Common Fixed Points for a Countable Family of Non-self Mappings in Cone Metric Spaces with the Convex Property

PIAO Yong-jie,LI Chun-hua

(College of Science,Yanbian University,Yanji 133002,China)

Common Fixed Points for a Countable Family of Non-self Mappings in Cone Metric Spaces with the Convex Property

PIAO Yong-jie,LI Chun-hua

(College of Science,Yanbian University,Yanji 133002,China)

A new common f i xed point result for a countable family of non-self mappings def i ned on a closed subset of a cone metric space with the convex property is obtained,and from which,a more general result is given.Our main results improve and generalize many known common f i xed point theorems.

common f i xed point;the convex property;cone metric space

§1.Introduction and Preliminaries

Huang and Zhang[1]recently have introduced the concept of cone metric spaces,where the set of real number is replaced by an ordered Banach space,and they have established some fi xed point theorems for a contractive type mapping on a normal cone metric space. Subsequently,some other authors[27]have generalized the results of Huang and Zhang[1]and have studied the existence of common fi xed points of a fi nite family of self mappings satisfying a contractive type condition in the framework of normal or non-normal cone metric spaces.In [8],the authors discussed some common fi xed point problems of a pair of non-self mappings de fi ned on a nonempty closed subset of a non-normal cone metric space.On the other hand, the authors recently have discussed and obtained some unique existence theorems of common fi xed points for a countable family of mappings on 2-metric spaces or metrically convex metric spaces respectively,see[9-12].

In this paper,we will give some common f i xed point theorems for a countable family of non-self mappings def i ned on a nonempty closed subset of a non-normal cone metric space with the convex property.

Let E be a real Banach space.A subset P0of E is called a cone if and only if

(i)P0is closed,nonempty and P0/={0};

(ii)a,b∈?,a,b≥0 and x,y∈P0implies ax+by∈P0;

(iii)P0∩(?P0)={0}.

Given a cone P0?E,we def i ne a partial ordering≤on E with respective to P0by x≤y if and only if y?x∈P0.We will write x＜y to indicate that x≤y but x/=y,while x?y will stand for y?x∈int P0(the interior of P0).

The cone P0is called normal if there is a number L＞0 such that for all x,y∈E,

The least positive number satisfying the above is then called the normal constant of P0.

Let X be a nonempty set.Suppose that the mapping d:X×X→E satisf i es

(d1)0≤d(x,y)for all x,y∈X and d(x,y)=0 if and only if x=y;

(d2)d(x,y)=d(y,x)for all x,y∈X;

(d3)d(x,y)≤d(x,z)+d(z,y),for all x,z,y∈X.

Then d is called a cone metric on X and(X,d)is called a cone metric space.

Let(X,d)be a cone metric space.We say that{xn}?X is

(e)Cauchy sequence if for every c∈E with 0?c,there is an N such that for all n,m＞N, d(xm,xn)?c;

(f)Convergent sequence if for every c∈E with 0?c,there is an N such that for all n＞N such that d(xn,x)?c for some x∈E.In this case,we say that x is the limit of{xn}and

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

A cone metric space X is said to have the convex property if for each nonempty closed subset C of X and each x∈C and y/∈C,there exists a point z∈?C such that

A metric space is said to be metrically convex[1314],if for any x,y∈with x/=y,there exists a point z∈X such that d(x,z)+d(z,y)=d(x,y).

Lemma 1[14]If K is a nonempty closed subset of a complete metrically convex space, then for any x∈K and y/∈K,there exists a point z∈?K such that

The above Lemma 1 shows that a complete metrically convex space is the example of a cone metric spaces with the convex property.

Lemma 2[8]Let(X,d)be a cone metric space.Then the following properties are often useful(particulary when dealing with cone metric spaces in which the cone needs not to be normal)

(P1)If u≤v and v?w,then u?w;

(P2)If 0≤u?c for each c∈intP0,then u=0;

(P3)If x≤y+c for each c∈intP0,then x≤y;

(P4)If 0≤x≤y and a∈? with a≥0,then 0≤ax≤ay;

(P5)If 0≤xn≤ynfor each n∈? and limn→∞xn=x,limn→∞yn=y,then 0≤x≤y;

(P6)If E is real Banach space with a cone P0and a≤λa where a∈P0and 0＜λ＜1, then a=0;

(P7)If c∈intP0,0≤anand an→0,then there exists n0such that for all n＞n0,we have an?c.

Remark 1It follows from(P7)that the sequence xnconverges to x∈X if d(xn,x)→0 as n→∞and xnis a Cauchy sequence if d(xn,xm)→0 as n,m→∞.In the case when the cone is not necessarily normal,we have only one half of the statements of Lemma 1 and Lemma 4 from[1].

The following is a particular form of the well-known result in[15].

Lemma 3Let(X,d)be a cone metric space with a cone P0,{xn}a sequence in X and {an}a sequence in P0and an→0.If for any m＞n＞1,d(xn,xm)≤an,then{xn}is a Cauchy sequence.

§2.Main Results

Theorem 1Let K be a nonempty complete and closed subset of a cone metric space X with the convex property,{Ti:K→X}i∈?a family of non-self mappings satisfying that there exists λ∈(0,)such that for each x,y∈K and i,j∈? with i/=j,

where

If Ti(x)∈K for all x∈?K and i∈?,then{Ti}i∈?have a unique common fi xed point in K.

ProofTake x0∈K.We will construct two sequences{xn}and{in the following manner.De fi ne=T1x0.If∈K,then put x1=∈/K,then by the convex property of X,there exists x1∈?K such that d(x0,x1)+d()=d().De fi ne=T2x1.If∈K,then put x2=∈/K,then by the convex property of X,there exists x2∈?K such that d(x1,x2)+d()=d().Continuing this way,we obtain{xn}and{x′n}

where

hence

hence

Case IIIf xn∈P and xn+1∈Q,then=Tn+1xn. Hence

where

un,n+1(xn?1,xn)

hence

hence

where

By Case II,we obtain that

hence we obtain that

By Case II again,we have

hence

But d(xn,xn+1)≤d(xn?1,+dn+1),hence we obtain that

and therefore

By Case II again,we have

or

and therefore for any n∈? with n≥2,

or

So,for any n∈? with n≥2,

Let δ=h?1[d(x2,x1)+d(x1,x0)]and K=h12,then K＜1,δ∈P0and for all m＞n≥2,

By the properties of P and Q,we can see that there are in fi nite elements xnk+1∈{xn}such that xnk+1∈P.

For any fi xed n∈?,there exists an enough large k∈? such that nk+1＞n and xnk+1∈P. And we can obtain the following

where

If un,nk+1(x?,xnk)=d(x?,xnk),then for any c∈intP0,there exists a large k0∈? such that for k≥k0,

Hence

and therefore,for any c∈intP,there exists a large k0∈? such that for k≥k0,

Hence we get that d(Tnx?,x?)?c for all c∈intP,so by Lemma 2(P2),Tnx?=x?for all n∈?.This means that x?is a common f i xed point of{Tn}n∈?.

Suppose that p and q are all common f i xed points of{Tn}n∈?,then

If u1,2(p,q)=d(p,q),then d(p,q)≤λd(p,q),hence d(p,q)=0 by Lemma 2(P6)and therefore p=q;

Hence x?is the unique common fi xed point of{Tn}n∈?.

Remark 2If K=X itself is complete,then the boundary condition is super fl ous.In this case,we can easily know that xn=x′n,hence the convex structure of X is also super fl ous.

Remark 3In fact,the condition“i/=j”in Theorem 1 can be replaced by the weaker condition“i＜j”.

Remark 4Many authors in the references and others obtained many common fi xed point theorems only for a fi nite family of mappings,but we fi rst introduced the concept of the convex property to discuss the existence of common fi xed point for a countable family of non-self-mappings on cone metric spaces in Theorem 1.Since we treat non-self-mappings,we need to consider the boundary condition of the given closed subset K of X in Theorem 1.The boundary condition in Theorem 1 is very weaker than that in[8]and very di ff erent from that in[8].So,we think that our technique is very di ff erent from the previous ones and our method is new.

Theorem 2Let K a nonempty complete and closed subset of a cone metric space(X,d) with the convex property,{Ti,j:X→X}i,j∈?a family ofmappings,{mi,j}i,j∈?a family of positive integral numbers such that there exists λ∈(0,)such that for each x,y∈X and i1,i2,j∈? with i1/=i2,

where

Furthermore,if(a)for each i,j∈?,(?K)?K,(b)for each i1,i2,μ,ν∈? withμ/=ν, Ti1,μTi2,ν=Ti2,νTi1,μ.Then{Ti,j}i,j∈?has a unique common fi xed point in K.

where

If ui,k,j(Ti,j(pj),Ti,j(pj))=0,then d(Ti,j(pj),Sk,j(Ti,j(pj)))≤λ0=0,hence d(Ti,j(pj), Sk,j(Ti,j(pj)))=0,i.e.,Ti,j(pj)=Sk,j(Ti,j(pj));

Hence in any situation,we have that Ti,j(pj)is a fi xed point of Sk,jfor each k with k/=i. So Ti,j(pj)is a common fi xed point of{Si,j}i∈?.By uniqueness of common fi xed points of {Si,j}i∈?,we haveTi,j(pj)=pjfor each i∈?.Hence pjis a common fi xed point of{Ti,j}i∈?.

If ujand vjare common fi xed points of{Ti,j}i∈?,then they are also common fi xed points of{Si,j}i∈?,hence ui=pj=vj.So j∈?,{Ti,j}i∈?has a unique common fi xed point pj.

Finally,we will prove that{Ti,j}i,j∈?has a unique common fi xed point.Now,we prove that for eachμ,ν∈?,pμ=pν.In fact,for any i1,i2,μ,ν∈? withμ/=ν,since Ti1,μ(pμ)=pμand Ti2,ν(pν)=pν,Ti1,μ(Ti2,ν(pν))=Ti1,μ(pν),hence Ti2,ν(Ti1,μ(pν))=Ti1,μ(Ti2,ν(pν))=Ti1,μ(pν) by(b).This means that Ti1,μ(pν)is a fi xed point of Ti2,νfor each i2,i.e.,Ti1,μ(pν)is a common fi xed point of{Ti2,ν}i2∈?.But{Ti2,ν}i2∈?has a unique common fi xe point pν,hence Ti1,μ(pν)=pνfor each i1and therefore pνis a common fi xed point of{Ti1,μ}i1∈?.But {Ti1,μ}i1∈?has a unique common fi xed point pμ,hence pμ=pν.Let p?=pj,then p?is thecommon f i xed point of{Ti,j}i,j∈?.The uniqueness of common f i xed points of{Ti,j}i,j∈?is obvious.

Remark 5In Theorem 2,the domain of{Ti,j}i,j∈?must be X.In fact,we can not be sure that Ti,j(pj)∈K even if pj∈K in the proof of Theorem 2.Hence we should not suppose that the domain of{Ti,j}i,j∈?is K.

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tion:47H05,47H10

CLC number:O189.1,O177.91Document code:A

1002–0462(2014)02–0221–10

date:2012-07-19

Supported by the National Natural Science Foundation of China(11361064)

Biographies:PIAO Yong-jie(1962-),male(Chaoxianzu),native of Jiutai,Jilin,a professor of Yanbian University,Ph.D.,engages in nonlinear theory and space theory;LI Chun-hua(1975-),female(Chaoxianzu),native of Baishan,Jilin,a lecturer of Yanbian University,Ph.D.,engages in functional analysis and space theory.