Fixed-Point Theorems for <span class="nowrap"><svg xmlns:xlink="http://www.w3.org/1999/xlink" xmlns="http://www.w3.org/2000/svg" style="vertical-align:-4.5269pt" id="M1" height="16.7875pt" version="1.1" viewBox="-0.0657574 -12.2606 36.2221 16.7875" width="36.2221pt"><g transform="matrix(.017,0,0,-0.017,0,0)"><path id="g113-230" d="M475 507C475 612 440 712 326 712C139 712 23 420 23 215C23 96 58 -12 180 -12C369 -12 475 293 475 507ZM391 522C391 486 387 448 379 394H126C155 538 222 677 310 677C386 677 391 571 391 522ZM373 346C344 193 283 22 189 22C126 22 106 114 106 196C106 243 111 293 118 346H373Z"/></g><g transform="matrix(.017,0,0,-0.017,12.383,0)"><path id="g117-33" d="M535 230V280H52V230H535Z"/></g><g transform="matrix(.017,0,0,-0.017,26.29,0)"><path id="g113-243" d="M542 242C542 369 469 429 379 468C370 472 372 496 376 513L416 681L400 691L344 648L219 51C167 69 113 128 113 217C113 324 166 387 266 428L252 459C124 418 23 336 23 202C23 95 90 13 205 -8L182 -132C165 -222 191 -254 209 -261L215 -258L270 -7C415 -4 542 98 542 242ZM469 232C469 126 403 33 279 39L355 409C399 394 469 331 469 232Z"/></g></svg>-</span>Contraction in Generalized Asymmetric Metric Spaces

Kari, Abdelkarim; Rossafi, Mohamed; Saffaj, Hamza; Marhrani, El Miloudi; Aamri, Mohamed

doi:https://doi.org/10.1155/2020/8867020

International Journal of Mathematics and Mathematical Sciences

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Research Article | Open Access

Volume 2020 | Article ID 8867020 | https://doi.org/10.1155/2020/8867020

Fixed-Point Theorems for -Contraction in Generalized Asymmetric Metric Spaces

Abdelkarim Kari,¹Mohamed Rossafi,²Hamza Saffaj,¹El Miloudi Marhrani,¹and Mohamed Aamri¹

Academic Editor: Sonia Perez Diaz

Received31 Aug 2020

Revised24 Sept 2020

Accepted21 Oct 2020

Published12 Nov 2020

Abstract

In the last few decades, a lot of generalizations of the Banach contraction principle had been introduced. In this paper, we present the notion of -contraction and -contraction in generalized asymmetric metric spaces to study the existence and uniqueness of the fixed point for them. We will also provide some illustrative examples. Our results improve many existing results.

1. Introduction

The problem of the existence of the solution of many mathematical models is equivalent to the existence of a fixed-point problem for a certain map. The study of fixed points, therefore, has a central role in many disciplines of applied sciences. The most essential and key part of the theory of fixed points is the existence of the solution of operator equations satisfying certain conditions, for example, Fredholm integral equations, Volterra integral equations, and two-point boundary-value problems in differential equations, as well as some eigenvalue problems [1–3]. A beautiful blend of analysis, topology, and geometry has laid down the foundation of the theory of fixed points.

The Banach contraction principle [4] has become a powerful tool in modern analysis, and it is an important tool for solving existence problems in mathematics and physics. Many authors have established the theory of fixed points particularly in two directions: one by stating the conditions on the mapping and second, taking the set as a more general structure [5–8].

Many generalizations of the concept of metric spaces are defined, and some fixed-point theorems are proved in these spaces. In particular, asymmetric metric spaces were introduced by Wilson [9] as metric spaces, but without the requirement that the asymmetric metric has to satisfy .

Asymmetric metric spaces have numerous recent applications both in pure and applied mathematics, for example, in rate-independent models for plasticity [10], shape-memory alloys [11], models for material failure [12], and the questions of the existence and uniqueness of Hamilton–Jacobi equations [13].

Many mathematicians worked on this interesting space. For more details, refer [14, 15].

A. Branciari in [16] initiated the notions of a generalized metric space as a generalization of a metric space, where the triangular inequality of metric spaces was replaced by . Various fixed-point results were established on such spaces, see [4, 17–21] and references therein.

Combining conditions used for definitions of asymmetric metric and generalized metric spaces, Piri et al. [22] announced the notions of the generalized asymmetric metric space.

In this paper, we introduce the notion of -contraction and -contraction and establish some new fixed-point theorems for mappings in the setting of complete generalized asymmetric metric spaces. Our result generalizes, improves, and extends the corresponding results due to Kannan and Reich. Moreover, illustrative examples are presented to support the obtained results.

2. Preliminaries

In the following, we recollect some definitions which will be useful in our main results.

Definition 1. (see [16]). Let be a nonempty set and be a mapping such that, for all and for all distinct points , each of them different from and , one has(i) if and only if (ii) for all distinct points (iii) (quadrilateral inequality)Then, is called a generalized metric space.

Definition 2. (see [22]). Let be a nonempty set and be a mapping such that, for all and for all distinct points , each of them different from and , one has(i) if and only if (ii)Then, is called a generalized asymmetric metric space.

Definition 3. (see [22]). Let be a generalized asymmetric metric space and be a sequence in X, and . Then,(i)We say that forward (backward) converges to if and only if(ii)We say that forward (backward) Cauchy if

Example 1. Let , where and and defined byfor all . Then, is a generalized asymmetric metric space. However, we have the following:(1) is not a metric space as , for all (2) is not a asymmetric metric space as (3) is not a rectangular metric space as , for all

Remark 1. Let be as in Example 1 and be a sequence in . However, we have the following:(i), and , . Then, the sequence forward converges to 2 and backward converges to 0, so the limit is not unique.(ii). So, forward (backward) convergence does not imply forward (backward) Cauchy.

Lemma 1. (see [22]). Let be a generalized asymmetric metric space and be a forward (or backward) Cauchy sequence with pairwise disjoint elements in X. If forward converges to and backward converges to , then x = y.

Definition 4. (see [22]). Let be a generalized asymmetric metric space. is said to be forward (backward) complete if every forward (backward) Cauchy sequence in X forward (backward) converges to .

Definition 5. (see [22]). Let be a generalized asymmetric metric space. is said to be complete if X is forward and backward complete.The following definition was given by Jleli and Samet in [23].

Definition 6. (see [23]). Let be the family of all functions : such that is increasing, for each sequence , is continuous.Recently, Zheng et al. [24] introduced the new type of contractive mappings as follows:

Definition 7. (see [24]). Let be the family of all functions : such that is increasing for each , is continuous

Lemma 2. (see [24]). If , then = 1, and for all .

Definition 8. (see [24]). Let be a metric space and be a mapping.
is said to be a -contraction if there exist and such that, for any ,where

Theorem 1. (see [22]). Let be a generalized asymmetric metric space and be a mapping. Suppose that there exist and such that, for any ,

Then, T has a unique fixed point.

3. Main Result

Motivated and inspired by Piri et al. [22] and Zheng et al. [24], we define the notions of -contraction and -contraction on the generalized asymmetric metric space, and we give some results on such space.

Definition 9. Let be a generalized asymmetric metric space and be a mapping.(1) is said to be a -contraction if there exist and such that, for any , we have where(2) is said to be a -contraction if there exist and such that, for any , we have where(3) is said to be a -Kannan-type contraction if there exist and such that for any , we have(4) is said to be a -Reich-type contraction if there exist and such that for any , we have

Theorem 2. Let be a complete generalized asymmetric metric space, and let be a -contraction, i.e., there exist and such that, for anyThen, has a unique fixed point.

Proof. Let be a fixed point, and define a sequence by , for all . If there exists such that or , then proof is finished.
We can suppose that and for all ; then, we haveSubstituting and , from (16), for all , we havewhereNow, we set .
Therefore,If , then from the assumption of the theorem, we havewhich is a contradiction. Hence, .
Thus,Repeating this step, we conclude thatFrom , we getTherefore, is a monotone strictly decreasing sequence of nonnegative real numbers. Consequently, there exists such thatNow, we claim that . Arguing by the contraction, we assume that . Since is a nonnegative decreasing sequence, we haveFrom the property of , we getBy letting in inequality (27), we obtainIt is a contradiction. Therefore,Next, we shall prove thatWe assume that for every . Indeed, suppose that for some with , so we have .
So, from the assumption of the theorem, we getSince is increasing, we haveContinuing this process, we can say thatIt is a contradiction. Therefore,for every , .
Substituting and ,Applying (16) with and , we havewhereSo, we getTake and . Thus, by (38), one can writeBy , we getBy (24), we havewhich implies thatTherefore, the sequence is monotone nonincreasing. Thus, there exists such thatBy (29), we assume that ; then, we getTaking in (38) and using the properties of , we obtainTherefore,By , we getIt is a contradiction. Therefore,Next, we shall prove that is a Cauchy sequence, i.e., , for all . Suppose to the contrary. Then, there is such that, for an integer , there exist two sequences and such thatNow, using (29), (48), (49), and the quadrilateral inequality, we findThen,Now, by the quadrilateral inequality, we haveLetting in the above inequalities and using (29), (48), and (51), we obtainTherefore, by (29) and (51), we get thatBy (53), there exists such thatTherefore,So,Applying with and , we haveLetting in the above inequality and using , (53), and (54), we obtainTherefore,Since is increasing, we getwhich is a contradiction. Then,Equivalently,Hence, is a forward and backward Cauchy sequence in X. By completeness of , there exists such thatSo, from Lemma 1, we get .
Now, we show that = . Arguing by contradiction, we assume thatTherefore,Now, by the quadrilateral inequality, we getBy letting in inequalities (67) and (68), we obtainTherefore,On the contrary,By letting in inequalities (71) and (72), we obtainTherefore,By (70) and from the definition of the limit, there exists such thatSimilarly, by (74), there exists such thatLet ; we concludeApplying (16) with and , we havewhereTherefore,By letting in inequality (80) and using , we obtainBy , we getwhich is a contradiction. Hence, .
Uniqueness: now, suppose that are two fixed points of such that . Therefore, we haveTherefore,Applying (16) with and , we havewhereTherefore, we haveTherefore,is a contradiction. Therefore, .

Corollary 1. (Theorem 2; see [22]). Let be a complete generalized asymmetric metric space and be a given mapping. Suppose that there exist and such that, for any , we have .

Then, has a unique fixed point.

Example 2. Consider . Let be a mapping defined by the following:(i) if and only (ii)(iii)(iv), and (v)Clearly, is not an asymmetric metric space. Indeed, .
However, it is a complete generalized asymmetric metric space.
Let be given bySuppose and . Therefore, and . First, observe that , , , , or .
For , we haveTherefore,So,For , we havewhich implies thatTherefore,For , we haveOn the contrary,Then,For , we haveTherefore,For , , we haveThen,Hence, satisfies the assumption of the theorem, and is the unique fixed point of .

Example 3. Let , where and .
Define as follows:Then, is a generalized asymmetric metric space. However, we have the following:(1) is not a metric space as (2) is not a generalized metric space as Define mapping byThen, . Let , and . It is obvious that and .
Consider the following possibilities: Case 1: with , and assume that . Therefore, On the contrary, which implies that Case 2: or . Therefore, , ; then, .In this case, consider two possibilities:(i): then, . Therefore, So, we have On the contrary, Since which implies that(ii): then, . Therefore,So, we haveOn the contrary,Since ,which implies thatHence, satisfies the assumption of the theorem, and is the unique fixed point of .

Theorem 3. Let be a complete generalized asymmetric metric space and be a mapping. Suppose that there exist and such that, for any ,where

Then, has a unique fixed point.

Proof. Let be a fixed point, and define a sequence byIf there exists such that or , then proof is finished.
We can suppose that and for all ; then, we haveSubstituting and , from (119), for all , we havewhereNow, we set .
Therefore,if for some , .
From (123), , and using Lemma 2, we getwhich implies thatwhich is contradiction. Hence, .
Therefore,As is increasing,Therefore, is a monotone strictly decreasing sequence of nonnegative real numbers. Consequently, there exists such thatNow, we claim that . Arguing by contradiction, we assume that . Since is a nonnegative decreasing sequence, we haveThus, we haveBy letting in inequality (132) and using and , we obtainIt is a contradiction. Therefore,Next, we shall prove thatWe assume that for every . Indeed, suppose that for some with , so we have .
So, from the assumption of the theorem, we getSince is increasing,Continuing this process, we can get thatwhich is a contradiction. Therefore,Substituting and ,Applying (119) with and , we havewhereTherefore,So, from Lemma 2, we haveTake and . Thus, from (144) and , we haveAgain by (137),Therefore,Then, the sequence is monotone nonincreasing, so it converges to some such thatBy (134) and assuming that , we haveTaking in (144) and using , we obtainTherefore,From , we getwhich is a contradiction. Therefore,Next, We shall prove that is a Cauchy sequence, i.e., , for all . Suppose to the contrary. Then, there is such that, for an integer , there exist two sequences and , , such thatAs in the proof of Theorem 2, we conclude thatHence, there exists such thatSince T is a -contraction, we derivewhereAs in the proof of Theorem 2, we haveApplying (119) with and , we haveBy letting in inequality (160) and using , and Lemma 2, we obtainwhich implies thatwhich is a contradiction. Therefore,Thus,Hence, is a forward and backward Cauchy sequence in X. By completeness of , there exists such thatSo, from Lemma 1, we get .
Now, we show that = . Arguing by contradiction, we assume thatTherefore,As in the proof of Theorem 2, we conclude thatBy (168) and (169), there exists such thatSince T is a -contraction, we derivewherewhich implies thatBy letting in inequality (172) and using , and Lemma 2, we obtainTherefore,which is a contradiction. Thus, . Thus, has a fixed point.
Uniqueness: let Fix, where . Then,Therefore,From the assumption of the theorem, we getwhereTherefore, we havewhich implies that . It is a contradiction. Therefore, .
From Theorem 3, we obtain the following fixed-point theorems for -Reich-type contraction and -Kannan-type contraction.

Theorem 4. Let be a complete generalized asymmetric space and be a -Kannan-type contraction; then, has a unique fixed point.

Proof. Since is a -Kannan-type contraction, there exist and such thatTherefore, T is a -contraction. As in the proof of Theorem 3, we conclude that T has a unique fixed point.

Theorem 5. Let be a complete generalized asymmetric space and be a -Reich-type contraction.

Then, T has a unique fixed point.

Proof. Since is a -Reich-type contraction, there exist and such thatTherefore, is a -contraction. As in the proof of Theorem 3, we conclude that T has a unique fixed point.

Corollary 2. Let be a complete metric space and be a Kannan-type mapping. i.e., there exists such that, for all ,

Then, T has a unique fixed point.

Proof. Let for all and for all . We prove that T is a -Kannan-type contraction. Indeed,Therefore, as in the proof of Theorem 4, has a unique fixed point .

Corollary 3. Let be a complete metric space and be a Reich-type mapping, i.e., there exists such that, for all ,and we have

Then, T has a unique fixed point.

Proof. Let for all and for all .
We prove that T is a -Reich-type contraction.Therefore, as in the proof of Theorem 5, has a unique fixed point .

Corollary 4 (Theorem 2). Let be a complete generalized asymmetric space and be a given mapping. Suppose that there exist and such that, for any ,where

Then, has a unique fixed point.

Proof. If , with , we prove that T is a -contraction. Therefore, as in the proof of Theorem 3, has a unique fixed point.

Example 4. Consider . Let be a mapping defined by the following:(i).(ii), .(iii).(iv).(v).Clearly, is not an asymmetric metric space, from .
However, it is a complete generalized asymmetric metric space. Let be given bySuppose and . Therefore, and .
First, observe that , , or ; for , we haveOn the contrary,Therefore,So,For , , we haveTherefore,So,For , , we haveTherefore,So,Hence, satisfies the assumption of the theorem, and is the unique fixed point of .

Example 5. Let and defined byClearly, is not metric, asymmetric metric, but it is a complete generalized asymmetric metric space.
Let be given byLet and . It is obvious that and .
Consider the following possibilities: Case 1: . We have Then, So, On the contrary, Since , which implies that Case 2: . Similar to case 1, we conclude thatHence, satisfies the assumption of the theorem, and is the unique fixed point of .

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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Copyright © 2020 Abdelkarim Kari 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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