Abstract

The aim of this work is to prove some coupled random coincidence theorems for a pair of compatible mixed monotone random operators satisfying weak contractive conditions. These results are some random versions and extensions of results of Karapınar et al. (2012). Our results generalize the results of Shatanawi and Mustafa (2012).

1. Introduction

Random coincidence point theorems are stochastic generalizations of classical coincidence point theorems and play an important role in the theory of random differential and integral equations. Random fixed point theorems for contractive mapping on complete separable metric space have been proved by several authors (see [1–8]). Fixed point theorems for monotone operators in ordered Banach spaces have been investigated and found various applications. Since then, fixed point theorems for mixed monotone mappings in partially ordered metric spaces are of great importance and have been utilized for matrix equations, ordinary differential equations, and the existence and uniqueness of solutions for some boundary value problems (see [9–17]).

Ćirić and Lakshmikantham [18] and Zhu and Xiao [19] proved some coupled random fixed point and coupled random coincidence results in partially ordered complete metric spaces. Moreover coupled random coincidence results in partially ordered complete metric spaces were considered in [20–22]. Following Karapınar et al. [17] and Shatanawi and Mustafa [21], we improve these results for a pair of compatible mixed monotone random mappings and , where and satisfy some weak contractive conditions. Presented results are also referred to the extensions and improve the corresponding results in [19, 21] and many other authors’ work.

2. Preliminaries

Let be a partially ordered set. The concept of a mixed monotone property of the mappings and has been introduced by Lakshmikantham and Ćirić in [16].

Definition 1 (see [16]). Let be a partially ordered set and a mapping. Then the map is said to have mixed -monotone property if is monotone -nondecreasing in and is monotone -nonincreasing in ; that is, for any ,

Definition 2 (see [16]). An element is called a coupled coincidence point of the mapping and if and .

Definition 3 (see [22]). The mappings and are said to be compatible if where and are sequences in such that and for all being satisfied.

Theorem 4 (see [17]). Let be a partially ordered set and suppose that there exists a metric on such that is a complete metric space. Let and be two mappings such that has the mixed -monotone property and satisfies for all with and , where and . Let and , , be continuous and let and be compatible mappings. If there exist such that and , then and have a coupled coincidence point in .
Denote as the set of functions satisfying the following:(i)is continuous,(ii) for all and if and only if .

Let be a measurable space with sigma algebra of subsets of and let be a metric space. A mapping is called -measurable if, for any open subset of , . In what follows, when we speak of measurability, we will mean -measurability. A mapping is called a random operator if, for any is measurable. A measurable mapping is called a random fixed point of a random function , if , for every . A measurable mapping is called a random coincidence of and if for each .

Definition 5 (see [22]). Let be a separable metric space and a measurable space. Then and are said to be compatible random operators if where and are sequences in such that and for all and for all being satisfied.

Theorem 6 (see [21]). Let be a partially ordered set, a complete separable metric space, and a measurable space. Let and be mappings such that there are two nonnegative real numbers and with such that for all with and for all . Assume that and satisfy the following conditions: are continuous, for all , are measurable, for all and , respectively,, for each , is continuous and commutes with and also suppose that either is continuous or has the following properties:if a nondecreasing sequence , then , for all ,if a nonincreasing sequence , then , for all .
If there exist measurable mappings such that and , then there are measurable mappings such that and for all ; that is, and have a coupled random coincidence.

Now, we state our main results as follows.

3. Main Results

In this section, we study coupled random coincidence and coupled random fixed point theorems for a pair of random mappings and . Then we will prove some results for random mixed monotone mappings, which are the extensions of corresponding results for deterministic mixed monotone mappings of Karapınar et al. [17].

Theorem 7. Let be a partially ordered set, a complete separable metric space, a measurable space, and and mappings such that for all with and for all , where and . Assume that and satisfy the following conditions: are continuous, for all ,, are measurable, for all and , respectively,, for each , is continuous and commutes with and also suppose that either is continuous or has the following properties:if a nondecreasing sequence , then , for all ,if a nonincreasing sequence , then , for all .
If there exist measurable mappings such that and , then there are measurable mappings such that and , for all ; that is, and have a coupled random coincidence.

Proof. Let be a family of measurable mappings. Define a function as . Since is continuous, for all , we conclude that is continuous, for all . Also, since is measurable, for all , we conclude that is measurable, for all (see [23, page 868]). Thus, is the Caratheodory function. Thus, if is measurable mapping, then is also measurable (see [24]). Also, for each , the function defined by is measurable; that is, .
Now we are going to construct two sequences of measurable mappings and in and two sequences and in as follows. Let be such that and , for all . Since , by a sort of Filippov measurable implicit function theorem (see [25, 26]), there is such that . Similarly, as , there is such that . Thus and are well defined now. Again, since there are such that Continuing this process we can construct sequences and in such that for all . Now, we use mathematical induction to prove that for all . Let , and by assumption we have Since we have Therefore, (10) holds for . Suppose (10) holds for some fixed number . Then, since and is monotone -nondecreasing in its first argument, we have Also, since and . and is monotone -nonincreasing in its second argument, we have Thus, from (9), we get Thus, by mathematical induction, we conclude that (10) holds for all . Now, we prove that and are Cauchy sequences. Let , and, by (6)–(10), we have which implies that Similarly, we have which implies that From (19) and (21), we get that Since , for all , by (22), we have Set , then is a nonincreasing sequence of positive real numbers. Thus, there is such that
Suppose that ; letting in two sides of (22) and using the properties of , we have which is a contradiction. Hence ; that is, We will show that and are Cauchy sequences. Suppose, to the contrary, that at least one of or is not a Cauchy sequence. This means that there exists an for which we can find subsequences of and of with such that Further, corresponding to , we can choose in such a way that it is the smallest integer with and satisfies (27). Then,
Using the triangle inequality and (28), we have By (27) and (29), we obtain Letting , in the inequalities above, we get By the triangle inequalities, we have
By the above inequalities and (27), we have Again, by the triangle inequality, we obtain Therefore, Taking in (33) and (35), we have Since , and . Then, from (6)–(10), we get Similarly, From (37) and (38), we arrive at Letting in the above inequality and using (26), (27), and the properties of , we have which is a contradiction. This means that and are Cauchy sequences.
Since is complete, for all , there exist the functions and such that Thus,
Since and are compatible mappings, we have
Suppose at first that assumption (a) holds. Taking the limit as in the following inequalities and using (9) and the continuity of , , we get This implies . Similarly, we can show that for each . The proof is complete.
Suppose now that (b) holds. From (9), we have Since and , we have Taking in the above inequality and using (46) and the properties of , we have Hence .
Similarly, one can show that .
The proof is complete.