Abstract
In this paper, we prove a fixed point theorem for generalized weak contractions satisfying rational expressions in partially ordered metric spaces. The result is a generalization of a recent result of Harjani et al. (Abstr. Appl. Anal, Vol.2010, 1-8, 2010). An example is also given to show that our result is a proper generalization of the existing one.
2000 Mathematics Subject Classification: 47H10, 54H25.
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1 Introduction and preliminaries
It is well known that the Banach contraction mapping principle is one of the pivotal results of analysis. Generalizations of this principle have been obtained in several directions. The following is an example of such generalizations. Jaggi in [1] proved the following theorem satisfying a contractive condition of rational type
Theorem 1.1. ([1]) Let T be a continuous self-map defined on a complete metric space (X, d). Suppose that T satisfies the following condition:
for all x, y ∈ X, x ≠ y and for some α, β ≥ 0 with α + β < 1, then T has a unique fixed point in X.
Another generalization of the contraction principle was suggested by Alber and Guerre-Delabriere [2] in Hilbert spaces. Rhoades [3] has shown that their result is still valid in complete metric spaces.
Definition 1.2. ([3]) Let (X, d) be a metric space. A mapping T : X → X is said to be φ-weak contraction if
for all x, y ∈ X, where φ : [0, ∞) → [0, ∞) is a continuous and non-decreasing function with φ(t) = 0 if and only if t = 0.
Theorem 1.3. ([3]) Let (X, d) be a complete metric space and T be a φ-weak contraction on X. Then, T has a unique fixed point.
In fact, while Alber and Guerre-Delabriere assumed an additional assumption limt→∞φ(t) = ∞ on φ, but Rhoades proved Theorem 1.3 without this particular condition. A number of extensions of Theorem 1.3 were presented in [4–9] and references therein. Some of these results were presented without the continuity and monotonicity of φ.
Recently, existence of fixed points in partially ordered sets has been considered, and first results were obtain by Ran and Reurings [10] and then by Nieto and Lopez [11]. The following fixed point theorem is the version of theorems, which were proved in those papers.
Theorem 1.4. ([10, 11]) Let (X, ≤) be a partially ordered set, and suppose that there is a metric d such that (X, d) be a complete metric space. Let T : X → X be a non-decreasing mapping satisfying the following inequality
where k ∈ (0, 1). Also, assume either
(i) T is continuous or
(ii) X has the property:
If there exists x0 ∈ X such that x0 ≤ Tx0, then T has a fixed point.
Besides, applications to matrix equations and ordinary differential equations were presented in [10, 11]. Afterward, coupled fixed point and common fixed point theorems and their applications to periodic boundary value problems and integral equations were given in [5–7, 12–19]. In particular, Harjani and Sadarangani [5] proved some fixed point theorems in the context of ordered metric spaces as the extensions of Theorem 1.3. We state one of their results.
Theorem 1.5. ([5]) Let (X, ≤) be a partially ordered set and suppose that there is a metric d such that (X, d) be a complete metric space. Let T : X → X be a non-decreasing mapping satisfying the following inequality
where φ : [0, ∞) → [0, ∞) is a continuous and non-decreasing function with φ(t) = 0 if and only if t = 0. Also, assume either
(i) T is continuous or
(ii) X has the property (1).
If there exists x0 ∈ X such that x0 ≤ Tx0, then T has a fixed point.
In addition, Harjani et al. in [12] proved the following theorem as a version of Theorem 1.1 in partially ordered metric spaces where they replaced the condition (1) by a stronger condition, that is
Theorem 1.6. ([12]) Let (X, ≤) be a partially ordered set and suppose that there is a metric d such that (X, d) be a complete metric space. Let T : X → X be a non-decreasing mapping such that
where 0 ≤ α, β and α + β < 1. Also, assume either
(i) T is continuous or
(ii) X has the property (2).
If there exists x0 ∈ X such that x0 ≤ Tx0, then T has a fixed point.
In this paper, we prove a fixed point theorem for generalized weak contractions satisfying rational expressions in partially metric spaces, which is a generalization of the result of Harjani et al. [12]. We also give an example to show that our result is a proper extension of the result in [12].
2 Main theorem
Theorem 2.1. Let (X, ≤) be a partially ordered set, and suppose that there is a metric d such that (X, d) be a complete metric space. Let T : X → X be a non-decreasing mapping satisfying the following inequality
where φ : [0, ∞) → [0, ∞) is a lower semi-continuous function with φ(t) = 0 if and only if t = 0, and
Also, assume either
(i) T is continuous or
(ii) X has the property (2).
If there exists x0 ∈ X such that x0 ≤ Tx0, then T has a fixed point.
Proof. Let x0 ∈ X be such that x0 ≤ Tx0, we construct the sequence {x n } in X as follows
Since T is a non-decreasing mapping, by induction, we can show that
If there exists n0 such that , then . This means that is a fixed point of T and the proof is finished. Thus, we can suppose that x n ≠ xn+1for all n.
Since x n > xn-1for all n ≥ 1, from (4), we have
Suppose that there exists m0 such that , from (7), we have
which is a contradiction. Hence, d (xn+1, x n ) ≤ d (x n , xn-1) for all n ≥ 1.
Since {d(xn+1, x n )} is a non-increasing sequence of positive real numbers, there exists δ ≥ 0 such that
We shall show that δ = 0. Assume, to the contray, that δ > 0. Taking the upper limit as n → ∞ in (7) and using the properties of the function φ, we get
which is a contradiction. Therefore, δ = 0, that is,
In what follows, we shall prove that {x n } is a Cauchy sequence. Suppose, to the contrary, that {x n } is not a Cauchy sequence. Then, there exists ε > 0 such that we can find subsequences {xm(k)}, {xn(k)} of {x n } with n(k) > m(k) ≥ k satisfying
Further, corresponding to m(k), we can choose n(k) in such way that it is the smallest integer with n(k) > m(k) ≥ k satisfying (9). Hence,
We have
Taking k → ∞ and using (8), we get
By the triangle inequality,
Taking k → ∞ in the above inequalities and using (7), (11), we obtain
Since m(k) < n(k), xn(k)-1> xm(k)-1, from (4), we have
Taking upper limit as k → ∞ in (13) and using (7), (11), (12) and the properties of the function φ, we have
which is a contradiction. Therefore, {x n } is a Cauchy sequence. Since X is a complete metric space, there exists x ∈ X such that limn→∞x n = x.
Now, suppose that the assumption (a) holds. The continuity of T implies
and this proved that x is a fixed point of T.
Finally, suppose that the assumption (b) holds. Since {x n } is a non-decreasing sequence and x n → x, then x = sup{x n }. Particularly, x n ≤ x for all n. Since T is non-decreasing, Tx n ≤ Tx for all n, that is, xn+1≤ Tx for all n. Moreover, as x n ≤ xn+1≤ Tx for all n and x = sup{x n }, we obtain x ≤ Tx. Consider the sequence {y n } that is constructed as follows
Since y0 ≤ Ty0, arguing like above part, we obtain that {y n } is a non-decreasing sequence and for certain y ∈ X. By the assumption (b), we have y = sup{y n }.
Since x n < x = y0 ≤ Tx = Ty0 ≤ y n ≤ y for all n, suppose that x ≠ y, from (4), we have
Taking upper limit as n → ∞ in the above inequality, we have
which is a contradiction. Hence, x = y. We have x ≤ Tx ≤ x, therefore Tx = x. That is, x is a fixed point of T.
The proof is complete. □
Corollary 2.2. Let (X, ≤) be a partially ordered set, and suppose that there is a metric d such that (X, d) be a complete metric space. Let T : X → X be a non-decreasing mapping such that
for all x, y ∈ X with x ≥ y, x ≠ y, where k ∈ (0, 1). Also, assume either
(i) T is continuous or
(ii) X has the property (2).
If there exists x0 ∈ X such that x0 ≤ Tx0, then T has a fixed point.
Proof. In Theorem 2.1, taking φ(t) = (1 - k)t, for all t ∈ [0, ∞), we get Corollary 2.2.
□
Remark 2.3. For α, β > 0, α + β < 1 and for all x, y ∈ X, x ≠ y, we have
where k = α + β ∈ (0,1). Therefore, Corollary 2.2 is a generalization of Theorem 1.6, so is Theorem 2.1.
Now, we shall prove the uniqueness of the fixed point.
Theorem 2.4. In addition to the hypotheses of Theorem 2.1, suppose that
then T has a unique fixed point.
Proof. From Theorem 2.1, the set of fixed points of T is non-empty. Suppose that x, y ∈ X are two fixed points of T. By the assumption, there exists z ∈ X that is comparable to x and y.
We define the sequence {z n } as follows
Since z is comparable with x, we may assume that z ≤ x. Using the mathematical induction, it is easy to show that z n ≤ x for all n.
Suppose that there exists n0 ≥ 1 such that , then z n = Tzn-1= Tx = x for all n ≥ n0 - 1. Hence, z n → x as n → ∞.
On the other hand, if z n ≠ x for all n, from (4), we have
It implies that d (x, z n ) < d (x, zn-1) for all n ≥ 1, that is, {d(x, z n )} is a decreasing sequence of positive real numbers. Therefore, there is an α ≥ 0 such that d(x, z n ) → α. We shall show that α = 0. Suppose, to the contrary, that α > 0. Taking the upper limit as n → ∞ in (16) and using the properties of φ, we have
which is a contradiction. Hence, α = 0, that is, z n → x as n →∞. Therefore, in both cases, we have
Similarly, we have
From (17) and (18), we get x = y. □
Example 2.5. Letwith the usual metric d (x, y) = |x - y|, ∀x, y ∈ X. Obviously, (X, d) is a complete metric space. We consider the ordered relation in X as follows
where ≤ be the usual ordering.
Let T : X → X be given by
It is easy to see that T is non-decreasing and X has the property (2). Also, there is x0 = 0 in X such that x0 = 0 ≼ 0 = Tx0.
Clearly, T has a fixed point that is 0. However, we cannot apply Theorem 1.6 because the condition (3) is not true. Indeed, suppose that the condition (3) holds. Taking y = 0 and x = 1/n, n = 2, 3, 4,... in (3), we have
This implies
or
Taking n → ∞ in the last inequality, we have 1 ≤ β and we obtain a contradiction.
We now show that T satisfies (4) with φ : [0, ∞) → [0, ∞) which is given by
We have x, y ∈ X, x ≽ y, x ≠ y if x = 1/n, y = 0 or x = 1/n, y = 1/m, m > n ≥ 2. So, we have two possible cases.
Case 1. x = 1/n, n ≥ 2 and y = 0, we have
Case 2. x = 1/n, y = 1/m, m > n ≥ 2, we have
For m > n ≥ 2, we have
is equivalent to
or
The last inequality holds since
Therefore,
We have
is equivalent to
or
or
or
We have
Thus, the inequality (20) holds, so does the inequality (19).
Therefore, all the conditions of Theorem 2.1 are satisfied. Applying Theorem 2.1, we conclude that T has a fixed point in X.
Notice that since T is not continuous, this example cannot apply to Theorem 1.1.
Moreover, since the condition (15) is not satisfied, the uniqueness of fixed point of T does not guarantee. In fact, T has two fixed points that are 0 and .
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Luong, N.V., Thuan, N.X. Fixed point theorem for generalized weak contractions satisfying rational expressions in ordered metric spaces. Fixed Point Theory Appl 2011, 46 (2011). https://doi.org/10.1186/1687-1812-2011-46
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DOI: https://doi.org/10.1186/1687-1812-2011-46