KANNAN–TYPE FIXED POINT THEOREM IN CONE PENTAGONAL METRIC SPACES

In this paper, we prove Kannan type fixed point theorem for two self mappings in non-normal cone pentagonal metric spaces. Our results extend and improve the recent results announced by many authors. AMS Subject Classification: 47H10, 54H25


Introduction
A mapping T : X → X on a metric space (X, d) is called Kannan contraction if there exists λ ∈ [0, 1/2) such that d(T x, T y) ≤ λ d(x, T x) + d(y, T y) , for all x, y ∈ X. (1.1) Kannan [10] proved that if X is complete, then every Kannan contraction has a fixed point.
The study of existence and uniqueness of fixed point of a mapping and common fixed points of two or more mappings has become a subject of great interest.Many authors proved the Kannan contraction principle in various generalized metric spaces (e.g., see [2,7,8,11]).
Long-Guang and Xian [7] introduced the concept of a cone metric space and proved some fixed point theorems for contractive type conditions in cone metric spaces.Later on many authors have (for e.g., [1,3,5,6,12]) proved some fixed point theorems for different contractive types conditions in cone metric spaces.
Recently, Garg and Agarwal [6] introduced the notion of cone pentagonal metric space and proved Banach contraction mapping principle in a normal cone pentagonal metric space setting.
Motivated and inspired by the results of [6,11], it is our purpose in this paper to continue the study of common fixed points for a two self mappings in non-normal cone pentagonal metric space setting.Our results extend and improve the results of [2,8,11], and many others.
Given a cone P ⊆ E, we defined a partial ordering ≤ with respect to P by x ≤ y if and only if y − x ∈ P. We shall write x < y to indicate that x ≤ y but x = y, while x ≪ y will stand for y − x ∈ int(P ), where int(P ) denotes the interior of P.
In this paper, we always suppose that E is a real Banach space and P is a cone in E with int(P ) = ∅ and ≤ is a partial ordering with respect to P. Definition 2.2.Let X be a nonempty set.Suppose the mapping ρ : X × X → E satisfies: (1) 0 < ρ(x, y) for all x, y ∈ X and ρ(x, y) = 0 if and only if x = y; (2) ρ(x, y) = ρ(y, x) for all x, y ∈ X; (3) ρ(x, y) ≤ ρ(x, z) + ρ(z, y) for all x, y, z ∈ X.
Then ρ is called a cone metric on X, and (X, ρ) is called a cone metric space.
Then ρ is called a cone rectangular metric on X, and (X, ρ) is called a cone rectangular metric space.
Then d is called a cone pentagonal metric on X, and (X, d) is called a cone pentagonal metric space.
Remark 2.6.Every cone rectangular metric space and so cone metric space is cone pentagonal metric space.The converse is not necessarily true (e.g., see [6]).
Let (X, d) be a cone pentagonal metric space.Let {x n } be a sequence in (X, d) and x ∈ X.If for every c ∈ E with 0 ≪ c there exist n 0 ∈ N and that for all n > n 0 , d(x n , x) ≪ c, then {x n } is said to be convergent and {x n } converges to x, and x is the limit of {x n }.We denote this by lim n→∞ Let T and S be self maps of a nonempty set X.If w = T x = Sx for some x ∈ X, then x is called a coincidence point of T and S and w is called a point of coincidence of T and S. Also, T and S are said to be weakly compatible if they commute at their coincidence points, that is, T x = Sx implies that T Sx = ST x.Lemma 2.7.Let T and S be weakly compatible self mappings of nonempty set X.If T and S have a unique point of coincidence w = T x = Sx, then w is the unique common fixed point of T and S.
Lemma 2.8.Let (X, d) be a cone metric space with cone P not necessary to be normal.Then for a, c, u, v, w ∈ E, we have Lemma 2.9.Let (X, d) be a complete cone pentagonal metric space.Let {x n } be a Cauchy sequence in X and suppose that there is natural number N such that: 1. x n = x m for all n, m > N ; 2. x n , x are distinct points in X for all n > N ; 3. x n , y are distinct points in X for all n > N ; 4. x n → x and x n → y as n → ∞.
Then x = y.

Main Results
In this section, we prove Kannan -type contraction principle in cone pentagonal metric spaces for two self mappings.We give an example to illustrate the result.Theorem 3.1.Let (X, d) be a cone pentagonal metric space.Suppose the mappings f, g : X → X satisfy the contractive condition: for all x, y ∈ X, where λ ∈ [0, 1/2).Suppose that f (X) ⊆ g(X) and g(X) or f (X) is a complete subspace of X, then the mappings f and g have a unique point of coincidence in X.Moreover, if f and g are weakly compatible then f and g have a unique common fixed point in X.
Proof.Let x 0 be an arbitrary point in X.Since f (X) ⊆ g(X), we can choose x 1 ∈ X such that f x 0 = gx 1 .Continuing this process, having chosen x n in X, we obtain x n+1 in X such that That is, f and g have a point of coincidence y k in X.We assume that y n = y n+1 , for all n ∈ N.Then, from (3.1), we have So that, Also from (3.1) and (3.2), we obtain That is, d(y n , y n+2 ) ≤ αr n−1 d(y 0 , y 1 ), ∀n ≥ 1, ( where α = λ(1 + r 2 ) > 0.
Therefore, {y n } is a Cauchy sequence in (X, d).Since g(X) is a complete subspace of X, there exists a points u, v ∈ g(X) such that lim n→∞ y n = v = gu.Now, we show that gu = f u.Given c ≫ 0, we choose a natural numbers 3(1+λ) , ∀n ≥ M 3 .Since x n = x m for n = m, by pentagonal property, we have that where By definition of cone we get that d(gu, f u) = 0, and so gu = f u = v.Hence, v is a point of coincidence of f and g.Similarly, if f (X) is a complete subspace of X the result holds.
Next, we show that v is unique.For suppose v ′ be another point of coincidence of f and g, that is gu Hence, v = v ′ .Since (f, g) is weakly compatible, by Lemma 2.7, v is the unique common fixed point of f and g.This completes the proof of the theorem.
To illustrate Theorem 3.1, we give the following example.Then (X, ρ) is a complete cone pentagonal metric space, but (X, ρ) is not a complete cone rectangular metric space because it lacks the rectangular property: Define a mapping f, g : X → X as follows: Clearly f (X) ⊆ g(X) and g(X) is a complete subspace of X.Also f and g are weakly compatible mappings.Thus, the conditions of Theorem 3.1 holds for all x, y ∈ X, where λ = 1 5 and d ∈ X is the unique common fixed point of the mappings f and g.Corollary 3.3.(see [2]) Let (X, d) be a complete cone pentagonal metric space and P be a normal cone with normal constant k.Suppose the mapping S : X → X satisfies the contractive condition: for all x, y ∈ X, where λ ∈ [0, 1/2).Then 1. S has a unique fixed point in X.
2. For any x ∈ X, the iterative sequence {S n x} converges to the fixed point.
Proof.Take g = I and P be a normal cone in Theorem 3.1.This completes the proof.Corollary 3.4.(see [11]) Let (X, d) be a cone rectangular metric space and P be a normal cone with normal constant k.Suppose the mappings S, g : X → X satisfies the contractive condition: d(Sx, Sy) ≤ λ d(gx, Sx) + d(gy, Sy) , for all x, y ∈ X, where λ ∈ [0, 1/2).Suppose that S(X) ⊆ g(X) and S(X) or g(X) is a complete subspace of X, then the mappings S and g have a unique coincidence point in X.Moreover, if S and g are weakly compatible then S and g have a unique common fixed point in X.
Proof.This follows from Remark 2.6 and Theorem 3.1, where P is a normal cone.
2. For any x ∈ X, the iterative sequence {S n x} converges to the fixed point.
Proof.Take g = I and P be a normal cone in Theorem 3.1 and Remark 2.6.This completes the proof.