R E S E A R C H
Open Access
Fixed point theorems on cone
-metric
spaces over Banach algebras and an
application
Tao Wang, Jiandong Yin
*and Qi Yan
*Correspondence: [email protected]
Department of Mathematics, Nanchang University, Nanchang, 330031, P.R. China
Abstract
In this paper, the concept of cone 2-metric spaces over Banach algebras is introduced and some existence and uniqueness theorems of fixed points for some mappings satisfying certain contractive conditions are established. As an application of one of the main results, an example is given at the end of the paper.
MSC: 54H25; 47H40
Keywords: cone 2-metric spaces over Banach algebras; fixed point theorems; spectral radius
1 Introduction
The notion of -metric spaces can be dated back to the work of Gähler, who established some fixed point theorems in -metric spaces in [–]. In , Sharmaet al.in [] investi-gated the existence and uniqueness of the fixed points of a family of mappings in -metric spaces as follows.
Theorem . Let(X,d)be a complete bounded-metric space(namely,there exists a con-stant K such that d(a,b,c)≤K for all a,b,c∈X)and let{Ti}∞i= be a family of mappings
from X to itself.Suppose that there exists a non-negative integer sequence{mi}∞i=such that
for all positive integers i,j and for all x,y,a∈X,we have dTmi
i x,T mj
j y,a
≤αdx,Tmi
i x,a
+dy,Tjmjy,a+βd(x,y,a)
for non-negative constantsα,βwithα+β< .Then{Ti}share a unique fixed point in X.
In , Rhoades [] obtained the following.
Theorem . Let(X,d)be a complete-metric space.Suppose the mapping T:X→X satisfies
d(Tx,Ty,a)≤hmaxd(x,y,a),d(x,Tx,a),d(y,Ty,a),d(x,Ty,a),d(y,Tx,a)
for≤h< .Then T has a unique fixed point x in X and for every x∈X, the
it-erative sequence {Tnx
} converges to x. Besides, the following error estimate formula
holds:
dTnx,x,a
≤ hn
–hd(x,Tx,a).
In , by defining the distanced(x,y) as a vector in an ordered Banach spaceE, Huang and Zhang [] introduced the concept of cone metric spaces and gave the new version of the Banach contraction principle in the setting of cone metric spaces. Since then, a few fixed point results have been proven on cone metric spaces (see [–] for more details). Moreover, Liu and Xu [] introduced the concept of cone metric spaces over Banach algebras and showed that cone metric spaces over Banach algebras are not equivalent to metric spaces. Based on the work of Liu and Xu [], the authors of [, ] obtained some interesting results on cone metric spaces over Banach algebras. By combining the concepts of -metric spaces and cone metric spaces, Singhet al.[] introduced a new space, which is called a cone -metric space and established some fixed point theorems of a contractive mapping on cone -metric spaces.
Enlightened by these thoughts, we define a new space in this paper, which is called the cone -metric space over a Banach algebra and prove some fixed point theorems of some mappings satisfying certain contractive conditions. Our main results extend Theo-rem . and TheoTheo-rem . to cone -metric spaces over Banach algebras. Moreover, some frequently used conditions as the normality of a cone and the boundedness of a metric space are not needed in our results. In addition, the proofs of our results are quite differ-ent from those of some known results since a given pair of elemdiffer-ents in a Banach algebra may not be compared.
2 Preliminaries
LetAalways be a real Banach algebra, that is,Ais a real Banach space in which an op-eration of multiplication is defined, subject to the following properties (for allx,y,z∈A,
a∈R):
(i) (xy)z=x(yz);
(ii) x(y+z) =xy+xzand(x+y)z=xz+yz; (iii) α(xy) = (αx)y=x(αy);
(iv) xy ≤ xy.
Throughout this paper, we shall assume that a Banach algebra has a unit (i.e., a mul-tiplicative identity)esuch thatex=xe=xfor allx∈A. An elementx∈Ais said to be invertible if there is an inverse elementy∈Asuch thatxy=yx=e. The inverse ofxis denoted byx–. For more details, we refer to [].
The following several lemmas about spectral radius are needed in this paper.
Lemma .(see []) LetAbe a Banach algebra with a unit e and x∈A.If the spectral radius r(x)of x is less than,i.e.,
r(x) = lim
n→∞x
nn=inf
n≥x
nn < ,
then e–x is invertible.Actually
(e–x)–=
∞
i=
Remark . Ifr(x) < , thenxn → (n→ ∞).
Lemma .(see []) LetAbe a Banach algebra and let x,y be vectors inA.If x and y
commute,then the following hold:
(i) r(xy)≤r(x)r(y); (ii) r(x+y)≤r(x) +r(y); (iii) |r(x) –r(y)| ≤r(x–y).
Lemma .(see []) LetAbe a Banach algebra and let k be a vector inA.If≤r(k) < ,
then we have
r(e–k)–≤ –r(k)–.
Now let us recall the concepts of cone and partial ordering for the Banach algebraA. A subsetPofAis called a cone ofAif:
(i) Pis non-empty closed and{θ,e} ⊂P;
(ii) αP+βP⊂Pfor all non-negative real numbersα,β; (iii) P=PP⊂P;
(iv) P∩(–P) ={θ},
whereθ denotes the null of the Banach algebraA. For a given coneP⊂A, we can define
a partial orderingwith respect toPbyxyif and only ify–x∈P.x≺ywill stand for
xyandx=y, whilexywill stand fory–x∈intP, whereintPdenotes the interior ofP. IfintP=∅, thenPis called a solid cone.
In the following we always assume thatPis a cone inAwithintP=∅andis the partial ordering with respect toP.
Definition .(see []) LetXbe a non-empty set. Suppose that the mappingd:X×X× X→R+satisfies:
(i) for every pair of distinct pointsx,y∈X, there exists a pointz∈Xsuch that
d(x,y,z)= ;
(ii) d(x,y,z) = if and only if at least two ofx,y,zare equal;
(iii) d(x,y,z) =d(p(x,y,z))for allx,y,z∈Xand for all permutationsp(x,y,z)ofx,y,z; (iv) d(x,y,z)≤d(x,y,w) +d(x,w,z) +d(w,y,z)for allx,y,z,w∈X.
Thendis called a -metric onX, and (X,d) is called a -metric space.
Definition . LetXbe a non-empty set. Suppose that the mappingd:X×X×X→A
satisfies:
(i) for every pair of distinct pointsx,y∈X, there exists a pointz∈Xsuch that
d(x,y,z)=θ;
(ii) θd(x,y,z)for allx,y,z∈Xandd(x,y,z) =θif and only if at least two ofx,y,zare equal;
(iii) d(x,y,z) =d(p(x,y,z))for allx,y,z∈Xand for all permutationsp(x,y,z)ofx,y,z; (iv) d(x,y,z)d(x,y,w) +d(x,w,z) +d(w,y,z)for allx,y,z,w∈X.
Thendis called a cone -metric onX, and (X,d) is called a cone -metric space over the Banach algebraA.
Definition . Let (X,d) be a cone -metric space over the Banach algebraA,x∈X, and let{xn}be a sequence inX. Then:
(i) {xn}converges toxwhenever for eachc∈Awithθcthere is a natural number
Nsuch thatd(xn,x,a)cfor alla∈Xand for alln≥N. We denote it by
limn→∞xn=xorxn→xasn→ ∞.
(ii) {xn}is a Cauchy sequence whenever for eachc∈Awithθcthere is a natural numberNsuch thatd(xn,xm,a)cfor alla∈Xand for alln,m≥N.
(iii) (X,d)is a complete cone-metric space if every Cauchy sequence is convergent inX.
Lemma .(see []) If E is a real Banach space with a solid cone P and ifθuc for
eachθc,then u=θ.
Lemma .(see []) IfAis a real Banach space with a solid cone P and ifxn →
(n→ ∞),then for anyθc,there exists N∈Nsuch that,for any n>N,we have xnc.
Now, we review some facts onc-sequence theories.
Definition .(see [, ]) LetPbe a solid cone in a Banach spaceA. A sequence{xn} ⊂
Pis ac-sequence if for eachθcthere existsN∈Nsuch thatxncfor alln>N.
Lemma .(see []) Let P be a solid cone in a Banach spaceAand let{xn}and{yn}
be sequences in P.If {xn}and {yn} are c-sequences andα,β > ,then{αxn+βyn} is a
c-sequence.
Lemma .(see []) Let P be a solid cone in a Banach spaceAand let{xn}be a sequence
in P.Suppose that k∈P is an arbitrarily given vector and{xn}is a c-sequence in P.Then
{kxn}is a c-sequence.
3 Main results and proofs
In this section, we will give some fixed point theorems in the setting of cone -metric spaces over Banach algebras.
Proposition . Let(X,d)be a complete cone-metric space over the Banach algebraA
and let P be the underlying solid cone inA.Let{xn}be a sequence in X.If{xn}converges to
x∈X,then we have:
(i) {d(xn,x,a)}is ac-sequence for alla∈X.
(ii) For anyp∈N,{d(xn,xn+p,a)}is ac-sequence for alla∈X.
Proof The proof of this proposition is easy, so we omit it here.
Proposition . Let P be a solid cone in the Banach algebraAand let x be a vector inA.
Suppose that k∈P is an arbitrarily given vector and xc for anyθ c,then we have
kxc for anyθc.
Proof Fixcθ, then c
mθ for allm∈N. It is clear thatx c
m for allm∈N. Sokx kc m.
Since kc
m →θ asm→ ∞, there existsM∈Nsuch thatkx kc
m cwhenm>M. This
Theorem . Let(X,d)be a complete cone-metric space over the Banach algebraAand
P be the underlying solid cone and{Ti}∞i=be a family of mappings from X to itself.Suppose
that there exists a non-negative integer sequence{mi}∞i=such that for all positive integers i,
j and for all x,y,a∈X,we have
dTmi
i x,T mj
j y,a
kd
x,Tmi
i x,a
+kd
y,Tjmjy,a+kd(x,y,a) (.)
for k,k,k∈P with r(k) +r(k) +r(k) < and k,k,and kcommute.Then{Ti}∞i=share
a unique fixed point in X.
Proof Writegi=Timi,i= , , . . . . Then for alli,j, by (.), we have
dgi(x),gj(y),a
kd
x,gi(x),a
+kd
y,gj(y),a
+kd(x,y,a).
Pickx∈Xand setxn=gn(xn–),n≥. Then
d(xn+,xn,a) =d
gn+(xn),gn(xn–),a
kd(xn,xn+,a) +kd(xn–,xn,a) +kd(xn,xn–,a),
which together with Lemma . yields
d(xn+,xn,a)kd(xn,xn–,a)
kd(xn–,xn–,a) ..
.
knd(x,x,a),
wherek= (e–k)–(k+k). In this case, for alll<n, one has
d(xn,xn–,xl)kd(xn–,xn–,xl)
.. .
kn–l–d(x
l+,xl,xl).
It means thatd(xn,xn–,xl) =θfor alll<n. Thus, forn>m, we have
d(xn,xm,a)d(xn,xm,xn–) +d(xn,xn–,a) +d(xn–,xm,a)
kn–d(x,x,a) +d(xn–,xm,xn–) +d(xn–,xn–,a) +d(xn–,xm,a)
kn–+kn–d(x,x,a) +d(xn–,xm,a)
.. .
kn–+kn–+· · ·+km+d(x
,x,a) +d(xm+,xm,a)
=e+k+· · ·+kn–m+kmd(x,x,a)
∞
i=
ki kmd(x,x,a)
= (e–k)–kmd(x,x,a).
On the other hand, it follows from Lemma . and Lemma . that
r(k) =r(e–k)–(k+k)≤r
(e–k)–
r(k+k)
≤r(k+k) –r(k)
≤r(k) +r(k) –r(k)
< .
Hence, from Lemma . and the fact that(e–k)–kmd(x
,x,a) → (n→ ∞), it follows that, for anyc∈Awithθc, there existsN∈Nsuch that for anyn>m>N, we have
d(xn,xm,a)(e–k)–kmd(x,x,a)c.
So{xn}is a Cauchy sequence inX. AsXis complete, there existsx∈Xsuch thatxn→x
(n→ ∞). We claim thatxis the unique fixed point ofTifor alli≥. In fact
dgn(x),x,a
dgn(x),x,gm+(xm)
+dgn(x),gm+(xm),a
+dgm+(xm),x,a
d(xm+,x,a) +kd
x,gn(x),x
+kd
xm,gm+(xm),x
+kd(x,xm,x)
+kd
xm,gm+(xm),a
+kd
x,gn(x),a
+kd(xm,x,a),
which implies
(e–k)d
gn(x),x,a
d(xm+,x,a) +kd(xm,xm+,x) +kd(xm,xm+,a) +kd(xm,x,a).
Therefore, it follows from Lemma ., Lemma ., and Proposition . that (e–k)d(gn(x),
x,a)ym, where{ym}is ac-sequence inP. In this case, we have (e–k)d(gn(x),x,a)cfor
anycθ, which together with Proposition . implies thatθd(gn(x),x,a)cfor any
a∈X,n∈N, andcθ ase–kis invertible. Therefore, by Lemma .,d(gn(x),x,a) =θ
for anyn∈N. Namely,gn(x) =xfor anyn∈N.
Suppose thatyis another fixed point ofgninX, then
d(x,y,a)kd(x,x,a) +kd(y,y,a) +kd(x,y,a).
That is (e–k)d(x,y,a)θ, which meansd(x,y,a) =θfor anya∈Xbecause of the invert-ibility ofe–k. Sox=y. That is, the fixed point ofgnis unique. Asx=gn(x) =Tnmn(x), we
haveTn(x) =Tn(Tnmn(x)) =Tnmn(Tn(x)) =gn(Tn(x)). It is easy to see thatTn(x) =x. Similarly,
the uniqueness of the fixed point ofTncan be proven.
this case, we do not need the bounded property of the underlying space (X,d), so Theo-rem . generalizes and extends TheoTheo-rem ..
Corollary . Let (X,d)be a complete cone -metric space over the Banach algebraA
and P be the underlying solid cone and let{Ti}∞i=be a family of mappings from X to itself.
Suppose that there exists a non-negative integer sequence{mi}∞i=such that for all positive
integers i,j and for all x,y,a∈X,we have dTmi
i x,T mj
j y,a
kd(x,y,a)
for some fixed k∈P with r(k) < .Then the{Ti}share a unique fixed point in X.
Corollary . Let(X,d)be a complete cone-metric space over the Banach algebraA
and P be the underlying solid cone and let{Ti}∞i=be a family of mappings from X to itself.
Suppose that there exists a non-negative integer sequence{mi}∞i=such that for all positive
integers i,j and for all x,y,a∈X,we have dTmi
i x,T mj
j y,a
kdx,Tmi
i x,a
+dy,Tjmjy,a
for some fixed k∈P with r(k) <.Then the{Ti}share a unique fixed point in X.
Proof Note thatr(k+k)≤r(k) < according to Lemma ., so the proof is rightward
from Theorem ..
In the following, we will present some notations. Letai,bi,x,y,k∈A. Write: (i) xi≥{ai} ⇔there existsi≥such thatxai.
(ii) i≥{ai}
j≥{bj} ⇔for anyi≥there existsj≥such thataibj. (iii) i≥{ai} x⇔for anyi≥, we haveaix.
(iv) ki≥{ai}=i≥{kai}.
Lemma .(see []) LetAbe a Banach algebra with a unit e,P be a cone inA,andbe
the partial ordering generated by P.Suppose that x is invertible and that x–θ implies
xθ.Letλ∈P.If the spectral radius r(λ)ofλis less than,then for any integer n≥,we
haveλnλe.
Lemma . Let(X,d)be a complete cone-metric space,P be a cone in the Banach algebra
Aandbe the partial ordering generated by P.Let{xn}and{yn}be two sequences in X
withlimn→∞xn=x andlimn→∞yn=y.Then for any cθ,there exists N∈Nsuch that
for any n>N,we have
–cd(xn,yn,a) –d(x,y,a)c
for all a∈X.
Proof The proof is easy, so we omit it here.
Lemma . Let(X,d)be a complete cone-metric space over the Banach algebraAand
() Suppose thatxis invertible and thatx–θ impliesxθ.
() Suppose the mappingT:X→Xsatisfies
d(Tx,Ty,a)kd(x,y,a),d(x,Tx,a),d(y,Ty,a)d(x,Ty,a),d(y,Tx,a) (.)
for an invertiblek∈Pwithr(k) < .
Then for any positive integers n and any x∈X,the following assertions hold:
(i) ≤i,j≤n{d(Tix,Tjx,a)} k
≤i,j≤n{d(Tix,Tjx,a)}.
(ii) ≤i,j≤n{d(Tix
,Tjx,a)} k
≤j≤n{d(x,Tjx,a)}.
(iii) For alli,j= , , . . . ,n,we haved(Tix,Tjx,x) =θ.
(iv) For anyn∈N,
≤i,j≤n
dTix,Tjx,a
(e–k)–(e–k)–+ (e–k)–d(x,Tx,a).
Proof According to (.), for alli,j, ≤i,j≤n, we have
dTix,Tjx,a
=dTTi–x,TTj–x,a
kdTi–x,Tj–x,a
,dTi–x,Tix,a
,dTj–x,Tjx,a
,
dTi–x,Tjx,a
,dTj–x,Tix,a
k
≤i,j≤n
dTix,Tjx,a
.
So (i) is true.
From (i), we know that
≤i,j≤n
dTix,Tjx,a
k
≤i,j≤n
dTix,Tjx,a
,
≤j≤n
dx,Tjx,a
.
We claim that≤i,j≤n{d(Tix
,Tjx,a)} k
≤j≤n{d(x,Tjx,a)}. In fact, for any ≤i,j≤
n, there exist ≤i(),j()≤nsuch thatd(Tix
,Tjx,a)kd(Ti ()
x,Tj ()
x,a) (otherwise
d(Tix
,Tjx,a)kd(x,Tjx,a) for some ≤j≤n, then the claim is true). Also, there exist ≤i(),j()≤nsuch thatd(Ti()x,Tj
()
x,a)kd(Ti ()
x,Tj ()
x,a) (otherwise, by Lemma .,d(Tix
,Tjx,a)k(kd(x,Tjx,a)) =kd(x,Tjx,a)kd(x,Tjx,a) for some ≤j≤n; so the claim is true). As the procedure continues, we obtain
dTix,Tjx,a
kdTi()x,Tj ()
x,a
kdTi()x,Tj ()
x,a
.. .
ktdTi(t)x,Tj (t)
x,a
.
Otherwise, the claim is true. Besides, there exists ≤s <t such that i(s) =i(t), j(s) =
j(t), where i() =i, j()=j. Then we have ksd(Ti(s)x
,Tj (s)
x,a)ktd(Ti (t)
x,Tj (t)
ktd(Ti(s)x
,Tj (s)
x,a), which implies
ks–ktdTi(s)x,Tj (s)
x,a
θ.
Sinceks–ktis invertible, we can obtaind(Ti(s)x
,Tj (s)
x,a) =θby multiplying (ks–kt)–in both sides. This meansd(Tix,Tjx,a) =θ for anyi,j,a, which contradicts the definition of cone -metric spaces over Banach algebras. Hence, our claim is true. Namely, (ii) is proven.
Now, takea=x, for any ≤i,j≤n, by (ii), there exists ≤m≤nsuch that
dTix,Tjx,x
kdx,Tmx,x
=θ,
which gives (iii).
According to (ii) and (iii), for fixed n∈Nand for any ≤i,j≤n, there exists ≤
j(),j(), . . . ,j(t)≤nsuch that
dTix,Tjx,a
kdx,Tj ()
x,a
kdTx,Tj ()
x,a
+d(x,Tx,a) +d
x,Tj ()
x,Tx
=kdTx,Tj ()
x,a
+d(x,Tx,a)
kdx,Tj ()
x,a
+kd(x,Tx,a)
kdTx,Tj ()
x,a
+d(x,Tx,a) +d
x,Tj ()
x,Tx
+kd(x,Tx,a)
=kdTx,Tj ()
x,a
+k+kd(x,Tx,a)
kdx,Tj ()
x,a
+k+kd(x,Tx,a)
.. .
ktdx
,Tj (t)
x,a
+kt–+· · ·+kd(x
,Tx,a)
.. .
There exists ≤s<tsuch thatj(s)=j(t), then
ksdx,Tj (s)
x,a
+ks–+· · ·+kd(x,Tx,a)
ktdx,Tj (t)
x,a
+kt–+· · ·+kd(x,Tx,a),
which implies
ks–ktdx,Tj (s)
x,a
kt–+· · ·+ksd(x,Tx,a).
Hence,
dTix,Tjx,a
ksks–kt–kt–+· · ·+ksd(x,Tx,a) +
On the other hand, by Lemma ., we have
ksks–kt–=kse–kt–sks–=e–kt–s–=
∞
i=
kt–si ∞
i=
ki= (e–k)–,
kt–+· · ·+ks=kse+k+· · ·+kt–s–ks(e–k)–(e–k)–.
Also, we can get (ks–+· · ·+k)(e–k)–in a similar way. So we have
dTix,Tjx,a
(e–k)–(e–k)–+ (e–k)–d(x,Tx,a).
In fact, we can obtaind(x,Tjx,a)((e–k)–(e–k)–+ (e–k)–)d(x,Tx,a) for any
≤j≤nby using the same approach. Then (iv) is given.
Theorem . Let(X,d)be a complete cone-metric space over the Banach algebraAand
P be the underlying solid cone.
(i) Suppose thatxis invertible and thatx–θimpliesxθ. (ii) Assume the mappingT:X→Xsatisfies
d(Tx,Ty,a)kd(x,y,a),d(x,Tx,a),d(y,Ty,a),d(x,Ty,a),d(y,Tx,a) (.)
for an invertiblek∈Pwithr(k) < .
Then T has a unique fixed point x in X and for every x∈X,the iterative sequence{Tnx}
converges to x.Besides,the following error estimate formula holds for any cθ:
dTnx
,x,a
kn(e–k)–(e–k)–+ (e–k)–d(x
,Tx,a) +c.
Proof Take x∈X. Set xn=Tnx, i= , , . . . . For any n, m, n<m, it follows from
Lemma . and (.) that
d(xn,xm,a) =d
Txn–,Tm–n+xn–,a
k
≤i,j≤m–n+
dTixn–,Tjxn–,a
=k
n–≤i,j≤m
dTix,Tjx,a
k
n–≤i,j≤m
dTix,Tjx,a
.. .
kn
≤i,j≤m
dTix,Tjx,a
kn(e–k)–(e–k)–+ (e–k)–d(x,Tx,a).
Hence, by Lemma . and the fact thatkn((e–k)–(e–k)–+ (e–k)–)d(x
m>n>N, we have
d(xn,xm,a)kn
(e–k)–(e–k)–+ (e–k)–d(x,Tx,a)c.
So{xn}is a Cauchy sequence inX. AsXis complete, there existsx∈Xsuch thatxn→x
(n→ ∞). Letm→ ∞and by Lemma ., we have
d(xn,x,a)d(xn,xm,a) +ckn
(e–k)–(e–k)–+ (e–k)–d(x,Tx,a) +c
for alln≥ and anycθ, which gives the error estimate formula. For alla∈X, we have
d(xn+,Tx,a)k
d(xn,x,a),d(xn,xn+,a),d(x,Tx,a),d(xn+,x,a),d(xn,Tx,x)
.
By Proposition ., for anyθcthere existsN∈Nsuch that for alln>N,
kd(xn,x,a)
c
, kd(xn,xn+,a)
c
,
kd(xn+,x,a)
c
, kd(xn,Tx,x)
c
.
Hence, we have d(xn+,Tx,a) c or d(xn+,Tx,a)kd(x,Tx,a) for alln>N. By Lem-ma ., it is easy to see thatd(x,Tx,a)cor (e–k)d(x,Tx,a)c, which together with Proposition . implies thatd(x,Tx,a) =θfor anya∈Xase–kis invertible. Furthermore,
Tx=x.
Suppose thatyis another fixed point ofT inX, then
d(x,y,a)kd(x,y,a),d(x,x,a),d(y,y,a),d(x,y,a),d(y,x,a),
which givesd(x,y,a)θord(x,y,a)kd(x,y,a). Sod(x,y,a) =θ. Thenx=y. Remark . If we replace the Banach algebraAwith the real numbers fieldRin The-orem ., then we get TheThe-orem ., so TheThe-orem . is just a corollary of TheThe-orem .. Moreover, it should be noted that, in a cone -metric space over the Banach algebraA, for any pair of given elements, they may not be comparable, which is the key difficulty for us in proving Theorem .. Fortunately, we found an effective way for it, and our way may be helpful for other researchers to deal with similar problems.
We conclude the paper with an example, which can illustrate the result of Theorem ..
Example . LetA=R. For each (x
,x)∈A,(x,x)=|x|+|x|. The multiplication is defined byxy= (x,x)(y,y) = (xy,xy+xy). ThenAis a Banach algebra with unit
e= (, ). LetP={(x,x)∈R|x,x≥}. ThenPis a cone inA.
Let X={(x, )∈R|≤x≤} ∪ {(,x)∈R|≤x≤}. The mapping is defined by
d(α,α,α) =d(β,β) whereα,α,α∈Xandβ,β∈ {α,α,α}are such thatβ– β=min{α–α,α–α,α–α}and
d
(x, ), (y, )=
|x–y|,|x–y|
d
(,x), (,y)=
|x–y|, |x–y|
,
d
(x, ), (,y)=d
(,y), (x, )=
x+y,x+ y
.
Then (X,d) is a complete cone -metric space over the Banach algebraA. Now we define mappingsTi:X→X(i≥) by
Ti
(x, )=
, i–
i–
i–
x
and
Ti
(,x)=
i–
–i
i–
x,
.
Then it is not difficult to verify that Tii–((x, )) = (,x) and Tii–((,x)) = (x, ). Hence, it follows from [] thatTisatisfies the contractive condition
dTmi
i x,T mj
j y,a
kd
x,Tmi
i x,a
+kd
y,Tjmjy,a+kd(x,y,a)
for allx,y,a∈Xandi≥, wheremi= i– ,k= (, ),k= (, ),k= (, ). Besides,
r(k) =r(k) =,r(k) =. By Theorem ., we see thatTihas a unique fixed point (, )
for alli≥.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Acknowledgements
The authors are grateful to the referees and the editors for valuable comments and suggestions, which have greatly improved the original manuscript.
Received: 18 May 2015 Accepted: 5 November 2015
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