On a Fixed Point Theorem with Application to Integral Equations

Abstract

We introduce the notion of dualistic Geraghty’s type contractions. We prove some fixed point theorems for ordered mappings satisfying the abovementioned contractions. We discuss an application of our fixed point results to show the existence of solution of integral equations.

1. Introduction and Preliminaries

In [1], Matthews introduced the concept of partial metric space as a suitable mathematical tool for program verification and proved an analogue of Banach fixed point theorem in complete partial metric spaces. O’Neill [2] introduced the concept of dualistic partial metric, which is more general than partial metric, and established a robust relationship between dualistic partial metric and quasi metric. In [3], Valero presented a Banach fixed point theorem on complete dualistic partial metric spaces. Valero also showed that the contractive condition in Banach fixed point theorem in complete dualistic partial metric spaces cannot be replaced by the contractive condition of Banach fixed point theorem for complete partial metric spaces. Later, Valero [3] generalized the main theorem of [4] using nonlinear contractive condition instead of Banach contractive condition.

For the sake of completeness, we recall Geraghty’s Theorem. For this purpose, we first recall the class of all functions which satisfies the condition

In [5], Geraghty presented a new class of mappings , called Geraghty contraction, which satisfies the following condition:for all , where .

For this new family of mappings, Geraghty in [5] proved a fixed point theorem, stated below.

Theorem 1 (see [5]). Let be a complete metric space and let be a mapping. Assume that there exists such that, for all , Then, has a unique fixed point and, for any choice of the initial point , the sequence defined by for each converges to the point .

Following [5], Amini-Harandi and Emami generalized Theorem 1 in context of ordered metric spaces (see [6]).

Theorem 2 (see [6]). Let be an ordered set and suppose that there exists a metric in such that is a complete metric space. Let be an increasing mapping such that there exists with . Suppose that there exists such thatAssume that either is continuous or is such that if an increasing sequence converges to , then for each .
Besides, if for all there exists which is comparable to and , then has a unique fixed point in .

In [7], La Rosa and Vetro have extended the notion of Geraghty contraction mappings to the context of partial metric spaces. Besides, they have yielded partial metric version of Theorem 1, stated below.

Theorem 3 (see [7]). Let be a complete partial metric space and let be a Geraghty contraction mapping. Then, has a unique fixed point and the Picard iterative sequence converges to with respect to , for any . Moreover, .

In this paper, we will present Theorems 1, 2, and 3 in dualistic partial metric spaces. We will show that our results generalize Theorems 1, 2, and 3 in many ways. In the section, we will apply our fixed point theorem to show the existence of solution of particular class of integral equations:

We need some mathematical basics of dualistic partial metric space and results to make this paper self-sufficient.

Throughout, in this paper, the letters , , and will represent the set of nonnegative real numbers, the set of real numbers, and the set of natural numbers, respectively.

According to O’Neill, a dualistic partial metric can be defined as follows.

Definition 4 (see [2]). Let be a nonempty set. If a function satisfies, for all , the following properties:  ⁡, ⁡, ⁡, ⁡, then is called dualistic partial metric and the pair is known as dualistic partial metric space.

If is a dualistic partial metric space, then defined byis called a dualistic quasi metric on such that   . Moreover, if is a dualistic quasi metric on , then is a metric on .

Remark 5. It is obvious that every partial metric is a dualistic partial metric but the converse is not true. To support this comment, define by It is easy to check that is a dualistic partial metric. Note that is not a partial metric, because . However, the restriction of to , , is a partial metric.

Example 6. If is a metric space and is an arbitrary constant, thendefines a dualistic partial metric on .

Following [2], each dualistic partial metric on generates a topology on . The topology consists of open balls of the form , where .

Definition 7 (see [2]). Let be a dualistic partial metric space:(1)A sequence in converges to a point if and only if .(2)A sequence in is called a Cauchy sequence if exists and is finite.(3)A dualistic partial metric space is said to be complete if every Cauchy sequence in converges, with respect to , to a point such that .

The following lemma will be helpful in the sequel.

Lemma 8 (see [2, 3]). (1)A dualistic partial metric is complete if and only if the metric space is complete.(2)A sequence in converges to a point , with respect to if and only if .(3)If such that , then for every .

Oltra and Valero, in [4], established a Banach fixed point theorem, stated as follows.

Theorem 9 (see [4]). Let be a complete dualistic partial metric space and let be a mapping such that there exists satisfying for all . Then, has a unique fixed point . Moreover, and the Picard iterative sequence converges to with respect to , for any .

2. The Results

In this section, we will prove the existence of fixed points of dualistic contractions in ordered dualistic partial metric spaces. To this end, we need to define the followings notions.

Definition 10. Let be a nonempty set. Then, is said to be an ordered dualistic partial metric space if(i) is a partially ordered set;(ii) is a dualistic partial metric space.

Definition 11. Let be a partially ordered set and suppose that is a complete dualistic partial metric space; a mapping is called dualistic Geraghty’s type contraction provided there exists such thatfor all comparable .

Now we present our main result.

Theorem 12. Let be a partially ordered set and suppose that is a complete dualistic partial metric space and let be a mapping such that (1) is a dominated mapping;(2) is a dualistic Geraghty’s type contraction;(3)either is continuous or if is a nonincreasing sequence in such that , then .Then, has a fixed point and the Picard iterative sequence converges to with respect to , for every . Moreover, .

Proof. Let be an initial element and for all ; if there exists a positive integer such that , then , so we are done. Suppose that ; then, since is dominated mapping, therefore ; that is and imply and implies ; continuing in a similar way, we get Since , from contractive condition (10) we have This implies that is a monotone and bounded above sequence; it is convergent and converges to a point ; that is,If , then we are done but if , then from (10) we haveThis implies thatTaking limit, we have Since , which entails .
Hence,Similarly, we can prove that Now since we deduce thatNow, we show that sequence is a Cauchy sequence . Suppose on the contrary that is not a Cauchy sequence. Then, given , we will construct a pair of subsequences and violating the following condition for least integer such that , where :In addition, upon choosing the smallest possible , we may assume thatBy the triangle inequality, we have That is, for all . In the view of (24) and (17), we have Again using triangle inequality, we have Taking limit as and using (17) and (25), we obtain Now from contractive condition (10), we have We conclude that By using (17), letting in the above inequality, we obtainSince , and hence which contradicts our assumption (21). Arguing like above, we can have . Hence, is a Cauchy sequence in ; that is, . Since is a complete metric space, converges to a point in ; that is, ; then, from Lemma 8, we get We are left to prove that is a fixed point of . For this purpose, we have to deal with two cases.
Case  1. If is continuous, then Hence, ; that is, is a fixed point of .
Case  2. If is a nonincreasing sequence in such that , then .
Using contractive condition (10) and (31), we get This shows that . So from and , we deduce that and hence is a fixed point of .

Note that in the above result fixed point may not be unique; in order to prove uniqueness of the fixed point, we need some more conditions and for this purpose we have following theorem.

Theorem 13. Let be an ordered complete dualistic partial metric space. Let be a mapping satisfying all the conditions of Theorem 12. Besides, if for each there exists which is comparable to and , then has a unique fixed point.

Proof. Following the proof of Theorem 12, we know that is a fixed point of . We are left to prove the uniqueness of the fixed point . Let be another fixed point of ; then, . Two cases arise; first if and are comparable, then from (10) it follows that . Secondly, if and are not comparable, then there exists which is comparable to both and ; that is, and . Since is dominated mapping, we deduce that and . Moreover, consider and by contractive condition (10), we obtainThis implies It shows that is a nonnegative and decreasing sequence, so for we get We claim that . Suppose on the contrary that .
By passing to subsequences, if necessary, we may assume that Then, by (34) we have . Since , we have Hence, Similarly, we can prove that Finally, by we have Taking limit, we get , since , which implies . Hence, . From and , we deduce that which proves the uniqueness of .

For monotone mappings, we present following result.

Theorem 14. Let be a partially ordered set and suppose that is a complete dualistic partial metric space and let be a mapping such that (1) is an increasing map with for some ;(2) is a dualistic Geraghty’s type contraction;(3)either is continuous or is such that if an increasing sequence then . Besides, if for each there exists which is comparable to and , then has a unique fixed point and the Picard iterative sequence converges to with respect to , for any . Moreover, .

Proof. We begin by defining a Picard iterative sequence in by for all . Given , then . Since is increasing, implies ; that is, ; this in turn gives which implies . Continuing in a similar way we getSince for each , from contractive condition (10) we have This implies that is a monotone and bounded below sequence; it is convergent and converges to a point ; that is, If , then we are done but if , then from (10) we have This implies that Taking limit, we have Since , which implies .
Hence,Similarly, we can prove thatAnd the desired conclusion follows arguing like in the proofs of Theorems 12 and 13.

Clearly, Theorem 3 is a particular case of Theorems 12 and 14.

A natural question that can be raised is whether the contractive condition in the statements of Theorems 12 and 14 can be replaced by the contractive condition in Theorem 3. The following easy example provides a negative answer to such a question.

Example 15. Consider the complete dualistic partial metric and the mapping defined by It is easy to check that the contractive condition in the statement of Theorem 3 holds: for all . However, does not have a fixed point. Observe that does not satisfy the contractive condition in the statements of Theorems 12 and 14. Indeed,

The analogues of Theorems 12 and 14 are given below without proofs as they can be obtained easily by following proofs of above theorems.

Theorem 16. Let be a partially ordered set and suppose that is a complete dualistic partial metric space and let be a mapping such that (1) is a dominating map;(2) is a dualistic Geraghty’s type contraction;(3)either is continuous or M is such that if an increasing sequence then . Besides, if for each there exists which is comparable to and , then, has a unique fixed point and the Picard iterative sequence converges to with respect to , for any . Moreover, .

Theorem 17. Let be a partially ordered set and suppose that is a complete dualistic partial metric space and let be a mapping such that (1) is a decreasing map with ;(2) is a dualistic Geraghty’s type contraction;(3)either is continuous or is such that if a decreasing sequence then . Besides, if for each there exists which is comparable to and , then has a unique fixed point and the Picard iterative sequence converges to with respect to , for any . Moreover, .

Observations. If we set in Theorem 14, we retrieve Theorem 2 as a particular case. If we set in Theorem 14, we retrieve Theorem 3 as a particular case.

3. Application to Integral Equations

In this section, we will show how Theorem 14 can be applied to prove the existence of solution of integral equation (53).

Let represent the class of functions with the following properties:(1) is increasing;(2)for each , ;(3);(4). For example, and are elements of .

Let us consider the following integral equation:To show the existence of solution of integral equation (53), we need following lemma.

Lemma 18. Let . Assume the following hypotheses are satisfied: (1);(2);(3), where and . Then, operator defined by satisfies .

Proof. We begin by defining the operator : Since , choose such thatThis implies that ; hence, .

Now we are in position to state our result regarding application.

Theorem 19. Assume that the following hypotheses are satisfied: (1)The conditions supposed in Lemma 18 hold.(2), comparable and for large . Then, integral equation (53) has a solution.

Proof. Let and   , where and is a sequence of real numbers satisfying for large . Suppose that is a mapping defined by We introduce a partial ordering on , settingThen, is a complete ordered dualistic partial metric space. Notice that is well-defined and (53) has a solution if and only if the operator has a fixed point. Precisely, we have to show that Theorem 14 is applicable to the operator . Then, for all comparable , we write It follows thatThis impliesand hence satisfies all the conditions of Theorem 14, so it has a fixed point and hence (53) has a solution.

Remark 20. The significance of the above results lies in the fact that these results are true for all real numbers whereas such results proved in partial metric spaces are only true for positive real numbers.

Competing Interests

The authors declare that they have no competing interests.