A Fixed Point Theorem for Mappings Satisfying a Contractive Condition of Rational Type on a Partially Ordered Metric Space

and Applied Analysis 3 Again, using induction d xn 1, xn ≤ ( β 1 − α )n d x1, x0 , 2.5 Put k β/ 1 − α < 1. Moreover, by the triangular inequality, we have, form ≥ n, d xm, xn ≤ d xm, xm−1 d xm−1, xm−2 · · · d xn 1, xn ≤ ( km−1 km−2 · · · k ) d x1, x0 ≤ ( k 1 − k ) d x1, x0 , 2.6 and this proves that d xm, xn → 0 as m,n → ∞. So, {xn} is a Cauchy sequence and, since X is a complete metric space, there exists z ∈ X such that limn→∞xn z. Further, the continuity of T implies Tz T ( lim n→∞ xn ) lim n→∞ Txn lim n→∞ xn 1 z, 2.7 and this proves that z is a fixed point. This finishes the proof. In what follows, we prove that Theorem 2.2 is still valid for T , not necessarily continuous, assuming the following hypothesis in X: if xn is a nondecreasing sequence in X such that xn −→ x, then x sup{xn}. 2.8 Theorem 2.3. Let X,≤ be a partially ordered set and suppose that there exists a metric d in X such that X, d is a complete metric space. Assume that X satisfies 2.8 . Let T : X → X be a nondecreasing mapping such that d ( Tx, Ty ) ≤ α x, Tx · d ( y, Ty ) d ( x, y ) βd ( x, y ) , for x, y ∈ X, x ≥ y, x / y, 2.9 with α β < 1. If there exists x0 ∈ X with x0 ≤ Tx0, then T has a fixed point. Proof. Following the proof of Theorem 2.2, we only have to check that Tz z. As {xn} is a nondecreasing sequence in X and xn → z, then, by 2.8 , z sup{xn}. Particularly, xn ≤ z for all n ∈ N. Since T is a nondecreasing mapping, then Txn ≤ Tz, for all n ∈ N or, equivalently, xn 1 ≤ Tz for all n ∈ N. Moreover, as x0 < x1 ≤ Tz and z sup{xn}, we get z ≤ Tz. Suppose that z < Tz. Using a similar argument that in the proof of Theorem 2.2 for x0 ≤ Tx0, we obtain that {Tnz} is a nondecreasing sequence and limn→∞Tz y for certain y ∈ X. 4 Abstract and Applied Analysis Again, using 2.8 , we have that y sup{Tnz}. Moreover, from x0 ≤ z, we get xn Tx0 ≤ Tz for n ≥ 1 and xn < Tz for n ≥ 1 because xn ≤ z < Tz ≤ Tz for n ≥ 1. As xn and Tz are comparable and distinct for n ≥ 1, applying the contractive condition we get d ( xn 1, T n 1z ) d Txn, T Tz ≤ α d xn, Txn · d ( Tz, T 1z ) d xn, Tnz βd xn, Tz α d xn, xn 1 · d ( Tz, T 1z ) d xn, Tnz βd xn, Tz . 2.10 Making n → ∞ in the last inequality, we obtain d ( z, y ) ≤ βdz, y. 2.11 As β < 1, d z, y 0, thus, z y. Particularly, z y sup{Tnz} and, consequently, Tz ≤ z and this is a contradiction. Hence, we conclude that z Tz. Now, we present an example where it can be appreciated that hypotheses in Theorem 2.2 do not guarantee uniqueness of the fixed point. This example appears in 8 . Let X { 1, 0 , 0, 1 } ⊂ R2 and consider the usual order ( x, y ) ≤ z, t ⇐⇒ x ≤ z, y ≤ t. 2.12 Thus, X,≤ is a partially ordered set whose different elements are not comparable. Besides, X, d2 is a complete metric space considering, d2, the Euclidean distance. The identity map T x, y x, y is trivially continuous and nondecreasing and assumption 2.1 of Theorem 2.2 is satisfied since elements in X are only comparable to themselves. Moreover, 1, 0 ≤ T 1, 0 and T has two fixed points in X. In what follows, we give a sufficient condition for the uniqueness of the fixed point in Theorems 2.2 and 2.3. This condition appears in 14 and for x, y ∈ X, there exists a lower bound or an upper bound. 2.13 In 8 , it is proved that the above-mentioned condition is equivalent, for x, y ∈ X, there exists z ∈ X which is comparable to x and y. 2.14 Theorem 2.4. Adding condition 2.14 to the hypotheses of Theorem 2.2 (or Theorem 2.3) one obtains uniqueness of the fixed point of T . Proof. Suppose that there exists z, y ∈ X which are fixed point. Abstract and Applied Analysis 5 We distinguish two cases. Case 1. If y and z are comparable and y / z, then using the contractive condition we haveand Applied Analysis 5 We distinguish two cases. Case 1. If y and z are comparable and y / z, then using the contractive condition we have d ( y, z ) d ( Ty, Tz ) ≤ α ( y, Ty ) · d z, Tz d ( y, z ) βd ( y, z ) α d ( y, y ) · d z, z d ( y, z ) βd ( y, z ) βd ( y, z ) . 2.15 As β < 1 is the last inequality, it is a contradiction. Thus, y z. Case 2. If y is not comparable to z, then by 2.14 there exists x ∈ X comparable to y and z. Monotonicity implies that Tx is comparable to Ty y and Tz z for n 0, 1, 2, . . .. If there exists n0 ≥ 1 such that T0x y, then as y is a fixed point, the sequence {Tnx : n ≥ n0} is constant, and, consequently, limn→∞Tx y. On the other hand, if Tx / y for n ≥ 1, using the contractive condition, we obtain, for n ≥ 2, d ( Tx, y ) d ( Tx, Ty ) ≤ α ( Tn−1x, Tx ) · dTn−1y, Tny d ( Tn−1x, Tn−1y ) βd ( Tn−1x, Tn−1y ) ≤ α ( Tn−1x, Tx ) · dy, y d ( Tn−1x, y ) βd ( Tn−1x, y ) βd ( Tn−1x, y ) . 2.16