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

In this paper, we establish some tripled coincidence point results for a mixed g-monotone mapping F : X³ → X satisfying (ψ, ϕ)-contractions in ordered generalized metric spaces. Also, an application and some examples are given to support our results.

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. [3-7]).

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. [9-11] 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 × X → X 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. [15-18]). For more results on G-metric spaces, one can refer to the papers [9-13,15-28].

Throughout the paper, ℕ* 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 → ℝ⁺ 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 x ≠ y,

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

(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

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

or

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 , that is, for any ε > 0, there exists N ∈ ℕ such that G(x, x_n, x_m) < ε, for all n, m ≥ N. We call x the limit of the sequence and write x_n → x or .

Proposition 4. (see [8]). Let (X,G) be a G-metric space. The following are equivalent:

(1) {x_n} is G-convergent to x,

(2) G(x_n, x_n, x) → 0 as n → +∞,

(3) G(x_n, x, x) → 0 as n → +∞,

(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 ∈ ℕ such that G(x_n, x_m, x_l) < ε for all m, n, l ≥ N, 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) the sequence {x_n} is G-Cauchy,

(2) for any ε > 0, there exists N ∈ ℕ such that G(x_n, x_m, x_m) < ε, for all m, n ≥ N.

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 × X → X 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 × X → X, g : X → X. The mapping F is said to has the mixed g-monotone property if for any x, y, z ∈ X

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

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

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

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

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) ψ is continuous and non-decreasing,

(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

(i) ϕ is lower-continuous and non-decreasing,

(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 [33-36]). In this paper, we establish tripled coincidence point results for mappings F : X³ → X and g : X → X involving nonlinear contractions in the setting of ordered G-metric spaces. Also, we present an application and some examples in support of our 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

Then, .

Proof. First, we have c_n ≤ max{a_n, b_n, c_n}, then

For all n ∈ ℕ, we have

which implies that . Having in mind that

so it follows that

By (3) and (4), we get that . □

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 : X³ → X and g : X → X. Assume there exist ψ ∈ Ψ and ϕ ∈ Φ such that for x, y, z, a, b, c, u, v, w ∈ X, with gx ≥ ga ≥ gu, gy ≤ gb ≤ gv and gz ≥ gc ≥ gw, we have

Assume that F and g satisfy the following conditions:

(1) F(X³) ⊆ g(X),

(2) F has the mixed g-monotone property,

(3) F is continuous,

(4) g is continuous and commutes with F.

Suppose 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 F and g have a tripled coincidence point in X, i.e., there exist x, y, z ∈ X such that

Proof. Suppose x₀, y₀, z₀ ∈ X are such that gx₀ ≤ F(x₀, y₀, z₀), gy₀ ≥ F(y₀, x₀, y₀), and gz₀ ≤ F(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 gx₀ ≤ gx₁, gy₀ ≥ gy₁ and gz₀ ≤ gz₁. 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 gx₀ ≤ gx₁ ≤ gx₂, gy₂ ≤ gy₁ ≤ gy₀ and gz₀ ≤ gz₁ ≤ gz₂. Continuing this process, we can construct three sequences {x_n}, {y_n} and {z_n} in X such that

and

If, for some integer n₀, we have , then , and ; i.e., 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 ∈ ℕ; i.e., we assume that either gx_n+1 ≠ gx_n or gy_n+1 ≠ gy_n or gz_n+1 ≠ gz_n. For any n ∈ ℕ*, we have from (5)

and

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

The fact that ψ is non-decreasing yields that

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

Having in mind that ϕ is non-decreasing, then

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

On the other hand,

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 ≤ r₁ ≤ r and subsequences {x_n(k)} of {x_n} and {y_n(k)} of {y_n} such that

We rewrite (12)

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

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

For any k ∈ ℕ, we rewrite (8) as

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

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

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.,

or . 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

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

By (G5) and (20), we have

Thus, by (18) we obtain

Similarly, we have

Again by (G5) and (20), we have

Letting k → +∞ and using (18), we get

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

Now, using inequality (5) we obtain

and

We deduce from (28)-(30) that

On the other hand, since

then from (27),

Therefore, the real sequence is bounded. Thus, up to a subsequence (still denoted the same), there exists ε₁ with 0 ≤ ε₁ ≤ ε such that

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

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

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

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

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

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

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

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 : X³ → X and g : X → X. Assume there exists k ∈ [0,1) such that for x, y, z, a, b, c, u, v, w ∈ X, with gx ≥ ga ≥ gu, gy ≤ gb ≤ gv and gz ≥ gc ≥ gw, we have

Assume that F and g satisfy the following conditions:

(1) F(X³) ⊂ g(X),

(2) F has the mixed g-monotone property,

(3) F is continuous,

(4) g is continuous and commutes with F.

Suppose 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 F and g have a tripled coincidence point in X, i.e., there exist x, y, z ∈ X such that

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 : X³ → X and g : X → X. Assume there exists k ∈ [0,1) such that for x, y, z, a, b, c, u, v, w ∈ X, with gx ≥ ga ≥ gu, gy ≤ gb ≤ gv and gz ≥ gc ≥ gw, we have

Assume that F and g satisfy the following conditions:

(1) F(X³) ⊆g(X),

(2) F has the mixed g-monotone property,

(3) F is continuous,

(4) g is continuous and commutes with F.

Suppose 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 F and g have a tripled coincidence point in X, i.e., there exist x, y, z ∈ X such that

Proof. It suffices to remark that

□

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:

(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 19. Let (X, ≤) be a partially ordered set and (X, G) be a G-metric space. Let F : X³ → X and g: X → X. Assume there exist ψ ∈ Ψ and ϕ ∈ Φ such that for x, y, z, a, b, c, u, v, w ∈ X, with gx ≥ ga ≥ gu, gy ≤ gb ≤ gv and gz ≥ gc ≥ gw, we have

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 gx₀ ≤ F(x₀, y₀, z₀), gy₀ ≥ F(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

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 gy ≤ gy_n for all n ≥ 0. Using (5), we get

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

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 : X³ → X and g: X → X. Assume there exists k ∈ [0, 1) such that for x, y, z, a, b, c, u, v, w ∈ X, with gx ≥ ga ≥ gu, gy ≤ gb ≤ gv and gz ≥ gc ≥ gw, we have

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 gx₀ ≤ F(x₀, y₀, z₀), gy ₀ ≥ F(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

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 : X³ → X and g : X → X. Assume there exists k ∈ [0, 1) such that for x, y, z, a, b, c, u, v, w ∈ X, with gx ≥ ga ≥ gu, gy ≤ gb ≤ gv and gz ≥ gc > gw, we have

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 gx₀ ≤ F(x₀, y₀, z₀), gy₀ ≥ F(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

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:

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

(ii) for all ε > 0, we have

Let N ∈ 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:

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 : X³ → X and g : X → X. Assume there exist ψ ∈ Ψ and ϕ ∈ Φ such that for x, y, z, a, b, c, u, v, w ∈ X, wth gx ≥ ga ≥ gu, gy ≤ gb ≤ gv and gz ≥ gc ≥ gw, we have

Assume that F and g satisfy the following conditions:

(1) F(X³) ⊆ g(X),

(2) F has the mixed g-monotone property,

(3) F is continuous,

(4) g is continuous and commutes with F.

Suppose 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 F and g have a tripled coincidence point in X, i.e., there exist x, y, z ∈ X such that

Proof. Take

It is easy to show that and . From (46), we have

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 : X³ → X and g: X → X. Assume there exist ψ ∈ Ψ and ϕ ∈ Φ such that for x, y, z, a, b, c, u, v, w ∈ X, with gx ≥ ga ≥ gu, gy ≤ gb ≤ gv and gz ≥ gc ≥ gw, we have

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 gx₀ ≤ F(x₀, y₀, z₀), gy₀ ≥ F(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

In this section, we study the existence of solutions to nonlinear integral equations using the results proved in section "Main results".

Consider the integral equations in the following system

We will analyze the system (48) under the following assumptions:

(i) f, k, h: [0, T] × ℝ → ℝ are continuous,

(ii) p : [0, T] → ℝ is continuous,

(iii) S: [0, T] × ℝ → [0, ∞) is continuous,

(iv) there exists q > 0 such that for all x, y ∈ ℝ, y ≥ x,

(v) We suppose that

(vi) There exist continuous functions α, β, γ : [0, T] → ℝ such that

We consider the space X = C([0,T],ℝ) of continuous functions defined on [0,T] endowed with the (G-complete) G-metric given by

We endow X with the partial ordered ≤ given by: x, y ∈ X, x ≤ y ⇔ x(t) ≤ y(t) for all t ∈ [0, T].

On the other hand, we may adjust as in [37] to prove that (X, G, ≤) is regular.

Our result is the following.

Theorem 25. Under assumptions (i)-(vi), the system (48) has a solution in X³ = (C([0, T], ℝ))³.

Proof. We consider the operators F : X³ → X and g : X → X defined by

for all x₁, x₂, x₃, x ∈ X.

First, we will prove that F has the mixed monotone property (since g is the identity on X).

In fact, for x₁ ≤ y₁ and t ∈ [0, T], we have

Taking into account that x₁(t) ≤ y₁(t) for all t ∈ [0, T], so by (iv), f(s, y₁ (s)) ≥ f(s, x₁ (s)). Then F(y₁, x₂, x₃)(t) ≥ F(x₁, x₂, x₃)(t) for all t ∈ [0,T],i.e.,

Similarly, for x₂ ≤ y₂ and t ∈ [0, T], we have

Having x₂(t) ≤ y₂(t), so by (iv), k(s, x₂(s)) ≥ k(s, y₂(s)). Then F(x₁, x₂, x₃)(t) ≥ F(x₁, y₂, x₃)(t) for all t ∈ [0,T], i.e.,

Now, for x₃ ≤ y₃ and t ∈ [0, T], we have

Taking into account that x₃(t) ≤ y₃(t) for all t ∈ [0, T], so by (iv), h(s, x₃(s)) ≥ h(s, y₃(s)). Then F(x₁, x₂, x₃)(t) ≥ F(x₁, x₂, y₃)(t) for all t ∈ [0, T], i.e.,

Therefore, F has the mixed monotone property.

In what follows we estimate the quantity G(F(x, y, z), F(a, b, c), F(u, v, w)) for all x, y, z, a, b, c, u, v, w ∈ X, with x ≥ a ≥ u, y ≤ b ≤ v and z ≥ c ≥ w. Since F has the mixed monotone property, we have

We obtain

Note that for all t ∈ [0, T], from (iv), we have

Thus,

Repeating this idea, we may get using definition of the the G-metric G

From (v), we have . This proves that the operator F satisfies the contractive condition appearing in Corollary 20.

Let α, β, γ be the functions appearing in assumption (vi), then by (vi), we get

Applying Corollary 20, we deduce the existence of x₁, x₂, x₃ ∈ X such that

i.e., (x₁, x₂, x₃) is a solution of the system (48). □

In this section, we state two examples to support the usability of our results. Before we present our first example we worth to mention the following remark.

Remark 26. All our results still valid if (u, v, w) = (a, b, c).

Example 27. Let X = [0, 1] with usual order. Define G : X × X × X → X by

Define F: X × X × X → X by

Also, define ψ, ϕ : [0, +∞) → [0, +∞) by ψ(t) = t and . Then:

a. (X, G, ≤) is a G-complete regular G-metric space.

b. For x, y, z, u, v, w ∈ X with x ≥ u ≥ u, y ≤ v ≤ v and z ≥ w ≥ w, we have

c. F has the mixed monotone property.

Proof. To prove (b), given x, y, z, u, v, w ∈ X with x ≥ u, y ≤ v and z ≥ w. Then:

Case 1: y > min{x, z} and v ≥ min{u, w}. Here, we have

Case 2: y ≥ min{x, z} and u ≥ w ≥ v. Here, we have y ≤ v ≤ w ≤ u ≤ x and y ≤ v ≤ w ≤ z. Hence y = v = w = u = x or y = v = w = z. Therefore

Case 3: y ≥ min{x, z} and w ≥ u ≥ v. Here, we have y ≤ v ≤ u ≤ w ≤ z and y ≤ v ≤ u ≤ x. Thus y = v = u = w = z or y = v = u = x. Therefore

Case 4: x ≥ z ≥ y and v ≥ min{u, w}.

Suppose w ≤ v, then w - y ≤ v - y and hence

Then

Suppose v < w, then u ≤ v < w and hence u ≤ v ≤ w ≤ z ≤ x. So

Therefore,

Case 5: z ≥ x ≥ y and v ≥ min{u, w}.

Suppose u ≤ v, then u - y ≤ v - y and hence

Therefore,

Suppose v < u, then w ≤ v < u and hence w ≤ v <u ≤ x ≤ z. So

Therefore,

Case 6: x ≥ z ≥ y and u ≥ w ≥ v. Here, we have

Case 7: x ≥ z ≥ y and w ≥ u ≥ v. Here, we have y ≤ v ≤ u ≤ w ≤ z ≤ x. Thus,

Case 8: z ≥ x ≥ y and u ≥ w ≥ v. Here, we have y ≤ v ≤ w ≤ u ≤ x ≤ z. Therefore, we have

Case 9: z ≥ x ≥ y, w ≥ u ≥ v. Here, we have y ≤ v ≤ u ≤ w ≤ z. Therefore, we have

To prove (c), let x, y, z ∈ X. To show that F(x, y, z) is monotone non-decreasing in x, let x₁, x₂ ∈ X with x₁ ≤ x₂. If y ≥ min{x₁, z}, then F(x₁, y, z) = 0 ≤ F(x₂, y, z).

If y < min{x₁, z}, then

Therefore, F(x, y, z) is monotone non-decreasing in x. Similarly, we may show that F(x, y, z) is monotone non-decreasing in z and monotone non-increasing in y. Thus, by Theorem 19 and Remark 26, F has a tripled fixed point. Here, (0, 0,0) is the unique tripled fixed point of F.

□

Now, we state our second example to support the usability of our results for non-symmetric G-metric spaces.

Example 28. Let X = {0, 1, 2, 3,...}. Define G : X × X × X → X by

F: X × X × X → X by

and g: X → X by gx = x². Also, define ψ, ϕ: [0, +∞) → [0, +∞) by ψ(t) = t² and ϕ(t) = t. Then

a. (X, G, ≤) is a complete nonsymmetric G-metric space.

b. For x, y, z, u, v, w ∈ X with gx ≥ gu ≥ gu, gy ≤ gv ≤ gv and gz ≥ gw ≥ gw, we have

c. F has the mixed g-monotone property.

d. F(X × X × X) ⊆ gX.

e. (X, G, ≤) is regular.

Proof. For (a) see Example 3.5 of Choudhury and Maity [17]. To prove (b), given x, y, z, u, v, w ∈ X with gx ≥ gu, gy ≤ gv and gz ≥ gw. Then: □

Case 1: (x ≤ 3 ⋁ z ≤ 3) ⋀ (u ≤ 3 ⋁ w ≤ 3). Here, we have F(x, y, z) = 0 and F(u, v, w) = 0. Thus

Case 2: (x ≤ 3 ⋁ z ≤ 3) ⋀ (u ≥ 4 ⋀ w ≥ 4). Here, x < u or z < w which is impossible because gx ≥ gu and gz ≥ gw.

Case 3: (x ≥ 4 ⋀ z ≥ 4) ⋀ (u ≤ 3 ⋁ w ≤ 3). Here, we have F(x, y, z) = 1 and F(u, v, w) = 0. Thus

Also,

In both cases, we have

Using the fact that if a, b ∈ ℕ with a ≤ b, then a² - a ≤ b² - b, we deduce that

Therefore,

Case 4: (x ≥ 3 ⋀ z ≥ 3) ⋀ (u ≥ 3 ⋀ w ≥ 3). Here, we have F(x, y, z) = 1 and F(u, v, w) = 1. Thus,

The proof of (c) and (d) are easy. To prove (e), let {x_n} be a non-decreasing sequences in X such that {x_n} G-converges to x. Then G(x_n, x, x) → 0. By definition of G, we conclude that x_n = x for all n except finitely many. Thus x_n ≤ x for all n ∈ ℕ. Similarly, we show that if {y_n} is a non-increasing sequence in X such that {y_n} G-converges to y, then y_n ≥ y for all n ∈ ℕ. Thus, (X, G, ≤) is regular.

By Theorem 19 and Remark 26, F and g have a tripled coincidence point in X. Here, (0, 0, 0) is the tripled coincidence point of F and g.

The authors declare that they have no competing interests.

All authors read and approved the final manuscript.

The authors thank the editor and the referees for their useful comments and suggestions.

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