Abstract
In this paper, we introduced a new type of a contractive condition defined on an ordered space, namely a -contraction, which generalizes the weak contraction. We also proved some fixed point theorems for such a condition in ordered metric spaces. A supporting example of our results is provided in the last part of our paper as well.
MSC:06A05, 06A06, 47H09, 47H10, 54H25.
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1 Introduction and preliminaries
It is well known that the Banach contraction principle has been improved in different directions in different spaces by mathematicians over the years. Even in the contemporary research, it remains a heavily investigated branch. Thus, several authors have generalized the principle in various ways (see, for example, [1–14]).
In 1997, Alber and Guerre-Delabriere [15] have introduced the concept of weak contraction in Hilbert spaces. Later, Rhoades [16] showed, in 2001, that these results are also valid in complete metric spaces. We state the result of Rhoades in the following.
A mapping , where is a metric space, is said to be weakly contractive if
for all and is a function satisfying:
-
(i)
φ is continuous and nondecreasing;
-
(ii)
if and only if ;
-
(iii)
.
Note that (1.1) reduces to an ordinary contraction when , where .
Theorem 1.1 ([16])
Let be a complete metric space and f be a weakly contractive mapping. Then f has a unique fixed point in X.
An interesting way to generalize this theorem is to consider it in case a partial ordering is defined on the space. Recall that a relation ⊑ is a partial ordering on a set X if it is reflexive, antisymmetric and transitive. By this meaning, we write instead of to emphasize some particular cases. Any are said to be comparable if or . If a set X has a partial ordering ⊑ defined on it, we say that it is a partially ordered set (w.r.t. ⊑) and denote it by . is said to be a totally ordered set if any two elements in X are comparable. Moreover, it is said to be a sequentially ordered set if each element of a convergent sequence in X is comparable with its limit. Yet, if is a metric space and is a partially ordered (totally ordered, sequentially ordered) set, we say that X is a partially ordered (totally ordered, sequentially ordered, respectively) metric space, and it will be denoted by .
In 2009, Harjani and Sadarangani [17] carried the work of Rhoades [16] into partially ordered metric spaces. We now state the result proved in [17] as follows.
Theorem 1.2 ([17])
Let be a complete partially ordered metric space and let be a continuous and nondecreasing mapping such that
for , where is a function satisfying:
-
(i)
φ is continuous and nondecreasing;
-
(ii)
if and only if ;
-
(iii)
.
If there exists such that , then f has a fixed point.
Harjini and Sadarangani [17] also proved fixed point theorems for noncontinuous mappings, nonincreasing mappings and even for non-monotonic mappings.
The aim of this paper is to introduce a weak condition which resulted in the concept called a -contraction.
2 -functions
In this section, we introduce our concept of a -function and some of its fundamental properties. Not to be ambiguous, we assume that ℝ represents the set of all real numbers while ℕ represents the set of all positive integers.
Definition 2.1 Let be a partially ordered metric space. A function is called a -function w.r.t. ⊑ in X if it satisfies the following conditions:
-
(i)
for every comparable ;
-
(ii)
for any sequences , in X such that and are comparable at each , if and , then ;
-
(iii)
for any sequences , in X such that and are comparable at each , if , then .
If, in addition, the following condition is also satisfied:
-
(A)
for any sequences , in X such that and are comparable at each , if the limit exists, then the limit also exists,
then ϱ is said to be a -function of type (A) w.r.t. ⊑ in X.
Example 2.2 Let be a partially ordered metric space. Suppose that the function is defined as in Theorem 1.2. Then is a -function of type (A) w.r.t. ⊑ in X.
Proposition 2.3 Let be a partially ordered metric space and be a -function w.r.t. ⊑ in X. If are comparable and , then .
Proof Let be comparable and . Define and to be two constant sequences in X such that and for all . It follows from the definition of a -function, since x and y are comparable, that . That is, . □
Corollary 2.4 Let be a totally ordered metric space and be a -function w.r.t. ⊑ in X. If and , then .
Proof Since X is totally ordered, any are comparable. The rest of the proof is straightforward. □
Example 2.5 Let . Define with and . If X is endowed with a usual ordering ≤, then is a totally ordered metric space with ϱ as a -function of type (A) w.r.t. ≤ in X. Note that for all , even when .
This example shows that the converse of Proposition 2.3 and that of Corollary 2.4 are not generally true.
Definition 2.6 Let be a partially ordered metric space, a mapping is called a -contraction w.r.t. ⊑ if there exists a -function w.r.t. ⊑ in X such that
for any comparable . Naturally, if there exists a -function of type (A) w.r.t. ⊑ in X such that the inequality (2.1) holds for any comparable , then f is said to be a -contraction of type (A) w.r.t. ⊑.
Remark 2.7 From Example 2.2, it follows that in partially ordered metric spaces, a weak contraction is also a -contraction.
3 Fixed point results
3.1 Fixed point theorems for monotonic mappings
Theorem 3.1 Let be a complete partially ordered metric space and be a continuous and nondecreasing -contraction of type (A) w.r.t. ⊑. If there exists with , then converges to a fixed point of f in X.
Proof For the existence of the fixed point, we choose such that . If , then the proof is finished. Suppose that . We define a sequence such that . Since and f is nondecreasing w.r.t. ⊑, we obtain
If there exists such that , then by the notion of -contractivity, the proof is finished. Therefore, we assume that for all . Also, assume that for all . Otherwise, we can find with , that is, , and the proof is finished. Hence, we consider only the case where for all .
Since for all , we have
for all . Therefore, we have nonincreasing. Since is bounded, there exists such that . Thus, there exists such that .
Assume that . Then, by the -contractivity of f, we have
Hence, , which implies that , a contradiction. Therefore, we have
Now we show that is a Cauchy sequence in X. Assume the contrary. Then there exists for which we can define subsequences and of such that is minimal in the sense that and . Therefore, . Observe that
Letting , we obtain and so
By the two following inequalities:
and
we can apply (3.1) and (3.2) to obtain
Furthermore, we deduce that the limit also exists. Now, by the -contractivity, we have
From (3.2) and (3.3), we may find that
which further implies that . Notice that at each . Consequently, we obtain that , which is a contradiction. So, is a Cauchy sequence. Since X is complete, there exists such that as . Finally, the continuity of f and imply that . Therefore, is a fixed point of f. □
Remark 3.2 In the setting of Remark 2.7, Theorem 3.1 reduces to Theorem 1.2 of [17].
Next, we drop the continuity of f in the Theorem 3.1, and find out that we can still guarantee a fixed point if we strengthen the condition of a partially ordered set to a sequentially ordered set.
Theorem 3.3 Let be a complete sequentially ordered metric space and be a nondecreasing -contraction of type (A) w.r.t. ⊑. If there exists with , then converges to a fixed point of f in X.
Proof If we take in the proof of Theorem 3.1, then we conclude that converges to a point in X.
Next, we prove that is a fixed point of f in X. Indeed, suppose that is not a fixed point of f, i.e., . Since is comparable with for all , we have
for all . By the definition of a convergent sequence, we have, for any , there exists such that for all with . Therefore, we have
As easily seen, is less than any nonnegative real number, so , which is a contradiction. Hence, is a fixed point of f. □
Corollary 3.4 Let be a complete totally ordered metric space and be a nondecreasing -contraction of type (A) w.r.t. ⊑. If there exists with , then converges to a unique fixed point of f in X.
Proof Take as in the proof of Theorem 3.1. Since the total ordering implies the partial ordering, we conclude that converges to a fixed point.
Next, we show that the fixed point of f is unique. Assume that u and v are two distinct fixed points of f, i.e., . Since X is totally ordered, u and v are comparable. Thus, we have
which is a contradiction. Therefore, and the fixed point of f is unique. □
We can still guarantee the uniqueness of a fixed point by weakening the total ordering condition as stated and proved in the next theorem.
Theorem 3.5 Let be a complete partially ordered metric space and be a continuous and nondecreasing -contraction of type (A) w.r.t. ⊑. Suppose that for each , there exists which is comparable to both x and y. If there exists with , then converges to a unique fixed point of f in X.
Proof If we take in the proof of Theorem 3.1, then we conclude that converges to a fixed point of f in X.
Next, we show that the fixed point of f is unique. Assume that u and v are two distinct fixed points of f, i.e., . Since , there exists such that w is comparable to both u and v. We will prove this part by showing that the sequence given by converges to both u and v. Therefore, we have
If we define a sequence and , we may obtain from (3.5) that is nonincreasing and there exist such that and .
Assume that . Then, by the -contractivity of f, we have
which is a contradiction. Hence, . In the same way, we can also show that . That is, converges to both u and v. Since the limit of a convergent sequence in a metric space is unique, we conclude that . Hence, this yields the uniqueness of the fixed point. □
Theorem 3.6 Let be a complete sequentially ordered metric space and be a nondecreasing -contraction of type (A) w.r.t. ⊑. Suppose that for each , there exists which is comparable to both x and y. If there exists with , then converges to a unique fixed point of f in X.
Proof If we take in the proof of Theorem 3.1, then we conclude that converges to a fixed point of f in X. The rest of the proof is similar to the proof of Theorem 3.5. □
Remark 3.7 In parallel with the study of Theorems 3.1, 3.3, 3.5 and 3.6, we can also prove in the same way that if the mapping f is nonincreasing, the above theorems still hold. However, we will omit the result for nonincreasing mappings.
3.2 Fixed point theorems for mappings with the lack of monotonicity
In this section, we drop the monotonicity conditions of f and find out that we can still apply our results to confirm the existence and uniqueness of a fixed point of f.
Theorem 3.8 Let be a complete partially ordered metric space and be a continuous -contraction of type (A) w.r.t. ⊑ such that the comparability of implies the comparability of . If there exists such that and are comparable, then converges to a fixed point of f in X.
Proof For the existence of the fixed point, we choose such that and are comparable. If , then the proof is finished. Suppose that . We define a sequence such that . Since and are comparable, we have and comparable for all .
If there exists such that , then by the notion of -contractivity, the proof is finished. Therefore, we assume that for all . Also, assume that for all . Otherwise, we can find with , that is, , and the proof is finished. Hence, we consider only the case where for all .
Since and are comparable for all , we have
for all . Therefore, we have is nonincreasing. Since is bounded, there exists such that . Thus, there exists such that .
Assume that . Then, by the -contractivity of f, we have
Hence, , which implies that , a contradiction. Hence, .
Now we show that is a Cauchy sequence in X. Assume the contrary. Then there exists for which we can define subsequences and of such that is minimal in the sense that and . Therefore, . Observe that
Letting , we obtain and so
By the two following inequalities:
and
we can apply the fact that and (3.6) to obtain
Furthermore, we deduce that the limit also exists. Now, by the -contractivity, we have
From (3.6) and (3.7), we may find that
which further implies that . Notice that at each . Consequently, we obtain that , which is a contradiction. So, is a Cauchy sequence. Since X is complete, there exists such that as . Finally, the continuity of f and imply that . Therefore, is a fixed point of f. □
Further results can be proved using the same plots as those of the earlier theorems in this paper, so we omit them.
Theorem 3.9 Let be a complete sequentially ordered metric space and be a -contraction of type (A) w.r.t. ⊑ such that the comparability of implies the comparability of . If there exists such that and are comparable, then converges to a fixed point of f in X.
Corollary 3.10 Let be a complete totally ordered metric space with and be a -contraction of type (A) w.r.t. ⊑ such that the comparability of implies the comparability of . If there exists such that and are comparable, then converges to a unique fixed point of f in X.
Theorem 3.11 Let be a complete partially ordered metric space and be a continuous -contraction of type (A) w.r.t. ⊑ such that the comparability of implies the comparability of . Suppose that for each , there exists which is comparable to both x and y. If there exists such that and are comparable, then converges to a unique fixed point of f in X.
Theorem 3.12 Let be a complete sequentially ordered metric space and be a -contraction of type (A) w.r.t. ⊑ such that the comparability of implies the comparability of . Suppose that for each , there exists which is comparable to both x and y. If there exists such that and are comparable, then converges to a unique fixed point of f in X.
4 Example
We give an example to ensure the applicability of our theorems.
Example 4.1 Let and suppose that we write and for .
Define by
and
Let ⊑ be an ordering in X such that for , if and only if and . Then is a partially ordered metric space with ϱ as a -function of type (A) w.r.t. ⊑ in X.
Now, let f be a self mapping on X defined by for all . It is obvious that f is continuous and nondecreasing w.r.t. ⊑.
Let be comparable w.r.t. ⊑. If , then they clearly satisfy the inequality (2.1). On the other hand, if , we have
Therefore, the inequality (2.1) is satisfied for every comparable . So, f is a continuous and nondecreasing -contraction of type (A) w.r.t. ⊑. Let , so we have . Now, applying Theorem 3.1, we conclude that f has a fixed point in X which is the point .
5 Conclusion
It is undeniable that Rhoades’s weak contraction is one of the earliest and the most important extensions of the contraction principle. The results in this paper give a new direction to expanding the framework of contractive type mappings in metric spaces. Still, there is a question to be raised from this paper onwards.
Question Are our results still true for any -contractions (not necessarily of type (A))?
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Acknowledgements
This work was partially supported by the Higher Education Research Promotion and National Research University Project of Thailand, Office of the Higher Education Commission (NRU-CSEC No. 55000613). The second author would like to thank the Research Professional Development Project under the Science Achievement Scholarship of Thailand (SAST). Also, the third author was supported by the Commission on Higher Education, the Thailand Research Fund and the King Mongkut’s University of Technology Thonburi (Grant No. MRG5580213).
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Chaipunya, P., Sintunavarat, W. & Kumam, P. On -contractions in ordered metric spaces. Fixed Point Theory Appl 2012, 219 (2012). https://doi.org/10.1186/1687-1812-2012-219
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DOI: https://doi.org/10.1186/1687-1812-2012-219