Some Coupled Fixed Point Results on Partial Metric Spaces

In this paper we give some coupled fixed point results for mappings satisfying different contractive conditions on complete partial metric spaces.


INTRODUCTION AND PRELIMINARIES
For a given partially ordered set X, Bhaskar and Lakshmikantham in [3] introduced the concept of coupled fixed point of a mapping F : X ×X → X . Later in [4], Lakshmikantham and Círíc investigated some more coupled fixed point theorems in partially ordered sets. The following is the corresponding definition of a coupled fixed point. Definition 1.1. [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.
F. Sabetghadam et al. [12] obtained the following 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 d(F (x, y), F (u, v)) ≤ kd(x, u) + ld(y, v) where k, l are nonnegative constants with k + l < 1. Then F has a unique coupled fixed point.
In this paper, we give the analogous of this result (and some others in [12]) on partial metric spaces, and we establish some coupled fixed point results. The concept of partial metric space (X, p) was introduced by Matthews in 1994. In such spaces, the distance of a point in the self may not be zero. First, we start with some preliminaries definitions on the partial metric spaces [1,2,5,6,7,8,9,10,11,13] Definition 1.3. ( [5,6,7]) 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).
A partial metric space is a pair (X, p) such that X is a nonempty set and p is a partial metric on X.
But if x = y, p(x, y) may not be 0.
If p is a partial metric on X, then the function p s : X × X −→ R + given by is a metric on X.
Definition 1.5. ( [5,6,7]) 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

MAIN RESULTS
Our first main result is the following 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 x 0 , y 0 ∈ X and set x 1 = F (x 0 , y 0 ) and y 1 = F (y 0 , x 0 ). 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 ≤kp(x n−1 , x n ) + lp(y n−1 , y n ), (2.2) and similarly p(y n , y n+1 ) =p(F (y n−1 , x n−1 ), F (y n , x n )) ≤kp(y n−1 , y n ) + lp(x n−1 , x n ). (2.3) Therefore, by letting (2.4) d n = p(x n , x n+1 ) + p(y n , y n+1 ), we have d n =p(x n , x n+1 ) + p(y n , y n+1 ) ≤kp(x n−1 , x n ) + lp(y n−1 , y n ) + kp(y n−1 , y n ) + lp(x n−1 , x n ) =(k + l)[p(y n−1 , y n ) + p(x n−1 , x n )] =(k + l)d n−1 . (2.5) Consequently, if we set δ = k + l then for each n ∈ N we have If d 0 = 0 then p(x 0 , x 1 ) + p(y 0 , y 1 ) = 0. Hence, from Remark 1.4, we get Similarly, we have (2.7) 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), Similarly, we get Therefore, we have using (2.1) and letting n → +∞, then from (2.10) and (2.11), we obtain It follows that It is worth noting that when the constants in Theorem 2.1 are equal we have the following Corollary 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. 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 (2.12) 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 (2.12) 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 can not 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 x n+1 = F (x n , y n ), y n+1 = F (y n , x n ) for any n ∈ N.
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 By the condition (p2) and (2.14), we have Therefore, by (2.13) and letting n → +∞, then from (2.16)-(2.19), 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, is another coupled fixed point of F , then in view of (2.13) 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 x n+1 = F (x n , y n ), y n+1 = F (y n , x n ) for any n ∈ N.
When the constants in Theorems 2.4 and 2.5 are equal, we get the following corollaries 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 (2.28) p(F (x, y), F (u, v)) ≤ k 2 (p (F (x, y), u) + p(F (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.
• Theorem 2.1 extends the Theorem 2.2 of [12] on the class of partial metric spaces.
• Theorem 2.4 extends the Theorem 2.5 of [12] 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 (2.13) for all x, y, u, v ∈ X, then F also satisfies the following contractive condition p(F (x, y), F (u, v)) =p(F (u, v), F (x, y)) ≤kp(F (u, v), u) + lp(F (x, y), x) (2.29) Consequently, by adding (2.13) and (2.29), F also satisfies the following: (2.30) p(F (x, y), F (u, v)) ≤ k + l 2 p(F (u, v), u) + k + l 2 p(F (x, y), x) which is a contractive condition of the type (2.27) in Corollary 2.6 with equal constants. Therefore, one can also reduce the proof of general case (2.13) in Theorem 2.4 to the special case of equal constants. A similar argument is valid for the contractive conditions (2.21) in Theorem 2.5 and (2.28) in Corollary 2.7.