#### Abstract

We present new results on the existence and uniqueness of tripled fixed points for nonlinear mappings in partially ordered complete metric spaces that extend the results in the previous works: Berinde and Borcut, 2011, Borcut and Berinde, 2012, and Borcut, 2012. An example and an application to support our new results are also included in the paper.

#### 1. Introduction

In some very recent papers, Berinde and Borcut , Borcut and Berinde , and Borcut  have introduced the concepts of tripled fixed point and tripled coincidence point, respectively, for nonlinear contractive mappings in partially ordered complete metric spaces and obtained existence and uniqueness theorems of tripled fixed points and tripled coincidence points, respectively, for some general classes of contractive type mappings.

The presented theorems in  extend several existing results in the literature . We recall the main concepts needed to present them.

Let be a partially ordered set and let be a metric on such that is a complete metric space. Consider on the product space the following partial order: for ,

Definition 1 (see ). Let be a partially ordered set and . We say that has the mixed monotone property if is nondecreasing in and and is nonincreasing in , that is, for any , , ,

Definition 2 (see ). An element is called a tripled fixed point of if

Let be a metric space. The mapping , given by defines a metric on , which will be denoted for convenience by , too.

Definition 3. Let , , be nonempty sets and , . We define the symmetric composition (or, the s-composition, for short) of and , by ,

For each nonempty set , denote by the projection mapping The symmetric composition has the following properties.

Proposition 4 (associativity). If , and

Proposition 5 (identity element). If , then

Proposition 6 (mixed monotonicity). If   , , are partially ordered sets and the mappings , are mixed monotone, then is mixed monotone.

Proposition 7. If   is a partially ordered set and is mixed monotone, then is mixed monotone for every .

The first main result in  is given by the following theorem.

Theorem 8 (see ). Let be a partially ordered set and suppose that there is a metric on such that is a complete metric space. Let be a continuous mapping having the mixed monotone property on . Assume that there exist the constants , , with for which for all , , . If there exist , , such that then there exist such that

Remark 9. If we take in Theorem 8, then the contraction condition (9) can be written in a slightly simplified form

Theorem 10 (see ). By adding to the hypotheses of Theorem 8 the condition, for every , , there exists a that is comparable to and ; then, the tripled fixed point of is unique.

Theorem 11 (see ). In addition to the hypotheses of Theorem 8, suppose that , , are comparable. Then .

#### 2. Main Results

Starting from the results presented in the first section, we will obtain new existence and uniqueness theorems for operators which verify a Kannan type contraction condition; see .

Denote

Theorem 12. Let be a partially ordered set and suppose that there is a metric on such that is a complete metric space. Let be a mapping having the mixed monotone property on . Assume that there exists a such that Also suppose that either(a)is continuous or(b) has the following property:(i)if a nondecreasing sequence , then for all ,(ii)if a nonincreasing sequence , then for all .If there exist , , such that then has a triple fixed point; that is, there exist , , such that

Proof. Let the sequences , , be defined by Since is mixed monotone for every , by Proposition 7, it follows by (15) that and are nondecreasing and is nonincreasing. Due to the mixed monotone property of , it is easy to show that and thus we obtain three sequences satisfying the following conditions: Now, for , denote Using (14), we get and so Similarly, we obtain
By (22) and (23), we get Therefore, for all , we have Because , , and , we have This implies that , , are Cauchy sequences in . Indeed, let ; then,
Similarly, we can verify that and are also Cauchy sequences.
Since is a complete metric space, there exist , , such that Finally, we claim that
Suppose first that assumption (a) holds. Hence is continuous at , and, therefore, for any given , there exists such that Since for , there exist , , such that, for , , , Hence, for , , This shows that . Similarly, one can show that
Suppose now that assumption (b) holds. Since , are nondecreasing and , , is nonincreasing and , by assumption (b), we have that , , and , for all . Then, by triangle inequality and (14), we get By summing (35), we obtain and by letting in the previous inequality, one obtains which proves that , , .

#### 3. Uniqueness of Tripled Fixed Points

In , the authors also considered some additional conditions that ensure the uniqueness of the tripled fixed point or that, for the tripled fixed point , we have .

Similarly, one can prove that the tripled fixed point in Theorem 12 is in fact unique, provided that the product space endowed with the partial order mentioned earlier has an additional property, as shown in the next theorem.

Theorem 13. By adding to the hypotheses of Theorem 12 the condition, for every , , there exists a which is comparable to and ; then, the tripled fixed point of is unique.

Proof. If is another tripled fixed point of , then we show that where are as in the proof of Theorem 12. We consider two cases.
Case  1. If are comparable to with respect to the ordering in , then, for every , Also, using the process of obtaining (26), we get Now letting , this implies that .
Case  2. If are not comparable to , then there exists an upper bound or a lower bound of and . Then, for all , We have

Theorem 14. In addition to the hypotheses of Theorem 12, suppose that , , are comparable. Then .

Proof. Recall that , , , are such that Now, if and , we claim that, for all , and . Indeed, by the mixed monotone property of , Assume that and for some . Now, consider similarly for . Hence, as . This implies that and hence we have .
Similarly, we obtain that and . The other remaining cases for , , are similar.

#### 4. An Example

Let be endowed with the usual metric and let be given by , for , and , for .

Then satisfies the Kannan type contractive condition (14) with but does not satisfy the Banach type contractive condition (9).

Let us first prove the first part of the assertion above. It suffices to completely cover the following limit case.

Case 1 (, , , , , ). In this case, condition (14) reduces to
Since , we have and hence the minimum value of the right hand side of (48) is greater or equal to .
Therefore, in order to have (48) satisfied for all and , , , , , with , , , that is, it suffices to take such that .

Note that, for the remaining cases to be discussed, the right hand side of (14) will be greater than the value obtained in Case 1.

Case 2 ( and , , , ). In this case, for example, the minimum value of the right hand side of (14) will be greater or equal to .
Note also that, in the cases , or , , the left hand side of (14) is always zero and so (14) is satisfied for all values of , , , .
This proves that, indeed, satisfies (14) with .
is not continuous but satisfies assumption (b) in Theorem 12. Moreover, by taking , , and , one can easily check that (15) is fulfilled.
Thus, the assumptions in Theorem 12 are satisfied and hence does admit tripled fixed points. By Theorem 13, we actually conclude that has a unique tripled fixed point, .
Now let us show that does not satisfy (9).
Assume the contrary, that is, that does satisfy (9) and let such that , and then take and , arbitrary in (9) to obtain the inequality Now by letting in (51), we reach a contradiction. This proves that, indeed, does not satisfy (9).

Remark 15. For various particular cases of the results established in this section and for possible further developments, we refer to [2, 46, 828].

#### 5. Applications

In this section, we present an application of tripled fixed point theorems for establishing existence and uniqueness results for the solutions of the nonlinear integral equation We consider the space of continuous real functions defined on the interval , endowed with metric Define the partial order “” on by Thus, is a partially ordered complete metric space.

For (52) we consider the following assumptions:(i) are continuous;(ii) is continuous;(iii) is continuous;(iv)there exist the constants , , , such that, for all , , we have (v)we suppose that ;(vi)there exist continuous functions such that

Theorem 16. Under assumptions (i)–(vi), (52) has a unique solution in .

Proof. We consider the operator defined by for any , , .
We prove that the operator fulfills the conditions of Theorem 8.
First, we prove that has the mixed-monotone property.
Let , , with and , then, we have Given that for all and based on our assumption (iv), we have That is, .
For , , with and , we have Given that for all and based on our assumption (iv), that is, That is, .
Similarly, one proves the same property for the third component and hence we have .
So, has the mixed-monotone property.
Now, we estimate for , , , and with having the mixed-monotone property, we get Now, for all , by using (iv), we have which implies where Using (v), we have . So, fulfills the conditions of Theorem 8.
Let , , be the functions appearing in assumption (vi); then, we have If , , , then all assumptions of Theorem 8 are fulfilled. So, there exists a tripled fixed point for the operator ; that is, Now, we show that, for any , , there exists which is comparable to both of them. Indeed, denote Then we have , , and , , . This implies that is comparable with and with , so Now, by Theorem 10, it follows that has a unique triple fixed point which is in fact the solution the (52).

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

The research was supported by the Grant PN-II-RU-TE-2011-3-0239 of the Romanian Ministry of Education and Research.