# Common Fixed Point Theorems Using the E.A. and CLR Properties in 2-Menger Spaces

Authors: Balbir Singh; Vishal Gupta; Sanjay Kumar
Journal: International Journal of Analysis
 (2013)

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## Abstract

First, we prove a common fixed point theorem using weakly compatible maps in 2-Menger space witht-norm of Hadzic type. Second, we prove a common fixed point theorem using the E.A. property along with weakly compatible maps. Further, we obtained a common fixed point theorem using the CLR property along with weakly compatible maps. At the end, we provide an application of our main theorem for four finite families of mappings.


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## Body

## 1. Introduction

The theory of probabilistic metric spaces is an important part of stochastic analysis, and so it is of interest to develop the fixed point theory in such spaces. The first result from the fixed point theory in probabilistic metric spaces was obtained by Sehgal and Bharucha-Reid [1]. Since then many fixed points theorems for single valued and multivalued mappings in probabilistic metric spaces have been proved in [2–8]. The study of 2-metric spaces was initiated by Gähler [9] and some fixed point theorems in 2-metric spaces were proved in [2, 8, 10–13]. In 1987, Zeng [14] gave the generalization of 2-metric to probabilistic 2-metric as follows.A probabilistic 2-metric space is an ordered pair(X,F), where X is an arbitrary set and F is a mapping from X3 into the set of distribution functions. The distribution function Fx,y,z(t) will denote the value of Fx,y,z at the positive real number t. The function Fx,y,z is assumed to satisfy the following conditions:(i)
Fx,y,z(0)=0 for all x,y,z∈X;(ii)
Fx,y,z(t)=1 for all t>0 if and only if at least two of the three points x,y,z are equal;(iii)
for distinct pointsx,y∈X, there exists a point z∈X such that Fx,y,z(t)≠1 for t>0;(iv)
Fx,y,z(t)=Fy,z,x(t)=Fz,x,y(t)=⋯ for all x,y,z∈X and t>0;(v)
ifFx,y,w(t1)=1, Fx,w,z(t2)=1, and  Fw,y,z(t3)=1, then Fx,y,z(t1+t2+t3)=1 for all x,y,z,w∈X  and t1,t2,t3>0.In 2003,  Shi et al. [15] gave the notion of nth order t-norm as follows.Definition 1.
A mappingΔ:∏i=1n[0,1]→[0,1] is called an nth order t-norm if the following conditions are satisfied:(i)
Δ(0,0,…,0)=0, Δ(a,1,1,…,1)=a for all a∈[0,1];(ii)
Δ(a1,a2,a3,…,an)=Δ(a2,a1,a3,…,an)=Δ(a2,a3,a1,…,an)=⋯=Δ(a2,a3,a4,…,an,a1);(iii)
ai≥bi,i=1,2,3…,n, implies Δ(a1,a2,a3,…,an)≥Δ(b1,b2,b3,…,bn);(iv)
Δ(Δ(a1,a2,a3,…,an),b2,b3,…,bn)=Δ(a1,Δ(a2,a3,…,an,b2),b3,…bn)=Δ(a1,a2,Δ(a3,a4,….,an,b2,b3),b4,…,bn)=⋯=Δ(a1,a2,a3,…,an-1,Δ(an,b2,b3,…,bn)).
Forn=2, we have a binary t-norm, which is commonly known as t-norm.
Basic examples oft-norm are the Lukasiewicz t-norm ΔL,ΔL(a,b)=max(a+b-1,0),t-norm ΔP, ΔP(a,b)=ab, and t-norm ΔM, ΔM(a,b)=min{a,b}.Definition 2 (see [6]).
A special class oft-norms (called a Hadzic-type t-norm) is introduced as follows.
LetΔ be a t-norm and let Δn:[0,1]→[0,1](n∈N) be defined in the following way:
(1)Δ1(x)=Δ(x,x),Δn+1(x)=Δ(Δn(x),x)(n∈N,x∈[0,1]).
We say that thet-norm Δ is of H type if Δ is continuous and the family {Δn(x),n∈N} is equicontinuous at x=1.
The family{Δn(x),n∈N} is equicontinuous at x=1 if for every λ∈(0,1) there  exists δ(λ)∈(0,1) such that the following implication holds:
(2)x>1-δ(λ)impliesΔn(x)>1-λ∀n∈N.
A trivial example oft-norm of H type is Δ=ΔM.Remark 3.
Everyt-norm ΔM is of Hadzic type but the converse need not be true; see [7].
There is a nice characterization of continuoust-norm.(i)
If there exists a strictly increasing sequence{bn}n∈N∈[0,1] such that limn→∞bn=1and Δ(bn,bn)=bn for all n∈N, then Δ is of Hadzic type.(ii)
IfΔ is continuous and Δ is of Hadzic type, then there exists a sequence {bn}n∈N as in (i).Definition 4 (see [7]).
IfΔ is a t-norm and (x1,x2,x3,…,xn)∈[0,1]n(n∈N), then Δi=1nxi  is defined recurrely by 1, if n=1 and Δi=1nxi=Δ(Δi=1n-1xi,xn)  for all n≥2. If {bn}n∈N is a sequence of numbers from [0,1], then Δi=1∞xi is defined as limn→∞Δi=1nxi (this limit always exists) and Δi=1∞xi as Δi=1∞xn+i.Definition 5.
LetX be any nonempty set and D the set of all left-continuous distribution functions. A triplet (X,F,Δ) is said to be a 2-Menger space if the probabilistic 2-metric space (X,F) satisfies the following condition:Fx,y,z(t)≥Δ(Fx,y,w(t1), Fx,w,z(t2), Fw,y,z(t3)), where t1,t2,t3>0, t1+t2+t3=t; x,y,z,w∈X;   and Δ is the 3rd order t-norm.Definition 6.
A sequence{xn} in a 2-Menger space (X,F,Δ) is said to be(i)
converge with limitx if limn→∞Fxn,x,a(t)=1 for all t>0 and for every a∈X,(ii)
Cauchy sequence inX, if given ϵ>0, λ>0, there exists a positive integer Nϵ,λ such that
(3)Fxn,xm,a(ϵ)>1-λ∀m,n>Nϵ,λ,foreverya∈X,(iii)
complete if every Cauchy sequence inX is convergent in X.In 1996, Jungck’s [16] introduced the notion of weakly compatible as follows.Definition 7.
Two mapsf and g are said to be weakly compatible if they commute at their coincidence points.In 2002, Aamri and Moutawakil [17] generalized the notion of noncompatible mapping to the E.A. property. It was pointed out in [17] that the property E.A. buys containment of ranges without any continuity requirements besides minimizing the commutativity conditions of the maps to the commutativity at their points of coincidence. Moreover, the E.A. property allows replacing the completeness requirement of the space with a more natural condition of closeness of the range. Recently, some common fixed point theorems in probabilistic metric spaces/fuzzy metric spaces by the E.A. property under weak compatibility have been obtained in [18–20].Definition 8 (see [17]).
Letf and g be two self-maps of a metric (X, d). The maps f and g  are said to satisfy the E.A. property if there exists a sequence {xn} in X such that
(4)limn→∞fxn=limn→∞gxn=uforsomeu∈X.
Now in a similar mode, we can state the E.A. property in 2-Menger space as follows.Definition 9.
A pair of self-mappings(f,g) of 2-Menger spaces (X,F,Δ) is said to hold the E.A. property if there exists a sequence {xn} in X such that
(5)limn→∞Ffxn,gxn,p(t)=1∀t>0,p∈X.Example 10.
LetX=[0,∞) be the usual metric space. Define f,g:X→X by fx=x/4 and gx=3x/4 for all x∈X. Consider the sequence {xn}=1/n. Since limn→∞fxn=limn→∞gxn=0, then f and g satisfy the E.A. property.
Although E.A property is generalization of the concept of noncompatible maps, yet it requires either completeness of the whole space or any of the range spaces or continuity of maps. But on the contrary, the new notion of the CLR property (common limit range property) recently given by Sintunavarat and Kumam [21] does not impose such conditions. The importance of the CLR property ensures that one does not require the closeness of range subspaces.Definition 11 (see [21]).
Two mapsf and g on 2-Menger spaces X satisfy the common limit in the range of g (CLRg) property if limn→∞fxn=limn→∞gxn=gx for some x∈X.Example 12.
LetX=[0,∞) be the usual metric space. Define f,g:X→X by fx=x+1 and gx=2x for all x∈X. Consider the sequence {xn}=1+(1/n). Since limn→∞fxn=limn→∞gxn=2=g1, f and g satisfy the CLRg property.Now we state a lemma which is useful in our study.Lemma 13 (see [22]).
Let(X,F,Δ) be a 2-Menger space. If there exists q∈(0,1) such that F(x,y,z,qt)≥F(x,y,z,t) for all x,y,z∈X with z≠x,z≠y, and t>0,  then  x=y.


## 2. Weakly Compatible Maps

Theorem 14.
Let(X,F,Δ) be a complete 2-Menger space with continuous t-norm Δ of H type. Let S and T be self-mappings of X. Then S and T have a unique common fixed point in X if and only if there exist two self-mappings A,B of X satisfying the following:(2.1)
A(X)⊂T(X) and B(X)⊂S(X);(2.2)
the pairs(A,S) and (B,T) are weakly compatible;(2.3)
there existsq∈(0,1) such that for every x,y,p∈X and t>0, F(Ax,By,p,qt)≥min{F(Sx,Ty,p,t),F(Ax,Sx,p,t),  F(By,Ty,p,t),F(Ax,Ty,p,t)};(2.4)
one of the subsetsA(X), B(X), S(X), or T(X) is a closed subset of X.Indeed, A,B,S, and T have a unique common fixed point in X.Proof.
Suppose thatS and T have a unique common fixed point, say z∈X.
DefineA:X→X by Ax=z for all x∈X and B:X→X by Bx=z for all x∈X.
Then one can see that (2.1)–(2.4) are satisfied.
Conversely, assume that there exist two self-mappingsA,B of X satisfying conditions (2.1)–(2.4). From condition (2.1) we can construct two sequences {xn} and {yn} of X such that
(6)y2n-1=Tx2n-1=Ax2n-2,y2n=Sx2n=Bx2n-1forn=1,2,….
Putting x=x2n and y=x2n+1 in (2.3), we have that for all p∈X and t>0(7)F(y2n+1,y2n+2,p,qt)=F(Ax2n,Bx2n+1,p,qt)≥min{F(Sx2n,Tx2n+1,p,t),F(Ax2n,Sx2n,p,t),F(Bx2n+1,Tx2n+1,p,t),F(Ax2n,Tx2n+1,p,t)}=min{F(y2n,y2n+1,p,qt),F(y2n+1,y2n+2,p,qt)}
implies F(y2n+1,y2n+2,p,qt)≥F(y2n,y2n+1,p,t), because F is nondecreasing.
Also, lettingx=x2n+2 and y=x2n+1 in (2.3), we have that
(8)F(y2n+2,y2n+3,p,qt)≥F(y2n+1,y2n+2,p,t)∀p∈X,t>0.
In general, we have
(9)F(yn,yn+1,p,qt)≥F(yn-1,yn,p,t).
Thus for all p∈X,t>0andn=1,2,3…(10)F(yn,yn+1,p,t)≥F(y0,y1,p,tqn).
We now show that{yn} is a Cauchy sequence in X.
Letϵ∈(0,1) be given. Since the t-norm Δ is of H type, there exists λ∈(0,1) such that for all m, n∈N with m>n(11)Δ2m-n(1-λ)>1-ϵ.
Since limn→∞F(y0,y1,p,t/qn)=1, there exists n0∈N such that for all p∈X and t>0,F(y0,y1,p,t/qn)>1-λ for all n≥n0.
From (10) we have for all p∈X and t>0, F(yn,yn+1,p,t)>1-λ for all n≥n0.
Letm>n≥n0. Then for all p∈X and t>0, we have
(12)F(ym,yn,p,t)=F(ym,yn,p,t3+t3+t3)≥Δ(F(ym,yn,yn+1,t3),F(ym,yn+1,p,t3),F(yn+1,yn,p,t3))≥Δ(F(yn+1,yn,p,t3),Δ(F(yn+1,yn,ym,t3),F(yn+1,ym,p,t3)))≥Δ(Δ2((1-λ),F(yn+1,ym,p,t3))).
Since(13)F(yn+1,ym,p,t3)≥Δ(F(yn+2,yn+1,p,t32),Δ(F(yn+2,yn+1,ym,t32),F(yn+2,ym,p,t32))).From (13), we get
(14)F(ym,yn,p,t)≥Δ(Δ22(1-λ),F(yn+2,ym,p,t32)).
Inductively, we obtain
(15)F(ym,yn,p,t)≥Δ(Δ2m-n(1-λ),F(ym,ym,p,t3m-n))=Δ2m-n(1-λ).
From (11) and (13) we get, for all p∈X and t>0,F(ym,yn,p,t)>1-ϵ for all m>n≥n0. Thus {yn} is a Cauchy sequence in X. Since X is complete, there exists a point z in X such that limn→∞yn=z and this gives
(16)limn→∞Sx2n=limn→∞Tx2n-1=limn→∞Ax2n-2=limn→∞Bx2n-1=z∀n∈N.
Without loss of generality, we assume that S(X) is a complete subspace of X; therefore, z=Su for some u∈X.
Subsequently, we have(17)limn→∞Sxn=limn→∞Txn=limn→∞Axn=limn→∞Bxn=z=Su.
Next, we claim that Au=Su.
For this purpose, we putx=u and y=xn in (2.3); we have
(18)F(Au,Bxn,p,qt)≥min{F(Su,Txn,p,t),F(Au,Su,p,t),F(Bxn,Txn,p,t),F(Au,Txn,p,t)}.
Taking limit asn→∞(19)F(Au,z,p,qt)≥min{F(z,z,p,t),F(Au,z,p,t),F(z,z,p,t),F(Au,z,p,t)}≥F(Au,z,p,t).
By Lemma 13, we have Au=z.
HenceAu=Su=z.
SinceA(X)⊂T(X), there exists a point v∈X such that Au=z=Tv.
Next we claim thatTv=Bv.
Puttingx=u and y=v in (2.3), we have
(20)F(Au,Bv,p,qt)≥min{F(Su,Tv,p,t),F(Au,Su,p,t),F(Bv,Tv,p,t),F(Au,Tv,p,t)}≥min{F(z,z,p,t),F(z,z,p,t),F(Bv,Tv,p,t),F(z,z,p,t)}≥F(Bv,Tv,p,t).
That is,F(Tv,Bv,p,qt)≥F(Bv,Tv,p,t).
By Lemma13, we have Tv=Bv.
ThusAu=Su=Tv=Bv=z.
Since the pairs(A,S) and (B,T) are weakly compatible and u and v are their points of coincidence, respectively, then
(21)Az=A(Su)=S(Au)=Sz,Bz=B(Tv)=T(Bv)=Tz.
Now we prove thatz is a common fixed point of A,B,S, and T.
For this purpose, putingx=z and  y=v in (2.3), we get
(22)F(Az,Bv,p,qt)≥min{F(Sz,Tv,p,t),F(Az,Sz,p,t),F(Bv,Tv,p,t),F(Az,Tv,p,t)}≥min{F(Az,Bv,p,t),F(Az,Az,p,t),F(Bv,Bv,p,t),F(Az,Bv,p,t)}≥F(Az,Bv,p,t).
By Lemma13, we have Az=Bv=z.
Hencez=Az=Sz, and z is a common fixed point of A and S. One can prove that Bv=z is also a common fixed point of B and T.
Uniqueness. Suppose z≠w is another fixed point of A,B,S, and T.
Then, for allp∈X with p≠z and p≠w and t>0,
(23)F(z,w,p,qt)=F(Az,Bw,p,qt)≥min{F(Sz,Tw,p,t),F(Az,Sz,p,t),F(Bw,Tw,p,t),F(Az,Tw,p,t)}≥min{F(z,w,p,t),F(z,z,p,t),F(w,w,p,t),F(z,w,p,t)}≥F(z,w,p,t),
which implies that F(z,w,p,qt)≥F(z,w,p,t).
Hencez=w is a unique common fixed point of A,B,S, and T.
This completes the proof of the theorem.Corollary 15.
Let(X,F,Δ) be a 2-Menger space with continuous t-norm Δ of H type. Let S and T be self-mappings of X. Then S and T have a unique common fixed point in X if and only if there exist two self-mappings A,B of X satisfying the following:(2.5)
A(X)⊂S(X);(2.6)
pair(A,S) is weakly compatible;(2.7)
there existsq∈(0,1) such that for every x,y,p∈X and t>0,
(24)F(Ax,Ay,p,qt)≥min{F(Sx,Sy,p,t),F(Ax,Sx,p,t),F(Ay,Sy,p,t),F(Ax,Sy,p,t)};(2.8)
one of the subspacesA(X) or S(X) is a closed subspace of X.Indeed, A and S have a unique common fixed point in X.Proof .
We can easily prove the theorem by settingB=A and T=S, in the proof of the Theorem 14.Corollary 16.
Let(X,F,Δ) be a 2-Menger space with continuous t-norm Δ of H type. Let S and T be self-mappings of X. Then S and T have a unique common fixed point in X if and only if there exist two self-mappings A,B of X satisfying (2.1), (2.2), (2.4), and the following:(2.9)
there existsq∈(0,1) such that for every x,y,p∈X and t>0, F(Ax,By,p,qt)≥F(Sx,Ty,p,t).Indeed, A,B,S, and T have a unique common fixed point in X.Proof.
Proof easily follows by settingmin{F(Sx,Ty,p,t), F(Ax,Sx,p,t), F(By,Ty,p,t), F(Ax,Ty,p,t)}=F(Sx,Ty,p,t) in the proof of Theorem 14.


## 3. E.A. Property and Weakly Compatible Maps

Now we prove our main result for weakly compatible maps along with the E.A. property as follows.Theorem 17.
Let(X,F,Δ) be a 2-Menger space with continuous t-norm Δ of H type. Let A, B, S, and T be self-mappings of X satisfying (2.1), (2.2), (2.3), and (2.4) and the following:(3.1)
pair(A,S) or pair (B,T) satisfies E.A. property.Then A,B,S, and T have a unique common fixed point in X.Proof.
If the pair(B,T) satisfies the property E.A., then there exists a sequence {xn} in X such that limn→∞Bxn=limn→∞Txn=z for some z∈X.
SinceB(X)⊂S(X), there exists a sequence {yn} in X such that Bxn=Syn. Hence limn→∞Syn=z. Also A(X)⊂T(X); there exists a sequence {yn′} in X such that Ayn′=Txn. Hence limn→∞Ayn′=z.
Suppose thatS(X) is a complete subspace of X. Then z=Su for some u∈X; subsequently; we have
(25)limn→∞Bxn=limn→∞Txn=z=limn→∞Ayn′=limn→∞Syn=z=Suforsomeu∈X.
Next, we claim thatAu=Su.
For this purpose, we putx=u and y=xn in (2.3); this gives
(26)F(Au,Bxn,p,qt)≥min{F(Su,Txn,p,t),F(Au,Su,p,t),F(Bxn,Txn,p,t),F(Au,Txn,p,t)}.
Taking limit asn→∞(27)F(Au,z,p,qt)≥min{F(z,z,p,t),F(Au,z,p,t),F(z,z,p,t),F(Au,z,p,t)}≥F(Au,z,p,t).
By Lemma13, we have Au=z.
HenceAu=Su=z.
SinceA(X)⊂T(X), there exists a point v∈X such that Au=z=Tv.
Next we claim thatTv=Bv.
Puttingx=u and y=v in (2.3), we have
(28)F(Au,Bv,p,qt)≥min{F(Su,Tv,p,t),F(Au,Su,p,t),F(Bv,Tv,p,t),F(Au,Tv,p,t)}≥min{F(z,z,p,t),F(z,z,p,t),F(Bv,Tv,p,t),F(z,z,p,t)}≥F(Bv,Tv,p,t).
That is, F(Tv,Bv,p,qt)≥F(Bv,Tv,p,t).
By Lemma13, we have Tv=Bv.
ThusAu=Su=Tv=Bv=z.
Since the pairs(A,S) and (B,T) are weakly compatible and u and v are their points of coincidence, respectively, then Az=A(Su)=S(Au)=Sz and Bz=B(Tv)=T(Bv)=Tz.
Now we prove thatz is a common fixed point of A,B,S, and T.
For this purpose, we putx=z and y=v in (2.3); we get
(29)F(Az,Bv,p,qt)≥min{F(Sz,Tv,p,t),F(Az,Sz,p,t),F(Bv,Tv,p,t),F(Az,Tv,p,t)}≥min{F(Az,Bv,p,t),F(Az,Az,p,t),F(Bv,Bv,p,t),F(Az,Bv,p,t)}≥F(Az,Bv,p,t).
By Lemma13, we have Az=Bv=z.
Hencez=Az=Sz and z is a common fixed point of A and S. One can prove that Bv=z is also a common fixed point of B and T.
Uniqueness. Suppose z≠w is another fixed point of A,B,S, and T.
Then, for allp∈X with p≠z and p≠w and t>0,
(30)F(z,w,p,qt)=F(Az,Bw,p,qt)≥min{F(Sz,Tw,p,t),F(Az,Sz,p,t),F(Bw,Tw,p,t),F(Az,Tw,p,t)}≥min{F(z,w,p,t),F(z,z,p,t),F(w,w,p,t),F(z,w,p,t)}≥F(z,w,p,t),
which implies that F(z,w,p,qt)≥F(z,w,p,t).
Hencez=w is a unique common fixed point of A,B,S, and T.Corollary 18.
Let(X,F,Δ) be a 2-Menger space with continuous t-norm Δ of H type. Let A,B,S, and T be self-mappings of X satisfying (2.1), (2.2), (2.4), and (2.9).
ThenA,B,S, and T have a unique common fixed point in X.


## 4. CLR Property and Weakly Compatible Maps

Now we prove our main result for weakly compatible maps along with the CLRS property as follows.Theorem 19.
Let(X,F,Δ) be a 2-Menger space with continuous t-norm Δ of H type. Let A,B,S, and T be self mapping of X satisfying (2.1), (2.2), (2.3), and the following:(4.1)
pair(A,S) or pair (B,T) satisfies CLRS  property;(4.2)
one of the subspacesA(X),B(X),S(X), or T(X) is a closed subspace of X.Then A,B,S, and T have a unique common fixed point in X.Proof.
If the pair(A,S) satisfies the CLRS property then there exists a sequence {xn} in X such that limn→∞Axn=limn→∞Sxn=z, where z∈S(X). Therefore, there exists a point u∈X such that Su=z. Since T(X) is a closed subspace of X and A(X)⊂T(X), for each {xn}⊂X there corresponds a sequence {yn}⊂X such that Axn=Tyn.
Therefore,limn→∞Txn=limn→∞Axn=z, where z∈S(X).
Thus in all, we have(31)limn→∞Axn=limn→∞Sxn=limn→∞Tyn=z.
Now we are required to show thatlimn→∞Byn=z.
Puttingx=xn and y=yn in (2.3), we get
(32)F(Axn,Byn,p,qt)≥min{F(Sxn,Tyn,p,t),F(Axn,Sxn,p,t),F(Byn,Tyn,p,t),F(Axn,Tyn,p,t)}.
Letlimn→∞Byn=l≠z for t>0.Then taking limit as n→∞, we have
(33)F(z,l,p,qt)≥min{F(z,z,p,t),F(z,z,p,t),F(l,z,p,t),F(z,z,p,t)}≥F(l,z,p,t).
By Lemma13, we  have  z=l, then   limn→∞Byn=z.
Hencelimn→∞Axn = limn→∞Sxn=limn→∞Tyn=limn→∞Byn=z=Su,  for  some  u∈X.
Uniqueness follows   from Theorem14.Corollary 20.
Let(X,F,Δ)   be  a   2-Menger  space  with  continuous t-norm Δ of  H type. Let A,B,S, and  T  be  self-mappings  of  X  satisfying (2.1), (2.2), (2.9), (4.1), and (4.2).
ThenA,B,S,  and T  have  a  unique  common  fixed  point  in  X.


## 5. Application

As an application of Theorem14, we prove a common fixed point theorem for four finite families of mappings which runs as follows.Theorem 21.
Let{A1,A2,…,Am}, {B1,B2,…,Bn}, {S1,S2,…,Sp}, and {T1,T2,…,Tq} be four finite families of self-mappings of a 2-Menger space (X,F,Δ) with continuous t-norm  of  Hadzic type such that A=A1A2⋯Am, B=B1B2⋯Bn, S=S1S2⋯Sp, and T=T1T2⋯Tq satisfy condition (2.1), (2.2), (2.3), and (2.4); then(i)
A andS have a point of coincidence;(ii)
B andT have a point of coincidence.Moreover, if AiAj=AjAi, BkBl=BlBk, SrSs=SsSr, TtTu=TuTt, AiSr=SrAi, and BkTt=TtBk for all i,j∈I1={1,2,…,m}, k, l∈I2={1,2,…,n}, r, s∈I3={1,2,…,p}, and t, u∈I4={1,2,…,q}, then (for all i∈I1,k∈I2,r∈I3, and t∈I4)Ai, Sr, Bk, and Tt have a common fixed point.Proof.
The conclusions (i) and (ii) are immediate asA,S,B, and T satisfy all the conditions of Theorem 14. Now appealing to component wise commutativity of various pairs, one can immediately prove that AS=SA and BT=TB, and hence, obviously both pairs (A,S) and (B,T) are coincidently commuting. Note that all the conditions of Theorem 14 (for mappings A,S,B, and T) are satisfied ensuring the existence of a unique common fixed point, say z. Now one needs to show that z remains the fixed point of all the component maps. For this consider
(34)A(Aiz)=((A1A2⋯Am)Ai)z=(A1A2⋯Am-1)((AmAi)z)=(A1A2⋯Am-1)(AiAmz)=(A1A2⋯Am-2)(Am-1Ai(Amz))=(A1A2⋯Am-2)(AiAm-1(Amz))=⋯=A1Ai(A2A3A4⋯Amz)=AiA1(A2A3A4⋯Amz)=Ai(Az)=Aiz.
Similarly, one can show that
(35)A(Srz)=Sr(Az)=Srz,S(Srz)=Sr(Sz)=Srz,S(Aiz)=Ai(Sz)=Aiz,B(Bkz)=Bk(Bz)=Bkz,B(Ttz)=Tt(Bz)=Ttz,T(Ttz)=Tt(Tz)=Ttz,T(Bkz)=Bk(Tz)=Bkz,
which show that (for all i,r,k, and t) AizandSrz are other fixed points of the pair (A,S) whereas BkzandTtz are other fixed points of the pair (B,T). Now appealing to the uniqueness of common fixed points of both pairs separately, we get
(36)z=Aiz=Srz=Bkz=Ttz,
which shows that z is a common fixed point of Ai,Sr,Bk, and Tt for all i,r,k, and t.By settingA=A1=A2=⋯=Am,B=B1=B2=⋯=Bn,S=S1=S2=⋯=Sp, and T=T1=T2=⋯=Tq, one deduces the following for certain iterates of maps, which run as follows.Corollary 22.
LetA,B,S, and T be four self-mappings of a 2-Menger space (X,F,Δ) such that Am,Bn,Sp, and Tq satisfy the conditions (2.1) and (2.3). If one of Am(X),Bn(X),Sp(X), or Tq(X) is a closed subset of X, then A,B,S, and T have a unique common fixed point provided (A,S) and (B,T) commute.



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*Source: 934738-2013-05-27.xml*