# Common fixed points of six mappings in partially ordered G-metric spaces

Vahid Parvaneh1*, Abdolrahman Razani2 and Jamal Rezaei Roshan3

Author Affiliations

1 Department of Mathematics, Gilan-E-Gharb Branch, Islamic Azad University, Gilan-E-Gharb 67871-54699, Iran

2 Department of Mathematics, Faculty of Science, Imam Khomeini International University, P.O. Box 34149-16818, Qazvin, Iran

3 Department of Mathematics, Qaemshahr Branch, Islamic Azad University, Qaemshahr 35953-47631, Iran

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Mathematical Sciences 2013, 7:18  doi:10.1186/2251-7456-7-18

 Received: 28 January 2013 Accepted: 3 March 2013 Published: 8 April 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

The aim of this paper is to present some common fixed point results for six selfmappings satisfying generalized weakly (ψ,φ)-contractive condition in the setup of partially ordered G-metric spaces. Our results extend and generalize the comparable results in the work of Abbas from the context of ordered metric spaces to the setup of ordered G-metric spaces. Also, our results are supported by an example.

##### Keywords:
Common fixed point; Generalized weakly contraction; Generalized metric space; Partially ordered set; Weak annihilator map; Dominating map

### Introduction and preliminaries

Alber and Guerre-Delabrere [1] defined weakly contractive mappings on Hilbert spaces as follows:

#### Definition 1.1

A mapping f:XX is said to be a weakly contractive mapping if

where x,yX and φ: [0,)→ [0,) is a continuous and nondecreasing function such that φ(t) = 0 if and only if t = 0.

#### Theorem 1.2

[2] Let (X,d) be a complete metric space and f:XX be a weakly contractive mapping. Then f has a unique fixed point.

Recently, Zhang and Song [3] have introduced the concept of a generalized φ-weak contractive condition and obtained a common fixed point for two maps.

#### Definition 1.3

Two mappings T,S:XX are called generalized φ-weak contractions if there exists a lower semicontinuous function φ: [ 0,)→ [ 0,) with φ(0)=0 and φ(t)>0 for all t>0 such that

for all x,yX, where

Zhang and Song proved the following theorem.

#### Theorem 1.4

Let (X,d) be a complete metric space and T,S:XX be generalized φ-weak contractive mappings where φ: [ 0,)→ [ 0,) is a lower semicontinuous function with φ(0) = 0 and φ(t)>0 for all t>0. Then, there exists a unique point uX such that u = Tu = Su.

Dorić [4], Moradi et al. [5], Abbas and Dorić [6], and Razani et al. [7] obtained some common fixed point theorems which are extensions of the result of Zhang and Song in the framework of complete metric spaces. Also, in these years many authors have focused on different contractive conditions in complete metric spaces with a partially order and have obtained some common fixed point theorems. For more details on fixed point theory, its applications, comparison of different contractive conditions and related results in ordered metric spaces we refer the reader to [8-15] and the references mentioned therein.

The concept of a generalized metric space, or a G-metric space, was introduced by Mustafa and Sims [16]. In recent years, many authors have obtained different fixed point theorems for mappings satisfying various contractive conditions on G-metric spaces (see e.g., [9,17-34]).

#### Definition 1.5

[16] (G-metric space) Let X be a nonempty set and G:X×X×XR+ be a function satisfying the following properties:

(G1) G(x,y,z)=0 iff x = y = z;

(G2) 0<G(x,x,y), for all x,yX with xy;

(G3) G(x,x,y)≤G(x,y,z), for all x,y,zX with zy;

(G4) G(x,y,z)=G(x,z,y)=G(y,z,x)=⋯, (symmetry

in all three variables);

(G5) G(x,y,z)≤G(x,a,a)+G(a,y,z), for all x,y,z,aX

(rectangle inequality).

Then the function G is called a G-metric on X and the pair (X,G) is called a G-metric space.

#### Definition 1.6

[16] Let (X,G) be a G-metric space and let {xn} be a sequence of points in X. A point xX is said to be the limit of the sequence {xn} and if and one says that the sequence {xn} is G-convergent to x. Thus, if xnx in a G-metric space (X,G), then for any ε>0, there exists a positive integer N such that G(x,xn,xm)<ε, for all n,mN.

#### Definition 1.7

[16] Let (X,G) be a G-metric space. A sequence {xn} is called G-Cauchy if for every ε>0, there is a positive integer N such that G(xn,xm,xl)<ε, for all n,m,lN, that is, if G(xn,xm,xl)→0, as n,m,l.

#### Lemma 1.8

[16] Let (X,G) be a G-metric space. Then the following are equivalent:

(1) {xn} is G-convergent to x.

(2) G(xn,xn,x)→0, as n.

(3) G(xn,x,x)→0, as n.

#### Lemma 1.9

[35] If (X,G) is a G-metric space, then {xn} is a G-Cauchy sequence if and only if for every ε>0, there exists a positive integer N such that G(xn,xm,xm)<ε, for all m>nN.

#### Definition 1.10

[16] A G-metric space (X,G) is said to be G-complete if every G-Cauchy sequence in (X,G) is convergent in X.

#### Definition 1.11

[16] Let (X,G) and (X,G) be two G-metric spaces. A function f:XX is G-continuous at a point xX if and only if it is G-sequentially continuous at x, that is, whenever {xn} is G-convergent to x, {f(xn)} is G-convergent to f(x).

#### Definition 1.12

A G-metric on X is said to be symmetric if G(x,y,y)=G(y,x,x), for all x,yX.

The concept of an altering distance function was introduced by Khan et al. [36] as follows.

#### Definition 1.13

The function ψ : [ 0,)→ [ 0,) is called an altering distance function if the following conditions hold:

1. ψ is continuous and nondecreasing.

2. ψ(t)=0 if and only if t = 0.

#### Definition 1.14

[8] Let (X,≼) be a partially ordered set. A mapping f is called a dominating map on X if xfx, for each x in X.

#### Example 1.15

[8] Let X = [ 0,1] be endowed with the usual ordering. Let f:XX be defined by Then, , for all xX. Thus, f is a dominating map.

#### Example 1.16

[8] Let X = [ 0,) be endowed with the usual ordering. Let f:XX be defined by for x∈ [ 0,1) and fx = xn for x∈ [ 1,), for any positive integer n. Then for all xX, xfx; that is, f is a dominating map.

A subset W of a partially ordered set X is said to be well ordered if every two elements of W be comparable [8].

#### Definition 1.17

[8] Let (X,≼) be a partially ordered set. A mapping f is called a weak annihilator of g if fgxx for all xX.

Jungck in [37] introduced the following definition.

#### Definition 1.18

[37] Let (X,d) be a metric space and f,g:XX be two mappings. The pair (f,g) is said to be compatible if and only if , whenever {xn} is a sequence in X such that , for some tX.

#### Definition 1.19

[38,39] Let (X,G) be a G-metric space and f,g:XX be two mappings. The pair (f,g) is said to be compatible if and only if , whenever {xn} is a sequence in X such that , for some tX.

#### Definition 1.20

[40] Let f and g be two self mappings of a metric space (X,d). The f and g are said to be weakly compatible if for all xX, the equality fx = gx implies fgx = gfx.

Let X be a non-empty set and f:XX be a given mapping. For every xX, let f−1(x)={uX:fu = x}.

#### Definition 1.21

Let (X,≼) be a partially ordered set and f,g,h:XX be mappings such that fXhX and gXhX. The ordered pair (f,g) is said to be partially weakly increasing with respect to h if for all xX, fxgy, for all yh−1(fx) [41].

Since we are motivated by the work in [8] in this paper, we prove some common fixed point theorems for nonlinear generalized (ψ,φ)-weakly contractive mappings in partially ordered G-metric spaces.

### Main results

Abbas et al. [8] proved the following theorem.

#### Theorem 2.1

Let (X,≼,d) be an ordered complete metric space. Let f, g, S and T be selfmaps on X, (T,f) and (S,g) be partially weakly increasing with f(X)⊆T(X) and g(X)⊆S(X), dominating maps f and g be weak annihilators of T and S, respectively. Suppose that there exist altering distance functions ψ and φ such that for every two comparable elements x,yX,

is satisfied where

If for a nondecreasing sequence {xn} with xnyn for all n, ynu implies that xnu and either of the following:

(a) (f,S) are compatible, f or S is continuous, and (g,T) are weakly compatible or

(b) (g,T) are compatible, g or T is continuous, and (f,S) are weakly compatible,

then f, g, S, and T have a common fixed point. Moreover, the set of common fixed points of f, g, S and T is well ordered if and only if f, g, S, and T have one and only one common fixed point.

Let (X,≼,G) be an ordered G-metric space and f,g,h,R,S,T:XX be six self mappings. In the rest of this paper, unless otherwise stated, for all x,y,zX, let

Our first result is the following.

#### Theorem 2.2

Let (X,≼,G) be a partially ordered complete G-metric space. Let f,g,h,R,S,T:XX be the six mappings such that f(X)⊆R(X), g(X)⊆S(X), h(X)⊆T(X) and dominating maps f, g, and h are weak annihilators of R, S, and T, respectively. Suppose that for every three comparable elements x,y,zX,

(1)

where ψ,φ: [ 0,)→ [ 0,) are altering distance functions. Then, f, g, h, R, S, and T have a common fixed point in X provided that for a nondecreasing sequence {xn} with xnyn for all n, ynu implies that xnu and either of the following:

(i) One of g or R and one of f or T are continuous, the pairs (f,T) and (g,R) are compatible, and the pair (h,S) is weakly compatible or

(ii) One of h or S and one of f or T are continuous, the pairs (f,T) and (h,S) are compatible, and the pair (g,R) is weakly compatible or

(iii) One of g or R and one of h or S are continuous, the pairs (g,R) and (h,S) are compatible, and the pair (f,T) is weakly compatible.

Moreover, the set of common fixed points of f, g, h, R, S, and T is well ordered if and only if f, g, h, R, S, and T have one and only one common fixed point.

#### Proof 2.3

Let x0X be an arbitrary point. Since f(X)⊆R(X), we can choose x1X such that fx0=Rx1. Since g(X)⊆S(X), we can choose x2X such that gx1=Sx2. Also, as h(X)⊆T(X), we can choose x3X such that hx2=Tx3.

Continuing this process, we can construct a sequence {zn} defined by

and

for all n≥0.

Now, since f, g and h are dominating and f, g, and h are weak annihilators of R, S and T, we obtain that

By continuing this process, we get

We will complete the proof in three steps.

Step I. We will prove that

Define Gk=G(zk,zk+1,zk+2). Suppose for some k0. Then, . Consequently, the sequence {zk} is constant, for kk0. Indeed, let k0=3n. Then z3n=z3n+1=z3n+2, and we obtain from (1),

(2)

where

Now from (2),

and so, φ(G(z3n+1,z3n+2,z3n+3))=0, that is, z3n+1=z3n+2=z3n+3.

Similarly, if k0=3n+1 or k0=3n+2, one can easily obtain that z3n+2=z3n+3=z3n+4 and z3n+3=z3n+4=z3n+5, and so the sequence {zk} is constant (for kk0), and is a common fixed point of R,S, T, f,g, and h.

Suppose

(3)

for all k. We prove that for each k = 1,2,3,⋯

(4)

Let k = 3n. Since xk−1xk, using (1) we obtain that

(5)

where

Since ψ is a nondecreasing function from (5), we get

(6)

If for an n≥0, G(z3n+1,z3n+2,z3n+3)>G(z3n,z3n+1, z3n+2)>0, then

Therefore, (5) implies that

which is only possible when G(z3n+1,z3n+2,z3n+3)=0. This is a contradiction to (3). Hence, G(z3n+1,z3n+2,z3n+3) ≤G(z3n,z3n+1,z3n+2) and

Therefore, (4) is proved for k = 3n. Similarly, it can be shown that

(7)

and

(8)

Hence, we conclude that {G(zk,zk+1,zk+2)} is a nondecreasing sequence of nonnegative real numbers. Thus, there is an r≥0 such that

(9)

Since

(10)

letting k in (10), we get

(11)

Letting n in (5) and using (9) and (11) and the continuity of ψ and φ, we get ψ(r)≤ψ(r)−φ(r)≤ψ(r) and hence φ(r)=0. This gives us

(12)

from our assumptions about φ. Also, from Definition 1.5, part (G3), we have

(13)

Step II. We will show that {zn} is a G-Cauchy sequence in X. Therefore, we will show that for every ε>0, there exists a positive integer k such that for all m,nk, G(zm,zn,zn)<ε. Suppose the above statement is false. Then there exists ε>0 for which we can find subsequences {zm(k)} and {zn(k)} of {zn} such that n(k)>m(k)≥k and

(a) m(k)=3t and n(k)=3t+1, where t and t are nonnegative integers.

(b)

(14)

(c) n(k) is the smallest number such that the condition (b) holds; i.e.,

(15)

From rectangle inequality and (15), we have

(16)

Making k in (16) from (12) and (15), we conclude that

(17)

Again, from rectangle inequality,

(18)

and

(19)

Hence, in (18) and (19), if k, using (12), (14), and (17), we have

(20)

On the other hand,

(21)

and

(22)

Hence, in (21) and (22), if k is from (13), (17), and (20), we have

(23)

In a similar way, we have

(24)

and

(25)

and therefore, from (24) and (25) by taking limit when k, using (13) and (20), we get that

(26)

Also,

(27)

and

(28)

Hence in (27) and (28), if k from (13), (23), and (25), we have

(29)

Also,

(30)

and

(31)

So from (13), (26), (29), and (30), we have

(32)

Finally,

(33)

and

(34)

Hence in (33) and (34), if k and by using (12) and (32), we have

(35)

Since xm(k)xn(k)xn(k)+1, putting x = xm(k), y = xn(k), and z = xn(k)+1 in (1) for all k≥0, we have

(36)

where

Now, from (13), (19), (26), and (35), if k in (36), we have

(37)

Hence, ε = 0, which is a contradiction. Consequently, {zn} is a G-Cauchy sequence.

Step III. We will show that f, g, h, R, S, and T have a common fixed point.

Since {zn} is a G-Cauchy sequence in the complete G-metric space X, there exists zX such that

(38)

(39)

and

(40)

Let (i) holds. Assume that R and T are continuous and let the pairs (f,T) and (g,R) are compatible. This implies that

(41)

and

(42)

Since

by using (1) we obtain that

(43)

where

as n.

On taking the limit as n in (43), we obtain that

(44)

and hence, Tz = Rz = z.

Since x3n+1x3n+2hx3n+2 and hx3n+2z, as n, we have x3n+1x3n+2z. Therefore, from (1),

(45)

where

as n.

If in (45) n, we obtain that

(46)

hence fz = z.

Since x3n+2hx3n+2 and hx3n+2z, as n, we have x3n+2z. Hence from(1),

(47)

where

as n.

Making n in (47), we obtain that

(48)

which implies that gz = z.

Since g(X)⊆S(X), there exists a point wX such that z = gz = Sw. Suppose that hwSw. Since zgz = SwgSww, we have zw. Hence, from (1), we obtain that

(49)

where

as n.

On taking the limit as n in (49), we obtain that

(50)

which yields that hw = z.

Now, Since h and S are weakly compatible, we have hz = hSw = Shw = Sz. Thus, z is a coincidence point of h and S.

Now, we are ready to show that hz = z.

Since x3nfx3n and fx3nz, as n, we have x3nz. Hence, from (1),

(51)

where

as n.

Letting n in (51), we obtain that

(52)

hence hz = z. Therefore, fz = gz = hz = Rz = Sz = Tz = z.

Similarly, the result follows when (ii) or (iii) hold.

Suppose that the set of common fixed points of f, g, h, R, S, and T is well ordered. We claim that common fixed point of f, g, h, R, S, and T is unique. Assume on contrary that fu = gu = hu = Ru = Su = Tu = u, fv = gv = hv = Rv = Sv = Tv = v, and uv. By using (1), we obtain

(53)

where

On the other hand, as v and u are comparable,

(54)

where

From (53) and (54),

(55)

Therefore, φ(max{G(u,v,v),G(v,u,u)})=0 which yields that u = v is a contradiction. Conversely, if f, g, h, R, S, and T have only one common fixed point then, clearly, the set of common fixed points of f, g, h, R, S, and T is well ordered.

We assume that

Taking f = g = h in Theorem 2.2, we obtain the following common fixed point result in corollary.

#### Corollary 2.4

Let (X,≼,G) be a partially ordered complete G-metric space. Let f,R,S,T:XX be four mappings such that f(X)⊆R(X)∪S(X)∪T(X) and dominating map f is a weak annihilator of R, S, and T. Suppose that for every three comparable elements x,y,zX,

(56)

where ψ,φ: [ 0,)→ [ 0,) are altering distance functions. Then, f, R, S, and T have a common fixed point in X provided that for a nondecreasing sequence {xn} with xnyn for all n, ynu implies that xnu and either of the following:

(i) One of f or R and one of f or T are continuous, the pairs (f,T), and (f,R) are compatible, and the pair (f,S) is weakly compatible or

(ii) One of f or S and one of f or T are continuous, the pairs (f,T), and (f,S) are compatible, and the pair (f,R) is weakly compatible or

(iii) One of f or R and one of f or S are continuous, the pairs (f,R), and (f,S) are compatible, and the pair (f,T) is weakly compatible.

Moreover, the set of common fixed points of f, R, S, and T is well ordered if and only if f, R, S, and T have one and only one common fixed point.

Let

Taking T = R = S in Theorem 2.2, we obtain the following common fixed point result.

#### Corollary 2.5

Let (X,≼,G) be a partially ordered complete G-metric space. Let f,g,h,T:XX be four mappings such that f(X)∪g(X)∪h(X)⊆T(X) and dominating maps f, g, and h are weak annihilators of T. Suppose that for every three comparable elements x,y,zX,

(57)

where ψ,φ: [ 0,)→ [ 0,) are altering distance functions. Then, f, g, h, and T have a common fixed point in X provided that for a nondecreasing sequence {xn}, with xnyn for all n, ynu implies that xnu and either of the following:

(i) One of f or T and one of g or T are continuous, the pairs (f,T) and (g,T) are compatible, and the pair (h,T) is weakly compatible or

(ii) One of f or T and one of h or T are continuous, the pairs (f,T) and (h,T) are compatible, and the pair (g,T) is weakly compatible or

(iii) One of g or T and one of h or T are continuous, the pairs (g,T) and (h,T) are compatible, and the pair (f,T) is weakly compatible.

Moreover, the set of common fixed points of f, g, h, and T is well ordered if and only if f, g, h, and T have one and only one common fixed point.

Let

Taking S = T and g = h in Theorem 2.2, we obtain the following common fixed point result.

#### Corollary 2.6

Let (X,≼,G) be a partially ordered complete G-metric space. Let f,g,R,S:XX be four mappings such that f(X)⊆R(X) and g(X)⊆S(X) and dominating maps f and g are weak annihilators of R and S, respectively. Suppose that for every three comparable elements x,y,zX,

(58)

where ψ,φ: [ 0,)→ [ 0,) are altering distance functions. Then, f, g, R, and S have a common fixed point in X provided that for a nondecreasing sequence {xn} with xnyn for all n, ynu implies that xnu and either of the following:

(i) One of g or R and one of f or S are continuous, the pairs (f,S) and (g,R) are compatible, and the pair (g,S) is weakly compatible or

(ii) One of g or S and one of f or S are continuous, the pairs (f,S) and (g,S) are compatible, and the pair (g,R) is weakly compatible or

(iii) One of g or R and one of g or S are continuous, the pairs (g,R) and (g,S) are compatible, and the pair (f,S) is weakly compatible.

Moreover, the set of common fixed points of f, g, R and S is well ordered if and only if f, g, R and S have one and only one common fixed point.

Let

Taking R = S = T and f = g = h in Theorem 2.2, we obtain the following common fixed point result:

#### Corollary 2.7

Let (X,≼,G) be a partially ordered complete G-metric space. Let f,T:XX be two mappings such that f(X)⊆T(X), dominating map f is a weak annihilator of T. Suppose that for every three comparable elements x,y,zX,

(59)

where ψ,φ: [ 0,)→ [ 0,) are altering distance functions. Then, f and T have a common fixed point in X provided that for a nondecreasing sequence {xn}, xnyn for all n, and ynu implies that xnu and one f or T is continuous and the pair (f,T) is compatible.

Moreover, the set of common fixed points of f and T is well ordered if and only if f and T have one and only one common fixed point.

#### Example 2.8

(see also [42]) Let X = [ 0,) and G on X be given by G(x,y,z)=|xy|+|yz|+|xz|, for all x,y,zX. We define an ordering ‘ ≼’ on X as follows:

(60)

Define self-maps f, g, h, S, T and R on X by

(61)

For each xX, we have 1+xex, and Hence, fx = ln(1+x)≤x, , and , which yields that xfx, xgx, and xhx, so f, g, and h are dominating.

Also, for each xX, we have fRx = ln(1+Rx)=3xx,

and since t6−3t+2≥0 for each t≥1, we have

Hence, fRxx, gSxx and hTxx. Thus f, g, and h are weak annihilators of S, T, and R, respectively.

Furthermore, fX = TX = gX = SX = hX = RX = [ 0,) and the pairs (f,T), (g,R), and (h,S) are compatible. For example, we will show that the pair (f,T) is compatible. Let {xn} is a sequence in X such that for some tX, and Therefore, we have

Since f and T are continuous, we have

On the other hand, |ln(1+t)−e6t+1|=0⇔t = 0.

Define control functions ψ,φ: [ 0,)→ [ 0,) with ψ(t)=bt and φ(t)=(b−1)t for all t∈ [ 0,), where 1<b≤6.

Now, we will show that f, g, h, R, S and T satisfy (1). Using the mean value theorem, we have

Thus, (1) is satisfied for all x,y,zX. Therefore, all the conditions of the Theorem 2.2 are satisfied. Moreover, 0 is the unique common fixed point of f, g, h, R, S, and T.

Denoted by Λ, the set of all functions μ: [ 0+)→ [ 0,+), verifying the following conditions:

(I) μ is a positive Lebesgue integrable mapping on each compact subset of [ 0,+).

(II) For all ε>0, .

Other consequences of the main theorem are the following results for mappings satisfying contractive conditions of integral type.

#### Corollary 2.9

We replaced the contractive condition (1) of Theorem 2.2 by the following condition: There exists a μΛ such that

(62)

Then, f, g, h, R, S, and T have a coincidence point, if the other conditions of Theorem 2.2 be satisfied.

#### Proof 2.10

Consider the function . Then (62) becomes

Taking ψ1=Γoψ and φ1=Γoφ and applying Theorem 2.2, we obtain the proof (it is easy to verify that ψ1 and φ1 are altering distance functions).

Similar to [43], let N be a fixed positive integer. Let {μi}1≤iN be a family of N functions which belong to Λ. For all t≥0, we define

We have the following result.

#### Corollary 2.11

We replaced the inequality (1) of Theorem 2.2 by the following condition:

(63)

Then, f, g, h, R, S, and T have a coincidence point, if the other conditions of Theorem 2.2 be satisfied.

#### Proof 2.12

We consider that and .

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

VP, AR, and JRR have worked together on each section of the paper such as the literature review, results and examples. All authors read and approved the final manuscript.

### Acknowledgments

The authors thank the referees for the extremely careful reading that contributed to the improvement of the manuscript.

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