# Fixed Point Theorems of Single-Valued Mapping forc-Distance in Cone Metric Spaces

Authors: Zaid Mohammed Fadail; Abd Ghafur Bin Ahmad; Zoran Golubović
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

A new concept of thec-distance in cone metric space has been introduced recently in 2011. The aim of this paper is to extend and generalize some fixed point theorems on c-distance in cone metric space.


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## Body

## 1. Introduction

The concept of cone metric spaces is a generalization of metric spaces, where each pair of points is assigned to a member of a real Banach space with a cone, for new results on cone metric spaces see [1–6]. This cone naturally induces a partial order in the Banach spaces. The concept of cone metric space was introduced in the work of Huang and Zhang [7], where they also established the Banach contraction mapping principle in this space. Then, several authors have studied fixed point problems in cone metric spaces. Some of these works are noted in [8–13].In [14], Cho et al. introduced a new concept of the c-distance in cone metric spaces and proved some fixed point theorems in ordered cone metric spaces. This is more general than the classical Banach contraction mapping principle.In [15], Sintunavarat et al. extended and developed the Banach contraction theorem on c-distance of Cho et al. [14]. They gave some illustrative examples of the main results. Their results improve, generalize, and unify the results of Cho et al. [14] and some results of the fundamental metrical fixed point theorems in the literature.In this paper we proved some fixed point theorems forc-distance in cone metric space. These theorems extend and develop some theorems in literature on c-distance of Cho et al. [14] in cone metric space.The following theorems are the main results given in [7, 14, 16].Theorem 1.1 (see [16]).
Let(X,d) be a complete cone metric space. Suppose that the mapping  f:X→X satisfies the contractive condition:
(1.1)d(fx,fy)⪯kd(x,y),
for all x,y∈X, where k∈[0,1) is a constant. Then f has a unique fixed point in X and for any x∈X, iterative sequence {fnx} converges to the fixed point.Theorem 1.2 (see [7]).
Let(X,d) be a complete cone metric space and  P be a normal cone with normal constant  K. Suppose that the mapping  f:X→X satisfies the contractive condition:
(1.2)d(fx,fy)⪯k(d(fx,x)+d(fy,y)),
for all x,y∈X, where k∈[0,1/2) is a constant. Then f has a unique fixed point in X and for any x∈X, iterative sequence {fnx} converges to the fixed point.Theorem 1.3 (see [7]).
Let(X,d) be a complete cone metric space and  P be a normal cone with normal constant  K. Suppose that the mapping  f:X→X satisfies the contractive condition:
(1.3)d(fx,fy)⪯k(d(fx,y)+d(fy,x)),
for all x,y∈X, where k∈[0,1/2) is a constant. Then f has a unique fixed point in X and for any x∈X, iterative sequence {fnx} converges to the fixed point.Theorem 1.4 (see [14]).
Let(X,⊑) be a partially ordered set and suppose that  (X,d) is a complete cone metric space. Let  q is a c-distance on  X and  f:X→X be a continous and nondecreasing mapping with respect to  ⊑. Suppose that the following two assertions hold: (1)
there existα,β,γ>0 with α+β+γ<1 such that
(1.4)q(fx,fy)⪯αq(x,y)+βq(x,fx)+γq(y,fy),
for all x,y∈X with y⊑x,(2)
there existsx0∈X such that x0⊑fx0.Then f has a fixed point x*∈X. If v=fv then q(v,v)=θ.


## 2. Preliminaries

LetE be a real Banach space and θ denote to the zero element in E. A cone P is a subset of E such that(1)
P is nonempty set closed and P≠{θ},(2)
ifa,b are nonnegative real numbers and x,y∈P then ax+by∈P,(3)
x∈P and -x∈P imply that x=θ.For any coneP⊂E, the partial ordering ⪯ with respect to P is defined by x⪯y if and only if y-x∈P. The notation of ≺ stands for x⪯y but x≠y. Also, we used x≪y to indicate that y-x∈intP, where intP denotes the interior of P. A cone P is called normal if there exists a number K such that
(2.1)θ⪯x⪯y⟹‖x‖≤K‖y‖,
for all x,y∈E. The least positive number K satisfying the above condition is called the normal constant of P.Definition 2.1 (see [7]).
LetX be a nonempty set and E be a real Banach space equipped with the partial ordering ⪯ with respect to the cone P. Suppose that the mapping d:X×X→E satisfies the following conditions: (1)
θ≺d(x,y) for all x,y∈X and d(x,y)=θ if and only if x=y,(2)
d(x,y)=d(y,x) for all x,y∈X,(3)
d(x,y)⪯d(x,y)+d(y,z) for all x,y,z∈X.Then d is called a cone metric on X and (X,d) is called a cone metric space.Definition 2.2 (see [7]).
Let(X,d) be a cone metric space, {xn} be a sequence in X, and x∈X.(1)
For allc∈E with θ≪c, if there exists a positive integer N such that d(xn,x)≪c for all n>N, then xn is said to be convergent and x is the limit of {xn}. We denote this by xn→x.(2)
For allc∈E with θ≪c, if there exists a positive integer N such that d(xn,xm)≪c for all n,m>N, then {xn} is called a Cauchy sequence in X.(3)
A cone metric space(X,d) is called complete if every Cauchy sequence in X is convergent.Lemma 2.3 (see [17]).
(1)
IfE is a real Banach space with a cone P and a⪯λa, where a∈P and 0<λ<1, then a=θ.(2)
Ifc∈intP, θ⪯an and an→θ, then there exists a positive integer N such that an≪c for all n≥N.Next we give the notation ofc-distance on a cone metric space which is a generalization of ω-distance of Kada et al. [18] with some properties.Definition 2.4 (see [14]).
Let(X,d) is a cone metric space. A function q:X×X→E is called a c-distance on X if the following conditions hold:(q1)
θ⪯q(x,y) for all x,y∈X,(q2)
q(x,y)⪯q(x,y)+q(y,z) for all x,y,z∈X,(q3)
for eachx∈X and n≥1, if q(x,yn)⪯u for some u=ux∈P, then q(x,y)⪯u whenever {yn} is a sequence in X converging to a point y∈X,(q4)
for allc∈E with θ≪c, there exists e∈E with θ≪e such that q(z,x)≪e and q(z,y)≪e imply d(x,y)≪c.Example 2.5 (see [14]).
LetE=ℝ and P=(x∈E:x≥0). let X=[0,∞) and define a mapping d:X×X→E by d(x,y)=|x-y| for all x,y∈X. then (X,d) is a cone metric space. define a mapping q:X×X→E by q(x,y)=y for all x,y∈X. Then q is a c-distance on X.Lemma 2.6 (see [14]).
Let(X,d) be a cone metric space and  q is a c-distance on  X. Let  {xn} and  {yn} be sequences in  X and  x,y,z∈X. Suppose that  un is a sequences in  P converging to  θ. Then the following hold: (1)
Ifq(xn,y)⪯un and q(xn,z)⪯un, then y=z,(2)
Ifq(xn,yn)⪯un and q(xn,z)⪯un, then {yn} converges to z,(3)
Ifq(xn,xm)⪯un for m>n, then {xn}  is a Cauchy sequence in X,(4)
Ifq(y,xn)⪯un, then {xn} is a Cauchy sequence in X.Remark 2.7 (see [14]).
(1)
q(x,y)=q(y,x) does not necessarily for all x,y∈X.(2)
q(x,y)=θ is not necessarily equivalent to x=y for all x,y∈X.


## 3. Main Results

In this section we prove some fixed point theorems usingc-distance in cone metric space. In whole paper cone metric space is over nonnormal cone with nonempty interior.Theorem 3.1.
Let(X,d) be a complete cone metric space and  q is a  c-distance on  X. Suppose that the mapping  f:X→X satisfies the contractive condition:
(3.1)q(fx,fy)⪯kq(x,y),
for all x,y∈X, where k∈[0,1) is a constant. Then f has a fixed point x*∈X and for any x∈X, iterative sequence {fnx} converges to the fixed point. If v=fv then q(v,v)=θ. The fixed point is unique.Proof.
Choosex0∈X. Set x1=fx0,x2=fx1=f2x0,…,xn+1=fxn=fn+1x0. We have:
(3.2)q(xn,xn+1)=q(fxn-1,fxn)⪯kq(xn-1,xn)⪯k2q(xn-2,xn-1)⪯⋯⪯knq(x0,x1).
Let m>n≥1. Then it follows that
(3.3)q(xn,xm)⪯q(xn,xn+1)+q(xn+1,xn+2)+⋯+q(xm-1,xm)⪯(kn+kn+1+⋯+km-1)q(x0,x1)⪯kn1-kq(x0,x1).
Thus, Lemma2.6 shows that {xn} is a Cauchy sequence in X. Since X is complete, there exists x*∈X such that xn→x* as n→∞. By (q3) we have:
(3.4)q(xn,x*)⪯kn1-kq(x0,x1).
On the other hand,(3.5)q(xn,fx*)=q(fxn-1,fx*)⪯kq(xn-1,x*)⪯kkn-11-kq(x0,x1)=kn1-kq(x0,x1).
By Lemma2.6 part 1, (3.4) and (3.5), we have x*=fx*. Thus, x* is a fixed point of f.
Suppose thatv=fv, then we have the following: q(v,v)=q(fv,fv)⪯kq(v,v). Since k<1, Lemma 2.3 show that q(v,v)=θ.
Finally suppose there is another fixed pointy* of f, then we have the following: q(x*,y*)=q(fx*,fy*)⪯kq(x*,y*). Since k<1, Lemma 2.3 show that q(x*,y*)=θ and also we have q(x*,x*)=θ. Hence by Lemma 2.6 part 1, x*=y*. Therefore the fixed point is unique.Corollary 3.2.
Let(X,d) be a complete cone metric space and  q is a c-distance on  X. Suppose that the mapping  f:X→X satisfies the contractive condition:
(3.6)q(fnx,fny)⪯kd(x,y),
for all x,y∈X, where k∈[0,1) is a constant. Then f has a unique fixed point x*∈X. If v=fv then q(v,v)=θ.Proof.
From Theorem3.1  fn has a unique fixed point x*. But fn(fx*)=f(fnx*)=f(x*), so f(x*) is also a fixed point of fn. Hence x*=fx*. Thus, x* is a fixed point of f. Since the fixed point of f is also fixed point of fn, the fixed point of f is unique.
Suppose thatv=fv. From above the fixed point of f is also fixed point of fn, then we have the following: q(v,v)=q(fv,fv)=q(fnv,fnv)⪯kq(v,v). Since k<1, Lemma 2.3 show that q(v,v)=θ.The following result is generalized from Theorem1.4. We prove a fixed point theorem and we do not require that X is a partially ordered set.Theorem 3.3.
Let(X,d) be a complete cone metric space and q is a c-distance on X. Suppose that the mapping f:X→X is continuous and satisfies the contractive condition:
(3.7)q(fx,fy)⪯kq(x,y)+lq(x,fx)+rq(y,fy),
for all x,y∈X, where k, l, r are nonnegative real numbers such that k+l+r<1. Then f has a fixed point x*∈X and for any x∈X, iterative sequence {fnx} converges to the fixed point. If v=fv then q(v,v)=θ. The fixed point is unique.Proof.
Choosex0∈X. Set x1=fx0,x2=fx1=f2x0,…,xn+1=fxn=fn+1x0. We have the following:
(3.8)q(xn,xn+1)=q(fxn-1,fxn)⪯kq(xn-1,xn)+lq(xn-1,fxn-1)+rq(xn,fxn)=kq(xn-1,xn)+lq(xn-1,xn)+rq(xn,xn+1).
So
(3.9)q(xn,xn+1)⪯k+l1-rq(xn-1,xn)=hq(xn-1,xn),
where h=(k+l)/(1-r)<1.
Letm>n≥1. Then it follows that
(3.10)q(xn,xm)⪯q(xn,xn+1)+q(xn+1,xn+2)+⋯+q(xm-1,xm)⪯(hn+hn+1+⋯+hm-1)q(x0,x1)⪯hn1-hq(x0,x1).
Thus, Lemma2.6 shows that {xn} is a Cauchy sequence in X. Since X is complete, there exists x*∈X such that xn→x* as n→∞. Since f is continuous, then x*=limxn+1=limf(xn)=f(limxn)=f(x*). Therefore x* is a fixed point of f.
Suppose thatv=fv, then we have
(3.11)q(v,v)=q(fv,fv)⪯kq(v,v)+lq(v,fv)+rq(v,fv)=(k+l+r)q(v,v),
since k+l+r<1, Lemma 2.3 show that q(v,v)=θ.
Finally, suppose that, there is another fixed pointy* of f, then we have the following:
(3.12)q(x*,y*)=q(fx*,fy*)⪯kq(x*,y*)+lq(x*,fx*)+rq(y*,fy*)=kq(x*,y*)+lq(x*,x*)+rq(y*,y*)=kq(x*,y*)⪯kq(x*,y*)+lq(x*,y*)+rq(x*,y*)=(k+l+r)q(x*,y*).
Since k+l+r<1<1, Lemma 2.3 shows that q(x*,y*)=θ and also we have q(x*,x*)=θ. Hence by Lemma 2.6 part 1, x*=y*. Therefore the fixed point is unique.Ifk=0 and r=l, then we have the following result.Corollary 3.4.
Let(X,d) be a complete cone metric space and  q is a c-distance on  X. Suppose that the mapping  f:X→X is continuous and satisfies the contractive condition:
(3.13)q(fx,fy)⪯l(q(x,fx)+q(y,fy)),
for all x,y∈X,where l∈[0,1/2) is a constant. Then f has a fixed point x*∈X and for any x∈X, iterative sequence {fnx} converges to the fixed point. If v=fv then q(v,v)=θ. The fixed point is unique.Finally, we provide another result and we do not require thatf is continuous.Theorem 3.5.
Let(X,d) be a complete cone metric space and  q is a  c-distance on  X. Suppose that the mapping  f:X→X satisfies the contractive condition:
(3.14)(1-r)q(fx,fy)⪯kq(x,fy)+lq(x,fx),
for all x,y∈X, where k,l,r are nonnegative real numbers such that 2k+l+r<1. Then f has a fixed point x*∈X and for any x∈X, iterative sequence {fnx} converges to the fixed point. If v=fv then q(v,v)=θ. The fixed point is unique.Proof.
Choosex0∈X. Set x1=fx0,x2=fx1=f2x0,…,xn+1=fxn=fn+1x0. Observe that
(3.15)(1-r)q(fx,fy)⪯kq(x,fy)+lq(x,fx),
equivalently
(3.16)q(fx,fy)⪯kq(x,fy)+lq(x,fx)+rq(fx,fy).
Then we have:
(3.17)q(xn,xn+1)=q(fxn-1,fxn)⪯kq(xn-1,fxn)+lq(xn-1,fxn-1)+rq(fxn-1,fxn)=kq(xn-1,xn+1)+lq(xn-1,xn)+rq(xn,xn+1)⪯kq(xn-1,xn)+kq(xn,xn+1)+lq(xn-1,xn)+rq(xn,xn+1).
So,
(3.18)q(xn,xn+1)⪯k+l1-k-rq(xn-1,xn)=hq(xn-1,xn),
where h=(k+l)/(1-k-r)<1.
Letm>n≥1. Then it follows that
(3.19)q(xn,xm)⪯q(xn,xn+1)+q(xn+1,xn+2)+⋯+q(xm-1,xm)⪯(hn+hn+1+⋯+hm-1)q(x0,x1)⪯hn1-hq(x0,x1).
Thus, Lemma2.6 shows that {xn} is a Cauchy sequence in X. Since X is complete, there exists x*∈X such that xn→x* as n→∞.
By (q3) we have:(3.20)q(xn,x*)⪯hn1-hq(x0,x1).
On the other hand,(3.21)q(xn,fx*)=q(fxn-1,fx*)⪯kq(xn-1,fx*)+lq(xn-1,fxn-1)+rq(fxn-1,fx*)=kq(xn-1,fx*)+lq(xn-1,xn)+rq(xn,xn+1)⪯kq(xn-1,xn)+kq(xn,fx*)+lq(xn-1,xn)+rq(xn,fx*).
So,
(3.22)q(xn,fx*)⪯k+l1-k-rq(xn-1,xn)⪯k+l1-k-rhn-1q(x0,x1)=hhn-1q(x0,x1)=hnq(x0,x1)⪯hn1-hq(x0,x1).
By Lemma2.6 part 1, (3.20) and (3.22), we have x*=fx*. Thus, x* is a fixed point of f.
Suppose thatv=fv, then we have
(3.23)q(v,v)=q(fv,fv)⪯kq(v,fv)+lq(v,fv)+rq(v,fv)=kq(v,v)+lq(v,v)+rq(v,v)⪯kq(v,v)+kq(v,v)+lq(v,v)+rq(v,v)=(2k+l+r)q(v,v).
Since 2k+l+r<1, Lemma 2.3 shows that q(v,v)=θ.
Finally, suppose that, there is another fixed pointy* of f, then we have
(3.24)q(x*,y*)=q(fx*,fy*)⪯kq(x*,fy*)+lq(x*,fx*)+rq(fx*,fy*)⪯kq(x*,fy*)+kq(x*,fy*)+lq(x*,fx*)+rq(fx*,fy*)=kq(x*,y*)+kq(x*,y*)+lq(x*,x*)+rq(x*,y*)=kq(x*,y*)+kq(x*,y*)+rq(x*,y*)⪯kq(x*,y*)+kq(x*,y*)+lq(x*,y*)+rq(x*,y*)=(2k+l+r)q(x*,y*).
Since2k+l+r<1, Lemma 2.3 shows that q(x*,y*)=θ and also we have q(x*,x*)=θ. Hence by Lemma 2.6 part 1, x*=y*. Therefore the fixed point is unique.Example 3.6.
Consider Example2.5. Define the mapping f:X→X by f(3/4)=1/4 and fx=x/2 for all x∈X with x≠3/4. Since d(f(1),f(3/4))=d(1,3/4), there is not k∈[0,1) such that d(fx,fy)⪯kd(x,y). Since Theorem  2.3 of Rezapour and Hamlbarani [16] cannot be applied to this example on cone metric space. To check this example on c-distance we have:(1)
Ifx=y=3/4, then we have the following.(3.25)q(f(34),f(34))=f(34)=14⪯k34=kq(34,34)withk=23.(2)
Ifx≠y≠3/4, then we have
(3.26)q(fx,fy)=y2⪯kq(x,y)withk=23.(3)
Ifx=3/4,y≠3/4, then we have
(3.27)q(f(34),fy)=y2⪯kq(34,y)withk=23.(4)
Ifx≠3/4,y=3/4, then we have
(3.28)q(fx,f(34))=f(34)=14⪯k34=kq(x,34)withk=23.Hence q(fx,fy)⪯kq(x,y) for all x∈X. Therefore, the condition of Theorem 3.1 are satisfied and then f has a unique fixed point x=0, f(0)=0 with q(0,0)=0.



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