# Related Results to Hybrid Pair of Mappings and Applications in Bipolar Metric Spaces

Authors: G. N. V. Kishore; K. P. R. Rao; A. Sombabu; R. V. N. S. Rao
Journal: Journal of Mathematics
 (2019)

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## Abstract

In this paper, we introduce the concept of multivalued contraction mappings in partially ordered bipolar metric spaces and establish the existence of unique coupled fixed point results for multivalued contractive mapping by using mixed monotone property in partially ordered bipolar metric spaces. Some interesting consequences of our results are obtained.


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## Body

## 1. Introduction and Preliminaries

Fixed point theory has been playing a vital role in the study of nonlinear phenomena. The Banach fixed point theorem or contraction mapping principle was proved by Banach [1] in 1922. Turinici [2] extended the Banach contraction principle in the setting of partially ordered sets and laid the foundation of a new trend in fixed point theory.The theory of mixed monotone multivalued mappings in ordered Banach spaces was extensively investigated by Y. Wu [3]. Existence of fixed points in ordered metric spaces was initiated by Ran and Reurings [4], and later on several authors studied the problem of existence and uniqueness of a fixed point for mappings satisfying different contractive conditions in the framework of partially ordered metric spaces ([1–27] and references therein).In [19], Bhaskar and Lakshmikantham introduced the concept of coupled fixed point and proved some coupled fixed point theorems in partially ordered metric spaces (see also [1–27] for more works). The study of fixed points for multivalued contraction mappings using the Hausdorff metric was initiated by Markin [20]. Later, many authors established hybrid fixed point theorems and gave applications of their results (see also [21–24]).Very recently, in 2016 Mutlu and Gu¨rdal [25] introduced the notion of bipolar metric spaces. Also they investigated some fixed point and coupled fixed point results on this space (see [25, 26]).This paper aims to introduce some coupled fixed point theorems for a multivalued mappings satisfying various contractive conditions defined on partially ordered bipolar metric spaces. We have illustrated the validity of the hypotheses of our results.First we recall some basic definitions and results.Definition 1 ([25]).
LetA and B be two nonempty sets. Suppose that d:A×B→[0,∞) is a mapping satisfying the following properties: (
B0)d(a,b)=0 if and only if a=b for all (a,b)∈A×B,(
B1)d(a,b)=d(b,a), for all a,b∈A∩B,(
B2)d(a1,b2)≤d(a1,b1)+d(a2,b1)+d(a2,b2), for all a1,a2∈A, b1,b2∈B.
Then the mappingd is called a bipolar metric on the pair (A,B) and the triple (A,B,d) is called a bipolar metric space.Definition 2 ([25]).
Assume(A1,B1) and (A2,B2) as two pairs of sets.
The functionF:A1∪B1→A2∪B2 is said to be a covariant map, if F(A1)⊆A2 and F(B1)⊆B2 and denote this as F:(A1,B1)⇉(A2,B2).
The mappingF:A1∪B1→A2∪B2 is said to be a contravariant map, if F(A1)⊆B2 and F(B1)⊆A2 and denote this as F:(A1,B1)⇋(A2,B2).
In particular,d1 and d2 are bipolar metrics in (A1,B1) and (A2,B2), respectively. Sometimes we use the notations F:(A1,B1,d1)⇉(A2,B2,d2) and F:(A1,B1,d1)⇋(A2,B2,d2).Definition 3 ([25]).
Let(A,B,d) be a bipolar metric space. A point v∈A∪B is said to be left point if v∈A, a right point if v∈B, and a central point if both.
Similarly, a sequence{an} on the set A and a sequence {bn} on the set B are called a left and right sequence, respectively.
In a bipolar metric space, sequence is the simple term for a left or right sequence.
A sequence{vn} is convergent to a point v if and only if {vn} is a left sequence, v is a right point, and limn→∞d(vn,v)=0; or {vn} is a right sequence, v is a left point, and limn→∞d(v,vn)=0.
A bisequence({an},{bn}) on (A,B,d) is sequence on the set A×B. If the sequences {an} and {bn} are convergent, then the bisequence ({an},{bn}) is said to be convergent. ({an},{bn}) is Cauchy sequence, if limn,m→∞d(an,bm)=0. In a bipolar metric space, every convergent Cauchy bisequence is biconvergent.
A bipolar metric space is called complete, if every Cauchy bisequence is convergent, hence biconvergent.Now we give our main results.


## 2. Main Results

The following definitions and results will be needed in the sequel.Let(A,B,d) be a bipolar metric space. For points a∈A,b∈B and the subsets X⊆A, Y⊆B, consider the bipolar metric d(a,Y)=infd(a,y)/y∈Y and d(X,b)=infd(x,b)/x∈X. We denote by CB(A) and CB(B) a class of all nonempty closed and bounded subsets of A and B, respectively. Also denote A2=A×A and B2=B×B. Let H be the Hausdorff bipolar metric induced by the bipolar metric d on (A,B); that is,(1)HX,Y=max⁡supx∈Xdx,B,supy∈YdA,y,for every X∈CB(A) and Y∈CB(B).Definition 4.
LetF:A2∪B2→CB(A∪B) be given the mapping; an element (a,b)∈A2∪B2 is called a coupled fixed point of a set valued mapping F if a∈F(a,b) and b∈F(b,a).Lemma 5 ([21]).
Letκ≥0. If X∈CB(A),Y∈CB(B) with H(X,Y)≤κ, then, for each x∈X, there exists an element y∈Y such that d(x,y)≤κ.Definition 6.
Let(A,B,≤) be a partially ordered set and let F:(A2,B2)⇉CB(A,B) be covariant map. We say that F has the mixed monotone property if F is monotone-nondecreasing in its first argument a and is monotone-nonincreasing in its second argument b, that is, for any (a,b)∈A2∪B2.(2)a1,a2∈A2,a1≤a2⇒Fa1,b≤Fa2,bb1,b2∈B2,b1≤b2⇒Fa,b1≥Fa,b2Note that ifa1≤a2, b1≥b2, and F has mixed monotone property, by Definition 6, we obtain F(a1,b1)≤F(a2,b2) and F(b1,a1)≥F(b2,a2).Theorem 7.
Let(A,B,≤) be a partially ordered set such that there exists a bipolar metric d on (A,B) with (A,B,d) being complete bipolar metric spaces. Consider the covariant mapping F:(A2,B2)⇉CB(A,B) satisfying the following condition:(3)HFa,b,Fp,q≤μda,p+κdb,qfor all a,b∈A, p,q∈B, and μ,κ are nonnegative constants with a≥p and b≤q. And μ+κ<1.(7.1)
F has a mixed monotone property(7.2)
There exists(a0,b0)∈A2∪B2 and, for some a1∈F(a0,b0), b1∈F(b0,a0), we have a0≤a1 and b0≥b1(7.3)
If a nondecreasing sequencean,pn is convergent to (p,a) for a∈A,p∈B, then an≤p, pn≤a for all n and if a nonincreasing sequence bn,qn is convergent to (q,b) for b∈A,q∈B, then bn≥q, qn≥b for all n
ThenF has a coupled fixed point.Proof.
Leta0,b0∈A and p0,q0∈B. Consider the sequences {an},{bn},{pn} and {qn} such that a1∈F(a0,b0), b1∈F(b0,a0) and p1∈F(p0,q0), q1∈F(q0,p0). By (7.2), we have that a0≤p1 and b0≥q1 and p0≤a1 and q0≥b1, where a1,b1∈A and p1,q1∈B.Applying this in inequality (3), we have(4)HFa0,b0,Fp1,q1≤μda0,p1+κdb0,q1and(5)HFb0,a0,Fq1,p1≤μdb0,q1+κda0,p1.On adding (4) and (5), we get(6)HFa0,b0,Fp1,q1+HFb0,a0,Fq1,p1≤μ+κda0,p1+db0,q1On the other hand(7)HFa1,b1,Fp0,q0≤μda1,p0+κdb1,q0and(8)HFb1,a1,Fq0,p0≤μdb1,q0+κda1,p0On adding (7) and (8), we get(9)HFa1,b1,Fp0,q0+HFb1,a1,Fq0,p0≤μ+κda1,p0+db1,q0Moreover,(10)HFa0,b0,Fp0,q0≤μda0,p0+κdb0,q0and(11)HFb0,a0,Fq0,p0≤μdb0,q0+κda0,p0On adding (10) and (11), we get(12)HFa0,b0,Fp0,q0+HFb0,a0,Fq0,p0≤μ+κda0,p0+db0,q0Also, if d(a0,p1)=d(b0,q1)=0, then a0=p1∈F(p0,q0),b0=q1∈F(q0,p0). If d(a1,p0)=d(b1,q0)=0, then p0=a1∈F(a0,b0),q0=b1∈F(b0,a0). If d(a0,p0)=d(b0,q0)=0, then a0=p0,b0=q0.It follows that(a0,b0) is a coupled fixed point of F.Assume that eitherd(a0,p1)≠0 or d(b0,q1)≠0 and d(a1,p0)≠0 or d(b1,q0)≠0; also d(a0,p0)≠0 or d(b0,q0)≠0.Sincea1∈F(a0,b0), b1∈F(b0,a0), then from (6) and Lemma 5 there exist p2∈F(p1,q1), q2∈F(q1,p1) such that(13)da1,p2+db1,q2≤μ+κda0,p1+db0,q1and since p1∈F(p0,q0), q1∈F(q0,p0), then from (9) and Lemma 5 there exist a2∈F(a1,b1), b2∈F(b1,a1) such that(14)da2,p1+db2,q1≤μ+κda1,p0+db1,q0Also since a1∈F(a0,b0), b1∈F(b0,a0) and p1∈F(p0,q0), q1∈F(q0,p0), from (12) and Lemma 5 then we have(15)da1,p1+db1,q1≤μ+κda0,p0+db0,q0Since, a0≤p1, b0≥q1 and p0≤a1, q0≥b1, a1∈Fa0,b0, p1∈F(p0,q0),b1∈F(b0,a0),q1∈F(q0,p0) and a2∈F(a1,b1),p2∈F(p1,q1), b2∈F(b1,a1),q2∈F(q1,p1), by assumption (7.1), we get(16)a1≤p2,b1≥q2andp1≤a2,q1≥b2.Similarly from (3) and above, we have(17)da2,p3+db2,q3≤μ+κda1,p2+db1,q2and(18)da3,p2+db3,q2≤μ+κda2,p1,db2,q1and also(19)da2,p2+db2,q2≤μ+κda1,p1,db1,q1Since, we havea1≤p2, b1≥q2 and p1≤a2, q1≥b2, a2∈F(a1,b1), p2∈F(p1,q1),b2∈F(b1,a1),q2∈F(q1,p1) and a3∈F(a2,b2),p3∈F(p2,q2), b3∈F(b2,a2),q3∈F(q2,p2). Again, applying our assumption (7.1), we get(20)a2≤p3,b2≥q3andp2≤a3,q2≥b3Continuing similarly this process, we have an+1∈F(an,bn), pn+1∈F(pn,qn),bn+1∈F(bn,an),qn+1∈F(qn,pn) with(21)an≤pn+1,bn≥qn+1andpn≤an+1,qn≥bn+1such that(22)dan,pn+1+dbn,qn+1≤μ+κdan-1,pn+dbn-1,qnand(23)dan+1,pn+dbn+1,qn≤μ+κdan,pn-1+dbn,qn-1and also(24)dan,pn+dbn,qn≤μ+κdan-1,pn-1+dbn-1,qn-1Put tn=dan,pn+1+dbn,qn+1 for any n∈N; then(25)tn≤μ+κtn-1Put sn=dan+1,pn+dbn+1,qn for any n∈N; then(26)sn≤μ+κsn-1Put rn=dan,pn+dbn,qn for any n∈N; then(27)rn≤μ+κrn-1Therefore tn, sn, and rn are nonincreasing sequences. From (25), (26), and (27) we have that(28)limn→∞tn=0,limn→∞sn=0andlimn→∞rn=0which implies that(29)limn→∞dan,pn+1=limn→∞dbn,qn+1=0,limn→∞dan+1,pn=limn→∞dbn+1,qn=0,limn→∞dan,pn=limn→∞dbn,qn=0.Using the property (B2), we have(30)dan,pm≤dan,pn+1+dan+1,pn+1+⋯+dam-1,pmdbn,qm≤dbn,qn+1+dbn+1,qn+1+⋯+dbm-1,qmand(31)dam,pn≤dam,pm-1+dam-1,pm-1+⋯+dan+1,pndbm,qn≤dbm),qm-1+dbm-1,qm-1+⋯+dbn+1,qnNext, we show that an,pn and bn,qn are Cauchy bisequence in A,B for each n;m∈N such that n<m. From (25), (26), (27), (30), and (31), we have(32)dan,pm+dbn,qm≤dan,pn+1+dbn,qn+1+dan+1,pn+1+dbn+1,qn+1+⋯+dam-1,pm-1+dbm-1,qm-1+dam-1,pm+dbm-1,qm≤tn+rn+1+⋯+tm-1+rm-1≤tn+tn+1+⋯+tm-1+rn+1+rn+2+⋯+rm-1≤∑k=nm-1tk+∑k=n+1m-1rk→0asm,n→∞and(33)dam,pn+dbm,qn≤dam,pm-1+dbm,qm-1+dam-1,pm-1+dbm-1,qm-1+⋯+dan+1,pn+1+dbn+1,qn+1+dan+1,pn+dbn+1,qn≤sm-1+rm-1+⋯+rn+1+sn≤sn+sn+1+⋯+sm-1+rn+1+rn+2+⋯+rm-1≤∑k=nm-1sk+∑k=n+1m-1rk→0asm,n→∞From above, it is clear thatan,pn and bn,qn are Cauchy bisequences in (A,B). Since, (A,B,d) is complete, a,b∈A and p,q∈B such that(34)limn→∞an+1=p,limn→∞bn+1=q,limn→∞pn+1=a,limn→∞qn+1=bNow we will show that a∈F(a,b),b∈F(b,a) and p∈F(p,q),q∈F(q,p). As an,pn is a nondecreasing bisequence and bn,qn is a nonincreasing bisequence in (A, B), (35)an→p,bn→qandpn→a,qn→b.By assumption (7.1), we get an≤p,pn≤a and bn≥q,qn≥b for all n. If an=p,pn=a and bn=q,qn=b for some n≥0, then p=an≤pn+1≤a=pnq=bn≥qn+1≥b=qn and a=pn≤an+1≤p=an, b=qn≥bn+1≥q=bn implies a=p and b=q; therefore, an=an+1∈F(an,bn) and bn=bn+1∈F(bn,an).So(an,bn) is coupled fixed point of F.Suppose that(an,pn)≠(p,a) and (bn,qn)≠(q,b) for all n≥0.From (3), we have(36)HFa,b,Fpn,qn≤μda,pn+κdb,qnand(37)HFb,a,Fqn,pn≤μdb,qn+κda,pnTherefore,(38)HFa,b,Fpn,qn+HFb,a,Fqn,pn≤μ+κda,pn+db,qnLetting n→∞, we have that(39)limn→∞HFa,b,Fpn,qn+HFb,a,Fqn,pn=0Since pn+1∈F(pn,qn) and limn→∞da,pn+1=0, we have a∈F(a,b) and since qn+1∈F(qn,pn) and limn→∞db,qn+1=0, we have b∈F(b,a).Similarly, we can provep∈F(p,q) and q∈F(q,b).On the other hand,(40)da,p=dlimn→∞pn,limn→∞an=limn→∞dan,pn=0and(41)db,q=dlimn→∞qn,limn→∞bn=limn→∞dbn,qn=0.Therefore, a=p and b=g and hence F has a coupled fixed point.Theorem 8.
Let(A,B,≤) be a partially ordered set such that there exists a bipolar metric d on (A,B) with (A,B,d) being complete bipolar metric spaces. Consider F:(A2,B2)⇉CB(A,B) a covariant set valued mapping, such that(42)HFa,b,Fp,q≤κ2da,p+db,qfor all a,b∈A, p,q∈B, and κ∈(0,1) with a≥p and b≤q. Suppose also that (8.1)
F has a mixed monotone property(8.2)
there exist(a0,b0)∈A2∪B2 and for some a1∈F(a0,b0), b1∈F(b0,a0) we have a0≤a1 and b0≥b1(8.3)
if a nondecreasing sequencean,pn is convergent to (p,a) for a∈A,p∈B, then an≤p, pn≤a for all n and if a nonincreasing sequence bn,qn is convergent to (q,b) for b∈A,q∈B, then bn≥q, qn≥b for all n
ThenF has a coupled fixed point; that is, there exist (a,b)∈A2∪B2 such that a∈F(a,b) and b∈F(b,a).Example 9.
LetA=Um(R)/Um(R) be upper triangular matrices over R and let B=Lm(R)/Lm(R) be lower triangular matrices over R with the bipolar metric dP,Q=∑i,j=1m|pij-qij| for all P=(pij)m×m∈Um(R) and Q=(qij)m×m∈Lm(R). On the set A,B, consider the following relation:(43)P,Q∈A2∪B2,P⪯Q⇔pij≤qij where ≤ is usual ordering. Then, clearly, A,B,d is a complete bipolar metric space and A,B,⪯ is a partially ordered set. Let F:A2,B2⇉CB(A,B) be defined as FP,Q=(pij+qij)/5m×m+(3/5)(Iij)m×m for all P=(pij)m×m,Q=(qij)m×m∈A2∪B2.
Then obviouslyF has mixed monotone property; also there exist P=(Oij)m×m and Q=(Iij)m×m such that(44)FOijm×m,Iijm×m=Oij+Iij5m×m+35Iijm×m⪰Oijm×mand(45)FIijm×m,Oijm×m=Oij+Iij5m×m+35Iijm×m⪯Iijm×m Taking P=(pij)m×m,Q=(qij)m×m,R=(rij)m×m,S=(sij)m×m∈A2∪B2 with P⪰R, Q⪯S, that is, pij≥rij, qij≤sij, we have(46)dFP,Q,FR,S=dpij+qij5+35Iij,rij+sij5+35Iij=15∑i,j=1mpij+qij-rij+sij≤15∑i,j=1mpij-rij+∑i,j=1mqij-sij≤15dP,R+dQ,S Therefore, all the conditions of Theorem 8 hold and (((3/5)Iij)m×m,((3/5)Iij)m×m) is a coupled fixed point of F.Definition 10.
Let(A,B,d) be bipolar metric spaces, a∈A,p∈B, and let F:(A×B)∪(B×A)→CB(A∪B) be a covariant multivalued map. An element (a,p) is called a coupled fixed point of F if a∈F(a,p) and p∈F(p,a).Theorem 11.
Let(A,B,≤) be a partially ordered set such that there exists a bipolar metric d on (A,B) with (A,B,d) being complete bipolar metric spaces. Consider F:(A×B,B×A)⇉CB(A,B) a covariant set valued mapping, such that(47)HFa,p,Fq,b≤μda,q+κdb,pfor all a,b∈A, p,q∈B and μ+κ<1 with a≥q and b≤p. Suppose also that (11.1)
F has a mixed monotone property(11.2)
there exista0∈A,p0∈B and for some a1∈F(a0,p0), p1∈F(p0,a0) we have a0≤a1 and p0≥p1(11.3)
if a nondecreasing sequencean,qn is convergent to (q,a) for a∈A,q∈B, then an≤q, qn≤a for all n and if a nonincreasing sequence bn,pn is convergent to (p,b) for b∈A,p∈B, then bn≥p, pn≥b for all n
ThenF has a coupled fixed point; that is, there exist a∈A,p∈B such that a∈F(a,p) and p∈F(p,a).Proof.
The proof will follow when we replaceA×B and B×A in place of A2, B2, respectively, in Theorem 7.Theorem 12.
Let(A,B,≤) be a partially ordered set such that there exists a bipolar metric d on (A,B) with (A,B,d) being complete bipolar metric spaces. Consider F:(A×B,B×A)⇉CB(A,B) a covariant set valued mapping, such that(48)HFa,p,Fq,b≤κ2da,q+db,pfor all a,b∈A, p,q∈B and κ∈(0,1) with a≥q and b≤p. Suppose also that (12.1)
F has a mixed monotone property(12.2)
there exista0∈A,p0∈B and for some a1∈F(a0,p0), p1∈F(p0,a0) we have a0≤a1 and p0≥p1(12.3)
if a nondecreasing sequencean,qn is convergent to (q,a) for a∈A,q∈B, then an≤q, qn≤a for all n and if a nonincreasing sequence bn,pn is convergent to (p,b) for b∈A,p∈B, then bn≥p, pn≥b for all n
ThenF has a coupled fixed point; that is, there exist a∈A,p∈B such that a∈F(a,p) and p∈F(p,a).


## 3. Conclusions

In the present research, we introduced and proved a coupled fixed point theorem for a multivalued mapping, satisfying various contractive conditions, defined on a partially ordered bipolar metric space, and gave suitable example that supports our main result.



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