Fixed point theory has fascinated hundreds of researchers since 1922 with the celebrated Banach's fixed point theorem. This theorem provides a technique for solving a variety of applied problems in mathematical sciences and engineering. There exists a last literature on the topic and this is a very active field of research at present. There are great number of generalizations of the Banach contraction principle. Bhaskar and Lakshmikantham 1 introduced the notion of coupled fixed point and proved some coupled fixed point results under certain conditions, in a complete metric space endowed with a partial order. Later, Lakshmikantham and Ćirić 2 extended these results by defining the mixed g-monotone property. More accurately, they proved coupled coincidence and coupled common fixed point theorems for a mixed g-monotone mapping in a complete metric space endowed with a partial order. Karapınar 34 generalized these results on a complete cone metric space endowed with a partial order. For other results on coupled fixed point theory, we address the readers to 5678910111213.

To make our exposition self contained, in this section we recall some previous notations and known results.

For simplicity, we denote from now on X × X ⋯ X × X ︸ k terms by X^k, where k ∈ ℕ and X be a non-empty set.

Let (X, ≤) be a partially ordered set. According to 1, the mapping F: X² → X is said to have mixed monotone property if F(x, y) is monotone non-decreasing in x and is monotone non-increasing in y, that is, for any x, y ∈ X,

x 1 ≤ x 2 ⇒ F ( x 1 , y ) ≤ F ( x 2 , y ) , for x 1 , x 2 ∈ X , y 1 ≤ y 2 ⇒ F ( x , y 2 ) ≤ F ( x , y 1 ) , for y 1 , y 2 ∈ X .

An element (x, y) ∈ X² is said to be a coupled fixed point of the mapping F: X² → X if

F ( x , y ) = x and F ( y , x ) = y .

Theorem 1.1. (1) Let (X, ≤) be an ordered set such that there exists a metric d on X such that (X, d) is complete. Let F: X² → X be a continuous mapping having the mixed monotone property on X. Assume that there exists k ∈ [0, 1) with

d ( F ( x , y ) , F ( u , v ) ) ≤ k 2 [ d ( x , u ) + d ( y , v ) ] , f o r a l l u ≤ x , y ≤ u .

If there exist x₀, y₀ ∈ X such that x₀ ≤ F(x₀, y₀) and F(y₀, x₀) ≤ y₀, then, there exist x, y ∈ X such that x = F(x, y) and y = F(y, x).

Recently, Samet and Vetro 14 introduced the notion of fixed point of N-order as natural extension of that of coupled fixed point and established some new coupled fixed point theorems in complete metric spaces, using a new concept of F-invariant set. Later, Berinde and Borcut 15 obtained existence and uniqueness of triple fixed point results in a complete metric space, endowed with a partial order.

Again, let (X, ≤) be a partially ordered set. In accordance with 15, the mapping F: X³ → X is said to have the mixed monotone property if for any x, y, z ∈ X

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

An element (x, y, z) ∈ X³ is called a tripled fixed point of F if

F ( x , y , z ) = x , F ( y , x , y ) = y and F ( z , y , x ) = z .

Berinde and Borcut 15 proved the following theorem.

Theorem 1.2. (15) Let (X, ≤) be a partially ordered set and (X, d) be a complete metric space. Let F: X³ → X be a mapping having the mixed monotone property on X. Assume that there exist constants a, b, c ∈ [0, 1) such that a + b + c < 1 for which

d ( F ( x , y , z ) , F ( u , v , w ) ) ≤ a d ( x , u ) + b d ( y , v ) + c d ( z , w )

for all x ≥ u, y ≤ υ, z ≥ w. Assume either

(I) F is continuous, or

(II) X has the following properties:

(i) if non-decreasing sequence x_n → x, then x_n ≤ x for all n,

(ii) if non-increasing sequence y_n → y, then y_n ≥ y for all n.

If there exist x₀, y₀, z₀ ∈ X such that

x 0 ≤ F ( x 0 , y 0 , z 0 ) , y 0 ≥ F ( y 0 , x 0 , y 0 ) , a n d z 0 ≤ F ( x 0 , y 0 , z 0 )

then there exist x, y, z ∈ X such that

F ( x , y , z ) = x , F ( y , x , y ) = y , a n d F ( z , y , x ) = z .

In this article, we establish tripled coincidence point theorems for F: X³ → X and g: X → X satisfying nonlinear contractive conditions, in partially ordered metric spaces. The presented theorems extend and improve some results in litterature.

We shall start this section by recalling the following basic notions, introduced by [Abbas, Aydi and Karapınar, Tripled common fixed point in partially ordered metric spaces, submitted]. In this respect, let us consider (X, ≤) a partially ordered set, F: X³ → X and g: X → X two mappings. The mapping F is said to have 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 ) .

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

while (gx, gy, gz) is said a tripled point of coincidence of mappings F and g. Moreover, (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 , and F ( z , y , x ) = g z = z .

At last, mappings F and g are called commutative if

g ( F ( x , y , z ) ) = F ( g x , g y , g z ) , ∀ x , y , z ∈ X .

In the same paper, they proved the following result.

Theorem 2.1. Let (X, ≤) be a partially ordered set and suppose there is a metric d on X such that (X, d) is a complete metric space. Assume there is a function φ: [0, +∞) → [0, +∞) such that φ(t) < t for each t > 0. Also suppose F: X³ → X and g: X → X are such that F has the mixed g-monotone property and suppose there exist p, q, r ∈ [0, 1) with p + 2q + r < 1 such that

d ( F ( x , y , z ) , F ( u , v , w ) ) ≤ φ p d ( g x , g u ) + q d ( g y , g v ) + r d ( g z , g w ) ,

for any x, y, z ∈ X for which gx > gu, gυ ≥ gy and gz ≥ gw.

Suppose F(X³) ⊂ g(X), g is continuous and commutes with F. Suppose either

(a) F is continuous, or

(b) X has the following properties:

(i) if non-decreasing sequence gx_n → x (respectively, gz_n → z), then gx_n ≤ x (respectively, gz_n ≤ z) for all n,

(ii) if non-increasing sequence gy_n → y, then gy_n ≥ y for all n.

If there exist x₀, y₀, z₀ ∈ X such that gx₀ ≤ F(x₀, y₀, z₀), gy₀ ≥ F(y₀, x₀, y₀) and gz₀ ≤ F(z₀, y₀, x₀), then there exist x, y, z ∈ X such that

F ( x , y , z ) = g x , F ( y , x , y ) = g y , a n d F ( z , y , x ) = g z ,

that is, F and g have a tripled coincidence point.

Before starting to introduce our results, let us consider the set of functions

Φ = φ : [ 0 , + ∞ ) → [ 0 , + ∞ ) | φ ( t ) < t and lim r → t + φ ( r ) < t , t > 0 .

Our first main result is the following:

Theorem 2.2. Let (X, ≤) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose F: X³ → X and g: X → X are such that F has the mixed g-monotone property and F(X³) ⊂ g(X). Assume there is a function φ ∈ Φ such that

d ( F ( x , y , z ) , F ( u , v , w ) ) + d ( F ( y , x , y ) , F ( v , u , v ) ) + d ( F ( z , y , x ) , F ( w , v , u ) ) ≤ 3 φ d ( g x , g u ) + d ( g y , g v ) + d ( g z , g w ) 3 ,

for any x, y, z, u, υ, w ∈ X for which gx ≥ gu, gυ ≥ gy and gz ≥ gw. Assume that F is continuous, g is continuous and commutes with F. If there exist x₀, y₀, z₀ ∈ X such that

g x 0 ≤ F ( x 0 , y 0 , z 0 ) , g y 0 ≥ F ( y 0 , x 0 , y 0 ) a n d g z 0 ≤ F ( z 0 , y 0 , x 0 ) ,

then there exist x, y, z ∈ X such that

F(x,y,z)=gx,F(y,x,y)=gy,andF(z,y,x)=gz,

that is, F and g have a tripled coincidence point.

Proof. Let x₀, y₀, z₀ ∈ X be such that gx₀ ≤ F(x₀, y₀, z₀), gy₀ ≥ F(y₀, x₀, y₀) and gz₀ ≤ F(z₀, y₀, x₀). We can choose x₁, y₁, z₁ ∈ X such that

g x 1 = F ( x 0 , y 0 , z 0 ) , g y 1 = F ( y 0 , x 0 , y 0 ) and g z 1 = F ( z 0 , y 0 , x 0 ) .

This can be done because F(X³) ⊂ g(X). Continuing this process, we construct sequences {x_n}, {y_n}, and {z_n} in X such that

g x n + 1 = F ( x n , y n , z n ) , g y n + 1 = F ( y n , x n , z n ) , and g z n + 1 = F ( z n , y n , x n ) .

By induction, we will prove that

g x n ≤ g x n + 1 , g y n + 1 ≤ g y n , and g z n ≤ g z n + 1 .

Since gx₀ ≤ F(x₀, y₀, z₀), gy₀ ≥ F(y₀, x₀, y₀), and gz₀ ≤ F(z₀, y₀, x₀), therefore by (2.4) we have

g x 0 ≤ g x 1 , g y 1 ≤ g y 0 , and g z 0 ≤ g z 1 .

Thus (2.6) is true for n = 0. We suppose that (2.6) is true for some n > 0. Since F has the mixed g-monotone property, by gx_n ≤ gx_n+1, gy_n+1 ≤ gy_n, and gz_n ≤ gz_n+1, we have that

g x n + 1 = F ( x n , y n , z n ) ≤ F ( x n + 1 , y n , z n ) ≤ F ( x n + 1 , y n , z n + 1 ) ≤ F ( x n + 1 , y n + 1 , z n + 1 ) = g x n + 2 ,

g y n + 2 = F ( y n + 1 , x n + 1 , y n + 1 ) ≤ F ( y n + 1 , x n , y n + 1 ) ≤ F ( y n , x n , y n + 1 ) ≤ F ( y n , x n , y n ) = g y n + 1 ,

and

g z n + 1 = F ( z n , y n , x n ) ≤ F ( z n + 1 , y n , x n ) ≤ F ( z n + 1 , y n + 1 , x n ) ≤ F ( z n + 1 , y n + 1 , x n + 1 ) = g z n + 2 .

That is, (2.6) is true for any n ∈ ℕ. If for some k ∈ ℕ

g x k = g x k + 1 , g y k = g y k + 1 , and g z k = g z k + 1 ,

then, by (2.5), (x_k, y_k, z_k) is a tripled coincidence point of F and g. From now on, we assume that at least

g x n ≠ g x n + 1 or g y n ≠ g y n + 1 or g z n ≠ g z n + 1

for any n ∈ ℕ. From (2.6) and the inequality (2.2)

d ( g x n + 1 , g x n ) + d ( g y n + 1 , g y n ) + d ( g z n + 1 , g z n )   = d ( F ( x n , y n , z n ) , F ( x n − 1 , y n − 1 , z n − 1 ) ) + d ( F ( y n , x n , y n ) , F ( y n − 1 , x n − 1 , y n − 1 ) )   + d ( F ( z n , y n , x n ) , F ( z n − 1 , y n − 1 , x n − 1 ) )   ≤ 3 φ ( 1 3 ( d ( g x n , g x n − 1 ) + d ( g y n , g y n − 1 ) + d ( g z n , g z n − 1 ) ) .

For each n ≥ 1, take

δ n : = 1 3 ( d ( g x n , g x n - 1 ) + d ( g y n , g y n - 1 ) + d ( g z n , g z n - 1 ) ) .

One can write

δ n + 1 ≤ φ ( δ n ) ∀ n ≥ 1 .

By (2.7), we have δ_n > 0. Having in mind φ(t) < t for each t > 0, so we have φ(δ_n) < δ_n. From (2.9), we get

δ n + 1 < δ n ∀ n ≥ 1 ,

that is, the sequence {δ_n} is non-negative and decreasing. Therefore, there exists some δ ≥ 0 such that

lim n → + ∞ δ n = lim n → + ∞ 1 3 d ( g x n , g x n - 1 ) + d ( g y n , g y n - 1 ) + d ( g z n , g z n - 1 ) = δ + .

We shall prove that δ = 0. Assume, on the contrary, that δ > 0. Then by letting n → +∞ in (2.9) we have

0 < δ = lim n → + ∞ δ n + 1 ≤ lim n → + ∞ φ ( δ n ) = lim r → δ + δ ( r ) < δ ,

which is a contradiction. Thus, δ = 0, and by (2.10), we get

lim n → + ∞ δ n = 0 .

We now prove that {gx_n}, {gy_n}, and {gz_n} are Cauchy sequences in (X,d).

Suppose, on the contrary, that at least one of {gx_n}, {gy_n}, and {gz_n} is not a Cauchy sequence. So, there exists ε > 0 for which we can find subsequences {gx_n(k)}, {gx_m(k)} of {gx_n}, {gy_n(k)}, {gy_m(k)} of {gy_n}, and {gz_n(k)}, {gz_m(k)} of {gz_n} with n(k) > m(k) ≥ k such that

d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) + d ( g z n ( k ) , g z m ( k ) ) ≥ ε .

Additionally, corresponding to m(k), we may choose n(k) such that it is the smallest integer satisfying (2.12) and n(k) > m(k) ≥ k. Thus,

d ( g x n ( x ) - 1 , g x m ( k ) ) + d ( g y n ( k ) - 1 , g y m ( k ) ) + d ( d z n ( k ) - 1 , g z m ( k ) ) < ε .

By using triangle inequality and having in mind (2.12) and (2.13)

ε ≤ t k = d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) + d ( g z n ( k ) , g z m ( k ) ) ≤ d ( g x n ( k ) , g x n ( k ) - 1 ) + d ( g x n ( k ) - 1 , g x m ( k ) ) + d ( g y n ( k ) , g y n ( k ) - 1 ) + d ( g y n ( k ) - 1 , g y m ( k ) ) + d ( g z n ( k ) , g z n ( k ) - 1 ) + d ( g z n ( k ) - 1 , g z m ( k ) ) < d ( g x n ( k ) , g x n ( k ) - 1 ) + d ( g y n ( k ) , g y n ( k ) - 1 ) + d ( g z n ( k ) , g z n ( k ) - 1 ) + ε .

Letting k → ∞ in (2.14) and using (2.11)

lim k → ∞ t k = lim k → ∞ d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) + d ( g z n ( k ) , g z m ( k ) ) = ε .

Again by triangle inequality,

t k = d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) + d ( g z n ( k ) , g z m ( k ) ) ≤ d ( g x n ( k ) , g x n ( k ) + 1 ) + d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g x m ( k ) + 1 , g x m ( k ) ) + d ( g y n ( k ) , g y n ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) + d ( g y m ( k ) + 1 , g y m ( k ) ) + d ( g z n ( k ) , g z n ( k ) + 1 ) + d ( g z n ( k ) + 1 , g y m ( k ) + 1 ) + d ( g z m ( k ) + 1 , g z m ( k ) ) ≤ δ n ( k ) + 1 + δ m ( k ) + 1 + d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) + d ( g z n ( k ) + 1 , g z m ( k ) + 1 ) .

Since n(k) > m(k), then

g x n ( k ) ≥ g x m ( k ) , g y n ( k ) ≤ g y m ( k ) , g z n ( k ) ≥ g z m ( k ) .

Take (2.17) in (2.2) to get

d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) + d ( g z n ( k ) + 1 , g z m ( k ) + 1 )     = d ( F ( x n ( k ) , y n ( k ) , z n ( k ) ) , F ( x m ( k ) , y m ( k ) , z m ( k ) )     + d ( F ( y n ( k ) , x n ( k ) , y n ( k ) ) , F ( y m ( k ) , x m ( k ) , y m ( k ) )     + d ( F ( z n ( k ) , y n ( k ) , x n ( k ) ) , F ( z m ( k ) , y m ( k ) , x m ( k ) ) )     ≤ 3 φ ( 1 3 [ d ( g x n ( k ) , g x m ( k ) ) + d ( g y n ( k ) , g y m ( k ) ) + d ( g z n ( k ) , g z m ( k ) ) ] )     = 3 φ ( t k 3 ) .

Combining this in (2.16), we obtain that

t k ≤ δ n ( k ) + 1 + δ m ( k ) + 1 + d ( g x n ( k ) + 1 , g x m ( k ) + 1 ) + d ( g y n ( k ) + 1 , g y m ( k ) + 1 ) + d ( g z n ( k ) + 1 , g z m ( k ) + 1 ) ≤ δ n ( k ) + 1 + δ m ( k ) + 1 + 3 φ t k 3 .

Letting k → ∞ and having in mind (2.11) and (2.15), we get

ε ≤ 3 lim k → + ∞ φ 1 3 t k = 3 lim r → 1 3 t + ϕ ( r ) < 3 1 3 ε = ε ,

which is a contradiction. This shows that {gx_n}, {gy_n}, and {gz_n} are Cauchy sequences in (X, d).

Since X is complete, there exist x, y, z ∈ X such that

lim n → + ∞ g x n = x , lim n → + ∞ g y n = y and lim n → + ∞ g z n = z .

From (2.18) and the continuity of g.

lim n → + ∞ g ( g x n ) = g x , lim n → + ∞ g ( g y n ) = g y , and lim n → + ∞ g ( g z n ) = g z .

From the commutativity of F and g, we have

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

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

Suppose that F is continuous. Letting n → +∞ in (2.20), therefore by (2.18) and (2.19), we obtain

g x = lim n → + ∞ g ( g x n + 1 ) = lim n → + ∞ F ( g x n , g y n , g z n ) = F lim n → + ∞ g x n , lim n → + ∞ g y n , lim n → + ∞ g z n = F ( x , y , z ) ,

g y = lim n → + ∞ g ( g y n + 1 ) = lim n → + ∞ F ( g y n , g x n , g y n ) = F lim n → + ∞ g y n , lim n → + ∞ g x n , lim n → + ∞ g y n = F ( y , x , y ) ,

and

g z = lim n → + ∞ g ( g z n + 1 ) = lim n → + ∞ F ( g z n , g y n , g x n ) = F lim n → + ∞ g z n , lim n → + ∞ g y n , lim n → + ∞ g x n = F ( z , y , x ) .

We have proved that F and g have a tripled coincidence point.

Corollary 2.3. Let (X, ≤) be a partially ordered set and suppose there is a metric d on X such that (X,d) is a complete metric space. Suppose F: X³ → X and g: X → X are such that F has the mixed g-monotone property and F(X³) ⊂ g(X). Assume there exists α ∈ [0,1) such that

d ( F ( x , y , z ) , F ( u , v , w ) ) + d ( F ( y , x , y ) , F ( v , u , v ) ) + d ( F ( z , y , x ) , F ( w , v , u ) ) ≤ α ( d ( g x , g u ) + d ( g y , g v ) + d ( g z , g w ) ) ,

for any x, y, z, u, υ, w ∈ X for which gx ≥ gu, gυ ≥ gy, and gz ≥ gw. Assume that F is continuous, g is continuous and commutes with F. If there exist x₀, y₀, z₀ ∈ X such that

g x 0 ≤ F ( x 0 , y 0 , z 0 ) , g y 0 ≥ F ( y 0 , x 0 , y 0 ) , and g z 0 ≤ F ( z 0 , y 0 , x 0 ) ,

then there exist x, y, z ∈ X such that

F ( x , y , z ) = g x , F ( y , x , y ) = g y , a n d F ( z , y , x ) = g z ,

that is, F and g have a tripled coincidence point.

Proof. It follows by taking φ(t) = αt in Theorem 2.2.

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

Definition 2.1. Let (X, ≤) be a partially ordered set and d be a metric on X. We say that (X, d,≤) is regular if the following conditions hold:

(i) if a non-decreasing sequence (x_n) is such that x_n → x, then x_n ≤ x for all n,

(ii) if a non-increasing sequence (y_n) is such that y_n → y, then y ≤ y_n for all n.

Theorem 2.4. Let (X, ≤) be a partially ordered set and d be a metric on X such that (X, d, ≤) is regular. Suppose that there exist φ ∈ Φ and mappings F: X³ → X and g: X → X such that (2.2) holds for any x, y, z, u, υ, w ∈ X for which gx ≥ gu, gυ ≥ gy and gz ≥ gw. Suppose also that (g(X), d) is complete, F has the mixed g-monotone property and F(X³) ⊂ g(X). If there exist x₀, y₀, z₀ ∈ X such that gx₀ ≤ F(x₀, y₀, z₀), gy₀ ≥ F(y₀, x₀, y₀), and gz₀ ≤ F(z₀, y₀, x₀), then there exist x, y, z ∈ X such that

F(x,y,z)=gx,F(y,x,y)=gy,andF(z,y,x)=gz,

that is, F and g have a tripled coincidence point.

Proof. Proceeding exactly as in Theorem 2.2, we have that (gx_n), (gy_n), and (gz_n) are Cauchy sequences in the complete metric space (g(X), d). 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, d, ≤), we have gx_n ≤ gx, gz_n ≤ gz, and gy ≤ gy_n for all n ≥ 0. If gx_n = gx, gy_n = gy, and gz_n = gz for some n ≥ 0, then gx = gx_n ≤ gx_n+1 ≤ gx = gx_n, gz = gz_n ≤ gz_n+1 ≤ gz = gz_n, and gy ≤ gy_n+1 ≤ gy_n = gy, which implies that gx_n = gx_n+1 = F(x_n, y_n, z_n), gy_n = gy_n+1 = F(y_n, x_n, y_n), and gz_n = gz_n+1 = F(z_n, y_n, x_n), that is, (x_n, y_n, z_n) is a tripled coincidence point of F and g. Then, we suppose that (gx_n, gy_n, gz_n) ≠ (gx, gy, gz) for all n ≥ 0. Using the triangle inequality, (2.2) and the property φ(t) < t for all t > 0,

d ( g x , F ( x , y , z ) ) ≤ d ( g x , g x n + 1 ) + d ( g x n + 1 , F ( x , y , z ) ) = d ( g x , g x n + 1 ) + d ( F ( x , y , z ) , F ( x n , y n , z n ) ) ≤ d ( g x , g x n + 1 ) + 3 φ 1 3 [ d ( g x n , g x ) + d ( g y n , g y ) + d ( d z n , g z ) ] < d ( g x , g x n + 1 ) + d ( g x n , g x ) + d ( g y n , g y ) + d ( g z n , g z ) .

Taking n → ∞ in the above inequality we obtain that d(gx,F(x, y, z)) = 0, so gx = F(x, y, z).

Analogously, we find that

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

thus, we have proved that F and g have a tripled coincidence point.

Corollary 2.5. Let (X, ≤) be a partially ordered set and suppose there is a metric d on X such that (X, ≤,d) is regular. Suppose F: X³ → X and g: X → X are such that F has the mixed g-monotone property and F(X³) ⊂ g(X). Assume there exists α ∈ [0,1) such that

d ( F ( x , y , z ) , F ( u , v , w ) ) + d ( F ( y , x , y ) , F ( v , u , v ) ) + d ( F ( z , y , x ) , F ( w , v , u ) ) ≤ α ( d ( g x , g u ) + d ( g y , g v ) + d ( g z , g w ) ) ,

for any x, y, z, u, υ, w ∈ X for which gx ≥ gu, gυ ≥ gy, and gz ≥ gw. Suppose also that (g(X), d) is complete. If there exist x₀, y₀, z₀ ∈ X such that

g x 0 ≤ F ( x 0 , y 0 , z 0 ) , g y 0 ≥ F ( y 0 , x 0 , y 0 ) and g z 0 ≤ F ( z 0 , y 0 , x 0 ) ,

then there exist x, y, z ∈ X such that

F ( x , y , z ) = g x , F ( y , x , y ) = g y , a n d F ( z , y , x ) = g z ,

that is, F and g have a tripled coincidence point.

Proof. It follows by taking φ(t) = αt in Theorem 2.4.

Now, we shall prove the existence and the uniqueness of a tripled common fixed point theorem. For a product X³ = X × X × X of a partial ordered set (X, ≤), we define a partial ordering in the following way: For all (x, y, z), (u, υ, r) ∈ X³

( x , y , z ) ≤ ( u , v , r ) ⇔ x ≤ u , y ≥ v and z ≤ r .

We say that (x, y, z) and (u, υ, w) are comparable if

( x , y , z ) ≤ ( u , v , r ) or ( u , v , r ) ≤ ( x , y , z ) .

Also, we say that (x, y, z) is equal to (u, υ, r) if and only if x = u, y = υ and z = r.

Theorem 2.6. In addition to hypothesis of Theorem 2.2, suppose that for all (x, y, z) and (u, υ, r) in X³, there exists (a, b, c) in X³ such that (F(a, b, c), F(b, a, b), F(c, b, a)) is comparable to (F(x, y, z), F(y, x, y), F(z, y, x)) and (F(u, υ, r), F(υ, u, υ), F(r, υ, u)). Also, assume that φ is non-decreasing. Then, F and g have a unique tripled common fixed point (x, y, z), that is

x = g x = F ( x , y , z ) , y = g y = F ( y , x , y ) , a n d z = g z = F ( z , y , x ) .

Proof. Due to Theorem 2.2, the set of tripled coincidence points of F and g is not empty. Assume now, that (x, y, z) and (u,υ,r) are two tripled coincidence points of F and g, that is,

F ( x , y , z ) = g x , F ( y , x , y ) = g y , and F ( z , y , x ) = g z , F ( u , v , r ) = g u , F ( v , u , v ) = g v , and F ( r , v , u ) = g r .

We shall show that (gx, gy, gz) and (gu, gυ, gr) are equal.

By assumption, there is (a, b, c) in X³ such that (F(a, b, c), F(b, a, b), F(c, b, a)) is comparable to (F(x, y, z), F(y, x, y), F(z, y, x)) and (F(u, υ, r), F(υ, u, v), F(r, υ, u)).

Define the sequences {ga_n},{gb_n}, and {gc_n} such that a = a₀, b = b₀, c = c₀ and

g a n = F ( a n - 1 , b n - 1 , c n - 1 ) , g b n = F ( b n - 1 , a n - 1 , b n - 1 ) , g c n = F ( c n - 1 , b n - 1 , a n - 1 ) ,