# Generalized Altering Distances and Common Fixed Points in Ordered Metric Spaces

Authors: Hemant Kumar Nashine; Hassen Aydi
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

Coincidence point and common fixed point results with the concept of generalized altering distance functions in complete ordered metric spaces are derived. These results generalize the existing fixed point results in the literature. To illustrate our results and to distinguish them from the existing ones, we equip the paper with examples. As an application, we study the existence of a common solution to a system of integral equations.


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## Body

## 1. Introduction and Preliminaries

The study of common fixed points of mappings satisfying certain contractive conditions has been researched extensively by many mathematicians since fixed point theory plays a major role in mathematics and applied sciences (see [1–40] and others). A new category of contractive fixed point problems was addressed by Khan et al. [1]. In this work, they introduced the notion of an altering distance function, which is a control function that alters distance between two points in a metric space.Definition 1.1 (see [1]).
A functionφ:[0,+∞)→[0,+∞) is called an altering distance function if and only if(i)
φis continuous,(ii)
φ is nondecreasing,(iii)
φ(t)=0⇔t=0.
Khan et al. [1] proved the following result.Theorem 1.2 (see [1]).
Let(X,d) be a complete metric space, φ:[0,+∞)→[0,+∞) an altering distance function, and T:X→X a self-mapping which satisfies the following inequality:
(1.1)φ(d(Tx,Ty))≤cφ(d(x,y))
for all x,y∈X and for some 0<c<1. Then T has a unique fixed point.Puttingφ(t)=t in Theorem 1.2, we retrieve immediately the Banach contraction principle.In 1997, Alber and Guerre-Delabriere [2] introduced the concept of weak contractions in Hilbert spaces.This concept was extended to metric spaces by Rhoades in [3].Definition 1.3.
A mappingT:X→X, where (X,d) is a metric space, is said to be weakly contractive if and only if
(1.2)d(Tx,Ty)≤d(x,y)-φ(d(x,y)),∀x,y∈X,
where φ:[0,+∞)→[0,+∞) is an altering distance function.Theorem 1.4 (see [3]).
Let(X,d) be a complete metric space and T:X→X a weakly contractive map. Then, T admits a unique fixed point.Weak inequalities of the above type have been used to establish fixed point results in a number of subsequent works, some of which are noted in [4–7, 20]. In [5], Choudhury introduced the concept of a generalized altering distance function for three variables.Definition 1.5 (see [5]).
A functionφ:[0,+∞)×[0,+∞)×[0,+∞)→[0,+∞) is said to be a generalized altering distance function if and only if(i)
φ is continuous,(ii)
φ is nondecreasing in all the three variables,(iii)
φ(x,y,z)=0⇔x=y=z=0.In [5], Choudhury proved the following common fixed point theorem using altering distances for three variables.Theorem 1.6 (see [5]).
Let(X,d) be a complete metric space and S,T:X→X two self mappings such that the following inequality is satisfied:
(1.3)Φ1(d(Sx,Ty))≤ψ1(d(x,y),d(x,Sx),d(y,Ty))-ψ2(d(x,y),d(x,Sx),d(y,Ty))
for all x,y∈X, where ψ1 and ψ2 are generalized altering distance functions and Φ1(x)=ψ1(x,x,x). Then S and T have a common fixed point.In [31], Rao et al. generalized Theorem 1.6 for four mappings satisfying a generalized contractive condition and used the following generalized altering distance function for four variables.Definition 1.7 (see [31]).
A functionφ:[0,+∞)×[0,+∞)×[0,+∞)×[0,+∞)→[0,+∞) is said to be a generalized altering distance function if and only if(i)
φ is continuous,(ii)
φ is nondecreasing in all the three variables,(iii)
φ(t1,t2,t3,t4)=0⇔t1=t2=t3=t4=0.
LetΨ4 denote the set of all functions φ satisfying (i)–(iii) in Definition 1.7.Recently, there have been so many exciting developments in the field of existence of fixed point in partially ordered sets (see [8–19, 21–36] and the references cited therein). The first result in this direction was given by Turinici [36], where he extended the Banach contraction principle in partially ordered sets. Ran and Reurings [33] presented some applications of Turinici’s theorem to matrix equations. The obtained result by Turinici was further extended and refined in [28–32]. Subsequently, Harjani and Sadarangani [19] generalized their own results [18] by considering pair of altering functions (ψ,φ). Nashine and Altun [21] and Nashine and Altun [22] generalized the results of Harjani and Sadarangani [18, 19]. Also, Nashine and Altun [22] and Shatanawi and Samet [41] worked for a pair (T,S) of weakly increasing mappings with respect to a third mapping R. In another paper, Nashine et al. [24] prove coincidence point and common fixed point results for mappings satisfying a contractive inequality which involves two generalized altering distance functions for three variables in ordered complete metric spaces. As application, they studied the existence of a common solution to a system of integral equations.The aim of this paper is to generalize the results of Nashine et al. [24] in the sense of four variables. We obtain coincidence point and common fixed point theorems in complete ordered metric spaces for mappings satisfying a contractive condition which involves two generalized altering distance functions in four variables. Presented theorems are ordered version of Theorem  2.1 of Rao et al. [31] for three mappings. In addition, an application to the study of the existence of a common solution to a system of integral equations is given.


## 2. Main Results

First we introduce some notations and definitions that will be used later.


### 2.1. Notations and Definitions

The following definition was introduced by Jungck in [37].Definition 2.1 (see [37]).
Let(X,d) be a metric space and f,g:X→X. If w=fx=gx, for some x∈X, then x is called a coincidence point of f and g, and w is called a point of coincidence of f and g. The pair {f,g} is said to be compatible if and only if limn→+∞d(fgxn,gfxn)=0, whenever {xn} is a sequence in X such that limn→+∞fxn=limn→+∞gxn=t for some t∈X.LetX be a nonempty set and R:X→X a given mapping. For every x∈X, we denote by R-1(x) the subset of X defined by
(2.1)R-1(x):={u∈X∣Ru=x}.Definition 2.2 (see [23]).
Let(X,⪯) be a partially ordered set and T,S,R:X→X are given mappings such that TX⊆RX and SX⊆RX. We say that S and T are weakly increasing with respect to R  if and only if, for all x∈X, we have
(2.2)Tx⪯Sy,∀y∈R-1(Tx),Sx⪯Ty,∀y∈R-1(Sx).Remark 2.3.
IfR:X→X is the identity mapping (Rx=x for all x∈X), then S and T are weakly increasing with respect to Rwhichimplies that S and T are weakly increasing mappings. Note that the notion of weakly increasing mappings was introduced in [9] (also see [17, 38]).Example 2.4.
LetX={1,2,3} be endowed with the partial order ⪯ given by
(2.3)⪯:={(1,1),(2,2),(3,3),(2,3),(3,1),(2,1)}.
Define the mappings T,S,R:X→X by
(2.4)T1=T3=1,T2=3,S1=S2=S3=1,R1=1,R2=R3=2.
We will show that the mappings S and T are weakly increasing with respect to R.
Letx,y∈X such that y∈R-1(Tx). By the definition of S, we have Sy=1. On the other hand, Tx∈{1,3} and (1,1),(3,1)∈⪯. Thus, we have Tx⪯Sy for all y∈R-1(Tx).
Letx,y∈X such that y∈R-1(Sx). By the definitions of S and R, we have R-1(Sx)=R-1(1)={1}. Then we have y=1. On the other hand, 1=Sx⪯Ty=T1=1. Then, Sx⪯Ty for all y∈R-1(Sx).
Thus, we proved thatS and T are weakly increasing with respect to R.Definition 2.5.
LetX be a nonempty set. Then (X,d,⪯) is called an ordered metric space if and only if:(i)
(X,d) is a metric space,(ii)
(X,⪯) is a partially ordered set.


### 2.2. Results

Our first result is the following.Theorem 2.6.
Let(X,d,⪯) be an ordered complete metric space. Let T,S,R:X→X be given mappings satisfying for every pair (x,y)∈X×X such that Rx and Ry are comparable,
(2.5)Φ1(d(Sx,Ty))≤ψ1(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)])-ψ2(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)]),
where ψ1 and ψ2 are generalized altering distance functions (in Ψ4) and Φ1(x)=ψ1(x,x,x,x). One assumes the following hypotheses:(i)
T, S, and R are continuous,(ii)
TX⊆RX, SX⊆RX,(iii)
T and S are weakly increasing with respect to R,(iv)
the pairs{T,R} and {S,R} are compatible.
Then,T, S, and R have a coincidence point, that is, there exists u∈X such that Ru=Tu=Su.Proof.
Letx0∈X be an arbitrary point. Since TX⊆RX, there exists x1∈X such that Rx1=Tx0. Since SX⊆RX, there exists x2∈X such that Rx2=Sx1. Continuing this process, we can construct a sequence {Rxn} in X defined by
(2.6)Rx2n+1=Tx2n,Rx2n+2=Sx2n+1,∀n∈N.
We claim that
(2.7)Rxn⪯Rxn+1,∀n∈N.
To this aim, we will use the increasing property of the mappings S and T with respect to R. From (2.6), we have
(2.8)Rx1=Tx0⪯Sy,∀y∈R-1(Tx0).
Since Rx1=Tx0, then x1∈R-1(Tx0), and we get
(2.9)Rx1=Tx0⪯Sx1=Rx2.
Again,
(2.10)Rx2=Sx1⪯Ty,∀y∈R-1(Sx1).
Since x2∈R-1(Sx1), we get
(2.11)Rx2=Sx1⪯Tx2=Rx3.
Hence, by induction, (2.7) holds.
Without loss of the generality, we can assume that(2.12)Rxn≠Rxn+1,∀n∈N.
Now, we will prove our result on three steps.
Step  1. We claim that
(2.13)lim⁡n→+∞d(Rxn+1,Rxn+2)=0.
Putting x=x2n+1 and y=x2n, from (2.7) and the considered contraction (2.5), we have
(2.14)Φ1(d(Rx2n+2,Rx2n+1))=Φ1(d(Sx2n+1,Tx2n))≤ψ1(12[d(Rx2n+1,Tx2n)+d(Rx2n,Sx2n+1)]d(Rx2n+1,Rx2n),d(Rx2n+1,Sx2n+1),d(Rx2n,Tx2n),12[d(Rx2n+1,Tx2n)+d(Rx2n,Sx2n+1)])-ψ2(12[d(Rx2n+1,Tx2n)+d(Rx2n,Sx2n+1)]d(Rx2n+1,Rx2n),d(Rx2n+1,Sx2n+1),d(Rx2n,Tx2n),12[d(Rx2n+1,Tx2n)+d(Rx2n,Sx2n+1)])=ψ1(12[d(Rx2n+1,Tx2n)+d(Rx2n,Sx2n+1)]d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2))-ψ2(12[d(Rx2n+1,Tx2n)+d(Rx2n,Sx2n+1)]d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2)).
Suppose, for some n∈ℕ, that
(2.15)d(Rx2n+1,Rx2n+2)>d(Rx2n,Rx2n+1).
Using (2.15) and a triangular inequality, we have
(2.16)12d(Rx2n,Rx2n+2)≤12(d(Rx2n,Rx2n+1)+d(Rx2n+1,Rx2n+2)<d(Rx2n+1,Rx2n+2).
Using this and (2.15) together with a property of the generalized altering function ψ1, we get
(2.17)ψ1(d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2))≤Φ1(d(Rx2n+2,Rx2n+1)).
Hence, we obtain
(2.18)Φ1(d(Rx2n+2,Rx2n+1))≤Φ1(d(Rx2n+2,Rx2n+1))-ψ2(d12d(Rx2n,Rx2n+2)(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2)).
This implies that
(2.19)ψ2(d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2))=0,
which yields that
(2.20)d(Rx2n+1,Rx2n)=0.
Hence, we obtain a contradiction with (2.12). We deduce that
(2.21)d(Rx2n+1,Rx2n+2)≤d(Rx2n,Rx2n+1),∀n∈N.
Similarly, putting x=x2n+1 and y=x2n+2, from (2.7) and the considered contraction (2.5), we have
(2.22)Φ1(d(Rx2n+2,Rx2n+3))≤ψ1(d(Rx2n+1,Rx2n+2),d(Rx2n+1,Rx2n+2),d(Rx2n+2,Rx2n+3),12d(Rx2n+1,Rx2n+3))-ψ2(d(Rx2n+1,Rx2n+2),d(Rx2n+1,Rx2n+2),d(Rx2n+2,Rx2n+3),12d(Rx2n+1,Rx2n+3)).
Suppose, for some n∈ℕ, that
(2.23)d(Rx2n+2,Rx2n+3)>d(Rx2n+1,Rx2n+2).
Then, by a triangular inequality, we have
(2.24)12d(Rx2n+1,Rx2n+3)≤12(d(Rx2n+1,Rx2n+2)+d(Rx2n+2,Rx2n+3))<d(Rx2n+2,Rx2n+3).
Hence, from this, (2.22), and (2.23), we obtain
(2.25)Φ1(d(Rx2n+2,Rx2n+3))≤Φ1(d(Rx2n+2,Rx2n+3))-ψ2(d(Rx2n+1,Rx2n+2),d(Rx2n+1,Rx2n+2),d(Rx2n+2,Rx2n+3),12d(Rx2n+1,Rx2n+3)).
This implies that
(2.26)ψ2(d(Rx2n+1,Rx2n+2),d(Rx2n+1,Rx2n+2),d(Rx2n+2,Rx2n+3),12d(Rx2n+1,Rx2n+3))=0,
which leads to
(2.27)d(Rx2n+1,Rx2n+2)=0.
Hence, we obtain a contradiction with (2.12). We deduce that
(2.28)d(Rx2n+1,Rx2n+2)≥d(Rx2n+2,Rx2n+3),∀n∈N.
Combining (2.21) and (2.28), we obtain
(2.29)d(Rxn+1,Rxn+2)≥d(Rxn+2,Rxn+3),∀n∈N.
Then, {d(Rxn+1,Rxn+2)} is a nonincreasing sequence of positive real numbers. This implies that there exists r≥0 such that
(2.30)lim⁡n→+∞d(Rxn+1,Rxn+2)=r.
By (2.14), we have
(2.31)Φ1(d(Rx2n+2,Rx2n+1))≤ψ1(12d(Rx2n,Rx2n+2)d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2))-ψ2(12d(Rx2n,Rx2n+2)d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2))≤ψ1(d(Rx2n+1,Rx2n),d(Rx2n,Rx2n+1),d(Rx2n,Rx2n+1),d(Rx2n,Rx2n+1))-ψ2(d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),0)=Φ1(d(Rx2n,Rx2n+1))-ψ2(d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),0).
Letting n→+∞ in (2.31) and using the continuities of Φ1 and ψ2, we obtain
(2.32)Φ1(r)≤Φ1(r)-ψ2(r,r,r,0),
which implies that ψ2(r,r,r,0)=0, so r=0. Hence
(2.33)lim⁡n→+∞d(Rxn+1,Rxn+2)=0.
Hence, (2.13) is proved.
Step  2. We claim that {Rxn} is a Cauchy sequence.
From (2.13), it will be sufficient to prove that {Rx2n} is a Cauchy sequence. We proceed by negation and suppose that {Rx2n} is not a Cauchy sequence. Then, there exists ε>0 for which we can find two sequences of positive integers {m(i)} and {n(i)} such that, for all positive integer i,
(2.34)n(i)>m(i)>i,d(Rx2m(i),Rx2n(i))≥ε,d(Rx2m(i),Rx2n(i)-2)<ε.
From (2.34) and using a triangular inequality, we get
(2.35)ε≤d(Rx2m(i),Rx2n(i))≤d(Rx2m(i),Rx2n(i)-2)+d(Rx2n(i)-2,Rx2n(i)-1)+d(Rx2n(i)-1,Rx2n(i))<ε+d(Rx2n(i)-2,Rx2n(i)-1)+d(Rx2n(i)-1,Rx2n(i)).
Letting i→+∞ in the previous inequality and using (2.13), we obtain
(2.36)lim⁡i→+∞d(Rx2m(i),Rx2n(i))=ε.
Again, a triangular inequality gives us
(2.37)|d(Rx2n(i),Rx2m(i)-1)-d(Rx2n(i),Rx2m(i))|≤d(Rx2m(i)-1,Rx2m(i)).
Letting i→+∞ in the above inequality and using (2.13) and (2.36), we get
(2.38)lim⁡i→+∞d(Rx2n(i),Rx2m(i)-1)=ε.
On the other hand, we have
(2.39)d(Rx2n(i),Rx2m(i))≤d(Rx2n(i),Rx2n(i)+1)+d(Rx2n(i)+1,Rx2m(i))=d(Rx2n(i),Rx2n(i)+1)+d(Tx2n(i),Sx2m(i)-1).
Then, from (2.13), (2.36), and the continuity of Φ1, we get by letting i→+∞ in the above inequality
(2.40)Φ1(ε)≤lim⁡i→+∞Φ1(d(Sx2m(i)-1,Tx2n(i))).
Now, using the considered contractive condition (2.5) for x=x2m(i)-1 and y=x2n(i), we have
(2.41)Φ1(d(Sx2m(i)-1,Tx2n(i)))≤ψ1(12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))]d(Rx2m(i)-1,Rx2n(i)),d(Rx2m(i)-1,Rx2m(i)),d(Rx2n(i),Rx2n(i)+1),12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))])-ψ2(12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))]d(Rx2m(i)-1,Rx2n(i)),d(Rx2m(i)-1,Rx2m(i)),d(Rx2n(i),Rx2n(i)+1),12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))]).
Then, from (2.13), (2.38), and the continuities of ψ1 and ψ2, we get by letting i→+∞ in the above inequality
(2.42)lim⁡i→+∞Φ1(d(Sx2m(i)-1,Tx2n(i)))≤ψ1(ε,0,0,ϵ)-ψ2(ε,0,0,ϵ)≤Φ1(ε)-ψ2(ε,0,0,ϵ).
Now, combining (2.40) with the previous inequality, we get
(2.43)Φ1(ε)≤Φ1(ε)-ψ2(ε,0,0,ϵ),
which implies that ψ2(ε,0,0,ϵ)=0, that is a contradiction since ε>0. We deduce that {Rxn} is a Cauchy sequence.
Step  3. We claimexistence of a coincidence point.
Since{Rxn} is a Cauchy sequence in the complete metric space (X,d), there exists u∈X such that
(2.44)lim⁡n→+∞Rxn=u.
From (2.44) and the continuity of R, we get
(2.45)lim⁡n→+∞R(Rxn)=Ru.
By the triangular inequality, we have
(2.46)d(Ru,Tu)≤d(Ru,R(Rx2n+1))+d(R(Tx2n),T(Rx2n))+d(T(Rx2n),Tu).
On the other hand, we have
(2.47)Rx2n⟶u,Tx2n⟶uasn→+∞.
Since R and T are compatible mappings, this implies that
(2.48)lim⁡n→+∞d(R(Tx2n),T(Rx2n))=0.
Now, from the continuity of T and (2.44), we have
(2.49)lim⁡n→+∞d(T(Rx2n),Tu)=0.
Combining (2.45), (2.48), and (2.49) and letting n→+∞ in (2.46), we obtain
(2.50)d(Ru,Tu)≤0,
that is,
(2.51)Ru=Tu.
Again, by a triangular inequality, we have
(2.52)d(Ru,Su)≤d(Ru,R(Rx2n+2))+d(R(Sx2n+1),S(Rx2n+1))+d(S(Rx2n+1),Su).
On the other hand, we have
(2.53)Rx2n+1⟶u,Sx2n+1⟶uasn⟶+∞.
Since R and S are compatible mappings, this implies that
(2.54)lim⁡n→+∞d(R(Sx2n+1),S(Rx2n+1))=0.
Now, from the continuity of S and (2.44), we have
(2.55)lim⁡n→+∞d(S(Rx2n+1),Su)=0.
Combining (2.45), (2.54), and (2.55) and letting n→+∞ in (2.52), we obtain
(2.56)d(Ru,Su)≤0,
that is,
(2.57)Ru=Su.
Finally, from (2.51) and (2.57), we have
(2.58)Tu=Ru=Su,
that is, u is a coincidence point of T, S, and R. This makes end to the proof.In the next theorem, we omit the continuity hypotheses satisfied byT, S, and R.Definition 2.7.
Let(X,d,⪯) be a partially ordered metric space. We say that X is regular if and only if the following hypothesis holds: if {zn} is a nondecreasing sequence in X with respect to ⪯such that zn→z∈X as n→+∞, then zn⪯z for all n∈ℕ.Now, our second result is the following.Theorem 2.8.
Let(X,d,⪯) be an ordered complete metric space. Let T,S,R:X→X be given mappings satisfying for every pair (x,y)∈X×X such that Rx and Ry are comparable,
(2.59)Φ1(d(Sx,Ty))≤ψ1(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)])-ψ2(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)]),
where ψ1 and ψ2 are generalized altering distance functions and Φ1(x)=ψ1(x,x,x,x). We assume the following hypotheses:(i)
X is regular,(ii)
T and S are weakly increasing with respect to R,(iii)
RX is a complete subspace of (X,d),(iv)
TX⊆RX, SX⊆RX.
Then,T, S, and R have a coincidence point.Proof.
Following the proof of Theorem2.6, we have {Rxn} is a Cauchy sequence in (RX,d). Since RX is a complete, there exists u=Rv, v∈X such that
(2.60)lim⁡n→+∞Rxn=u=Rv.
Since {Rxn} is a nondecreasing sequence and X is regular, it follows from (2.60) that Rxn⪯Rv for all n∈ℕ. Hence, we can apply the contractive condition (2.5). Then, for x=v and y=x2n, we obtain
(2.61)Φ1(d(Sv,Rx2n+1))=Φ1(d(Sv,Tx2n))≤ψ1(12[d(Rv,Tx2n)+d(Rx2n,Sv)]d(Rv,Rx2n),d(Rv,Sv),d(Rx2n,Rx2n+1),12[d(Rv,Tx2n)+d(Rx2n,Sv)])-ψ2(12[d(Rv,Tx2n)+d(Rx2n,Sv)]d(Rv,Rx2n),d(Rv,Sv),d(Rx2n,Rx2n+1),12[d(Rv,Tx2n)+d(Rx2n,Sv)]).
Letting n→+∞ in the above inequality and using (2.13), (2.60), and the properties of ψ1 and ψ2, we obtain
(2.62)Φ1(d(Sv,Rv))≤ψ1(0,d(Rv,Sv),0,12d(Rv,Sv))-ψ2(0,d(Rv,Sv),0,12d(Rv,Sv))≤Φ1(d(Sv,Rv))-ψ2(0,d(Rv,Sv),0,12d(Rv,Sv)).
This implies that ψ2(0,d(Rv,Sv),0,(1/2)d(Rv,Sv))=0, which gives us that d(Rv,Sv)=0, that is,
(2.63)Rv=Sv.
Similarly, for x=x2n+1 and y=v, we obtain
(2.64)Φ1(d(Rx2n+2,Tv))=Φ1(d(Sx2n+1,Tv))≤ψ1(12[d(Rx2n+1,Tv)+d(Rv,Sx2n+1)]d(Rx2n+2,Rv),d(Rx2n+1,Rx2n+2),d(Rv,Tv),12[d(Rx2n+1,Tv)+d(Rv,Sx2n+1)])-ψ2(12[d(Rx2n+1,Tv)+d(Rv,Sx2n+1)]d(Rx2n+2,Rv),d(Rx2n+1,Rx2n+2),d(Rv,Tv),12[d(Rx2n+1,Tv)+d(Rv,Sx2n+1)]).
Letting n→+∞ in the above inequality, we get
(2.65)Φ1(d(Rv,Tv))≤ψ1(0,0,d(Rv,Tv),12d(Rv,Tv))-ψ2(0,0,d(Rv,Tv),12d(Rv,Tv))≤Φ1(d(Rv,Tv))-ψ2(0,0,d(Rv,Tv),12d(Rv,Tv)).
This implies that ψ2(0,0,d(Rv,Tv),(1/2)d(Rv,Tv))=0, and then,
(2.66)Rv=Tv.
Now, combining (2.63) and (2.66), we obtain
(2.67)Rv=Tv=Sv.
Hence, v is a coincidence point of T, S, and R. This makes end to the proof.Now, we give a sufficient condition that assures the uniqueness of the common coincidence point of{T,R} and {S,R}.Theorem 2.9.
Under the hypotheses of Theorem2.6 (resp., Theorem 2.8) and suppose that (SX,⪯) is a totally ordered set and S is one-to-one mapping, then one obtains a unique common coincidence point of {T,R} and {S,R}.Proof.
Following the proof of Theorem2.6 (resp., Theorem 2.8), the set of common coincidence points of {T,R} and {S,R} is nonempty. Let μ,ν∈Xbe two common coincidence points of {T,R} and {S,R}, that is,
(2.68)Tμ=Rμ=Sμ,Tν=Rν=Sν.
This implies that Rμ,Rν∈SX, and then, Rμ and Rν are comparable with respect to ⪯. Then, we can apply the contractive condition (2.5). We have
(2.69)Φ1(d(Sμ,Tν))≤ψ1(d(Rμ,Rν),d(Rμ,Sμ),d(Rν,Tν),12[d(Rμ,Tν)+d(Rν,Sμ)])-ψ2(d(Rμ,Rν),d(Rμ,Sμ),d(Rν,Tν),12[d(Rμ,Tν)+d(Rν,Sμ)])=ψ1(d(Sμ,Tν),0,0,d(Sμ,Tν))-ψ2(d(Sμ,Tν),0,0,d(Sμ,Tν))≤Φ1(d(Sμ,Tν))-ψ2(d(Sμ,Tν),0,0,d(Sμ,Tν)).
This implies that ψ2(d(Sμ,Tν),0,0,d(Sμ,Tν)), which gives us that d(Sμ,Tν)=0, that is, Sμ=Tν. From (2.68), we get Sμ=Sν. Since S is one-to-one, we have μ=ν. This makes end to the proof.Now, it is easy to state a corollary of Theorem2.6 or Theorem 2.8 involving contractions of integral type.Corollary 2.10.
LetR,S and T satisfy the conditions of Theorem 2.6 or Theorem 2.8, except that condition (2.5) is replaced by the following: there exists a positive Lebesgue integrable function u on ℝ+ such that ∫0ε‍u(t)dt>0 for each ε>0 and that
(2.70)∫0Φ1(d(Sx,Ty))u(t)dt≤∫0ψ1(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),(1/2)[d(Rx,Ty)+d(Ry,Sx)])u(t)dt-∫0ψ2(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),(1/2)[d(Rx,Ty)+d(Ry,Sx)])‍u(t)dt.
Then, T, S, and R have a coincidence point.IfR:X→X is the identity mapping, we can deduce easily the following common fixed point results.The next result is an immediate consequence of Theorem2.6.Corollary 2.11.
Let(X,d,⪯) be an ordered complete metric space. Let T,S:X→X be given mappings satisfying for every pair (x,y)∈X×X such that x and y are comparable,
(2.71)Φ1(d(Sx,Ty))≤ψ1(d(x,y),d(x,Sx),d(y,Ty),12[d(x,Ty)+d(y,Sx)])-ψ2(d(x,y),d(x,Sx),d(y,Ty),12[d(x,Ty)+d(y,Sx)]),
where ψ1 and ψ2 are generalized altering distance functions and Φ1(x)=ψ1(x,x,x,x). One assumes the following hypotheses:(i)
T and S are continuous,(ii)
T and S are weakly increasing.
Then,T and S have a common fixed point, that is, there exists u∈X such that u=Tu=Su.The following result is an immediate consequence of Theorem2.8.Corollary 2.12.
Let(X,d,⪯) be an ordered complete metric space. Let T,S:X→X be given mappings satisfying for every pair (x,y)∈X×X such that x and y are comparable,
(2.72)Φ1(d(Sx,Ty))≤ψ1(d(x,y),d(x,Sx),d(y,Ty),12[d(x,Ty)+d(y,Sx)])-ψ2(d(x,y),d(x,Sx),d(y,Ty),12[d(x,Ty)+d(y,Sx)]),
where ψ1 and ψ2 are generalized altering distance functions and Φ1(x)=ψ1(x,x,x,x). One assume the following hypotheses:(i)
X is regular,(ii)
T and S are weakly increasing.
Then,T and S have a common fixed point.Now, we give some examples to support our results.Example 2.13.
LetX={4,5,6} be endowed with the usual metric d(x,y)=|x-y| for all x,y∈X, and ⪯:={(4,4),(5,5),(6,6),(6,4)}. Consider the mappings
(2.73)S=T=(456464),R=(456456).
We define the functions ψ1,ψ2:[0,+∞)4→[0,+∞) by
(2.74)ψ1(t1,t2,t3,t4)=14(t1+t2+t3+t4),ψ2(t1,t2,t3,t4)=116(t1+t2+t3+t4).
Clearly, Φ1(t)=t for all t≥0. Now, we will check that all the hypotheses required by Theorem 2.8 are satisfied.
(i)X is regular.
Let{zn} be a nondecreasing sequence in X with respect to ⪯ such that zn→z∈X as n→+∞. We have zn⪯zn+1 for all n∈ℕ.(a)
Ifz0=4, then z0=4⪯z1. From the definition of ⪯, we have z1=4. By induction, we get zn=4 for all n∈ℕ and z=4. Then, zn⪯z for all n∈ℕ.(b)
Ifz0=5, then z0=5⪯z1. From the definition of ⪯, we have z1=5. By induction, we get zn=5 for all n∈ℕ and z=5. Then, zn⪯z for all n∈ℕ.(c)
Ifz0=6, then z0=6⪯z1. From the definition of ⪯, we have z1∈{6,4}. By induction, we get zn∈{6,4} for all n∈ℕ. Suppose that there exists p≥1 such that zp=4. From the definition of ⪯, we get zn=zp=4 for all n≥p. Thus, we have z=4 and zn⪯z for all n∈ℕ. Now, suppose that zn=6 for all n∈ℕ. In this case, we get z=6 and zn⪯z for all n∈ℕ.
Thus, we proved that, in all cases, we havezn⪯z for all n∈ℕ. Then, X is regular.
(ii)T and S are weakly increasing.
SinceS=T, we have to check that Tx⪯T(Tx) for all x∈X.
Forx=4, we have
(2.75)T4=4⪯T(T4)=T4=4.
For x=5, we have
(2.76)T5=6⪯T(T5)=T6=4.
For x=6, we have
(2.77)T6=4⪯T(T6)=T4=4.
Thus, we proved that T and S are weakly increasing.
On the other hand, it is very easy to show that (2.5) is satisfied for all x,y∈X such that Rx⪯Ry.
Now, all the hypotheses of Theorem2.8 are satisfied. Then R, S, and T have a coincidence point u=4.
Note that inequality (2.5) is not satisfied for x=4 and y=5. Indeed,
(2.78)ψ1(d(R4,R5),d(R4,S4),d(R5,T5),12[d(R4,T5)+d(R5,S4)])-ψ2(d(R4,R5),d(R4,S4),d(R5,T5),12[d(R4,T5)+d(R5,S4)])=ψ1(1,0,1,32)-ψ2(1,0,1,32)=2132<Φ1(d(S4,T5))=Φ1(2)=2.
Then, Theorem  2.1 of Rao et al. [31] cannot be applied (for three maps) in this case.Example 2.14.
LetX={(0,1),(1,0),(1,1)}⊂ℝ2 with the Euclidean distance d2. (X,d2) is, obviously, a complete metric space. Moreover, we consider the order ⪯ in X given by ℛ={(x,x),x∈X}. Notice that the elements in X are only comparable to themselves, so (X,⪯) is regular. Also we consider R,S,T:X→X given by
(2.79)T(1,0)=S(1,0)=R(1,0)=(1,0),T(0,1)=S(0,1)=(1,0),R(0,1)=(0,1),T(1,1)=S(1,1)=R(1,1)=(1,1).
It is easy that, for all x∈X, TSx⪯Sx, and STx⪯Tx, so the pair {S,T} is weakly increasing.
Defineψ1,ψ2:[0,+∞)4→[0,+∞) by
(2.80)ψ1(t1,t2,t3,t4)=max⁡{t1,t2,t3,t4},ψ2(t1,t2,t3,t4)=12max⁡{t1,t2,t3,t4}.
Then, ψ1(t)=t for all t≥0.
As the elements inX are only comparable to themselves, condition (2.5) appearing in Theorem 2.8 is, obviously, satisfied. Now, all the hypotheses of Theorem 2.8 are satisfied. (1,1) and (1,0) are the coincidence points of the mappings R, S, and T.
On the other hand, the inequality (2.5) is not satisfied for x=(1,0) and y=(1,1). Indeed,
(2.81)ψ1(12[d2(R(1,0),T(1,1))+d2(R(1,1),S(1,0))]d2(R(1,0),R(1,1)),d2(R(1,0),S(1,0)),d2(R(1,1),T(1,1)),12[d2(R(1,0),T(1,1))+d2(R(1,1),S(1,0))])-ψ2(12[d2(R(1,0),T(1,1))+d2(R(1,1),S(1,0))]d2(R(1,0),R(1,1)),d2(R(1,0),S(1,0)),d2(R(1,1),T(1,1)),12[d2(R(1,0),T(1,1))+d2(R(1,1),S(1,0))])=ψ1(1,0,0,1)-ψ2(1,0,0,1)=12<Φ1(d2(S(1,0),T(1,1)))=Φ1(1)=1.
Then, again Theorem  2.1 of Rao et al. [31] cannot be applied (for three maps) in this case.A number of fixed point results may be obtained by assuming different forms for the functionsψ1 and ψ2. In particular, fixed point results under various contractive conditions may be derived from the above theorems. For example, if we consider
(2.82)ψ1(x,y,z,t)=k1xs+k2ys+k3zs+k4t4,ψ2(x,y,z,t)=(1-k)[k1xs+k2ys+k3zs+k4t4],
where s>0 and 0<k=k1+k2+k3+k4<1, we obtain the following results.The next result is an immediate consequence of Corollaries2.11 and 2.12.Corollary 2.15.
Let(X,d,⪯) be an ordered complete metric space. Let T,S:X→X be given mappings satisfying for every pair (x,y)∈X×X such that x and y are comparable,
(2.83)[d(Sx,Ty)]s≤k1[d(x,y)]s+k2[d(x,Sx)]s+k3[d(y,Ty)]s+k4[12(d(x,Ty)+d(y,Sx))]s,
where s>0 and 0<k=k1+k2+k3+k4<1. One assumes the following hypotheses:(i)
T and S are continuous or X is regular,(ii)
T and S are weakly increasing.
Then,T and S have a common fixed point, that is, there exists u∈X such that u=Tu=Su.Remark 2.16.
Other fixed point results may also be obtained under specific choices ofψ1 and ψ2.


## 2.1. Notations and Definitions

The following definition was introduced by Jungck in [37].Definition 2.1 (see [37]).
Let(X,d) be a metric space and f,g:X→X. If w=fx=gx, for some x∈X, then x is called a coincidence point of f and g, and w is called a point of coincidence of f and g. The pair {f,g} is said to be compatible if and only if limn→+∞d(fgxn,gfxn)=0, whenever {xn} is a sequence in X such that limn→+∞fxn=limn→+∞gxn=t for some t∈X.LetX be a nonempty set and R:X→X a given mapping. For every x∈X, we denote by R-1(x) the subset of X defined by
(2.1)R-1(x):={u∈X∣Ru=x}.Definition 2.2 (see [23]).
Let(X,⪯) be a partially ordered set and T,S,R:X→X are given mappings such that TX⊆RX and SX⊆RX. We say that S and T are weakly increasing with respect to R  if and only if, for all x∈X, we have
(2.2)Tx⪯Sy,∀y∈R-1(Tx),Sx⪯Ty,∀y∈R-1(Sx).Remark 2.3.
IfR:X→X is the identity mapping (Rx=x for all x∈X), then S and T are weakly increasing with respect to Rwhichimplies that S and T are weakly increasing mappings. Note that the notion of weakly increasing mappings was introduced in [9] (also see [17, 38]).Example 2.4.
LetX={1,2,3} be endowed with the partial order ⪯ given by
(2.3)⪯:={(1,1),(2,2),(3,3),(2,3),(3,1),(2,1)}.
Define the mappings T,S,R:X→X by
(2.4)T1=T3=1,T2=3,S1=S2=S3=1,R1=1,R2=R3=2.
We will show that the mappings S and T are weakly increasing with respect to R.
Letx,y∈X such that y∈R-1(Tx). By the definition of S, we have Sy=1. On the other hand, Tx∈{1,3} and (1,1),(3,1)∈⪯. Thus, we have Tx⪯Sy for all y∈R-1(Tx).
Letx,y∈X such that y∈R-1(Sx). By the definitions of S and R, we have R-1(Sx)=R-1(1)={1}. Then we have y=1. On the other hand, 1=Sx⪯Ty=T1=1. Then, Sx⪯Ty for all y∈R-1(Sx).
Thus, we proved thatS and T are weakly increasing with respect to R.Definition 2.5.
LetX be a nonempty set. Then (X,d,⪯) is called an ordered metric space if and only if:(i)
(X,d) is a metric space,(ii)
(X,⪯) is a partially ordered set.


## 2.2. Results

Our first result is the following.Theorem 2.6.
Let(X,d,⪯) be an ordered complete metric space. Let T,S,R:X→X be given mappings satisfying for every pair (x,y)∈X×X such that Rx and Ry are comparable,
(2.5)Φ1(d(Sx,Ty))≤ψ1(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)])-ψ2(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)]),
where ψ1 and ψ2 are generalized altering distance functions (in Ψ4) and Φ1(x)=ψ1(x,x,x,x). One assumes the following hypotheses:(i)
T, S, and R are continuous,(ii)
TX⊆RX, SX⊆RX,(iii)
T and S are weakly increasing with respect to R,(iv)
the pairs{T,R} and {S,R} are compatible.
Then,T, S, and R have a coincidence point, that is, there exists u∈X such that Ru=Tu=Su.Proof.
Letx0∈X be an arbitrary point. Since TX⊆RX, there exists x1∈X such that Rx1=Tx0. Since SX⊆RX, there exists x2∈X such that Rx2=Sx1. Continuing this process, we can construct a sequence {Rxn} in X defined by
(2.6)Rx2n+1=Tx2n,Rx2n+2=Sx2n+1,∀n∈N.
We claim that
(2.7)Rxn⪯Rxn+1,∀n∈N.
To this aim, we will use the increasing property of the mappings S and T with respect to R. From (2.6), we have
(2.8)Rx1=Tx0⪯Sy,∀y∈R-1(Tx0).
Since Rx1=Tx0, then x1∈R-1(Tx0), and we get
(2.9)Rx1=Tx0⪯Sx1=Rx2.
Again,
(2.10)Rx2=Sx1⪯Ty,∀y∈R-1(Sx1).
Since x2∈R-1(Sx1), we get
(2.11)Rx2=Sx1⪯Tx2=Rx3.
Hence, by induction, (2.7) holds.
Without loss of the generality, we can assume that(2.12)Rxn≠Rxn+1,∀n∈N.
Now, we will prove our result on three steps.
Step  1. We claim that
(2.13)lim⁡n→+∞d(Rxn+1,Rxn+2)=0.
Putting x=x2n+1 and y=x2n, from (2.7) and the considered contraction (2.5), we have
(2.14)Φ1(d(Rx2n+2,Rx2n+1))=Φ1(d(Sx2n+1,Tx2n))≤ψ1(12[d(Rx2n+1,Tx2n)+d(Rx2n,Sx2n+1)]d(Rx2n+1,Rx2n),d(Rx2n+1,Sx2n+1),d(Rx2n,Tx2n),12[d(Rx2n+1,Tx2n)+d(Rx2n,Sx2n+1)])-ψ2(12[d(Rx2n+1,Tx2n)+d(Rx2n,Sx2n+1)]d(Rx2n+1,Rx2n),d(Rx2n+1,Sx2n+1),d(Rx2n,Tx2n),12[d(Rx2n+1,Tx2n)+d(Rx2n,Sx2n+1)])=ψ1(12[d(Rx2n+1,Tx2n)+d(Rx2n,Sx2n+1)]d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2))-ψ2(12[d(Rx2n+1,Tx2n)+d(Rx2n,Sx2n+1)]d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2)).
Suppose, for some n∈ℕ, that
(2.15)d(Rx2n+1,Rx2n+2)>d(Rx2n,Rx2n+1).
Using (2.15) and a triangular inequality, we have
(2.16)12d(Rx2n,Rx2n+2)≤12(d(Rx2n,Rx2n+1)+d(Rx2n+1,Rx2n+2)<d(Rx2n+1,Rx2n+2).
Using this and (2.15) together with a property of the generalized altering function ψ1, we get
(2.17)ψ1(d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2))≤Φ1(d(Rx2n+2,Rx2n+1)).
Hence, we obtain
(2.18)Φ1(d(Rx2n+2,Rx2n+1))≤Φ1(d(Rx2n+2,Rx2n+1))-ψ2(d12d(Rx2n,Rx2n+2)(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2)).
This implies that
(2.19)ψ2(d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2))=0,
which yields that
(2.20)d(Rx2n+1,Rx2n)=0.
Hence, we obtain a contradiction with (2.12). We deduce that
(2.21)d(Rx2n+1,Rx2n+2)≤d(Rx2n,Rx2n+1),∀n∈N.
Similarly, putting x=x2n+1 and y=x2n+2, from (2.7) and the considered contraction (2.5), we have
(2.22)Φ1(d(Rx2n+2,Rx2n+3))≤ψ1(d(Rx2n+1,Rx2n+2),d(Rx2n+1,Rx2n+2),d(Rx2n+2,Rx2n+3),12d(Rx2n+1,Rx2n+3))-ψ2(d(Rx2n+1,Rx2n+2),d(Rx2n+1,Rx2n+2),d(Rx2n+2,Rx2n+3),12d(Rx2n+1,Rx2n+3)).
Suppose, for some n∈ℕ, that
(2.23)d(Rx2n+2,Rx2n+3)>d(Rx2n+1,Rx2n+2).
Then, by a triangular inequality, we have
(2.24)12d(Rx2n+1,Rx2n+3)≤12(d(Rx2n+1,Rx2n+2)+d(Rx2n+2,Rx2n+3))<d(Rx2n+2,Rx2n+3).
Hence, from this, (2.22), and (2.23), we obtain
(2.25)Φ1(d(Rx2n+2,Rx2n+3))≤Φ1(d(Rx2n+2,Rx2n+3))-ψ2(d(Rx2n+1,Rx2n+2),d(Rx2n+1,Rx2n+2),d(Rx2n+2,Rx2n+3),12d(Rx2n+1,Rx2n+3)).
This implies that
(2.26)ψ2(d(Rx2n+1,Rx2n+2),d(Rx2n+1,Rx2n+2),d(Rx2n+2,Rx2n+3),12d(Rx2n+1,Rx2n+3))=0,
which leads to
(2.27)d(Rx2n+1,Rx2n+2)=0.
Hence, we obtain a contradiction with (2.12). We deduce that
(2.28)d(Rx2n+1,Rx2n+2)≥d(Rx2n+2,Rx2n+3),∀n∈N.
Combining (2.21) and (2.28), we obtain
(2.29)d(Rxn+1,Rxn+2)≥d(Rxn+2,Rxn+3),∀n∈N.
Then, {d(Rxn+1,Rxn+2)} is a nonincreasing sequence of positive real numbers. This implies that there exists r≥0 such that
(2.30)lim⁡n→+∞d(Rxn+1,Rxn+2)=r.
By (2.14), we have
(2.31)Φ1(d(Rx2n+2,Rx2n+1))≤ψ1(12d(Rx2n,Rx2n+2)d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2))-ψ2(12d(Rx2n,Rx2n+2)d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),12d(Rx2n,Rx2n+2))≤ψ1(d(Rx2n+1,Rx2n),d(Rx2n,Rx2n+1),d(Rx2n,Rx2n+1),d(Rx2n,Rx2n+1))-ψ2(d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),0)=Φ1(d(Rx2n,Rx2n+1))-ψ2(d(Rx2n+1,Rx2n),d(Rx2n+1,Rx2n+2),d(Rx2n,Rx2n+1),0).
Letting n→+∞ in (2.31) and using the continuities of Φ1 and ψ2, we obtain
(2.32)Φ1(r)≤Φ1(r)-ψ2(r,r,r,0),
which implies that ψ2(r,r,r,0)=0, so r=0. Hence
(2.33)lim⁡n→+∞d(Rxn+1,Rxn+2)=0.
Hence, (2.13) is proved.
Step  2. We claim that {Rxn} is a Cauchy sequence.
From (2.13), it will be sufficient to prove that {Rx2n} is a Cauchy sequence. We proceed by negation and suppose that {Rx2n} is not a Cauchy sequence. Then, there exists ε>0 for which we can find two sequences of positive integers {m(i)} and {n(i)} such that, for all positive integer i,
(2.34)n(i)>m(i)>i,d(Rx2m(i),Rx2n(i))≥ε,d(Rx2m(i),Rx2n(i)-2)<ε.
From (2.34) and using a triangular inequality, we get
(2.35)ε≤d(Rx2m(i),Rx2n(i))≤d(Rx2m(i),Rx2n(i)-2)+d(Rx2n(i)-2,Rx2n(i)-1)+d(Rx2n(i)-1,Rx2n(i))<ε+d(Rx2n(i)-2,Rx2n(i)-1)+d(Rx2n(i)-1,Rx2n(i)).
Letting i→+∞ in the previous inequality and using (2.13), we obtain
(2.36)lim⁡i→+∞d(Rx2m(i),Rx2n(i))=ε.
Again, a triangular inequality gives us
(2.37)|d(Rx2n(i),Rx2m(i)-1)-d(Rx2n(i),Rx2m(i))|≤d(Rx2m(i)-1,Rx2m(i)).
Letting i→+∞ in the above inequality and using (2.13) and (2.36), we get
(2.38)lim⁡i→+∞d(Rx2n(i),Rx2m(i)-1)=ε.
On the other hand, we have
(2.39)d(Rx2n(i),Rx2m(i))≤d(Rx2n(i),Rx2n(i)+1)+d(Rx2n(i)+1,Rx2m(i))=d(Rx2n(i),Rx2n(i)+1)+d(Tx2n(i),Sx2m(i)-1).
Then, from (2.13), (2.36), and the continuity of Φ1, we get by letting i→+∞ in the above inequality
(2.40)Φ1(ε)≤lim⁡i→+∞Φ1(d(Sx2m(i)-1,Tx2n(i))).
Now, using the considered contractive condition (2.5) for x=x2m(i)-1 and y=x2n(i), we have
(2.41)Φ1(d(Sx2m(i)-1,Tx2n(i)))≤ψ1(12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))]d(Rx2m(i)-1,Rx2n(i)),d(Rx2m(i)-1,Rx2m(i)),d(Rx2n(i),Rx2n(i)+1),12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))])-ψ2(12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))]d(Rx2m(i)-1,Rx2n(i)),d(Rx2m(i)-1,Rx2m(i)),d(Rx2n(i),Rx2n(i)+1),12[d(Rx2m(i)-1,Rx2n(i)+1)+d(Rx2n(i),Rx2m(i))]).
Then, from (2.13), (2.38), and the continuities of ψ1 and ψ2, we get by letting i→+∞ in the above inequality
(2.42)lim⁡i→+∞Φ1(d(Sx2m(i)-1,Tx2n(i)))≤ψ1(ε,0,0,ϵ)-ψ2(ε,0,0,ϵ)≤Φ1(ε)-ψ2(ε,0,0,ϵ).
Now, combining (2.40) with the previous inequality, we get
(2.43)Φ1(ε)≤Φ1(ε)-ψ2(ε,0,0,ϵ),
which implies that ψ2(ε,0,0,ϵ)=0, that is a contradiction since ε>0. We deduce that {Rxn} is a Cauchy sequence.
Step  3. We claimexistence of a coincidence point.
Since{Rxn} is a Cauchy sequence in the complete metric space (X,d), there exists u∈X such that
(2.44)lim⁡n→+∞Rxn=u.
From (2.44) and the continuity of R, we get
(2.45)lim⁡n→+∞R(Rxn)=Ru.
By the triangular inequality, we have
(2.46)d(Ru,Tu)≤d(Ru,R(Rx2n+1))+d(R(Tx2n),T(Rx2n))+d(T(Rx2n),Tu).
On the other hand, we have
(2.47)Rx2n⟶u,Tx2n⟶uasn→+∞.
Since R and T are compatible mappings, this implies that
(2.48)lim⁡n→+∞d(R(Tx2n),T(Rx2n))=0.
Now, from the continuity of T and (2.44), we have
(2.49)lim⁡n→+∞d(T(Rx2n),Tu)=0.
Combining (2.45), (2.48), and (2.49) and letting n→+∞ in (2.46), we obtain
(2.50)d(Ru,Tu)≤0,
that is,
(2.51)Ru=Tu.
Again, by a triangular inequality, we have
(2.52)d(Ru,Su)≤d(Ru,R(Rx2n+2))+d(R(Sx2n+1),S(Rx2n+1))+d(S(Rx2n+1),Su).
On the other hand, we have
(2.53)Rx2n+1⟶u,Sx2n+1⟶uasn⟶+∞.
Since R and S are compatible mappings, this implies that
(2.54)lim⁡n→+∞d(R(Sx2n+1),S(Rx2n+1))=0.
Now, from the continuity of S and (2.44), we have
(2.55)lim⁡n→+∞d(S(Rx2n+1),Su)=0.
Combining (2.45), (2.54), and (2.55) and letting n→+∞ in (2.52), we obtain
(2.56)d(Ru,Su)≤0,
that is,
(2.57)Ru=Su.
Finally, from (2.51) and (2.57), we have
(2.58)Tu=Ru=Su,
that is, u is a coincidence point of T, S, and R. This makes end to the proof.In the next theorem, we omit the continuity hypotheses satisfied byT, S, and R.Definition 2.7.
Let(X,d,⪯) be a partially ordered metric space. We say that X is regular if and only if the following hypothesis holds: if {zn} is a nondecreasing sequence in X with respect to ⪯such that zn→z∈X as n→+∞, then zn⪯z for all n∈ℕ.Now, our second result is the following.Theorem 2.8.
Let(X,d,⪯) be an ordered complete metric space. Let T,S,R:X→X be given mappings satisfying for every pair (x,y)∈X×X such that Rx and Ry are comparable,
(2.59)Φ1(d(Sx,Ty))≤ψ1(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)])-ψ2(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),12[d(Rx,Ty)+d(Ry,Sx)]),
where ψ1 and ψ2 are generalized altering distance functions and Φ1(x)=ψ1(x,x,x,x). We assume the following hypotheses:(i)
X is regular,(ii)
T and S are weakly increasing with respect to R,(iii)
RX is a complete subspace of (X,d),(iv)
TX⊆RX, SX⊆RX.
Then,T, S, and R have a coincidence point.Proof.
Following the proof of Theorem2.6, we have {Rxn} is a Cauchy sequence in (RX,d). Since RX is a complete, there exists u=Rv, v∈X such that
(2.60)lim⁡n→+∞Rxn=u=Rv.
Since {Rxn} is a nondecreasing sequence and X is regular, it follows from (2.60) that Rxn⪯Rv for all n∈ℕ. Hence, we can apply the contractive condition (2.5). Then, for x=v and y=x2n, we obtain
(2.61)Φ1(d(Sv,Rx2n+1))=Φ1(d(Sv,Tx2n))≤ψ1(12[d(Rv,Tx2n)+d(Rx2n,Sv)]d(Rv,Rx2n),d(Rv,Sv),d(Rx2n,Rx2n+1),12[d(Rv,Tx2n)+d(Rx2n,Sv)])-ψ2(12[d(Rv,Tx2n)+d(Rx2n,Sv)]d(Rv,Rx2n),d(Rv,Sv),d(Rx2n,Rx2n+1),12[d(Rv,Tx2n)+d(Rx2n,Sv)]).
Letting n→+∞ in the above inequality and using (2.13), (2.60), and the properties of ψ1 and ψ2, we obtain
(2.62)Φ1(d(Sv,Rv))≤ψ1(0,d(Rv,Sv),0,12d(Rv,Sv))-ψ2(0,d(Rv,Sv),0,12d(Rv,Sv))≤Φ1(d(Sv,Rv))-ψ2(0,d(Rv,Sv),0,12d(Rv,Sv)).
This implies that ψ2(0,d(Rv,Sv),0,(1/2)d(Rv,Sv))=0, which gives us that d(Rv,Sv)=0, that is,
(2.63)Rv=Sv.
Similarly, for x=x2n+1 and y=v, we obtain
(2.64)Φ1(d(Rx2n+2,Tv))=Φ1(d(Sx2n+1,Tv))≤ψ1(12[d(Rx2n+1,Tv)+d(Rv,Sx2n+1)]d(Rx2n+2,Rv),d(Rx2n+1,Rx2n+2),d(Rv,Tv),12[d(Rx2n+1,Tv)+d(Rv,Sx2n+1)])-ψ2(12[d(Rx2n+1,Tv)+d(Rv,Sx2n+1)]d(Rx2n+2,Rv),d(Rx2n+1,Rx2n+2),d(Rv,Tv),12[d(Rx2n+1,Tv)+d(Rv,Sx2n+1)]).
Letting n→+∞ in the above inequality, we get
(2.65)Φ1(d(Rv,Tv))≤ψ1(0,0,d(Rv,Tv),12d(Rv,Tv))-ψ2(0,0,d(Rv,Tv),12d(Rv,Tv))≤Φ1(d(Rv,Tv))-ψ2(0,0,d(Rv,Tv),12d(Rv,Tv)).
This implies that ψ2(0,0,d(Rv,Tv),(1/2)d(Rv,Tv))=0, and then,
(2.66)Rv=Tv.
Now, combining (2.63) and (2.66), we obtain
(2.67)Rv=Tv=Sv.
Hence, v is a coincidence point of T, S, and R. This makes end to the proof.Now, we give a sufficient condition that assures the uniqueness of the common coincidence point of{T,R} and {S,R}.Theorem 2.9.
Under the hypotheses of Theorem2.6 (resp., Theorem 2.8) and suppose that (SX,⪯) is a totally ordered set and S is one-to-one mapping, then one obtains a unique common coincidence point of {T,R} and {S,R}.Proof.
Following the proof of Theorem2.6 (resp., Theorem 2.8), the set of common coincidence points of {T,R} and {S,R} is nonempty. Let μ,ν∈Xbe two common coincidence points of {T,R} and {S,R}, that is,
(2.68)Tμ=Rμ=Sμ,Tν=Rν=Sν.
This implies that Rμ,Rν∈SX, and then, Rμ and Rν are comparable with respect to ⪯. Then, we can apply the contractive condition (2.5). We have
(2.69)Φ1(d(Sμ,Tν))≤ψ1(d(Rμ,Rν),d(Rμ,Sμ),d(Rν,Tν),12[d(Rμ,Tν)+d(Rν,Sμ)])-ψ2(d(Rμ,Rν),d(Rμ,Sμ),d(Rν,Tν),12[d(Rμ,Tν)+d(Rν,Sμ)])=ψ1(d(Sμ,Tν),0,0,d(Sμ,Tν))-ψ2(d(Sμ,Tν),0,0,d(Sμ,Tν))≤Φ1(d(Sμ,Tν))-ψ2(d(Sμ,Tν),0,0,d(Sμ,Tν)).
This implies that ψ2(d(Sμ,Tν),0,0,d(Sμ,Tν)), which gives us that d(Sμ,Tν)=0, that is, Sμ=Tν. From (2.68), we get Sμ=Sν. Since S is one-to-one, we have μ=ν. This makes end to the proof.Now, it is easy to state a corollary of Theorem2.6 or Theorem 2.8 involving contractions of integral type.Corollary 2.10.
LetR,S and T satisfy the conditions of Theorem 2.6 or Theorem 2.8, except that condition (2.5) is replaced by the following: there exists a positive Lebesgue integrable function u on ℝ+ such that ∫0ε‍u(t)dt>0 for each ε>0 and that
(2.70)∫0Φ1(d(Sx,Ty))u(t)dt≤∫0ψ1(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),(1/2)[d(Rx,Ty)+d(Ry,Sx)])u(t)dt-∫0ψ2(d(Rx,Ry),d(Rx,Sx),d(Ry,Ty),(1/2)[d(Rx,Ty)+d(Ry,Sx)])‍u(t)dt.
Then, T, S, and R have a coincidence point.IfR:X→X is the identity mapping, we can deduce easily the following common fixed point results.The next result is an immediate consequence of Theorem2.6.Corollary 2.11.
Let(X,d,⪯) be an ordered complete metric space. Let T,S:X→X be given mappings satisfying for every pair (x,y)∈X×X such that x and y are comparable,
(2.71)Φ1(d(Sx,Ty))≤ψ1(d(x,y),d(x,Sx),d(y,Ty),12[d(x,Ty)+d(y,Sx)])-ψ2(d(x,y),d(x,Sx),d(y,Ty),12[d(x,Ty)+d(y,Sx)]),
where ψ1 and ψ2 are generalized altering distance functions and Φ1(x)=ψ1(x,x,x,x). One assumes the following hypotheses:(i)
T and S are continuous,(ii)
T and S are weakly increasing.
Then,T and S have a common fixed point, that is, there exists u∈X such that u=Tu=Su.The following result is an immediate consequence of Theorem2.8.Corollary 2.12.
Let(X,d,⪯) be an ordered complete metric space. Let T,S:X→X be given mappings satisfying for every pair (x,y)∈X×X such that x and y are comparable,
(2.72)Φ1(d(Sx,Ty))≤ψ1(d(x,y),d(x,Sx),d(y,Ty),12[d(x,Ty)+d(y,Sx)])-ψ2(d(x,y),d(x,Sx),d(y,Ty),12[d(x,Ty)+d(y,Sx)]),
where ψ1 and ψ2 are generalized altering distance functions and Φ1(x)=ψ1(x,x,x,x). One assume the following hypotheses:(i)
X is regular,(ii)
T and S are weakly increasing.
Then,T and S have a common fixed point.Now, we give some examples to support our results.Example 2.13.
LetX={4,5,6} be endowed with the usual metric d(x,y)=|x-y| for all x,y∈X, and ⪯:={(4,4),(5,5),(6,6),(6,4)}. Consider the mappings
(2.73)S=T=(456464),R=(456456).
We define the functions ψ1,ψ2:[0,+∞)4→[0,+∞) by
(2.74)ψ1(t1,t2,t3,t4)=14(t1+t2+t3+t4),ψ2(t1,t2,t3,t4)=116(t1+t2+t3+t4).
Clearly, Φ1(t)=t for all t≥0. Now, we will check that all the hypotheses required by Theorem 2.8 are satisfied.
(i)X is regular.
Let{zn} be a nondecreasing sequence in X with respect to ⪯ such that zn→z∈X as n→+∞. We have zn⪯zn+1 for all n∈ℕ.(a)
Ifz0=4, then z0=4⪯z1. From the definition of ⪯, we have z1=4. By induction, we get zn=4 for all n∈ℕ and z=4. Then, zn⪯z for all n∈ℕ.(b)
Ifz0=5, then z0=5⪯z1. From the definition of ⪯, we have z1=5. By induction, we get zn=5 for all n∈ℕ and z=5. Then, zn⪯z for all n∈ℕ.(c)
Ifz0=6, then z0=6⪯z1. From the definition of ⪯, we have z1∈{6,4}. By induction, we get zn∈{6,4} for all n∈ℕ. Suppose that there exists p≥1 such that zp=4. From the definition of ⪯, we get zn=zp=4 for all n≥p. Thus, we have z=4 and zn⪯z for all n∈ℕ. Now, suppose that zn=6 for all n∈ℕ. In this case, we get z=6 and zn⪯z for all n∈ℕ.
Thus, we proved that, in all cases, we havezn⪯z for all n∈ℕ. Then, X is regular.
(ii)T and S are weakly increasing.
SinceS=T, we have to check that Tx⪯T(Tx) for all x∈X.
Forx=4, we have
(2.75)T4=4⪯T(T4)=T4=4.
For x=5, we have
(2.76)T5=6⪯T(T5)=T6=4.
For x=6, we have
(2.77)T6=4⪯T(T6)=T4=4.
Thus, we proved that T and S are weakly increasing.
On the other hand, it is very easy to show that (2.5) is satisfied for all x,y∈X such that Rx⪯Ry.
Now, all the hypotheses of Theorem2.8 are satisfied. Then R, S, and T have a coincidence point u=4.
Note that inequality (2.5) is not satisfied for x=4 and y=5. Indeed,
(2.78)ψ1(d(R4,R5),d(R4,S4),d(R5,T5),12[d(R4,T5)+d(R5,S4)])-ψ2(d(R4,R5),d(R4,S4),d(R5,T5),12[d(R4,T5)+d(R5,S4)])=ψ1(1,0,1,32)-ψ2(1,0,1,32)=2132<Φ1(d(S4,T5))=Φ1(2)=2.
Then, Theorem  2.1 of Rao et al. [31] cannot be applied (for three maps) in this case.Example 2.14.
LetX={(0,1),(1,0),(1,1)}⊂ℝ2 with the Euclidean distance d2. (X,d2) is, obviously, a complete metric space. Moreover, we consider the order ⪯ in X given by ℛ={(x,x),x∈X}. Notice that the elements in X are only comparable to themselves, so (X,⪯) is regular. Also we consider R,S,T:X→X given by
(2.79)T(1,0)=S(1,0)=R(1,0)=(1,0),T(0,1)=S(0,1)=(1,0),R(0,1)=(0,1),T(1,1)=S(1,1)=R(1,1)=(1,1).
It is easy that, for all x∈X, TSx⪯Sx, and STx⪯Tx, so the pair {S,T} is weakly increasing.
Defineψ1,ψ2:[0,+∞)4→[0,+∞) by
(2.80)ψ1(t1,t2,t3,t4)=max⁡{t1,t2,t3,t4},ψ2(t1,t2,t3,t4)=12max⁡{t1,t2,t3,t4}.
Then, ψ1(t)=t for all t≥0.
As the elements inX are only comparable to themselves, condition (2.5) appearing in Theorem 2.8 is, obviously, satisfied. Now, all the hypotheses of Theorem 2.8 are satisfied. (1,1) and (1,0) are the coincidence points of the mappings R, S, and T.
On the other hand, the inequality (2.5) is not satisfied for x=(1,0) and y=(1,1). Indeed,
(2.81)ψ1(12[d2(R(1,0),T(1,1))+d2(R(1,1),S(1,0))]d2(R(1,0),R(1,1)),d2(R(1,0),S(1,0)),d2(R(1,1),T(1,1)),12[d2(R(1,0),T(1,1))+d2(R(1,1),S(1,0))])-ψ2(12[d2(R(1,0),T(1,1))+d2(R(1,1),S(1,0))]d2(R(1,0),R(1,1)),d2(R(1,0),S(1,0)),d2(R(1,1),T(1,1)),12[d2(R(1,0),T(1,1))+d2(R(1,1),S(1,0))])=ψ1(1,0,0,1)-ψ2(1,0,0,1)=12<Φ1(d2(S(1,0),T(1,1)))=Φ1(1)=1.
Then, again Theorem  2.1 of Rao et al. [31] cannot be applied (for three maps) in this case.A number of fixed point results may be obtained by assuming different forms for the functionsψ1 and ψ2. In particular, fixed point results under various contractive conditions may be derived from the above theorems. For example, if we consider
(2.82)ψ1(x,y,z,t)=k1xs+k2ys+k3zs+k4t4,ψ2(x,y,z,t)=(1-k)[k1xs+k2ys+k3zs+k4t4],
where s>0 and 0<k=k1+k2+k3+k4<1, we obtain the following results.The next result is an immediate consequence of Corollaries2.11 and 2.12.Corollary 2.15.
Let(X,d,⪯) be an ordered complete metric space. Let T,S:X→X be given mappings satisfying for every pair (x,y)∈X×X such that x and y are comparable,
(2.83)[d(Sx,Ty)]s≤k1[d(x,y)]s+k2[d(x,Sx)]s+k3[d(y,Ty)]s+k4[12(d(x,Ty)+d(y,Sx))]s,
where s>0 and 0<k=k1+k2+k3+k4<1. One assumes the following hypotheses:(i)
T and S are continuous or X is regular,(ii)
T and S are weakly increasing.
Then,T and S have a common fixed point, that is, there exists u∈X such that u=Tu=Su.Remark 2.16.
Other fixed point results may also be obtained under specific choices ofψ1 and ψ2.


## 3. Application

Consider the integral equations:(3.1)u(t)=∫0T‍K1(t,s,u(s))ds+g(t),t∈[0,T],u(t)=∫0T‍K2(t,s,u(s))ds+g(t),t∈[0,T],
where T>0.The purpose of this section is to give an existence theorem for common solution of (3.1) using Corollary 2.15. This application is inspired by [9].Previously, we consider the spaceX=C(I)(I=[0,T]) of continuous functions defined on I. Obviously, this space with the metric given by
(3.2)d(x,y)=sup⁡t∈I|x(t)-y(t)|,∀x,y∈C(I),
is a complete metric space. C(I) can also be equipped with the partial order ⪯ given by
(3.3)x,y∈C(I),x⪯y⟺x(t)≤y(t),∀t∈I.
Moreover, in [28], it is proved that (C(I),⪯) is regular.Now, we will prove the following result.Theorem 3.1.
Suppose that the following hypotheses hold:
(i)K1,K2:I×I×ℝ→ℝ and g:ℝ→ℝ are continuous,
(ii) for allt,s∈I,
(3.4)K1(t,s,u(t))≤K2(t,s,∫0TK1(s,τ,u(τ))dτ+g(s)),K2(t,s,u(t))≤K1(t,s,∫0T‍K2(s,τ,u(τ))dτ+g(s)),
(iii) there existk1,k2,k3≥0 such that
(3.5)|Gx(t)-Fy(t)|≤k1(y(t)-x(t))+k2|x(t)-Gx(t)-g(t)|+k3|y(t)-Fy(t)-g(t)|,
where
(3.6)Fx(t)=∫0TK1(t,s,x(s))ds,t∈I,Gx(t)=∫0T‍K2(t,s,x(s))ds,t∈I,
andk1+k2+k3<1, for every x,y∈X and x⪯y and t∈I.
Then, the integral equations (3.1) have a solution u*∈C(I).Proof.
DefineT,S:C(I)→C(I) by
(3.7)Tx(t)=Fx(t)+g(t),t∈I,Sx(t)=Gx(t)+g(t),t∈I.
Now, we will prove that T and S are weakly increasing. From (ii), for all t∈I, we have
(3.8)Tx(t)=∫0TK1(t,s,x(s))ds+g(t)≤∫0T‍K2(t,s,∫0T‍K1(s,τ,x(τ))dτ+g(s))ds+g(t)=∫0T‍K2(t,s,Tx(s))ds+g(t)=STx(t).
Similarly,
(3.9)Sx(t)=∫0TK2(t,s,x(s))ds+g(t)≤∫0TK1(t,s,∫0T‍K2(s,τ,x(τ))dτ+g(s))ds+g(t)=∫0TK1(t,s,Sx(s))ds+g(t)=TSx(t).
Then, we have Tx⪯STx and Sx⪯TSx for all x∈C(I). This implies that T and S are weakly increasing.
Now, for allx,y∈C(I) such that x⪯y, by (iii), we have
(3.10)|Sx(t)-Ty(t)|=|Gx(t)-Fy(t)|≤k1(y(t)-x(t)|+k2|x(t)-Gx(t)-g(t)|+k3|y(t)-Fy(t)-g(t)|.
Hence
(3.11)d(Sx,Ty)=sup⁡t∈[0,T]|Sx(t)-Ty(t)|≤k1sup⁡t∈[0,T]|x(t)-y(t)|+k2sup⁡t∈[0,T]|x(t)-Gx(t)-g(t)|+k3sup⁡t∈[0,T]|y(t)-Fy(t)-g(t)|=k1sup⁡t∈[0,T]|x(t)-y(t)|+k2sup⁡t∈[0,T]|x(t)-Sx(t)|+k3sup⁡t∈[0,T]|y(t)-Ty(t)|.
Then
(3.12)d(Sx,Ty)≤k1d(x,y)+k2d(x,Sx)+k3d(y,Ty)
for all x,y∈X such that y⪯x.
This implies that, for allx,y∈C(I) such that x⪯y,
(3.13)d(Sx,Ty)≤k1d(x,y)+k2d(x,Sx)+k3d(y,Ty).
Hence the contractive condition required by Corollary 2.15 is satisfied with s=1, k4=0, and k1+k2+k3<1.
Now, all the required hypotheses of Corollary2.15 are satisfied. Then, there exists u*∈C(I), a common fixed point of T and S, that is, u* is a solution to (3.1).



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*Source: 736367-2012-07-29.xml*