Fixed Points for Contractive Mappings of Integral Type Involving <span class="nowrap"><svg xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg" style="vertical-align:-0.2724395pt" id="M1" height="8.14084pt" version="1.1" viewBox="-0.0657574 -7.8684 11.1206 8.14084" width="11.1206pt"><g transform="matrix(.017,0,0,-0.017,0,0)"><path id="g113-246" d="M615 290C615 392 563 448 529 448C511 448 491 441 484 434L483 425C518 400 546 349 546 273C546 158 504 46 417 46C381 46 356 77 356 142C356 184 380 293 392 334L387 340L307 323C305 288 296 228 287 188C265 94 221 52 180 52C144 52 112 89 112 171C112 278 164 372 241 430L226 454C106 394 23 288 23 155C23 40 79 -12 139 -12C184 -12 245 32 285 85C292 31 322 -12 376 -12C423 -12 470 18 501 45C570 104 615 184 615 290Z"/></g></svg>-</span>Distance and Applications

Gao, Haiyan; Liu, Na; Zhao, Liangshi

doi:https://doi.org/10.1155/2021/6648527

Journal of Function Spaces

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Abstract Introduction Preliminaries Applications Conclusion Data Availability Conflicts of Interest Authors’ Contributions Acknowledgments References Copyright Related Articles

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Fixed Point Theory and Applications for Function Spaces

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Research Article | Open Access

Volume 2021 | Article ID 6648527 | https://doi.org/10.1155/2021/6648527

Fixed Points for Contractive Mappings of Integral Type Involving -Distance and Applications

Haiyan Gao,^1,2Na Liu,³and Liangshi Zhao⁴

Academic Editor: Zoran Mitrovic

Received14 Nov 2020

Revised14 Dec 2020

Accepted11 Jan 2021

Published23 Jan 2021

Abstract

In this paper, we use -distance to prove the existence, uniqueness, and iterative approximations of fixed points for a few contractive mappings of integral type in complete metric spaces. The proved results are used to investigate the solvability of certain nonlinear integral equations. Four examples are given.

1. Introduction

The researchers [1–17] attained various generalizations of the well-known Banach contraction principle. In 2001, Rhoades [15] certified several fixed point results for the weakly contractive mappings. Branciari [1] introduced the notion of integral type contraction and established a nice fixed point result for the mapping. Many authors investigated the existence of fixed points for a lot of contractive mappings of integral type, for example, see [11–14]. Particularly, Liu et al. [14] obtained several fixed point theorems for contractive mappings of integral type in complete metric spaces.

Kada et al. [8] introduced the concept of -distance in metric spaces and proved a few fixed point theorems for some contractive mappings by using -distance. It is clear that the results in [8] extended the Caristi’s fixed point theorem, Ekeland’s -variational’s principle, and the nonconvex minimization theorem. The researchers in [3, 5–7, 9, 10, 17] got several fixed point results for certain contractive mappings with respect to -distance.

In this paper, we prove the existence, uniqueness, and iterative approximations of fixed points for several kinds of mappings, which satisfy some contractive conditions of integral type with respect to -distance in complete metric spaces. We also construct four illustrative examples and give applications of the obtained results in nonlinear Fredholm and Volterra integral equations, respectively. Our results generalize or differ from the corresponding fixed point theorems in [1, 14, 15].

2. Preliminaries

Let denotes the set of all positive integers, , , and

Recall that a self-mapping in a metric space is called orbitally continuous if implies for each and .

3. Fixed Point Results with respect to -Distance

In this section, using -distance, we give four fixed point theorems for the contractive mappings (2), (38), (82), and (114) below.

Theorem 1. Let be a -distance in a complete metric space and let satisfy thatHere, . Then, possesses a unique fixed point such that ,

Proof. Firstly, we claim the existence of fixed points of in . Put and for each . Now, we need to think over two situations as follows:

Case 2. for some . Clearly, is a fixed point of and . Suppose that . Making use of (2) and , we obtain thatwhich is ridiculous. Hence, , which means that

Case 3. for all . Suppose that(2), (6), and ensure thatthat is,

The above equation and give that

Combining (6), (9), and , we know thatthat is,

Because of Lemma 1 in [8], (6), and (11), we deduce that , which is contradictive, and, hence,

By means of , (2) and (12), we havewhich together with (12) and ensures that

We see from (14) that is a positive and strictly decreasing sequence. It follows that

Now, we claim that . Otherwise, . In view of Lemma 2.1 in [12], (2), (15), and , we see that

It is ridiculous. Therefore, . Consequently,

In the same way, we have

Now, we proceed to show that

Suppose that there exists a real number such that for every , there exist satisfying

For each , denotes the least integer exceeding and satisfying (20). Obviously,

On account of and (21), we obtain that

Letting tend to infinity in (22) and taking advantage of (17), (18), and (21), we have

In light of Lemma 2.1 in [12], (2), (23), and , we deduct that

It is ridiculous. Of course, (19) is true.

Assume that and denotes the real number appearing in (3) of [8]. By means of (19), we infer that there is satisfyingwhich ensures that

So is a Cauchy sequence. Completeness of means that

According to (19), we are aware of the fact that for any there is withwhich together with (27) gives that

It follows that

Taking account of (2), (30), , and Lemma 2.1 in [12], we infer that

It follows that

Lemma 2.2 in [12] and the above equation give that

We get from (1) in [8] and (17) that

Clearly,

Applying (30), (35), and Lemma 1 in [8], we gain that .

Secondly, we prove that . Suppose that . In view of and (2), we receive thatwhich is not possible. Hence, .

Thirdly, we assert the uniqueness of fixed points of in . Assume that possesses two fixed points . We know, analogous to the proof of (36), that . Assume that . Due to and (2), we deduce thatwhich is ridiculous. Consequently, . Using Lemma 1 in [8] and , we obtain that .

Theorem 4. Let be a -distance in a complete metric space and let satisfy thatHere, . Then, possesses a unique fixed point such that ,

Proof. Firstly, we show that possesses fixed points in . Let and for each . Now, we divide the proof into two steps.

Step 5. Put for some . In addition, is a fixed point of and . Suppose that . Owing to (38) and , we acquire thatwhich is ridiculous. Hence, , which means that

Step 6. for all . Assume that

Using , (38), and (42), we conclude that

It becomes that

Thus, and the above equation guarantee that

As a result of (42), (45), and (1) in [8], we deduce thatin other words,

From (45), (47), and Lemma 1 in [8], we obtain that , which is impossible, and, hence,

Suppose that there exists with

In light of (38), (48), (49), and , we infer that

It is ridiculous. By means of (48), we get that

Thus, (51) means that the sequence is both positive and decreasing. Consequently,

Assume that

In terms of (38), (53), and , we attain thatwhich is ridiculous. Hence,

We get from (55) that the sequence is both nonnegative and nonincreasing. Thus,

Assume that . In view of (38), (52), (56), , and Lemma 2.1 in [12], we gain that

It is impossible. Thus, (18) is true. Suppose that (6) holds. Taking advantage of (6), (38), and , we have

It follows that

Thus, the above equation and give (9). Using (6), (9), and (1) in [8], we conclude thatthat is, (11) holds. By virtue of (6), (11), and Lemma 1 in [8], we obtain that , which is impossible. As a result, (12) holds. Assume that there exists with

We know from (12), (38), (61), and thatwhich is ridiculous. By means of (12), we obtain that

With the help of (63), there is a real number satisfying (15). Assume that . Let . Clearly, there exists a subsequence of with

Note that (38) and infer that

Letting in (65) and utilizing (18), and Lemma 2.2 in [12], we make a conclusion that

It follows that

In view of (18), (38), (64), (65), and (67), and Lemma 2.2 in [12], we deduct thatwhich together with yields that . It is obvious that

Using Lemma 2.2 in [12], (15), (38), (69), and , we find thatwhich together with implies that . Thus, (17) holds.

Now, we assert that (19) holds. Otherwise, there is a real number such that for arbitrary , there are with (20) and (21). On account of (1) in [8] and (3.12), we gain that

Letting in (71) and making use of (17), (18), and (21), we require that

Taking notice of (38), (72), , and Lemma 2.1 in [12], we receive that

It is ridiculous. That is, (19) is true.

We deduce, similar to the proof of Theorem 1, that (27) holds. It follows from (19) that for every real number there is withwhich together with (2) in [8] and (27) gives thatthat is, (30) holds. In terms of (30), (38), , and Lemma 2.1 in [12], we get that

It follows that

Thus, Lemma 2.2 in [12] and the above equation ensure that

In light of (1) in [8] and (17), we arrive atthat is to say, (35) holds. By virtue of (30), (35) and Lemma 1 in [8], we have .

Secondly, we assert that . Assume that . Owing to (38) and , we deduce that

It is ridiculous. Hence, .

Thirdly, we show the uniqueness of fixed points of in . Assume that possesses two fixed points . We get, similar to the proof of (80), that . Assume that . Taking account of and (38), we get thatwhich is ridiculous. Therefore, . It follows from and Lemma 1 in [8] that .

Theorem 7. Let be a -distance in a complete metric space and let satisfy thathere . Then, possesses a unique fixed point satisfying ,

Proof. Firstly, we demonstrate that possesses fixed points in . Let and for each . Now, we consider two cases below:

Case 8. for some . Since is a fixed point of , it follows that . Assume that . Due to (82) and , we havewhich is absurd. Hence, and

Case 9. for all . Assume that (6) holds. In view of (6), (82), and , we obtain thatwhich means that

Combining and the above equation, we get (3.3). We gain from (6), (9), and thatin other words, (11) sets up. In terms of (6), (11), and Lemma 1 in [8], we know immediately that , which is absurd, and, hence, (12) is true. Assume that there is satisfying (61). We conclude from (12), (61), (82), and thatwhich is impossible. Hence, (63) is true. It follows from (63) that there is a real number with (15). Assume that there is with

In terms of (82), (90), and , we infer that

It is absurd, and, hence,

(92) means that is both nonnegative and nonincreasing. Consequently,

Assume that . Owing to (15), (82), (93), , and Lemma 2.1 in [12], we get that

It is contradictive. Thus, (17) is true. Suppose that (42) holds. We infer from (42), (82), and thatthat is,

Thus, (45) follows from the above equation and . We deduct from (42), (45), and (1) in [8] thatthat is to say, (47) holds. Thus, is easily obtained from (45), (47), and Lemma 1 in [8], which is ridiculous. As a result, (48) holds. Suppose that there exists satisfying (49). By virtue of (48), (49), (82), and , we know thatwhich is ridiculous. By means of (48), we have (51). It follows from (51) that the sequence is both positive and decreasing, which yields (52) for some a constant . Suppose that . Put . Obviously, there is a subsequence of with (64). Using (82), and , we deduce that

Letting in (99) and using (17), and Lemma 2.2 in [12], we find that

It follows that

We attain from Lemma 2.2 in [12], (17), (64), (82), (99), (101), and that

It follows that . By virtue of , we have and . It means that (69) holds. In view of (52), (69), (82), , and Lemma 2.2 in [12], we infer thatwhich yields that . Using , we obtain that and . Thus, (3.9) holds.

Now, we prove that (19) holds. Suppose that there is an such that for arbitrary , (20), (21), and (22) hold for some . As in (3.13) and by virtue of (17), (18), and (21), we acquire that

In terms of Lemma 2.1 in [12], (82), (104), and , we are aware of the fact thatwhich is absurd. Thus, (19) is true.

We infer, similar to the proof of Theorem 1, that (27) holds. It follows from (19) that for each there is withwhich together with (2) in [8] and (27) gets thatthat is, (30) holds. On account of Lemma 2.1 in [12], (30), (82), and , we deduct thatin other words,

Lemma 2.2 in [12] and the above equation give that

In light of (1) in [8] and (17), we attain thatthat is to say, (35) holds. Using (30), (35), and Lemma 1 in [8], we have .

Secondly, we prove that . Assume that . Because of (82) and , we deduce thatwhich is impossible. Hence, .

Thirdly, we assert the uniqueness of fixed points of in . Assume that possesses two fixed points . We deduce, similar to the proof of (112), that . Assume that . On account of (82) and , we get thatwhich is ridiculous. Therefore, . Using and Lemma 1 in [8], we infer immediately that .

We have, similar to the proof of Theorem 1, the result below and omit its proof.

Theorem 10. Let be a -distance in a complete metric space and let satisfy thatHere, . Then, possesses a unique fixed point such that ,

4. Four Examples

Now, we give four examples to explain the fixed point results obtained in Section 3.

Remark 11. Letting , we deduce that Theorem 1 reduces to Theorem 2.1 in [14], which generalizes Theorem 1 in [15]. On the other hand, the example below proves that Theorem 1 extends indeed these results in [14, 15] and differs from Theorem 2.1 in [1].

Example 12. Let and . Let , and be defined by, respectively,and

It follows that is a -distance in and . Put . In order to check (2), we consider two cases below:

Case 13. . It follows that

Case 14. . Note that

That is, (2) is true. Hence, the conditions of Theorem 1 are fulfilled. Thus, Theorem 1 ensures that possesses a unique fixed point in . Now, we need to prove that Theorem 2.1 in [1], Theorem 2.1 in [14], and Theorem 1 in [15] are useless in checking the existence of fixed points for the mapping in .

If there is satisfying the conditions of Theorem 1 in [15], we know thatwhich is absurd.

If there are and satisfying the conditions of Theorem 2.1 in [1], we attain thatwhich is ridiculous.

If there is satisfying the conditions of Theorem 2.1 in [14], we conclude thatwhich is impossible.

Remark 15. Examples 16, 21, and 26 explain that Theorems 4, 7, and 10 are different from Theorem 2.1 in [14].

Example 16. Let and . Let , and be defined by, respectively,and

Evidently, is a -distance in and . Let . For the sake of verifying (38), we take into account the following four possible cases:

Case 17. . Note that

Case 18. . Obviously,

Case 19. . Notice that

Case 20. . It follows that

That is to say, (38) is true. Therefore, the conditions of Theorem 4 are fulfilled. Consequently, Theorem 4 means that possesses a unique fixed point in . However, we cannot use Theorem 2.1 in [14] to show the existence of fixed points for the mapping in . Or else, there is withwhich is absurd.

Example 21. Let and . Let , and be defined by, respectively,and

Obviously, is a -distance in and . Let . To demonstrate (82), we consider four cases below:

Case 22. . Evidently,

Case 23. . Clearly,

Case 24. . It is obvious that

Case 25. . Clearly

In other words, (82) is true, and consequently, the conditions of Theorem 7 are fulfilled. Thus, Theorem 7 yields that possesses a unique fixed point in . Next, we testify that Theorem 2.1 in [14] is unapplicable in ensuring the existence of fixed points for the mapping in .

If there is satisfying the conditions of Theorem 2.1 in [14], we havewhich is ridiculous.

Example 26. Let and . Let , and be defined by, respectively,and

It is easy to see that is a -distance in and . Let . To prove (114), we have to consider two cases below:

Case 27. . Apparently,

Case 28. . It is easy to demonstrate that

Hence, (114) is true, and the conditions of Theorem 10 are satisfied. Thus, Theorem 10 guarantees that possesses a unique fixed point in . Then, we certify that Theorem 2.1 in [14] is unfulfilled in showing the existence of fixed points for the mapping in . Otherwise, there is satisfying the conditions Theorem 2.1 in [14]. It means thatwhich is ridiculous.

5. Applications

In this section, we utilize Theorems 1 and 7 to investigate the solvability of the nonlinear Fredholm and Volterra integral equations below, respectively,where and are constants in with , and are given functions.

We assume that denotes the Banach space of all continuous functions with the norm . Let and

Obviously, is a complete metric space. Define two mappings and as follows:

Theorem 29. Let and satisfy that
(a1) and are continuous;
(a2) there is with

Then, Eq. (142) possesses a unique solution in .

Proof. Define two functions and byObviously, is a -distance and . It follows from and (145) that for arbitrary , is continuous in , which means that maps into itself. Taking account of (145) and , we get thatIt follows thatThat is, (2) and (82) hold. It follows from each of Theorems 1 and 7 that possesses a unique fixed point , that is, Eq. (142) has a unique solution .
We get, similar to the proof of Theorem 29, the following result and omit its proof.

Theorem 30. Let and satisfy and . Then, Eq. (143) possesses a unique solution in .

6. Conclusion

By using -distance, we prove several fixed point results for a few contractive mappings of integral type, some of which are used to investigate the existence and uniqueness of solutions for certain nonlinear Fredholm and Volterra integral equations, respectively. Four examples are provided to testify that our results extend or differ from some known results in the literature.

Data Availability

The data used to support the findings of this study are included within the article.

Conflicts of Interest

The authors declare that they have no competing interests.

Authors’ Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Acknowledgments

The authors thank the referees for their useful comments and suggestions. This work was supported by the National Natural Science Foundation of China (No. 41701616).

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Copyright © 2021 Haiyan Gao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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