Near-Common Fixed Point Result in Cone Interval b -Metric Spaces over Banach Algebras

: In this article, we proposed the concept of cone interval b -metric space over Banach algebras. Furthermore, some near-ﬁxed point and near-common ﬁxed point results are proved in the context of cone interval b -metric space and normed interval spaces for self-mappings under different types of generalized contractions. An example is presented to validate our main outcome.


Introduction and Preliminaries
Fixed point theory is without a doubt one of the most powerful techniques from nonlinear analysis. This provides a best method to study various problems in economics, science and mathematics; particularly the initial and boundary value problems involving ordinary, partial and fractional differential equations, difference equation, integral equations, functional equations, variational inequality, etc.; in particular, this provides the existence and uniqueness of solutions of such problems. If T : X → X is a mapping of a space X into itself, then the question arises whether some points are mapped onto itself, i.e., does the equation T(x) = x have a solution? If so, x is called a fixed point of T. The Banach contraction principle was considered as the most fundamental result in metric fixed point theory. Several authors generalized the Banach contraction principle, and this generalization goes in two different directions or ways, as recorded below: i To generalize the contraction condition of Banach, ii Replacing complete metric space with some generalized metric space.
Some authors used both ways or directions to generalize the Banach contraction principle. In this paper, we make an attempt to cover both ways or directions to generalize the Banach contraction principle, i.e., we make an effort to generalize the contraction condition and also metric space and its axioms.
By swapping the real numbers with ordered Banach space, Huang and Zhang [1] generalized the notion of a metric space and demonstrated some fixed point results of contractive maps using the normality condition in such spaces. Rezapour and Hamlbarani [2] subsequently ignored the normality assumption and obtained some generalizations of the Huang and Zhang [1] results. However, it should be noted that in recent research, some scholars established an equivalence between cone metric spaces and metric spaces in the sense of the existence of fixed points of the mappings involved. See, for instance, [3][4][5]. Liu and Xu [6] proposed the concept of a cone metric space over Banach algebra in order to solve these shortcomings by replacing Banach space with Banach algebra, which became an interesting discovery in the study of fixed point theory since it can be shown that cone metric spaces over Banach algebra are not equal to metric spaces in terms of the presence of the fixed points of mappings. Among these generalizations, Wu [7] introduced the concept of metric interval spaces and normed interval spaces as a generalization of metric spaces, and also studied some near-fixed point results in such spaces.
In 1973, Hardy and Rogers [8] proposed a new definition of mappings called the contraction of Hardy-Rogers that generalizes the theory of Banach contraction and the theorem of Reich [9] in metric space setting. For other related work about the concept of Hardy-Rogers contractions, see for instance ( [10][11][12] and the references therein).
We recollect certain essential notes, definitions required and primary results consistent with the literature.
Interval Space 1. [7] Let I denote the set of all closed and bounded intervals [ , ε], where , ε ∈ R and ≤ ε, we consider ∈ R as the element [ , ] ∈ I. The addition is given by and the scalar multiplication is calculated as follows: It is evident that I is not a (conventional) vector space under the aforementioned addition and scalar multiplication. The main reason for this is that, as the following explanation will show, the inverse element does not exist for any non-degenerated closed interval.
It is clear to see that the zero element is [0, 0] ∈ I. However, for any [ , ε] ∈ I, the subtraction is not a zero element. In other words, there is no inverse element of [ , ε].
By considering the zero element, the null set is defined as follows: It is easy to see that It may also be demonstrated that [−1, 1] generates Ω via non-negative scalar multiplication, as demonstrated below: We call [−1, 1] a generator of the null set Ω in this situation. The following observations in interval space are noteworthy.
• The distributive law for scalar addition does not hold in general; that is to say, for any [ , ε] ∈ I and α, β ∈ R. • For positive scalar addition, the distributive law is true; that is, for any [ , ε] ∈ I and α, β > 0.
• For negative scalar addition, the distributive law is valid; that is, It is clear to see that The family of all classes [ , ε] for [ , ε] ∈ I is denoted by I .
According to the preceding Proposition, the classes defined in (2) create the equivalence classes. The family I is referred to as the quotient set of I in this case. It is also seen In other words, the whole set I is partitioned by the family of all equivalence classes.
Metric Interval Space 1. [7] Let I denote the set of all closed and bounded intervals in R with the null set Ω, and d denote a mapping from I × I to non-negative real numbers that fulfills the following axioms: If the following condition iv is satisfied, then we say that d satisfies the null equalities: iv. for any ω 1 , ω 2 ∈ Ω and [ , ε], [ρ, ] ∈ I, the following holds true: Example 1. [7] Let I be the set of all closed bounded intervals in R, and d denote the function from Then, (I, d) is a metric interval space in which d fulfills the null equalities.
Then, (I, · ) is a normed interval space in which the null equality is satisfied by the norm · .
i Given the normed interval space (I, · ) such that · satisfies the null equality. For any Given the normed interval space (I, · ). For any The concept of convergence and Cauchy sequence in normed interval spaces is given below. Given the normed interval space (I, · ).
Definition 1. [7] Given the normed interval space (I, · for all n, m > N with n = m. If every Cauchy sequence in I is convergent, then I is complete. iv If I is complete, then it is also called a Banach interval space.
Unless otherwise stated, we will assume in this article that B is a real Banach algebra. If ∈ B occurs, we call e the unit of B, so that e = e = . We call B a unital in this case. If an inverse element ε ∈ B exists, the element ∈ B is said to be invertible, so that ε = ε = e. The inverse of in such case is unique and is denoted by −1 . We require the following propositions in the sequel. Lemma 1. [13] Consider the unital Banach algebra B with unit e and let ∈ B be an arbitrary element. Then, lim n→∞ n 1 n exists, and the spectral radius satisfies r( ) = lim n→∞ n 1 n = in f n 1 n .

Definition 3. [6]
Consider the Banach algebra B with unit element e, zero element θ and C = ∅.
If there is M > 0 such that for all , ε ∈ C, we have Then, (ℵ, d) over Banach algebra B with cone metric d is a cone metric space.
In [16], over Banach algebra with constant b ≥ 1 the cone b-metric space is introduced as a generalization of cone metric space over Banach algebra. Definition 5. [16] Let mapping d : ℵ × ℵ → B and ℵ = ∅: (e 1 ). for all , ε ∈ ℵ, θ d( , ε) and d( , ε) = θ if and only if = ε, Then, (ℵ, d) over Banach algebra B with cone b-metric d is cone b-metric space. Note that if we take b = 1, then it reduces to cone metric space over Banach algebra B.
Lemma 4. [18] Consider the Banach algebra B and intC = ∅. Furthermore, consider { n } a c-sequence in B and k ∈ C where k is arbitrary, then {k n } is a c-sequence.
Lemma 5. [14] Consider the Banach algebra B, e their unit element and C = ∅. Let L ∈ B and n = L n . If r(L) < 1, then { n } is a c-sequence.

Lemma 6.
[18] Consider the Banach algebra B and intC = ∅. Let { n } and {ε n } be c-sequences in B. Then, for arbitrary η, ζ ∈ C we have {η n + ζε n } is also a c-sequence.

Results and Discussion
In this section, we introduce the concept of so called cone interval b-metric space over Banach algebra in short CIbMS over BA as a generalization of metric interval space. In the rest of the below discussion, we consider θ as the zero element of the Banach algebra B, C as the cone in B and e as the unit element of B. Definition 7. Let I be the collection of all closed bounded intervals in R, s ≥ 1 a constant and B with a non-normal cone C as a Banach algebra. Suppose that the mapping d : Then, the pair (I, d) is CIbMS over BA B with parameter s ≥ 1. We claim that d satisfies the null equality if the following are satisfied: iv.

Remark 2.
In above definition, if we take the Banach algebra B = R + , we can obtain an interval b-metric space (IbMS) with parameter s ≥ 1. Furthermore, by taking B = R + and s = 1, then we can obtain metric interval space (MIS). By taking s = 1 in the above definition, we can obtain cone interval metric space (CI MS) over BA B.
Remark 3. The class of CIbMS over BA B is larger than the class of MIS and class of (CI MS) over BA B since the latter must be the former, but the converse is not true. We can present an example, as follows, which shows that introducing a CIbMS over BA B instead of a MIS and CI MS over BA B is very meaningful since there exists CIbMS over BA B, which is not MIS and not a CI MS over BA B.
] be the set of continuous functions on the closed interval [a, b] with supremum norm. Define multiplication in the usual way. Then, B is a Banach algebra with a unit 1.
]} and X = I, where I is the collection of all closed bounded intervals in R. Define a mapping d : We are going to claim that [ , ε] Furthermore, we have from [ , ε] and [ρ, ] that ≤ ε and ρ ≤ , and so we have from these that Using (3) in (4), we have From (4) and (5), we can form two identical intervals as and Let Therefore, from (6) and (7), we obtain We have  2]), which implies that 5 2 2 2 + 3 2 for p = 2, that is, 13 − 25 = −12 ∈ C, that implies that the triangle inequality does not hold true, but for the parameter s = 2 p−1 it is a CIbMS over BA B. (iv) Furthermore, d satisfies the null equality, that is, for any [ , ε], [ρ, ] ∈ I and k 1 , That completes the verification. It is easy to verify that (I , d) is a CIbMS over the BA B with parameter s ≥ 5 2 , but it is not a The concept of convergence, Cauchy sequence and completeness in CIbMS over BA B is defined as follows: Since it is not difficult to verify that for ω 1 = [−1, 1], ω 2 = [−2, 2] ∈ Ω we have which shows that [ , ε] is a near-fixed point of T.

Definition 13. Let T, F be two self-mappings of I into itself. A point [ , ε] ∈ I is called a near-common fixed point of T and F if and only if T([ , ε])
Example 6. Let us consider the mappings T, F : I → I defined by To show that [ , ε] is a near-common fixed point of T and F, we have to show that T  Clearly, by previous example, it is seen that T ([ , ε]) . Hence T and F are weakly compatible. Theorem 1. Let (I, d) be a CIbMS over BA B with the parameter s ≥ 1 and C be the underlying solid cone in B. Let a i ∈ C(i = 1, . . . , 5) be constants with 2sr(a 1 ) + (s + 1)r(a 2 + a 3 + sa 4 + sa 5 ) < 2. Suppose that a 1 commutes with a 2 + a 3 + sa 4 + sa 5 and the mappings T, F : On the other hand, we have Add up (9) and (10) yields that Denote a 2 + a 3 + sa 4 + sa 5 = a, then (11) becomes Note that for s ≥ 1, we have 2r(a) ≤ (s + 1)r(a) ≤ 2sr(a 1 ) + (s + 1)r(a) < 2, therefore r(a) < 1 < 2, then by Lemma 1 it follows that 2e − a is invertible. Furthermore, Multiply both sides of (12) by (2e − a) −1 , we arrive at Denote B = (2e − a) −1 (2a 1 + a), then by (13) we get Since a 1 commutes with a, it follows that that is to say, (2e − a) −1 commutes with 2a 1 + a. Then, by Lemmas 2 and 3, we gain [2r(a 1 ) + r(a)] < 1 s , which shows that e − sB is invertible and also by using Remark 1 B n → 0 as m → ∞. Hence, for any n ≥ 1, p ≥ 1 and B ∈ C with r(B) < 1 s , we have that However, since e − sB is invertible, therefore we have which implies that For another thing, As we can see that therefore, by combining (14) and (15), we have Now, we can see r(sa) = sr(a) ≤ (s + 1)r(a) ≤ 2sr(a) + (s + 1)r(a) < 2, thus by Lemma 1, it concludes that 2e − sa is invertible. As a result, it follows from (16) that