# Common Fixed Points of Generalized Rational Type Cocyclic Mappings in Multiplicative Metric Spaces

Authors: Mujahid Abbas; Manuel De la Sen; Talat Nazir
Journal: Discrete Dynamics in Nature and Society
 (2015)

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## Abstract

The aim of this paper is to present fixed point result of mappings satisfying a generalized rational contractive condition in the setup of multiplicative metric spaces. As an application, we obtain a common fixed point of a pair of weakly compatible mappings. Some common fixed point results of pair of rational contractive types mappings involved in cocyclic representation of a nonempty subset of a multiplicative metric space are also obtained. Some examples are presented to support the results proved herein. Our results generalize and extend various results in the existing literature.


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## Body

## 1. Introduction and Preliminaries

Fixed point theory provides the most important and traditional tools for proving the existence of solutions of many problems in both pure and applied mathematics. In metric fixed point theory, the interplay between contractive condition and the existence and uniqueness of a fixed point has been very strong and fruitful. The study of fixed points of mappings which satisfy certain contractive conditions has primary applications in the solution of differential and integral equations (see, e.g., [1–5]). Theorems dealing with fixed point of certain mappings inspired and motivated the development of many other important kinds of points like coincidence points, intersection points, sectional points, and so forth.One of the basic and most widely applied fixed point theorems in mathematical analysis is “Banach Contraction Mapping Principle” (or Banach’s Fixed Point Theorem). Kirk et al. [4] obtained some fixed point results for mappings satisfying cyclical contractive conditions. The Banach contraction mappings are continuous mappings, while cyclic contraction mappings need not be continuous. Păcurar and Rus [6] studied some fixed point results for cyclic weak contractions. Piątek [7] obtained some results on cyclic Meir-Keeler contractions in metric spaces. Using fixed point result of weakly contractive map, Karapinar [3] established some interesting fixed point results for cyclic ϕ-weak contraction mappings. Recently, Derafshpour et al. [8] obtained results on the existence of best proximity points of cyclic contractions. For more results in this direction, we refer to [9–15].Banach contraction principle has been generalized either by extending the domain of the mapping or by considering a more general contractive condition on the mappings. Ozavsar and Cevikel [16] proved an analogous result to Banach contraction principle in the framework of multiplicative metric spaces. They also studied some topological properties of the relevant multiplicative metric space. Bashirov et al. [17] studied the concept of multiplicative calculus and proved a fundamental theorem of multiplicative calculus. They also illustrated the usefulness of multiplicative calculus with some interesting applications. Multiplicative calculus provides natural and straightforward way to compute the derivative of product and quotient of two functions [18]. It was shown that the multiplicative differential equations are more suitable than the ordinary differential equations in investigating some problems in economics and finance. Due to its operational simplicity and support to Newtonian calculus, it has attracted the attention of several researchers in the recent years. Furthermore, based on the definition of multiplicative absolute value function, they defined the multiplicative distance between two nonnegative real numbers and between two positive square matrices. This provided the basis for multiplicative metric spaces. Florack and van Assen [19] gave applications of multiplicative calculus in biomedical image analysis. He et al. [20] studied common fixed points for weak commutative mappings on a multiplicative metric space (see also [21]). Recently, Yamaod and Sintunavarat [22] obtained some fixed point results for generalized contraction mappings with cyclic (α,β)-admissible mapping in multiplicative metric spaces.In this paper, we obtain fixed point result of a generalized rational contractive mapping in the framework of multiplicative metric spaces. Employing this result, a common fixed point of a pair of weakly compatible mappings is obtained. We study the sufficient conditions for the existence of common fixed points of pair of rational contractive types mappings involved in cocyclic representation of a nonempty subset of a multiplicative metric space. Our results generalize and extend comparable results in [3].ByR, R+, Rn+, and N we denote the set of all real numbers, the set of all nonnegative real numbers, the set of all n-tuples of positive real numbers, and the set of all natural numbers, respectively.Consistent with [16, 17], the following definitions and results will be needed in the sequel.Definition 1 (see [17]).
LetX be a nonempty set. A mapping d:X×X→R+ is said to be a multiplicative metric on X if for any x,y,z∈X, the following conditions hold:(i)
d(x,y)≥1 and d(x,y)=1 if and only if x=y.(ii)
d(x,y)=d(y,x).(iii)
d(x,y)≤d(x,z)·d(z,y).
The pair(X,d) is called a multiplicative metric space.We define absolute valued function that includes the negative real numbers in its domain.Definition 2.
A multiplicative absolute value function·:R→R+ is defined as(1)x=xifx≥1,1xifx∈0,1,1ifx=0,-1xifx∈-1,0,-xifx≤-1.Using the definition of multiplicative absolute value function, we can prove the following proposition.Proposition 3.
For arbitraryx,y∈R+, the multiplicative absolute value function ·:R+→R+ satisfies the following:(1)
x≥1.(2)
x≤x.(3)
1/x≤x if x>0 and x≤1/x if x≤0.(4)
x·y≤xy.Example 4 (see [16]).
LetX=Rn+. Then(2)d1x,y=x1y1·x2y2·⋯·xnyn,d2x,y=max⁡x1y1,x2y2,…,xnyndefine multiplicative metrices on X, where x=(x1,x2,…,xn) and y=(y1,y2,…,yn)∈Rn+.Definition 5 (see [16]).
Let(X,d) be a multiplicative metric space, x0 an arbitrary point in X, and ε>1. A multiplicative open ball B(x0,ε) of radius ε centered at x0 is the set {z∈X:d(z,x0)<ε}.
A sequence{xn} in a multiplicative metric space (X,d) is said to be multiplicative convergent to some point x∈X if, for any given ε>1, there exists n0∈N such that xn∈B(x,ε) for all n≥n0. If {xn} converges to x, we write xn→x as n→∞.Definition 6 (see [16]).
A sequence{xn} in a multiplicative metric space (X,d) is multiplicative convergent to x in X if and only if d(xn,x)→1 as n→∞.Definition 7.
Let(X,dX) and (Y,dY) be two multiplicative metric spaces and x0 an arbitrary but fixed element of X. A mapping f:X→Y is said to be multiplicative continuous at x0 if and only if xn→x0 in (X,dX) implies that f(xn)→f(x0) in (Y,dY). That is, given arbitrary ε>1, there exists δ>1 which depends on x0 and ε such that dY(fx,fx0)<ε for all those x in X for which dX(x,x0)<δ.Definition 8 (see [16]).
A sequence{xn} in a multiplicative metric space (X,d) is said to be multiplicative Cauchy sequence if, for any ε>1, there exists n0∈N such that d(xn,xm)<ε for all m,n≥n0.
A multiplicative metric space(X,d) is said to be complete if every multiplicative Cauchy sequence {xn} in X is multiplicative convergent in X.Theorem 9 (see [16]).
A sequence{xn} in a multiplicative metric space (X,d) is multiplicative Cauchy if and only if d(xn,xm)→1 as n,m→∞.Example 10.
LetX=C∗[a,b] be the collection of all real-valued multiplicative continuous functions over [a,b]⊆R+ with the multiplicative metric d defined by(3)df,g=supx∈a,b⁡fxgxfor arbitraryf,g∈X,where ·:R+→R+ is a multiplicative absolute value function. Then (C∗[a,b],d) is complete.Recall that iff and g are two self-maps on a set X and u=fx=gx for some x in X, then x is called acoincidence point of f and g, and u is called apoint of coincidence of f and g.Definition 11 (see [23]).
Two self-mapsf and g on a nonempty set X are called weakly compatible if they commute at their coincidence point.We will also need the following proposition from [23].Proposition 12.
Letf and g be weakly compatible self-maps on a set X. If f and g have a unique point of coincidence w=fx=gx, then w is the unique common fixed point of f and g.Definition 13 (see [12]).
Let{Xi:i=1,2,…,p} be a finite collection of nonempty subsets of a set X, where p is some positive integer and f,g:X→X. The set X is said to have a cocyclic representation with respect to the collection {Xi:i=1,2,…,p} and a pair (f,g) if(1)
X=∪i=1pXi;(2)
f(X1)⊆g(X2),…,f(Xp-1)⊆g(Xp), f(Xp)⊆g(X1).Definition 14.
The control functionsψ and φ are defined as follows:(i)
ψ:[1,∞)→[1,∞) is a continuous nondecreasing function with ψ(t)=1 if and only if t=1.(ii)
φ:[1,∞)→[1,∞) is a lower semicontinuous function with φ(t)=1 if and only if t=1.


## 2. Main Results

In this section, we obtain several fixed and common fixed point results of self-maps satisfying certain generalized contractive conditions in the framework of multiplicative metric space.We start with the following result.Theorem 15.
LetX,d be a complete multiplicative metric space and f:X→X. Suppose that there exist control functions ψ and φ such that(4)ψdfx,fy≤ψMfx,yφMfx,y,for any x,y∈X, where(5)Mfx,y=max⁡dx,y,dfx,x,dfy,y,dfx,y·dfy,x1/2.Then f has a unique fixed point.Proof.
Letx0 be a given point in X. Define a sequence {xn} in X as xn=fnx0 or equivalently as xn+1=fxn for n∈N. If xk=xk+1 for some k, then we have fxk=xk+1=xk and the result follows. Assume that xn+1≠xn, for all n∈N; that is, Mf(xn+1,xn)>1 for all n∈N. From (4), we have(6)ψdxn+2,xn+1=ψdfxn+1,fxn≤ψMfxn+1,xnφMfxn+1,xn,where(7)Mfxn+1,xn=max⁡dxn+1,xn,dfxn+1,xn+1,dfxn,xn,dfxn+1,xn·dfxn,xn+11/2=max⁡dxn+1,xn,dxn+2,xn+1,dxn+1,xn,dxn+2,xn·dxn+1,xn+11/2=max⁡dxn+1,xn,dxn+2,xn+1,dxn+2,xn1/2≤max⁡dxn+1,xn,dxn+2,xn+1,dxn+2,xn+1·dxn+1,xn1/2≤max⁡dxn+1,xn,dxn+2,xn+1,dxn+2,xn+1+dxn+1,xn2=max⁡dxn+1,xn,dxn+2,xn+1.If Mf(xk+1,xk)≤d(xk+2,xk+1) for some k∈N, then ψ(d(xk+2,xk+1))≤ψ(d(xk+2,xk+1))/φ(Mf(xk+1,xk)) implies φ(Mf(xk+1,xk))≤1, a contradiction. Hence Mf(xn+1,xn)≤d(xn+1,xn) for all n∈N. Also, d(xn+1,xn)≤Mf(xn+1,xn) for all n∈N and hence Mf(xn+1,xn)=d(xn+1,xn) for all n∈N. Thus, we have(8)ψdxn+2,xn+1≤ψdxn+1,xnφdxn+1,xn<ψdxn+1,xnwhich implies that(9)dxn+2,xn+1≤dxn+1,xnfor all n∈N. Thus {d(xn+1,xn)} is (strictly) decreasing of positive real numbers. Consequently, there exists c≥1 such that {d(xn+1,xn)} converges to c. Suppose that c>1. Now(10)ψdxn+2,xn+1≤ψMfxn+1,xnφMfxn+1,xn,and lower semicontinuity of φ gives that(11)limsupn→∞⁡ψdxn+2,xn+1≤limsupn→∞⁡ψMfxn+1,xnliminfn→∞⁡φMfxn+1,xn,which implies that ψ(c)≤ψ(c)/φc<ψ(c), a contradiction as φ(c)>1. Therefore, c=1; that is, limn→∞⁡d(xn+1,xn)=1.
Now, we claim thatlimn,m→∞⁡d(xn,xm)=1. If not, then there exist ε>1 and sequences {nk}, {mk} in N, with nk>mk≥k, such that d(xnk,xmk)≥ε for all k∈N. Without any loss of generality, we can assume that d(xnk,xmk-1)<ε. Since {d(xmk-1,xmk)} is a subsequence of convergent sequence {d(xn-1,xn)}→1 as n→∞, then {d(xmk-1,xmk)}→1 as k→∞. Now(12)ε≤dxmk,xnk≤dxnk,xmk-1·dxmk-1,xmkimplies that(13)limk→∞⁡dxmk,xnk=ε.From (13) and inequality d(xmk,xnk)≤d(xmk,xmk-1)·d(xmk-1,xnk), it follows that ε≤limk→∞⁡d(xmk-1,xnk). Also, the inequality d(xmk-1,xnk)≤d(xmk-1,xmk)·d(xmk,xnk) and (13) give that limk→∞⁡d(xmk-1,xnk)≤ε, and hence we have(14)limk→∞⁡dxmk-1,xnk=ε.Equation (14) and inequality d(xmk-1,xnk)≤d(xmk-1,xnk+1)·d(xnk+1,xnk) imply that ε≤limk→∞d(xmk-1,xnk+1), while inequality d(xmk-1,xnk+1)≤d(xmk-1,xnk)·d(xnk,xnk+1) and (14) imply that limk→∞d(xmk-1,xnk+1)≤ε, and hence we have(15)limk→∞⁡dxmk-1,xnk+1=ε.From (13) and d(xmk,xnk)≤d(xmk,xnk+1)·d(xnk+1,xnk), we have ε≤limk→∞d(xmk,xnk+1), and the inequality d(xmk,xnk+1)≤d(xmk,xnk)·d(xnk,xnk+1) and (13) give that limk→∞d(xmk,xnk+1)≤ε. So(16)limk→∞⁡dxmk,xnk+1=ε.As(17)Mfxnk,xmk-1=max⁡dxnk,xmk-1,dfxnk,xnk,dfxmk-1,xmk-1,dxnk,fxmk-1·dxmk-1,fxnk1/2=max⁡dxnk,xmk-1,dxnk+1,xnk,dxmk,xmk-1,dxnk,xmk·xmk-1,xnk+11/2,we have limk→∞Mf(xnk,xmk-1)=max⁡{ε,1,1,ε}=ε. From (4), it follows that(18)ψdxnk+1,xmk=ψdfxnk,fxmk-1≤ψMfxnk,xmk-1φMfxnk,xmk-1.Taking upper limit as k→∞ implies that ψ(ε)≤ψ(ε)/φ(ε)<ψ(ε), a contradiction as ε>1. Thus, we obtain that limn,m→∞d(xn,xm)=1, and hence {xn} is a multiplicative Cauchy sequence in (X,d). Next, we assume there exists a point u∈X such that limn→∞d(u,xn)=1; equivalently,(19)limn,m→∞⁡dxn,xm=limn→∞⁡dxn,u=1.Note that(20)dfu,u≤Mfu,xn=max⁡du,xn,dfu,u,dfxn,xn,du,fxn·dxn,fu1/2=max⁡du,xn,dfu,u,dxn+1,xn,du,xn+1·dxn,fu1/2≤max⁡du,xn,dfu,u,dxn+1,xn,du,xn+1·dxn,u·du,fu1/2.Taking limit as n→∞, we conclude that limn→∞Mf(u,xn)=d(fu,u). Hence(21)ψdfu,fxn+1≤ψMfu,xnφMfu,xnon taking upper limit as n→∞ implies that(22)ψdfu,u≤ψdfu,uφdfu,uwhich further implies that fu=u.
To prove the uniqueness of fixed point off, assume on the contrary that fv=v and fw=w with v≠w. Note that(23)ψdv,w=ψdfv,fw≤ψMfv,wφMfv,w,where(24)Mfv,w=max⁡dv,w,dfv,v,dfw,w,dfv,w·dfw,v1/2=max⁡dv,w,dv,v,dw,w,dv,w·dw,v1/2=dv,w.From (23) it follows that(25)ψdv,w=ψdfv,fw≤ψdv,wφdv,w<ψdv,w,a contradiction as d(v,w)>1. Hence v=w.Corollary 16.
LetX,d be a complete multiplicative metric space and f:X→X. Suppose that there exist control functions ψ and φ such that(26)ψdfnx,fny≤ψMfnx,yφMfnx,y,for any x,y∈X and n∈N where(27)Mfnx,y=max⁡dx,y,dfnx,x,dfny,y,dfnx,y·dfny,x1/2.Then f has a unique fixed point u and fnu=fu.Proof.
SetT=fn. From Theorem 15, T has a unique fixed point u. Now f(u)=f(Tu)=fn+1(u)=fn(fu)=Tfu which implies that fu is also a fixed point of T. By the uniqueness of fixed point of T, we have fu=u.Now, we recall the following lemma from [24].Lemma 17.
LetX be a nonempty set and f:X→X. Then there exists a subset E⊆X such that f(E)=f(X) and f:E→X is one-to-one.Theorem 18.
LetX,d be a multiplicative metric space and f,g:X→X. Suppose that there exist control functions ψ and φ such that(28)ψdfx,fy≤ψMgx,gyφMgx,gyholds for any x,y∈X, where(29)Mx,y=max⁡dx,y,dfx,gx,dfy,gy,dfx,gy·dfy,gx1/2.If gX is a complete subspace of X, then f and g have a unique coincidence point in X. Moreover, if f and g are weakly compatible, then f and g have at most one common fixed point.Proof.
By Lemma17, there exists E⊆X such that g(E)=g(X) and g:E→X is one-to-one. Define a map h:g(E)→g(E) by h(gx)=fx. Since g is one-to-one on E,  h is well defined. Note that(30)ψdhgx,hgy≤ψMgx,gyφMgx,gy,for all gx,gy∈g(E) where(31)Mgx,gy=max⁡dgx,gy,dhgx,gx,dhgy,gy,dhgx,gy·dhgy,gx.Since g(E)=g(X) is complete, by Theorem 15, there exists x0∈X such that h(gx0)=gx0=fx0. Hence, f and g have a unique point of coincidence. From Proposition 12, f and g have a unique common fixed point.


## 3. Cocyclic Contractions

Now we obtain common fixed point results for self-maps satisfying certain cocyclic contractions defined on a multiplicative metric space. We start with the following.Theorem 19.
LetX,d be a multiplicative metric space, A1,A2,…,Ap nonempty closed subsets of X, and Y=∪i=1pAi. Suppose that f,g:Y→Y are such that(a)
Y has a cocyclic representation with respect to pair (f,g) and the collection {Ai:i=1,2,…,p};(b)
there exists control functionsψ and φ such that, for any x,y∈Ai×Ai+1, i=1,2,…,p, the following,(32)ψdfx,fy≤ψMf,gx,yφMf,gx,y,holds where(33)Mf,gx,y=max⁡dgx,gy,dfx,gx,dfy,gy,dfx,gy·dfy,gx1/2withAp+1=A1.
IfgAi is complete subspace of X for each i∈{1,2,…,p}, then f and g have a unique coincidence point z∈∩i=1pgAi=Z provided that fZ⊆gZ⊆Z and gZ is closed. Moreover, if f and g are weakly compatible, then f and g have at most one common fixed point.Proof.
Letx0 be a given point in ∪i=1pAi. Then there exists i0∈{1,2,…,p} such that x0∈Ai0. Choose a point x1 in Ai0+1 such that f(x0)=g(x1). This can be done because f(Ai0)⊆g(Ai0+1). Continuing this process, for n>0, there exists in∈{1,2,…,p} such that having chosen xn in Ain, we obtain xn+1 in Ain+1 such that f(xn)=g(xn+1). If, for some n0≥0, we have fxn0=fxn0+1, then fxn0=gxn0 implies that xn0 is the coincidence point of f and g. Assume that d(fxn,fxn+1)>1 for all n. From (b), we have(34)ψdgxn+1,gxn+2=ψdfxn,fxn+1≤ψMf,gxn,xn+1φMf,gxn,xn+1≤ψMf,gxn,xn+1,where(35)Mf,gxn,xn+1=max⁡dgxn,gxn+1,dfxn,gxn,dfxn+1,gxn+1,dfxn,gxn+1·dfxn+1,gxn1/2=max⁡dgxn,gxn+1,dgxn+1,gxn,dgxn+2,gxn+1,dgxn+1,gxn+1·dgxn+2,gxn1/2≤max⁡dgxn,gxn+1,dgxn+2,gxn+1,dgxn+2,gxn+1·dgxn+1,gxn1/2≤max⁡dgxn,gxn+1,dgxn+2,gxn+1,dgxn+2,gxn+1+dgxn+1,gxn2=max⁡dgxn,gxn+1,dgxn+2,gxn+1.If, for some n, max{d(gxn,gxn+1),d(gxn+2,gxn+1)}=d(gxn+2,gxn+1), then Mf,g(xn,xn+1)=d(gxn+2,gxn+1) and(36)ψdgxn+1,gxn+2≤ψMf,gxn,xn+1φMf,gxn,xn+1≤ψdgxn+2,gxn+1φdgxn+2,gxn+1imply that φd(gxn+2,gxn+1)=1, a contradiction as d(gxn+2,gxn+1)>1. Hence d(gxn+2,gxn+1)≤d(gxn,gxn+1) for all n. Thus Mf,g(xn,xn+1)≤d(gxn,gxn+1). On the other hand, we have Mf,g(xn,xn+1)≥d(gxn,gxn+1). Hence Mf,g(xn,xn+1)=d(gxn,gxn+1).
Similarly we obtain thatd(gxn+1,gxn)≤d(gxn,gxn-1). Thus the sequence d(gxn+1,gxn) is nonincreasing. Consequently, there exists δ≥1 such that limn→∞⁡d(gxn+1,gxn)=δ. Suppose that δ>1. Now(37)ψdgxn+2,gxn+1≤ψMf,gxn+1,xnφMf,gxn+1,xn,and lower semicontinuity of φ gives that(38)limsupn→∞⁡ψdgxn+2,gxn+1≤limsupn→∞⁡ψMf,gxn+1,x2nliminfn→∞⁡φMf,gx2n+1,x2n,which implies that ψ(δ)≤ψ(δ)/φ(δ), a contradiction. Therefore δ=1. That is,(39)limn→∞⁡dgxn+1,gxn=1.Assume that {gxn} is not a multiplicative Cauchy sequence. Then, there is ε>1, and there are even integers nk and mk with mk>nk>k such that(40)dgxmk,gxnk≥ε,and d(gxmk-1,gxnk)<ε. Note that(41)ε≤dgxmk,gxnk≤dgxnk,gxmk-1·dgxmk-1,gxmk.From (39) and (40), it follows that(42)limk→∞⁡dgxmk,gxnk=ε.By (42) and(43)dgxnk+1,gxmk≤dgxnk+1,gxnk·dgxnk,gxmk,we have limk→∞⁡d(gxnk+1,gxmk)≤ε. Also, by (42) and(44)dgxmk,gxnk≤dgxnk+1,gxnk·dgxnk+1,gxmkwe obtain that ε≤limk→∞⁡d(gxnk+1,gxmk). Hence(45)limk→∞⁡dgxnk+1,gxmk=ε.From (39) and(46)dgxnk+1,gxmk-1≤dgxnk+1,gxnk·dgxnk,gxmk-1,we have limk→∞⁡d(gxnk+1,gxmk-1)≤ε. By (45) and the inequality(47)dgxnk+1,gxmk≤dgxnk+1,gxmk-1·dgxmk-1,gxmk,we have ε≤limk→∞⁡d(gxnk+1,gxmk-1). Thus(48)limk→∞⁡dgxnk+1,gxmk-1=ε.Note that(49)Mf,gxnk,xmk-1=max⁡dfxnk,fxmk-1,dfxnk,gxnk,dfxmk-1,gxmk-1,dfxnk,gxmk-1·dfxmk-1,gxnk1/2=max⁡dgxnk+1,gxmk,dgxnk+1,gxnk,dgxmk,gxmk-1,dgxnk+1,gxmk-1·dgxmk,gxnk1/2.Consequently, limk→∞⁡Mf,g(xnk,xmk-1)=max⁡{ε,1,1,ε}=ε. Since xnk and xmk-1 lie in different but adjacent labelled sets Ai and Ai+1 for some 1≤i≤p, we have(50)ψdgxnk+1,gxmk=ψdfxnk,gxmk-1≤ψMf,gxnk,xmk-1φMf,gxnk,xmk-1.Taking upper limit as k→∞ implies that ψ(ε)≤ψ(ε)/φ(ε), a contradiction as ε>1, and hence {gxn} is a multiplicative Cauchy sequence in ∪i=1pgAi. Since ∪i=1pgAi is complete, there exists y∈∪i=1pgAi such that(51)limk→∞⁡dgxn,y=1.Consequently, we can find a point z in Y such that g(z)=y.
Now we show thaty∈∩i=1pgAi=Z. From condition (a) and x0∈Ai0 for some i0∈{1,2,…,p}, we can choose a subsequence {xnk} in Ai0 out of the sequence {xn}. Obviously, {gxnk}⊆g(Ai0). As g(Ai0) is closed, y∈g(Ai0). Similarly, we can choose a subsequence {xnk+1} in Ai0+1 out of the sequence {xn}. Obviously, {gxnk+1}⊆g(Ai0+1). As g(Ai0+1) is closed, y∈g(Ai0+1). Continuing this way, we obtain that y∈∩i=1pgAi and hence ∩i=1pg(Ai)≠∅.
Now we show thatf(z)=g(z). Since z∈Y, there exists some i in {1,2,…,p} such that z∈Ai. Choose a subsequence {xnk} of {xn} with xnk∈Ai-1. From (b), we have(52)ψdgxnk+1,fz=ψdfxnk,fz≤ψMf,gxnk,zφMf,gxnk,z,where(53)Mf,gxnk,z=max⁡dgxnk,gz,dfxnk,gxnk,dfz,gz,dfxnk,gz·dfz,gxnk1/2=max⁡dgxnk,gz,dgxnk+1,gxnk,dfz,gz,dgxnk+1,gz·dfz,gxnk1/2.On taking upper limit as k→∞ we obtain that(54)ψdgz,fz≤ψdfz,gzφdfz,gz,and hence f(z)=g(z). Thus z is the coincidence point of f and g.
Note thatZ being a finite intersection of closed sets is closed and hence complete. Consider the restrictions of f and g on Z; that is, f|Z,  g|Z:Z→Z. Obviously, f|Z(Z)⊆g|Z(Z)⊆Z. Also g|Z(Z) is closed and hence complete. From Theorem 18, it follows that f|Z and g|Z have a unique coincidence point in Z. As f and g are weakly compatible, from Proposition 12, it follows that f|Z and g|Z have a unique common fixed point.Corollary 20.
LetX,d be a multiplicative metric space, A1,A2,…,Ap nonempty closed subsets of X, and Y=∪i=1pAi. Suppose that f,g:Y→Y are such that(a)
Y has a cocyclic representation with respect to pair (fm,gn) and the collection {Ai:i=1,2,…,p} for some m,n∈N;(b)
there exists two control functionsψ and φ such that, for any x,y∈Ai×Ai+1, i=1,2,…,p,(55)ψdfmx,fmy≤ψMf,gx,yφMf,gx,yholds for somem,n∈N, where (56)Mf,gx,y=max⁡dgnx,gny,dfmx,gnx,dfmy,gny,dfmx,gny·dfmy,gnx1/2withAp+1=A1.
IfgnAi is complete subspace of X for each i∈{1,2,…,p}, then fm and gn have a unique coincidence point z∈∩i=1pgnAi=Z provided that fmZ⊆gnZ⊆Z and gnZ is closed. Moreover, if fm and gn are weakly compatible, then fm and gn have a unique common fixed point. Furthermore, f and g have a unique common fixed point provided that f and g are commuting.Proof.
SetS=fm and T=gn. From Theorem 19, S and T have a unique common fixed point u. Now S(fu)=fSu=fu and Tfu=fTu=fu imply that fu is the common fixed point of S and T. Also, T(gu)=gTu=gu and Sgu=gSu=gu imply that gu is also the common fixed point of S and T. By uniqueness of common fixed point of S and T, we have fu=gu=u.Example 21.
LetX=R and let d be a multiplicative metric on X defined by d(x,y)=ax-y, where a>1 is a real number. For some c>1, set A1=[-c,0], A2=[0,c], and A3=A1. Define f,g:∪i=12Ai→∪i=12Ai by(57)fx=αxc,gx=-βxc,where α,β≥1 with 6α≤5β and β≤c. Note that f(A1)=[-α,0]⊆[-β,0]=g(A2) and f(A2)=[0,α]⊆[0,β]=g(A1). Y=A1∪A2 has a cocyclic representation with respect to pair (f,g) and the collection {A1,A2}.
Defineψ,φ:[1,∞)→[1,∞) by(58)ψt=t3,φt=t1/2,ift∈1,2β,t2,ift>2β.Clearly ψ is continuous and nondecreasing, φ is a lower semicontinuous, and ψ(t)=φ(t)=1 if and only if t=1.
We show that condition (b) is satisfied. Now, forx∈A1, y∈A2 implies(59)ψdfx,fy=ψdαxc,αyc=ψaα/cy-x=a3α/cy-x≤a5β/2cy-x=d-βxc,-βyc5/2=dgx,gy5/2≤Mf,gx,y5/2=Mf,gx,y3Mf,gx,y1/2=ψMf,gx,yφMf,gx,y.When x∈A2 and y∈A1, (60)ψdfx,fy=ψdαxc,αyc=ψaα/cx-y=a3α/cx-y≤a5β/2cx-y=d-βxc,-βyc5/2=dgx,gy5/2≤Mf,gx,y5/2=Mf,gx,y3Mf,gx,y1/2=ψMf,gx,yφMf,gx,y.Thus, f and g satisfy all the conditions of Theorem 19. Moreover, f and g have at most one common fixed point.Example 22.
LetX=R and d:X×X→R+ be the multiplicative metric defined by d(x,y)=ex-y. Suppose A1=[-c,0], A2=[-c/2,c/2], and A3=[-c/4,c] and A4=A1, where c>0. Define f,g:∪i=13Ai→∪i=13Ai by(61)fx=-xcifx>0,0otherwise,gx=2xc.Note that f(A1)={0}⊆[-1,1]=g(A2), f(A2)=[-1/2,0]⊆[-1/2,2]=g(A3), and f(A3)=[-1,0]⊆[-2,0]=g(A1). Hence Y=A1∪A2 is a cocyclic representation between f and g.
Letψ,φ:[1,∞)→[1,∞) be as defined in Example 21. To check condition (b), we consider the following cases:(1)
Letx∈A1 and y∈A2. If y∈[-c/2,0], then d(fx,fy)=e0=1 and hence (b) holds. If y∈(0,c/2], then we have(62)dfx,fy=d0,-yc=e1/cy≤e5/cy-x=d2xc,2yc5/2=dgx,gy5/2≤Mf,gx,y5/2=Mf,gx,y3Mf,gx,y1/2=ψMf,gx,yφMf,gx,y.(2)
Letx∈A2 and y∈A3. If x∈[-c/2,0] and y∈[-c/4,0], then d(fx,fy)=e0=1 and hence (b) is satisfied. If x∈[-c/2,0] and y∈(0,c], then we obtain that(63)dfx,fy=d0,-yc=e1/cy<e15/2cy=d-yc,2yc5/2=dfy,gy5/2≤Mf,gx,y5/2=Mf,gx,y3Mf,gx,y1/2=ψMf,gx,yφMf,gx,y.Whenx∈(0,c/2] and y∈(-c/4,0], we have(64)dfx,fy=d-xc,0=e1/cx<e15/2cx=d-xc,2xc5/2=dfx,gx5/2≤Mf,gx,y5/2=Mf,gx,y3Mf,gx,y1/2=ψMf,gx,yφMf,gx,y.Forx∈(0,c/2] and y∈(0,c], we obtain that(65)dfx,fy=d-xc,-yc=e1/cx-y<e5/cx-y=d2xc,2yc5/2=dgx,gy5/2≤Mf,gx,y5/2=Mf,gx,y3Mf,gx,y1/2=ψMf,gx,yφMf,gx,y.(3)
Letx∈A3 and y∈A1. If x∈[-c/4,0], then d(fx,fy)=1 and (b) is satisfied. If x∈(0,c], then we have(66)dfx,fy=d-xc,0=e1/cx<e15/2cx=d-xc,2xc5/2=dfx,gx5/2≤Mf,gx,y5/2=Mf,gx,y3Mf,gx,y1/2=ψMf,gx,yφMf,gx,y.Hence f and g satisfy all the conditions of Theorem 19. Moreover, 0 is a unique common fixed point of f and g in ∩i=13Ai.DenoteΥ={ϕ:[1,∞)→[1,∞); ϕ is a Lebesgue integrable with finite integral on each compact subset of [1,∞) and for each ε>1, ∫1ε‍φ(t)dt>1}.As a consequence of Theorem19, we obtain following common fixed point result for mappings satisfying cocyclic contractive condition of integral type in a multiplicative metric space.Corollary 23.
LetX,d be a multiplicative metric space, A1,A2,…,Ap, p nonempty closed subsets of X, and Y=∪i=1pAi. Suppose that f,g:Y→Y are mappings such that(a)
Y has a cocyclic representation with respect to pair (f,g) and the collection {Ai:i=1,2,…,p};(b)
there exists control functionsψ and φ such that, for any x,y∈Ai×Ai+1, i=1,2,…,p, the following,(67)∫1ψdfx,fyϕtdt≤∫1ψMf,gx,yϕtdt∫1φMf,gx,yϕtdt,is satisfied, whereϕ∈Υ, and(68)Mf,gx,y=max⁡dgx,gy,dfx,gx,dfy,gy,dfx,gy·dfy,gx1/2withAp+1=A1.
IfgAi is a complete subspace of X for each i∈{1,2,…,p}, then f and g have a unique coincidence point z∈∩i=1pgAi=Z provided that fZ⊆gZ⊆Z and gZ is closed. Moreover, if f and g are weakly compatible, then f and g have at most one common fixed point.Proof.
DefineΨ:[1,∞)→[1,∞) by Ψ(x)=∫1x‍ϕ(t)dt. From (67), it follows that(69)Ψψdfx,fy≤ΨψMf,gx,yΨφMf,gx,y,which can be written as(70)ψ1dfx,fy≤ψ1Mf,gx,yφ1Mx,y,where ψ1=Ψ∘ψ and φ1=Ψ∘φ. Clearly, ψ1,φ1:[1,∞)→[1,∞), ψ1 is continuous and nondecreasing, φ1 is a lower semicontinuous, and ψ1(t)=φ1(t)=1 if and only if t=1. Hence by Theorem 19, f and g have at most one common fixed point.



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