Contraction on Complete Rectangular Metric Spaces

Abstract

In this paper, we present some fixed point results for generalized contraction in the framework of compete rectangular metric spaces. Further, we establish some fixed point theorems for this type of mappings defined on such spaces. Our results generalize and improve many of the well-known results. Moreover, to support our main results, we give an illustrative example.

1. Introduction

The well-known Banach contraction theory is one of the methods used, which states that if is a complete metric space and is self-mapping with contraction, then has a unique fixed point [1].

In 2000, Branciari [2] introduced the notion of generalized metric spaces, for example, the triangle inequality is replaced by the inequality for all pairwise distinct points , , , and . Since then, several results have been proposed by many mathematicians on such spaces (see [3–8]).

The concept of metric space, as an ambient space in fixed point theory, has been generalized in several directions, such as, metric spaces [9] and generalized metric spaces.

Combining conditions are used for definitions of b-metric and generalized metric spaces. Roshan et al. [10] announced the notion of rectangular b-metric space.

Hussain et al. [11] introduced the concept of complete rectangular metric space and proved certain results of fixed point theory on such spaces.

In this paper, we provide some fixed point results for generalized contraction in the framework of compete rectangular metric spaces, and also we give two examples to support our results.

2. Preliminaries

Definition 1. (see[10]). Let be a nonempty set, be a given real number, and let d: be a mapping such that for all and all distinct points , each distinct from and :(1), if only if (2)(3)Then, is called a rectangular metric space.

Lemma 1. (see [10]). Let be a rectangular b-metric space.(a)Suppose that sequences and in are such that and as , with , , and for all . Then, we have(b)If and is a Cauchy sequence in with for any , , converging to , thenfor all .

Zheng et al. [12] introduced a new type of contractions called -contractions in metric spaces and proved a new fixed point theorems for such mapping.

Definition 2. (see [13]). We denote by the set of functions , satisfying the following conditions:  is increasing  for each sequence   is continuous on

Definition 3. (see [12]). We denote by the set of functions : satisfying the following conditions:  : is nondecreasing  for each ,   is continuous on

Lemma 2. (see [12]). If , then and for each .
In 2014, Hussain et al. [14] introduced a weaker notion than the concept of completeness and called it -completeness for metric spaces.

Definition 4. (see [14]). Let and :. We say that is a triangular admissible mapping if ,  ,   for all   for all

Definition 5. (see [14]). Let be a b-rectangular metric space and let : be two mappings. The space is said to be as follows:(a)T is continuous mapping on , if for given point and sequence in , and for all imply that .(b)T is subcontinuous mapping on , if for given point and sequence in , and for all imply that T.(c)T is continuous mapping on , if for given point and sequence in , and or for all imply that .The following definitions were given by Hussain et al. [11].

Definition 6. (see [11]). Let be a rectangular b-metric space and let : be two mappings. The space is said to be(a)complete, if every Cauchy sequence in with for all converges in (b), if every Cauchy sequence in with for all converges in (c)complete, if every Cauchy sequence in with or for all converges in

Definition 7. (see [11]). Let be a rectangular b-metric space and let : be two mappings. The space is said to be(a) is -regular, if , where for all implies for all (b) is subregular, if , where for all implies for all (c) is -regular, if , where or for all implies that or for all .

3. Main Results

Definition 8. Let be a -rectangular b-metric space with parameter and let be a self-mapping on . Suppose that are two functions. We say that is an contraction, if for all with and , we havewhere , for , , and .

Definition 9. Let be a -complete rectangular metric space and be a mapping.(1) is said to be a Kannan-type contraction if there exist and with or for any , , we have(2) is said to be a Reich-type contraction if there exist and with or for any , , we have(3) is said to be a Kannan-type mapping, that is, if there exists with or for any , , we have(4) is said to be a Reich-type mapping, that is, if there exists with or for any , , we have

Theorem 1. Let be a complete rectangular b-metric and let be two functions. Let be a self-mapping satisfying the following conditions:(i) is a triangular admissible mapping(ii) is an contraction(iii)There exists such that or (iv) is a continuous.Then T has a fixed point. Moreover, has a unique fixed point when or for all .

Proof. Let such that or .
Define a sequence by . Since T is a triangular admissible mapping, then or .
Continuing this process, we have or , for all . By and , one hasSuppose that there exists such that . Then, is a fixed point of and the proof is finished. Hence, we assume that , i.e., for all . We haveIndeed, suppose that for some , so we haveDenote . Then, (3) and Lemma 2 imply thatAs is increasing, soHence,SinceThusContinuing this process, we can prove that , which is a contradiction. Thus, in the following, we can assume that (8) and (9) hold.
We shall prove thatSince is contraction, we getSince is increasing, we deduce that , and thusSince , thenTherefore, is monotone strictly decreasing sequence of nonnegative real numbers. Consequently, there exists , such thatwhich again by (3) and (19) and property of , we haveBy taking the limit as in (21) and using , we haveThen, , by , we obtainOn the other hand,By and Lemma 2, we obtainTherefore,Taking the limit as in (28) and using (23), since , we haveHence, (16) is proved.
Next, we show that is an Cauchy sequence in , if otherwise there exists an for which we can find sequences of positive integers and of such that, for all positive integers , ,From (30) and using the rectangular inequality, we getTaking the upper limit as in (32) and using (16), we getMoreover,Then, from (23) and (31), we getOn the other hand, we haveThen, from (23) and 31 we getApplying (3) with and , we haveNow taking the upper limit as in (38) and using , (23), (33), (35), (37), and Lemma 2, we haveTherefore, implies , which is a contradiction.
Consequently, is a Cauchy sequence in complete rectangular b-metric space . Since or , for all .
This implies that the sequence converges to some . Suppose that . Then, we have all the assumption of Lemma 1 and is continuous, then as . Therefore,Hence, we have and so . Thus, is a fixed point of .

3.1. Uniqueness

Let Fix where and or .

Applying (3) with and , we have

Since is increasing, thereforewhich is a contradiction. Hence, and have a unique fixed point.

Recall that a self-mapping is said to have the property , if for every .

Theorem 2. Let : be two functions and let be an complete rectangular b-metric space. Let be a mapping satisfying the following conditions:(i) is a triangular admissible mapping(ii) is an contraction(iii) or , for all Fix Then has the property .

Proof. Let Fix for some fixed . As or and is a triangular -admissible mapping, thenContinuing this process, we havefor all . By and , we getAssume that Fix , i.e., .
Applying (3) with and , we getwhich implies thatSince is increasing, therefore,which is a contradiction as and .
Assuming the following conditions, we prove that Theorem 2 still holds for not necessarily continuous.

Theorem 3. Let : be two functions and let be an complete rectangular b-metric space.
Let be a mapping satisfying the following assertions:(i) is triangular admissible(ii) is contraction(iii)There exists such that or (iv) is an -regular rectangular b-metric spaceThen has a fixed point. Moreover, has a unique fixed point whenever or for all .

Proof. Let such that or . Similar to the proof of Theorem 3, we can conclude thatwhere .
From holds for .
Suppose that for some . From Theorem 3, we know that the members of the sequence are distinct. Hence, we have , i.e., for all . Thus, we can apply (3), to and for all to getBy Lemma 2 and , we obtainBy taking the limit as in (51), we haveAssume that . Then, from Lemma 1,By assumption , we have and so . Thus, is a fixed point of .
The proof of the uniqueness is similarly to that of Theorem 3.
above theorems, if we take , for some fixed , where . We obtain the following extension of Jamshaid et al. result (Theorem 1) [13] of complete rectangular -metric space.

Corollary 1. Let be two functions and be an complete rectangular -metric space and let be self-mapping. Suppose for all with or and , we havewhere and . If the mapping satisfies the following assertions: point, if(i) is a triangular admissible mapping(ii)There exists such that or (iii) is -continuous or(iv)is an regular rectangular metric spaceThen has a fixed point. Moreover, has a unique fixed point whenever or for all .

Proof. Let , we prove that is an contraction, Hence, T satisfies in assumption of Theorem 3 or 2 and is the unique fixed point of .
It follows from Theorem 3; we obtain the following fixed point theorems for -Kannan-type contraction and -Reich-type contraction.

Theorem 4. Let be a complete rectangular -metric space and let be two functions. Let be a self-mapping satisfying the following conditions:(i) is a triangular admissible mapping(ii) is a -Kannan-type contraction(iii)There exists such that or (iv)is a continuousThen T has a fixed point. Moreover, has a unique fixed point when or for all .

Proof. If is a -Kannan-type contraction, thus there exist and with or for any , , we haveTherefore,where , , which implies that is a contraction Therefore, from Theorem 2, has a unique fixed point.

Theorem 5. Let be a complete rectangular -metric space and let be two functions. Let be a self-mapping satisfying the following conditions:(i) is a triangular admissible mapping(ii) is a Reich-type contraction(iii)There exists such that or (iv) is a continuousThen T has a fixed point. Moreover, has a unique fixed point when or for all .

Proof. If is a Reich-type contraction, thus there exist and with or for any , , we haveTherefore,where , which implies that is a contraction. Therefore, from Theorem 3, has a unique fixed point.

Corollary 2. Let be a complete rectangular -metric space and let be two functions. Let be a self-mapping satisfying the following conditions:(i) is a triangular admissible mapping(ii) is a Kannan-type mapping(iii)There exists such that or (iv)is a continuousThen T has a fixed point. Moreover, has a unique fixed point when or for all . Then has a unique fixed point .

Proof. Let for all , and for all .
It is obvious that and . We prove that is a -Kannan-type contraction:Therefore, from Theorem 3, has a unique fixed point .

Corollary 3. Let be a complete rectangular -metric space and let be two functions. Let be a self-mapping satisfying the following conditions:(i) is a triangular admissible mapping(ii) is a Reich-type mapping(iii)There exists such that or (iv)is a continuousThen T has a fixed point. Moreover, has a unique fixed point when or for all .

Proof. Let for all , and for all t.
We prove that T is a -Reich-type contraction:Therefore, from Theorem 3, has a unique fixed point .

Example 1. Consider the set . It is easy to check that the mapping given by(i), for all (ii), , (iii), , and Clearly is a rectangular metric space with parameter .
Define mapping and byThen, is an continuous triangular admissible mapping.
Let , , and , , , and . It is obvious that and . Evidently, and are when, or . Consider the following four possibilities:
For and , thenThenFor and , thenThenFor and , thenThenHence, satisfying the assumption of Theorems 3 and 1 is the unique fixed point of .

Example 2. Let , where and . Define as follows:Then, is a rectangular b-metric space with coefficient s = 3. However, we have the following:(1) is not a metric space, as (2) is not a b-metric space for s = 3, as (3) is not a rectangular metric space, as Define mapping and byThen, is an continuous triangular admissible mapping.
Let , and taking and . It is obvious that and . Evidently, and are when with .
Consider two cases: Case 1: : On the other hand Since , then which implies that Case 2: :Similarly for Case 2, we conclude thatHence, condition (3) is satisfied. Therefore, has a unique fixed point .

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.