Common Fixed Point of Four Mapping With Contractive Modulus on
Cone Banach Space
R.Krishnakumar
a, D.Dhamodharan
b,∗aDepartment of Mathematics, Urumu Dhanalakshmi College, Tiruchirappalli-620019, Tamil Nadu, India.
bDepartment of Mathematics, Jamal Mohamed College (Autonomous), Tiruchirappalli-620020, Tamil Nadu, India.
Abstract
In this paper, we have proved the existence of unique common fixed point of four contractive maps on cone Banach space through an upper semi continuous contractive modulus and weakly compatible maps.
Keywords: Cone Banach Space, Common Fixed Point, Contractive Modulus, Weakly Compatible.
2010 MSC:65D30, 65D32. c2012 MJM. All rights reserved.
1
Introduction
The notion of cone metric space is initiated by Huang and Zhang [3] and also they discussed some properties of the convergence of sequences and proved the fixed point theorems of a contraction mapping for cone metric
spaces; Any mappingTof a complete cone metric spaceXinto itself that satisfies, for some 0 ≤ k < 1, the
inequalityd(Tx,Ty)≤kd(x,y),∀x,y∈ Xhas a unique fixed point. Some fixed theorems in cone Banach space
are proved by Karapinar[5].
In this paper, we investigate the common fixed point theorems with the assumption of weakly compatible and coincidence point of four maps on an upper semi continuous contractive modulus in cone Banach space
Definition 1.1. Let E be the real Banach space. A subset P of E is called a cone if and only if:
i. P is closed, non empty and P6=0
ii. ax+by∈ P for all x,y∈P and non negative real numbers a,b
iii. P∩(−P) ={0}.
Given a coneP⊂E, we define a partial ordering≤with respect to P byx≤yif and only ify−x∈P. We
will writex <yto indicate thatx≤ybutx6=y, whilex,ywill stand fory−x∈intP, whereintPdenotes the
interior ofP. The conePis called normal if there is a numberK>0 such that 0≤x ≤yimplieskxk ≤Kkyk
for allx,y∈E. The least positive number satisfying the above is called the normal constant.
∗Corresponding author.
Example 1.1. [12] Let K>1. be given. Consider the real vector space with
E={ax+b:a,b∈R;x∈[1−1
k, 1]}
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. Let X be a nonempty set. Suppose the mapping d:X×X→E satisfies
i. d(x,y)≥0, and d(x,y) =0if and only if x=y∀x,y∈X,
ii. d(x,y) =d(y,x), ∀x,y∈X,
iii. d(x,y)≤d(x,z) +d(z,y), ∀x,y,z∈X,
Then(X,d)is called a cone metric space(CMS).
Example 1.2. 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|)
whereα≥0is a constant. Then(X,d)is a cone metric space.
Definition 1.3. [5] Let X be a vector space over R. Suppose the mappingk.kc:X→E satisfies
i. kxkc≥0for all x∈X,
ii. kxkc=0if and only if x=0,
iii. kx+ykc≤ kxkc+kykcfor all x,y∈X,
iv. kkxkc=|k|kxkcfor all k∈ R and for all x∈ X, thenk.kcis called a cone norm on X, and the pair(X,k.kc)is
called a cone normed space(CNS).
Remark 1.1. Each Cone normed space is Cone metric space with metric defined by d(x,y) =kx−ykc
Example 1.3. Let X = R2, P = {(x,y) : x ≥ 0,y ≥ 0} ⊂ R2andk(x,y)kc = (a|x|,b|y|),a > 0,b > 0.Then
(X,k.kc)is a cone normed space over R2
Definition 1.4. Let(X,k.kc) be a CNS, x ∈ X and {xn}n≥0 be a sequence in X. Then {xn}n≥0 converges to x whenever for every c∈ E with0E, there is a natural number N∈N such thatkxn−xkcc for all n≥N. It is
denoted bylimn→∞xn=x or xn →x
Definition 1.5. Let (X,k.kc)be a CNS, x ∈ X and{xn}n≥0 be a sequence in X. {xn}n≥0 is a Cauchy sequence whenever for every c∈E with0c, there is a natural number N∈N, such thatkxn−xmkcc for all n,m≥N
Definition 1.6. Let(X,k.kc)be a CNS, x∈X and{xn}n≥0be a sequence in X.(X,k.kc)is a complete cone normed
space if every Cauchy sequence is convergent. Complete cone normed spaces will be called cone Banach spaces.
Lemma 1.1. [5] Let(X,k.kc)be a CNS, P be a normal cone with normal constant K, and{xn}be a sequence in X.
Then
ii. the sequence{xn}is Cauchy if and only ifkxn−xmkc→0as n,m→∞,
iii. the sequence{xn}converges to x and the sequence{yn}converges to y, thenkxn−ynkc→ kx−ykc.
Definition 1.7. Let f and g be two self maps defined on a set X maps f and g are said to be commuting of f gx=g f x for all x∈X
Definition 1.8. Let f and g be two self maps defined on a set X maps f and g are said to be weakly compatible if they commute at coincidence points. that is if f x=gx forall x∈X then f gx=g f x
Definition 1.9. Let f and g be two self maps on set X.If f x=gx,for some x∈X then x is called coincidence point of f and g
Lemma 1.2. Let f and g be weakly compatible self mapping of a set X.If f and g have a unique point of coincidence, that is w= f x=gx then w is the unique common fixed point of f and g.
2
Main Result
Theorem 2.1. Let(X,k.kc)be a Cone Banch space with the norm d(x, 0) =kxkc. Suppose that the mappings P,Q,S
and T are four self maps of(X,k.kc)such that T(X)⊆P(X)and S(X)⊆Q(X)and satisfying
kTy−Sxkc≤akPx−Qykc+b{kPx−Sxkc+kQy−Tykc}+c{kPx−Tykc+kQy−Sxkc} (2.1)
for all x,y∈X, where a,b,c≥0and a+2b+2c<1.suppose that the pairs{P,S}and{Q,T}are weakly compatible, then P,Q,S and T have a unique common fixed point.
Proof. Supposex0is an arbitrary initial point ofXand define the sequence{yn}inXsuch that
y2n =Sx2n=Qx2n+1 y2n+1=Tx2n+1=Px2n+2 By (2.1) implies that
ky2n+1−y2nkc=kTx2n+1−Sx2nkc
≤akPx2n−Qx2n+1kc+b{kPx2n−Sx2nkc+kQx2n−Tx2n+1kc}
+c{kPx2n−Tx2n+1kc+kQx2n+1−Sx2nkc}
≤aky2n−1−y2nkc+b{ky2n−1−y2nkc+ky2n−y2n+1kc}
+c{ky2n−1−y2n+1kc+ky2n−y2nkc}
≤aky2n−1−y2nkc+b{ky2n−1−y2nkc+ky2n−y2n+1kc}
+cky2n−1−y2n+1kc
≤(a+b+c)ky2n−1−y2nkc+ (b+c)ky2n−y2n+1kc
ky2n+1−y2nkc≤ a
+b+c
1−(b+c)ky2n−y2n−1kc
ky2n+1−y2nkc≤hky2n−y2n−1kc
whereh= 1a−+(bb++cc) <1 for alln∈ N
ky2n−y2n+1kc≤hky2n−1−y2nkc
≤h2ky2n−2−y2n−1kc
.. .
For allm>n
kyn−ymkc≤ kyn−yn+1kc+kyn+1−yn+2kc+· · ·+kym−1−ymkc
≤(hn+hn+1+· · ·hm−1)ky0−y1kc
≤hn(1+h+h2+· · ·+hm−1−n)ky0−y1kc
≤ h
n
1−hky0−y1kc
⇒ kyn−ymkc0 asn,m→∞.
Hence{yn}is a Cauchy sequence.
There exists a pointlin(X,k.kc)such that
lim
n→∞{yn}=l, limn→∞S2n =nlim→∞Q2n+1=land limn→∞Tx2n+1=nlim→∞Px2n+2=l that is,
lim
n→∞S2n=nlim→∞Q2n+1=nlim→∞Tx2n+1=nlim→∞Px2n+2=x
∗
SinceT(X)⊆P(X), there exists a pointzinXSuch thatx∗=Pzthen by(1)
kSz−x∗kc≤ kSz−Tx2n−1kc+kTx2n−1−x∗kc
≤akPz−Qx2n−1kc+b{kPz−Szkc+kQx2n−1−Tx2n−1kc}
+c{kPz−Tx2n−1kc+kQx2n−1−Szkc}+kTx2n−1−x∗kc
Taking the limit asn→∞
kSz−x∗kc≤akx∗−x∗kc+b{kx∗−x∗kc+kx∗−Szkc}
+c{kx∗−x∗kc+kx∗−Szkc}+kx∗−x∗kc
≤0+b{kx∗−Szkc+0}+c{0+kx∗−Szkc}+0+ (b+c)kx∗−Szkc
Which is a contraction sincea+2b+2c<1.
therefore Sz=Pz=x∗
SinceS(X)⊆Q(X)there exists a pointw∈Xsuch thatx∗=Qw.
by(1)
kSz−x∗kc≤ kSz−Twkc
≤akPz−Qwkc+b{kPz−Szkc+kQw−Twkc}+c{kPz−Twkc+kQw−Swkc}
≤akx∗−x∗kc+b{kx∗−x∗kc+kx∗−Twkc}+c{kx∗−Twkc+kx∗−x∗kc}
≤0+ +b{0+kx∗−Twkc}+c{kx∗−Twkc+0}
kx∗−Twkc≤(b+c)kx∗−Twkc
which is a contradiction sincea+2b+2c<1.
therefore Tw=Qw=x∗
ThusSz=Pz=Tw=Qw=x∗
SincePandSare weakly compatible maps,
To prove thatx∗is a fixed point ofS
SupposeSx∗ 6=x∗then by (2.1)
kSx∗−x∗kc≤ kSx∗−Tx∗kc
≤akPx∗−Qwkc+b{kPx∗−Sx∗kc+kQw−Twkc}+
+c{kPx∗−Twkc+kQw−Sx∗kc}
≤akSx∗−x∗kc+b{kSx∗−Sx∗kc+kx∗−x∗kc}+
+c{kSx∗−x∗kc+kx∗−Sx∗kc}
≤akSx∗−x∗kc+b{0+0}+2ckSx∗−x∗kc
kSx∗−x∗kc≤(a+2c)kSx∗−x∗kc
Which is a contradiction, Sincea+2b+2c<1.
Sx∗=x∗
HenceSx∗ = Px∗ =x∗Similarly,QandTare weakly compatible maps thenTQw=QTw, that isTx∗ =
Qx∗
To prove thatx∗is a fixed point ofT.
SupposeTx∗ 6=x∗by (2.1)
kTx∗−x∗kc≤ kSx∗−Tx∗kc
≤akPx∗−Qx∗kc+b{kPx∗−Sx∗kc+kQx∗−Tx∗kc}+
+c{kPx∗−Tx∗kc+kQx∗−Sx∗kc}
≤akx∗−Tx∗kc+b{kx∗−x∗kc+kTx∗−Tx∗kc}+
+c{kx∗−Tx∗kc+kTx∗−x∗kc}
≤akTx∗−x∗kc+b{0+0}+2ckTx∗−x∗kc
kTx∗−x∗kc≤(a+2c)kTx∗−x∗kc
which is a contradiction sincea+2b+2c<1.
Tx∗=x∗.
Hence.Tx∗=Qx∗=x∗
ThusSx∗=Px∗=Tx∗=Qx∗=x∗
That is,x∗is a common fixed point ofP,Q,SandT
To prove that the uniqueness ofx∗
Suppose thatx∗andy∗,x∗6=y∗are common fixed points ofP,Q,SandTrespectively, by (2.1) we have,
kx∗−y∗kc≤ kSx∗−Ty∗kc
≤akPx∗−Qy∗kc+b{kPx∗−Sx∗kc+kQy∗−Ty∗kc}+
+c{kPx∗−Ty∗kc+kQy∗−Sx∗kc}
≤akx∗−y∗kc+b{kx∗−x∗kc+ky∗−y∗kc}+c{kx∗−y∗kc+ky∗−x∗kc}
≤akx∗−y∗kc+b{0+0}+c{kx∗−y∗kc+ky∗−x∗kc}
≤(a+2c)kx∗−y∗kc
which is a contradiction. Sincea+2b+2c<1.
therefore x∗=y∗.
Corollary 2.1. Let(X,k.kc)be a Cone Banach space with the norm d(x, 0) =kxkc.. Suppose that the mappings P,S
and T are three self maps of(X,k.kc)such that T(X)⊆P(X)and S(X)⊆P(X)and satisfying
kSx−Tykc≤akPx−Pykc+b{kPx−Sykc+kPx−Tykc}+c{kPx−Tykc+kPy−Sxkc} (1)
for all x,y∈ X, where a,b,c≥0and a+2b+2c<1.suppose that the pairs{P,S}and{P,T}are weakly compatible, then P,S and T have a unique common fixed point.
Proof. The proof of the corollary immediate by takingP=Qin the above theorem (2.1).
Definition 2.10. A function Φ : [0,∞) → [0,∞) is said to be contractive modulus ifΦ : [0,∞) → [0,∞) and
Φ(t)<t for t>0
Definition 2.11. A real valued functionΦdefined on X⊆R is said to be upper semi continuous if lim
n→∞supΦ(tn)≤
Φ(t),for every sequence{tn} ∈X with tn →t as n→∞.
Remark 2.2. It is clear that every continuous function is upper semi continuous but converse may not true.
Theorem 2.2. Let(X,k.kc)be a Cone Banch space with the norm d(x, 0) =kxkc.. Suppose that the mappings P,Q,S
and T are four self maps of(X,k.kc)such that T(X)⊆P(X)and S(X)⊆Q(X)satisfying
kSx−Tykc≤Φ(λ(x,y)), (2.2)
whereΦis an upper semi continuous contractive modulus and
λ(x,y) =max{kPx−Qykc,kPx−Sxkc,kQy−Tykc,1
2{kPx−Tykc+kQy−Sxkc}}.
The pair{S,P}and{T,Q}are weakly compatible. Then P,Q,S and T have a unique common fixed point.
Proof. Let us takex0is an arbitrary point ofXand define a sequence{y2n}inXsuch that
y2n =Sx2n=Qx2n+1 y2n+1=Tx2n+1=Px2n+2 By (2.2) implies that
ky2n−y2n+1kc=kSx2n−Tx2n+1kc
≤Φ(λ(x2n,x2n+1))
≤λ(x2n,x2n+1)
=max{kPx2n−Qx2n+1kc,kPx2n−Sx2nkc,kQx2n+1−Tx2n+1kc,
1
2{kPx2n−Tx2n+1kc+kQx2n+1−Sx2nkc}}
=max{kTx2n−1−Sx2nkc,kTx2n−1−Sx2nkc,kSx2n−Tx2n+1kc,
1
2{kTx2n−1−Tx2n+1kc+kSx2n−Sx2nkc}}
=max{kTx2n−1−Sx2nkc,kTx2n−1−Sx2nkc,kSx2n−Tx2n+1kc,
1
2kTx2n−1−Tx2n+1kc}
=max{ky2n−y2n−1kc,ky2n−y2n+1kc,1
2ky2n−1−y2n+1kc}
SinceΦis an contractive modulus,λ(x2n−x2n+1) =ky2n−y2n+1kcis not possible. Thus,
ky2n−y2n+1kc≤Φ(ky2n−1−y2nkc) (2.3)
SinceΦis an upper semi continuous, contractive modulus. Equation (2.3) implies that the sequence{ky2n+1−
y2nkc}is monotonic decreasing and continuous. There exists a real number, sayr≥0 such that
lim
n→∞ky2n+1−y2nkc=r,
asn→∞equation (2.3)⇒
r≤Φ(r)
which is only possible ifr=0 becauseΦis a contractive modulus. Thus
lim
n→∞ky2n+1−y2nkc=0.
Claim:{y2n}is a Cauchy sequence.
Suppose{y2n}is not a Cauchy sequence.
Then there exists ane>0 and sub sequence{ni}and{mi}such thatmi <ni<mi+1
kymi−ynikc≥e and kymi −yni−1kc≤e (2.4)
e≤ kymi−ynikc≤ kymi−yni−1kc+kyni−1−ynikc
therefore lim
i→∞kymi −ynikc=e
now
e≤ kymi−1−yni−1kc≤ kymi−1−ymikc+kymi−yni−1kc
by taking limiti→∞we get,
lim
i→∞kymi−1−yni−1kc=e from (2.3) and (2.4)
e≤ kymi−ynikc=kSxmi−Txnikc≤Φ(λ(xmi,xni))
where implies
e≤Φ(λ(xmi,xni)) (2.5)
λ(xmi,xni) =max{kPxmi −Qxnikc,kPxmi−Sxmikc,kQxni−Txnikc,
1
2(kPxmi−Txnikc+kQxni−Sxmikc)}
=max{kTxmi−1−Sxni−1kc,kTxmi−1−Sxmikc,kSxni−1−Txnikc,
1
2(kTxmi−1 −Txnikc+kSxni−1−Sxmikc)}
=max{kymi−1−yni−1kc,kymi−1−ymikc,kyni−1−ynikc,
1
2(kymi−1−ynikc+kyni−1−ymikc)}
Taking limit asi→∞, we get
lim
i→∞λ(xmi,xni) =max{e, 0, 0,
1 2(e,e)} lim
i→∞λ(xmi,xni) =e
Therefore from (2.5) we have,e≤Φ(e)
Therefore{y2n}is Cauchy sequence inX
There exits a pointzinXsuch that lim
n→∞y2n =z Thus,
lim
n→∞Sx2n =nlim→∞Qx2n+1=z and lim
n→∞Tx2n+1=nlim→∞Px2n+2=z (i.e) lim
n→∞Sx2n=nlim→∞Qx2n+1=nlim→∞Tx2n+1=nlim→∞Px2n+2=z
T(X)⊆P(X), there exists a pointu∈ Xsuch thatz=Pu
kSu−zkc≤ kSu−Tx2n+1kc+kTx2n+1−zkc
≤Φ(λ(u,x2n+1)) +kTx2n+1−zkc
where
λ(u,x2n+1) =max{kPu−Qx2n+1kc,kPu−Sukc,kQx2n+1−Tx2n+1kc,
1
2(kPu−Tx2n+1kc+kQx2n+1−Sukc)}
=max{kz−Sx2nkc,kz−Sukc,kSx2n−Tx2n+1kc,
1
2(kz−Tx2n+1kc+kSx2n−Sukc)}.
Now taking the limit asn→∞we have,
λ(u,x2n+1) =max{kz−Sukc,kz−Sukc,kSu−Tukc,1
2(kz−Tukc+kz−Sukc)}
=max{kz−Sukc,kz−Sukc,kSu−zkc,1
2(kz−zkc+kz−Sukc)}
=kz−Sukc
Thus
kSu−zkc≤Φ(kSu−zkc) +kz−zkc
=Φ(kSu−zkc)
IfSu6=zthenkSu−zkc>0 and hence asΦis contracive modulus
Φ(kSu−zkc)<kSu−zkcWhich is a contradiction,Su=zso,Pu=Su=z
Souis a coincidence point ifPandS. The pair of mapsSandPare weakly compatibleSPu = PSuthat is
Sz=Pz.
S(X)⊆Q(X), there exists a pointv∈Xsuch thatz=Qv.
Then we have
kz−Tvkc=kSu−Tvkc
≤Φ(λ(u,v))
≤λ(u,v)
=max{kPu−Qvkc,kPu−Sukc,kQv−Tvkc,
1
2(kPu−Tvkc+kQv−Sukc)}
=max{kz−zkc,kz−zkc,kz−Tvkc,
1
2(kz−Tvkc+kz−zkc)}
Thuskz−Tvkc≤Φ(kz−Tvkc).
IfTv∈zthenkz−Tvkc≥0 and hence asΦis contractive modulus
Φ(kz−Tvkc)<kz−Tvkc
Thereforekz−Tvkc<kz−Tvkc
which is a contradiction. ThereforeTv=Qv=z
So,vis a coincidence point ofQandT.
Since the pair of mapsQandTare weakly compatible,QTv=TQv
(i.e)Qz=Tz.
Now show thatzis a fixed point ofS.
We have
kSz−zk=kSz−Tvkc
≤Φ(λ(z,v))
≤λ(z,v)
=max{kPz−Qvkc,kPz−Szkc,kQv−Tvkc,1
2(kPz−Tvkc+kQv−Szkc)}
=max{kSz−zkc,kSz−Szkc,kz−zkc,1
2(kSz−zkc+kz−Szkc)}
=kSz−zkc
ThuskSz−zkc≤Φ(kSz−zkc).
IfSz6=zthenkSz−zkc>0 and hence asΦis contractive modulusΦ(kSz−zkc)<kSz−zkc
which is a contradiction. There exitsSz=z. HenceSz=Pz=z
Show thatzis a fixed point ofT.
We have
kz−Tzkc=kSz−Tzkc
≤Φ(λ(z,z))
≤λ(z,z)
=max{kPz−Qzkc,kPz−Szkc,kQz−Tzkc,1
2(kPz−Tzkc+kQz−Szkc)}
=max{kz−Tzkc,kz−zkc,kTz−Tzkc,1
2(kz−Tzkc+kTz−zkc)}
=kz−Tzkc
Thuskz−Tzkc≤Φ(kz−Tzkc).
Ifz6=Tzthenkz−Tzkc>0 and hence asΦis contractive modulus
Φ(kz−Tzkc)<kz−Tzkc.
which is a contradiction. Hencez=Tz.
ThereforeTz=Qz=z.
ThereforeSz=Pz=Tz=Qz=z.
That iszis common fixed point ofP,Q,SandT.
Uniqueness
we have
kz−wkc=kSz−Twkc
≤Φ(λ(z,w))
≤λ(z,w)
=max{kPz−Qwkc,kPz−Szkc,kQw−Twkc,1
2(kPz−Twkc+kQw−Szkc)}
=max{kz−wkc,kz−zkc,kw−wkc,1
2(kz−wkc+kw−zkc)}
=kz−wkc
Thus,kz−wkc≤Φ(kz−wkc)
Sincez6=w, thenkz−wk>0 and hence asΦis contractive modulus.
Φ(kz−wkc)<kz−wkc
therefore kz−wkc<kz−wkc
which is a contradiction,
therefore z=w
Thuszis the unique common fixed point ofP,Q,SandT.
Corollary 2.2. Let(X,k.kc)be a Cone Banch space with the norm d(x, 0) = kxkc. Suppose that the mappings P,S
and T are three self maps of(X,k.kc)such that T(X)⊆P(X)and S(X)⊆P(X)satisfying
kSx−Tykc≤Φ(λ(x,y)), (2.6)
whereΦis an upper semi continuous contractive modulus and
λ(x,y) =max{kPx−Pykc,kPx−Sxkc,kPy−Tykc,1
2{kPx−Tykc+kPy−Sxkc}}.
The pair{S,P}and{T,P}are weakly compatible. Then P,S and T have a unique common fixed point.
Proof. The proof of the corollary immediate by takingP=Qin the above theorem (2.2).
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Received: July 10, 2016;Accepted: December 23, 2016
UNIVERSITY PRESS
