Fixed Point Theorems for Generalized Expansive Mapping in Rectangular Metric Spaces

Abstract

In this paper, we present the notion of expansive mapping in complete rectangular metric spaces and study various fixed point theorems for such mappings. The presented theorems extend, generalize, and improve many existing results in the literature.

1. Introduction

The problem of fixed points of mapping with an adequate contractive condition has been the focal point of a rigorous research activity. It has an extensive applications in different areas such as nonlinear and adaptive control systems, parametrized estimation problems, fractal image decoding, and convergence of recurrent networks.

In 1922, the Banach contraction principle [1] came into existence, which is a very famous theorem in nonlinear analysis and has many useful applications.

Several mathematicians generalized the Banach contraction principle in two major directions, one by stating the conditions on the mapping and second allowing the set to have a more general structure [2–7].

In 1984, Wang et al. [8] presented some interesting work on expansion mappings in metric spaces. Recently, Kumar et al. [9] introduced a new notion of generalized expansive mappings and established some fixed point theorems for such mappings in complete generalized metric spaces.

In this paper, inspired by the idea of contraction introduced by Zheng et al. [10] in metric spaces, we presented expansive mapping and establish various fixed point theorems for such mappings in complete rectangular metric spaces. Our theorems extend, generalize, and improve many existing results.

2. Preliminaries

In this section, we recall some definitions and results that will be used to prove our main results.

By an expansion mapping [8] on a metric space , we understand a mapping satisfying for all : where is a real in.

In 2000 Branciari [2] introduced the concept of rectangular metric spaces as follows:

Definition 1 (see [11]). 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) (the rectangular inequality)Then, is called a rectangular metric space.

Definition 2 (see [12]). Let and We say that is a triangular -admissible mapping if
(T₁)
(T₂)
(T₃) for all
(T₄) for all

Definition 3 (see [12]). Let be a rectangular metric space and let be two mappings. (a) 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 Recently, Hussain et al. [13] gave the following definitions.

Definition 4 (see [13]). Let be a rectangular 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)complete, if every Cauchy sequence in with for all converges in (c)complete, if every Cauchy sequence in with or for all converges in

Definition 5 (see [13]). Let be a rectangular 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 The following definitions were given by Jleli et al. in [14].

Definition 6 (see [14]). Let be the family of all functions such that
(θ₁) is increasing, i.e., for all such that
For each sequence is continuous

Definition 7 (see [14]). Let be a rectangular metric space and be a mapping.
is said to be a contraction if there exist and such that for any Recently, Zheng et al. [10] considered the following family of functions and introduced the notion of contraction of metric spaces:

Definition 8 (see [10]). Let be the family of all functions , such that
(θ₁) is increasing
(θ₂) For each , ;
(θ₃) is continuous.

Lemma 9 (see [10]). If . Then, and for all

Definition 10 (see [10]). Let be a metric space and be a mapping.
is said to be a contraction if there exist and such that for any , where Now, we recall the definition of generalized expansive mapping.
Let be the family of functions which satisfy the following: (i) is upper semicontinuous(ii) converges to as , for all (iii), for any

Definition 11 (see [10]). Let be a generalized metric space and be a given mapping. is said to be a generalized expansive mapping of type if there exists two functions and such that for all , where

Definition 12 (see [10]). Let be a generalized metric space and be a given mapping. is said to be a generalized expansive mapping of type II if there exists two functions and such that for all , where

3. Fixed Point Theorem on Rectangular Metric Spaces

We introduce a new notion of generalized expansive mapping in the context of rectangular metric spaces as follows.

Definition 13. Let be a rectangular metric space and be a given mapping. is said to be generalized expansive mapping if there exists two functions and such that where

Theorem 14. Let be a -complete generalized metric space, and be a bijective generalized expansive mapping satisfying the following conditions: (i) is a triangular admissible mapping(ii)There exists such that or (iii)is a continuousThen, has a fixed point. Moreover, has a unique fixed point when or for all

Proof. Let such that or We define the sequence in by , for all
Since is a triangular admissible mapping, then or Continuing this process, we have or for all By (T₃) and (T₄) one has. Suppose that there exists such that Then, is a fixed point of and the proof is finished. Hence, we assume that , i.e., for all
Step 1: We shall prove Applying inequality (11) with and , we obtain where If for some , then the inequality (19), we get It is a contradiction. Hence, Therefore, Thus, Continuing this process, we get Now, by (23) and (2.9), we deduce that By the property of and , it is evident that Step 2. Now, we shall prove On the contrary, assume that for some . Indeed, suppose that , so we have Denote By inequality (25), we have Continuing this process, we get which is a contradiction. Thus, (27) hold.
Step 3. We shall prove Applying inequality (11) with , , we obtain where Take and Thus, by (32), one can write Therefore, Again, by (25), which implies that Then, the sequence is monotone nonincreasing. Thus, there exists such that Assume that . By (26), we have Taking the in (32), and using , , and Lemma 9, we obtain which implies that Therefore, By (θ₁), we get It is a contradiction, then Step 4. We shall prove that is a Cauchy sequence in , that is, If otherwise there exists an for which we can find sequence of positive integers and of such that, for all positive integers , Now, using (26), (44), (46), and the rectangular inequality, we find Then, Now, by rectangular inequality, we have Letting in the above inequalities, using (26), (46), and (49), we obtain On the other hand, Letting in the above inequalities and using (26), (49), and (52), we get that By (51), let , from the definition of the limit, there exists such that This implies that and by (54), let , from the definition of the limit, there exists such that Applying (11) with and , we obtain Letting , the above inequality and using (θ₃) and , we obtain Therefore, It is a contradiction with Lemma 9, according to which Then, It follows that is a Cauchy sequence in . Since is -complete and for all , the there exists such that Step 5. We show that arguing by contradiction, we assume that By rectangular inequality, we get By letting in inequality (66) and (67), we obtain Therefore, Since is -continuous, then i.e . Hence, , so .
Step 6. Uniqueness. Now, suppose that are two fixed points of such that and or . Therefore, we have Applying (11) with and , we have where Therefore, we have It is a contradiction. Therefore, .

Theorem 15. Let : be two functions and let be a complete rectangular metric space. Let be a bijective, mapping satisfying the following conditions: (i) is a triangular -admissible mapping(ii) is a generalized -expansive mapping(iii) or for all Then, has the property

Proof. Let for some fixed . As or and is a triangular -admissible mapping, then Continuing this process, we have for all . By and , we get Since is a bijective mapping, then for all and . Therefore, Assume that , i.e., Then, we have Applying (11) with and , we obtain where Letting in (80), we obtain Now, using (θ₃) and , we get It is a contradiction. Then, .

Example 16. Let and defined by Then, is a metric space and rectangular metric space. Define mapping and by Then, is an continuous triangular admissible mapping and is a bijective mapping.
Let
Evidently, ( and are when .

Consider two cases:

Case 1. .

Thus,

We have

On the other hand,

Since , then

which implies that

Case 2.

Thus,

We have

On the other hand

Since , then

which implies that

Hence, condition (11) is satisfied. Therefore, has a unique fixed point .

Theorem 17. Let be two functions and let be a complete rectangular metric space. Let be a bijective mapping satisfying the following assertions: (i) is triangular admissible(ii) is a generalized expansive mapping(iii)There exists such that or(iv)is a-regular rectangular metric space

Then, has a fixed point. Moreover, has a unique fixed point whenever or for all

Proof. Let x such that or . Similar to the proof of Theorem 14, we can conclude that and from inequality (69), we have From (iv) hold for
Suppose that for some From Theorem 14, we know that the members of the sequence are distinct. Hence, we have i.e., for all Thus, we can apply (11), to and for all to get where Therefore, By letting in inequality (101), we obtain Since and are continuous functions and , we conclude that which implies that It is a contradiction. Hence, .
The proof of the uniqueness is similarly to that of Theorem 14.

Corollary 18. Let be two functions, be a -complete rectangular metric space and be a bijective mapping. Suppose that for all with or and we have where and , . Then, has a fixed point, if (i) is a triangular admissible mapping(ii)there exists such that or (iii)is a continuous or(iv) is an -regular rectangular metric spaceMoreover, has a unique fixed point when or for all

4. Fixed Point Theorem on Rectangular Metric Spaces Endowed with a Partial Order

Definition 19 (see [4]). Let be an ordered rectangular metric space and be a mapping. (1) is said to be -complete, if every Cauchy in with for all or for all converges in (2) is said to be -regular, if for each sequence in , and for all or for all imply that or , respectively(3) is said to be -continuous, if for given and sequence with or for all ,

Definition 20. Let be an ordered rectangular metric spaces and be a mapping. We say that aids to be an ordered expansive mapping, if for all withor such that where

Theorem 21. Let be an -complete partially ordered rectangular metric space. Let be a bijective self mapping on satisfying the following assertions: (i) is monotone(ii) is an ordered expansive mapping(iii)there exists such that or (iv)either is -continuous or(v) is -regularThen, has a fixed point. Moreover, has a unique fixed point whenever or for all

Proof. Define the mapping by and the mapping by Using condition (iii), we have Owing to the monotonicity of , we get Therefore, and hold.
On the other hand, if or Since be an -complete partially ordered rectangular metric space, we conclude that Thus, (T₃) and (T₄) hold. This shows that is a triangular admissible mapping then Now, if is -continuous, then
or or and as with
The existence and uniqueness of a fixed point follows from Theorem 14.
Now, suppose that follow is -regular. Let be a sequence such that which implies that for all and as This shows that is regular. Thus, the existence and uniqueness of fixed point from Theorem 17.

Corollary 22. Let be an -complete partially ordered rectangular metric spaces.
Let be a bijective self mapping on be such that is a monotone mapping. Suppose that there exists such that for all with or . Suppose also the following conditions hold: (i)There exists such that or (ii)Either is -continuous or(iii) is -regularThen, has a fixed point. Moreover, has a unique fixed point whenever or for all

Proof. By taking and , obviously that and and is a -expansive mapping. As in the proof of Theorem 21, we get the result.

5. Conclusion

In this article, we have proved some fixed point theorems for the expansive mapping in the settings of complete rectangular metric spaces. These results are extensions of the related results announced in [6, 7, 9].

Data Availability

No data were used to support this study.

Conflicts of Interest

On behalf of all authors, the corresponding author states that there is no conflict of interest.