Abstract

We study the existence and uniqueness of coincidence point for nonlinear mappings of any number of arguments under a weak ()-contractivity condition in partial metric spaces. The results we obtain generalize, extend, and unify several classical and very recent related results in the literature in metric spaces (see Aydi et al. (2011), Berinde and Borcut (2011), Gnana Bhaskar and Lakshmikantham (2006), Berzig and Samet (2012), Borcut and Berinde (2012), Choudhury et al. (2011), Karapınar and Luong (2012), Lakshmikantham and Ćirić (2009), Luong and Thuan (2011), and Roldán et al. (2012)) and in partial metric spaces (see Shatanawi et al. (2012)).

1. Introduction

The notion of coupled fixed point was introduced by Guo and Lakshmikantham [1] in 1987. In a recent paper, Gnana Bhaskar and Lakshmikantham [2] introduced the concept mixed monotone property for contractive operators of the form , where is a partially ordered metric space, and then established some coupled fixed-point theorems. After that, many results appeared on coupled fixed-point theory in different contexts (see, e.g., [36]). Later, Berinde and Borcut [7] introduced the concept of tripled fixed point and proved tripled fixed-point theorems using mixed monotone mappings (see also [810]).

Very recently, Roldán et al. [11] proposed the notion of coincidence point between mappings in any number of variables and showed some existence and uniqueness theorems that extended the mentioned previous results for this kind of nonlinear mappings, not necessarily permuted or ordered, in the framework of partially ordered complete metric spaces, using a weaker contraction condition, that also generalized other works by Berzig and Samet [12], Karapınar and Berinde [13].

Partial metric spaces were firstly introduced by Matthews in [14] as an attempt to generalize the metric spaces by establishing the condition that the distance between a point to itself (which is not necessarily zero) is less or equal than the distance between that point and another point of the space. In the mentioned papers, Matthews studied topological properties of partial metric spaces and stated a modified version of a Banach contraction mapping principle on this kind of spaces. After Matthews' pioneering work, the theory of partial metric spaces and particularly the field of fixed-point theorems have expansively been developed due to the increasing interest in this area and motivated by its possible applications (see [15, 16] and references therein).

In this paper, our main aim is to study a weaker contractivity condition for nonlinear mappings of any number of arguments. This condition can be particularized in a variety of forms that let us extend the previously mentioned results and other recent ones in this field (see [2, 5, 7, 9, 11, 12, 1620]). We also notice that our results cannot be obtained by the very recent paper of Haghi et al. [21] (for more details see Remark 26).

2. Preliminaries

Preliminaries and notation about coincidence points can also be found in [11]. Let be a positive integer. Henceforth, will denote a nonempty set, and will denote the product space . Throughout this paper, and will denote nonnegative integers and . Unless otherwise stated, “for all ” will mean “for all ”, and “for all ” will mean “for all ”. Let .

A metric on is a mapping satisfying, for all : (i) if, and only if, ;(ii).

From these properties, we can easily deduce that and for all . The last requirement is called the triangle inequality. If is a metric on , we say that is a metric space (for short, an MS).

Definition 1 (see [22]). A triple is called a partially ordered metric space if is a MS and is a partial order on .

Definition 2 (see [2]). An ordered MS is said to have the sequential -monotone property if it verifies(i)if is a nondecreasing sequence and , then for all ;(ii)if is a nonincreasing sequence and , then for all .

If is the identity mapping, then is said to have the sequential monotone property.

Henceforth, fix a partition of two non-empty subsets of ; that is, and We will denote If is a partially ordered space, , and , we will use the following notation:

Let and be two mappings.

Definition 3 (see [11]). One says that and are commuting if for all .

Definition 4 (see [11]). Let be a partially ordered space. One says that has the mixed -monotone property (with respect to ) if is -monotone nondecreasing in arguments of and -monotone nonincreasing in arguments of ; that is, for all and all ,
Henceforth, let be mappings from into itself, and let be the -tuple .

Definition 5 (see [11]). A point is called a -coincidence point of the mappings and if If is the identity mapping on , then is called a -fixed point of the mapping .

Remark 6. If and are commuting and is a -coincidence point of and , then also is a -coincidence point of and .

Definition 7 (see [14]). A partial metric on is a mapping verifying, for all :  ;  ;  ;  .
In this case, is a partial metric space (for short, a PMS).

Example 8 (see, e.g., [14]). Let , and define on by for all . Then, is a partial metric space.

Example 9 (see [14]). Let , and define = . Then, is a partial metric space.

Example 10 (see [14]). Let , and define by Then, is a partial metric space.

Example 11 (see, e.g., [23, 24]). Let and be a metric space and a partial metric space, respectively. Functions given by define partial metrics on , where is an arbitrary function and .
Obviously, if is a MS and we define , then is a PMS. Indeed, a partial metric on verifies(i); (ii); (iii),
but the condition does not necessarily hold. For a partial metric on , the mappings given by for all , are (usual) metrics on . On a PMS, the concepts of convergence, Cauchy sequences, completeness, and continuity are defined as follows.

Definition 12 (see [14, 25, 26]). Let be a sequence on a PMS .(i)  -converges to (and one will write ) if .(ii) is called -Cauchy if exists (and it is finite).(iii) is said to be -complete if every -Cauchy sequence in   -converges to a point such that .(iv)A mapping is said to be -continuous at if, for every , there exists such that .

We have used the previous notation because we need to distinguish between -convergence and -convergence on and usual convergence for real sequences.

Lemma 13 (see [14, 25, 26]). Let be a sequence on a PMS .(1) is -Cauchy if, and only if, it is -Cauchy.(2) if, and only if, and ; that is, (3) is complete if, and only if, the MS is complete.(4)If and , then for all .

3. Auxiliary Results

We will use the following results about real sequences in the proof of our main theorems.

Lemma 14. Let be real lower bounded sequences such that . Then, there exists and a subsequence such that .

Proof. Let for all . As is convergent, it is bounded. As for all and , then every is bounded. As is a real bounded sequence, it has a convergent subsequence . Consider the subsequences ; that are real bounded sequences and the sequence that also converges to . As is a real bounded sequence, it has a convergent subsequence . Then, the sequences , , also are real bounded sequences, , and . Repeating this process times, we can find subsequences , , (where ) such that for all . And . But so , and there exists such that . Therefore, there exists and a subsequence such that .

Lemma 15. Let be a sequence of nonnegative real numbers which has not any subsequence converging to zero. Then, for all , there exist and such that for all .

Proof. Suppose that the conclusion is not true. Then, there exists such that, for all , there exists verifying . Let be such that . For all , take . Then, there exists verifying . Taking limit when , we deduce that . Then, has a subsequence converging to zero (maybe, reordering ), but this is a contradiction.

Lemma 16. If is a sequence in a MS that is not Cauchy, then there exist and two subsequences and such that, for all ,

Proof. We know that If this condition is not true, then Let . Then, there exists such that and . Let , and consider the numbers Since , between the previous numbers there exists a first nonnegative integer such that but for all . In particular, .
Now, let . Then, there exists such that and . Let , and consider the numbers Since , between the previous numbers there exists a first nonnegative integer such that but for all . In particular, .
Repeating this process, we can find two subsequences and such that, for all :

Definition 17. Let be the family of all continuous, nondecreasing mappings such that if, and only if, .

These mappings are known as altering distance functions (see [27]). Note that every selected commutes with ; that is, for all .

Lemma 18. If and , then .

Proof. As there exists , then . If the conclusion is not true, there exists such that, for all , there exists verifying . This means that has a subsequence such that . As is nondecreasing, for all . Therefore, has a subsequence lower bounded by , but this is impossible since .

With regards to coincidence points, it is possible to consider the following simplification. If is a permutation of , and we reorder (4), then we deduce that every coincidence point may be seen as a coincidence point associated to the identity mapping on (see, for instance, [28]).

Lemma 19. Let be a permutation of , and let and = . Then, a point is a -coincidence point of the mappings and if, and only if, is a -coincidence point of the mappings and .

Therefore, in the sequel, without loss of generality, we will only consider -coincidence points where , that is, that verify for all . We also show some preliminary results on PMS.

Lemma 20. Let be a sequence on a PMS , and let .(1)If and , then , and for all .(2)If and , then .

Proof. Since and , then = . Therefore, = = , so . Since is continuous, then for all , and item 4 of Lemma 13 implies that .
Item 2 of Lemma 13 shows that .

Remark 21. Although the limit in a MS is unique, the -limit in a PMS is not necessarily unique. For instance, let as in Example 10. Then, is a complete PMS (see [14]). Consider for all . Then, but whenever .

Definition 22. Let , let be a PMS, let be a mapping, and let . We will say that is -continuous at if, for all sequences on such that for all , for all and , we have that and . One will say that is -continuous if it is continuous at every point .

Lemma 23. If is a PMS, and is -continuous at , then is -continuous at .

Proof. Let sequences on such that for all , for all , and . Item 1 of Lemma 20 implies that for all . Since is -continuous at , then . Item 2 of Lemma 13 assures us that and Then, is -continuous at .

4. Main Results

In the following result, we show sufficient conditions to ensure the existence of -coincidence points, where .

Theorem 24. Let be a complete PMS, and let a partial order on . Let be an -tuple of mappings from into itself verifying if and if . Let and be two mappings such that has the mixed -monotone property on , and is -continuous and commuting with . Assume that there exist such that for which for all . Suppose either is -continuous or has the sequential -monotone property. If there exist verifying for all , then and have, at least, one -coincidence point.

Proof. The proof is divided into seven steps. The first two steps are the same as in the proof of Theorem 9 in [11], since the contractivity condition does not play any role in these parts of the proof.
Step  1. There exist sequences such that = for all and all .
Step  2. for all and all .
Step  3. We claim that for all (i.e., ).
Indeed, define for all . As for all and all , then condition (17) implies that, for all and all :Therefore, for all , . This means that the sequence is nonincreasing and lower bounded. Hence, it is convergent; that is, there exists such that . We are going to show that . Since Lemma 14 assures that there exist and a subsequence such that . Repeating (18), for all , Consider the sequence Suppose that this sequence has no subsequence converging to zero. Using , Lemma 15 assures us that there exists and such that for all . It follows that Then, (20) says to us Taking limit in , we deduce that , which is impossible. Therefore, the sequence in (21) must have a subsequence converging to zero. Since and are continuous, taking limit when in (20) using this subsequence, we deduce that , so . Then, we have just proved that . Therefore, , and Lemma 18 assures that , which means that for all since for all and all .
Step  4. for all (i.e., ). It is the same proof of Step  3.
Since for all and , joining Steps 3 and 4, it follows that Step  5. Every sequence is -Cauchy. We reason by contradiction. Suppose that are not -Cauchy () and are -Cauchy, being . By Lemma 16, for all , there exists and subsequences and such that Now, let and . Since are -Cauchy, for all , there exists such that if , then . Since by Step  4, there exists such that if , then . Define . If , then Therefore, we have proved that there exists such that if , then
Next, let such that . Let such that , and define . Consider the numbers until finding the first positive integer verifying Now let such that , and define . Consider the numbers until finding the first positive integer verifying Repeating this process, we can find sequences such that, for all , Note that by (27), for all , so for all and all . Furthermore, for all , Therefore, for all and all , Next, for all , let be an index such that Then, for all , Applying the contractivity condition (17), it follows, for all , Consider the sequence: If this sequence has a subsequence that converges to zero, then we can take limit when in (36) using this subsequence, so that we would have , which is impossible since . Therefore, the sequence (37) has no subsequence converging to zero. In this case, taking in Lemma 15, there exist and such that , for all . It follows that, for all , . Thus, by (36), Fix any and we are going to prove that . Indeed, by Step  3 and (24), since are sequences converging to zero, we can find such that if , then Therefore, (33) implies that, for all and for all such that , Then, (38) guarantees that . This means that for all . If we take (where ), we deduce that for all . Since is continuous, we have that , which is impossible since . This contradiction finally proves that every sequence is -Cauchy.
Since is -complete, then is -complete (item 3 of Lemma 13). Then, there exist such that for all . Furthermore, = = = for all . Since is -continuous, then and for all . Item 1 of Lemma 20 shows that for all . Therefore, for all , . Moreover, for all and all , .
Step  6. Suppose that is -continuous. In this case, we know that and for all and which implies that and , for all . Item 1 of Lemma 20 assures us that, for all , Since the limit in a MS is unique, we deduce that for all , so is a -coincidence point of and .
Step  7. Suppose that   has the sequential -monotone property. In this case, by Step  2, we know that for all and all . This means that the sequence is monotone. As , we deduce that for all and all. This condition implies that, for all and all , (the first case occurs when and the second one when ). Then, by (17), for all , Since for all , then Therefore, . Taking limit when in (45), we deduce that for all . As , Lemma 18 guarantees that Finally, for all , Using (46) and (47), we conclude that for all .

Remark 25. In the previous theorem, if the image of the metric is not the whole set , then and can only be defined on , and we can consider a wider range of mappings since it is only necessary to impose that they are continuous and nondecreasing on .

Remark 26. We notice also that our paper cannot be deduced from the recent interesting paper of Haghi et al. [21] on partial metric space. In fact, we use a partial order . Then, we only suppose (17) for which for all (not necessarily on points which are not comparable). Further, we use a self-map which implies that is not necessarily a partial metric on . For instance, let provided with its usual partial order and the partial metric . Consider Then, is continuous, but but . Then, does not verify the axiom . Therefore, we cannot apply Theorem 2.4 on Haghi et al. [21].
As a result, we cannot use Theorem 2.7 in [21] since has an influence in , and our mapping has not a role in the left side of (17).

5. Consequences

Remark 27. Theorem 9 in [11] is an easy consequence of Theorem 24 if we take , , and for all .
In the next result, let be the family of all nondecreasing on each argument, continuous mappings verifying if, and only if, . Examples of such mappings are the following, where , and for all . (i).(ii).(iii).

Lemma 28. Let , and define as for all , where is the usual basis of . Then, and for all .

Proof. First part is clear. If , then = = . Therefore, = .

Corollary 29. Thesis of Theorem 24 also holds if one replaces the contractivity condition (17) by any of the following list (for which for all ).(A) This condition can be found in [11] and [12], there exist and such that (B)In [17], there exist and such that and (C) There exist and such that (D)In [2, 7, 9], there exist , , and such that and (E)In [5], there exist and such that (F)In [19, 20], there exist such that is subadditive for all and

Of course, it is also interesting to particularize all the previous items to the following cases: (where ), (where ), or for all .

Proof. (A) By Lemma 28, there exists such that for all , so (52) implies (17). (B) It is obvious that , so (53) implies (17). (C) We only take in item (A). (D) It is a mixture of (B) and (C). (E) It is a particular case of (D) where for all . (F) If is subadditive, then for all , so we may choose for all in (B).

6. Uniqueness of -Coincidence Points

Consider on the product space the following partial order: for , We say that and are comparable if or .

Theorem 30. Under the hypothesis of Theorem 24, assume that for all -coincidence points , of and there exists such that is comparable, at the same time, to and to .
Then, and have a unique -coincidence point such that for all .

Proof. From Theorem 24, the set of -coincidence points of and is nonempty. The proof is divided into two steps.
Step  1. We claim that if are two -coincidence points of and , then Let be two -coincidence points of and , and let be a point such that is comparable, at the same time, to and to . Using , define the following sequences. Let for all . Reasoning as in Theorem 24, we can determine sequences , such that for all and all . We are going to prove that for all , so (59) will be true.
Firstly, we reason with and , and the same argument holds for and . As and are comparable, we can suppose that (the other case is similar); that is, for all . Using that has the mixed -monotone property and reasoning as in Theorem 24, it is possible to prove that for all and all . This condition implies that, for all and all Define for all . Reasoning as in Theorem 24, it is not difficult to prove that which means that . As for all and all , we deduce that for all ; that is, for all . Item 1 of Lemma 20 shows that If we had supposed that , we would have obtained the same property (61). And as also is comparable to , we can reason in the same way to prove that for all . Since the limit in a MS is unique, for all .
Let be a -coincidence point of and , and define for all . As , Remark 6 assures us that also is a -coincidence point of and .
Step  2. We claim that is the unique -coincidence point of and such that for all . It is similar to Step 2 in Theorem 11 in [11].

It is natural to say that is injective on the set of all -coincidence points of and when for all implies for all when are two -coincidence points of and . For example, this is true is is injective on .

Corollary 31. In addition to the hypotheses of Theorem 30, suppose that is injective on the set of all -coincidence points of and . Then, and have a unique -coincidence point.

Proof. If and are two -coincidence points of and , we have proved in (59) that for all . As is injective on these points, then, for all .

Corollary 32. In addition to the hypotheses of Theorem 30, suppose that is comparable to for all . Then, .
In particular, there exists a unique such that , which verifies .

Proof. Let , let such that , and let Fix . As is comparable to , then either for all or for all . Since for all , we know that either for all or for all . In any case, applying (17), If , then , so , which is impossible. Then, , and (63) implies that , so . Therefore, for all and .

Example 33. Let provided with its usual partial order and the partial metric . Let , and let real numbers such that there exist verifying . Let , and consider and , for all . Then, is monotone nondecreasing in those arguments for which and monotone nonincreasing in those arguments for which . Furthermore, taking , it follows that Therefore, If and , all conditions of Theorems 24 and 30 (and Corollaries 31 and 32) are satisfied. Indeed, it is clear that is the unique fixed point of .

The following example is based on Examples 1.9 and 2.2 in [29].

Example 34. Let , and let be the partial metric on given by for all . Then, is complete, and generates the discrete topology on (indeed, is the Euclidean metric on ). Consider on the following partial order: Consider and defined by It is not difficult to prove the following statements.(1) and are -continuous mappings (since generates the discrete topology on ).(2) and are commuting.(3) If verify , then either or . Then, has the mixed -monotone property on .(4) If verify for all , then = . In particular, (17) holds (whatever and ; for instance, and for all ).For simplicity, henceforth, suppose that is even, and let (resp., ) be the set of all odd (resp., even) numbers in .(5) For a mapping , we use the notation and consider Then, if is odd, and if is even. Let .(6) Take if is odd and if is even. Then, for all .(7) has the sequential -monotone property.
Therefore, we can apply Theorems 24 and 30, and Corollaries 31 and 32, to conclude that and have a unique -coincidence point, which is .

Acknowledgments

This work has been partially supported by Junta de Andaluca, by projects FQM-268, FQM-178 and FQM-235 of the Andalusian CICYE.