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

d ( fx , fy ) d ( x , y ) φ ( d ( x , y ) ) ,

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

d ( Tx , Sy ) N ( x , y ) φ ( N ( x , y ) ) ,

for all x,yX, where

N ( x , y ) = max { d ( x , y ) , d ( x , Tx ) , d ( y , Sy ) , 1 2 [ d ( x , Sy ) + d ( y , Tx ) ] } .

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 = T u = S u.

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 [815] 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, 1734]).

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

Definition 1.7

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

Lemma 1.8

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

  1. (1)

    {x n } is G-convergent to x.

  2. (2)

    G(x n ,x n ,x)→0, as n.

  3. (3)

    G(x n ,x,x)→0, as n.

Lemma 1.9

[35] If (X,G) is a G-metric space, then {x n } is a G-Cauchy sequence if and only if for every ε>0, there exists a positive integer N such that G(x n ,x m ,x m )<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 {x n } is G-convergent to x, {f(x n )} 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. 1.

    ψ is continuous and nondecreasing.

  2. 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 xf x, 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 fx= x 1 3 . Then, x x 1 3 =fx, 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 fx= x n for x∈ [ 0,1) and f x = xn for x∈ [ 1,), for any positive integer n. Then for all xX, xf x; 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 f g xx 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 lim n d(fg x n ,gf x n )=0, whenever {x n } is a sequence in X such that lim n f x n = lim n g x n =t, 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 lim n G(fg x n ,fg x n ,gf x n )=0, whenever {x n } is a sequence in X such that lim n f x n = lim n g x n =t, 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 f x = g x implies f g x = g f x.

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

Definition 1.21

Let (X,≼) be a partially ordered set and f,g,h:XX be mappings such that f Xh X and g Xh X. The ordered pair (f,g) is said to be partially weakly increasing with respect to h if for all xX, f xg y, for all yh−1(f x) [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,

ψ ( d ( fx , gy ) ) ψ ( M ( x , y ) ) φ ( M ( x , y ) ) ,

is satisfied where

M ( x , y ) = max { d ( Sx , Ty ) , d ( fx , Sx ) , d ( gy , Ty ) , d ( Sx , gy ) + d ( fx , Ty ) 2 } .

If for a nondecreasing sequence {x n } with x n y n for all n, y n u implies that x n u and either of the following:

  1. (a)

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

  2. (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

M ( x , y , z ) = max { G ( Tx , Ry , Sz ) , G ( Tx , Ry , gy ) , G ( Ry , Sz , hz ) , G ( Sz , Tx , fx ) , G ( Tx , fx , gy ) + G ( Ry , gy , hz ) + G ( Sz , hz , fx ) 3 } .

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,

ψ G ( fx , gy , hz ) ψ M ( x , y , z ) φ M ( x , y , z ) ,
(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 {x n } with x n y n for all n, y n u implies that x n u and either of the following:

  1. (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

  2. (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

  3. (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 f x0=R x1. Since g(X)⊆S(X), we can choose x2X such that g x1=S x2. Also, as h(X)⊆T(X), we can choose x3X such that h x2=T x3.

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

z 3 n + 1 = R x 3 n + 1 = f x 3 n , z 3 n + 2 = S x 3 n + 2 = g x 3 n + 1 ,

and

z 3 n + 3 = T x 3 n + 3 = h x 3 n + 2 ,

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

x 0 f x 0 = R x 1 fR x 1 x 1 g x 1 = S x 2 gS x 2 x 2 h x 2 = T x 3 hT x 3 x 3 .

By continuing this process, we get

x 1 x 2 x 3 x k x k + 1 .

We will complete the proof in three steps.

Step I. We will prove that

lim k G( z k , z k + 1 , z k + 2 )=0.

Define G k =G(z k ,zk+1,zk+2). Suppose G k 0 =0 for some k0. Then, z k 0 = z k 0 + 1 = z k 0 + 2 . Consequently, the sequence {z k } is constant, for kk0. Indeed, let k0=3n. Then z3n=z3n+1=z3n+2, and we obtain from (1),

ψ G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) = ψ G ( f x 3 n , g x 3 n + 1 , h x 3 n + 2 ) ψ M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) φ M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) ,
(2)

where

M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) = max { G ( T x 3 n , R x 3 n + 1 , S x 3 n + 2 ) , G ( T x 3 n , R x 3 n + 1 , g x 3 n + 1 ) , G ( R x 3 n + 1 , S x 3 n + 2 , h x 3 n + 2 ) , G ( S x 3 n + 2 , T x 3 n , f x 3 n ) , G ( T x 3 n , f x 3 n , g x 3 n + 1 ) + G ( R x 3 n + 1 , g x 3 n + 1 , h x 3 n + 2 ) 3 + G ( S x 3 n + 2 , h x 3 n + 2 , f x 3 n ) 3 } = max { G ( z 3 n , z 3 n + 1 , z 3 n + 2 ) , G ( z 3 n , z 3 n + 1 , z 3 n + 2 ) , G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) , G ( z 3 n + 2 , z 3 n , z 3 n + 1 ) , G ( z 3 n , z 3 n + 1 , z 3 n + 2 ) + G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) 3 + G ( z 3 n + 2 , z 3 n + 3 , z 3 n + 1 ) 3 } = max { 0 , 0 , G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) , 0 , 0 + G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) + G ( z 3 n + 2 , z 3 n + 3 , z 3 n + 1 ) 3 } .

Now from (2),

ψ ( G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) ψ ( G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) φ ( G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) ,

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 {z k } is constant (for kk0), and z k 0 is a common fixed point of R,S, T, f,g, and h.

Suppose

G k =G( z k , z k + 1 , z k + 2 )>0
(3)

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

G ( z k + 1 , z k + 2 , z k + 3 ) M ( x k , x k + 1 , x k + 2 ) = G ( z k , z k + 1 , z k + 2 ) .
(4)

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

ψ G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) = ψ G ( f x 3 n , g x 3 n + 1 , h x 3 n + 2 ) ψ M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) φ M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) ψ M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) ,
(5)

where

M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) = max { G ( T x 3 n , R x 3 n + 1 , S x 3 n + 2 ) , G ( T x 3 n , R x 3 n + 1 , g x 3 n + 1 ) , G ( R x 3 n + 1 , S x 3 n + 2 , h x 3 n + 2 ) , G ( S x 3 n + 2 , T x 3 n , f x 3 n ) , G ( T x 3 n , f x 3 n , g x 3 n + 1 ) + G ( R x 3 n + 1 , g x 3 n + 1 , h x 3 n + 2 ) 3 + G ( S x 3 n + 2 , h x 3 n + 2 , f x 3 n ) 3 } = max { G ( z 3 n , z 3 n + 1 , z 3 n + 2 ) , G ( z 3 n , z 3 n + 1 , z 3 n + 2 ) , G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) , G ( z 3 n + 2 , z 3 n , z 3 n + 1 ) , G ( z 3 n , z 3 n + 1 , z 3 n + 2 ) + G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) 3 + G ( z 3 n + 2 , z 3 n + 3 , z 3 n + 1 ) 3 } .

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

G( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 )M( x 3 n , x 3 n + 1 , x 3 n + 2 ).
(6)

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

M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) = G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) .

Therefore, (5) implies that

ψ ( G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) ψ ( G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) φ ( G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ) ,

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

M ( x 3 n , x 3 n + 1 , x 3 n + 2 ) = G ( z 3 n , z 3 n + 1 , z 3 n + 2 ) .

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

G ( z 3 n + 2 , z 3 n + 3 , z 3 n + 4 ) M ( x 3 n + 1 , x 3 n + 2 , x 3 n + 3 ) = G ( z 3 n + 1 , z 3 n + 2 , z 3 n + 3 ) ,
(7)

and

G ( z 3 n + 3 , z 3 n + 4 , z 3 n + 5 ) M ( x 3 n + 2 , x 3 n + 3 , x 3 n + 4 ) = G ( z 3 n + 2 , z 3 n + 3 , z 3 n + 4 ) .
(8)

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

lim k G( z k , z k + 1 , z k + 2 )=r.
(9)

Since

G ( z k + 1 , z k + 2 , z k + 3 ) M ( x k , x k + 1 , x k + 2 ) G ( z k , z k + 1 , z k + 2 ) ,
(10)

letting k in (10), we get

lim k M( x k , x k + 1 , x k + 2 )=r.
(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

lim k G( x k , x k + 1 , x k + 2 )=0,
(12)

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

lim k G( x k , x k + 1 , x k + 1 )=0.
(13)

Step II. We will show that {z n } 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(z m ,z n ,z n )<ε. Suppose the above statement is false. Then there exists ε>0 for which we can find subsequences {zm(k)} and {zn(k)} of {z n } such that n(k)>m(k)≥k and

  1. (a)

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

  2. (b)
    G( z m ( k ) , z n ( k ) , z n ( k ) )ε.
    (14)
  3. (c)

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

    G( z m ( k ) , z n ( k ) 1 , z n ( k ) 1 )<ε.
    (15)

From rectangle inequality and (15), we have

G ( z m ( k ) , z n ( k ) , z n ( k ) ) G ( z m ( k ) , z n ( k ) 1 , z n ( k ) 1 ) + G ( z n ( k ) 1 , z n ( k ) , z n ( k ) ) < ε + G ( z n ( k ) 1 , z n ( k ) , z n ( k ) + 1 ) .
(16)

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

lim k G( z m ( k ) , z n ( k ) , z n ( k ) )=ε.
(17)

Again, from rectangle inequality,

G ( z m ( k ) , z n ( k ) , z n ( k ) + 1 ) G ( z m ( k ) , z n ( k ) , z n ( k ) ) + G ( z n ( k ) , z n ( k ) , z n ( k ) + 1 ) G ( z m ( k ) , z n ( k ) , z n ( k ) ) + G ( z n ( k ) , z n ( k ) + 1 , z n ( k ) + 2 ) ,
(18)

and

G( z m ( k ) , z n ( k ) , z n ( k ) )G( z m ( k ) , z n ( k ) , z n ( k ) + 1 ).
(19)

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

lim k G( z m ( k ) , z n ( k ) , z n ( k ) + 1 )=ε.
(20)

On the other hand,

G ( z m ( k ) , z n ( k ) + 1 , z n ( k ) + 1 ) G ( z m ( k ) , z n ( k ) , z n ( k ) ) + G ( z n ( k ) , z n ( k ) + 1 , z n ( k ) + 1 ) ,
(21)

and

G ( z n ( k ) , z n ( k ) + 1 , z m ( k ) ) G ( z n ( k ) , z n ( k ) + 1 , z n ( k ) + 1 ) + G ( z n ( k ) + 1 , z n ( k ) + 1 , z m ( k ) ) .
(22)

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

lim k G( z m ( k ) , z n ( k ) + 1 , z n ( k ) + 1 )=ε.
(23)

In a similar way, we have

G ( z m ( k ) + 1 , z n ( k ) , z n ( k ) + 1 ) G ( z m ( k ) + 1 , z m ( k ) , z m ( k ) ) + G ( z m ( k ) , z n ( k ) , z n ( k ) + 1 ) 2 G ( z m ( k ) , z m ( k ) + 1 , z m ( k ) + 1 ) + G ( z m ( k ) , z n ( k ) , z n ( k ) + 1 ) ,
(24)

and

G ( z m ( k ) , z n ( k ) , z n ( k ) + 1 ) G ( z m ( k ) , z m ( k ) + 1 , z m ( k ) + 1 ) + G ( z m ( k ) + 1 , z n ( k ) , z n ( k ) + 1 ) ,
(25)

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

lim k G( z m ( k ) + 1 , z n ( k ) , z n ( k ) + 1 )=ε.
(26)

Also,

G( z m ( k ) , z n ( k ) + 1 , z n ( k ) + 1 )G( z m ( k ) , z m ( k ) + 1 , z n ( k ) + 1 ),
(27)

and

G ( z m ( k ) , z m ( k ) + 1 , z n ( k ) + 1 ) G ( z m ( k ) , z m ( k ) + 1 , z m ( k ) + 1 ) + G ( z m ( k ) + 1 , z m ( k ) + 1 , z n ( k ) + 1 ) G ( z m ( k ) , z m ( k ) + 1 , z m ( k ) + 1 ) + G ( z m ( k ) + 1 , z n ( k ) , z n ( k ) + 1 ) .
(28)

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

lim k G( z m ( k ) , z m ( k ) + 1 , z n ( k ) + 1 )=ε.
(29)

Also,

G ( z m ( k ) + 1 , z n ( k ) + 1 , z n ( k ) + 1 ) G ( z m ( k ) + 1 , z n ( k ) + 1 , z n ( k ) )
(30)

and

G ( z m ( k ) + 1 , z n ( k ) , z n ( k ) + 1 ) G ( z m ( k ) + 1 , z n ( k ) + 1 , z n ( k ) + 1 ) + G ( z n ( k ) + 1 , z n ( k ) + 1 , z n ( k ) ) .
(31)

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

lim k G( z m ( k ) + 1 , z n ( k ) + 1 , z n ( k ) + 1 )=ε.
(32)

Finally,

G ( z m ( k ) + 1 , z n ( k ) + 1 , z n ( k ) + 2 ) G ( z m ( k ) + 1 , z n ( k ) + 1 , z n ( k ) + 1 ) + G ( z n ( k ) + 1 , z n ( k ) + 1 , z n ( k ) + 2 ) G ( z m ( k ) + 1 , z n ( k ) + 1 , z n ( k ) + 1 ) + G ( z n ( k ) , z n ( k ) + 1 , z n ( k ) + 2 ) ,
(33)

and

G( z m ( k ) + 1 , z n ( k ) + 1 , z n ( k ) + 1 )G( z m ( k ) + 1 , z n ( k ) + 1 , z n ( k ) + 2 ).
(34)

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

lim k G( z m ( k ) + 1 , z n ( k ) + 1 , z n ( k ) + 2 )=ε.
(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

ψ G ( z m ( k ) + 1 , z n ( k ) + 1 , z n ( k ) + 2 ) = ψ G ( f x m ( k ) , g x n ( k ) , h x n ( k ) + 1 ) ψ M ( x m ( k ) , x n ( k ) , x n ( k ) + 1 ) φ M ( x m ( k ) , x n ( k ) , x n ( k ) + 1 ) ,
(36)

where

M ( x m ( k ) , x n ( k ) , x n ( k ) + 1 ) = max { G ( T x m ( k ) , R x n ( k ) , S x n ( k ) + 1 ) , G ( T x m ( k ) , R x n ( k ) , g x n ( k ) ) , G ( R x n ( k ) , S x n ( k ) + 1 , h x n ( k ) + 1 ) , G ( S x n ( k ) + 1 , T x m ( k ) , f x m ( k ) ) , G ( T x m ( k ) , f x m ( k ) , g x n ( k ) ) + G ( R x n ( k ) , g x n ( k ) , h x n ( k ) + 1 ) + G ( S x n ( k ) + 1 , h x n ( k ) + 1 , f x m ( k ) ) 3 } = max { G ( z m ( k ) , z n ( k ) , z n ( k ) + 1 ) , G ( z m ( k ) , z n ( k ) , z n ( k ) + 1 ) , G ( z n ( k ) , z n ( k ) + 1 , z n ( k ) + 2 ) , G ( z n ( k ) + 1 , z m ( k ) , z m ( k ) + 1 ) , G ( z m ( k ) , z m ( k ) + 1 , z n ( k ) + 1 ) + G ( z n ( k ) , z n ( k ) + 1 , z n ( k ) + 2 ) + G ( z n ( k ) + 1 , z n ( k ) + 2 , z m ( k ) + 1 ) 3 } .

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

ψ(ε)ψ(ε)φ(ε).
(37)

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

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

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

lim n G ( z 3 n + 1 , z 3 n + 1 , z ) = lim n G ( R x 3 n + 1 , R x 3 n + 1 , z ) = lim n G ( f x 3 n , f x 3 n , z ) = 0 ,
(38)
lim n G ( z 3 n + 2 , z 3 n + 2 , z ) = lim n G ( S x 3 n + 2 , S x 3 n + 2 , z ) = lim n G ( g x 3 n + 1 , g x 3 n + 1 , z ) = 0 ,
(39)

and

lim n G ( z 3 n + 3 , z 3 n + 3 , z ) = lim n G ( T x 3 n + 3 , T x 3 n + 3 , z ) = lim n G ( h x 3 n + 2 , h x 3 n + 2 , z ) = 0 .
(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

lim n G ( Tf x 3 n , fT x 3 n , fT x 3 n ) = lim n G ( Tz , fT x 3 n , fT x 3 n ) = 0 ,
(41)

and

lim n G ( Rg x 3 n + 1 , gR x 3 n + 1 , gR x 3 n + 1 ) = lim n G ( Rz , gR x 3 n + 1 , gR x 3 n + 1 ) = 0 .
(42)

Since

R x 3 n + 1 fR x 3 n + 1 x 3 n + 1 g x 3 n + 1 = S x 3 n + 2 gS x 3 n + 2 x 3 n + 2 h x 3 n + 2 = T x 3 n + 3 ,

by using (1) we obtain that

ψ G ( fT x 3 n + 3 , gR x 3 n + 1 , h x 3 n + 2 ) ψ M ( T x 3 n + 3 , R x 3 n + 1 , x 3 n + 2 ) φ M ( T x 3 n + 3 , R x 3 n + 1 , x 3 n + 2 ) ,
(43)

where

M ( T x 3 n + 3 , R x 3 n + 1 , x 3 n + 2 ) = max { G ( TT x 3 n + 3 , RR x 3 n + 1 , S x 3 n + 2 ) , G ( TT x 3 n + 3 , RR x 3 n + 1 , gR x 3 n + 1 ) , G ( RR x 3 n + 1 , S x 3 n + 2 , h x 3 n + 2 ) , G ( S x 3 n + 2 , TT x 3 n + 3 , fT x 3 n + 3 ) , G ( TT x 3 n + 3 , fT x 3 n + 3 , gR x 3 n + 1 ) 3 + G ( RR x 3 n + 1 , gR x 3 n + 1 , h x 3 n + 2 ) 3 + G ( S x 3 n + 2 , h x 3 n + 2 , fT x 3 n + 3 ) 3 } max { G ( Tz , Rz , z ) , G ( Tz , Rz , Rz ) , G ( Rz , z , z ) , G ( z , Tz , Tz ) , G ( Tz , Tz , Rz ) + G ( Rz , Rz , z ) + G ( z , z , Tz ) 3 } ,

as n.

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

ψ G ( Tz , Rz , z ) ψ G ( Tz , Rz , z ) φ G ( Tz , Rz , z ) ,
(44)

and hence, T z = R z = z.

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

ψ G ( fz , g x 3 n + 1 , h x 3 n + 2 ) ψ M ( z , x 3 n + 1 , x 3 n + 2 ) φ M ( z , x 3 n + 1 , x 3 n + 2 ) ,
(45)

where

M ( z , x 3 n + 1 , x 3 n + 2 ) = max { G ( Tz , R x 3 n + 1 , S x 3 n + 2 ) , G ( Tz , R x 3 n + 1 , g x 3 n + 1 ) , G ( R x 3 n + 1 , S x 3 n + 2 , h x 3 n + 2 ) , G ( S x 3 n + 2 , Tz , fz ) , G ( Tz , fz , g x 3 n + 1 ) + G ( R x 3 n + 1 , g x 3 n + 1 , h x 3 n + 2 ) 3 } + G ( S x 3 n + 2 , h x 3 n + 2 , fz ) 3 max { G ( Tz , z , z ) , G ( Tz , z , z ) , G ( z , z , z ) , G ( z , Tz , fz ) , G ( Tz , fz , z ) + G ( z , z , z ) + G ( z , z , fz ) 3 } = G ( z , z , fz ) ,

as n.

If in (45) n, we obtain that

ψ G ( fz , z , z ) ψ G ( fz , z , z ) φ G ( fz , z , z ) ,
(46)

hence f z = z.

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

ψ G ( fz , gz , h x 3 n + 2 ) ψ M ( z , z , x 3 n + 2 ) φ M ( z , z , x 3 n + 2 ) ,
(47)

where

M ( z , z , x 3 n + 2 ) = max { G ( Tz , Rz , S x 3 n + 2 ) , G ( Tz , Rz , gz ) , G ( Rz , S x 3 n + 2 , h x 3 n + 2 ) , G ( S x 3 n + 2 , Tz , fz ) , G ( Tz , fz , gz ) + G ( Rz , gz , h x 3 n + 2 ) + G ( S x 3 n + 2 , h x 3 n + 2 , fz ) 3 } max { G ( Tz , Rz , z ) , G ( Tz , Rz , gz ) , G ( Rz , z , z ) , G ( z , Tz , fz ) , G ( Tz , fz , gz ) + G ( Rz , gz , z ) + G ( z , z , fz ) 3 } = G ( z , z , gz )

as n.

Making n in (47), we obtain that

ψ G ( z , gz , z ) ψ G ( z , z , gz ) φ G ( z , z , gz ) ,
(48)

which implies that g z = z.

Since g(X)⊆S(X), there exists a point wX such that z = g z = S w. Suppose that h wS w. Since zg z = S wg S ww, we have zw. Hence, from (1), we obtain that

ψ G ( fz , gz , hw ) ψ M ( z , z , w ) φ M ( z , z , w ) ,
(49)

where

M ( z , z , w ) = max { G ( Tz , Rz , Sw ) , G ( Tz , Rz , gz ) , G ( Rz , Sw , hw ) , G ( Sw , Tz , fz ) , G ( Tz , fz , gz ) + G ( Rz , gz , hw ) + G ( Sw , hw , fz ) 3 } max { G ( z , z , z ) , G ( z , z , z ) , G ( z , z , hw ) , G ( z , z , z ) , G ( z , z , z ) + G ( z , z , hw ) + G ( z , hw , z ) 3 } = G ( z , z , hw ) ,

as n.

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

ψ G ( z , z , hw ) ψ G ( z , z , hw ) φ G ( z , z , hw ) ,
(50)

which yields that h w = z.

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

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

Since x3nf x3n and f x3nz, as n, we have x3nz. Hence, from (1),

ψ G ( f x 3 n , gz , hz ) ψ M ( x 3 n , z , z ) φ M ( x 3 n , z , z ) ,
(51)

where

M ( x 3 n , z , z ) = max { G ( T x 3 n , Rz , Sz ) , G ( T x 3 n , Rz , gz ) , G ( Rz , Sz , hz ) , G ( Sz , T x 3 n , f x 3 n ) , G ( T x 3 n , f x 3 n , gz ) + G ( Rz , gz , hz ) + G ( Sz , hz , f x 3 n ) 3 } max { G ( z , z , z ) , G ( z , z , z ) , G ( z , z , hz ) , G ( z , z , z ) , G ( z , z , z ) + G ( z , z , hz ) + G ( z , hz , z ) 3 } = G ( z , z , hz ) ,

as n.

Letting n in (51), we obtain that

ψ G ( z , z , hz ) ψ G ( z , z , hz ) φ G ( z , z , hz ) ,
(52)

hence h z = z. Therefore, f z = g z = h z = R z = S z = T z = 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 f u = g u = h u = R u = S u = T u = u, f v = g v = h v = R v = S v = T v = v, and uv. By using (1), we obtain

ψ G ( fu , gv , hv ) ψ M ( u , v , v ) φ M ( u , v , v ) ,
(53)

where

M ( u , v , v ) = max { G ( Tu , Rv , Sv ) , G ( Tu , Rv , gv ) , G ( Rv , Sv , hv ) , G ( Sv , Tu , fu ) , G ( Tu , fu , gv ) + G ( Rv , gv , hv ) + G ( Sv , hv , fu ) 3 } = max { G ( u , v , v ) , G ( v , u , u ) } .

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

ψ G ( fv , gu , hu ) ψ M ( v , u , u ) φ M ( v , u , u ) ,
(54)

where

M ( v , u , u ) = max { G ( Tv , Ru , Su ) , G ( Tv , Ru , gu ) , G ( Ru , Su , hu ) , G ( Su , Tv , fv ) , G ( Tv , fv , gu ) + G ( Ru , gu , hu ) + G ( Su , hu , fv ) 3 } = max { G ( v , u , u ) , G ( u , v , v ) } .

From (53) and (54),

ψ max { G ( u , v , v ) , G ( v , u , u ) } = max { ψ G ( u , v , v ) , ψ G ( v , u , u ) } ψ max { G ( u , v , v ) , G ( v , u , u ) } φ max { G ( u , v , v ) , G ( v , u , u ) } .
(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

M 1 ( x , y , z ) = max { G ( Tx , Ry , Sz ) , G ( Tx , Ry , fy ) , G ( Ry , Sz , fz ) , G ( Sz , Tx , fx ) , G ( Tx , fx , fy ) + G ( Ry , fy , fz ) + G ( Sz , fz , fx ) 3 } .

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,

ψ G ( fx , fy , fz ) ψ M 1 ( x , y , z ) φ M 1 ( x , y , z ) ,
(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 {x n } with x n y n for all n, y n u implies that x n u and either of the following:

  1. (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

  2. (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

  3. (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

M 2 ( x , y , z ) = max { G ( Tx , Ty , Tz ) , G ( Tx , Ty , gy ) , G ( Ty , Tz , hz ) , G ( Tz , Tx , fx ) , G ( Tx , fx , gy ) + G ( Ty , gy , hz ) + G ( Tz , hz , fx ) 3 } .

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,

ψ G ( fx , gy , hz ) ψ M 2 ( x , y , z ) φ M 2 ( x , y , z ) ,
(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 {x n }, with x n y n for all n, y n u implies that x n u and either of the following:

  1. (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

  2. (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

  3. (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

M 3 ( x , y , z ) = max { G ( Sx , Ry , Sz ) , G ( Sx , Ry , gy ) , G ( Ry , Sz , gz ) , G ( Sz , Sx , fx ) , G ( Sx , fx , gy ) + G ( Ry , gy , gz ) + G ( Sz , gz , fx ) 3 } .

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,

ψ G ( fx , gy , gz ) ψ M 3 ( x , y , z ) φ M 3 ( x , y , z ) ,
(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 {x n } with x n y n for all n, y n u implies that x n u and either of the following:

  1. (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

  2. (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

  3. (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

M 4 ( x , y , z ) = max { G ( Tx , Ty , Tz ) , G ( Tx , Ty , fy ) , G ( Ty , Tz , fz ) , G ( Tz , Tx , fx ) , G ( Tx , fx , fy ) + G ( Ty , fy , fz ) + G ( Tz , fz , fx ) 3 } .

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,

ψ G ( fx , fy , fz ) ψ M 4 ( x , y , z ) φ M 4 ( x , y , z ) ,
(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 {x n }, x n y n for all n, and y n u implies that x n u 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:

xyyx,x,yX.
(60)

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

fx = ln ( 1 + x ) , Rx = e 3 x 1 , gx = ln ( 1 + x 2 ) , Sx = e 2 x 1 , hx = ln ( 1 + x 3 ) , Tx = e 6 x 1 .
(61)

For each xX, we have 1+xex,1+ x 2 e x and 1+ x 3 e x . Hence, f x = ln(1+x)≤x, gx=ln(1+ x 2 )x, and hx=ln(1+ x 3 )x, which yields that xf x, xg x, and xh x, so f, g, and h are dominating.

Also, for each xX, we have f R x = ln(1+R x)=3xx,

gSx = ln ( 1 + Sx 2 ) = ln ( 1 + e 2 x 1 2 ) = ln ( 1 + e 2 x 2 ) = ln ( e x e x + e x 2 ) = x + ln ( e x + e x 2 ) x ,

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

hTx = ln ( 1 + Tx 3 ) = ln ( 1 + e 6 x 1 3 ) = ln ( 2 + e 6 x 3 ) x.

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

Furthermore, f X = T X = g X = S X = h X = R X = [ 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 {x n } is a sequence in X such that for some tX, lim n G(t,f x n ,f x n )=0 and lim n G(t,T x n ,T x n )=0. Therefore, we have

lim n f x n t = lim n T x n t = 0 .

Since f and T are continuous, we have

lim n G ( fT x n , fT x n , Tf x n ) = 2 lim n fT x n Tf x n = 2 ft Tt = 2 ln ( 1 + t ) e 6 t + 1 .

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

Define control functions ψ,φ: [ 0,)→ [ 0,) with ψ(t)=b t 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

ψ ( G ( fx , gy , hz ) ) = b ( fx gy + fx hz + gy hz ) = b ( ln ( x + 1 ) ln ( y 2 + 1 ) + ln ( x + 1 ) ln ( z 3 + 1 ) + ln ( y 2 + 1 ) ln ( z 3 + 1 ) ) b ( 1 2 2 x y + 1 3 3 x z + 1 6 3 y 2 z ) = b ( 6 x 3 y + 6 x 2 z + 3 y 2 z ) 6 b 6 ( e 6 x e 3 y + e 3 y e 2 z + e 2 z e 6 x ) Tx Ry + Ry Sz + Sz Tx = G ( Tx , Ry , Sz ) M ( x , y , z ) = ψ ( M ( x , y , z ) ) φ ( M ( x , y , z ) ) .

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:

  1. (I)

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

  2. (II)

    For all ε>0, 0 ε μ(t)dt>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

0 ψ G ( fx , gy , hz ) μ ( t ) dt 0 ψ M ( x , y , z ) μ ( t ) dt 0 φ M ( x , y , z ) μ ( t ) dt.
(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 Γ(x)= 0 x μ(t)dt. Then (62) becomes

Γ ψ ( G ( fx , gy , hz ) ) Γ ψ ( M ( x , y , z ) ) Γ φ ( M ( x , y , z ) ) .

Taking ψ1=Γ o ψ and φ1=Γ o φ 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 {μ i }1≤iN be a family of N functions which belong to Λ. For all t≥0, we define

I 1 ( t ) = 0 t μ 1 ( s ) ds , I 2 ( t ) = 0 I 1 t μ 2 ( s ) ds = 0 0 t μ 1 ( s ) ds μ 2 ( s ) ds , I 3 ( t ) = 0 I 2 t μ 3 ( s ) ds = 0 0 0 t μ 1 ( s ) ds μ 2 ( s ) ds μ 3 ( s ) ds , , I N ( t ) = 0 I ( N 1 ) t μ N ( s ) ds.

We have the following result.

Corollary 2.11

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

I N ψ G ( fx , gy , hz ) I N ψ M ( x , y , z ) I N φ M ( x , y , z ) .
(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 Ψ ̂ = I N and Φ ̂ = I N .