Non-Unique Fixed Point Results in Extended B-Metric Space †

In this paper, we investigate the existence of fixed points that are not necessarily unique in the setting of extended b-metric space. We state some examples to illustrate our results.


Introduction and Preliminaries
Metric fixed point theory was initiated by the elegant results of Banach, the contraction mapping principle, and all researchers in this area agree on this.He formulated that every contraction in a complete metric space possesses a unique fixed point.Researchers have generalized this result by refining the contraction condition and/or by changing the metric space with a refined abstract space.One interesting generalization of metric space is b-metric space, formulated recently by Czerwik [1].Following this result on b-metric space, several authors have reported a number of fixed point results in the framework of b-metric space (see, e.g., [2][3][4][5][6][7][8][9] and related references therein).
Definition 1 (Czerwik [1]).For a non-empty set X, a function m b : X × X → R + 0 is said to be b-metric if it satisfies the following conditions: In addition, the pair (X, m b ) is called a b-metric space, in short, bMS.

One of the standard examples of b-metric is the following:
Example 1.Let X = R be the set of real numbers and m b (x, y) = (x − y) 2 .Then m b is a b-metric on R with s = 2.It is clear that m b is not a metric on R.
Remark 1.It is worth mentioning that, for s = 1, the b-metric becomes a usual metric.
Recently, Kamran et al. [10] introduced a new type of generalized b-metric space.Furthermore, they observed the analog of a Banach contraction mapping principle in the framework of this new space.Definition 2. [10] Let θ : X × X → [1, ∞) be a mapping.For a non-empty set X, a function d θ : X × X → [0, ∞) is said to be an extended b-metric if it satisfies the following state of affairs for all ξ, η, ζ ∈ X.In addition, the pair (X, d θ ) is called an extended b-metric space, in short extended-bMS.
Obviously, (d θ 1) and (d θ 2) hold.For (d θ 3), we have In conclusion, for any ξ, η, ζ ∈ X, the third axiom is satisfied.Accordingly, (X, d θ ) is an extended b-metric space.Notice also that the standard triangle inequality is not satisfied for the following case Hence, (X, d) does not form a standard metric space.
In an extended-bMS, it is possible to obtain an analogy of basic topological notions, such as convergence, Cauchy sequences, and completeness.For more details, see, e.g., [10].Definition 3. [10] Let (X, d θ ) be an extended-bMS.

(i)
We say that a sequence ξ n in X converges to ξ ∈ X, if for every > 0 there exists N = N( ) ∈ N such that d θ (ξ n , ξ) < , for all n ≥ N, and it is denoted as lim n→∞ ξ n = ξ.

(ii)
We say that a sequence ξ n in X is Cauchy if, for every > 0, there exists N = N( ) In what follows, we recollect the notion of completeness: Definition 4. [10].An extended-bmetric space (X, d θ ) is complete if every Cauchy sequence in X is convergent.
Lemma 1. [10] Suppose that the pair (X, d θ ) is a complete extended-bMS, where d θ is continuous.Then every convergent sequence has a unique limit.
We say that the set O(ξ; ∞) is the orbit of T. Definition 6. Suppose that the pair (X, Remark 3. It is evident that the orbital continuity of T yields orbital continuity of any iterative power of T, that is, orbital continuity of T m for any m ∈ N. Definition 7. [12] Suppose that T is a self-mapping on a non-empty set X. Let α : X × X → [0, ∞) be a mapping.Then T is called an α-orbital admissible if, for all ξ ∈ X, we have ( Remark 4. We note that any α-admissible mapping is also an α-orbital admissible mapping.(see, e.g., [12]).

Main Results
Throughout the paper, we shall assume that d θ is a continuous functional.
Lemma 2. [13] Let (X, d θ ) be an extended b-metric space.If there exists q ∈ [0, 1) such that the sequence {x n }, for any n ∈ N, then the sequence {x n } is Cauchy in X.
Proof.Let {x n } n∈N be a given sequence.By employing Inequality (3), recursively, we derive that Since q ∈ [0, 1), we find that lim On the other hand, by (d θ 3), together with triangular inequality, for p ≥ 1, we derive that Notice that the inequality above is dominated by On the other hand, by employing the ratio test, we conclude that the series which is why we obtain the desired result.Thus, we have Consequently, we observe for n ≤ 1, p ≤ 1 that Letting n → ∞ in Equation ( 7), we conclude that the constructive sequence {x n } is Cauchy in the extended b-metric space (X, d θ ).Lemma 3. Let T : X → X be an α-orbital admissible mapping and x n = Tx n−1 , n ∈ N. If there exists x 0 ∈ X such that α(x 0 , Tx 0 ) ≥ 1, then we have Proof.By assumption, there exists a point x 0 ∈ X such that α(x 0 , Tx 0 ) ≥ 1.On account of the definition of {x n } ⊂ X and owing to the fact that T is α-orbital admissible, we derive Recursively, we have Theorem 2. Suppose that T is an orbitally continuous self-mapping on the T-orbitally complete extended-bMS (X, d θ ).Assume that there exists k ∈ [0, 1) and a ≥ 1 such that α(x, y) min{d θ (Tx, Ty), d θ (x, Tx), d θ (y, Ty)} − a min{d θ (x, Ty), d θ (Tx, y)} ≤ kd θ (x, y)) for all x, y ∈ X.Furthermore, we presume that (i) T is α-orbital admissible; (ii) there exists x 0 ∈ X such that α(x 0 , Tx 0 ) ≥ 1; (iii) Then, for each x 0 ∈ X, the sequence {T n x 0 } n∈N converges to a fixed point of T.

Proof.
By assumption (ii), there exists a point x 0 ∈ X such that α(x 0 , Tx 0 ) ≥ 1.We construct the sequence {x n } in X such that If x n 0 = x n 0 +1 = Tx n 0 for some n 0 ∈ N 0 , then x * = x n 0 forms a fixed point for T that the proof finishes.Hence, from now on, we assume that On account of the assumptions (i) and (ii), together with Lemma (3), Inequality ( 8) is yielded, that is, By replacing x = x n−1 and y = x n in Inequality ( 9) and taking Equation ( 12) into account, we find that min {d θ (Tx or, min {d θ (x n , x n+1 ), Since k ∈ [0, 1), the case d θ (x n−1 , x n ) ≤ kd θ (x n−1 , x n ) is impossible.Thus, we conclude that On account of Lemma 2, we find that the sequence {x n } is a Cauchy sequence.By completeness of (X, d θ ), the sequence x n converges to some point u ∈ X as n → ∞.Owing to the construction x n = T n x 0 and the fact that (X, d θ ) is T-orbitally complete, there is u ∈ X such that x n → u.Since T, is orbital continuity, we deduce that x n → Tu.Accordingly, we conclude that u = Tu.
Let us first notice that for any x ∈ {1, 2, 3, 4}, the sequence {T n x} tends to 1 when n → ∞.For this reason, we can conclude that the mapping T is orbitally continuous and lim n,m→∞ θ(T n x, T m x) = 3 < 4 = 1 k , so (iii) is satisfied.It can also be easily verified that T is orbital admissible.If x = 1 or y = 1, then d(1, T1) = 0 so Inequality (9) holds.We have to consider the following cases.
Therefore, all the conditions of Theorem 2 are satisfied and T has a fixed point, x = 1.
In Theorem 2, if we presume that α(x, y) = 1 and θ(x, y) = 1, then we deduce the renowned non-unique fixed point theorem of Ćirić [14] as follows: Corollary 1. [ Ćirić [14]] Suppose that T is an orbitally continuous self-map on the T-orbitally complete standard metric space (X, d).We presume that there is a k ∈ [0, 1) such that min{d(Tx, Ty), d(x, Tx), d(y, Ty)} − min{d(x, Ty), d(Tx, y)} ≤ kd(x, y) for all x, y ∈ X.Then, for each x 0 ∈ X, the sequence {T n x 0 } n∈N converges to a fixed point of T. Theorem 3. Suppose that T is an orbitally continuous self-map on the T-orbitally complete extended-bMS (X, d).We presume that there exists k ∈ [0, 1) such that α(x, y)Γ(x, y) ≤ kd θ (x, y) for all x, y ∈ X, where where R(x, y) = 0. Furthermore, we assume that

(i)
T is α-orbital admissible; (ii) there exists x 0 ∈ X such that α(x 0 , Tx 0 ) ≥ 1; (iii) Then, for each x 0 ∈ X, the sequence {T n x 0 } n∈N converges to a fixed point of T.
Proof.As a first step, we construct an iterative sequence {x n } as in the proof of Theorem 2. For this purpose, we take an arbitrary initial value x ∈ X and define the following recursion: x 0 := x and x n = Tx n−1 for all n ∈ N.
We also suppose that as is discussed in the proof of Theorem 2. For x = x n−1 and y = x n , Inequality ( 16) becomes (taking into account Lemma (3)) where We obtain that which is a contraction, since k ∈ [0, 1).Consequently, we deduce that Applying Equation ( 22) recurrently, we find that The rest of the proof is a verbatim restatement of the related lines in the proof of Theorem 2.
Then, for each x 0 ∈ X, the sequence {T n x 0 } n∈N converges to a fixed point of T.
Proof.Basically, we shall use the same technique that was used in the proof of Theorem 2. We built a recursive {x n }, x 0 := x and x n = Tx n−1 for all n ∈ N (25) for an arbitrary initial value x ∈ X. Regarding the discussion in the proof of Theorem 2, we presume that For x = x n−1 and y = x n , Inequality (24) becomes (taking into account Lemma 3) where a contraction, since k ∈ [0, 1).Accordingly, we conclude that Recursively, we derive that By following the related lines in the proof of Theorem 2, we complete the proof.
Then, for each x 0 ∈ X, the sequence {T n x 0 } n∈N converges to a fixed point of T.
Proof.As a first step, we shall construct an recursive sequence {x n = Tx n−1 } n∈N , for an arbitrary initial value x 0 := x ∈ X, as in the proof of Theorem 2. By following the same steps in the proof of Theorem 2, we deduce that adjacent terms of the sequence {x n } should be chosen distinct, that is, x n = x n−1 for all n ∈ N.

(m b 1 )
m b (x, y) = 0 if and only if x = y.(m b 2) m b (x, y) = m b (y, x) for all x, y ∈ X. (m b 3) m b (x, y) ≤ s[m b (x, z) + m b (z, y)]for all x, y, z ∈ X, where s ≥ 1.