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# Tripled coincidence point results for generalized contractions in ordered generalized metric spaces

## Abstract

In this paper, we establish some tripled coincidence point results for a mixed g-monotone mapping F : X3X satisfying (ψ, ϕ)-contractions in ordered generalized metric spaces. Also, an application and some examples are given to support our results.

## Introduction and preliminaries

Banach contraction principle is one of the core subject that has been studied. It has so many different generalizations with different approaches. One of the remarkable generalizations, known as Φ-contraction, was given by Boyd and Wong [1] in 1969. In 1997, Alber and Guerre-Delabriere [2], introduced the notion of a weak ϕ-contraction which generalizes Boyd and Wong results, so Banach's result. Very recently, inspired from the notion of weak ϕ-contractions, a new concept of ( ψ, ϕ)-contractions was introduced (see e.g. [37]).

Mustafa and Sims [8] introduced the notion of generalized metric spaces or simply G-metric spaces as a generalization of the concept of a metric space. Based on the concept of G-metric space, Mustafa et al. [911] proved several fixed point theorems for mapping satisfying different contractive condition. The study of common fixed point was initiated by Abbas and Rhoades [12]. The first result for contractive mappings in ordered G-metric spaces was obtained by Saadati et al. [13]. Bhashkar and Lakshmikantham [14] introduced the concept of a coupled fixed point of a mapping F : X × XX and proved some coupled fixed point theorems in ordered metric spaces. Some authors obtained some interesting coupled fixed point theorems in G-metric spaces (see e.g. [1518]). For more results on G-metric spaces, one can refer to the papers [913, 1528].

Throughout the paper, * is the set of positive integers. In 2004, Mustafa and Sims [8] introduced the concept of G-metric spaces as follows:

Definition 1. (see [8]). Let X be a non-empty set, G : X × X × X+ be a function satisfying the following properties

(G1) G(x, y, z) = 0 if x = y = z,

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

(G3) G(x, x, y) ≤ G(x, y, z) for all x, y, z X with yz,

(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, a X (rectangle inequality).

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

Every G-metric on X defines a metric d G on X by

$d G ( x , y ) = G ( x , y , y ) + G ( y , x , x ) , for all x , y ∈ X .$
(1)

Example 2. Let (X, d) be a metric space. The function G : X × X × X → [0, +∞), defined by

$G ( x , y , z ) = max { d ( x , y ) , d ( y , z ) , d ( z , x ) } ,$

or

$G ( x , y , z ) = d ( x , y ) + d ( y , z ) + d ( z , x ) ,$

for all x, y, z X, is a G-metric on X.

Definition 3. (see [8]). Let (X, G) be a G-metric space, and let {x n } be a sequence of points of X, therefore, we say that {x n } is G-convergent to x X if $lim n , m → + ∞ G ( x , x n , x m ) = 0$, that is, for any ε > 0, there exists N such that G(x, x n , x m ) < ε, for all n, mN. We call x the limit of the sequence and write x n x or $lim n → + ∞ x n =x$.

Proposition 4. (see [8]). Let (X,G) be a G-metric space. 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 → +∞,

4. (4)

G(x n , x m , x) → 0 as n, m → +∞.

Definition 5. (see [8]). Let (X, G) be a G-metric space. A sequence {x n } is is called a G-Cauchy sequence if, for any ε > 0, there is N such that G(x n , x m , x l ) < ε for all m, n, lN, i.e., G(x n , x m , x l ) → 0 as n, m, l → ∞.

Proposition 6. (see [8]). Let (X, G) be a G-metric space. Then the following are equivalent:

1. (1)

the sequence {x n } is G-Cauchy,

2. (2)

for any ε > 0, there exists N such that G(x n , x m , x m ) < ε, for all m, nN.

Definition 7. (see [8]). A G-metric space (X, G) is called G-complete if every G-Cauchy sequence is G-convergent in (X, G).

Definition 8. Let (X, G) be a G-metric space. A mapping F : X × X × XX is said to be continuous if for any three G-convergent sequences {x n }, {y n } and {z n } converging to x, y and z respectively, {F(x n , y n , z n )} is G-convergent to F(x, y, z).

Following the paper of Berinde and Borcut [29], Aydi, Karapınar and Postolache [30] introduced the following definitions:

Definition 9. (see [30]). Let (X, ≤) be a partially ordered set and F : X × X × XX, g : XX. The mapping F is said to has the mixed g-monotone property if for any x, y, z X

$x 1 , x 2 ∈ X , g x 1 ≤ g x 2 ⇒ F ( x 1 , y , z ) ≤ F ( x 2 , y , z ) , y 1 , y 2 ∈ X , g y 1 ≤ g y 2 ⇒ F ( x , y 1 , z ) ≥ F ( x , y 2 , z ) , z 1 , z 2 ∈ X , g z 1 ≤ g z 2 ⇒ F ( x , y , z 1 ) ≤ F ( x , y , z 2 ) .$
(2)

Definition 10. (see [30]). Let F : X × X × XX and g : XX. An element (x, y, z) is called a tripled coincidence point of F and g if

$F ( x , y , z ) = g x , F ( y , x , y ) = g y , F ( z , y , x ) = g z .$

The point (gx, gy, gz) is called a point of coincidence of F and g.

Definition 11. (see [30]). Let F : X × X × XX and g : XX. An element (x, y, z) is called a tripled common fixed point of F and g if

$F ( x , y , z ) = g x = x , F ( y , x , y ) = g y = y , F ( z , y , x ) = g z = z .$

Definition 12. (see [30]). Let X be a non-empty set. Let F: X × X × XX and g: XX are such that

$g ( F ( x , y , z ) ) = F ( g x , g y , g z )$

whenever x, y, z X, then F and g are said to be commutative.

Khan et al. [31] introduced the concept of altering distance function as follows:

Definition 13. (altering distance function, [31]) The function ψ : [0, + ∞) → [0, + ∞) is called an altering distance function if the following properties are satisfied:

1. (1)

ψ is continuous and non-decreasing,

2. (2)

ψ(t) = 0 if and only if t = 0.

Let Ψ be the set of altering distances. Again, we denote by Φ the set of functions ϕ : [0, +∞) → [0, +∞) such that

1. (i)

ϕ is lower-continuous and non-decreasing,

2. (ii)

ϕ(t) = 0 if and only if t = 0.

The notion of a fixed point of N-order was first introduced by Samet and Vetro [32]. Later, Berinde and Borcut [29] proved some tripled fixed point results (N = 3) in partially ordered metric spaces (see also [3336]). In this paper, we establish tripled coincidence point results for mappings F : X3X and g : XX involving nonlinear contractions in the setting of ordered G-metric spaces. Also, we present an application and some examples in support of our results.

## Main results

Before stating our results, we give the following useful lemma.

Lemma 14. Consider three non-negative real sequences {a n }, {b n } and {c n }. Suppose there exists α ≥ 0 such that

$lim n → + ∞ max { a n , b n } = 0 a n d lim n → + ∞ max { a n , b n , c n } = α .$

Then, $lim sup n → + ∞ c n = α$.

Proof. First, we have c n ≤ max{a n , b n , c n }, then

$lim sup n → + ∞ c n ≤ α .$
(3)

For all n , we have

$0 ≤ max { a n , b n , c n } - c n ≤ max { a n , b n } + c n - c n = max { a n , b n } ,$

which implies that $lim sup n → + ∞ ( max { a n , b n , c n } - c n ) ≤ 0$. Having in mind that

$lim sup n → + ∞ ( max { a n , b n , c n } - c n ) ≥ lim sup n → + ∞ max { a n , b n , c n } - lim sup n → + ∞ c n = α - lim sup n → + ∞ c n ,$

so it follows that

$lim sup n → + ∞ c n ≥ α .$
(4)

By (3) and (4), we get that $lim sup n → + ∞ c n = α$. □

The aim of this paper is to prove the following theorem.

Theorem 15. Let (X, ≤) be a partially ordered set and (X, G) be a G-metric space such that (X, G) is G-complete. Let F : X3X and g : XX. Assume there exist ψ Ψ and ϕ Φ such that for x, y, z, a, b, c, u, v, w X, with gxgagu, gygbgv and gzgcgw, we have

$ψ ( G ( F ( x , y , z ) , F ( a , b , c ) , F ( u , v , w ) ) ) ≤ ψ ( m a x { G ( g x , g a , g u ) , G ( g y , g b , g v ) , G ( g z , g c , g w ) } ) - ϕ ( m a x { G ( g x , g a , g u ) , G ( g y , g b , g v ) , G ( g z , g c , g w ) } ) .$
(5)

Assume that F and g satisfy the following conditions:

1. (1)

F(X3) g(X),

2. (2)

F has the mixed g-monotone property,

3. (3)

F is continuous,

4. (4)

g is continuous and commutes with F.

Suppose there exist x0, y0, z0 X such that gx0F(x0, y0, z0), gy0F(y0, x0, y0) and gz0F(z0, y0, x0), then F and g have a tripled coincidence point in X, i.e., there exist x, y, z X such that

$F ( x , y , z ) = g x , F ( y , x , y ) = g y , F ( z , y , x ) = g z .$

Proof. Suppose x0, y0, z0 X are such that gx0F(x0, y0, z0), gy0F(y0, x0, y0), and gz0F(z0, y0, x0). Since F(X3) g(X), we can choose gx1 = F(x0, y0, z0), gy1 = F(y0, x0, y0) and gz1 = F(z0, y0, x0). Then gx0gx1, gy0gy1 and gz0gz1. Similarly, define gx2 = F(x1, y1, z1), gy2 = F(y1, x1, y1) and gz2 = F(z1, y1, x1). Since F has the mixed g-monotone property, we have gx0gx1gx2, gy2gy1gy0 and gz0gz1gz2. Continuing this process, we can construct three sequences {x n }, {y n } and {z n } in X such that

$g x n = F ( x n - 1 , y n - 1 , z n - 1 ) ≤ g x n + 1 = F ( x n , y n , z n ) , g y n + 1 = F ( y n , x n , y n ) ≤ g y n = F ( y n - 1 , x n - 1 , y n - 1 ) ,$

and

$g z n = F ( z n - 1 , y n - 1 , x n - 1 ) ≤ g z n + 1 = F ( z n , y n , x n ) .$

If, for some integer n0, we have $( g x n 0 + 1 , g y n 0 + 1 , g z n 0 + 1 ) = ( g x n 0 , g y n 0 , g z n 0 )$, then $F ( x n 0 , y n 0 , z n 0 ) = g x n 0 , F ( y n 0 , x n 0 , y n 0 ) = g y n 0$, and $F ( z n 0 , y n 0 , x n 0 ) =g z n 0$; i.e., $( x n 0 , y n 0 , z n 0 )$ is a tripled coincidence point of F and g. Thus we shall assume that (gx n +1, gy n +1, gz n +1) ≠ (gx n , gy n , gz n ) for all n ; i.e., we assume that either gx n +1gx n or gy n +1gy n or gz n +1gz n . For any n *, we have from (5)

$ψ ( G ( g x n + 1 , g x n , g x n ) ) : = ψ ( G ( F ( x n , y n , z n ) , F ( x n − 1 , y n − 1 , z n − 1 ) , F ( x n − 1 , y n − 1 , z n − 1 ) ) ) ≤ ψ ( max { G ( g x n , g x n − 1 , g x n − 1 ) , G ( g y n , g y n − 1 , g y n − 1 ) , G ( g z n − 1 , g z n − 1 ) } ) − ϕ ( max { G ( g x n , g x n − 1 , g x n − 1 ) , G ( g y n , g y n − 1 , g y n − 1 ) , G ( g z n , g z n − 1 , g z n − 1 ) } ) ≤ ψ ( max { G ( g x n , g x n − 1 , g x n − 1 ) , G ( g y n , g y n − 1 ) g y n − 1 ) , G ( g z n , g z n − 1 , g z n − 1 ) } ) ,$
(6)
$ψ ( G ( g y n , g y n , g y n + 1 ) ) : = ψ ( G ( F ( y n − 1 , x n − 1 , y n − 1 ) , F ( y n − 1 , x n − 1 , y n − 1 ) , F ( y n , x n , y n ) ) ) ≤ ψ ( max { G ( g y n − 1 , g y n − 1 , g y n ) , G ( g x n − 1 , g x n − 1 , g x n ) } − ϕ ( max { G ( g y n − 1 , g y n − 1 , g y n ) , G ( g x n − 1 , g x n − 1 , g x n ) } ) ≤ ψ ( max { G ( g y n , g y n − 1 , g y n − 1 ) , G ( g x n , g x n − 1 , g x n − 1 ) , G ( g z n , g z n − 1 , g z n − 1 ) } ) ,$
(7)

and

$ψ ( G ( g z n + 1 , g z n , g z n ) ) : = ψ ( G ( F ( z n , y n , x n ) , F ( z n - 1 , y n - 1 , x n - 1 ) , F ( z n - 1 , y n - 1 , x n - 1 ) ) ) ≤ ψ ( max { G ( g z n , g z n - 1 , g z n - 1 ) , G ( g y n , g y n - 1 , g y n - 1 ) , G ( g x n , g x n - 1 , g x n - 1 ) } ) - ϕ ( max { G ( F ( z n , y n , x n ) , F ( z n - 1 , y n - 1 , x n - 1 ) , F ( z n - 1 , y n - 1 , x n - 1 ) ) } ) ≤ ψ ( max { G ( g z n , g z n - 1 , g z n - 1 ) , G ( g y n , g y n - 1 , g y n - 1 ) , G ( g x n , g x n - 1 , g x n - 1 ) } ) .$
(8)

Since ψ:[0,+∞) → [0, +∞) is a non-decreasing function, for a, b, c [0,+∞), we have ψ(max{a, b, c}) = max{ψ(a), ψ(b), ψ(c)}. Then, from (6), (7), and (8), it follows that

$ψ ( max { G ( g x n + 1 , g x n , g x n ) , G ( g y n , g y n , g y n + 1 ) , G ( g z n + 1 , g z n , g z n ) } ) = max ( { ψ ( G ( g x n + 1 , g x n , g x n ) ) , ψ ( G ( g y n , g y n , g y n + 1 ) ) , ψ ( G ( g z n + 1 , g z n , g z n ) ) } ) ≤ ψ ( max { G ( g x n , g x n - 1 , g x n - 1 ) , G ( g y n , g y n - 1 , g y n - 1 ) , G ( g z n , g z n - 1 , g z n - 1 ) } ) .$

The fact that ψ is non-decreasing yields that

$max { G ( g x n + 1 , g x n , g x n ) , G ( g y n , g y n , g y n + 1 ) , G ( g z n + 1 , g z n , g z n ) } ≤ max { G ( g x n , g x n - 1 , g x n - 1 ) , G ( g y n , g y n - 1 , g y n - 1 ) , G ( g z n , g z n - 1 , g z n - 1 ) } .$
(9)

Thus, {max{G(gx n +1, gx n , gx n ),G(gy n , gy n , gy n +1),G(gz n +1, gz n , gz n )}} is a positive non-increasing sequence. Hence there exists r ≥ 0 such that

$lim n → + ∞ max { G ( g x n + 1 , g x n , g x n ) , G ( g y n , g y n , g y n + 1 ) , G ( g z n + 1 , g z n , g z n ) } = r .$
(10)

Having in mind that ϕ is non-decreasing, then

$ϕ ( max { G ( g x n , g x n - 1 , g x n - 1 ) , G ( g y n , g y n - 1 , g y n - 1 ) , G ( g z n , g z n - 1 , g z n - 1 ) } ) ≥ ϕ ( max { G ( g x n , g x n - 1 , g x n - 1 ) , G ( g y n , g y n - 1 , g y n - 1 ) } ) ,$
(11)

so from (6)-(8), we get that

$ψ ( max { G ( g x n + 1 , g x n , g x n ) , G ( g y n , g y n , g y n + 1 ) , G ( g z n + 1 , g z n , g z n ) } ) = max ( { ψ ( G ( g x n + 1 , g x n , g x n ) ) , ψ ( G ( g y n , g y n , g y n + 1 ) ) , ψ ( G ( g z n + 1 , g z n , g z n ) ) } ) ≤ ψ ( max { G ( g x n , g x n - 1 , g x n - 1 ) , G ( g y n , g y n - 1 , g y n - 1 ) , G ( g z n , g z n - 1 , g z n - 1 ) } ) - ϕ ( max { G ( g x n , g x n - 1 , g x n - 1 ) , G ( g y n , g y n - 1 , g y n - 1 ) } ) .$
(12)

On the other hand,

$0 ≤ max { G ( g x n , g x n - 1 , g x n - 1 ) , G ( g y n - 1 , g y n - 1 , g y n ) } ≤ max { G ( g x n , g x n - 1 , g x n - 1 ) , G ( g y n - 1 , g y n - 1 , g y n ) , G ( g z n , g z n - 1 , g z n - 1 ) } ,$
(13)

so by (10), the real sequence {max{G(gx n , gx n -1, gx n -1),G(gy n -1, gy n -1, gy n )}} is bounded. Thus, there exists a real number r1 with 0 ≤ r1r and subsequences {x n ( k )} of {x n } and {y n ( k )} of {y n } such that

$lim k → + ∞ max { G ( g x n ( k ) + 1 , g x n ( k ) , g x n ( k ) ) , G ( g y n ( k ) , g y n ( k ) , g y n ( k ) + 1 ) } = r 1 .$
(14)

We rewrite (12)

$ψ ( max { G ( g x n ( k ) + 1 , g x n ( k ) , g x n ( k ) ) , G ( g y n ( k ) , g y n ( k ) , g y n ( k ) + 1 ) , G ( g z n ( k ) + 1 , g z n ( k ) , g z n ( k ) } ) ≤ ψ ( max { G ( g x n ( k ) , g x n ( k ) − 1 , g x n ( k ) − 1 ) , G ( g y n ( k ) , g y n ( k ) − 1 , g y n ( k ) − 1 ) , G ( g z n ( k ) , g z n ( k ) − 1 , g z n ( k ) − 1 ) } ) − ϕ ( max { G ( g x n ( k ) , g x n ( k ) − 1 , g x n ( k ) − 1 ) , G ( g y n ( k ) , g y n ( k ) − 1 , g y n ( k ) − 1 ) } ) .$
(15)

Letting k → +∞ in (15), having in mind (10), (14), the continuity of ψ and the lower semi-continuity of ϕ, we obtain

$ψ ( r ) = lim sup k → + ∞ ψ ( max { G ( g x n ( k ) + 1 , g x n ( k ) , g x n ( k ) ) , G ( g y n ( k ) , g y n ( k ) , g y n ( k ) + 1 ) , G ( g z n ( k ) + 1 , g z n ( k ) , g z n ( k ) ) } ) ≤ lim sup k → + ∞ ψ ( max { G ( g x n ( k ) , g x n ( k ) - 1 , g x n ( k ) - 1 ) , G ( g y n ( k ) , g y n ( k ) - 1 , g y n ( k ) - 1 ) , G ( g z n ( k ) , g z n ( k ) - 1 , g z n ( k ) - 1 ) } ) - lim inf k → + ∞ ϕ ( max { G ( g x n ( k ) , g x n ( k ) - 1 , g x n ( k ) - 1 ) , G ( g y n ( k ) , g y n ( k ) - 1 , g y n ( k ) - 1 ) } ) ≤ ψ ( r ) - ϕ ( r 1 ) ,$

which implies that ϕ(r1) = 0, and using a property of ϕ, we find r1 = 0. Thanks to Lemma 14 together with (10) and (14), it yields that

$r = : lim k → + ∞ max { G ( g x n ( k ) , g x n ( k ) - 1 , g x n ( k ) - 1 ) , G ( g y n ( k ) , g y n ( k ) - 1 , g y n ( k ) - 1 ) , G ( g z n ( k ) , g z n ( k ) - 1 , g z n ( k ) - 1 ) } = lim sup k → + ∞ G ( g z n ( k ) , g z n ( k ) - 1 , g z n ( k ) - 1 ) .$
(16)

For any k , we rewrite (8) as

$ψ ( G ( g z n ( k ) + 1 , g z n ( k ) , g z n ( k ) ) ) ≤ ψ ( max { G ( g z n ( k ) , g z n ( k ) - 1 , g z n ( k ) - 1 ) , G ( g y n ( k ) , g y n ( k ) - 1 , g y n ( k ) - 1 ) , G ( g x n ( k ) , g x n ( k ) - 1 , g x n ( k ) - 1 ) } ) - ϕ ( max { G ( g z n ( k ) , g z n ( k ) - 1 , g z n ( k ) - 1 ) , G ( g y n ( k ) , g y n ( k ) - 1 , g y n ( k ) - 1 ) , G ( g x n ( k ) , g x n ( k ) - 1 , g x n ( k ) - 1 ) } ) .$
(17)

Again, letting k → +∞ in (17), having in mind (10), (16) and by the properties of ψ, ϕ, we obtain

$ψ ( r ) = lim sup k → + ∞ ψ ( G ( g z n ( k ) + 1 , g z n ( k ) , g z n ( k ) ) ) ≤ lim sup k → + ∞ ψ ( max { g z n ( k ) , g z n ( k ) − 1 , g z n ( k ) − 1 ) , G ( g y n ( k ) , g y n ( k ) − 1 , g y n ( k ) − 1 ) , G ( g x n ( k ) , g x n ( k ) − 1 , g x n ( k ) − 1 ) } ) − lim inf k → + ∞ ϕ ( max { G ( g z n ( k ) , g z n ( k ) − 1 , g z n ( k ) − 1 ) , G ( g y n ( k ) , g y n ( k ) − 1 , g y n ( k ) − 1 ) , G ( g x n ( k ) , g x n ( k ) − 1 , g x n ( k ) − 1 ) } ) ≤ ψ ( r ) − ϕ ( r ) ,$

which gives that ϕ(r) = 0, so r = 0, i.e., by (10),

$lim n → + ∞ max { G ( g x n + 1 , g x n , g x n ) , G ( g y n , g y n , g y n + 1 ) , G ( g z n + 1 , g z n , g z n ) } = 0 .$
(18)

Our next step is to show that {gx n }, {gy n } and {gz n } are G-Cauchy sequences.

Assume the contrary, i.e., {gx n }, {gy n } or {gz n } is not a G-Cauchy sequence, i.e.,

$lim n , m → + ∞ G ( g x m , g x n g x n ) ≠ 0 , or lim n , m → + ∞ G ( g y m , g y n , g y n ) ≠ 0 ,$

or $lim n , m → + ∞ G ( g z m , g z n , g z n ) ≠ 0$. This means that there exists ε > 0 for which we can find subsequences of integers {m k } and {n k } with n k > m k > k such that

$max { G ( g x m k , g x n k , g x n k ) , G ( g y m k , g y n k , g y n k ) , G ( g z m k , g z n k , g z n k ) } ≥ ε .$
(19)

Further, corresponding to m k we can choose n k in such a way that it is the smallest integer with n k > m k and satisfying (19). Then

$max { G ( g x m k , g x n k - 1 , g x n k - 1 ) , G ( g y m k , g y n k - 1 , g y n k - 1 ) , G ( g z m k , g z n k - 1 , g z n k - 1 ) } < ε .$
(20)

By (G5) and (20), we have

$G ( g x m k , g x n k , g x n k ) ≤ G ( g x m k , g x n k - 1 , g x n k - 1 ) + G ( g x n k - 1 , g x n k , g x n k ) < ε + G ( g x n k - 1 , g x n k , g x n k ) .$

Thus, by (18) we obtain

$lim k → + ∞ G ( g x m k , g x n k , g x n k ) ≤ lim k → + ∞ G ( g x m k , g x n k - 1 , g x n k - 1 ) ≤ ε .$
(21)

Similarly, we have

$lim k → + ∞ G ( g y m k , g y n k , g y n k ) ≤ lim k → + ∞ G ( g y m k , g y n k - 1 , g y n k - 1 ) ≤ ε .$
(22)
$lim k → + ∞ G ( g z m k , g z n k , g z n k ) ≤ lim k → + ∞ G ( g z m k , g z n k - 1 , g z n k - 1 ) ≤ ε .$
(23)

Again by (G 5) and (20), we have

$G ( g x m k , g x n k , g x n k ) ≤ G ( g x m k , g x m k - 1 , g x m k - 1 ) + G ( g x m k - 1 , g x n k - 1 , g x n k - 1 ) + G ( g x n k - 1 , g x n k , g x n k ) ≤ G ( g x m k , g x m k - 1 , g x m k - 1 ) + G ( g x m k - 1 , g x m k , g x m k ) + G ( g x m k , g x n k - 1 , g x n k - 1 ) + G ( g x n k - 1 , g x n k , g x n k ) < G ( g x m k , g x m k - 1 , g x m k - 1 ) + G ( g x m k - 1 , g x m k , g x m k ) + ε + G ( g x n k - 1 , g x n k , g x n k ) .$

Letting k → +∞ and using (18), we get

$lim k → + ∞ G ( g x m k , g x n k , g x n k ) ≤ lim k → + ∞ G ( g x m k - 1 , g x n k - 1 , g x n k - 1 ) ≤ ε ,$
(24)
$lim k → + ∞ G ( g y m k , g y n k , g y n k ) ≤ lim k → + ∞ G ( g y m k - 1 , g y n k - 1 , g y n k - 1 ) ≤ ε ,$
(25)
$lim k → + ∞ G ( g z m k , g z n k , g z n k ) ≤ lim k → + ∞ G ( g z m k - 1 , g z n k - 1 , g z n k - 1 ) ≤ ε .$
(26)

Using (19) and (24)-(26), we have

$lim k → + ∞ max { G ( g x m k , g x n k , g x n k ) , G ( g y m k , g y n k , g y n k ) , G ( g z m k , g z n k , g z n k ) } = lim k → + ∞ max { G ( g x m k − 1 , g x n k − 1 , g x n k − 1 ) , G ( g y m k − 1 , g y n k − 1 , g y n k − 1 ) , G ( g z m k − 1 , g z n k − 1 , g z n k − 1 ) = ε .$
(27)

Now, using inequality (5) we obtain

$ψ ( G ( g x m k , g x n k , g x n k ) ) = ψ ( G ( F ( x m k - 1 , y m k - 1 , z m k - 1 ) , F ( x n k - 1 , y n k - 1 , z n k - 1 ) , F ( x n k - 1 , y n k - 1 , z n k - 1 ) ) ) ≤ ψ ( max { G ( x m k - 1 , x n k - 1 , x n k - 1 ) , G ( y m k - 1 , y n k - 1 , y n k - 1 ) , G ( z m k - 1 , z n k - 1 , z n k - 1 ) } ) - ϕ ( max { G ( x m k - 1 , x n k - 1 , x n k - 1 ) , G ( y m k - 1 , y n k - 1 , y n k - 1 ) , G ( z m k - 1 , z n k - 1 , z n k - 1 ) } )$
(28)
$ψ ( G ( g y m k , g y n k , g y n k ) ) = ψ ( G ( F ( y m k - 1 , x m k - 1 , y m k - 1 ) , F ( y n k - 1 , x n k - 1 , y n k - 1 ) , F ( y n k - 1 , x n k - 1 , y n k - 1 ) ) ) ≤ ψ ( max { G ( x m k - 1 , x n k - 1 , x n k - 1 ) , G ( y m k - 1 , y n k - 1 , y n k - 1 ) } ) - ϕ ( max { G ( x m k - 1 , x n k - 1 , x n k - 1 ) , G ( y m k - 1 , y n k - 1 , y n k - 1 ) } )$
(29)

and

$ψ ( G ( g z m k , g z n k , g z n k ) ) = ψ ( G ( F ( z m k - 1 , y m k - 1 , x m k - 1 ) , F ( z n k - 1 , y n k - 1 , x n k - 1 ) , F ( z n k - 1 , y n k - 1 , x n k - 1 ) ) ) ≤ ψ ( max { G ( x m k - 1 , x n k - 1 , x n k - 1 ) , G ( y m k - 1 , y n k - 1 , y n k - 1 ) , G ( z m k - 1 , z n k - 1 , z n k - 1 ) } ) - ϕ ( max { G ( x m k - 1 , x n k - 1 , x n k - 1 ) , G ( y m k - 1 , y n k - 1 , y n k - 1 ) , G ( z m k - 1 , z n k - 1 , z n k - 1 ) } )$
(30)

We deduce from (28)-(30) that

$ψ ( max { G ( g x m k , g x n k , g x n k ) , G ( g y m k , g y n k , g y n k ) , G ( g z m k , g z n k , g z n k ) } = max { ψ ( G ( g x m k , g x n k , g x n k ) ) , ψ ( G ( g y m k , g y n k , g y n k ) ) , ψ ( G ( g z m k , g z n k , g z n k ) ) } ≤ ψ ( max { G ( x m k − 1 , x n k − 1 , x n k − 1 ) , G ( y m k − 1 , y n k − 1 , y n k − 1 ) , G ( z m k − 1 , z n k − 1 , z n k − 1 ) } ) − ϕ ( max { G ( x m k − 1 , x n k − 1 , x n k − 1 ) , G ( y m k − 1 , y n k − 1 , y n k − 1 ) } ) .$
(31)

On the other hand, since

$max { G ( x m k - 1 , x n k - 1 , x n k - 1 ) , G ( y m k - 1 , y n k - 1 , y n k - 1 ) } ≤ max { G ( x m k - 1 , x n k - 1 , x n k - 1 ) , G ( y m k - 1 , y n k - 1 , y n k - 1 ) , G ( z m k - 1 , z n k - 1 , z n k - 1 ) } ,$
(32)

then from (27),

$lim sup k → + ∞ max { G ( x m k - 1 , x n k - 1 , x n k - 1 ) , G ( y m k - 1 , y n k - 1 , y n k - 1 ) } ≤ ε .$

Therefore, the real sequence ${ max { G ( x m k − 1 , x n k − 1 , x n k − 1 ) , G ( y m k − 1 , y n k − 1 , y n k − 1 ) } }$ is bounded. Thus, up to a subsequence (still denoted the same), there exists ε1 with 0 ≤ ε1ε such that

$lim k → + ∞ max { G ( x m k - 1 , x n k - 1 , x n k - 1 ) , G ( y m k - 1 , y n k - 1 , y n k - 1 ) } = ε 1 .$
(33)

Inserting this in (31) and using (27), (33) together with the properties of ψ, ϕ, we get that

$ψ ( ε ) = lim sup k → + ∞ ψ ( max { G ( g x m k , g x n k , g x n k ) , G ( g y m k , g y n k , g y n k ) , G ( g z m k , g z n k , g z n k ) } ≤ lim sup k → + ∞ ψ ( max { G ( x m k − 1 , x n k − 1 , x n k − 1 ) , G ( y m k − 1 , y n k − 1 , y n k − 1 ) , G ( z m k − 1 , z n k − 1 , z n k − 1 ) } ) − lim inf k → + ∞ ϕ ( max { G ( x m k − 1 , x n k − 1 , x n k − 1 ) , G ( y m k − 1 , y n k − 1 , y n k − 1 ) } ) ≤ ψ ( ε ) − ϕ ( ε 1 ) ,$

which leads to ϕ(ε1) = 0, so ε1 = 0. By this and (27), due to Lemma 14, we obtain

$lim sup k → + ∞ G ( z m k - 1 , z n k - 1 , z n k - 1 ) = ε .$

Combining this to (19) and (26), we find

$lim sup k → + ∞ G ( z m k , z n k , z n k ) = ε .$

Letting k → ∞ in (30) and using (27), we deduce

$ψ ( ε ) ≤ ψ ( ε ) - ϕ ( ε ) ,$

i.e., ε = 0, it is a contradiction. We conclude that {gx n }, {gy n } and {gz n } are G-Cauchy sequences in the G-metric space (X, G), which is G-complete. Then, there are x, y, z X such that {gx n }, {gy n } and {gz n } are respectively G-convergent to x, y and z, i.e., from Proposition 4, we have

$lim n → + ∞ G ( g x n , g x n , x ) = lim n → + ∞ G ( g x n , x , x ) = 0 ,$
(34)
$lim n → + ∞ G ( g y n , g y n , y ) = lim n → + ∞ G ( g y n , y , y ) = 0 ,$
(35)
$lim n → + ∞ G ( g z n , g z n , z ) = lim n → + ∞ G ( g z n , z , z ) = 0 .$
(36)

From (34)-(36) and the continuity of g, we get thanks to Proposition 8

$lim n → + ∞ G ( g ( g x n ) , g ( g x n ) , g x ) = lim n → + ∞ G ( g ( g x n ) , g x , g x ) = 0 ,$
(37)
$lim n → + ∞ G ( g ( g y n ) , g ( g y n ) , g y ) = lim n → + ∞ G ( g ( g y n ) , g y , g y ) = 0 ,$
(38)
$lim n → + ∞ G ( g ( g z n ) , g ( g z n ) , g z ) = lim n → + ∞ G ( g ( g z n ) , g z , g z ) = 0 .$
(39)

Since gx n +1 = F(x n , y n , z n ), gy n +1 = F(y n , x n , y n ) and gz n +1 = F(z n , y n , x n ), so the commutativity of F and g yields that

$g ( g x n + 1 ) = g ( F ( x n , y n , z n ) ) = F ( g x n , g y n , g z n ) ,$
(40)
$g ( g y n + 1 ) = g ( F ( y n , x n , y n ) ) = F ( g y n , g x n , g y n ) ,$
(41)
$g ( g z n + 1 ) = g ( F ( z n , y n , x n ) ) = F ( g z n , g y n , g x n ) .$
(42)

Now we show that F(x, y, z) = gx, F(y, x, y) = gy and F(z, y, x) = gz.

The mapping F is continuous, so since the sequences {gx n }, {gy n } and {gz n } are, respectively, G-convergent to x, y and z, hence using Definition 8, the sequence {F(gx n , gy n , gz n )} is G-convergent to F(x, y, z). Therefore, from (40), {g(gx n +1)} is G-convergent to F(x, y, z). By uniqueness of the limit, from (37) we have F(x, y, z) = gx.

Similarly, one finds F(y, x, y) = gy and F(z, y, x) = gz, and this makes end to the proof. □

Corollary 16. Let (X, ≤) be a partially ordered set and (X, G) be a G-metric space such that (X, G) is G-complete. Let F : X3X and g : XX. Assume there exists k [0,1) such that for x, y, z, a, b, c, u, v, w X, with gxgagu, gygbgv and gzgcgw, we have

$G ( F ( x , y , z ) , F ( a , b , c ) , F ( u , v , w ) ) ≤ k max { G ( g x , g a , g u ) , G ( g y , g b , g v ) , G ( g z , g c , g w ) } .$

Assume that F and g satisfy the following conditions:

1. (1)

F(X3) g(X),

2. (2)

F has the mixed g-monotone property,

3. (3)

F is continuous,

4. (4)

g is continuous and commutes with F.

Suppose there exist x0, y0, z0 X such that gx0F(x0, y0, z0), gy 0 F(y0, x0, y0) and gz0F(z0, y0, x0), then F and g have a tripled coincidence point in X, i.e., there exist x, y, z X such that

$F ( x , y , z ) = g x , F ( y , x , y ) = g y , F ( z , y , x ) = g z .$

Proof. If follows by taking ψ(t) = t and ϕ(t) = (1 - k)t for all t ≥ 0. □

Corollary 17. Let (X, ≤) be a partially ordered set and (X, G) be a G-metric space such that (X, G) is G-complete. Let F : X3X and g : XX. Assume there exists k [0,1) such that for x, y, z, a, b, c, u, v, w X, with gxgagu, gygbgv and gzgcgw, we have

$G ( F ( x , y , z ) , F ( a , b , c ) , F ( u , v , w ) ) ≤ k 3 ( G ( g x , g a , g u ) + G ( g y , g b , g v ) + G ( g z , g c , g w ) ) .$

Assume that F and g satisfy the following conditions:

1. (1)

F(X3) g(X),

2. (2)

F has the mixed g-monotone property,

3. (3)

F is continuous,

4. (4)

g is continuous and commutes with F.

Suppose there exist x0, y0, z0 X such that gx0F(x0, y0, z0), gy0F(y0, x0, y0) and gz0F(z0, y0, x0), then F and g have a tripled coincidence point in X, i.e., there exist x, y, z X such that

$F ( x , y , z ) = g x , F ( y , x , y ) = g y , F ( z , y , x ) = g z .$

Proof. It suffices to remark that

$k 3 ( G ( g x , g a , g u ) + G ( g y , g b , g v ) + G ( g z , g c , g w ) ) ≤ k max { G ( g x , g a , g u ) , G ( g y , g b , g v ) , G ( g z , g c , g w ) } .$
(43)

In the next theorem, we omit the continuity hypothesis of F. We need the following definition.

Definition 18. Let (X, ) be a partially ordered set and G be a G-metric on X. We say that (X, G, ≤) is regular if the following conditions hold:

1. (i)

if a non-decreasing sequence {x n } is such that x n x, then x n x for all n,

2. (ii)

if a non-increasing sequence {y n } is such that y n y, then yy n for all n.

Theorem 19. Let (X, ≤) be a partially ordered set and (X, G) be a G-metric space. Let F : X3X and g: XX. Assume there exist ψ Ψ and ϕ Φ such that for x, y, z, a, b, c, u, v, w X, with gxgagu, gygbgv and gzgcgw, we have

$ψ ( G ( F ( x , y , z ) , F ( a , b , c ) , F ( u , v , w ) ) ) ≤ ψ ( max { G ( g x , g a , g u ) , G ( g y , g b , g v ) , G ( g z , g c , g w ) } ) - ϕ ( max { G ( g x , g a , g u ) , G ( g y , g b , g v ) , G ( g z , g c , g w ) } ) .$
(44)

Assume that (X, G, ≤) is regular. Suppose that (g(X),G) is G-complete, F has the mixed g-monotone property and F(X × X × X) g(X). Also, assume there exist x0, y0, z0 X such that gx0F(x0, y0, z0), gy0F(y0, x0, y0) and gz 0 F(z0, y0, x0), then F and g have a tripled coincidence point in X, i.e., there exist x, y, z X such that

$F ( x , y , z ) = g x , F ( y , x , y ) = g y , F ( z , y , x ) = g z .$

Proof. Proceeding exactly as in Theorem 15, we have that {gx n }, {gy n } and {gz n } are G-Cauchy sequences in the G-complete G-metric space (g(X), G). Then, there exist x, y, z X such that gx n gx, gy n gy and gz n gz. Since {gx n } and {gz n } are non-decreasing and {gy n } is non-increasing, using the regularity of (X, G, ≤), we have gx n gx, gz n gz and gygy n for all n ≥ 0. Using (5), we get

$ψ ( G ( F ( x , y , z ) , g x n + 1 , g x n + 1 ) = ψ ( G ( F ( x , y , z ) , F ( x n , y n , z n ) , F ( x n , y n , z n ) ) ) ≤ ψ ( max { G ( g x , g x n , g x n ) , G ( g y , g y n , g y n ) , G ( g z , g z n , g z n ) } ) − ϕ ( max { G ( g x , g x n , g x n ) , G ( g y , g y n , g y n ) , G ( g z , g z n , g z n ) } ) .$
(45)

Letting n → +∞ in the above inequality, we obtain that

$ψ ( G ( F ( x , y , z ) , g x , g x ) ) ≤ ψ ( 0 ) - ϕ ( 0 ) = 0 ,$

which implies that G(F(x, y, z), gx, gx) = 0, i.e., gx = F(x, y, z).

Similarly, one can show that gy = F(y, x, y) and gz = F(z, y, x). Thus we proved that (x, y, z) is a tripled coincidence point of F and g. □

Similarly, we can state the following corollary.

Corollary 20. Let (X, ≤) be a partially ordered set and (X, G) be a G-metric space. Let F : X3X and g: XX. Assume there exists k [0, 1) such that for x, y, z, a, b, c, u, v, w X, with gxgagu, gygbgv and gzgcgw, we have

$G ( F ( x , y , z ) , F ( a , b , c ) , F ( u , v , w ) ) ≤ k max { G ( g x , g a , g y ) , G ( g y , g b , g v ) , G ( g z , g c , g w ) } .$

Assume that (X, G, ≤) is regular. Suppose that (g(X),G) is G-complete, F has the mixed g-monotone property and F(X × X × X) g(X). Also, assume there exist x0, y0, z0 X such that gx0F(x0, y0, z0), gy 0F(y0, x0, y0) and gz0F(z0, y0, x0), then F and g have a tripled coincidence point in X, i.e., there exist x, y, z X such that

$F ( x , y , z ) = g x , F ( y , x , y ) = g y , F ( z , y , x ) = g z .$

Corollary 21. Let (X, ≤) be a partially ordered set and (X, G) be a G-metric space. Assume that (X, G, ≤) is regular. Let F : X3X and g : XX. Assume there exists k [0, 1) such that for x, y, z, a, b, c, u, v, w X, with gxgagu, gygbgv and gzgc > gw, we have

$G ( F ( x , y , z ) , F ( a , b , c ) , F ( u , v , w ) ) ≤ k 3 ( G ( g x , g a , g u ) + G ( g y , g b , g v ) + G ( g z , g c , g w ) ) .$

Suppose that (g(X),G) is G-complete, F has the mixed g-monotone property and F(X × X × X) g(X). Also, assume there exist x0, y0, z0 X such that gx0F(x0, y0, z0), gy0F(y0, x0, y0) and gz0 < F(z0, y0, x0), then F and g have a tripled coincidence point in X, i.e., there exist x, y, z X such that

$F ( x , y , z ) = g x , F ( y , x , y ) = g y , F ( z , y , x ) = g z .$

Remark 22. Other corollaries could be derived from Theorems 15 and 19 by taking g = Id x .

Now, form previous obtained results, we will deduce some tripled coincidence point results for mappings satisfying a contraction of integral type in G-metric space. Let us introduce first some notations.

We denote by Γ the set of functions α: [0, +∞) → [0, +∞) satisfying the following conditions:

1. (i)

α is a Lebesgue integrable mapping on each compact subset of [0, +∞),

2. (ii)

for all ε > 0, we have

$∫ 0 ε α ( s ) d s > 0 .$

Let N * be fixed. Let {α i }1 ≤ i N be a family of N functions that belong to Γ. For all t ≥ 0, we denote (I i )i =1,..., N as follows:

$I 1 ( t ) = ∫ 0 t α 1 ( s ) d s , I 2 ( t ) = ∫ 0 I 1 ( t ) α 2 ( s ) d s = ∫ 0 ∫ 0 t α 1 ( s ) d s α 2 ( s ) d s , ⋮ I N ( t ) = ∫ 0 I N - 1 ( t ) α N ( s ) d s .$

We have the following result.

Theorem 23. Let (X, ≤) be a partially ordered set and (X, G) be a G-metric space such that (X, G) is G-complete. Let F : X3X and g : XX. Assume there exist ψ Ψ and ϕ Φ such that for x, y, z, a, b, c, u, v, w X, wth gxgagu, gygbgv and gzgcgw, we have

$I N ( ψ ( G ( F ( x , y , z ) , F ( a , b , c ) , F ( u , v , w ) ) ) ) ≤ I N ( ψ ( m a x { G ( g x , g a , g u ) , G ( g y , g b , g v ) , G ( g z , g c , g w ) } ) ) - I N ( ϕ ( m a x { G ( g x , g a , g u ) , G ( g y , g b , g v ) , G ( g z , g c , g w ) } ) ) .$
(46)

Assume that F and g satisfy the following conditions:

1. (1)

F(X3) g(X),

2. (2)

F has the mixed g-monotone property,

3. (3)

F is continuous,

4. (4)

g is continuous and commutes with F.

Suppose there exist x0, y0, z0 X such that gx0F(x0, y0, z0), gy0F(y0, x0, y0) and gz0F(z0, y0, x0), then F and g have a tripled coincidence point in X, i.e., there exist x, y, z X such that

$F ( x , y , z ) = g x , F ( y , x , y ) = g y , F ( z , y , x ) = g z .$

Proof. Take

$ϕ ̃ = I N ∘ φ and ψ ̃ = I N ∘ ψ .$

It is easy to show that $ψ ̃ ∈Ψ$ and $ϕ ̃ ∈Φ$. From (46), we have

$ψ ˜ ( G ( F ( x , y , z ) , F ( a , b , c ) , F ( u , v , w ) ) ≤ ψ ˜ ( max { G ( g x , g a , g u ) , G ( g y , g b , g v ) , G ( g z , g c , g w ) } ) − ϕ ˜ ( max { G ( g x , g a , g u ) , G ( g y , g b , g v ) , G ( g z , g c , g w ) } ) .$
(47)

Now, applying Theorem 15, we obtain the desired result. □

Similarly, we have

Theorem 24. Let (X, ≤) be partially ordered set and (X, G) be a G-metric space. Let F : X3X and g: XX. Assume there exist ψ Ψ and ϕ Φ such that for x, y, z, a, b, c, u, v, w X, with gxgagu, gygbgv and gzgcgw, we have

$I N ( ψ ( G ( F ( x , y , z ) , F$