Fixed Point Results for α-ψ-Contractive Mappings Including Almost Contractions and Applications

Abstract

In the recent paper (B. Samet, C. Vetro, and P. Vetro, Fixed point theorems for --contractive type mappings, Nonlinear Analysis. Theory, Methods and Applications, 75 (2012), 2154-2165.), the authors introduced the concept of -admissible maps on metric spaces. Using this new concept, they presented some nice fixed point results. Also, they gave an existence theorem for integral equation to show the usability of their result. Then, many authors focused on this new concept and obtained a lot of fixed point results, which are used for existence theorems. In this paper, we not only extend some of the recent results about this direction but also generalize them. Then, we give some examples to show our results are proper extensions. Furthermore, we use our results to obtain the existence and uniqueness result for a solution of fourth order two-point boundary value problem.

1. Introduction and Preliminaries

Fixed point theory contains many different fields of mathematics, such as nonlinear functional analysis, mathematical analysis, operator theory, and general topology. Historically, the study of fixed point theory has developed in two major branches: the first is fixed point theory for contraction or contraction type mappings on complete metric spaces and the second is fixed point theory for continuous operators on compact and convex subsets of a normed space. Recently, there has been a lot of activities in the first branch and several fundamental fixed point results have been extended and generalized by many authors in different directions. In this paper, we mention some important of them and give some new fixed point results. Also, we support our results by giving a lot of nontrivial examples. First, we give some notations, which will be used in this paper. Let be a function. For convenience, we consider the following properties of this function: is nondecreasing, for all , for ,, is continuous, is upper semicontinuous from the right, for any .

In the light of the above properties, the following hold: if and satisfied, then holds. If and are satisfied, then holds. If and are satisfied, then and hold.

Denoted by the family of functions satisfying and , which is called comparison functions in the literature (see [1]), by the family of functions satisfying and , which is called -comparison functions in the literature (see [1]), and by the family of functions satisfying and . Now, we give some examples showing the relations between the sets , and . First, it is clear that .

Example 1. Let be defined by , where , and then .

Example 2. Let be defined by , and then , but .

Example 3. Let be defined by and then , but .

Example 4. Let be defined by and then , but .

Example 5. Let be defined by and then , but .

In their recent paper, Samet et al. [2] introduced the notions of -admissible and --contractive mappings and then gave some fixed point results for such mappings. Their results are closely related to some ordered fixed point results. Then, using their idea, some authors presented fixed point results for single and multivalued mappings (see, e.g., [2–6]).

Definition 6 (see [2]). Let be a metric space, let be a self-map on , , and let be a function. Then is called --contractive whenever for all .

Note that every Banach contraction mapping is an --contractive mapping with and for some .

Definition 7 (see [2]). is called -admissible whenever implies .

There exist some examples for -admissible mappings in [2]. For convenience, we mention here one of them. Let . Define and by for all and for and for . Then is -admissible.

Theorem 8 (see [2]). Let be a complete metric space and let be an -admissible and --contractive mapping. If there exists such that and is continuous, then has a fixed point.

The following theorems are a generalization of Theorem 8.

Theorem 9 (see [3]). Let be a complete metric space and let be an -admissible mapping satisfying for all , where and If there exists such that and is continuous, then has a fixed point.

Theorem 10 (see [4]). Let be a complete metric space and let be an -admissible mapping satisfying. Assume that where and If there exists such that and is continuous or is regular, then has a fixed point.

For the sake of brevity, we will say that is regular whenever, for any sequence in with for all and as , we have for all .

The aim of this paper is to extend and generalize the above results. Note that, in these theorems, the function belongs to the class ; that is, is -comparison function. Also, in Theorem 10, the contractive condition is used with . In this paper, we give three existence results. In the first result, the contractive condition (7) will be generalized to almost contraction case. Here we take , but we use instead of . In the second result, we take in , which is a wider class of . And in the third result, we take in , which is a different class of . Also, we present some uniqueness theorems and some supporting examples.

2. Existence Results

Our first result is almost contraction version of fixed points of -admissible mapping. We can find detailed information about almost contractions in [7–12].

Theorem 11. Let be a complete metric space and let be an -admissible mapping. Assume that holds for all , where , , and as in Theorem 9. Also, suppose is continuous or is regular and there exists such that . Then has a fixed point.

Proof. Let such that . Define a sequence in by for all . If for some , then is a fixed point for and result is proved. So, we suppose that for all . Since is -admissible mapping and , we deduce that . Continuing this process, we get for all . Now by (9) with , we get where If for some , then from (11), we have which is a contradiction. Thus for all and so from (11), we have By induction, we have
for all . Now, for each , , we have Therefore, is a Cauchy sequence in . Since is complete, there exists such that . If is continuous, then we have So, is a fixed point of . Now, suppose is regular. Since for all and as , then we have for all . From (9) we have where Now, suppose . Taking into account (15) and , there exists such that and for all . Therefore we have for all . Now, from (20), we obtain for all . Letting in the last equality, we get that which is a contradiction. Therefore and so has a fixed point.

Remark 12. In Theorem 11, if we take , then we obtain an extension of Theorem 10. Even if , we can extend Theorem 10 by taking the function from and taking instead of as follows.

Theorem 13. Let be a complete metric space and let be an -admissible mapping. Assume that holds for all , where and as in Theorem 11. Also, suppose is continuous or is regular and there exists such that . Then has a fixed point.

Proof. Let such that . As in the proof of Theorem 11, we can construct a sequence in and we can obtain (15) if the consecutive terms are different (otherwise, has a fixed point). Now we show that is a Cauchy sequence. Let . Taking into account (15), there exists , such that . Therefore we have and so from (25) Again using (25) we have where If , then, from (28), we have and so Therefore, we have which is a contradiction. Thus, and so, from (28), we have By continuing this way, we can obtain for all . Now let with , and then that is, is a Cauchy sequence in . The rest of the proof can be made as in the proof of Theorem 11.

In the following theorem we take the function from instead of .

Theorem 14. Let be a complete metric space and let be an -admissible mapping. Assume that holds for all , where , and as in Theorem 11. Also, suppose is continuous or is regular and there exists such that . Then has a fixed point.

Proof. Let such that . As in the proof of Theorem 11, we can construct a sequence in and we may assume that for all . For the sake of brevity we put . Since is an -admissible mapping, we can obtain as in the proof of Theorem 11 that Therefore, it should be for all integer and so, from (37), we have Consequently, the sequence of positive numbers is decreasing and bounded below. So, there exists such that . We claim that . Suppose to the contrary that . Using the fact that is upper semicontinuous from the right function, we get from (38) which is a contradiction. Hence, we conclude that ; that is, Now, we prove that the sequence is Cauchy in . Suppose, to the contrary, there exists such that where and are subsequences of with for all . Moreover, is chosen as the smallest integer satisfying (41). Thus, we have By the triangle inequality, we get Letting in above inequality and using (40), we get that Now let be such that and for all . Then for all . Using (40) and (44) and letting in (45), we get Since for all and is upper semicontinuous from the right function, we deduce that On the other hand, for each , we have so which is a contradiction. Thus is a Cauchy sequence in . Since is complete, there exists such that . If is continuous, then we have So is a fixed point of . Now, suppose is regular and . We first note that hence by the upper semicontinuity of , we get On the other hand, since is regular, for all integer and as , then we have for all integer . Thus, from (36), we have and taking limit supremum, we get which is a contradiction. Therefore and so has a fixed point.

3. Uniqueness Results

In this section, we consider some properties to obtain the uniqueness of the fixed point in the above theorems. For this, we denote the set of fixed points of by Fix.

Theorem 15. Assume that all hypotheses of Theorem 13 hold. Also suppose holds, then the fixed point of is unique.

Proof. Suppose and are two fixed points of , and then there exists such that and . Since is an -admissible mapping, then for all . Therefore Without loss of generality, we can assume for all . Therefore from (59) we have (note it should be . Otherwise we obtain a contradiction from (59)) Letting in the above inequality, we have . Similarly, we can obtain and so .

Remark 16. The condition (57) is not sufficient to obtain the uniqueness of the fixed point in Theorem 11.

Example 17. Let with the usual metric. Define and by and . Then, it is clear that is -admissible mapping and is regular. Also, there exists such that . Now, we show that (9) is satisfied with and . For this, we consider the following cases.

Case 1. If , then

Case 2. If , then .

Case 3. If , then .

Case 4. If and , then

Case 5. If and , then

Case 6. If and , then Therefore, all conditions of Theorem 11 are satisfied and so has a fixed point. Although the condition (57) of Theorem 15 is satisfied, the fixed point of is not unique.

In the following, we give a uniqueness theorem by adding some conditions in Theorem 11.

Theorem 18. Assume that all hypotheses of Theorem 11 hold. Also suppose for all , there exists such that and and holds for all , where and , then the fixed point of is unique.

Proof. Suppose and are two fixed points of , and then there exists such that and . Since is an -admissible mapping, then for all . Therefore from (66) Without loss of generality, we can assume for all . Therefore we have Letting in the above inequality, we have . Similarly, we can obtain and so .

4. Some Corollaries and Example

Corollary 19. Let be a complete metric space and be -admissible. Assume that holds for all , where and . Also, suppose is continuous and there exists such that . Then has a fixed point.

Corollary 20. Let be a complete metric space and let be self-map of . Assume that holds for all , where , , and as in Theorem 9. Then has a fixed point.

Corollary 21. Let be a complete metric space and let be self-map of . Assume that holds for all , where and . Then has a fixed point.

Corollary 22. Let be a complete metric space and let be -admissible. Assume that holds for all , where (or ). Also, suppose is continuous and there exists such that . Then has a fixed point.

Corollary 23. Let be a complete metric space and let be a self-map of . Assume that holds for all , where (or ), and as in Theorem 9. Then has a unique fixed point.

Corollary 24. Let be a complete metric space and let be a self-map of . Assume that holds for all , where (or ). Then has a unique fixed point.

Example 25. Let with the usual metric. Define and by for all . Then, it is clear that is -admissible mapping and is continuous. Also, . But, although , we cannot find any satisfying Therefore, Theorems 10 and 9 cannot be applied to this example. Now, we show that (9) is satisfied with and . Let , and then . We have to consider the following cases.

Case 1. If , then

Case 2. If and , then Therefore, all conditions of Theorem 11 are satisfied and so has a fixed point in .

5. Applications

In this section, we apply Corollary 22 to the following fourth order two-point boundary value problem: which describes the bending of an elastic beam clamped at both endpoints. In [13], using an ordered version of Geraghty’s fixed point result, an existence theorem for a nonnegative solution of (80) is given. The boundary value problem can be written as the integral equation (see [14]) where is the Green’s function given by Then it is clear that, is continuous on , , and for and .

Here we will consider with the uniform metric, that is, for . Then is a Banach space. For the convenience we consider the operator defined on by

Theorem 26. Under the following assumptions, Problem (80) has a solution.(A) is bounded,(B)there exist and such that if for , then for we have (C)if , then ,(D)there exists such that ,(E)if is a sequence in such that for all and , then for all .

Proof. Let with the metric . First, we show that . Let and , and then, by the continuity of and the boundedness of , we have where such that .
Now, define by Then, it is clear that is -admissible. Also, for all with , we have for all . Therefore hold. Finally, by the condition (E), is regular. Therefore all hypotheses of Corollary 22 are satisfied and so has a fixed point in . Thus, the problem (80) has a solution in .

In the following, we also give an existence and uniqueness theorem for (80) under slight different conditions.

Theorem 27. Under the following assumptions, problem (80) has a unique solution.(F) is bounded, nondecreasing with respect to second variable and for all ,(G)there exists such that, for with and , we have where for all .

Proof. As in the proof of Theorem 26, we can show that . Now define by Therefore, if , then and for all . Since is nondecreasing with respect to second variable, we have for all . Thus, we have and so . It shows that is -admissible. Also, for all with , we have for all . Therefore hold. Also, since , we have and so . Now, let for all and in . Then, we have and so for all . Therefore, for all and . That is, for all and so . Hence is regular. Therefore all existence hypotheses of Corollary 22 are satisfied and so has a fixed point in . Thus, the problem (80) has a solution in . Finally, let and be two solutions of (80), and then and and . Thus, and . Therefore, by Corollary 22 the solution is unique.

Example 28. Consider the nonlinear fourth order two point boundary value problem where . In this case, . It is easy to see that is bounded and nondecreasing with respect to the second variable. Also, for all . Now let with then, for all , where . Therefore, the conditions (F) and (G) of Theorem 27 are satisfied and so the boundary value problem (94) has a unique solution.

Conflict of Interests

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