Kannan Prequasi Contraction Maps on Nakano Sequence Spaces

Bakery, Awad A.; Mohamed, O. M. Kalthum S. K.

doi:https://doi.org/10.1155/2020/8871563

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Fixed Point Theory and Applications for Function Spaces

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Research Article | Open Access

Volume 2020 | Article ID 8871563 | https://doi.org/10.1155/2020/8871563

Kannan Prequasi Contraction Maps on Nakano Sequence Spaces

Awad A. Bakery^1,2and O. M. Kalthum S. K. Mohamed^1,3

Academic Editor: Nawab Hussain

Received29 Sept 2020

Revised28 Nov 2020

Accepted19 Dec 2020

Published31 Dec 2020

Abstract

In this article, we explore the concept of the prequasi norm on Nakano special space of sequences (sss) such that its variable exponent in . We evaluate the sufficient setting on it with the definite prequasi norm to configuration prequasi Banach and closed (sss). The Fatou property of different prequasi norms on this (sss) has been investigated. Moreover, the existence of a fixed point of Kannan prequasi norm contraction maps on the prequasi Banach (sss) and the prequasi Banach operator ideal constructed by this (sss) and numbers have been examined.

1. Introduction

Ideal maps and summability theorems [1–6] are extremely significant in mathematical models and have more achievements, such as ideal transformations, normal series, fixed point theory, geometry of Banach spaces, and approximation theory. By , we mark the spaces of all sequences of real numbers. We denote the space of all bounded linear maps from a Banach space into a Banach space by , and if , we indicate , the -th number by [7], the -th approximation number by , and , where 1 shows at the place, for every .

Notations 1. The sets , , and , (cf. [8]) denote

Let , the Nakano sequence space defined and studied in [9–11] is denoted by: where And is a Banach space, however, Faried and Bakery [8] assumed the hypothesis of prequasi operator ideal that is more established than the quasi operator ideal. Bakery and Abou Elmatty [9] demonstrated the strictly inclusion of the prequasi operator ideal , for inconsistent powers. It was a small prequasi operator ideal. As the literature of the Banach fixed point theorem [12], many mathematicians created on many actions. Haghi et al. [13, 14] showed that some generalizations in fixed point theory are not real generalizations and investigated some fixed point generalizations to partial metric spaces, which are obtained from the corresponding results in metric spaces. Kannan [15] presented a representation of a class of operators with the same fixed point actions as contractions nevertheless that fails to be continuous. They only try to illustrate Kannan maps [16] in modular vector spaces. The target of this paper is to appraise the concept of prequasi norm on . The Fatou property of different prequasi norms on this (sss) has been examined. We are delving the sufficient set-up on equipped with the definite prequasi norm to pattern prequasi Banach and closed (sss). The existence of a fixed point of Kannan prequasi norm contraction mapping on the prequasi Banach (sss) has been given. Finally, the existence of a fixed point of Kannan prequasi norm contraction mapping on the prequasi Banach operator ideal has been made current.

2. Definitions and Preliminaries

Definition 2 (see [2]). The linear space of sequences is detailed as a special space of sequences (sss), if (1) is solid, i.e., let , , and , for every , then (2), where marks the integral part of , if

Definition 3 (see [8]). A subclass of is definite a premodular (sss), if there is verifying the set-up: (i)For , with , where is the zero vector of (ii)For every and , we have for which ,(iii), for each , for some (iv)For and , then (v)The inequality, holds, for some (vi)Assume be the space of finite sequences, then (vii)There is such that for every

Definition 4 (see [17]). Suppose be a (sss). The function is called prequasi norm on , if it provides the conditions (i), (ii), and (iii) of Definition 3.

Theorem 5 (see [17]). Pick up be a premodular (sss), then it is prequasi normed (sss).

Theorem 6 (see [17]). is a prequasi normed (sss), if it is quasinormed (sss).

Definition 7 (see [3]). Let be the class of all bounded linear operators between any two arbitrary Banach spaces. A subclass of is named an operator ideal, if every vector verifies the next setting: (i) where denotes Banach space of one dimension(ii)The space is linear over (iii)Assume , , and , then, , where and are normed spaces (see [18, 19])

The theory of prequasi operator ideal, which is more general than the quasi operator ideal.

Definition 8 (see [8]). A function is named a prequasi norm on the ideal if the following setting includes (1)Assume , , and (2)There is so as to , for every and (3)There is such that , for each ,(4)There is for to if , and , then

Theorem 9 (see [20]). Pick up be a premodular (sss), then be a prequasi norm on .

Theorem 10 (see [9]). Suppose and be Banach spaces, and be a premodular (sss), then be a prequasi Banach operator ideal, such that .

Theorem 11 (see [8]). is a prequasi norm on the ideal , if is a quasinorm on the ideal .

The agreeable inequality [21] will be used in the consequence: Suppose and , for every , then

3. Main Results

3.1. Prequasi Normed (sss)

We illustrate the adequate set-up on equipped with a prequasi norm to generate prequasi Banach and closed (sss).

Definition 12. (a) is -convergent to If the -limit exists, then it is unique
(b) is -Cauchy, if
(c) is -closed, if for all -converging to , then

Theorem 13. , where , for all , is a premodular (sss), if is an increasing.

Proof. First, we have to prove is a (sss):
(1) Suppose . Since , we have so .
(2) Assume and . As , one has Hence, . So, by using Parts (1) and (2), we get is linear. Also for all since
(3) Let , for every and . One can see we have . This implies the sequence space is solid.
(4) Suppose and be an increasing sequence, one has then . Secondly, we show that the functional on is a premodular: (i)Evidently, and (ii)We have such that , for every and . For , there is such that , for every (iii)We have so that , for every (iv)Clearly, since is solid(v)From (49), we have (vi)Clearly, (vii)There is , for or , for such that

Theorem 14. Assume be an increasing, then be a prequasi Banach (sss), where , for every .

Proof. Let the set-up be verified. From Theorem 13, the space is a premodular (sss). By Theorem 5, the space is a prequasi normed (sss). To prove that is a prequasi Banach (sss), assume be a Cauchy sequence in . Hence, for every , we have such that for all , one has

Therefore, for and , we get So is a Cauchy sequence in , for constant . Which implies , for fixed . Hence, , for every . Then, to show that , we have So . This explains that is a prequasi Banach (sss).

Theorem 15. Pick up be an increasing, then be a prequasi closed (sss), where , for every .

Proof. Assume the conditions be verified. From Theorem 13, the space be a premodular (sss). By Theorem 5, the space is a prequasi normed (sss). To show that is a prequasi closed (sss), suppose and , then for all , we have so that for all , we have Hence, for and , one has Therefore, is a convergent sequence in , for fixed . Hence, , for constant . Finally, to prove that , we obtain hence, . This gives that is a prequasi closed (sss).

Example 16. The functional is a prequasi norm (not a quasinorm) on Nakano special space of sequences .

Example 17. The functional is a prequasi norm (not a quasinorm) on Nakano special space of sequences .

Example 18. The functional is a prequasi norm (not a norm) on -absolutely summable sequences of real numbers , for all .

Example 19. For , the functional is a prequasi norm (a quasinorm and a norm) on Nakano special space of sequences .

4. The Fatou Property

We investigate here the Fatou property of different prequasi norms on .

Definition 20. A prequasi norm on provides the Fatou property, if for all sequence with and any then

Theorem 21. The function provides the Fatou property, if is an increasing, for all .

Proof. Let the set-up be satisfied and with Since the space is a prequasi closed space, then . So for every , one has

Theorem 22. The function does not fulfill the Fatou property, for all , if with .

Proof. Suppose the set-up be confirmed and with Since the space is a prequasi closed space, then . Then, for each , we get

So, does not indulge the Fatou property.

5. Kannan Prequasi -Contraction Operator

Now, we explain the definition of Kannan -contraction mapping on the prequasi normed (sss). We study the sufficient setting on constructed with definite prequasi norm so that there is one and only one fixed point of Kannan prequasi norm contraction mapping.

Definition 23. An operator is called a Kannan -contraction, if there is , so that for all .
An element is named a fixed point of , if

Theorem 24. Assume be an increasing, and be Kannan -contraction mapping, where , for all , then has one fixed point.

Proof. Let the setting be satisfied. For each , then . As is a Kannan -contraction operator, one has

So, for all with , one can see

Therefore, is a Cauchy sequence in . As the space is prequasi Banach space. Hence, there is so that . To prove that . Since has the Fatou property, we have hence, . Then, is a fixed point of . To show that the fixed point is unique. Let we have two distinctive fixed points of . So, we have

Therefore,

Corollary 25. Let be an increasing, and be Kannan -contraction mapping, where , for all , then has unique fixed point with

Proof. Pick up the conditions be satisfied. By Theorem 24, we have a unique fixed point of . Hence, one has

Definition 26. Suppose be a prequasi normed (sss), and The operator is called -sequentially continuous at , if and only if, when then .

Theorem 27. Pick up with , and , where , for all . The point is the only fixed point of , if the following conditions are satisfied: (a) is Kannan -contraction mapping(b) is -sequentially continuous at (c)There is so that the sequence of iterates has a subsequence converging to

Proof. Let the set-up be verified. Suppose be not a fixed point of , then . By the set-up (b) and (c), we have

As the operator is Kannan -contraction, one has

As , we have a contradiction. Therefore, is a fixed point of . To prove that the fixed point is unique. Assume we have two different fixed points of . So, one can see

Therefore,

Example 28. Let , where , for all and

Since for all with , we have

For all with , we have

For all with and , we have

Therefore, the map is Kannan -contraction mapping. Since satisfies the Fatou property. By Theorem 24, the map has a unique fixed point

Let be such that where with . Since the prequasi norm is continuous, we have

Hence, is not -sequentially continuous at . So, the map is not continuous at .

If , for all . Since for all with , we have

For all with , we have

For all with and , we have

Therefore, the map is Kannan -contraction mapping and

It is clear that is -sequentially continuous at and has a subsequence converging to . By Theorem 27, the point is the only fixed point of .

Example 29. Let, where , for all and

Since for all with , we have

For all with , then for any , we have

For all with and , we have

Therefore, the map is Kannan -contraction mapping. It is clear that is -sequentially continuous at and there is with such that the sequence of iterates has a subsequence converging to . Then, has one fixed point . Note that is not continuous at .

6. Kannan Contraction Maps on Prequasi Ideal

We account the being present of a fixed point of Kannan prequasi norm contraction operator on the prequasi Banach operator ideal investigated by and numbers.

Theorem 30. Let and be Banach spaces, and be an increasing, then , where be a prequasi Banach operator ideal.

Proof. Pick up the conditions be verified. By Theorem 13, the space is a premodular (sss). Therefore, from Theorem 9, one has is a prequasi norm on . So, from Theorem 10, we obtain the space is a prequasi Banach operator ideal.

Theorem 31. Pick up and be Banach spaces, and be an increasing, then , where be a prequasi closed operator ideal.

Proof. By Theorem 13, the space is a premodular (sss). Therefore, from Theorem 9, we have is a prequasi norm on . Assume , for each and . Hence, we have and since , we have

So is convergent in , i.e., and as , for every and is a premodular (sss). Therefore, we get

we have , so .

Definition 32. A prequasi normon the ideal, where, provides the Fatou property if for every sequencewithand all, then

Theorem 33. The prequasi norm, for alldoes not satisfy the Fatou property, ifis increasing.

Proof. Let the setting be provided and with Since the space is a prequasi closed ideal, so . Therefore, for every , we have

Hence, does not support the Fatou property.

Now, we introduce the definition of Kannan -contraction operator on the prequasi operator ideal.

Definition 34. For the prequasi normon the ideal, where. An operatoris called a Kannan-contraction, if we haveso thatfor all.

Definition 35. For the prequasi normon the ideal, where, andThe operatoris called-sequentially continuous at, if and only if, whenthen.

Theorem 36. Set upbe an increasing and, where, for every. The pointis the unique fixed point of, if the following set up are satisfied:(a) is Kannan -contraction mapping(b) is -sequentially continuous at a point (c)There is such that the sequence of iterates has a subsequence converging to

Proof. Let the conditions be verified. If is not a fixed point of , then . From the setting (b) and (c), we have

Since is Kannan -contraction mapping, one can see

As , this implies a contradiction. Therefore, is a fixed point of . To show that the fixed point is unique. Let we have two different fixed points of . Hence, one has

Therefore,

Example 37. Letandbe Banach spaces, , where, for everyand

Since for all with , we have

For all with , we have

For all with and , we have

Therefore, the map is Kannan -contraction mapping and

It is clear that is -sequentially continuous at the zero operator and has a subsequence converging to . By Theorem 36, the zero operator is the only fixed point of . Let be such that where with . Since the prequasi norm is continuous, we have

Hence is not -sequentially continuous at . So, the map is not continuous at .

7. Application to the Existence of Solutions of Summable Equations

Summable equations like (45) were studied by Salimi et al. [22], Agarwal et al. [23], and Hussain et al. [24]. In this section, we search for a solution to (45) in where be an increasing and , for all . Consider the summable equations and let defined by

Theorem 38. The summable equations ((45)) has a solution inif and for all, there is, so that

Proof. Let the conditions be verified. Consider the mapping defined by (46). We have

Then, from Theorem 24, we have a solution of equation (45) in

Example 39. Given the sequence space, where, for all. Consider the summable equationswhere and let defined by

It is easy to see that

By Theorem 38, the summable equations (49) has a solution in .

Data Availability

Not applicable.

Ethical Approval

This article does not contain any studies with human participants or animals performed by any of the authors.

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

This work was funded by the University of Jeddah, Saudi Arabia, under grant No. (UJ-20-078-DR). The authors, therefore, acknowledge with thanks the University’s technical and financial support. Also, the authors are extremely grateful to the reviewers for their valuable suggestions and leading a crucial role for a better presentation of this manuscript.

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Copyright

Copyright © 2020 Awad A. Bakery and O. M. Kalthum S. K. Mohamed. 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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