Some Coupled Fixed Point Results on Partial Metric Spaces

Definition 1.1 (see [3].)An element (x, y) ∈ X × X is said to be a coupled fixed point of the mapping F : X × X → X if F(x, y) = x and F(y, x) = y.

Theorem 1.2. Let (X, d) be a complete cone metric space. Suppose that the mapping F : X × X → X satisfies the following contractive condition for all x, y, u, v ∈ X

where k, l are nonnegative constants with k + l < 1. Then, F has a unique coupled fixed point.

Definition 1.3 (see [6]–[8].)A partial metric on a nonempty set X is a function p : X × X → ℝ⁺ such that for all x, y, z ∈ X:

• (p1)

  x = y⇔p(x, x) = p(x, y) = p(y, y),

• (p2)

  p(x, x) ≤ p(x, y),

• (p3)

  p(x, y) = p(y, x),

• (p4)

  p(x, y) ≤ p(x, z) + p(z, y) − p(z, z).

Remark 1.4. It is clear that if p(x, y) = 0, then from (p1), (p2), and (p3), x = y. But if x = y, p(x, y) may not be 0.

Definition 1.5 (see [6]–[8].)Let (X, p) be a partial metric space. Then,

• (i)

  a sequence {x_n} in a partial metric space (X, p) converges to a point x ∈ X if and only if p(x, x) = lim _n→+∞p(x, x_n);

• (ii)

  a sequence {x_n} in a partial metric space (X, p) is called a Cauchy sequence if there exists (and is finite) lim _n,m→+∞p(x_n, x_m);

• (iii)

  a partial metric space (X, p) is said to be complete if every Cauchy sequence {x_n} in X converges to a point x ∈ X, that is, p(x, x) = lim _n,m→+∞p(x_n, x_m).

Lemma 1.6 (see [6], [7], [9].)Let (X, p) be a partial metric space;

• (a)

  {x_n} is a Cauchy sequence in (X, p) if and only if it is a Cauchy sequence in the metric space (X, p^s),

• (b)

  a partial metric space (X, p) is complete if and only if the metric space (X, p^s) is complete; furthermore, lim _n→+∞p^s(x_n, x) = 0 if and only if

Theorem 2.1. Let (X, p) be a complete partial metric space. Suppose that the mapping F : X × X → X satisfies the following contractive condition for all x, y, u, v ∈ X

where k, l are nonnegative constants with k + l < 1. Then, F has a unique coupled fixed point.

Proof. Choose x₀, y₀ ∈ X and set x₁ = F(x₀, y₀) and y₁ = F(y₀, x₀). Repeating this process, set x_n+1 = F(x_n, y_n) and y_n+1 = F(y_n, x_n). Then, by (2.1), we have

and similarly Therefore, by letting we have Consequently, if we set δ = k + l, then, for each n ∈ ℕ, we have If d₀ = 0 then p(x₀, x₁) + p(y₀, y₁) = 0. Hence, from Remark 1.4, we get x₀ = x₁ = F(x₀, y₀) and y₀ = y₁ = F(y₀, x₀), meaning that (x₀, y₀) is a coupled fixed point of F. Now, let d₀ > 0. For each n ≥ m, we have, in view of the condition (p4) Similarly, we have Thus, By definition of p^s, we have p^s(x, y) ≤ 2p(x, y), so, for any n ≥ m which implies that {x_n} and {y_n} are Cauchy sequences in (X, p^s) because of 0 ≤ δ = k + l < 1. Since the partial metric space (X, p) is complete, hence thanks to Lemma 1.6, the metric space (X, p^s) is complete, so there exist u^*, v^* ∈ X such that Again, from Lemma 1.6, we get But, from condition (p2) and (2.6), so since δ ∈ [0,1[, hence letting n → +∞, we get lim _n→+∞p(x_n, x_n) = 0. It follows that Similarly, we get Therefore, we have, using (2.1), and letting n → +∞, then from (2.14) and (2.15), we obtain p(F(u^*, v^*), u^*)) = 0, so F(u^*, v^*) = u^*. Similarly, we have F(v^*, u^*) = v^*, meaning that (u^*, v^*) is a coupled fixed point of F.

Now, if (u′, v′) is another coupled fixed point of F, then

It follows that In view of k + l < 1, this implies that p(u′, u^*) + p(v′, v^*) = 0, so u^* = u′ and v^* = v′. The proof of Theorem 2.1 is completed.

Corollary 2.2. Let (X, p) be a complete partial metric space. Suppose that the mapping F : X × X → X satisfies the following contractive condition for all x, y, u, v ∈ X

where 0 ≤ k < 1. Then, F has a unique coupled fixed point.

Example 2.3. Let X = [0, +∞[ endowed with the usual partial metric p defined by p : X × X → [0, +∞[ with p(x, y) = max {x, y}. The partial metric space (X, p) is complete because (X, p^s) is complete. Indeed, for any x, y ∈ X,

Thus, (X, p^s) is the Euclidean metric space which is complete. Consider the mapping F : X × X → X defined by F(x, y) = (x + y)/6. For any x, y, u, v ∈ X, we have which is the contractive condition (2.19) for k = 1/3. Therefore, by Corollary 2.2, F has a unique coupled fixed point, which is (0,0). Note that if the mapping F : X × X → X is given by F(x, y) = (x + y)/2, then F satisfies the contractive condition (2.19) for k = 1, that is, In this case, (0,0) and (1,1) are both coupled fixed points of F, and, hence, the coupled fixed point of F is not unique. This shows that the condition k < 1 in Corollary 2.2, and hence k + l < 1 in Theorem 2.1 cannot be omitted in the statement of the aforesaid results.

Theorem 2.4. Let (X, p) be a complete partial metric space. Suppose that the mapping F : X × X → X satisfies the following contractive condition for all x, y, u, v ∈ X

where k, l are nonnegative constants with k + l < 1. Then, F has a unique coupled fixed point.

Proof. We take the same sequences {x_n} and {y_n} given in the proof of Theorem 2.1 by

Applying (2.23), we get where δ = k/(1 − l). By definition of p^s, we have Since k + l < 1, hence δ < 1, so the sequences {x_n} and {y_n} are Cauchy sequences in the metric space (X, p^s). The partial metric space (X, p) is complete, hence from Lemma 1.6, (X, p^s) is complete, so there exist u^*, v^* ∈ X such that From Lemma 1.6, we get By the condition (p2) and (2.25), we have so lim _n→+∞p(x_n, x_n) = 0. It follows that Similarly, we find Therefore, by (2.23), and letting n → +∞, then from (2.27)–(2.32), we obtain From the preceding inequality, we can deduce a contradiction if we assume that p(F(u^*, v^*), u^*) ≠ 0, because in that case we conclude that 1 ≤ k and now this inequality is, in fact, a contradiction, so p(F(u^*, v^*), u^*) = 0, that is, F(u^*, v^*) = u^*. Similarly, we have F(v^*, u^*) = v^*, meaning that (u^*, v^*) is a coupled fixed point of F. Now, if (u′, v′) is another coupled fixed point of F, then, in view of (2.23), that is, p(u′, u^*) = 0 since (k + l) < 1. It follows that u^* = u′. Similarly, we can have v^* = v′, and the proof of Theorem 2.4 is completed.

Theorem 2.5. Let (X, p) be a complete partial metric space. Suppose that the mapping F : X × X → X satisfies the following contractive condition for all x, y, u, v ∈ X

where k, l are nonnegative constants with k + 2l < 1. Then, F has a unique coupled fixed point.

Proof. Since, k + 2l < 1, hence k + l < 1, and as a consequence the proof of the uniqueness in this theorem is as trivial as in the other results. To prove the existence of the fixed point, choose the sequences {x_n} and {y_n} like in the proof of Theorem 2.1, that is

Applying again (2.37), we have It follows that for any n ∈ ℕ^* Let us take δ = l/(1 − l − k). Hence, we deduce Under the condition 0 ≤ k + 2l < 1, we get 0 ≤ δ < 1. From this fact, we immediately obtain that {x_n} is Cauchy in the complete metric space (X, p^s). Of course, similar arguments apply to the case of the sequence {y_n} in order to prove that and, thus, that the sequence {y_n} is Cauchy in (X, p^s). Therefore, there exist u^*, v^* ∈ X such that Thanks to Lemma 1.6, we have The condition (p2) together with (2.41) yield that hence letting n → +∞, we get lim _n→+∞p(x_n, x_n) = 0. It follows that Similarly, we have Therefore, we have, using (2.37), Letting n → +∞ yields, using (2.46), and since k < 1, we have p(F(u^*, v^*), u^*) = 0, that is, F(u^*, v^*) = u^*. Similarly, thanks to (2.47), we get F(v^*, u^*) = v^*, and hence (u^*, v^*) is a coupled fixed point of F.

Corollary 2.6. Let (X, p) be a complete partial metric space. Suppose that the mapping F : X × X → X satisfies the following contractive condition for all x, y, u, v ∈ X

where 0 ≤ k < 1. Then, F has a unique coupled fixed point.

Corollary 2.7. Let (X, p) be a complete partial metric space. Suppose that the mapping F : X × X → X satisfies the following contractive condition for all x, y, u, v ∈ X

where 0 ≤ k < 2/3. Then, F has a unique coupled fixed point.

Proof. The condition 0 ≤ k < 2/3 follows from the hypothesis on k and l given in Theorem 2.5.

Remark 2.8. (i) Theorem 2.1 extends the Theorem  2.2 of [4] on the class of partial metric spaces.

(ii) Theorem 2.4 extends the Theorem  2.5 of [4] on the class of partial metric spaces.

Remark 2.9. Note that in Theorem 2.4, if the mapping F : X × X → X satisfies the contractive condition (2.23) for all x, y, u, v ∈ X, then F also satisfies the following contractive condition:

Consequently, by adding (2.23) and (2.52), F also satisfies the following: which is a contractive condition of the type (2.50) in Corollary 2.6 with equal constants. Therefore, one can also reduce the proof of general case (2.23) in Theorem 2.4 to the special case of equal constants. A similar argument is valid for the contractive conditions (2.37) in Theorem 2.5 and (2.51) in Corollary 2.7.