# A Common Fixed Point Theorem in Metric Space under General Contractive Condition

Authors: Sunny Chauhan; Zoran Kadelburg; Sumitra Dalal
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We prove a common fixed point theorem for two pairs of compatible and subsequentially continuous (alternately subcompatible and reciprocally continuous) mappings satisfying a general contractive condition in a metric space. Some illustrative examples are furnished to highlight the realized improvements. Our result improves the main result of Sedghi and Shobe (2007).


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## Body

## 1. Introduction

Fixed point theory is one of the most fruitful and effective tools in mathematics which has enormous applications within as well as outside mathematics. Starting from the celebrated Banach contraction principle [1], many authors have obtained its several generalizations in different ways (see, e.g., [2–4]).A metrical common fixed point theorem generally involves conditions on commutativity, continuity, and contraction of the given mappings, as well as completeness (or closedness) of the underlying space (or subspaces), along with conditions on suitable containment amongst the ranges of involved mappings. Hence, in order to prove a new metrical common fixed point theorem, one is always required to weaken one or more of these conditions. In order to weaken commutativity conditions in common fixed point theorems, Sessa [5] introduced the concept of weakly commuting mappings. Jungck [6] defined the notion of compatible mappings in order to generalize the concept of weak commutativity and showed that weakly commuting mappings are compatible, but the converse is not true. Afterwards Jungck and Rhoades [7] introduced the concept of weak compatibility to the setting of single-valued and multivalued mappings which is more general than compatibility. However, the study of common fixed points of noncompatible mappings is also equally interesting, and it was initiated by Pant [8] in metric spaces. Researchers of this domain introduced several definitions of weak commutativity such as compatible mappings, compatibility of type (A), (B), (C), and (P), and several others, whose systematic comparisons and illustrations are available in Murthy [9] and Singh and Tomar [10].In 2002, Aamri and El Moutawakil [11] introduced the notion of property (E.A). Further, Al-Thagafi and Shahzad [12] introduced the notion of occasionally weakly compatible mappings. It was used further by many authors (see, e.g., [13]). However, in [14], Đorić et al. observed that the condition of occasionally weak compatibility reduces to weak compatibility in the presence of a unique point of coincidence of the given pair of mappings. Thus, no generalization can be obtained by replacing weak compatibility with occasionally weak compatibility (see also [15–17]).In 2009, Bouhadjera and Godet-Thobie [18] further enlarged the class of compatible (reciprocally continuous) pairs by introducing the concept of subcompatibility (subsequential continuity) of pairs of mappings, which is substantially weaker than compatibility (reciprocal continuity). Since then, Imdad et al. [19] improved the results of Bouhadjera and Godet-Thobie and showed that these results can easily be recovered by replacing subcompatibility with compatibility or subsequential continuity with reciprocal continuity. Very recently, Chauhan et al. [20] obtained some results of this kind using integral type contractive conditions. Many authors established a number of other fixed point results in metric and related spaces (see, e.g., [21–31]).In 2007, Sedghi and Shobe [32] proved a common fixed point theorem for weakly compatible mappings satisfying a new general contractive type condition. The aim of this paper is to prove a common fixed point theorem for two pairs of self-mappings by using the notions of compatibility and subsequential continuity (alternately subcompatibility and reciprocal continuity) satisfying general contractive condition in a metric space. Some examples are furnished which demonstrate the validity of our result.


## 2. Preliminaries

Definition 1.
LetA,S:X→X be two self-mappings on a metric space (X,d). The mappings A and S are said to be (1)
weakly commuting ifd(ASx,SAx)≤d(Ax,Sx) for all x∈X [5],(2)
compatible iflimn→∞d(ASxn,SAxn)=0 for each sequence {xn} in X such that limn→∞Axn=limn→∞Sxn [6],(3)
noncompatible if there exists a sequence{xn} in X such that limn→∞Axn=limn→∞Sxn, but limn→∞d(ASxn,SAxn) is either nonzero or nonexistent [8],(4)
weakly compatible if they commute at their coincidence points; that is,ASu=SAu whenever Au=Su, for some u∈X [7],(5)
occasionally weakly compatible if there is a pointx∈X which is a coincidence point of A and S at which A and S commute [12],(6)
with the property (E.A) if there exists a sequence{xn} in X and some z∈X such that limn→∞Axn=limn→∞Sxn=z [11].
It can be noticed that arbitrary noncompatible self-mappings satisfy the property (E.A) but two mappings satisfying the property (E.A), need not be noncompatible (see [33, Example 1]). Also, weak compatibility and property (E.A) are independent of each other (see [31, Examples 2.1 and 2.2]).Definition 2 (see [34]).
A pair(A,S) of self-mappings on a metric space (X,d) is called reciprocally continuous if for a sequence {xn} in X, limn→∞ASxn=Az and limn→∞SAxn=Sz, whenever limn→∞Axn=limn→∞Sxn=z, for some z∈X.It is easy to see that if two self-mappings are continuous, then they are obviously reciprocally continuous, but the converse is not true. Moreover, in the setting of common fixed point theorems for compatible pairs of self-mappings satisfying contractive conditions, continuity of one of the mappings implies their reciprocal continuity but not conversely (see [35]).Definition 3 (see [18]).
A pair(A,S) of self-mappings on a metric space (X,d) is said to be subcompatible if there exists a sequence {xn} such that limn→∞Axn=limn→∞Sxn=z, for some z∈X and limn→∞d(ASxn,SAxn)=0.A pair of subcompatible mappings satisfies the property (E.A). Obviously, compatible mappings which satisfy the property (E.A) are subcompatible, but the converse statement does not hold in general (see [36, Example 2.3]). Two occasionally weakly compatible mappings are subcompatible; however, the converse is not true in general (see [18, Example 1.2]).Definition 4 (see [18]).
A pair(A,S) of self-mappings on a metric space (X,d) is called subsequentially continuous if there exists a sequence {xn} in X such that limn→∞Axn=limn→∞Sxn=z, for some z∈X such that limn→∞ASxn=Az and limn→∞SAxn=Sz.One can easily check that if two self-mappingsA and S are both continuous, hence also reciprocally continuous mappings but A and S are not subsequentially continuous (see [35, Example 1]).Definition 5 (see [32]).
By⋄:ℝ+×ℝ+→ℝ+, a binary operation will be denoted, satisfying the following conditions: (1)
⋄ is associative and commutative,(2)
⋄ is continuous.
Some typical examples of⋄ are a⋄b=max{a,b},  a⋄b=a+b,  a⋄b=ab,  a⋄b=ab+a+b, and a⋄b=ab/max{a,b,1}, for each a,b∈ℝ+.Definition 6 (see [32]).
The binary operation⋄ is said to satisfy α-property if there exists a positive real number α such that
(1)a⋄b≤αmax{a,b},
for all a,b∈ℝ+.Example 7 (see [32]).
(1) Ifa⋄b=a+b, for each a,b∈ℝ+, then for α≥2, we have a⋄b≤αmax{a,b}.
(2) Ifa⋄b=ab/max{a,b,1}, for each a,b∈ℝ+, then for α≥1, we have a⋄b≤αmax{a,b}.


## 3. Main Results

In 2007, Sedghi and Shobe [32] proved the following result.Theorem 8 (see [32, Theorem 2.1]).
Let(X,d) be a complete metric space such that ⋄ satisfies the α-property with α>0. Let A, B, S, and T be self-mappings on X satisfying the following conditions: (1)
A(X)⊆T(X), B(X)⊆S(X), and T(X) or S(X) is a closed subset of X,(2)
the pairs(A,S) and (B,T) are weakly compatible,(3)
for allx,y∈X,
(2)d(Ax,By)≤k1(d(Sx,Ty)⋄d(Ax,Sx))+k2(d(Sx,Ty)⋄d(By,Ty))+k3(d(Sx,Ty)⋄d(Sx,By)+d(Ax,Ty)2),wherek1,k2,k3>0 and 0<α(k1+k2+k3)<1. Then, A, B, S, and T have a unique common fixed point in X.Now we prove our main result.Theorem 9.
LetA, B, S, and T be four self-mappings on a metric space (X,d), and let the operation ⋄ satisfy the α-property with α>0. Suppose that the pairs (A,S) and (B,T) are compatible and subsequentially continuous (alternately subcompatible and reciprocally continuous), satisfying inequality (2) of Theorem 8. Then A, B, S, and T have a unique common fixed point in X.Proof.
If the pair of mappings(A,S) is subsequentially continuous and compatible, there exists a sequence {xn} in X such that
(3)limn→∞Axn=limn→∞Sxn=z,
for some z∈X, and
(4)limn→∞d(ASxn,SAxn)=d(Az,Sz)=0;
that is, Az=Sz. Similarly, with respect to the pair (B,T), there exists a sequence {yn} in X such that
(5)limn→∞Byn=limn→∞Tyn=w,
for some w∈X, and
(6)limn→∞d(BTyn,TByn)=d(Bw,Tw)=0;
that is, Bw=Tw. Hence z is a coincidence point of the pair (A,S) whereas w is a coincidence point of the pair (B,T).
Now we assert thatz=w. If z≠w then using inequality (2) with x=xn and y=yn, we have
(7)d(Axn,Byn)≤k1(d(Sxn,Tyn)⋄d(Axn,Sxn))+k2(d(Sxn,Tyn)⋄d(Byn,Tyn))+k3(d(Sxn,Tyn)⋄d(Sxn,Byn)+d(Axn,Tyn)2).
Letting n→∞, we get
(8)d(z,w)≤k1(d(z,w)⋄d(z,z))+k2(d(z,w)⋄d(w,w))+k3(d(z,w)⋄d(z,w)+d(z,w)2).
Since ⋄ satisfies the α-property, we obtain
(9)d(z,w)≤k1αmax{d(z,w),d(z,z)}+k2αmax{d(z,w),d(w,w)}+k3αmax{d(z,w),d(z,w)+d(z,w)2}=k1αmax{d(z,w),0}+k2αmax{d(z,w),0}+k3αmax{d(z,w),d(z,w)}=α(k1+k2+k3)d(z,w)<d(z,w),
which is a contradiction. Hence z=w. Now we prove that Az=z. If we suppose that Az≠z, then from inequality (2) with x=z and y=yn, we have
(10)d(Az,Byn)≤k1(d(Sz,Tyn)⋄d(Az,Sz))+k2(d(Sz,Tyn)⋄d(Byn,Tyn))+k3(d(Sz,Tyn)⋄d(Sz,Byn)+d(Az,Tyn)2).
Taking the limit as n→∞, we get
(11)d(Az,w)≤k1(d(Sz,w)⋄d(Az,Az))+k2(d(Az,w)⋄d(w,w))+k3(d(Az,w)⋄d(Az,w)+d(Az,w)2);
that is,
(12)d(Az,z)≤k1(d(Sz,z)⋄d(Az,Az))+k2(d(Az,z)⋄d(z,z))+k3(d(Az,z)⋄d(Az,z)+d(Az,z)2)≤k1αmax{d(Sz,z),d(Az,Az)}+k2αmax{d(Az,z),d(z,z)}+k3αmax{d(Az,z),d(Az,z)+d(Az,z)2}.
Then, simplifying, we obtain
(13)d(Az,z)≤α(k1+k2+k3)d(Az,z)<d(Az,z),
a contradiction. Hence, Az=z. Therefore, Az=Sz=z. Now we show that Bz=z. If Bz≠z then using (2) with x=xn and y=z, we have
(14)d(Axn,Bz)≤k1(d(Sxn,Tz)⋄d(Axn,Sxn))+k2(d(Sxn,Tz)⋄d(Bz,Tz))+k3(d(Sxn,Tz)⋄d(Sxn,Bz)+d(Axn,Tz)2).
As n→∞, we get
(15)d(z,Bz)≤k1(d(z,Bz)⋄d(z,z))+k2(d(z,Bz)⋄d(Bz,Bz))+k3(d(z,Bz)⋄d(z,Bz)+d(z,Bz)2)≤k1αmax{d(z,Bz),d(z,z)}+k2αmax{d(z,Bz),d(Bz,Bz)}+k3αmax{d(z,Bz),d(z,Bz)+d(z,Bz)2}.
Then, simplifying, we obtain
(16)d(z,Bz)≤α(k1+k2+k3)d(z,Bz)<d(z,Bz),
a contradiction. Hence, Bz=z. Therefore, z=Az=Sz=Bz=Tz; that is, z is a common fixed point of A, B, S, and T. The uniqueness of common fixed point is an easy consequence of inequality (2).
Now suppose that the mappings(A,S) (as well as (B,T)) are subcompatible and reciprocally continuous. Then there exists a sequence {xn} in X such that
(17)limn→∞Axn=limn→∞Sxn=z,
for some z∈X, and
(18)limn→∞d(ASxn,SAxn)=d(Az,Sz)=0,
whereas in respect of the pair (B,T), there exists a sequence {yn} in X with
(19)limn→∞Byn=limn→∞Tyn=w,
for some w∈X, and
(20)limn→∞d(BTyn,TByn)=d(Bw,Tw)=0.
Therefore, Az=Sz and Bw=Tw; that is, z is a coincidence point of the pair (A,S) whereas w is a coincidence point of the pair (B,T). The rest of the proof can be completed easily.Example 10.
LetX=[0,∞), and let d be the usual metric on X. Define self-mappings A, B, S, and T by
(21)Ax=Bx={x7,ifx∈[0,1];x+67,ifx∈(1,∞),Sx=Tx={x6,ifx∈[0,1];x+56,ifx∈(1,∞).
Consider the sequence {xn}={1/n}n∈ℕ in X. Then
(22)limn→∞Axn=limn→∞17n=0=limn→∞16n=limn→∞Sxn.
Next,
(23)limn→∞ASxn=limn→∞A(16n)=limn→∞142n=0=A0,limn→∞SAxn=limn→∞S(17n)=limn→∞142n=0=S0,limn→∞d(ASxn,SAxn)=0.
Consider another sequence {xn}={1+1/n}n∈ℕ in X. Then
(24)limn→∞Axn=limn→∞(1+17n)=1=limn→∞(1+16n)=limn→∞Sxn.
However,
(25)limn→∞ASxn=limn→∞A(1+16n)=limn→∞(1+142n)=1≠A1,limn→∞SAxn=limn→∞S(1+17n)=limn→∞(1+142n)=1≠S1,
but limn→∞d(ASxn,SAxn)=0. Thus, the pair (A,S) is compatible as well as subsequentially continuous but not reciprocally continuous (the same for the pair (B,T)). It is easy to check that condition (2) is satisfied with ⋄=max, α=1, and k1+k2+k3=6/7. Therefore, all the conditions of Theorem 9 are satisfied. Here, 0 is a coincidence as well as the unique common fixed point of the mappings A, B, S, and T.
It can be noted that this example cannot be covered by those fixed point theorems which assume both compatibility and reciprocal continuity or by involving conditions of closedness of respective ranges. Indeed, in this exampleA(X)=[0,1/7]∪(1,∞) and S(X)=[0,1/6]∪(1,∞); hence, neither of A(X) and S(X) is closed.Example 11.
LetX=ℝ (set of real numbers), and let d be the usual metric on X. Define self-mappings A, B, S, and T by
(26)Ax=Bx={x4,ifx∈(-∞,1);4x-3,ifx∈[1,∞),Sx=Tx={x+3,ifx∈(-∞,1);5x-4,ifx∈[1,∞).
Consider the sequence {xn}={1+1/n}n∈ℕ in X. Then
(27)limn→∞Axn=limn→∞(1+4n)=1=limn→∞(1+5n)=limn→∞Sxn.
Also,
(28)limn→∞ASxn=limn→∞A(1+5n)=limn→∞(1+20n)=1=A1,limn→∞SAxn=limn→∞S(1+4n)=limn→∞(1+20n)=1=S1,limn→∞d(ASxn,SAxn)=0.
Consider another sequence {xn}={(1/n)-4}n∈ℕ in X. Then
(29)limn→∞A(xn)=limn→∞(14n-1)=-1=limn→∞(1n-1)=limn→∞S(xn).
Next,
(30)limn→∞ASxn=limn→∞A(1n-1)=limn→∞(14n-14)=-14=A(-1),limn→∞SAxn=limn→∞S(14n-1)=limn→∞(14n-1+3)=2=S(-1),
and limn→∞d(ASxn,SAxn)≠0. Thus, the pair (A,S) is reciprocally continuous as well as subcompatible but not compatible (the same for the pair (B,T)). It is easy to check that condition (2) is satisfied with ⋄=max, α=1, and k1+k2+k3=4/5. Therefore, all the conditions of Theorem 9 are satisfied. Thus, 1 is a coincidence as well as the unique common fixed point of the pair (A,S).
It can be noted that this example cannot be covered by those fixed point theorems which involve both compatibility and reciprocal continuity. Again,A(X)=(-∞,1/4)∪[1,+∞) is not closed. Note also that the mappings A and S have two points of coincidence (-1 and 1), which are occasionally weakly compatible but not weakly compatible.In the next example (taken from [18, Example 1.4]), we demonstrate the situation when conditions of Theorem 9 are not satisfied, and the given pairs have no common fixed points.Example 12.
LetX=[0,+∞) with the standard metric d, and let A,B,S,T:X→X be given by
(31)Ax=Bx={x+1,0≤x≤1,2x-1,x>1,Sx=Tx={1-x,0≤x<1,3x-2,x≥1.
Then, as it was shown in [18, Example 1.4], the pairs (A,S) and (B,T) are subsequentially continuous and subcompatible. However, they are neither reciprocally continuous nor compatible (not even occasionally weakly compatible). We note that (A,S) has no common fixed points, although it has a unique point of coincidence z=1.We present an example of different kind, inspired by [20, Example 3].Example 13.
LetX={0,1,2,…,10}, and define a metric d on X by
(32)d(x,y)={0,x=y,max{x,y},x≠y.
Consider the mappings A,B,S,T:X→X given by
(33)Ax=Bx={0,x=0,x-1,x≥1;Sx=Tx={0,x=0,x+1,1≤x≤9,10,x=10.
Take ⋄=max (which satisfies α-condition with α=1). Then,(1)
the pair(A,S) (as well as (B,T)) is compatible and subsequentially continuous,(2)
condition (2) is satisfied with k1=k2=k3=0.3. Indeed, in order to prove (1), take xn=0 for all n but finitely many (which is the only possibility to obtain the same limit for (Axn) and (Sxn)). Then d(Axn,0)→0 and d(Sxn,0)→0; also SAxn→0=S0 and ASxn→0=A0. Hence, the pair (A,S) is compatible and subsequentially continuous.
In order to prove (2), suppose thatx,y∈X, x≠y (the case x=y is trivial). Since we have A=B, S=T, and k1=k2, condition (2) is symmetric in x,y; hence, without loss of generality, we can suppose that x≥y. Consider the following possible cases.
Case
1. One has y=0 and x=1. Then Ax=By=0, d(Ax,By)=0, and (2) is satisfied.
Case
2. One has y∈{0,1} and 2≤x≤9. Then Ax=x-1, By=0, d(Ax,By)=x-1. The right-hand side of (2) becomes (t∈{0,2})
(34)R=k1max{x+1,t}+k2max{x+1,x+1}+k3max{x+1,12[max{Sx,By}+x+1]}=(k1+k2+k3)(x+1)=0.9(x+1)≥0.9·108(x-1)>x-1=d(Ax,By).Case3. One has y∈{0,1} and x=10. Then
(35)d(Ax,By)=9=(k1+k2+k3)·10=R.Case4. One has 2≤y<x≤9. Then d(Ax,By)=d(x-1,y-1)=x-1 and
(36)R=k1max{x+1,x+1}+k2max{x+1,y+1}+k3max{x+1,12[x+1+max{Ax,Ty}]}=(k1+k2+k3)(x+1)=0.9(x+1)≥0.9·108(x-1)>x-1=d(Ax,By).Case5. One has 2≤y<x=10. Then (2) again reduces to 9=(k1+k2+k3)·10.
All the conditions of Theorem9 are satisfied, and A, B, S, and T have a unique common fixed point (which is z=0).
By choosingA, B, S, and T suitably in Theorem 9, we can deduce corollaries for two or three self-mappings. As a sample, we deduce the following corollary for two self-mappings.Corollary 14.
LetA and S be two self-mappings on a metric space (X,d) such that ⋄ satisfies the α-property with α>0. If the pair (A,S) is compatible and subsequentially continuous (alternately subcompatible and reciprocally continuous) satisfying
(37)d(Ax,Ay)≤k1(d(Sx,Sy)⋄d(Ax,Sx))+k2(d(Sx,Sy)⋄d(Ay,Sy))+k3(d(Sx,Sy)⋄d(Sx,Ay)+d(Ax,Sy)2),
for all x,y∈X, where k1,k2,k3>0 and 0<α(k1+k2+k3)<1, then A and S have a unique common fixed point in X.Remark 15.
The conclusion of Theorem9 remains true if we replace inequality (2) by the following:
(38)d(Ax,By)≤k1(d(Sx,Ty)+d(Ax,Sx))+k2(d(Sx,Ty)+d(By,Ty))+k3(d(Sx,Ty)+d(Sx,By)+d(Ax,Ty)2),
for all x,y∈X, where k1,k2,k3>0 and 0<k1+k2+k3<1/2.
Similarly, other variants of contractive condition can be obtained by specifying operation⋄.Remark 16.
Similar results can be obtained if condition (2) is replaced by the following one:
(39)d(Ax,By)≤ψ(u)forsomeu∈{d(Sx,Ty),d(Ax,Sx),forsomeu∈2d(By,Ty),d(Sx,By),forsomeu∈2d(Ax,Ty)},
for a suitable function ψ:[0,+∞)→[0,+∞).


## 4. Conclusion

Theorem9 is proved for two pairs of compatible and subsequentially continuous (alternately subcompatible and reciprocally continuous) mappings satisfying a general contractive condition. Theorem 9 improves the main result of Sedghi and Shobe [32, Theorem 2.1] as we do not require any condition on the containment of ranges of involved mappings and completeness (or closedness) of the whole space (or any subspace). A natural result is defined in the form of a corollary (see Corollary 14). On the other hand, Remark 15 is developed for a particular case, a⋄b=a+b, which also improves the result of Sedghi and Shobe [32, Corollary 2.2].



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