Remarks on Some Recent Coupled Coincidence Point Results in Symmetric G-Metric Spaces

Definition 1. Let X be a nonempty set, and let G : X³ → ℝ⁺ be a function satisfying the following properties:

• (a)

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

• (b)

  0 < G(x, y, z) for all x, y, and z ∈ X with x ≠ y;

• (c)

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

• (d)

  G(x, y, z) = G(x, z, y) = G(y, z, x) = ⋯ (symmetry in all three variables); and

• (e)

  G(x, y, z) ≤ G(x, a, a)  +  G(a, y, z) for all x, y, z, and a ∈ X.

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

Definition 2. Let (X, G) be a G-metric space and let {x_n} be a sequence of points in X.

• (i)

  A point x ∈ X is said to be the limit of a sequence {x_n} if lim _n,m→∞G(x, x_n, x_m) = 0, and one says that the sequence {x_n} is G-convergent to x.

• (ii)

  The sequence {x_n} is said to be a G-Cauchy sequence 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, l ≥ N; that is, G(x_n, x_m, x_l) → 0, as n, m, l → ∞.

• (iii)

  (X, G) is said to be G-complete (or a complete G-metric space) if every G-Cauchy sequence in (X, G) is G-convergent in X.

Proposition 3 (see [1].)Let (X, G) be a G-metric space, and let {x_n} be a sequence of points in X. Then the following are equivalent.

• (1)

  The sequence {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 4 (see [1], [10].)A G-metric on X is said to be symmetric if G(x, y, y) = G(y, x, x) for all x, y ∈ X. Every G-metric on X defines a metric d_G on X by

For a symmetric G-metric space, one obtains However, for an arbitrary G-metric G on X, just the following inequality holds: (3/2) G (x, y, y) ≤ d_G (x, y) ≤ 3G (x, y, y), for all x, y ∈ X.

Definition 5. In this work, one will consider the following three classes of mappings [3, 21]:

•  

  Ψ = {ψ∣ψ : [0, ∞)→[0, ∞) is continuous and nondecreasing and ψ⁻¹({0}) = {0}},

•  

  Φ = {ϕ∣ϕ : [0, ∞)→[0, ∞) is lower semicontinuous and ϕ⁻¹({0}) = {0}}, and

•  

  $$.

For weak ϕ-contractions in the frame of metric spaces see [19, 21].

At first we need the following well-known definitions and results (see, e.g., [9, 15, 22]).

Definition 6. Let (X, ⪯) be a partially ordered nonempty set, and let F : X² → X,    g : X → X be two mappings. The mapping F has the mixed g-monotone property if for any x, y ∈ X the following hold:

Note that, if g = i_X, the identity mapping, then F is said to have the mixed monotone property.

Remark 7. If (X, ⪯) is a partially ordered set, then (X², ⊑) is also partially ordered with

Definition 8. Let F : X² → X and g : X → X. An element (x, y) ∈ X² is called a coupled coincidence point of F and g if

while (gx, gy) ∈ X² is called a coupled point of coincidence of mappings F and g. Moreover, (x, y) is called a coupled common fixed point of F and g if

Remark 9. Otherwise, (x, y) is a coupled coincidence point of F and g if and only if (x, y) is a coincidence point of the mappings T_F : X² → X² and T_g : X² → X² which are defined by

Definition 10. Mappings f,  g : X → X are said to be compatible in a G-metric space (X, G) if

whenever {x_n} is a sequence in X such that lim _n→∞fx_n = lim _n→∞gx_n in (X, G).

It is easy to prove that f and g are compatible in (X, G) if and only if they are compatible in the associated metric space (X, d_G).

Definition 11. Let (X, G) be a G-metric space, and let F : X² → X and g : X → X be two mappings. One says that F and g are compatible if

whenever x_n and y_n are such that The proof of the following lemma is immediate (for (2) see [18]).

Lemma 12. (1) Let (X, G) be a symmetric G-metric space. Define G₁ : X² × X² × X² → ℝ⁺ by

Then (X², G₁) is a symmetric G-metric space.

(2) Let X = {a, b}. Define

and extend G to X³ by using the symmetry in the variables. Then it is clear that (X, G) is an asymmetric G-metric space. It is not hard to see that (X³, G₁) is not a G-metric space.

Remark 13. One can prove that mappings F and g are compatible in (X, G) if and only if T_F and T_g are compatible in (X², G₁).

Definition 14. (1) Let X be a nonempty set. Then (X, G, ⪯) is called an ordered G-metric space if

• (i)

  (X, G) is a G-metric space and

• (ii)

  (X, ⪯) is a partially ordered set.

(2) Let (X, G, ⪯) be a partially ordered G-metric space. One says that (X, G, ⪯) is regular if the following hypotheses hold:

• (i)

  if a nondecreasing sequence {x_n} is such that x_n → x as n → ∞, then x_n⪯x⪯x for all n ∈ ℕ,

• (ii)

  if a nonincreasing sequence {y_n} is such that y_n → y as n → ∞, then y_n⪰y⪰y for all n ∈ ℕ.

Theorem 15. Let (X, G, ⪯) be a partially ordered symmetric G-metric space, F : X² → X, and g : X → X. Assume that there exist ψ ∈ Ψ and ϕ ∈ Φ such that

for all x, y, u, v, a, and b ∈ X for which gx⪯gu⪯ga∧gy⪰gv⪰gb or gx⪰gu⪰ga∧gy⪯gv⪯gb. Assume that F and g satisfy the following conditions:

•  

  (1) F(X²) ⊂ g(X);

•  

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

•  

  (3) F and g are continuous and compatible and (X, G) is G-complete, or

•  

  (3^′) (X, G, ⪯) is regular and one of F(X²) or g(X) is G-complete;

•  

  (4) there exist x₀, y₀ ∈ X such that

  $$ ()

 Then F and g have a coupled coincidence point.

Remark 16. (a) Obviously, the condition (1) from [23] is equivalent to the condition (13). Hence, by using a new symmetric G-metric space (X², G₁) we have obtained a method of reducing coupled coincidence and coupled fixed point results in (ordered) symmetric G-metric spaces to the respective results for mappings with one variable, even obtaining (in some cases) more general theorems. We note that this method cannot be applied in the case of asymmetric G-metric spaces (see (2) of Lemma 12). For other details of coupled case in ordered metric spaces see also [22].

(b) Also, we note that Theorem 3.1. from [23] holds if F and g are compatible instead of commuting (see Step 3 in [23]). Indeed, since

then because F and g are compatible.

Therefore, now we have

that is, F(x, y) = gx. Similarly, we obtain that F(y, x) = gy.

Lemma 17. Let (X, G) be a G-metric space, and let {x_n} be a sequence in X such that lim _n→∞G(x_n, x_n+1, x_n+1) = 0. If {x_n} is not a G-Cauchy sequence in (X, G), then there exist ε > 0 and two sequences {m_k} and {n_k} of positive integers such that the following four sequences tend to ε⁺ when k → ∞:

Lemma 18. Let (X, G, ⪯) be a partially ordered G-metric space, and let f and g be two self-mappings on X. Assume that there exist ψ ∈ Ψ and ϕ ∈ Φ such that

for all x, y, z, u, v, w, a, b, and c ∈ X for which gx⪯gu⪯ga∧gy⪰gv ≥ gb∧gz⪯gw⪯gc or gx⪰gu⪰ga∧gy⪯gv⪯gb∧gz⪰gw⪰gc. If the following conditions hold:

• (i)

  f is a g-nondecreasing with respect to ⪯ and fX ⊂ gX;

• (ii)

  there exists x₀ ∈ X such that gx₀⪯fx₀;

• (iii)

  f and g are continuous and compatible, and (X, G) is G-complete or

•  

  (iii^′) (X, G, ⪯) is regular, and one of fX or gX is G-complete.

Then f and g have a coincidence point in X.

Proof. If gx₀ = fx₀, then x₀ is a coincidence point of f and g. Therefore, let gx₀≺fx₀. Since fX ⊂ gX, we obtain a Jungck sequence y_n = fx_n = gx_n+1 for all n = 0,1, 2, …, where x_n ∈ X, and by induction we get that y_n⪯y_n+1. If y_n = y_n+1 for some n ∈ ℕ, then x_n+1 is a coincidence point of f and g. Therefore, suppose that y_n ≠ y_n+1 for each n. Now, we will prove the following:

• (1)

  G(y_n, y_n+1, y_n+1) → 0 as n → ∞;

• (2)

  {y_n} is a G-Cauchy sequence.

Indeed, by putting x = u = a = x_n, y = v = b = x_n+1, and z = u = c = x_n+1 in (19) we get

and since the function ψ is nondecreasing, it follows that G(y_n, y_n+1, y_n+1) ≤ G(y_n−1, y_n, y_n); that is, there exists lim _n→∞G(y_n, y_n+1, y_n+1) = G^* ≥ 0. If G^* > 0, we get from the previous relation ψ(G^*) ≤ ψ(G^*) − ϕ(G^*); that is, G^* = 0 which is a contradiction. Hence, we obtain that lim _n→∞G(y_n, y_n+1, y_n+1) = 0.

Further, using Lemma 17 we shall prove that {y_n} is a G-Cauchy sequence. Suppose this is not the case. Then, by Lemma 17 there exist ε > 0 and two sequences {m_k} and {n_k} of positive integers such that the following sequences tend to ε⁺ when k → ∞:

Putting $$, $$, and$$ in (19) we have that is, Letting k → ∞, we get ψ(ε) ≤ ψ(ε) − ϕ(ε); that is, ϕ(ε) = 0. Since ϕ ∈ Φ, we get ε = 0, which is a contradiction. We have proved that {y_n} is a G-Cauchy sequence in (X, G).

In case (iii), since (X, G) is G-complete, there exists z ∈ X such that y_n → z. Then we have

Further, according to Definition 1 (e) and since f and g are continuous and compatible, we get It follows that z is a coincidence point for f and g.

In case (iii^′), it follows that y_n = fx_n = gx_n+1 → gz, z ∈ X (in both cases when fX or gX is G-complete), and then gx_n⪯gz⪯gz, and by the contractive condition (19) we have

By taking limit as n → ∞ in the above inequality, we obtain and hence fz = gz.

Proof of Theorem 15. Firstly, (13) implies

for all Y = (x, y),   V = (u, v), and A = (a, b) from X² for which T_g(Y)⊑T_g(V)⊑T_g(A) or T_g(A)⊑T_g(V)⊑T_g(Y).

Further,

• (1)

  implies that T_F(X²) ⊂ T_g(X²);

• (2)

  implies that T_F is T_g-nondecreasing with respect to ⊑ and T_F(X²) ⊂ T_g(X²);

• (3)

  implies that T_F and T_g are continuous and compatible and (X², G₁) is G-complete,

or

•  

  (3^′) implies that (X², G₁, ⊑) is regular and one of T_F(X²) or T_g(X²) is G-complete;

• (4)

  implies that there exists Y₀ = (x₀, y₀) ∈ X² such that T_g(Y₀)⊑T_F(Y₀) or T_F(Y₀)⊑T_g(Y₀).

All conditions of Lemma 18 for the ordered G-metric space (X², G₁) are satisfied. Therefore, the mappings T_F and T_g have a coincidence point in X². According to Remark 9 the mappings F and g have a coupled coincidence point. The proof of Theorem 15 is complete.

Theorem 19. Let (X, G, ⪯) be a partially ordered symmetric G-metric space, F : X² → X, and g : X → X. Assume that there exists φ ∈ Θ such that

for all x, y, u, v, a, and b ∈ X for which gx⪯gu⪯ga∧gy⪰gv ≥ gb or gx⪰gu⪰ga∧gy⪯gv⪯gb. Assume that F and g satisfy the following conditions:

• (1)

  F(X²) ⊂ g(X);

• (2)

  F has the mixed g-monotone property;

• (3)

  F and g are continuous and compatible and (X, G) is G-complete, or

•  

  (3^′) (X, G, ⪯) is regular and one of F(X²) or g(X) is G-complete;

• (4)

  there exist x₀, y₀ ∈ X such that

  $$ ()

Then F and g have a coupled coincidence point.

Lemma 20. Let (X, G, ⪯) be a partially ordered G-metric space, and let f and g be two self-mappings on X. Assume that there exists φ ∈ Θ such that

for all x, y, z, u, v, w, a, b, and c ∈ X for which gx⪯gu⪯ga∧gy⪰gv ≥ gb∧gz⪯gw⪯gc or gx⪰gu⪰ga∧gy⪯gv⪯gb∧gz⪰gw⪰gc. If the following conditions hold:

• (i)

  f is g-nondecreasing with respect to ⪯ and fX ⊂ gX;

• (ii)

  there exists x₀ ∈ X such that gx₀⪯fx₀;

• (iii)

  f and g are continuous and compatible, and (X, G) is G-complete or

•  

  (iii^′) (X, G, ⪯) is regular and one of fX or gX is G-complete,

Then f and g have a coincidence point in X.

Proof. If gx₀ = fx₀, then x₀ is a coincidence point of f and g. Therefore, let gx₀≺fx₀. Since fX ⊂ gX, we obtain a Jungck sequence y_n = fx_n = gx_n+1 for all n = 0,1, 2, …, where x_n ∈ X, and by induction we get that y_n⪯y_n+1. If y_n = y_n+1 for some n ∈ ℕ, then x_n+1 is a coincidence point of f and g. Therefore, suppose that y_n ≠ y_n+1 for each n. Now, we will prove that G(y_n, y_n+1, y_n+1) → 0 as n → ∞.

Indeed, by putting x = u = a = x_n, y = v = b = x_n+1, and z = u = c = x_n+1 in (31) we get

That is, there exists lim _n→∞G(y_n, y_n+1, y_n+1) = G^* ≥ 0. If G^* > 0, we get which is a contradiction. Hence, we obtain that lim _n→∞G(y_n, y_n+1, y_n+1) = 0.

Further, by using Lemma 17, we shall prove that {y_n} is a G-Cauchy sequence. Suppose this is not the case. Then, by Lemma 17 there exist ε > 0 and two sequences {m_k} and {n_k} of positive integers such that the following sequences tend to ε⁺ when k → ∞:

Putting $$, $$, and $$ in (31) we have that is, Letting k → ∞, we obtain Hence, we get ε < ε, which is a contradiction. We have proved that {y_n} is a G-Cauchy sequence in (X, G).

Now, in case (iii^′), since (X, G) is G-complete, there exists z ∈ X such that y_n → z. Then we have

and because it follows that z is a coincidence point for f and g.

In case  (iii^′), it follows that y_n = fx_n = gx_n+1 → gz, z ∈ X (in both cases when fX or gX is G-complete), and then gx_n⪯gz⪯gz, and by the contractive condition (31) we have

By taking the limit as n → ∞ in the above inequality we obtain and hence fz = gz.

Proof of Theorem 19. The proof is very similar to the proof of Theorem 15. Namely, the contractive condition (29) for the mappings F and g is equivalent to the following condition:

for the mappings T_F and T_g. The proof is further an immediate consequence of Lemma 20.

Remark 21. We have obtained that, in case of symmetric G-metric spaces, Theorem 19 generalizes both results (Theorems 3.1 and 3.2) from [3]. Also, Example 22 shows that this generalization is proper. It is clear that our new method with ordered symmetric G-metric spaces (X², G₁) and with mappings T_F and T_g implies that all results from [3] can be reduced to known results with one variable.

The following example supports both of our theorems, the first with ψ(t) = t and ϕ(t) = (1/2)t and the second with φ(t) = (1/2)t.

Example 22. Let X = ℝ be endowed with the complete G-metric

for all x, y, and z ∈ X and with the usual order. Consider the mappings F(x, y) = (x³ − 2y³)/8 and g(x) = x³. All the conditions of Theorems 15 and 19 are satisfied. In particular, the mapping F has the mixed g-monotone property, and we will check that F and g are compatible.

Let {x_n} and {y_n} be two sequences in X such that

Then (A − 2B)/8 = A and (B − 2A)/8 = B, wherefrom it follows that A = B = 0. Then and similarly Also, F and g do not commute, and therefore a coupled coincidence point of F and g cannot be obtained by Theorem 3.1 from [3].

The contractive condition (13) is satisfied with ψ(t) = t and ϕ(t) = (1/2)t which follows from

for all x, y, u, v, a, and b ∈ X for which gx ≤ gu ≤ ga∧gy ≥ gv ≥ gb or gx ≥ gu ≥ ga∧gy ≤ gv ≤ gb.

Hence,

Similarly, the condition (29) is satisfied with φ(t) = (1/2)t; that is, for all x, y, u, v, a, and b ∈ X for which gx ≤ gu ≤ ga∧gy ≥ gv ≥ gb or gx ≥ gu ≥ ga∧gy ≤ gv ≤ gb. There exists a coupled coincidence point (0,0) of the mappings F and g.