Fixed Point Results on Closed Ball in Convex Rectangular <svg xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg" style="vertical-align:-0.4618998pt" id="M1" height="12.7225pt" version="1.1" viewBox="-0.0657574 -12.2606 18.3204 12.7225" width="18.3204pt"><g transform="matrix(.017,0,0,-0.017,0,0)"><path id="g113-99" d="M452 333C452 398 425 448 375 448C329 448 235 404 140 292H138L229 683C233 701 234 712 225 712C208 712 149 686 66 677L64 652H97C135 652 140 651 129 598L38 167C23 96 29 57 44 33C60 7 98 -12 132 -12C169 -12 232 3 290 41C383 102 452 223 452 333ZM365 316C365 199 308 87 239 49C221 39 200 35 183 35C137 35 99 70 112 163C116 191 123 215 129 233C190 316 290 392 330 392C348 392 365 372 365 316Z"/></g><g transform="matrix(.017,0,0,-0.017,8.053,0)"><path id="g117-33" d="M535 230V280H52V230H535Z"/></g></svg>Metric Spaces and Applications

Li, Chaobo; Cui, Yunan; Chen, Lili

doi:https://doi.org/10.1155/2022/8840964

Journal of Function Spaces

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Fixed Point Theory and Applications for Function Spaces 2021

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Research Article | Open Access

Volume 2022 | Article ID 8840964 | https://doi.org/10.1155/2022/8840964

Fixed Point Results on Closed Ball in Convex Rectangular Metric Spaces and Applications

Chaobo Li,¹Yunan Cui,¹and Lili Chen²

Academic Editor: Mohamed A. Taoudi

Received10 Nov 2021

Revised04 Jan 2022

Accepted07 Apr 2022

Published28 Apr 2022

Abstract

In this paper, the concept of convex rectangular metric spaces is introduced as a generalization of both convex metric spaces and rectangular metric spaces. The purpose of this study is to indicate a way of generalizing Mann’s iteration algorithm and a series of fixed point results in rectangular metric spaces. Furthermore, certain examples are given to support the results. We also study well posedness of fixed point problems of some mappings in convex rectangular metric spaces, and an application to the dynamic programming is entrusted to manifest the viability of the obtained results. Our results extend comparable results in the existing literature.

1. Introduction and Preliminaries

It is well known that fixed point theory has become an important field of mathematics due to its high degree of unity and wide application. No doubt that the most significant fundamental result of this theory is Banach contraction principle [1] which was published in 1922. Banach contraction principle proposes for the first time to use Picard iteration to approximate a fixed point, which not only proves the existence of the fixed point but also proves the uniqueness of the fixed point. Later in 1968, Kannan [2] studied a new type of contractive mapping. Since then, there have been many results related to mappings satisfy various types of contractive inequality, see for example [3–9].

In 2000, Branciari [10] developed the notion of a rectangular space as a generalization of normal metric space via substituting the triangle inequality with the quadrilateral inequality and extended Banach contraction principle to this space. Successively, George et al. [11] introduced the notion of a rectangular metric space as a generalization of rectangular metric space and they also proved some fixed point results for contractive mappings. The concept of a convex structure and a convex metric space was introduced by Takahashi [12]. Later, Goebel and Kirk [13] studied some iterative processes for nonexpansive mappings in a hyperbolic metric space, and in 1988, Ding [14] found fixed points of quasicontraction mappings in convex metric spaces by Ishikawa’s iteration scheme. However, iterative methods have received vast investigation for finding fixed points of nonexpansive mappings, see [15–17]. Particularly, in the process of the research on some fixed point problem, one of the most famous fixed point method is the Mann iteration [18, 19] as follows: for some suitably chosen scalars . Due to [20], Mann iterative sequence converges weakly to a fixed point of if the sequence satisfies following conditions: .

Very recently, Chen et al. [21] introduced the notion of a convex metric space and extend Mann’s algorithms directly to metric spaces. After that, Asif et al. [22] investigate fixed point of single-valued Hardy-Roger’s type contraction globally as well as locally in a convex metric space. Along the line, Chen et al. [23] introduce the concept of a convex graphical rectangular metric space and prove some fixed point theorems in this space. New some fixed point results on a closed ball can be seen in [22, 24–26].

In this work, we firstly introduce the concept of the convex rectangular metric spaces which is a combination of properties of rectangular metric spaces and convex metric spaces. However, we prove some fixed point theorems using generalized Mann’s iteration algorithm and show concrete examples supporting our main results. In addition, we claim that fixed point problem is well posed and as an application, we apply our main results to solve the dynamic programming problem.

Some fundamental definitions related to our work are given below:

Definition 1. (See [11]). Let be a nonempty set and the mapping satisfy if and only if for all for all (3)There exists a real number such that for all and all distinct points Then, is called a rectangular metric on , and is called a rectangular metric space with coefficient .

Remark 2. Note that every metric space is a rectangular metric space (see [11]), and every is a with coefficient .

Definition 3. (See [11]). Let be a , be a sequence in and . Then, (a)The sequence is said to be convergent in to , if for every , there exists such that for all , and this fact is represented by (b)The sequence is said to be a Cauchy sequence in if for every , there exists such that for all , and this fact is represented by is said to be a complete if every Cauchy sequence in converges to some

Definition 4. (See [12]). Let be a metric space and . A continuous function is said to be a convex structure on if for each and , for all . A metric space with a convex structure is called a convex metric space.

2. Main Results

In this section, we introduce a generalization of both convex metric spaces and rectangular metric spaces, which we call convex rectangular metric spaces. We also establish some fixed point theorems arising from this metric space.

Definition 5. Let be a with constant . If a mapping satisfies for all and , then is said to be a convex rectangular metric space .

Definition 6. Let be a and be a mapping. Let be the sequence generated by Mann’s iterative procedure involving the mapping , as follows: where and are the initial value.

Definition 7. If in Definition 5, we call the resultant space to be a convex rectangular metric space , which is, indeed, the with a convex structure .

Next, we see some specific examples of .

Example 8. Let . For any , we define the metric by and . Notice that, for any and , then the convex of the function implies that Then, for any distinct points , we have Hence, is a with . For any , let be a mapping defined by Now, we verify that satisfies inequality (3). In fact, for any , we can see that Therefore, is a with . Note that is not a metric space as follows: for we take . Moreover, is a when we let , and it shows that reduces to a for .

Example 9. Let . For any , we define the metric by . From Example 8, it follows that is a with . For any , let be a mapping defined by For any , we obtain that Therefore, is a with , but not a .

Example 10. Let , , such that and where is a constant. Then, is a with coefficient . The mapping is defined by , , and then

So, is a with coefficient , but not a .

Definition 11. Let be a with constant , is some element in , and , and then the set is called a closed ball in .

In the paper [3], George et al. proved Banach contraction principle in complete by means of Picard iteration. Now, we will show Banach contraction principle for complete using generalized Mann’s iteration algorithm.

Theorem 12. Let be a complete with constant and be a mapping satisfying for all , where . Let the sequence generated by the Mann iterative process and such that . If and ( is an arbitrary positive real number and ), then has a unique fixed point in . Moreover, the sequence and as , if the following inequality holds:

Proof. Without loss of generality, we suppose that for all . Indeed, If , then . We conclude that and it shows ; then, is a fixed point of , and the proof is finished. It follows from Definition 5 and Definition 6,

Now, we consider the following two cases:

Case 13. If for all , we have Let , with the assumption and , and we obtain that Hence,

Case 14. If for some , we have Denote that , and it follows from (20) and (21) that which implies that is a decreasing sequence of nonnegative reals. Hence, there exists such that

We will show that . Suppose that , letting in inequality (22), we obtain a contradiction. Hence, we get that . Furthermore, we have which shows that . Also, we can assume for any . Indeed, if , then using the inequality (21), we have in which shows that and , and then is a fixed point, and the proof is finished. Next, we shall prove for all . In order to do it, we will consider the following two cases:

Case 15. If for all , then we have which establishes that

Case 16. If there exist some such that , then It follows from (28) and (29) that Next, we claim that is a Cauchy sequence by contradiction. Assume that there exists and the subsequences and of such for with , . On the one hand, taking the limit superior in above inequality as , and we conclude

On the other hand, let and , and we have by taking the limit superior on both sides of above the inequality as , and we get a contradiction. Thus, is a Cauchy sequence in . Since the space is complete, there exists such that . We shall show that is a fixed point of . Applying the rectangular inequality, we obtain that since , and then letting , and we deduce which implies . Thus, is a fixed point of . Suppose that are two distinct fixed points of , that is, and . Then, which is a contradiction. Therefore, we must have , i.e., . Thus, has a unique fixed point. Next, we proceed to show that the sequence . In order to complete it, we will use mathematical induction. Thanks to Definition 5 and Definition 6, we obtain which implies ; therefore, . Suppose , observe from above proof, we get for all . It is easy to see that . Now, we can assume that . If , then

We also need to distinguish the following four cases:

Case 17. If , then we have

Case 18. if , then we have

Case 19. if , then we have

Case 20. if , then we have which implies

Finally, by above cases, we prove that which show that . Hence, by induction , therefore, we conclude that for all . As every closed ball in a complete metric space is complete, so , as .

The following example illustrates the above theorem.

Example 21. Let and for all . For any , we define by . The mapping is defined by Set and . If , then and have a unique fixed point in .

Proof. It is easy to see that is a with . In addition, for any , we have So, is a with . It is not difficult to see that satisfies for . According to , we have , since , and we obtain that is, , then And we obtain while , getting and . Hence, 0 is a fixed point of in . Suppose are two distinct fixed points of , then we have which shows that , that is, . Thus, has a unique fixed point in . Let , then . For all , we obtain , and this means that the sequence . Furthermore, , that is, .

Now, we prove the Kannan type fixed point theorem for a complete , which extends the results in the paper [3], replacing Picard’s iteration algorithm by Mann’s iteration algorithm.

Theorem 22. Let be a with constant and the mapping be defined by for all , and . Let the sequence generated by the Mann iterative process and such that . If and ( is an arbitrary real number and ), then has a unique fixed point in . Moreover, the sequence and as , if the following inequality holds:

Proof. Without loss of generality, we suppose that for all . Indeed, If , that is, . Then, we have and it shows that and , which means that is a fixed point of , and the proof is finished. Thanks to Definition 5 and Definition 6, we have

Now, we have the following two cases:

Case 23. If for all , we have which establishes that Notice that , then we have Since , we conclude that

Case 24. If for some , then and this implies that Since , then we get Noticing that

Let , it is clear that , and for any , we obtain the following inequality: and it implies that is a decreasing sequence of nonnegative reals. Hence, there exists such that

We will show that . Suppose that . Letting in inequality (63), we obtain a contradiction. Hence, we get that . Moreover, we have which shows that . Next, we shall prove that . In order to do it, we will consider the following two cases:

Case 25. if for all , then we obtain Hence,

Case 26. If there exist some such that , then we get Hence, It follows from (68) and (70) that Next, we will claim that is a Cauchy sequence by contradiction. Assume there exists and the subsequences and of such for with , . On the one hand, taking the limit superior in above inequality as , and we get On the other hand, let and , and we have We obtain a contradiction. Thus, is a Cauchy sequence in . Since the space is complete, there exists such that . We shall show that is a fixed point of . Applying the rectangular inequality, we obtain that since , and then letting , and we deduce which implies . Thus, is a fixed point of . Suppose that are two distinct fixed points of , that is, . Then, which is a contradiction. Therefore, we must have , that is, . Thus, has a unique fixed point. Finally, we will prove the iteration sequence . In order to complete it, we will use mathematical induction. Choose , and we have which implies ; therefore, . Suppose . It is easy to see that . Without loss of generality, we can assume that . If , then

We also need to distinguish the following four cases:

Case 27. If , then we have

Case 28. If , then we have

Case 29. If , then we have

Case 30. If , then we have which implies

Finally, by above cases, we prove that , which show that . Hence, by induction . Therefore, we conclude that for all . As every closed ball in a complete metric space is complete, so , as .

Next, we give the following example to illustrate above theorem.

Example 31. Let and the mapping such that for any . Let us define the metric by the formula as well as the mapping by the formula . Choose to be the initial value and , where . If , then , and has a unique fixed point in .

Proof. It is easy to see that is a with . We claim that satisfies inequality for any . Next, we will consider the four cases: (a)If , then it is easy to see that inequality (87) holds(b)If and , thenwhich implies that holds for any and . (c)If and , then, similarly to case (b), we can also get that inequality (87) holds(d)If , thenwhich shows that holds for all .
Summarizing, inequality, (87) holds for all . Next, we will show that has a unique fixed point in . In order to do it, we will consider the following two cases: (i)If , thenObviously, as , (ii)If , thenIf , then . From the case (i), it follows that as . If , then . From the above procedure, without loss of generality, we can assume that . Then, we obtain which implies that .
Hence, , where 0 is a fixed point of . Actually, 0 is a unique fixed point of in . Indeed, suppose that is a fixed point of , then , that is, , which implies , a contradiction. Thus, has a unique fixed point in . Let , then . For all , from above proof, we can obtain , then , and this means that the sequence . Furthermore, , that is, , and the proof is finished.

The concept of well posedness is very important in many fields of mathematics and has evoked much interest to several researchers [27–29].

Definition 32. (see [26]). Let be a metric space and be a self-map. The fixed point problem of is said to be well posed if has a unique fixed point (2)For any sequence in with , we have

We next study the well posedness of the fixed point problem of in complete .

Theorem 33. Let be a with constant and all the hypotheses of Theorem 12 hold. If the constant , then fixed point problem of is well posed.

Proof. Let is a unique fixed point of and assume be a sequence in such that . Because of uniqueness of the fixed point of , for all , we can assume that . If for some , , then since , and we get . Due to , it is not difficult to see that , indeed, if not, a contradiction. Therefore, let us assume that , and then combining with , and we obtain which implies , which completes the proof.

Theorem 34. Let be a with constant and all the hypotheses of Theorem 22 hold. If the constant , then fixed point problem of is well posed.

Proof. Let be a unique fixed point of and a sequence in sequence in such that . Without loss of generality, let , for all . By the help of uniqueness of the fixed point of , then we have . If for some , , then Hence, and we conclude that . Due to , it is not difficult to see that , indeed, if not, a contradiction. Therefore, let us assume that , and then combining with , and we obtain which implies , which completes the proof.

3. Applications

In this section, we will apply our result to solving the following functional equation arising in dynamic programming: for all , where , , and . We assume that and are Banach spaces, is a state space, and is a decision space. Precisely, see also [30, 31]. Let denote the set of all bounded real-valued functions on and the norm defined as for all . Clearly, is a Banach space. Moreover, we can define a rectangular metric by for all . Since is complete, we deduce that is a complete rectangular metric space with . In order to show the existence of a solution of equation (104), we consider the operator of the form for all and . We will prove the following theorem.

Theorem 35. Let be given by (106). Suppose that the following hypotheses hold:
(A1) and are bounded functions;
(A2) There exists , for all , and , such that

Then, the functional equation (104) has a bounded solution.

Proof. Obviously, is well defined, since and are bounded. That is, and operator are well defined. Then, from (A2), we have Let ; thus, all the conditions of Theorem 12 are fulfilled, and there exists a fixed point of such that . In other words, for all . This completes the proof.

Example 36. Consider the functional equation for . We let , . is defined by , is defined by , and is defined by for . It is not difficult to see that and are bounded functions. Moreover,

Thus, all the conditions of Theorem 35 are fulfilled. Hence, functional equation (110) has a solution .

Data Availability

No data were used to support this study.

Conflicts of Interest

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

Acknowledgments

The authors would like to thank the referees for valuable comments and suggestions for improving this work. This work was supported by the National Natural Science Foundation of China under Grant 11871181.

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Copyright © 2022 Chaobo Li 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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