

Corollary 3.1Letbe a partially ordered complete cone metric space and P a regular cone such that d(x,y)∈intP for x,y∈X withSuppose that F,G,H:X→X are three mappings such that F(X)⊆H(X),G(X)⊆H(X)and H(X)is a closed subset in X.If the following conditions are satis fied.

(1)There exists u(x,y)∈ MF,G,Hsuch that ψ(d(Fx,Gy))≤ ψ(u(x,y))−ϕ(d(Hx,Hy))for x,y∈X such that Hx and Hy are comparable,where

ψ :P → P is an altering distance function and ϕ :intP ∪{θ} → intP ∪ {θ}is a continuous function with the properties as in Theorem 3.1.

(2)F and G are weakly increasing with respect to H;

(3)H has the property(A);

(4){F,H}and{G,H}are weakly compatible pairs.

Then F,G and H have a common fixed point.

ProofBy setting S=T=H in Theorem 3.1,we know that the conclusions in the Corollary are true.

Corollary 3.2Letbe a partially ordered complete cone metric space and P a regular cone such that d(x,y)∈intP for x,y∈X withSuppose that F,S,T:X→X are three mappings such that F(X)⊆S(X),F(X)⊆T(X)and,both S(X)and T(X)are closed subsets in X.If the following conditions are satis fied.

(1)There exists u(x,y)∈ MF,S,Tsuch that ψ(d(Fx,Fy))≤ ψ(u(x,y))−ϕ(d(Sx,Ty))for x,y∈X such that Sx and Ty are comparable,where

ψ :P → P is an altering distance function and ϕ :intP ∪{θ} → intP ∪ {θ}is a continuous function with the properties as in Theorem 3.1.

(2)F is weakly increasing with respect to S as well as T;

(3)S and T have the property(A);

(4){F,S}and{F,T}are weakly compatible pairs.

Then then F,S and T have a common fixed point.

ProofBy setting F=G in Theorem 3.1,we know that the conclusions in the corollary are true.

Example 3.1Let E=R2with usual norm,then E is a Banach space with the zero element θ=(0,0).Let P={(x,y):x,y ≥ 0},then P is a regular cone in E.Let X=[0,1]be a partially ordered set with orderingwhich is de fined by

for all x,y∈X,where“≤”is the usual ordering.We de fine d:X×X→E as follows

for x,y∈X,then(X,d)is a complete cone metric space and d(x,y)∈intP for x,y∈X withLet ψ :P → P and ϕ :intP ∪{θ} → intP ∪{θ}be two functions de fined by

for t∈ P and s∈ intP ∪{θ},where t=(t1,t2),s=(s1,s2)and v=min{s1,s2}.

Then ψ and ϕ are continuous functions, ψ(t)= θ if and only if t= θ and ψ is strongly increasing.If ϕ(s)= θ,then v=0.Note that s=(s1,s2) ∈ intP ∪ {θ}.So s1=s2=0,therefore s=θ.

For(s1,s2)∈ intP ∪ {θ}and x,y ∈ X,we suppose that ϕ(s)≤ d(x,y)is not true,thenthat is,ϕ(s)∈ intP andwhich implies that(|x−y|,|x−y|)¿that is,d(x,y)¿ ϕ(s).

Now,we de fine the four mappings from X to X as follows:for all x∈X and

Then S(X)and T(X)are closed in X,F(X)⊆T(X)and G(X)⊆S(X).Clearly,{F,S}and{G,T}are weakly compatible pairs of mappings.

Let{Sxn}be a nondecreasing sequence,converging to Sz.Ifthenas n→ ∞,that isas n→ ∞,it implies thatas n→∞in R which is the real number space.Since{Sxn}is a nondecreasing sequence,we haveand sofor all n=1,2,···.Clearly,.If,also we haveandHence,S has the property(A).Similarly,also T has the property(A).Since

It is easy to check that,for all x,y∈ X,ψ(d(Fx,Gy))≤ψ(u(x,y))−ϕ(d(Sx,Ty))holds.By Theorem 3.1,F,G,S and T have a common fixed point

Lemma 3.1Letbe a partially ordered complete cone metric space and P a regular cone such that d(x,y)∈ intP for x,y ∈ X withLet ϕ :intP ∪{θ}→ intP ∪{θ}be a continuous function with the properties as follows.

(i)ϕ(t)= θ if and only if t= θ;

(ii)ϕ(t)¿t for t∈ intP;

(iii)Either ϕ(t)≤ d(x,y)or d(x,y)≤ ϕ(t)for t∈ intP ∪{θ}and x,y∈ X.

Suppose that w is a point and{yn}is a sequence in X such that{d(w,yn)}is monotonic decreasing.Then{d(w,yn)}is convergent to either r=θ or r∈intP.

ProofSince P is a regular cone,and{d(w,yn)}is a decreasing sequence which is bounded from the below,there exists r∈P such that d(w,yn)→r as n→∞.If there exists n0such that d(w,yn0)= θ,then d(w,yn)= θ for n>n0,and so r= θ.Therefore,we can suppose that d(w,yn)∈ intP for n=1,2,···.

LetSince P is a regular cone,it is also a normal cone.Let B={t∈ intP:ktkm0.Let n→ ∞ in the inequality and by using Remark2.1(vi),we have r≤ ϕ(t0).From the condition(ii)of the lemma,we get r¿t0,which implies thatThis contradicts that t0∈B.

Hence,from the condition(iii)of the lemma,we have ϕ(t) ≤ d(w,yn)for t ∈ B and n=1,2,···.Let n → ∞ in the inequality,we get ϕ(t) ≤ r for t∈ B.Since ϕ(t) ∈ intP for t∈ intP,we havethat is,r∈ intP.

Theorem 3.2Letbe a partially ordered complete cone metric space,and P a regular cone such that d(x,y)∈intP for x,y∈X withLet F,G,S and T:X→X be four mappings satisfying conditions(1)∼(4)in Theorem 3.1.If following condition(5)is also satis fied.

(5)For x,y∈S(X)∩T(X),there exists z0∈X such thatandwhere-sequence with initial point z0andis a(G,F)−(S,T)-sequence with initial point z0.

Then the common fixed point of F,G,S and T is unique.

ProofLet w1and w2be two common fixed points of F,G,S and T.

If w1and w2are comparable,that is,Sw1and Tw2are comparable,then there existssuch that

where

Thenwhich implies that ϕ(d(w1,w2))=0.Hence d(w1,w2)=0,that is,w1=w2.

If w1and w2are not comparable,then there exists a point z0∈X such thatandandwhere{yn}is a(F,G)−(T,S)-sequence with initial point z0andis a(G,F)−(S,T)-sequence with initial point z0.There exist two sequenceswithsuch that

and

where n=0,1,2,···.As we do it in the proof of Theorem 3.1,we can prove that

Since,and,we get.From(3.1),we have

where

and n=1,2,···.For every n,there are three possible cases.

Case 1From(3.14),we haveBy using the strongly monotone property of ψ,we have

Case 2ThenBy using the strongly monotone property of ψ and the triangular inequality,we haveIt follows thatd(w1,y2n).

Case 3Also we can obtain that

By a similar argument to the above,we can obtain

In view of(3.16)and(3.17),{d(w1,yn)}is a monotonic decreasing sequence.From the lemma3.1,{d(w1,yn)}is convergent to either r=θ or r∈intP.

From(3.14)and(3.15),there is a sequence{nk}of positive integer numbers with nk

Case 1for k=1,2,···.ThenBy taking k → ∞ and using the properties of ψ and ϕ,we get ψ(r)≤ ψ(r)−ϕ(r).Hence ϕ(r)= θ and so r= θ.

Case 2for k=1,2,···.By the triangular inequality,

By taking k → ∞,we get ψ(r)≤ ψ(r)− ϕ(r).Hence ϕ(r)= θ,and so r= θ.

Case 3.for k=1,2,···.Then

Clearly,also we have r=θ.Therefore,

as n→∞.By a similar argument to the above,we can obtain that

as n→∞.Now,by using the triangular inequality,we have

where n=1,2,···.By taking n → ∞ in the inequality above and using(3.18),(3.19),we obtain d(w1,w2)=θ and so w1=w2.

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