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Research Article

Fixed point theorems for generalized contraction mappings in multiplicative metric spaces

Afrah A. N. Abdou

Departement of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia.

Communicated by Y. J. Cho

Abstract

The purpose of this paper is to study and discuss the existence of common fixed points for weakly compatible mappings satisfying the generalized contractiveness and the (CLR)-property. Our results improve the corresponding results given in He et al. [X. He, M. Song, D. Chen, Fixed Point Theory Appl., 2014 (2014), 9 pages]. Moreover, we give some examples to illustrate for the main results. 2016 All rightsc reserved.

Keywords: Multiplicative metric space, common fixed point, compatible mappings, weakly compatible mappings, the (CLR)-property.

2010 MSC: 47H10, 54H25, 54E50.

1. Introduction

In 1922, Banach [5] proved the theorem which is well known as “Banach’s Fixed Point Theorem” to establish the existence of solutions for nonlinear operator equations and integral equations. It is widely considered as a source of metric fixed point theory and also its significance lies in its vast applications. The study on the existence of fixed points of some mappings satisfying certain contractions has many applications and has been the center various research activities. In the past years, many authors generalized Banach’s Fixed Point Theorem in various spaces such as quasi-metric spaces, fuzzy metric spaces, 2-metric spaces, cone metric spaces, partial metric spaces, probabilistic metric spaces and generalized metric spaces (see, for instance, [2, 3, 4, 7, 9, 15, 16, 17, 20, 22, 23, 24, 25, 26, 27, 28, 29, 30]).

On the other hand, in 2008, Bashirov et al. [6] defined a new distance so called a multiplicative distance by using the concepts of multiplicative absolute value. After then, in 2012, by using the same idea of

Email address: [email protected](Afrah A. N. Abdou) Received 2015-12-22

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multiplicative distance, ¨Ozav¸sar and C¸ evikel [18] investigate multiplicative metric spaces by remarking its topological properties and introduced the concept of a multiplicative contraction mapping and proved some fixed point theorems for multiplicative contraction mappings on multiplicative spaces. In 2012, He et al. [8]

proved a common fixed point theorem for four self-mappings in multiplicative metric space.

Recently, motivated by the concepts of compatible mappings and compatible mappings of types (A), (B) in metric spaces given by Jungck [10], [14], Jungck et al. [12] and Pathak and Khan [19], in 2015, Kang et al. [11] introduced the concepts of compatible mappings and its variants in multiplicative metric spaces, that is compatible mappings of types (A), (B) and others, and prove some common fixed point theorems for these mappings.

Especially, in 2002, Aamri and Moutawakil [1] introduced the concept of the (E.A)-property. Afterward, in 2011, Sintunaravat and Kumam [27] obtained that the notion of the (E.A)-property always requires a completeness of underlying subspaces for the existence of common fixed points for single-valued mappings and hence they coined the idea ofcommon limit in the range (shortly, the (CLR) property), which relaxes the completeness of the underlying of the subspaces.

Motivated by the above results, in this paper, we prove some common fixed point theorems for weakly compatible mappings satisfying some generalized contractions and the common limit range with respect to the value of given mappings in multiplicative metric spaces. Also, we give some examples to illustrate for our main results.

2. Preliminaries

Now, we present some necessary definitions and results in multiplicative metric spaces, which will be needed in the sequel.

Definition 2.1 ([6]). Let X be a nonempty set. A multiplicative metric is a mapping d: X×X → R+ satisfying the following conditions:

(M1) d(x, y)≥1 for allx, y∈X andd(x, y) = 1⇐⇒x=y;

(M2) d(x, y) =d(y, x) for allx, y∈X;

(M3) d(x, y)≤d(x, z)·d(z, y) for allx, y, z ∈X (: multiplicative triangle inequality).

The pair (X, d) is called amultiplicative metric space.

Proposition 2.2([18]). Let (X, d)be a multiplicative metric space, {xn}be a sequence inX and letx∈X.

Then

xn→x (n→ ∞) if and only if d(xn, x)→1 (n→ ∞).

Definition 2.3 ([18]). Let (X, d) be a multiplicative metric space and {xn} be a sequence in X. The sequence {xn} is called a multiplicative Cauchy sequence if for each > 0 there exists a positive integer N ∈Nsuch thatd(xn, xm)< for all n, m≥N.

Proposition 2.4 ([18]). Let (X, d) be a multiplicative metric space and {xn} be a sequence in X. Then {xn} is a multiplicative Cauchy sequence if and only if d(xn, xm)→1 as n, m→ ∞.

Definition 2.5 ([18]). A multiplicative metric space (X, d) is said to be multiplicative complete if every multiplicative Cauchy sequence in (X, d) is multiplicative convergent inX.

Note that R+ is not complete under the ordinary metric, of course, under the multiplicative metric,R+ is a complete multiplicative and the convergence of a sequence in R+ in both multiplicative and ordinary metric space are equivalent. But they may be different in more general cases.

Proposition 2.6 ([18]). Let (X, dX) and (Y, dY) be two multiplicative metric spaces, f : X → Y be a mapping and {xn} be a sequence in X. Then f is multiplicative continuous at x ∈ X if and only if f(xn)→f(x) for every sequence {xn} with xn→x as n→ ∞.

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Proposition 2.7 ([18]). Let (X, dX) be a multiplicative metric spaces, {xn} and {yn} be two sequences in X such that xn→x∈X and yn→y∈X as n→ ∞. Then

d(xn, yn)→d(x, y) (n→ ∞).

Definition 2.8. The self-mappingsf and gof a set X are said to be:

(1) commutativeorcommutingon X [10] if f gx=gf xfor allx∈X;

(2) weakly commutativeor weakly commutingon X [21] if d(f gx, gf x)≤d(f x, gx) for allx∈X;

(3) compatible on X [11] if limn→∞d(f gxn, gf xn) = 1 whenever {xn} is a sequence in X such that limn→∞f xn= limn→∞gxn=tfor somet∈X;

(4) weakly compatible on X [13] if f x =gx for all x ∈X implies f gx= gf x, that is, d(f x, gx) = 1 =⇒ d(f gx, gf x) = 1.

Remark 2.9. Weakly commutative mappings are compatible and compatible mappings are be weakly com- patible, but the converses are not true (see [11, 13]).

Example 2.10. LetX= [0,+∞) and define a mapping as follows: for allx, y∈X, d(x, y) =e|x−y|.

Then dsatisfies all the conditions of a multiplicative metric and so (X, d) is a multiplicative metric space.

Letf and g be two self-mappings ofX defined byf x=x3 and gx= 2−x for all x∈X. Then we have d(f xn, gxn) =e|xn−1|·|x2n+xn+2|→1 if and only if xn→1

and

n→∞lim d(f gxn, gf xn) = lim

n→∞e6|xn−1|2 = 1 if xn→1 asn→ ∞. Thusf and g are compatible. Note that

d(f g(0), gf(0)) =d(8,2) =e6> e2 =d(0,2) =d(f(0), g(0)) and so the pair (f, g) is not weakly commuting.

Example 2.11. LetX= [0,+∞), (X, d) be a multiplicative metric space defined byd(x, y) =e|x−y|for all x, yin X. Let f and gbe two self-mappings of X defined by

f x=

x, if 0≤x <2, 2, if x= 2,

4, if 2< x <+∞,

gx=

4−x, if 0≤x <2, 2, if x= 2,

7, if 2< x <+∞.

By the definition of the mappingsf and g, only for x= 2, f x=gx= 2 and sof gx=gf x= 2. Thus the pair (f, g) is weakly compatible.

For xn= 2−n1 ∈(0,2), from the definition of the mappings f and g, we have

n→∞lim f xn= lim

n→∞gxn= 2, but

n→∞lim d(f gxn, gf xn) = lim

n→∞exn =e2 6= 1 and so the pair (f, g) is not compatible.

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Definition 2.12. Two pairs S, A and T, B of self-mappings of a multiplicative metric spaces (X, d) are said to have the common limit range with respect to the value of the mapping A (or B) (shortly, the (CLR)-property) if there exists two sequences {xn}and {yn}inX such that

n→∞lim Axn= lim

n→∞Sxn= lim

n→∞Byn= lim

n→∞T yn=Az for somez∈X.

Example 2.13. Let X = [0,∞) be the usual metric space and define a mapping d : X ×X −→ R by d(x, d) = e|x−y| for all x, y ∈ X. Then (X, d) is a complete multiplicative metric space. Define mappings S, T, A, B :X−→X by

Sx= 1

64x, T x= 1

32x, Ax=x, Bx= 2x

for allx∈X. Then, for the sequences{xn} and{yn}inX defined by xn= n1 and yn=−1n for each n≥1, clearly, we can see that

n→∞lim Axn= lim

n→∞Sxn= lim

n→∞Byn= lim

n→∞T yn=A(0) = 0

orB(0) = 0. This show that the pairs (S, A) and (T, B) have the common limit range with respect to the value of the mappingA or the mappingB.

Definition 2.14 ([18]). Let (X, d) be a multiplicative metric space. A mapping f : X → X is called a multiplicative contraction if there exists a real constant λ ∈ (0,1] such that d(f x, f y) ≤ [d(x, y)]λ for all x, y∈X.

The following is Banach’s Fixed Point Theorem in multiplicative metric spaces, which was proved by Ozav¸sar and C¨ ¸ evikel.

Theorem 2.15 ([18]). Let (X, d) be a multiplicative metric space and f : X → X be a multiplicative contraction. If(X, d) is complete, then f has a unique fixed point in X.

Recently, He et al. [8] proved the following.

Theorem 2.16 ([8]). Let S, T, Aand B be four self-mappings of a multiplicative metric space X satisfying the following conditions:

(a) S(X)⊂B(X) and T(X)⊂A(X);

(b) the pairs(A, S) and (B, T) are weak commuting on X;

(c) one of S, T, A andB is continuous;

(d) there exists a number λ∈(0,12) such that, for all x, y∈X,

d(Sx, T y)≤[max{d(Ax, By), d(Ax, Sx), d(By, T y), d(Sx, By), d(Ax, T y)}]λ. ThenS, T, A and B have a unique common fixed point in X.

3. Common fixed points for weakly compatible mappings

In this section we prove some common fixed point theorems for weakly compatible mappings in multi- plicative metric spaces.

Theorem 3.1. Let (X, d)be a complete multiplicative metric space. LetS, T, A, B :X →X be single-valued mappings such thatS(X)⊂B(X), T(X)⊂A(X) and there exists λ∈(0,12) such that

dp(Sx, T y)≤h ϕ

maxn

dp(Ax, By),dp(Ax, Sx)dp(By, T y)

1 +dp(Ax, By) ,dp(Ax, T y)dp(By, Ax) 1 +dp(Ax, By)

oiλ

(3.1)

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for allx, y∈X and p≥1, where ϕ: [0,∞)→[0,∞) is a monotone increasing function such thatϕ(0) = 0 and ϕ(t)< t for all t >0.

Suppose that one of the following conditions is satisfied:

(a) eitherA or S is continuous, the pair (S, A) is compatible and the pair (T, B) is weakly compatible;

(b) eitherB or T is continuous, the pair (T, B) is compatible and the pair (S, A) is weakly compatible.

ThenS, T, A and B have a unique common fixed point in X.

Proof. Let x0 ∈X. Since S(X) ⊂B(X) and T(X) ⊂A(X), there exist x1, x2 ∈X such that y0 =Sx0 = Bx1 and y1=T x1 =Ax2. By induction, we can define the sequences{xn} and {yn} inX such that

y2n=Sx2n=Bx2n+1, y2n+1 =T x2n+1 =Ax2n+2 (3.2) for all n≥0.

Now, we prove that{yn}is a multiplicative Cauchy sequence inX. From (3.1) and (3.2), it follows that, for all n≥1,

dp(y2n, y2n+1) =dp(Sx2n, T x2n+1)

≤h ϕ

max

n

dp(Ax2n, Bx2n+1)),dp(Ax2n, Sx2n)dp(Bx2n+1, T x2n+1) 1 +dp(Ax2n, Bx2n+1) , dp(Ax2n, T x2n+1)dp(Bx2n+1, Ax2n)

1 +dp(Ax2n, Bx2n+1)

oiλ

≤h ϕ

max

n

dp(y2n−1, y2n)), dp(y2n, y2n+1), dp(y2n−1, y2n+1) oiλ

≤h ϕ

dp(y2n−1, y2n)·dp(y2n, y2n+1) iλ

≤[d(y2n−1, y2n)]·[d(y2n, y2n+1)], which implies that

d(y2n, y2n+1)≤[d(y2n−1, y2n)]p−pλ = [d(y2n−1, y2n)]1−λλ = [d(y2n−1, y2n)]h, (3.3) whereh= 1−λλ ∈(0,1). Similarly, we have

dp(y2n+2, y2n+1) =dp(Sx2n+2, T x2n+1)

≤h ϕ

max

n

dp(Ax2n+2, Bx2n+1)),dp(Ax2n+2, Sx2n+2)dp(Bx2n+1, T x2n+1) 1 +dp(Ax2n+2, Bx2n+1) , dp(Ax2n+2, T x2n+1)dp(Bx2n+1, Ax2n+2)

1 +dp(A2n+2, Bx2n+1)

oiλ

≤h ϕ

max

n

dp(y2n, y2n+1), dp(y2n+2, y2n+1), dp(y2n, y2n+1) oiλ

≤h ϕ

max

n

dp(y2n, y2n+1)·dp(y2n+2, y2n+1) oiλ

≤[d(y2n, y2n+1)]·[d(y2n+2, y2n+1)], which implies that

d(y2n+1, y2n+2)≤[d(y2n, y2n+1)]1−λλ = [d(y2n, y2n+1)]h. (3.4) It follows from (3.3) and (3.4) that, for all n∈N,

d(yn, yn+1)≤[d(yn−1, yn)]h≤[d(yn−2, yn−1)]h2 ≤ · · · ≤[d(y0, y1)]hn.

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Therefore, for alln, m∈Nwithn < m, by the multiplicative triangle inequality, we obtain d(yn, ym)≤d(yn, yn+1)·d(yn+1, yn+2)· · ·d(ym−1, ym)

≤[d(y0, y1)]hn·[d(y0, y1)]hn+1· · ·[d(y0, y1)]hm−1

≤[d(y0, y1)]1−hhn .

This means that d(yn, ym) → 1 as n, m → ∞. Hence {yn} is a multiplicative Cauchy sequence in X. By the completeness ofX, there existsz∈X such that yn→zas n→ ∞. Moreover, since {y2n}={Sx2n}= {Bx2n+1}and {y2n+1}={T x2n+1}={Ax2n+2}are subsequences of {yn}, we obtain

n→∞lim Sx2n= lim

n→∞Bx2n+1= lim

n→∞T x2n+1 = lim

n→∞Ax2n+2=z. (3.5)

Next, we show that z is a common fixed point ofS,T,A andB under the condition (a).

Case 1. Suppose that A is continuous. Then it follows that limn→∞ASx2n = limn→∞A2x2n = Az.

Since the pair (S, A) is compatible, it follows from (3.5) that

n→∞lim d(SAx2n, ASx2n) = lim

n→∞d(SAx2n, Az) = 1, this is, limn→∞SAx2n=Az. Using (3.1), we have

dp(SAx2n, T x2n+1)≤h ϕ

maxn

dp(A2x2n, Bx2n+1),dp(A2x2n, SAx2n)dp(Bx2n+1, T x2n+1) 1 +dp(A2x2n, Bx2n+1) , dp(Ax2x2n, T x2n+1)dp(Bx2n+1, A2x2n)

1 +dp(A22n, Bx2n+1)

oiλ

.

Takingn→ ∞on the two sides of the above inequality and using (3.5), we can obtain dp(Az, z)≤h

ϕ

max n

dp(Az, z),dp(Az, Az)dp(z, z)

1 +dp(Az, z) ,dp(Az, z)dp(z, Az) 1 +dp(Az, z)

oiλ

≤h ϕ

max

n

dp(Az, z), 1

dp(Az, z), dp(Az, z) oiλ

=h ϕ

dp(Az, z)iλ

≤[d(Az, z)].

This means that d(Az, z) = 1, this is,Az=z. Again, applying (3.1), we obtain dp(Sz, T x2n+1)≤h

ϕ maxn

dp(Az, Bx2n+1)),dp(Az, Sz)dp(Bx2n+1, T x2n+1) 1 +dp(Az, Bx2n+1) , dp(Az, T x2n+1)dp(Bx2n+1, Az)

1 +dp(Az, Bx2n+1)

oiλ

.

Lettingn→ ∞on both sides in the above inequality and using Az=zand (3.4), we can obtain dp(Sz, z)≤h

ϕ maxn

dp(Az, z),dp(z, Sz)dp(z, z)

1 +dp(Az, z) ,dp(Az, z)dp(z, Az) 1 +dp(z, z)

oiλ

≤h ϕ

dp(Sz, z) iλ

≤[d(Sz, z)].

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This implies that d(Sz, z) = 1, that is, Sz = z. On the other hand, since z = Sz ∈S(X) ⊂B(X), there exist z∈X such that z=Sz=Bz. By using (3.1) andz=Sz=Az=Bz, we can obtain

dp(z, T z) =dp(Sz, T z)

≤h ϕ

max

n

dp(Az, Bz),dp(Az, Sz)dp(Bz, T z)

1 +dp(Az, Bz) ,dp(Az, T z)dp(Bz, Az) 1 +dp(Az, Bz)

oiλ

≤h ϕ

dp(z, T z)iλ

≤[d(z, T z)].

This implies thatd(z, T z) = 1 and so T z=z=Bz. Since the pairT, B is weakly compatible, we have T z =T Bz =BT z=Bz.

Now, we prove that T z=z. From (3.1), we have dp(z, T z) =dp(Sz, T z)

≤h ϕ

max

n

dp(Az, Bz),dp(Az, Sz)dp(Bz, T z)

1 +dp(Az, Bz) ,dp(Az, T z)dp(Bz, Az) 1 +dp(Az, Bz)

oiλ

≤h ϕ

maxn

dp(z, T z), 1

dp(z, T z), dp(z, T z)oiλ

= h

ϕ

dp(z, T z) iλ

≤[d(z, T z)].

This implies thatd(z, T z) = 1 and soz=T z. Therefore, we obtainz=Sz=Az =T z =Bz and so zis a common fixed point ofS, T, Aand B.

Case 2. Suppose that S is continuous. Then limn→∞SAx2n = limn→∞S2x2n = Sz. Since the pair (S, A) is compatible, it follows from (3.5) that

n→∞lim dp(SAx2n, ASx2n) = lim

n→∞dp(Sz, ASx2n) = 1, this is, limn→∞ASx2n=Sz. From (3.1), we obtain

dp(S2x2n, T x2n+1)≤h ϕ

maxn

dp(ASx2n, Bx2n+1)),dp(ASx2n, S2x2n)dp(Bx2n+1, T x2n+1) 1 +dp(ASx2n, Bx2n+1) , dp(ASx2n, T x2n+1)dp(Bx2n+1, ASx2n)

1 +dp(ASx2n, Bx2n+1)

oiλ

. Takingn→ ∞on the both sides of the above inequality and using (3.4), we can obtain

dp(Sz, z)≤h ϕ

max

n

dp(Sz, z),dp(Sz, Sz)dp(z, z)

1 +dp(Sz, z) ,dp(Sz, z)dp(z, Sz) 1 +dp(Sz, z)

oiλ

≤h ϕ

maxn

dp(Sz, z), 1

dp(Sz, z), dp(z, Sz)oiλ

= h

ϕ

dp(z, T z) iλ

≤[d(Sz, z)].

This means thatd(Sz, z) = 1, this is,Sz=z. Sincez =Sz∈S(X)⊂B(X), there exist z ∈X such that z=Sz=Bz. From (3.1), we have

dp(S2x2n, T z)≤h ϕ

max

n

dp(ASx2n, Bz),dp(ASx2n, S2x2n)dp(Bz, T z) 1 +dp(ASx2n, Bz) , dp(ASx2n, T z)dp(Bz, ASx)

1 +dp(ASx2n, Bz)

oiλ

.

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Lettingn→ ∞on both sides in the above inequality and using Az=zand (3.5), we can obtain dp(Sz, T z)≤h

ϕ maxn

dp(Sz, z),dp(Sz, Sz)dp(z, T z)

1 +dp(Sz, z) ,dp(Sz, T z)dp(z, Sz) 1 +dp(Sz, z)

oiλ

=h ϕ

dp(z, T z)iλ

≤[d(z, T z)],

which implies that d(z, T z) = 1 and so T z = z =Bz. Since the pair (T, B) is weakly compatible, we obtain

T z =T Bz=BT z =Bz and so T z=Bz. By (3.1), we have

dp(Sx2n, T z)≤h ϕ

maxn

dp(Ax2n, Bz)),dp(Ax2n, Sx2n)dp(Bz, T z) 1 +dp(Ax2n, Bz) , dp(Ax2n, T z)dp(Bz, Ax2n)

1 +dp(Ax2n, Bz)

oiλ

.

Takingn→ ∞on the both sides of the above inequality and using Bz=T z, we can obtain dp(z, T z)≤h

ϕ maxn

dp(z, T z),dp(z, z)dp(T z, T z)

1 +dp(z, T z) ,dp(z, T z)dp(T z, z) 1 +dp(z, T z)

oiλ

≤h ϕ

maxn

dp(z, T z), 1

dp(z, T z), dp(T z, z)oiλ

= h

ϕ

dp(z, T z) iλ

≤[d(z, T z)].

This implies that d(z, T z) = 1 and so z = T z =Bz. On the other hand, since z = T z ∈ T(X) ⊂ A(X), there existz∗∗∈X such that z=T z =Az∗∗. By (3.1), usingT z =Bz=z, we can obtain

dp(Sz∗∗, z) =dp(Sz∗∗, T z)

≤h ϕ

maxn

dp(Az∗∗, Bz),dp(Az∗∗, Sz∗∗)dp(Bz, T z)

1 +dp(Az∗∗, Bz) ,d(Az∗∗, T z)dp(Bz, Az∗∗) 1 +dp(Az∗∗, Bz)

oiλ

= h

ϕ

dp(Sz∗∗, z) iλ

≤[d(Sz∗∗, z)].

This implies thatd(Sz∗∗, z) = 1 and so Sz∗∗=z=Az∗∗. Since the pairS, Ais compatible, we have d(Az, Sz) =d(SAz∗∗, ASz∗∗) =d(z, z) = 1

and so Az=Sz. Hence z=Sz=Az=T z=Bz.

Next, we prove that S, T, Aand B have a unique common fixed point. Suppose that w∈X is another common fixed point ofS, T, Aand B. Then we have

dp(z, w) =dp(Sz, T w)

≤h ϕ

max

n

dp(Az, Bw),dp(Az, Sz)dp(Bw, T w)

1 +dp(Az, Bw) ,dp(Az, T w)dp(Bz, Az) 1 +dp(Az, Bw)

oiλ

=h ϕ

dp(z, w)iλ

≤[d(z, w)],

which implies thatd(z, w) = 1 and sow=z. Therefore,z is a unique common fixed point ofS, T, AandB. Finally, if the condition (b) holds, then we obtain the same result. This completes the proof.

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Example 3.2. Let X = [0,2] and (X, d) be a complete multiplicative metric space, where dis defined by d(x, y) =e|x−y| for allx, y inX. LetS,T,A and B be four self-mappings of X defined by

Sx= 5

4, if x∈[0,2]; T x= 7

4, if x∈[0,1],

5

4, if x∈(1,2], Ax=

1, if x∈[0,1],

5

4, if x∈(1,2),

7

4, if x= 2,

Bx=

1

4, if x∈[0,1],

5

4, if x∈(1,2), 1, if x= 2.

Note that S is multiplicative continuous inX, but T,Aand B are not multiplicative continuous mappings inX. Also, we have the following:

(1) Clearly, S(X)⊂B(X) and T(X)⊂A(X).

(2) If {xn} ⊂(1,2), then we have

n→∞lim Sxn= lim

n→∞Axn=t= 5 4 and so

n→∞lim d(SAxn, ASnx) =d 5

4,5 4

= 1,

which means that the pair (S, A) is compatible. By the definition of the mappingsT,B, for any x∈(1,2), we have T x=Bx= 54 and so

T Bx=T(5 4) = 5

4 =B(5

4) =BT x.

ThusT Bx=BT x, which implies that the pair T, Bis weakly compatible.

(3) Now, we prove that the mappings S, T, A and B satisfy the condition (3.1) of Theorem 3.1 with λ= 23 andp= 1. For this, we consider the following cases:

Case 1. If x, y∈[0,1], then we have

d(Sx, T y) =d 5

4,7 4

=e12 and, sinceϕ(t)< t for all t >0, we have

h ϕ

maxn

dp(Ax, By),dp(Ax, Sx)dp(By, T y)

1 +dp(Ax, By) ,dp(Ax, T y)dp(By, Ax) 1 +dp(Ax, By)

oiλ

≤maxn d23

1,1 4

, d23 1,5

4

d231 4,7

4

, d23 1,7

4

d231 4,1o

= max{e12, e16e, e34e34} ≤e.

Thus we have

d(Sx, T y) =e12

<maxn

d23(Ax, By),d23(Ax, Sx)d23(By, T y) 1 +d23(Ax, By)

,d23(Ax, T y)d23(By, Ax) 1 +d23(Ax, By)

o .

Case 2. If x∈[0,1] andy∈(1,2], then we obtain d(Sx, T y) =d

5 4,5

4

= 1≤maxn

d23(Ax, By),d23(Ax, Sx)d23(By, T y) 1 +d23(Ax, By)

,d23(Ax, T y)d23(By, Ax) 1 +d23(Ax, By)

o .

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Case 3. If x∈(1,2) and y∈[0,1], then we obtain d(Sx, T y) =d

5 4,7

4

=e12 and

h ϕ

maxn

dp(Ax, By),dp(Ax, Sx)dp(By, T y)

1 +dp(Ax, By) ,dp(Ax, T y)dp(By, Ax) 1 +dp(Ax, By)

oiλ

≤max n

d23(5 4,1

4),d23(54,54)d23(14,74)

1 +d23(54,14) ,d23(54,74)d23(14,54) 1 +d23(54,14)

o

= max

e23, e, e

< e.

Hence we have

d(Sx, T y) =e12 <

h ϕ

max

n

dp(Ax, By),dp(Ax, Sx)dp(By, T y)

1 +dp(Ax, By) ,dp(Ax, T y)dp(By, Ax) 1 +dp(Ax, By)

oiλ

.

Case 4. If x= 2 and y∈[0,1], then we have d(Sx, T y) =d

5 4,7

4

=e12. Hence we have

d(Sx, T y) =e12 <

h ϕ

max

n

dp(Ax, By),dp(Ax, Sx)dp(By, T y)

1 +dp(Ax, By) ,dp(Ax, T y)dp(By, Ax) 1 +dp(Ax, By)

oiλ

.

Case 5. If x, y∈(1,2], then we have d(Sx, T y) =d

5 4,5

4

= 1≤h ϕ

maxn

dp(Ax, By),dp(Ax, Sx)dp(By, T y)

1 +dp(Ax, By) ,dp(Ax, T y)dp(By, Ax) 1 +dp(Ax, By)

oiλ

.

Then, as in all the above cases, the mappingsS,T,AandB satisfy the condition (3.1) of Theorem 3.1. So, all the conditions of Theorem 3.1 are satisfied. Moreover, 54 is the unique common fixed point for all of the mappings S,T,A and B.

Theorem 3.3. Let (X, d)be a complete multiplicative metric space. LetS, T, A, B :X →X be single-valued mappings such thatS(X)⊂B(X), T X ⊂AX and there existsλ∈(0,12) such that

dp(Smx, Tqy)≤h ϕ

max

n

dp(Ax, By),dp(Ax, Smx)dp(By, Tqy)

1 +dp(Ax, By) ,dp(Ax, Tqy)dp(By, Ax) 1 +dp(Ax, By)

oiλ

(3.6) for all x, y ∈ X, p ≥1 and m, q ∈ Z+, where ϕ: [0,∞) → [0,∞) is a monotone increasing function such thatϕ(0) = 0 andϕ(t)< t for all t >0.

Assume the following conditions are satisfied:

(a) the pairs(S, A) and (T, B) are commutative mappings;

(b) one of S, T,A andB is continuous.

ThenS, T, A and B have a unique common fixed point in X.

Proof. FromS(X)⊂B(X) andT(X)⊂A(X) we have

Sm(X)⊂Sm−1(X)⊂ · · · ⊂S2(X)⊂S(X)⊂B(X)

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and

Tq(X)⊂Tq−1(X)⊂ · · · ⊂T2(X)⊂T(X)⊂A(X).

Since the pairs (S, A) and (T, B) are commutative mappings, we have

SmA=Sp−1SA=Sm−1AS =Sm−2(SA)S =Sm−2AS2 =· · ·=ASm and

TqB =Tq−1T B =Tq−1BT =Tq−2(T B)T =Tq−2BT2 =· · ·=BTq,

that is,SmA =ASm and TqB =BTq. It follows from Remark 2.9 that the pairs (Sp, A) and (Tq, B) are compatible and also weakly compatible. Therefore, by Theorem 3.1, we can obtain thatSm, Tq, A and B have a unique common fixed pointz∈X.

In addition, we prove thatS,T,A and B have a unique common fixed point. From (3.6), we have dp(Sz, z) =d(Sm(Sz), Tqz)

≤h ϕ

max

dp(ASz, Bz),dp(ASz, SmSz)dp(Bz, Tqz)

1 +dp(ASz, Bz) ,dp(ASz, Tqz)dp(Bz, ASz) 1 +dp(ASz, Bz)

o iλ

=h ϕ

max

dp(Sz, z),dp(Sz, Sz)dp(z, z)

1 +dp(Sz, z) ,dp(Sz, z)dp(z, Sz) 1 +dp(Sz, z)

iλ

≤h

ϕ(dp(Sz, z)) iλ

≤[d(Sz, z)].

This implies thatd(Sz, z) = 1 and so Sz=z. On the other hand, we have d(z, T z) =d(Sm(z), Tq(T z))

≤h ϕ

max

dp(Az, BT z),dp(Az, Smz)dp(BT z, Tqz)

1 +dp(Az, BT z) ,dp(Az, Tq(T z))dp(BT z, Az) 1 +dp(Az, BT z)

o iλ

=h ϕ

max

dp(z, z),dp(z, z)dp(T z, z)

1 +dp(z, T z) ,dp(z, T z)dp(z, T z) 1 +dp(z, T z)

iλ

≤h

ϕ(dp(z, T z))iλ

≤[d(z, T z)].

This implies that d(z, T z) = 1, i.e., T z =z. Therefore, we obtain Sz =T z =Az =Bz =z and so z is a common fixed point ofS,T,A and B.

Finally, we prove that S, T, Aand B have a unique common fixed point z. Suppose that w∈X is also a common fixed point ofS, T, Aand B. Then we have

d(z, w) =d(Sm(z), Tq(w))

≤h ϕ

max

dp(Az, Bz),dp(Az, Smz)dp(Bw, Tqw)

1 +dp(Az, Bz) ,dp(Az, Tq(w))dp(Bw, Az) 1 +dp(Az, Bz)

iλ

max

dp(z, T z),dp(z, z)dp(T z, T z)

1 +dp(z, T z) ,dp(z, T z)dp(T z, z) 1 +dp(z, T z)

iλ

≤h

ϕ(dp(z, T z)) iλ

≤[d(z, T z)].

This implies thatd(z, w) = 1 and sow=z. Therefore,z is a unique common fixed point ofS, T, Aand B. This completes the proof.

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Now, if we take ϕ(t) =tand p= 1 in Theorem 3.1, then we have the following result.

Corollary 3.4. Let(X, d)be a complete multiplicative metric space. LetS, T, A, B :X→Xbe single-valued mappings such thatS(X)⊂B(X), T(X)⊂A(X) and there exists λ∈(0,12) such that, for all x, y∈X,

d(Sx, T y) ≤maxn

dλ(Ax, By),dλ(Ax, Sx)dλ(By, T y)

1 +dλ(Ax, By) ,dλ(Ax, T y)dλ(By, Ax) 1 +dλ(Ax, By)

o

. (3.7)

Suppose that one of the following conditions is satisfied:

(a) eitherA or S is continuous, the pair (S, A) is compatible and the pair (T, B) is weakly compatible;

(b) eitherB or T is continuous, the pair (T, B) is compatible and the pair (S, A) is weakly compatible.

ThenS, T, A and B have a unique common fixed point in X.

Now, if we take ϕ(t) =tand p= 1 in Theorem 3.3, then we have the following result.

Corollary 3.5. Let(X, d)be a complete multiplicative metric space. LetS, T, A, B :X→Xbe single-valued mappings such thatS(X)⊂B(X), T(X)⊂A(X) and there exists λ∈(0,12) such that

d(Smx, Tqy)≤max n

dλ(Ax, By),dλ(Ax, Smx)dλ(By, Tqy)

1 +dλ(Ax, By) ,dλ(Ax, Tqy)dλ(By, Ax) 1 +dλ(Ax, By)

o

(3.8) for allx, y∈X and m, q∈Z+. Assume that the following conditions are satisfied:

(a) the pairs(S, A) and (T, B) are commutative mappings;

(b) one of S, T,A andB is continuous.

ThenS, T, A and B have a unique common fixed point in X.

Now, we give an example to illustrate Corollary 3.4.

Example 3.6. LetX = [0,2] and (X, d) be a multiplicative metric space defined byd(x, y) =e|x−y| for all x, yin X. Let S,T,A and B be four self mappings defined by

Sx= 7

6, if x∈[0,2], T x= 3

2, if x∈[0,1],

7

6, if x∈(1,2], Ax=

1, if x∈[0,1],

7

6, if x∈(1,2),

3

2, if x= 2,

Bx=

1

6, if x∈[0,1],

7

6, if x∈(1,2), 1, if x= 2.

Clearly, we can get S(X) ⊂ B(X) and T(X) ⊂ A(X). Note that T, A and B are not multiplicative continuous mappings and S is multiplicative continuous in X. By the definition of the mappings S and A, we have

d(SAx, ASx) =d 7

6,7 6

= 1≤d(Sx, Ax),

which implies that the pairS, A is weak commuting. Therefore, the pair (S, A) must be compatible.

Clearly, only for x ∈(1,2), T x =Bx= 76 and T Bx = T(76) = 76 =B(76) =BT x and so T Bx =BT x.

Thus the pair (T, B) is also weakly compatible.

Now, we prove that the mappingsS,T,AandB satisfy the Condition (3.7) of Corollary 3.4 withλ= 23. Let

M(x, y) = maxn

d(Ax, By),d(Ax, Sx)d(By, T y)

1 +d(Ax, By) ,d(Ax, T y)d(By, Ax) 1 +d(Ax, By)

oλ

. Now, we have the following 4 cases:

Case 1. If x, y∈[0,1], then we have d(Sx, T y) =d

7 6,3

2

=e13 < e56·23 =d23

1,1 6

=d23(Ax, By)≤M(x, y).

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Case 2. If x∈[0,1] andy∈(1,2], then we have d(Sx, T y) =d

7 6,7

6

= 1≤M(x, y).

Case 3. If x∈(1,2] andy∈[0,1], then we have d(Sx, T y) =d

7 6,3

2

=e13 <1≤M(x, y).

Case 4. If x, y∈(1,2], then we have

d(Sx, T y) =d 7

6,7 6

= 1≤M(x, y).

Then, in all the above cases, the mappings S,T,A and B satisfy the Condition (3.7) of Corollary 3.4 with λ= 23. So, all the conditions of Corollary 3.4 are satisfied. Moreover, 76 is the unique common fixed point for the mappings S,T,Aand B.

4. Common fixed points for mappings with the (CLR)-property

In this section, we prove some common fixed point theorems for weakly compatible mappings satisfying the (CLR) property without completeness of multiplicative metric space.

Theorem 4.1. Let(X, d)be a multiplicative metric space. LetS, T, A, B :X→Xbe single-valued mappings such thatS(X)⊂B(X),T(X)⊂A(X) and there exists λ∈(0,12) such that

dp(Sx, T y)≤h ϕ

maxn

dp(Ax, By),dp(Ax, Sx)dp(By, T y)

1 +dp(Ax, By) ,dp(Ax, T y)dp(By, Ax) 1 +dp(Ax, By)

oiλ

(4.1) for allx, y∈X and p≥1, where ϕ: [0,∞)→[0,∞) is a monotone increasing function such thatϕ(0) = 0 and ϕ(t)< t for all t >0. Assume the following conditions are satisfied:

(a) the pairs(S, A) and (T, B) are weakly compatible;

(b) the pairs (B, T) and (T, B) have the common limit with respect to the value of the mapping A (or B).

ThenS, T, A and B have a unique common fixed point in X.

Proof. Since two pairs (S, A) and (T, B) have the common limit with respect to the value ofA, there exists two sequence {xn} and {yn}inX such that

n→∞lim Axn= lim

n→∞Sxn= lim

n→∞Byn= lim

n→∞T yn=Az for somez∈X.

Now, we show that Sz=Az. By (4.1), we have dp(Sz, T yn)≤h

ϕ

max n

dp(Az, Byn),dp(Az, Sz)dp(Byn, T yn)

1 +dp(Az, Byn) ,dp(Az, T yn)dp(Byn, Az) 1 +dp(Az, Byn)

oiλ

.

Takingn→ ∞in the above inequality, we obtain dp(Sz, Az)≤h

ϕ maxn

dp(Az, Az),dp(Az, Sz)dp(Az, Az)

1 +dp(Az, Az) ,dp(Az, Az)dp(Az, Az) 1 +dp(Az, Az)

oiλ

≤[d(Sz, Az)],

which implies that Sz=Az. Since S(X)⊂B(X), there existsv∈X such thatSz=Bz.

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Next, we showT v=Bv. By (4.1), we have dp(Bv, T v) =d(Sz, T v)≤h

ϕ maxn

dp(Az, Bv),dp(Az, Sz)dp(Bv, T v)

1 +dp(Az, Bv) ,dp(Az, T v)dp(Bv, Az) 1 +dp(Az, Bv)

oiλ

≤[d(Bv, T v)] and so Bv=T v. Therefore, we have

Sz=Az=Bv=T v.

Since the pairsA, S and B, T are weakly compatible, Sz=Az,T v=Bv and

SAz=ASz=AAz =SSz, T Bv=BT v =T T v=BBv. (4.2) Now, we show that Sz is a common fixed point ofS, T, A andB. By (4.1), we obtain

dp(S2z, Sz) =d(S2z, T v)

≤h ϕ

maxn

dp(ASz, Bv),dp(ASz, S2z)dp(Bv, T v)

1 +dp(ASz, Bv) ,dp(ASz, T v)dp(Bv, Az) 1 +dp(ASz, Bv)

oiλ

≤[d(S2z, Sz)].

This implies thatS2z=Sz. Therefore,SSz=ASz=Sz.By (4.1), we obtain dp(T v, T2v) =d(Sz, T2v)

≤h ϕ

maxn

dp(Az, BT v),dp(Az, Sz)dp(BT v, T2v)

1 +dp(Az, BT v) ,dp(Az, T2v)dp(BT v, Az) 1 +dp(ASz, BT v)

oiλ

≤[d(T v, T2v)].

This implies thatT v=T T v. Therefore,T Bv=BT v=T T v, that is, Bv is a common fixed point ofB and T. SinceSv=T v, we have

SSz=ASz=T Sz =BSz

and soSz is a common fixed point of S, T, Aand B. We can obtain the uniqueness of common fixed point z, similarly, in Theorem 3.1. This completes the proof.

If we take p= 1 and ϕ(t) =tin Theorem 4.1, we have the following result.

Corollary 4.2. Let (X, d) be a multiplicative metric space. Let S, T, A, B : X → X be single-valued mappings such thatS(X)⊂B(X), T(X)⊂A(X) and there exists λ∈(0,12) such that

d(Sx, T y)≤h maxn

d(Ax, By),d(Ax, Sx)d(By, T y)

1 +d(Ax, By) ,d(Ax, T y)d(By, Ax) 1 +d(Ax, By)

oiλ

for allx, y∈X. Assume the following conditions are satisfied:

(a) the pairs(S, A) and (T, B) are weakly compatible;

(b) the pairs (B, T) and (T, B) have the common limit with respect to the value of the mapping A (or B).

ThenS, T, A and B have a unique common fixed point in X.

Now, we give an example to illustrate Corollary 4.2.

Example 4.3. LetX = [0,64) be a usual metric space. Define a mappingd:X×X −→Rbyd(x, d) =e|x−y|

for all x, y ∈ X. Then (X, d) is a complete multiplicative metric space. Define the mappings S, T, A, B : X−→X by

Sx= 1

64x, T x= 1

32x, Ax=x, Bx= 2x.

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Then we have have the following:

(1) T(X) = [0,2)⊂[0,64) =A(X);

(2) Clearly, the pairs (S, A) and (T, B) have the common limit of the value of the mapping B;

(3) Clearly, the pairs (S, A) and (T, B) are weakly compatible mappings. For allx, y∈X, d(Sx, T y) =e|641x−321y|≤h

maxn

e|x−2y|,e|x−641x|e|2y−321y|

1 +e|x−2y| ,e|x−321y|e|2y−x|

1 +e|x−2y|

oi23 .

Therefore, all the conditions of Corollary 4.2 are satisfied and, further,S(0) =T(0) =A(0) =B(0) = 0 and so 0 is a unique common fixed point of the mapsS, T, Aand B.

Note that (X, d), in Example 4.3, is not complete. Therefore, Theorem 4.1 cannot be applied.

As a consequence of Theorem 4.1, by putting A=B =Ix, we obtain the following result.

Corollary 4.4. Let (X, d) be a multiplicative metric space. Let S, T, A, B : X → X be single-valued mappings such thatS(X)⊂B(X), T(X)⊂A(X) and there exists λ∈(0,12) such that

dp(Sx, T y)≤h ϕ

maxn

dp(x, y),dp(x, Sx)dp(y, T y)

1 +dp(x, y) ,dp(x, T y)dp(y, Ax) 1 +dp(x, y)

oiλ

for allx, y∈X and p≥1, where ϕ: [0,∞)→[0,∞) is a monotone increasing function such thatϕ(0) = 0 and ϕ(t)< t for all t >0. Assume the following conditions are satisfied:

(a) the pairs(S, A) and (T, B) are weakly compatible;

(b) the pairs (B, T) and (T, B) have the common limit with respect to the value of the mapping A (or B).

ThenS, T, A and B have a unique common fixed point in X.

If we take ϕ(t) =t in Theorem 4.1, we have the following result.

Corollary 4.5. Let (X, d) be a multiplicative metric space. Let S, T, A, B : X → X be single-valued mappings such thatS(X)⊂B(X), T(X)⊂A(X) and there exists λ∈(0,12) such that

dp(Sx, T y)≤h max

n

dp(Ax, By),dp(Ax, Sx)dp(By, T y)

1 +dp(Ax, By) ,dp(Ax, T y)dp(By, Ax) 1 +dp(Ax, By)

oiλ

for allx, y∈X and p≥1. Assume the following conditions are satisfied:

(a) the pairs(S, A) and (T, B) are weakly compatible;

(b) the pairs (B, T) and (T, B) have the common limit with respect to the value of the mapping A (or B).

ThenS, T, A and B have a unique common fixed point in X.

If we take p= 1,A=B and S=T in Corollary 4.5, we have the following result.

Corollary 4.6. Let (X, d) be a multiplicative metric space. Let T, A : X → X be single-valued mappings such thatT(X)⊂A(X) and there exists λ∈(0,12) such that

d(T x, T y)≤h maxn

d(Ax, Ay),d(Ax, T x)d(Ay, T y)

1 +d(Ax, Ay) ,d(Ax, T y)d(Ay, Ax) 1 +d(Ax, Ay)

oiλ

for allx, y∈X. Assume the following conditions are satisfied:

(a) the pair (T, A) is weakly compatible;

(b) the pair(T, A) has the common limit with respect to the value of the mapping A (or B).

ThenT and A have a unique common fixed point in X.

Remark 4.7. In all results in this section, we don’t need the completeness of a multiplicative metric space.

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Acknowledgements

This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah.

The author, therefore, acknowledge with thanks DSR technical and financial support.

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