Tripled best proximity point in complete metric spaces

Definition 2.1

[3] Let (X, d) be a metric space and A ≠ ϕ, B ≠ ϕ are subsets of X. Let T : A → B be a mapping. Then a ∈ A is said to be a best proximity point if and only if d(a, Ta) = d(A, B).

Definition 2.2

[2] Let F : X × X × X → X. An element (a, b, c) is called a tripled fixed point of F if F(a, b, c) = a, F(b, a, b) = b and F(c, b, a) = c.

Definition 2.3

[7] Let (A, B) be a pair of nonempty subsets of a metric space (X, d) with A₀ ≠ Φ. Then, the pair (A, B) has P-property if and only if

where x₁, x₂ ∈ A and y₁, y₂ ∈ B.

Definition 2.4

[5] A map ϕ : ℝ⁺ → ℝ⁺ is said to be a comparison function if

1. x < y ⇒ ϕ(x) ≤ ϕ(y) ∀ x, y ∈ ℝ⁺;

2. lim_n→+∞ ϕⁿ(t) = 0.

If ϕ is a comparison function, we have ϕ(0) = 0 and ϕ(t) < t for all t > 0. Here [0, +∞)⁶ denote [0, +∞) × [0, +∞) × [0, +∞) × [0, +∞) × [0, +∞) × [0, +∞). Let Θ denote collection of continuous functions θ : [0, +∞)⁶ → [0, +∞) such that

Definition 2.5

[4] Let θ be a continuous function in Θ and ϕ be a comparison function. A mapping T : A → B is said to be a generalized almost (ϕ, θ)-contraction if

Definition 2.6

[4] Let (X, d) be a metric space with A ≠ ϕ and B ≠ ϕ are closed subsets. Let F : X × X → X be a mapping such that d(u, F(u, v)) = d(A, B) and d(v, F(v, u)) = d(A, B). Then F has a coupled best proximity point (u, v).

Definition 2.7

Let (X, d) be a complete metric space and A ≠ ϕ, B ≠ ϕ are closed subsets. An element (u, v, w) ∈ X × X × X is said to be a tripled best proximity point of F : X × X × X → X if u, w ∈ A and y ∈ B such that d(u, F(u, v, w)) = d(A, B), d(v, F(v, u, w)) = d(A, B) and d(w, F(w, v, u)) = d(A, B).

Theorem 3.1

Let (X, d) be a complete metric space. Let A ≠ ϕ, B ≠ ϕ are closed subsets such that A₀ and B₀ are nonempty. Let F : X × X × X → X be a continuous mapping which satisfies

1. F(A₀ × B₀ × A₀) ⊆ B₀;

2. F(B₀ × A₀ × B₀) ⊆ A₀;

3. Pair (A, B) has the (P)-property.

Let θ be a continuous function in Θ and ϕ be a comparison function satisfying

for all x, y, z, u, v, w ∈ X.

Then (u, u, u) is the unique tripled best proximity point of F.

Proof

Choose x₀, z₀ ∈ A₀ and y₀ ∈ B₀. Since F(x₀, y₀, z₀), F(z₀, y₀, x₀) ∈ B₀, F(y₀, x₀, y₀) ∈ A₀, there exists x₁, z₁ ∈ A and y₁ ∈ B such that d(x₁, F(x₀, y₀, z₀)) = d(y₁, F(y₀, x₀, y₀)) = d(z₁, F(z₀, y₀, x₀)) = d(A, B).

Continuing this way, there exist sequences {x_n}, {z_n} in A and {y_n} in B such that

If d(x_n, x_n+1) = d(y_n, y_n+1) = d(z_n, z_n+1) = 0 for all n ∈ ℕ, then we are done.

Suppose d(x_n, x_n+1) > 0 or d(y_n, y_n+1) > 0 or d(z_n, z_n+1) > 0.

Now, by condition (c), d(x_n, F(x_n–1, y_n–1, z_n–1)) = d(A, B), d(x_n+1, F(x_n, y_n, z_n)) = d(A, B), and using (3.1), we have

Similarly, from (c), d(y_n, F(y_n–1, x_n–1, y_n–1)) = d(A, B), d(y_n+1, F(y_n, x_n, y_n)) = d(A, B), and d(z_n, F(z_n–1, y_n–1, x_n–1)) = d(A, B), d(z_n+1, F(z_n, y_n, x_n)) = d(A, B) respectively and using (3.1), we obtain

From (3.5), (3.6) and (3.7), we get

Repeating (3.8) n-times, we obtain

Hence

Now,

which gives

Similarly,

Let ϵ > 0. When n → +∞, ϕⁿ(max{d(x₀, x₁), d(y₀, y₁), d(z₀, z₁)}) → 0 then there exists n ∈ ℕ, such that

Now, we have to prove

Suppose (3.9) is true for m = k. Now,

From (c), d(x_n+1, F(x_n, y_n, z_n)) = d(x_k+1, F(x_k, y_k, z_k)) = d(A, B), and using (3.1), we have

Similarly,

and

By using the properties of θ, lim_n→+∞d(x_n, F(x_n, y_n, z_n)) = d(A, B), lim_n→+∞d(y_n, F(y_n, x_n, y_n)) = d(A, B) and lim_n→+∞d(z_n, F(z_n, y_n, x_n)) = d(A, B), we have

and

When taking n₀ large enough, we have

and

From the relations (3.9)-(3.16), we get

Thus (3.9) is true for all m ≥ n ≥ n₀. Hence, {x_n} and {z_n} are Cauchy sequences in A and {y_n} in B. Since (X, d) is complete, there exist u, v, w ∈ X such that

Since A and B are closed, we get u, w ∈ A and v ∈ B. Since F is continuous,

Similarly, d(v, F(v, u, v)) = d(A, B) and d(w, F(w, v, u)) = d(A, B).

Thus, (u, v, w) is a tripled best proximity point of F. Now, we show that u = v = w.

Lastly, from (c) and using (3.1), we have

Therefore, u = v = w.

To prove the uniqueness, let t be another tripled best proximity point. Now,

This completes the proof.□

Theorem 3.2

Let (X, d) be a complete metric space. Let A ≠ ϕ, B ≠ ϕ are closed subsets such that A₀ and B₀ are nonempty. Let F : X × X × X → X be a continuous mapping which satisfies

1. F(A₀ × A₀ × A₀) ⊆ B₀ or F(B₀ × B₀ × B₀) ⊆ A₀;

2. Pair (A, B) has the (P)-property.

Let θ be a continuous function in Θ and ϕ be a comparison function satisfying

for all x, y, z, u, v, w ∈ X.

Then (u, u, u) is the unique tripled best proximity point of F.

Proof

Choose x₀, y₀, z₀ ∈ A₀. Since F(A₀ × A₀ × A₀) ⊆ B₀, we get F(x₀, y₀, z₀), F(y₀, x₀, y₀), F(z₀, y₀, x₀) ∈ B₀. Then by following Theorem 3.1, we get that (u, u, u) is the unique tripled best proximity point.□

Theorem 3.3

Let (X, d) be a complete metric space. Let A ≠ ϕ be a closed subset. Let F : X × X × X → X be a continuous mapping such that F(A × A × A) ⊆ A, θ be a continuous function in Θ and ϕ be a comparison function satisfying

Then (u, u, u) is the unique tripled point of F.

Example 3.4

Consider X = {0, 2, 3, 4, 5} and d(x, y) = x−y2 for all x, y ∈ X. Let U = {2, 5} and V = {2, 4} be subsets of X. Let F : X × X × X → X be a continuous mapping given by F(x, y, z) = x + y – z for all x, y, z ∈ X, θ : [0; +∞)⁶ → [0, +∞) given by θ(r, s, t, u, v, w) = min{r, s, t, u, v, w} and ϕ : [0, +∞) → [0, +∞) be given by ϕ(t) = t1+t .

Proof

Here, A₀ = {2}, B₀ = {2}, d(A, B) = 0. Take x, z ∈ A₀ and y ∈ B₀, then clearly F(A₀ × B₀ × A₀) ⊆ B₀, F(B₀ × A₀ × B₀) ⊆ A₀, condition (c) of Theorem 3.1 is true, satisfying (3.1) and also all the conditions of Theorem 3.2. Hence, from Theorem 3.1 and Theorem 3.2, (2, 2, 2) is the unique tripled best proximity point.□

Example 3.5

Consider (X, d) = ℝ, d(x, y) = |x – y| for all x, y ∈ ℝ. Let U = [1, 2] and V = [–2, –1] be subsets of X. Let F : X × X × X → X be a continuous mapping given by F(x, y, z) = x+y−z3 for all x, y, z ∈ X, θ : [0; +∞)⁶ → [0, +∞) given by θ(r, s, t, u, v, w) = min{r, s, t, u, v, w} and ϕ : [0, +∞) → [0, +∞) be given by ϕ(t) = t1+t .

Proof

Here, A₀ = {1}, B₀ = {–1}, d(A, B) = 2. Take x, z ∈ A₀ and y ∈ B₀, then clearly F(A₀ × B₀ × A₀) ⊆ B₀, F(B₀ × A₀ × B₀) ⊆ A₀, the pair (A, B) has the (P)-property and (1, -1, 1) is the unique tripled best proximity point but not of the form (u, u, u). This is because (3.1) is not satisfied. Therefore, by Theorem 3.1, we cannot get the results.□