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Largest and least fixed point theorems of increasing mappings in partially ordered metric spaces
Fixed Point Theory and Applications volume 2013, Article number: 74 (2013)
Abstract
In this paper, some largest and least fixed point theorems of increasing mappings in partially ordered metric spaces are proved, which extends and improves essentially many recent results since the additivity of η has been removed. In particular, the partial order used in this paper is not confined to that introduced by a functional.
MSC:06A06, 47H10.
1 Introduction
For improving Caristi’s fixed point theorem [1, 2], Feng and Liu [3] defined the following partial order on a metric space.
Lemma 1 (see [[3], Lemma 4.1])
Let be a metric space, let be a functional, and let be a nondecreasing and subadditive (i.e., , ) function with . Define a relation ⪯ on X by
Then ⪯ is a partial order on X.
This partial order is a generalized notion of the partial order defined by Caristi [1] as follows:
Since then the existence of fixed points in partially ordered metric spaces has been considered by many authors, and many satisfactory results have been obtained for Caristi-type mappings [2–7], mappings satisfying some monotone conditions with respect to the partial order introduced by a functional [8, 9], and mappings with some contractive conditions [10–18]. Recently, Li [9] proved the existence of maximal and minimal fixed points of increasing mappings by using the partial order introduced by (1).
It is worth mentioning that in [9], the function η is necessarily assumed to be subadditive for ensuring that the relation defined by (1) is a partial order. While it is well known that the additivity of η is no longer necessary for the study of fixed point theorems for a Caristi-type mapping (see [4–7]), naturally, one may wonder whether the additivity of η in [9] could be omitted.
In this paper we show how the additivity of η could be removed. Without the additivity of η, we prove not only the existence of maximal and minimal fixed points, but also the existence of largest and least fixed points of increasing mappings in a partially ordered metric space. In particular, the partial order used in this paper is not confined to that introduced by (1).
2 Fixed point theorems
In this section, let be a complete metric space, let be a function, let be a functional, and let ⪯ be a partial order on X such that
and
where and .
Remark 1 It is easy to see from Lemma 1 that the partial order introduced by (1) is certainly such that (3) is satisfied, but the converse is not true. In fact, a partial order such that (3) is satisfied is not necessarily confined to that introduced by (1). The following example shows that there does exist some partial order on X such that (3) is satisfied even though the relation defined by (1) is not a partial order on X.
Example 1 Let , , and ≤ is the usual order of reals. Let for each and for each . Define a relation ⪯ on X by
Clearly, ⪯ is an order on X. Direct calculation gives that
which implies (3) is satisfied. However, the relation defined by (1) is not a partial order on X since η is not subadditive.
Theorem 1 Let be a complete metric space, let be a bounded below functional, let be a nondecreasing function with , and let ⪯ be a partial order on X such that (3) and (4) are satisfied. Let be an increasing mapping. Assume that there exists such that . Then
-
(i)
T has a maximal fixed point , i.e., let be a fixed point of T, then implies ;
-
(ii)
T has a least fixed point , i.e., let be a fixed point of T, then .
Proof
-
(i)
Set
Clearly, Q is nonempty since . Let be an increasing chain, where Γ is a directed set. From (3) we find that is a decreasing net of reals. Since φ is bounded below, then exists. Let be an increasing sequence of elements from Γ such that
We claim that is a Cauchy sequence. If otherwise, there exist an increasing subsequence and such that
Since η is nondecreasing, then
which together with (3) implies that
So, we have
Let , then by and , we get
which is a contradiction, and hence is a Cauchy sequence. By the completeness of X, there exists some such that
For arbitrary , we have for each , and hence since is closed by (4). So, we have . Moreover, the arbitrary property of forces that
Since T is increasing and , then
Let , then since is closed by (4). This together with (6) indicates .
In the following, we show that has an upper bound in Q. For each , if there exists some such that , then by (6) we have for each , i.e., is an upper bound of . If there exists some such that for each n, by (3), we have for each n. Let , then we have by (4) and . We claim that
Otherwise, there exists some such that and . Then by (3) and , we have , i.e., . This contradicts , and hence for each , i.e., is an upper bound of .
By Zorn’s lemma, has a maximal element, denote it by . Since and T is increasing, then , and hence . Moreover, the maximality of in Q forces that . Therefore is a maximal fixed point of T in .
(ii) Set
From (i) we find that is nonempty. Set
Clearly, since . Define a relation on S by
It is easy to check that the relation is a partial order on S.
Let be a decreasing chain of S, where . From (3), (7), and (8), we find that is an increasing chain of M, where
Clearly, . Following the proof of (i), there exist and an increasing sequence of elements from Γ with such that (5) and (6) are satisfied. Since , then for each and each n. So, the increasing property of T implies that
Let , then
since is closed by (4). Therefore by and (9). In analogy to the proof of (i), we can prove has an upper bound in M, denote it by . Set . By and (7), we have . Note that is an upper bound of in M, then
which together with (8) implies that
i.e., is a lower bound of in S. By Zorn’s lemma, has a minimal element, denote it by . By (7) we have and
Moreover, by the increasing property of T, we have and for each . Set . Clearly, and by (7). So, by (8). Finally, the minimality of in S forces that , which implies that . Hence is a least fixed point of T in by (10). The proof is complete. □
Theorem 2 Let be a complete metric space, let be a bounded above functional, let be a nondecreasing function with , and let ⪯ be a partial order on X such that (3) and (4) are satisfied. Let be an increasing mapping. Assume that there exists such that . Then
-
(i)
T has a minimal fixed point , i.e., let be a fixed point of T, then implies ;
-
(ii)
T has a largest fixed point , i.e., let be a fixed point of T, then .
Proof Let ⪯1 be the inverse partial order of ⪯ and . Clearly, is bounded below on X since φ is bounded above, and by . It is easy to check that (3) is satisfied for ⪯1 and , and T is increasing with respect to ⪯1. Set and . Then and , and and are closed for each by (4). Applying Theorem 1 on , we find that T has a maximal fixed point and a least fixed point corresponding to ⪯1. Let be a fixed point of T. If , then , and hence by the maximality of corresponding to ⪯1, i.e., is a minimal fixed point of T corresponding to ⪯. By the least property of corresponding to ⪯1, we have , and hence , i.e., is a largest fixed point of T corresponding to ⪯. The proof is complete. □
Remark 2 From the proof of Theorem 1 (resp. Theorem 2), we find that it is only necessarily assumed in Theorem 1 (resp. Theorem 2) that the functional φ is bounded below (resp. above) on (resp. ) and T is increasing on (resp. ).
Theorem 3 Let be a complete metric space, let be a functional, let be a nondecreasing function with , and let ⪯ be a partial order on X such that (3) and (4) are satisfied. Let be a mapping. Assume that there exist with such that
and T is increasing on . Then T has a largest fixed point and a least fixed point in .
Proof Note that for each by (3), i.e., φ is bounded on . Then the conclusion follows from Remark 2, Theorem 2, and Theorem 3. The proof is complete. □
Remark 3 In our Theorems 1-3, the continuity and additivity of η necessarily assumed in [9] has been removed.
In analogy to the proof of [[7], Lemma 1], we can prove the following lemma.
Lemma 2 Let be a metric space, let be a continuous, nondecreasing, and subadditive function with , let be a continuous functional, and let ⪯ be the partial order introduced by (1). Then for each , and are closed.
It follows from Remark 1 and Lemma 2 that if η is a continuous, nondecreasing, and subadditive function with and φ is a continuous functional, then the relation ⪯ defined by (1) is a partial order on X such that (3) and (4) are satisfied. Therefore by Theorem 1 and Theorem 3, we have the following corollaries.
Corollary 1 Let be a complete metric space, let be a continuous and bounded below functional, let be a continuous, nondecreasing, and subadditive function with , and let ⪯ be the partial order introduced by (1). Let be an increasing mapping. If there exists such that , then T has a maximal fixed point and a least fixed point in .
Corollary 2 Let be a complete metric space, let be a continuous functional, let be a continuous, nondecreasing, and subadditive function with , and let ⪯ be the partial order introduced by (1). Let be a mapping. Assume that there exist with such that (11) is satisfied and T is increasing on . Then T has a largest fixed point and a least fixed point in .
Remark 4 It is clear that [[8], Theorem 3] is exactly a special case of Corollary 1 with . In addition, the existence of least fixed points has also been obtained in Theorem 1 and Corollary 1. Therefore both Theorem 1 and Corollary 1 indeed extend [[8], Theorem 3] and [[9], Theorem 2].
Remark 5 Note that each largest (resp. least) fixed point of T must be a maximal (resp. minimal) fixed point of T, but the converse is not true. Therefore both Theorem 3 and Corollary 2 improve essentially [[8], Theorem 6] and [[9], Theorems 5].
Example 2 Let X, d, φ, η, and ⪯ be the same as the ones appearing in Example 1 and
Clearly, is a complete metric space, φ is continuous, , and , . From Example 1 we know that ⪯ is a partial order such that (3) is satisfied. For each , we have
and
Note that , X, () and () are closed sets. Then, for each , and are closed, i.e., (4) is satisfied. By (12) we have
which implies that for each with , i.e., T is increasing on X. Therefore it follows from Theorem 3 that T has a largest fixed point and a least fixed point in X. In fact, 0 is the largest fixed point and is the least fixed point in .
Remark 6 (i) The existence of fixed points in Example 2 could not be obtained by [[9], Theorem 2 and Theorem 5] since η is not subadditive.
(ii) For each , , and each , , we have
Clearly, T is not a contractive mapping and hence the existence of fixed points in Example 2 could not be obtained by the fixed point theorems of contractive mappings in partially ordered metric spaces.
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Acknowledgements
This study was supported by the Natural Science Foundation of China (11161022), the Natural Science Foundation of Jiangxi Province (20114BAB211006, 20122BAB201015), the Educational Department of Jiangxi Province (GJJ12280) and the Program for Excellent Youth Talents of JXUFE(201201).
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Jiang, S., Li, Z. Largest and least fixed point theorems of increasing mappings in partially ordered metric spaces. Fixed Point Theory Appl 2013, 74 (2013). https://doi.org/10.1186/1687-1812-2013-74
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DOI: https://doi.org/10.1186/1687-1812-2013-74