Unique Common Fixed Points for Two Weakly C∗-contractive Mappings on Partially Ordered 2-metric Spaces

PIAO YONG-JIE

(Department of Mathematics,College of Science,Yanbian University,Yanji,Jilin,133002)

Communicated by Ji You-qing

1 Introduction and Preliminaries

Definition 1.1[1]–[3]A2-metric space(X,d)consists of a nonempty set X and a function d:X×X×X→[0,+∞)such that

(i)for distant elements x,y∈X,there exists a u∈X such that d(x,y,u)/=0;

(ii)d(x,y,z)=0if and only if at least two elements in{x,y,z}are equal;

(iii)d(x,y,z)=d(u,v,w),where{u,v,w}is any permutation of{x,y,z};

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

Choudhury[17]introduced the next definition in a real metric space:

Definition 1.3[17]Let(X,d)be a metric space and T:X→X be a map.T is said to be weak C-contraction if there exists a continuous function φ:[0,+∞)2→[0,+∞)with φ(s,t)=0⇐⇒s=t=0such that

Choudhury[17]also proved that any map satisfying the weakC-contraction has a unique fixed point on a complete metric space(see[17],Theorem 2.1).Later,the above result was extended to the case in a complete ordered metric spaces(see[18],Theorems 2.1,2.3 and 3.1).

In 2013,Definition 1.3 was extended to the case in a 2-metric space by Dung and Hang[10]as follows:

Definition 1.4[10]Let(X,≼,d)be a ordered2-metric space,T:X→X a map.T is said to be weak C-contraction if there exists a continuous function φ:[0,+∞)2→[0,+∞)with φ(s,t)=0⇐⇒s=t=0such that for any x,y,a∈X with x≼y or y≼x,

Dung and Hang[10]proved that any weaklyC-contractive map has fixed points on complete ordered 2-metric spaces(see[10],Theorems 2.3,2.4 and 2.5).The results generalized and improved the corresponding conclusions in[17]–[18].

Definition 1.5Let(X,≼,d)be a ordered2-metric space and S,T:X→X be two maps.S,T are said to be weakly C∗-contractive maps if there exists a continuous function φ:[0,+∞)2→[0,+∞)with φ(s,t)=0⇐⇒s=t=0such that for any x,y,a∈X with x≼y or y≼x,

d(Sx,Ty,a)≤kd(x,y,a)+l[d(x,Ty,a)+d(y,Sx,a)]−φ(d(x,Ty,a),d(y,Sx,a)),where k and l are two real numbers satisfying l＞0and0＜k+l≤1−l.

Definition 1.6[10]Let(X,d)be a2-metric space and a,b∈X,r＞0.The set

is said to be a2-ball with centers a and b and radius r.Each2-metric d on X generalizes a topology τ on X whose base is the family of2-balls.τ is said to be a2-metric topology.

Lemma 1.1[13]–[14]If a sequence{xn}n∈N+in a2-metric space(X,d)is convergent to x,then

Lemma 1.5[6]Each2-metric space is T2-space.

The purpose of this paper is to use the method in[10]to discuss and study the existence problems of common fixed points for two maps satisfying weaklyC∗-contractive condition on ordered 2-metric spaces and give a sufficient condition under which there exists a unique common fixed point.

2 Unique Common Fixed Points

Now,we discuss the existence problems of unique common fixed point for two maps on non-complete 2-metric spaces without ordered relation.

Theorem 2.1Let(X,d)be a2-metric space and S,T:X→X be two maps.Suppose that

where k,l are two real numbers such that l＞0and0＜k+l≤1−l.If S(X)or T(X)is complete,then S and T have a unique common fixed point.

Proof.Take any elementx0∈Xand construct a sequence{xn}satisfying

For anyn=0,1,2,···anda∈X,by(2.1),we can get

Takea=x2nin(2.2),we obtain

By using(2.3)and Definition 1.1(iv),we obtain from(2.2)that

which implies

Similarly,for anyn=0,1,2,···anda∈X,by(2.1),we have

Takea=x2n+1in(2.5),we obtain

By using(2.6)and Definition 1.1(iv),we obtain from(2.5)that

which implies

Combining(2.3),(2.4),(2.6)and(2.7),we have

It is easy to obtain

Letn→∞.Then from the first to third line,fifth line,sixth line in(2.9),we obtain

Letn→∞again.Then from(2.9),we obtain

Hence we have

Soξ(a)=0 by the property ofφ,that is,

By Definition 1.1(ii),

which implies that

by(2.8).Hence,by the mathematical induction,

And for any fixed pointm≥1,

Hence,by(2.8)and the mathematical induction,we have

For 0≤n＜m−1,sincem−1≥n+1,using(2.12),we obtain

So,by the mathematical induction,we have

Combining(2.11),(2.12)and(2.14),we obtain

For anyi,j,k=0,1,2,···(we can assumei＜j),by(2.8)and(2.15),we have

Suppose that{xn}is not Cauchy,then by Lemma 1.2,there exists ab∈Xand anϵ＞0 such that for any natural numberk,there exist two natural numbersm(k),n(k)satisfyingm(k)＞n(k)＞ksuch that the following holds

By(2.16)and(2.17),we have

Letk→∞.Then by(2.10)and from the above,we obtain

By Definition 1.1(iv)and(2.16),we have

Lettingk→∞in(2.19)and(2.20),and using(2.10)and(2.18),we have

On the other hand,it is easy to know that

Lettingk→∞in the above two inequalities,and using(2.10)and(2.21),we obtain

Using(2.10),we can assume that the parity ofm(k)andn(k)is different.Letm(k)be odd andn(k)be even.We obtain

Letk→∞in the above inequality.Then by(2.21)and(2.22),we have

which implies thatφ(ϵ,ϵ)=0,i.e.,ϵ=0.This is a contradiction.Hence{xn}is a Cauchy sequence.

Suppose thatSXis complete.Sincex2n+1=Sx2n∈SXfor alln=0,1,2,···,there exists au∈SXsuch thatx2n+1→uasn→∞.And since{xn}is a Cauchy sequence and the following holds

sox2n+2→uasn→∞.

By Lemma 1.1 and(2.1),for anya∈X,one has

Similarly,we have

ThereforeSu=u.So we haveTu=Su=u,that is,uis a common fixed point ofSandT.

Ifvis also a common fixed point ofSandTandu/=v,then there exists ana∗∈Xsuch thatd(u,v,a∗)＞0.By(2.1),we have

which implies that

by the property ofφ.This is a contradiction to the choice ofa∗.Souis the unique common fixed point ofSandT.

Similarly,we can prove the same result forTXbeing complete.The proof is completed.

From now,we discuss the existence problems of common fixed points for two mappings on non-complete ordered 2-metric spaces.

Theorem 2.2Let(X,≼,d)be an ordered2-metric space and S,T:X→X be two maps.Suppose that for each comparable elements x,y∈X,

where k,l are two real numbers satisfying l＞0and0＜k+l≤1−l.If S and T satisfy the following conditions:

(i)for each x∈X,x≼Sx and x≼Tx;

(ii)S and T are both continuous;

(iii)S(X)or T(X)is complete,

then S and T have a common fixed point.

Proof.Take an elementx0∈X.Using(i),we have

Hence we obtain a sequence{xn}satisfying

For eachm,n=0,1,2,···,xnandxmare comparable by(2.24),hence modifying the derivation process of Theorem 2.1,we can prove that{xn}is a Cauchy sequence.

Suppose thatSXis complete.Sincex2n+1=Sx2n∈SXfor alln=0,1,2,···,there exists au∈SXsuch thatx2n+1→uasn→∞.And since{xn}is Cauchy and

Therefore,uis a common fixed point ofSandT.

Similarly,we can prove the same result forTXbeing complete.The proof is completed.The following result is the non-continuous version of Theorem 2.2.

Theorem 2.3Let(X,≼,d)be a ordered2-metric space and S,T:X→X be two maps.Suppose that(2.23)holds.If S and T satisfy:

(i)for each x∈X,x≼Sx and x≼Tx;

(iii)S(X)or T(X)is complete,

then S and T have a common fixed point.

Similarly,Sinceuandx2n+1are comparable,we have

SoSu=u.ThereforeTu=Su=u,i.e.,uis a common fixed point ofSandT.The proof is completed.

Now,we give a sufficient condition under which there exists a unique common fixed point for two mappings in Theorems 2.2 and 2.3.

Theorem 2.4Suppose that all of the conditions in Theorem2.2or Theorem2.3hold.Furthermore,if

(I)for each x,y∈X,there exists a z∈X such that z and x are comparable,z and y are comparable;

(II)u≺v implies that Snu≼v and Tnu≼v for all n=1,2,···,then S and T have a unique common fixed point.

Proof.From Theorems 2.2 and 2.3 we know thatSandThave a common fixed pointu.Suppose thatvis another common fixed point ofS.Thenu/=v.

Case 1.uandvare comparable.

which impliesd(u,v,a∗)=0 by the property ofφ.This is a contradiction to the choice ofa∗.Therefore,uis the unique common fixed point ofSandT.

Case 2.uandvare not comparable.

which means thatSnuandTnware comparable.By(2.23),for each fixeda∈X,we have

This shows that{d(u,Tnw,a)}∞n=1is a non-increasing non-negative real number sequence.Hence there existsM(a)≥0 such that

Lettingn→∞in(2.25)and using(2.26),we obtain

which impliesM(a)=0,i.e.,

Henceu=vby Lemma 1.5,which is a contradiction.Souis the unique common fixed point ofSandT.

Ifuin the above derivation process is replaced byv,then we similarly obtain

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