Cone 2-metric space nad Fixed point theorem of contractive mappings

B. Singh, Shishir Jain, Prakash Bhagat

Cone 2-metric space nad Fixed point theorem of contractive mappings

Abstract. In this paper we introduce cone 2-metric space and prove some fixed point theorems of a contractive mapping on a cone 2-metric space.

2000 Mathematics Subject Classification: 47H10, 54H25.

Key words and phrases: Cone 2-metric space, Fixed point, Cone metric space, 2- metric space, Ordered Banach space.

The concept of 2-metric space has been investigated by S. Gahler in series of papers [1],[2],[3]. A 2-metric space is (X, d), where X is a non-empty set and d is a real valued function with domain X × X × X, which satisfies the following properties,

a. d(x, y, z) = 0 if at least two of x, y, z are equal,

b. d(x, y, z) = d(p(x, y, z)) for all x, y, z ∈ X and for all permutations p(x, y, z) of x, y, z.

c. d(x, y, z) ¬ d(x, y, w) + d(x, w, z) + d(w, y, z) for all x, y, z, w ∈ X.

Recently, L.G. Huang and X. Zhang [4] introduced cone metric space by generalized the concept of a metric space, replacing the set of real numbers, by an ordered Banach space and obtained some fixed point theorems for contractive mappings. By using both of the concepts, we define a new space, Cone 2-metric space by replacing real number in 2-metric space by an ordered Banach space.

1. Cone 2-metric space. Let E be a real Banach space and P a subset of E.

P is called a cone if :

1. P is closed, nonempty and P 6= {0}.

2. a, b ∈ R, a, b ­ 0, x, y ∈ P ⇒ ax + by ∈ P .

3. x ∈ P and −x ∈ P ⇒ x = 0.

Given a cone P ⊂ E, we define a partial ordering ¬ in E 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 6= y, while x  y will stand for y − x ∈ intP , intP denotes the interior of P .

The cone is called normal if there exists a number K > 0 such that ∀x, y ∈ P . x ¬ y implies kxk ¬ Kkyk.

The least number satisfying above is called the normal constant of P .

The cone P is called regular if every non-decreasing sequence in P , which is bounded from above is convergent. That is, if {x ⁿ } is sequence such that

x 1 ¬ x ² ¬ · · · ¬ x ⁿ ¬ · · · ¬ y

for some y in E, there exists x ∈ P such that kx ⁿ − xk → 0 (n → ∞). Equivalently the cone P is regular if and only if every decreasing sequence which is bounded from below is convergent. It can be easily proved that a regular cone is a normal cone.

Throughout this paper, we suppose that E is a real Banach space, P is cone in E with intP 6= φ and ¬ is partial ordering in E with respect to P .

Remark 1.1 [5] If E is a real Banach space with cone P and if a ¬ λa where a ∈ P and 0 < λ < 1, then a = 0.

Definition 1.2 : Let X be a nonempty set. Suppose the mapping d : X×X×X → P satisfies:

1. 0 ¬ d(x, y, z), for all x, y, z ∈ X and d(x, y, z) = 0, if and only if at least two of x, y, z are equal.

2. d(x, y, z) = d(p(x, y, z)) for all x, y, z ∈ X and for all permutations p(x, y, z) of x, y, z.

3. d(x, y, z) ¬ d(x, y, w) + d(x, w, z) + d(w, y, z) for all x, y, z, w ∈ X.

Then d is called a cone 2-metric on X, and (X, d) will be called a cone 2-metric space. Cone 2-metric space will be called normal, if the cone P is normal cone.

Example 1.3 Let E = R ² , P = {(x, y) ∈ E | x, y ­ 0} ⊂ R ² , X = R and d : X × X ×X → E and such that d(x, y, z) = (ρ ⁿ , αρ), where ρ = min( |x−y|, |y−z|, |z−x|) and, α and n are some fixed positive integers. Then (X, d) is a cone 2-metric space.

Definition 1.4 Let (X, d) be a cone 2-metric space with respect to a cone P in a real Banach space E. Let {x ⁿ } be a sequence in X and x ∈ X. If for every c ∈ E with 0  c i.e. c ∈ intP there is N such that d(x ⁿ , x, a)  c for all a ∈ X and for all n > N. Then {x ⁿ } is said to be convergent to x. We denote it by

n lim →∞ x n = x or x n → x as n → ∞

Lemma 1.5 Let (X, d) be a cone 2-metric space, P be a normal cone with normal constant K. Let {x ⁿ } be sequence in X. Then {x ⁿ } converges to x ∈ X if and only if d(x n , x, a) → 0 as n → ∞ for all a ∈ X.

Proof Suppose that {x ⁿ } is convergent to x. For every ε > 0 we can choose c ∈ E with 0  c and Kkck < ε. Then there is N, s.t. d(x ⁿ , x, a)  c for all a ∈ X and for all n ­ N. Hence kd(x ⁿ , x, a) k ¬ Kkck < ε for n > N. This means d(x n , x, a) → 0 (n → ∞).

Conversely, suppose that d(x n , x, a) → 0 (n → ∞). For c ∈ intP , there is ε > 0 and N such that kd(x ⁿ , x, a) k < ε, for all a ∈ X and for all n > N. So c − d(x ⁿ , x, a) ∈ intP , i.e. d(x ⁿ , x, a)  c. Hence {x ⁿ } converges to x. 

Lemma 1.6 Let (X, d) be a cone 2-metric space, P be a normal cone with normal constant K. Let {x ⁿ } be a sequence in X. Then limit of {x ⁿ } is unique if it exists.

Proof Assume that sequence {x ⁿ } converges to two points x and y. For any c ∈ intP , there is N such that for all n > N, d(x n , x, a)  c and d(x ⁿ , y, a)  c for all a ∈ X. Now,

d (x, y, a) ¬ d(x, y, x

) + d(x, x

, a) + d(x

, y, a ) ¬ 3c.

Hence kd(x, y, a)k ¬ 3Kkck. Since c is arbitrary d(x, y, a) = 0 for all a ∈ X, 

Definition 1.7 Let (X, d) be a cone 2-metric space. Let {x ⁿ } be a sequence in X. If for every c ∈ intP there is a N such that d(x ⁿ , x m , a)  c for all a ∈ X and for all m, n > N, then {x ⁿ } is said to be a Cauchy sequence in X.

Definition 1.8 Let (X, d) be a cone 2-metric space, if every Cauchy sequence is convergent in X, the X is said to be a complete cone 2-metric space.

Lemma 1.9 Let (X, d) be a cone 2-metric space, {x ⁿ } be a sequence in X. If {x ⁿ } converges to x, then {x ⁿ } is a Cauchy sequence.

Proof For any c ∈ intP , there is N such that d(x ^m , x, a)  c/3 and d(x ⁿ , x, a)  c/3 for all a ∈ X and for all m, n > N. Now for m, n > N,

d(x

, x

, a ) ¬ d(x

, x

, x) + d(x

, x, a) + d(x, x

, a )  c.

Hence {x ⁿ } is a Cauchy sequence. 

Lemma 1.10 Let (X, d) be a cone 2-metric space, P be a normal cone with normal

constant K. Let {x ⁿ } be a sequence in X. Then {x ⁿ } is a Cauchy sequence if and

only if d(x n , x m , a) → 0, (n, m → ∞), for all a ∈ X.

Proof Suppose that {x ⁿ } is a Cauchy sequence. For every ε > 0, choose c ∈ intP and Kkck < ε. Then there is N, such that m, n > N, d(x ^m , x n , a)  c, for all a ∈ X.

So that when m, n, > N, kd(x ^m , x n , a) k ¬ Kkck < ε. That is d(x ⁿ , x m , a) → 0, (n, m → ∞), for all a ∈ X.

Conversely, suppose that d(x _n , x m , a) → 0, (n, m → ∞), for all a ∈ X. For every c ∈ intP , there exists ε > 0 and an integral N, such that for all m, n > N, kd(x ^m , x n , a) k < ε. Hence c − d(x ^m , x n , a) ∈ intP , i.e. d(x ^m , x n , a)  c. Hence

{x ⁿ } is a Cauchy sequence. 

Lemma 1.11 Let (X, d) be a cone 2-metric space, P be a normal cone with normal constant K. Let {x n } and {y ⁿ } be two sequences in X, x ⁿ → x, y ⁿ → y (n → ∞).

Then

d(x n , y n , a) → d(x, y, a) (n → ∞) for all a ∈ X.

Proof For every ε > 0, choose c ∈ intP , such that kck < _8K+4 ^ε . As x n → x and y n → y, there is N such that for all n > N, d(x ⁿ , x, a)  c and d(y ⁿ , y, a)  c, for all a ∈ X. We have

scriptsize

d(x n , y n , a) ¬ d(x ⁿ , y n , x) + d(x n , x, a) + d(x, y n , a)

¬ d(x ⁿ , y n , x) + d(x n , x, a) + d(x, y n , y) + d(x, y, a) + d(y, y n , a)

¬ d(x, y, a) + 4c.

Similarly

d(x, y, a) ¬ d(x, y, x ⁿ ) + d(x, x n , a) + d(x n , y, a)

¬ d(x, y, x ⁿ ) + d(x, x n , a) + d(x n , y, y n ) + d(x n , y n , a) + d(y n , y, a)

¬ d(x ⁿ , y n , a) + 4c.

Hence

0 ¬ d(x, y, a) + 4c − d(x ⁿ , y n , a) ¬ 8c and

kd(x, y, a)−d(x ⁿ , y n , a) k ¬ kd(x, y, a)+4c−d(x ⁿ , y n , a) k+k4ck ¬ (8K +4)kck < ε.

Therefore d(x n , y n , a) → d(x, y, a) (n → ∞) for all a ∈ X. 

Definition 1.12 Let (X, d) be a cone 2-metric space. If for every sequence {x ⁿ } in X, there is a subsequence {x ⁿ

} of {x ⁿ } conveging in X. Then X is called a sequentially compact cone 2-metric space.

2. Main Results.

Theorem 2.1 Let (X, d) be a complete cone 2-metric space, P be a normal cone with normal constant K. Suppose the mapping T : X → X satisfies.

d (T x, T y, a) ¬ κd(x, y, a) + λd(T x, x, a) + µd(T y, y, a),

for all x, y, a ∈ X, (1)

for some fixed κ, λ, µ ∈ [0, 1) with κ + λ + µ < 1. Then T has a unique fixed point in X and for every x ∈ X, the iterative sequence {T ⁿ x } converges to the fixed point.

Proof Choose x 0 ∈ X. Set x ¹ = T x 0 , x 2 = T x 1 = T ² x 0 , · · · , x ⁿ⁺¹ = T x n = T ⁿ⁺¹ x 0 , · · · .

We have

d(x n+1 , x n , a) = d(T x n , T x n −1 , a)

¬ κd(x ⁿ , x n −1 , a) + λd(T x n , x n , a) + µd(T x n −1 , x n −1 , a)

¬ κd(x ⁿ , x n −1 , a) + λd(x n+1 , x n , a) + µd(x n , x n −1 , a)

This implies

(1 − λ)d(x ⁿ⁺¹ , x n , a) ¬ (κ + µ)d(x ⁿ , x n −1 , a)

⇒ d(x n+1 , x n , a) ¬ (κ + µ)

(1 − λ) d(x n , x n −1 , a), for ∀ n ­ 1

= ρd(x n −1 , x n −2 , a), where ρ = (κ + µ)

(1 − λ) < 1 (2)

¬ ρ ² d(x n −1 , x n −2 , a) (3)

.. .

¬ ρ ⁿ d(x 1 , x 0 , a).

Also for k > t, we have

d(x k , x _k−1 , x t ) ¬ ρ d(x k−1 , x _k−2 , x t )

¬ ρ ² d(x k −2 , x k −3 , x t ) .. .

¬ ρ ^{k −t−1} d(x t+1 , x t , x t )

= 0 (4)

Now for any n > m, with using (3) and (4), we have

d(x n , x m , a) ¬ d(x ⁿ , x m , x n −1 ) + d(x n , x n −1 , a) + d(x n −1 , x m , a)

¬ ρ ^{n −1} d(x 1 , x 0 , a) + d(x n −1 , x m , x n −2 ) + d(x n −1 , x n −2 , a) +d(x n −2 , x m , a)

¬ (ρ ^{n −1} + ρ ^{n −2} ) d(x 1 , x 0 , a) + d(x n −2 , x m , a) .. .

¬ (ρ ^{n −1} + ρ ^{n −2} + · · · + ρ ^m+1 ) d(x 1 , x 0 , a) + d(x m+1 , x m , a)

¬ (ρ ^{n −1} + ρ ^{n −2} + · · · + ρ ^m+1 + ρ ^m ) d(x 1 , x 0 , a)

= ρ ^m (1 + ρ + ρ ² + · · · + ρ ^{n −m−1} )d(x 1 , x 0 , a)

¬ _1−ρ ^ρ

d(x 1 , x 0 , a), as ρ < 1 and P is closed

Thus we have kd(x ⁿ , x m , a) k ¬ _1−ρ ^ρ

K kd(x ¹ , x 0 , a) k. This implies d(x ⁿ , x m , a) → 0, (n, m → ∞), for all a ∈ X. Hence {x ⁿ } is a Cauchy sequence. As X is complete, there is x ∈ X such that x ⁿ → x (n → ∞). Now for any a ∈ X, we have

d(T x, x, a) ¬ d(T x, x, T x

) + d(T x, T x

, a) + d(T x

, x, a)

¬ κd(x, x

, x) + λd(T x, x, x) + µd(T x

, x

, x) + κd(x, x

, a) +λd(T x, x, a) + µd(T x

, x

, a) + d(x

, x, a), by (1)

⇒ (1 − λ)d(T x, x, a) ¬ µd(x

, x

, x) + κd(x, x

, a) + µd(x

, x

, a) + d(x

, x, a)

⇒ d(T x, x, a) ¬

{µd(x

, x

, x) + κd(x, x

, a) + µd(x

, x

, a) +d(x

, x, a )}

On taking limit as n → ∞ and by using Lemma 1.10, we obtain that d(T x, x, a) = 0

This implies T x = x. So x is a fixed point for T in X.

Now it check its uniqueness let y be another fixed point of T in X, d(x, y, a) = d(T x, T y, a)

¬ κd(x, y, z) + λd(T x, x, a) + µd(T y, y, a)

⇒ d(x, y, a) ¬ κd(x, y, a)

By using remark 1.1, we obtain that d(x, y, a) = 0 . Therefore fixed point of T in

X is unique. _

On taking λ = µ = 0 in Theorem 2.1, we get the usual Banach contraction principal in the setting of a cone 2-metric space.

Corollary 2.2 Let (X, d) be a complete cone 2-metric space, P be a normal cone with normal constant K. Suppose the mapping T : X → X satisfies the contractive condition

d(T x, T y, a) ¬ kd(x, y, a), for all x, y, a ∈ X,

where k ∈ [0, 1) is a constant. Then T has a unique fixed point in X. Moreover for

any x ∈ X, the iterative sequence {T ⁿ x } converges to the fixed point.

Corollary 2.3 Let (X, d) be a complete cone 2-metric space, P be a normal cone with normal constant K. For c ∈ intP and x ⁰ ∈ X, set B(x ⁰ , c) = {x ∈ X |d(x ⁰ , x, a) ¬ c}. Suppose the mapping T : X → X satisfies the contractive condi- tion

d(T x, T y, a) ¬ kd(x, y, a), for all x, y ∈ B(x ⁰ , c) and for all a ∈ X, where k ∈ [0, 1) is a constant. Then T has a unique fixed point in B(x ⁰ , c).

Proof In view of Corollary 2.2, we only need to prove that B(x 0 , c) is complete and T x ∈ B(x 0 , c) for all x ∈ B(x 0 , c).

Let {x ⁿ } be a Cauchy sequence in B(x 0 , c). Then this sequence is also a Cauchy sequence in X. As X is complete, ∃ x ∈ X such that x ⁿ → x as n → ∞. For any a ∈ X we have

d(x 0 , x, a) ¬ d(x ⁰ , x, x n ) + d(x 0 , x n , a) + d(x n , x, a)

¬ d(x ⁰ , x, x n ) + d(x n , x, a) + c , for all a ∈ X

As x n → x, we have d(x 0 , x, x n ) = d(x n , x, a) = 0. Therefore d(x 0 , x, a) ¬ c and

x ∈ B(x 0 , c). Thus B(x 0 , c) is complete. _

Corollary 2.4 Let (X, d) be a complete cone 2-metric space, P be a normal cone with normal constant K. Suppose the mapping T : X → X with some positive integer n satisfies the contractive condition, i.e.

d(T ⁿ x, T ⁿ y, a) ¬ kd(x, y, a), for all x, y, a ∈ X, where k ∈ [0, 1) is a constant. Then T has a unique fixed point in X.

Proof In view of Corollary 2.2, T ⁿ has a unique fixed point x. Now T ⁿ (T x) = T (T ⁿ x) = T x, so T x is also a fixed point of T ⁿ . As the fixed point of T ⁿ is unique, there fore T x = x. Thus x is a fixed point of T . Uniqueness of fixed point of T is followed by the uniqueness of fixed point of T ⁿ . _

Corollary 2.5 Let (X, d) be a complete cone 2-metric space, P be a normal cone with normal constant K. Suppose the mapping T : X → X satisfies the contractive condition

d(T x, T y, a) ¬ k(d(T x, x, a) + d(T y, y, a)), for all x, y, a ∈ X,

where k ∈ [0, ¹ 2 ) is a constant. Then T has a unique fixed point in X, and for any x ∈ X, iterative sequence {T ⁿ x } converges to the fixed point.

Proof Taking κ = 0 and λ = µ = k in Theorem 2.1 the result follows. _

Theorem 2.6 Let (X, d) be a sequentially compact cone 2-metric space, P be a regular cone. Suppose the mapping T : X → X satisfies the contractive condition

d(T x, T y, a) < d(x, y, a), for all x, y, a ∈ X and x 6= y.

Then T has a unique fixed point in X.

Proof Choose x ₀ ∈ X. Set x 1 = T x ₀ , x 2 = T x ₁ = T ² x 0 , · · · , x n+1 = T x n = T ⁿ⁺¹ x 0 , · · · . If for some n, x ⁿ⁺¹ = x n , then x n is a fixed point of T , and the result follows. So we assume that for all n, x n+1 6= x ⁿ . Taking d n = d(x n , x n+1 , a), for all a ∈ X, we have

d n+1 = d(x n+1 , x n+2 , a) = d(T x n , T x n+1 , a) < d(x n x n+1 , a) = d n for all a ∈ X Therefore d n is a decreasing sequence bounded below by 0. Since P is regular, there is a d ∈ P such that d ⁿ → d (n → ∞.) From the sequence compactness of X, there are subsequence {x ⁿ

} of {x ⁿ } and x ∈ X such that x ⁿ

→ x as i → ∞. We have

d(T x n

, T x, a) < d(x n

, x, a), i = 1, 2, · · · for all a ∈ X.

So

kd(T x ⁿ

, T x, a) k ¬ Kkd(x ⁿ

, x, a) k → 0 (i → ∞), for all a ∈ X

where K is the normal constant of E. Hence T x n

→ T x (i → ∞). Simi- larly T ² x n

→ T x (i → ∞). By using Lemma 1.11, we have d(T x ⁿ

, x n

, a) → d(T x, x, a) (i → ∞) and d(T ² x n

, T x n

, a) → d(T ² x, T x, a) (i → ∞), for all a ∈ X. It is obvious that d(T x ⁿ

, T x, a) = d n

→ d = d(T x, x, .a) (i → ∞) for all a ∈ X. Now we shall prove that T x = x. If T x 6= x, then d 6= 0. Then have

d = d(T x, x, a) > d(T ² x, T x, a) = lim

i→∞ d(T ² x n

, T x n

, a) = lim

i→∞ d n

= d.

We have a contradiction, therefore T x = x. That is x is a fixed point of T . Uniqu-

eness follows easily. _

To illustrate the main result i.e. Theorem 2.1, of this paper an example is given.

Let E = R ² , and P = {(x, y) ∈ R ² |x, y ­ 0} be a normal cone in E. Let X = {(x, 0) ∈ R ² |0 ¬ x ¬ 1}∪{(0, x) ∈ R ² |0 ¬ x ¬ 1}. The mapping d : X ×X ×X → E is defined by

d(α 1 , α 2 , α 3 ) = d 1 (β ₁ , β ₂ ) where α 1 , α 2 , α 3 ∈ X and β 1 , β ₂ ∈ {α ¹ , α 2 , α 3 } are such that ||β 1 − β 2 || = min{||α ¹ − α ² ||, ||α ² − α ³ ||, ||α ³ − α ¹ ||} and

d 1 ((x, 0), (y, 0)) = ⁵ ₄ |x − y|, |x − y|
d 1 ((0, x), (0, y)) = |x − y|, ³ 4 |x − y|

d 1 ((x, 0), (0, y)) = d 1 ((0, y), (x, 0)) = ⁵ ₄ x + y, x + ³ ₄ y)
Then (X, d) is a complete cone 2-metric space.

Let mapping T : X → X with

T ((x, 0)) =

 0, 1

12 x



and T ((0, x)) =

 1 18 x, 0



Then T satisfies the contractive condition

d(T x, T y, a) ¬ κd(x, y, a) + λd(T x, x, a) + µd(T y, y, a), for all x, y, a ∈ X, with constant κ = 1/4, λ = µ = 1/3. It is obvious that T has a fixed point (0, 0) ∈ E.

On the other hand, we see that T is not a contractive mapping in the 2-metric space on X.

References

[1] S.G¨ahler, 2-metricsche R¨aume und ihre topologische strukture, Math. Nachr. 26 (1963), 115- 148.

[2] S.G¨ahler, ¨ Uber die Uniformisierbarkeit 2-metricsche R¨aume, Math. Nachr. 28 (1965), 235- 244.

[3] S.G¨ahler, Zur geometric 2-metricsche R¨aume, Revne Roumaine der Mathem. Pures et Ap- pliques, 11 (1966), 665-667.

[4] L. G. Huang, X. Zhang, Cone metric spaces and fixed point theorems of contractive mappings, J. Math. Anal. Appl. 332 (2007), 1468-1476.

[5] D. ˜Ilic, V. Rakocevic, Quasi-contraction on a cone metric space, Applied Mathematics Letters 22 (2009), 728- 731.

[6] Shobha Jain, Shishir Jain and Lal Bahadur, Compatibility and weak compatibility for four self maps in cone metric spaces, Bulletin of Mathematical Analysis and Application 2 (2010), 15-24.

[7] Shobha Jain, Shishir Jain and Lal Bahadur, Weakly compatibile maps in cone metric spaces, Rendiconti Del Semnario matematico 3 (2010), 13-23.

B. Singh

S. S. in Mathematics, Vikram University Ujjain (M.P.), India

Shishir Jain

Shri Vaishnav Institute of Technology and Science Indore (M.P.), India

E-mail: jainshishir11@rediffmail.com Prakash Bhagat

Shri Vaishnav Institute of Technology and Science Indore (M.P.), India

E-mail: prakash svits@ymail.com

(Received: 11.02.2012)