fixed point theorems in normal cone metric space

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FIXED POINT THEOREMS IN NORMAL CONE METRIC SPACE Article · December 2016

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International J. of Math. Sci. & Engg. Appls. (IJMSEA) ISSN 0973-9424, Vol. 10 No. III (December, 2016), pp. 213-224

FIXED POINT THEOREMS IN NORMAL CONE METRIC SPACE R. KRISHNAKUMAR1 AND D. DHAMODHARAN2 1 Department of Mathematics, Urumu Dhanalakshmi College, Tiruchirappalli-620019, India 2 Department of Mathematics, Jamal Mohamed College (Autonomous), Tiruchirappalli-620020, India

Abstract In this paper, we proved some fixed point theorems in complete normal cone metric spaces, which are the generalization of some existing results in the literature.

1. Introduction There exist a number of generalizations of metric spaces, and one of them is the cone metric spaces. The notion of cone metric space is initiated by Huang and Zhang [2] and also they discussed some properties of the convergence of sequences and proved the fixed point theorems of a contraction mappings cone metric spaces. Many authors have studied the existence and uniqueness of strict fixed points for single valued mappings and multivalued mappings in metric spaces [1, 5, 6, 10]. In this paper discuss existence and unique fixed point in complete normal cone metric spaces, which are the generalization of some existing Contraction principle. −−−−−−−−−−−−−−−−−−−−−−−−−−−−

Key Words : Normal cone, Cone metric space, Fixed point. 2000 AMS Subject Classification : Primary 54H25; Secondary 47H09, 47H10. c http: //www.ascent-journals.com

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R. KRISHNAKUMAR & D. DHAMODHARAN

Definition 1.1 : A subset P of E is called a cone if and only if: 1. P is closed, nonempty and P 6= 0 2. ax + by ∈ P for all x, y ∈ P and nonnegative real numbers a, b 3. P ∩ P − = {0}. Given a cone P ⊂ E, we define a partial ordering ≤ with respect to P by x ≤ y if and only if y − x ∈ P . We will write x < y to indicate that x ≤ y but x 6= y, while x, y will stand for y − x ∈ intP , where intP denotes the interior of P . The cone P is called normal if there is a number K > 0 such that 0 ≤ x ≤ y implies kxk ≤ Kkyk for all x, y ∈ E. The least positive number satisfying the above is called the normal constant. The cone P is called regular if every increasing sequence which is bounded from above is convergent. That is, if {xn } is sequence such that x1 ≤ x2 ≤ · · · ≤ xn · · · ≤ y for some y ∈ E, then there is x ∈ E such that kxn − xk → 0 as n → 0. Equivalently the cone P is regular if and only if every decreasing sequence which is bounded from below is convergent. It is well known that a regular cone is a normal cone. Suppose E is a Banach space, P is a cone in E with intP 6= 0 and ≤ is partial ordering with respect to P. Example 1 : Let K > 1 be given. Consider the real vector space with 1 E = {ax + b : a, b ∈ R; x ∈ [1 − , 1]} k with supremum norm and the cone P = {ax + b : a ≥ 0, b ≤ 0} in E. The cone P is regular and so normal. Definition 1.2 : Suppose that E is a real Banach space, then P is a cone in E with intP 6= ∅, and ≤ is partial ordering with respect to P . Let X be a nonempty set, a function d : X × X → E is called a cone metric on X if it satisfies the following conditions with 1. d(x, y) ≥ 0, and d(x, y) = 0 if and only if x = y ∀x, y ∈ X, 2. d(x, y) = d(y, x), ∀x, y ∈ X,

FIXED POINT THEOREMS IN NORMAL CONE METRIC SPACE

215

3. d(x, y) ≤ d(x, z) + d(z, y), ∀x, y, z ∈ X, Then (X, d) is called a cone metric space (CMS). Example 2 : Let E = R2 P = {(x, y) : x, y ≥ 0} X = R and d : X × X → E such that d(x, y) = (|x − y|, α|x − y|) where α ≥ 0 is a constant. Then (X, d) is a cone metric space. Definition 1.3 : Let (X, d) be a CMS and {xn }n≥0 be a sequence in X. Then {xn }n≥0 converges to x in X whenever for every c ∈ E with 0 c, there is a natural number N ∈ N such that d(xn , x) c for all n ≥ N . It is denoted by limn→∞ xn = x or xn → x. Definition 1.4 : Let (X, d) be a CMS and {xn }n≥0 be a sequence in X. {xn }n≥0 is a Cauchy sequence whenever for every c ∈ E with 0 c, there is a natural number N ∈ N, such that d(xn , xm ) c for all n, m ≥ N. Lemma 1.5 : Let (X, d) be a cone metric space, P be a normal cone with normal constant K. Let {xn } be a sequence in X. If {xn } converges to x and {xn } converges to y, then x = y. That is the limit of {xn } is unique. Definition 1.6 : Let (X, d) be a cone metric space, if every Cauchy sequence is convergent in X, then X is called a complete cone metric space. Lemma 1.7 : Let (X, d) be a cone metric space, P be a normal cone with normal constant K. Let {xn } be a sequence in X. Then {xn } is a Cauchy sequence if and only if d(xn , xm ) → 0 (n, m → ∞).

2. Main Result Theorem 2.1 : Let (X, d) be a complete cone metric space and P be a normal cone with normal constant K. Suppose the mapping T : X → X satisfies the following conditions: d(T x, T y) ≤

d(x, T x) + d(y, T y) d(x, y) d(x, T x) + d(y, T y) + l

for all x, y ∈ X, where l ≥ 1. Then

(1)

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(i) T has unique fixed point in X.

(ii) T n x0 converges to a fixed point, for all x0 ∈ X.

Proof : (i) Let x0 ∈ X be arbitrary and choose a sequence {xn } such that xn+1 =T xn . d(xn+1 , xn ) = d(T xn , T xn−1 ) d(x , T x ) + d(x n n n−1 , T xn−1 ) ≤ d(xn , xn−1 ) d(xn , T xn ) + d(xn−1 , T xn−1 ) + l d(x , x n n+1 ) + d(xn−1 , xn ) ≤ d(xn , xn−1 ) d(xn , xn+1 ) + d(xn−1 , xn ) + l Take λn =

d(xn , xn+1 ) + d(xn−1 , xn ) , d(xn , xn+1 ) + d(xn−1 , xn ) + l

we have d(xn+1 , xn ) ≤ λn d(xn , xn−1 ) ≤ (λn λn−1 )d(xn−1 , xn−2 ) .. . ≤ (λn λn−1 · · · λ1 )d(x1 , x0 ). Observe that (λn ) is non increasing, with positive terms. So, λ1 ...λn ≤ λn1 and λn1 → 0. It follows that

lim (λ1 λ2 · · · λn ) = 0.

n→∞

Thus, it is verified that

lim d(xn+1 , xn ) = 0

n→∞


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Now for all m, n ∈ N and m > n we have d(xm , xn ) ≤ d(xn , xn+1 ) + d(xn+1 , xn+2 ) + · · · + d(xm−1 , xm ) ≤ [(λn λn−1 · · · λ1 ) + (λn+1 λn · · · λ1 ) + · · · + (λm−1 λm−2 · · · λ1 )]d(x1 , x0 ) =

m−1 X

(λk λk−1 · · · λ1 )d(x1 , x0 ) k=n m−1 X

kd(xm , xn )k ≤ Kk kd(xm , xn )k ≤ K kd(xm , xn )k ≤ K

(λk λk−1 · · · λ1 )d(x1 , x0 )k

k=n m−1 X

(λk λk−1 · · · λ1 )kd(x1 , x0 )k

k=n m−1 X

ak kd(x1 , x0 )k,

k=n

where ak = (λk λk−1 · · · λ1 ) and K is normal constant of P . ∞ m−1 P P a < 1 and a is finite , and (λk λk−1 · · · λ1 ) → 0, as m, n → ∞. Now lim k+1 k ak k→∞

k=1

k=n

Hence {ak } is convergent by D’Alembert’s ratio test, Therefore {xn } is a Cauchy sequence. There is x0 ∈ X such that xn → x0 as n → ∞. d(T x0 , x0 ) ≤ d(T x0 , T xn ) + d(T xn , x0 ) d(x0 , T x0 ) + d(x , T x ) n n d(xn , x0 ) + d(T xn , x0 ) ≤ d(x0 , T x0 ) + d(xn , T xn ) + l d(x0 , T x0 ) + d(x , x n n+1 ) ≤ d(xn , x0 ) + d(xn+1 , x0 ) d(x0 , T x0 ) + d(xn , xn+1 ) + l d(T x0 , x0 ) ≤ 0 as n → ∞ Therefore kd(x0 , T x0 )k = 0. Thus, T x0 = x0 . Uniqueness Suppose x0 and y 0 are two fixed points of T . d(x0 , y 0 ) = d(T x0 , T y 0 ) d(x0 , T x0 ) + d(y 0 , T y 0 ) ≤ d(x0 , y 0 ) d(x0 , T x0 ) + d(y 0 , T y 0 ) + l ≤0 Therefore kd(x0 , y 0 )k = 0. Thus x0 = y 0 . Hence x0 is an unique fixed point of T.

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(ii) Now d(T n x0 , x0 ) = d(T n−1 (T x0 ), x0 ) = d(T n−1 x0 , x0 ) = d(T n−2 (T x0 ), x0 ) · · · = d(T x0 , x0 ) = 0 Hence T n x0 converges to a fixed point, for all x0 ∈ X.

2

Corollary 2.2 : Let (X, d) be a complete cone metric space and P be a normal cone with normal constant K. Suppose the mapping T : X → X satisfies the following conditions: d(T x, T y) ≤

d(x, T x) + d(y, T y) d(x, y) d(x, T x) + d(y, T y) + 1

(2)

for all x, y ∈ X. Then (i) T has unique fixed point in X. (ii) T n x0 converges to a fixed point, for all x0 ∈ X. Proof : The proof of the corollary immediate by taking l = 1 in the above theorem. 2 Theorem 2.3 : Let (X, d) be a complete metric space and let T be a mapping from X into itself. Suppose that T satisfies the following condition: d(y, T y) d(x, y) d(T x, T y) ≤ d(x, T x) + d(y, T y) + l

(3)

for all x, y ∈ X, where l ≥ 1. Then (i) T has unique fixed point in X. (ii) T n x0 converges to a fixed point, for all x0 ∈ X. Proof : (i) Let x0 ∈ X be arbitrary and choose a sequence {xn } such that xn+1 =T xn . We have d(xn+1 , xn ) = d(T xn , T xn−1 ) d(xn−1 , T xn−1 ) ≤ d(xn , xn−1 ) d(xn , xn+1 ) + d(xn−1 , xn ) + l d(xn−1 , xn ) ≤ d(xn , xn−1 ) d(xn , xn+1 ) + d(xn−1 , xn ) + l d(xn−1 , xn ) ≤ d(xn , xn−1 ). d(xn , xn+1 ) + d(xn−1 , xn ) + l


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Take λn =

d(xn−1 , xn ) , d(xn , xn+1 ) + d(xn−1 , xn ) + l

we have d(xn+1 , xn ) ≤ λn d(xn , xn−1 ) ≤ (λn λn−1 )d(xn−1 , xn−2 ) .. . ≤ (λn λn−1 · · · λ1 )d(x1 , x0 ). Observe that {λn } is non increasing, with positive terms. So, λ1 ...λn ≤ λn1 and λn1 → 0. It follows that lim (λ1 λ2 · · · λn ) = 0.

n→∞

Thus, it is verified that lim d(xn+1 , xn ) = 0.

n→∞

Now for all m, n ∈ N we have d(xm , xn ) ≤ d(xn , xn+1 ) + d(xn+1 , xn+2 ) + · · · + d(xm−1 , xm ) ≤ [(λn λn−1 · · · λ1 ) + (λn+1 λn · · · λ1 ) + · · · + (λm−1 λm−2 · · · λ1 )]d(x1 , x0 ) =

m−1 X

(λk λk−1 · · · λ1 )d(x1 , x0 ) k=n m−1 X

kd(xm , xn )k ≤ Kk kd(xm , xn )k ≤ K

(λk λk−1 · · · λ1 )d(x1 , x0 )k

k=n m−1 X

ak kd(x1 , x0 )k

k=n

where ak = λk λk−1 · · · λ1 and K is normal constant of P . ∞ m−1 P P a ak is finite , and (λk λk−1 · · · λ1 ) → 0, as m, → ∞. Now lim k+1 ak < 1 and k→∞

k=1

k=n

Hence {ak } is convergent by D’Alembert’s ratio test, Therefore {xn } is a Cauchy se-

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quence. There is x0 ∈ X such that xn → x0 d(T x0 , x0 ) ≤ d(T x0 , T xn ) + d(T xn , x0 ) d(xn , T xn ) ≤ d(xn , x0 ) + d(T xn , x0 ) d(x0 , T x0 ) + d(xn , T xn ) + l d(xn , xn+1 ) d(xn , x0 ) + d(xn+1 , x0 ) ≤ d(x0 , T x0 ) + d(xn , xn+1 ) + l d(T x0 , x0 ) ≤ 0 as n → ∞ Therefore kd(x0 , T x0 )k = 0. Thus, T x0 = x0 . Uniqueness Suppose x0 and y 0 are two fixed points of T . d(x0 , y 0 ) = d(T x0 , T y 0 ) d(y 0 , T y 0 ) ≤ d(x0 , y 0 ) d(x0 , T x0 ) + d(y 0 , T y 0 ) + l ≤0 Therefore kd(x0 , y 0 )k = 0. Thus x0 = y 0 . Hence x0 is an unique fixed point of T. (ii) Now d(T n x0 , x0 ) = d(T n−1 (T x0 ), x0 ) = d(T n−1 x0 , x0 ) = d(T n−2 (T x0 ), x0 ) · · · = d(T x0 , x0 ) = 0 Hence T n x0 converges to a fixed point, for all x0 ∈ X.

2

Corolary 2.4 : Let (X, d) be a complete metric space and let T be a mapping from X into itself. Suppose that T satisfies the following condition: d(T x, T y) ≤

d(y, T y) d(x, y) d(x, T x) + d(y, T y) + 1

(4)

for all x, y ∈ X. Then (i) T has unique fixed point in X. (ii) T n x0 converges to a fixed point, for all x0 ∈ X. Proof : The proof of the corollary immediate by taking l = 1 in the above theorem. 2 Theorem 2.5 : Let (X, d) be a complete cone metric space and P be a normal cone with normal constant K. Suppose the mapping T : X → X satisfies the following


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conditions: d(T x, T y) ≤

d(x, T y) + d(y, T x) (d(x, T x) + d(y, T y)) d(x, T x) + d(y, T y) + l

(5)

for all x, y ∈ X, where l ≥ 1. Then (i) T has unique fixed point in X. (ii) T n x0 converges to a fixed point, for all x0 ∈ X. Proof :(i) Let x0 ∈ X be arbitrary and choose a sequence {xn } such that xn+1 =T xn . d(xn , xn+1 ) = d(T xn , T xn−1 ) d(x , T x n n−1 ) + d(xn−1 , T xn ) (d(xn , T xn ) + d(xn−1 , T xn−1 )) ≤ d(xn , T xn ) + d(xn−1 , T xn−1 ) + l d(x , x ) + d(x n n n−1 , xn+1 ) ≤ (d(xn , xn+1 ) + d(xn , xn−1 )) d(xn , xn+1 ) + d(xn−1 , xn ) + l d(xn−1 , xn+1 ) ≤ )(d(xn , xn+1 ) + d(xn , xn−1 )) d(xn , xn+1 ) + d(xn−1 , xn ) + l d(x n−1 , xn ) + d(xn , xn+1 ) ≤ (d(xn , xn+1 ) + d(xn , xn−1 )) d(xn , xn+1 ) + d(xn−1 , xn ) + l Take λn =

d(xn−1 , xn ) + d(xn , xn+1 ) , d(xn , xn+1 ) + d(xn−1 , xn ) + l

we have d(xn+1 , xn ) ≤ λn (d(xn , xn+1 ) + d(xn , xn−1 )) (1 − λn )d(xn+1 , xn ) ≤ λn d(xn , xn−1 ) λn d(xn+1 , xn ) ≤ d(xn , xn−1 ) (1 − λn ) λn λn−1 ≤ d(xn−1 , xn−2 ) (1 − λn )(1 − λn−1 ) .. . λn λn−1 · · · λ1 ≤ d(x1 , x0 ). (1 − λn )(1 − λn−1 ) · · · (1 − λ1 ) ≤ γn d(x1 , x0 ) where γn =

λn λn−1 · · · λ1 (1 − λn )(1 − λn−1 ) · · · (1 − λ1 )

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Observe that {λn } is non increasing, with positive terms. So, λ1 ...λn ≤ λn1 and λn1 → 0. It follows that lim (λ1 λ2 · · · λn ) = 0.

n→∞

Therefore lim γn = 0

n→∞

Thus, it is verified that lim d(xn+1 , xn ) = 0

n→∞

Now for all m, n ∈ N we have d(xm , xn ) ≤ d(xn , xn+1 ) + d(xn+1 , xn+2 ) + · · · + d(xm−1 , xm ) ≤ [γn + γn+1 + · · · + γm−1 ]d(x1 , x0 ) ≤

m−1 X

γk d(x1 , x0 )

k=n m−1 X

kd(xm , xn )k ≤ Kk kd(xm , xn )k ≤ K

γk d(x1 , x0 )k

k=n m−1 X

γk kd(x1 , x0 )k,

k=n

where ak = γk and K is normal constant of P . ∞ m−1 P P a < 0 and Now lim k+1 a is finite. Since γk is convergent by D’Alembert’s ratio k ak k→∞

test, as m → ∞.

k=1

k=n

Therefore {xn } is a Cauchy sequence. There is x0 ∈ X such that xn → x0 as n → ∞. d(T x0 , x0 ) ≤ d(T x0 , T xn ) + d(T xn , x0 ) d(x0 , T x ) + d(x , T x0 ) n n (d(xn , x0 ) + d(T xn , x0 )) ≤ 0 d(x , T xn ) + d(xn , T x0 ) + l d(x0 , x 0 n+1 ) + d(xn , T x ) ≤ (d(xn , x0 ) + d(xn+1 , x0 )) d(x0 , xn+1 ) + d(xn , T x0 ) + l d(T x0 , x0 ) ≤ 0 as n → ∞ Therefore kd(x0 , T x0 )k = 0. Thus, T x0 = x0 .


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Uniqueness Suppose x0 and y 0 are two fixed points of T . d(x0 , y 0 ) = d(T x0 , T y 0 ) d(x0 , T y 0 ) + d(y 0 , T x0 ) ≤ (d(x0 , T x0 ) + d(y 0 , T y 0 )) d(x0 , T x0 ) + d(y 0 , T y 0 ) + l ≤0 Therefore kd(x0 , y 0 )k = 0. Thus x0 = y 0 . Hence x0 is an unique fixed point of T. (ii) Now d(T n x0 , x0 ) = d(T n−1 (T x0 ), x0 ) = d(T n−1 x0 , x0 ) = d(T n−2 (T x0 ), x0 ) · · · = d(T x0 , x0 ) = 0 Hence T n x0 converges to a fixed point, for all x0 ∈ X.

2

Corollary 2.6 : Let (X, d) be a complete cone metric space and P be a normal cone with normal constant K. Suppose the mapping T : X → X satisfies the following conditions: d(T x, T y) ≤

d(x, T y) + d(y, T x) (d(x, T x) + d(y, T y)) d(x, T x) + d(y, T y) + 1

(6)

for all x, y ∈ X. Then (i) T has unique fixed point in X. (ii) T n x0 converges to a fixed point, for all x0 ∈ X. Proof : The proof of the corollary immediate by taking l = 1 in the above theorem. 2

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