Abstract

The aim of this paper is to introduce a notion of -contraction defined on a metric space with -distance. Moreover, fixed-point theorems are given in this framework. As an application, we prove the existence and uniqueness of a solution for the nonlinear Fredholm integral equations. Some illustrative examples are provided to advocate the usability of our results.

1. Introduction

By a contraction on a metric space , we understand a mapping satisfying for all , where is a real in .

In 1922, Banach proved the following theorem.

Theorem 1 (see [1]). Let be a complete metric space. Let be a contraction. Then,(i) has a unique fixed point (ii)For every , the sequence , where , converges to (iii)We have the following estimate: for every , ,

As a result of its intelligibility and profitableness, the previous theorem has become a very celebrated and popular tool in solving the existence problems in many branches of mathematical analysis.

Many mathematicians extended the Banach contraction principle in two major directions, one by stating the conditions on the mapping and second by taking the set as more general structure.

Recently, Kari et al. [2] give some fixed-point results for generalized -contraction in the framework of -compete rectangular -metric spaces.

In 2012, Wardowski [3] introduced the concept of -contraction, using this concept, he proved the existence and uniqueness of a fixed point in complete metric spaces. This direction has been studied and generalized in different spaces, and various fixed-point theorems are developed [4, 5]. Cosentino and Vetro [6] presented some fixed-point results of Hardy-Rogers type for self-mappings on complete metric spaces or complete ordered metric spaces. In 2016, Piri and Kumam [7] introduced the modified generalized -contractions, by combining the ideas of Dung and Hang [8], Piri and Kumam [9], Wardowski [3], and Wardowski and Van Dung [10], and gave some fixed-point result for these type mappings on complete metric space.

In 1996, Kada et al. initiated the notion of -distance on a metric space; then, many authors used this concept to prove some results of fixed-point theory [11, 12].

Recently, Wongyat and Sintunavarat [13] introduced a special -distance called ceiling distance and proved some fixed point for generalized contraction mappings with respect to this distance.

Later, Wardowski [14] studied a new type of contractions called nonlinear -contraction.

In this paper, we shall obtain a fixed-point theorem for -contraction with respect to -distance on complete metric spaces. Various examples are constructed to illustrate our results. As an application, we prove the existence and uniqueness of a solution for the nonlinear Fredholm integral equations.

The paper is structured as follows:

In Section 2, we briefly recall some definitions and basic properties used to prove our main results.

In Section 3, we present our results.

Section 4 is devoted to the application of the result in nonlinear integral equations.

2. Preliminaries

Kada et al. [15] introduced the concept of -distance on a metric space as follows:

Definition 2 (see [15]). Let be a metric space. A function is called a -distance on , if it satisfies the following three conditions for all
(W1)
(W2) is lower semicontinuous on for all
(W3) For each , there exists such that and imply

Remark 3. Each metric on a nonempty set is a -distance on .

Example 1 (see [13]). Let be a metric space. The function defined by for every is a -distance on , where is a positive real number. But is not a metric since for any

The following lemma is a useful tool for proving our results.

Lemma 4 (see [15]). Let be a metric space, be a -distance on , and be two sequences in , and .(i)If then . In particular, if , then (ii)If and for all , where and are sequences in converging to , then converges to (iii)If for each , there exists such that implies , then is a Cauchy sequence

Definition 5 (see [13]). A -distance on a metric space is said to be a ceiling distance of if and only iffor all .

Example 2 (see [13]). Let with the metric defined by for all , and let . Define the function byfor all . Then, is a ceiling distance of .

The following definition was introduced by Wardowski.

Definition 6 (see [3]). Let be the family of all functions such that(i) is strictly increasing(ii)For each sequence of positive numbers,(iii)There exists such that

Recently, Piri and Kuman [9] extended the result of Wardowski [3] by changing the condition (iii) in Definition 6 as follows.

Definition 7 (see [9]). Let be the family of all functions such that(i) is strictly increasing(ii)For each sequence of positive numbers,(iii) is continuous

The following definition introduced by Wardowski [14] will be used to prove our result.

Definition 8 (see [14]). Let be the family of all functions and be the family of all functions satisfy the following conditions:(i) is strictly increasing(ii)For each sequence of positive numbers,(iii) for all (iv)There exists such that

By replacing the condition (iii) in Definition 8, we introduce a new class of -contraction.

Definition 9. Let be the family of all functions and be the family of all functions satisfy the following conditions:(i) is strictly increasing(ii) for all (iii)For each sequence of positive numbers,(iv) is continuous

Example 3. (i)Let , . Then, (ii)Let , , and . Then, and

Definition 10 (see [14]). Let be a metric space. A mapping is called a -contraction on , if there exist and such thatfor all for which

3. Main Result

In this paper, using the idea introduced by Wongyat and Sintunavarat [13], we presented the concept of -contraction on a complete metric space with -distance.

Definition 11. Let be a -distance on a metric space . A mapping is said to be a -generalized -contraction of type on if there exist and such thatfor all for which

Theorem 12. Let be a complete metric space and be a -distance on and a ceiling distance of , supposing that is a -generalized -contraction of type . Then, has a unique fixed point on .

Proof. Let be an arbitrary point in ; define a sequence byfor all If there exists such that , then the proof is finished.
We can suppose that for all
Since is a ceiling distance of , we obtain for all
Substituting and , from (9), for all , we haveImply thatSince is increasing, then . Therefore, is monotone strictly decreasing sequence of nonnegative real numbers. Consequently, there exists such thatThe inequality (11) impliesSince , we have ; then from the definition of the limit, there exists and such that for all , , hencefor all . Taking the limit as in the above inequality, we getthat is, ; then, from the condition (ii) of Definition 8, we conclude thatNext, we shall prove that is a Cauchy sequence, i.e., , for all .
The condition (iv) of Definition 8 implies that there exists such thatFrom the inequality (12), we getHence,Taking limit in the above inequality, we conclude thatThen, there exists , such that for all ,Therefore, for , we haveSince , thenBy Lemma 4, we can conclude that is a Cauchy sequence in . By the completeness of , there exists such thatNow, we show that ; arguing by contradiction, we assume thatFrom (24), for each , there is such thatfor all . Since is lower semicontinuous and as , we getimplying thatNow, by triangular inequality, we get,By letting in inequality (30) and (31), we obtainTherefore,Let , from the definition of the limit, there exists such thatwhich implies thatApplying (9) with and , we havewhich implies thatSince is increasing, we getBy letting in the above inequality, we obtainWhich is a contradiction, then , so .
For uniqueness, now, suppose that are two fixed points of such that . Therefore, we haveApplying (9) with and , we haveimplyingwhich is a contradiction. Therefore, .

Example 4. Let with the metric defined byfor all . Define a mapping bySuppose that and , clearly and . Also, we define a -distance byfor all . It is easy to see that is a ceiling distance of . Now, we will show that satisfies the condition (9).

Case 1. If , then , , and . Thus,We prove that is a contraction mapping of type . Indeed,Therefore,

Case 2. If , then , , and . Thus,Therefore,Hence, 0 is the unique fixed point of .

Example 5. Consider the sequence defined as follows:Let the metric defined byfor all .
Define a mapping byClearly, the Banach contraction is not satisfied. In fact, we can check easily thatSuppose that and , clearly and . Also, we define a -distance byfor all . It is easy to see that is a ceiling distance of .
Now, we will show that satisfies the condition (9).

Case 1. and . In this case, we haveThus,On the other hand,Therefore,

Case 2. . In this case, we haveThus,On the other hand,Therefore,Thus, the inequality (9) that is satisfied implies that has a unique fixed point. In this example, is the unique fixed point of .

Definition 13. Let be a -distance on a metric space . A mapping is said to be a -generalized -contraction of type on if there exist and such thatfor all for which

Theorem 14. Let be a complete metric space and be a -distance on and a ceiling distance of . Suppose that is a -contraction of type . Then, has a unique fixed point on .

Proof. As in the proof of Theorem 12, we can conclude thatNext, we show that is a Cauchy sequence, i.e.,Now, we claim that Arguing by contradiction, we assume that there exists we can find and sequences and of positive integers such that for all positive integers, ,Again by triangular inequality and using (65), (67), and (68), we getSo,Again by the triangular inequality, for all , we have the following two inequalitiesLetting in the above inequalities, using (65) and (70), we obtainHence, from the definition of the limit, there exists such thatApplying (64) with and , we obtainLetting the above inequality, we obtainSince is a continuous and , we conclude thatwhich is a contradiction. Then,By the condition (iii) of Lemma 4, we can conclude that is a Cauchy sequence. Since is a complete metric space, there exists such that as .
As in the proof of Theorem 12, we conclude that for each , there exists such that , , and for all .
Now applying (64) with and , we getwhich implies thatTherefore, . Hence, , so .
Following the proof of Theorem 12, we know that is a unique fixed point of . This complete the proof.

Example 6. Let with the metric defined byfor all . Define a mapping bySuppose that and , clearly and . Also, we define a -distance byfor all . It is easy to see that is a ceiling distance of . Now, we will show that satisfies the condition (64).
We prove that is a -contraction mapping of type .

Case 1. If , then , , and . Thus,On the other hand,Therefore,

Case 2. If , then , , and . Thus,On the other hand,Therefore,Hence, 0 is the unique fixed point of .

Example 7. Let be the set defined bywhereLet the metric defined byfor all . Define a mapping byClearly, the Banach contraction is not satisfied. In fact, we can check easily thatSuppose that and , clearly and . Also, we define a -distance byfor all . It is easy to see that is a ceiling distance of . Now, we will show that satisfies the condition (64).

Case 1. and . In this case, we haveThus,On the other hand,Therefore,

Case 2. . In this case, we haveThus,On the other hand,Therefore,Thus, the inequality (64) that is satisfied implies that has a unique fixed point. In this example, is the unique fixed point of .

Taking in Theorems 12 and 14, we obtain the following result.

Corollary 15. Let be a complete metric space and be a -contraction of type . Then, has a unique fixed point.

Corollary 16. Let be a complete metric space and be a -contraction of type . Then, has a unique fixed point.

4. Application to Nonlinear Integral Equations

In this section, we endeavor to apply Theorems 12 and 14 to prove the existence and uniqueness of the integral equation of Fredholm type:where , , and is a given continuous function.

Theorem 17. Consider the Fredholm integral equation (103) and assume that the kernel function satisfies the condition for all and . Then, the equation (4.1) has a unique solution for some constant depending on the constants .

Proof. Let and defined byfor all . Clearly, with the metric given byfor all , is a complete metric space. Next, define the function byfor all .
Clearly, is a -distance on and a ceiling distance of . We will find the condition on under which the operator has a unique fixed point which will be the solution of the integral equation (103). Assume that and . Then, we getwhich implies thatSince by the definition of the -distance on and a ceiling distance of , we have and for any , then we can take natural logarithm sides and getprovided that , which implies thatHence,for all . It follows that satisfies the condition (9). Therefore, there exists a unique solution of the nonlinear Fredholm inequality (103).

Example 8. Let . Consider the equationHere, and we haveThen, the condition (9) holds. From Theorem 17, the nonlinear integral equation (103) has a unique solution for .
By using direct computation, letThen, we have and hence,which implies thatTherefore,we obtain that this equation has a unique solution when

Data Availability

No data were used to support this study.

Conflicts of Interest

On behalf of all authors, the corresponding author states that there is no conflict of interest.