#### Abstract

We prove the existence and uniqueness of a fixed point of certain type mapping, extension of Suzuki-Edelstein mapping, in a partially ordered complete metric space. Our results extend, improve, and generalize the existence results on the topic in the literature. We state some examples to illustrate our results.

#### 1. Introduction

Fixed point theory lies in the center of nonlinear functional analysis because it has a broad range of applications in fields such as economics, computer science, and many others (see, e.g., [17]). In particular, fixed point theory is quite useful in finding the solutions of inverse problems and structural optimizations in science and engineering [810]. Banach contraction mapping principle [11] is considered to be the fundamental result in this theory. It states that each contraction defined on a complete metric space has a unique fixed point. The strength of Banach's result in applications comes from two remarkable implications of this principle. The first one is that it guarantees the existence and uniqueness of a fixed point of a contraction. The second and most effective one is that it provides a technique to evaluate the fixed point. The importance of these two properties of Banach contraction mapping principle evidently has attracted many prominent mathematicians (see, e.g., [1216]) interested in the fixed point theory and its applications as the use of this principle has widened considerably since its first appearance [1721]. In particular, in this paper, we will focus on one of the most remarkable generalizations of Banach contraction mapping principle developed by Edelstein [22]. This theorem can be stated as follows.

Theorem 1. Let be a compact metric space and let be a mapping on . Assume for all with . Then has a unique fixed point.

Inspired by Edelstein's theorem, Suzuki [23, 24] further improved Banach's principle through the two theorems below.

Theorem 2. Define a nonincreasing function from onto by Then for a metric space , the following is equivalent. (1)is complete. (2)Every mapping on satisfying the following has a fixed point: there exists such that implies for all .

Theorem 3. Let be a compact metric space and let be a mapping on . Assume that implies for all . Then has a unique fixed point.

Theorems 2 and 3 are extensively studied by many authors (see, e.g., [2529]). It is worth to point out that the studies mentioned above can be classified as the extensions of Banach's principle on compact/complete metric spaces, which are totally ordered.

On the other hand, Turinici [30], Ran and Reurings [31], and Nieto and Rodríguez-López [32] proved an analog of the desired Banach's principle in the context of partially ordered complete metric spaces for a certain class of maps. In particular, Ran and Reurings [31] applied their result to solve a matrix equation and Nieto and Rodríguez-López [32] applied their theorem to guarantee the existence and uniqueness of the solution of some boundary value problems.

Motivated by all these developments, we shall prove new fixed point theorems extending Edelstein-Suzuki type contraction results in the setting of partially ordered complete metric spaces.

#### 2. Main Results

We denote by the set of functions satisfying the following condition:

We denote by the set of nondecreasing functions such that

We have the following lemma.

Lemma 4 (see [33, 34]). If , then for all .

Definition 5 (see [35]). Let and . We say that is -admissible if

Theorem 6. Let be an ordered metric space such that is complete. Let be a self-mapping. Assume that there exists such that for all comparable , where . If the following conditions hold: (i) is non-decreasing, continuous, and -admissible, (ii)there exists such that and , then has a fixed point in .

Proof. Let . If , then the theorem follows. Hence we suppose that . Define a sequence by for all . Since is non-decreasing and then and hence is an increasing sequence. If for some , then the result is proved as is a fixed point of . In what follows, we will suppose that . Hence, is a strictly increasing sequence. Since is -admissible and there exists such that , we find . By continuing this process, we get for all . Now we obtain that and and are comparable for all . Then by (5), we have
Then By induction, we have
By taking the limit as in the inequality above, we deduce For a fixed , there exists such that As is nondecreasing, we get Then By continuing this process, we get for all and . Consequently . Hence is a Cauchy sequence. Since is complete, there is such that . Now, since is continuous, we have So is a fixed point of .

Example 7. Let . Define the metric on by . Define by and by
Let . Let if and only if . Hence the conditions of Theorem 6 hold; that is, has a fixed point.

Proof. Let . First we assume that . Then . On the other hand, for all , we have . Hence . That is, is -admissible. Clearly, and . Let , and . Then we derive Next we assume that . As a result we find Then the conditions of Theorem 6 hold; therefore, has a fixed point.

Theorem 8. Let be an ordered metric space such that is complete. Let be a self-mapping. Assume that there exists such that for all comparable , where . If the following conditions hold: (i) is non-decreasing and -admissible, (ii)there exists such that and , (iii)for a strictly increasing sequence , converging to one has for all , (iv)if is a sequence in such that for all and as , then for all , then has a fixed point.

Proof. Proceeding as in the proof of Theorem 6, we know that there is a point such that and by condition (iii). We prove that . From (iv), we have for all . Now, we suppose that the following inequalities hold for some .
Since, for all , we get for all . Now by Lemma 4 and (9), we have . As a result, we obtain This is a contradiction. Thus, for all , either or holds. By (21), we have or
Now since , we have for all , that is,
Consequently, there exists an infinite subset such that one of these inequalities holds for every . If we take the limit as and , then we get , that is, , which is a contradiction. Hence it is absurd to suppose . So is a fixed point of .

Example 9. Let . Define the metric on by . Define by and by
Let , and . Let if and only if . Hence the conditions of Theorem 8 hold. That is, has a fixed point.

Proof. Let . If , then . On the other hand, for all , we derive . Hence . That is, is -admissible.
Clearly, there exists such that .
If is a sequence in such that for all and as . Then and . That is, for all . We need to consider the following cases.(i)If and then (ii)If and , then But we see that
If it is the case that , then . Hence,
Then the conditions of Theorem 8 hold. Consequently, has a fixed point.

The following notions will be used to show the uniqueness of a fixed point. (A)For all that are not comparable, there exists comparable with and such that and . (B)For all that are comparable, there exists such that and .

Theorem 10. Adding conditions (A) and (B) to the hypotheses of Theorem 8 one obtains uniqueness of the fixed point of .

Proof. Suppose that and are two fixed points of such that . If and are not comparable, by condition (A), there exists comparable with and , such that and . Since is -admissible, we have and . Since we have and and are comparable, by (21), we derive By induction, we get If , then . That is, and are comparable, which is a contradiction. Hence we suppose that .
Taking the as in the first of the inequalities above, we have Similarly, we obtain that From the inequality we get by taking the limit as , that is, .
Similarly, if and are comparable, then from (21) and condition (B), it follows that .

We denote by the set of functions satisfying the following condition:

Theorem 11. Let be an ordered metric space such that is complete. Let be a self-mapping. Assume that there exist and such that for all comparable , where and . If the following conditions hold: (i) is non-decreasing, (ii)there exists such that , (iii)if a non-decreasing sequence is such that as , then then has a fixed point. Moreover, has a unique fixed point if: (iv)for all that are not comparable there exists comparable with and .

Proof. Let . If , then the conclusion follows. Hence we suppose that . Define a sequence by for all . Since is non-decreasing and , we have Hence is a non-decreasing sequence. Since and and are comparable for all , by (42), we have
Then we find for all . This implies that the sequence is decreasing and so there is a real number such that
Now, we show that must be equal to . Let . Then by taking the limit as in (46), we get which is a contradiction. Hence we conclude that
Next, we prove that is a Cauchy sequence. Suppose, to the contrary, that is not a Cauchy sequence. Then there exists an such that for all positive integer , for some sequences and . For all , we have Taking the limit as in the inequality above and using (50), we get Again, from the expressions (50), and (53), we deduce by taking the limit as . From the facts and , we deduce that there is such that for all
Since and and are comparable, by (5), we get Taking the limit as in the inequality above, we have which is a contradiction. Hence is a Cauchy sequence. Since is complete, there is a point such that . Furthermore, by condition (iii), .
Now, we suppose that the following inequalities hold: for some . By (47), we have a contradiction. Thus, for all , either or holds. By (42), it follows that either or
Consequently, there exists an infinite subset such that one of these inequalities holds for every . If we take the limit as and , then we get that , that is, which contradicts with the assumption . Hence is a fixed point of .
Now, assume that condition (iv) holds. To prove the uniqueness of , suppose that is another fixed point of . It is easy to see from (5) that , if and are comparable. If and are not comparable, by condition (iv), there exists an comparable with and . Now we suppose that First we note that for each comparable with , by (5) with and , we have Taking the as in the inequalities above, we have which is a contradiction. Hence . Similarly, we obtain that From the inequality we get , that is, , by taking the limit as .

Example 12. Let . We define a partial order by
Define the metric on by . Clearly, is a complete metric space. Also define by
Let , , , and , where . Hence the conditions of Theorem 11 hold; that is, has a unique fixed point. But in [36] cannot be applied to in this example.

Proof. Let , with . Then for all elements of we have We examine each of the following cases: (i)for , we have; (ii)for , we have (iii)for , we have That is, in [36, Theorem  ] cannot be applied to . But for , Hence the conditions of Theorem 11 hold; that is, has a unique fixed point in .

#### 3. Application to Existence of Solutions of Integral Equations

Let be the set of real continuous functions defined on and be defined by for all . Let be the partial order on defined by if and only if for all . Then is a complete partially ordered metric space.

Consider the following integral equation: where(A) is continuous, (B) is continuous, (C) is continuous and (D)there exists a nonempty and closed subset such that for all with , we have where and for all with , we have (E)for all , we have (F)for a strictly increasing sequence converging to , we have for all . We have the following result of existence of solutions for integral equations.

Theorem 13. Under assumptions (A)–(E), the integral equation (78) has a solution in .

Proof. Let be defined by
First, we will prove that is a nondecreasing mapping with respect to . Let . By , we have for all . On the other hand by definition of , we have
Then , that is, is a non-decreasing mapping with respect to . Now suppose that with . Then by , , and the definition of , we get
Define by
Clearly, holds for all with . If then and . Since, is increasing, we obtain . Similarly, since , we get . Therefore, is an -admissible mapping. Also, assume that is a sequence such that as and . Then . Since is a closed set, we have . That is, . Then conditions of Theorem 8 hold and has a solution in .

#### Acknowledgment

The first author is supported by Islamic Azad University, Astara Branch.