# Fixed Point Theorems of Quasicontractions on Cone Metric Spaces with Banach Algebras

Authors: Hao Liu; Shaoyuan Xu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We introduce the concept of quasicontractions on cone metric spaces with Banach algebras, and by a new method of proof, we will prove the existence and uniqueness of fixed points of such mappings. The main result generalizes the well-known theorem of Ćirić (Ćirić 1974).


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## Body

## 1. Introduction

Let(X,d)be a complete metric space. Recall that a mappingT:X→Xis called a quasicontraction if, for somek∈(0,1)and for allx,y∈X, one has
(1)d(Tx,Ty)⩽kmax{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)}.
Ćirić [1] introduced and studied quasicontractions as one of the most general classes of contractive-type mappings. He proved the well-known theorem that any quasicontractionThas a unique fixed point. Recently, scholars obtained various similar results on cone metric spaces. See, for instance, [2–5].In this paper, we study the quasicontractions on metric spaces with Banach algebras, which are introduced in [6] and turn out to be an interesting generalization of classic metric spaces. By a new method of proof, we generalize Ćirić theorem.LetAalways be a real Banach algebra with a multiplication unite; that is,ex=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-1. For more details, we refer to [7].The following proposition is well known (see [7]).Proposition 1 (see [7]).
LetAbe a Banach algebra with a unite, and let x∈A. If the spectral radiusρ(x)ofxis less than 1, that is,
(2)ρ(x)=limn→∞∥xn∥1/n=infn⩾1∥xn∥1/n<1,
thene-xis invertible. Actually,
(3)(e-x)-1=∑i=0∞‍xi.A subsetPofAis called a cone if (1)
Pis nonempty closed and{0,e}⊂P;(2)
αP+βP⊂Pfor all nonnegative real numbersα,β;(3)
P2=PP⊂P;(4)
P∩(-P)={0}.For a given coneP⊂A, we can define a partial ordering⩽with respect toPbyx⩽yif and only if y-x∈P. And x≨y will stand forx⩽yandx≠y, whilex<ywill stand for y-x∈intP, where intP denotes the interior ofP.Remark 2.
In the literature on cone metric spaces, authors usex<yto meanx⩽yandx≠y andx≪yto meany-x∈intP. To our knowledge, and from a topological point of view, the order relationy-x∈intPplays a very similar role in cone metric spaces asx<ydoes inℝ.The conePis called normal if there is a numberM>0such that for allx,y∈A,
(4)0⩽x⩽y⟹∥x∥⩽M∥y∥.
The least positive number satisfying above is called the normal constant ofP(see [8]).In the following, we always assume thatPis a cone inAwith intP≠∅ and⩽is partial ordering with respect toP.Definition 3 (see [8]).
LetXbe a nonempty set. Suppose the mappingd:X×X→Asatisfies (1)
0⩽d(x,y)for allx,y∈Xandd(x,y)=0if and only ifx=y;(2)
d(x,y)=d(y,x)for allx,y∈X;(3)
d(x,y)⩽d(x,z)+d(z,x)for allx,y,z∈X.Then, d is called a cone metric onX, and(X,d)is called a cone metric space (with Banach algebraA).For more details about cone metric spaces with Banach algebras, we refer the readers to [6].Definition 4 (see [8]).
Let(X,d)be a cone metric space, and let x∈Xand{xn} be a sequence inX. Then, (1)
{xn}converges toxwhenever for each c∈A with 0<c there is a natural numberNsuch that d(xn,x)<c for all n⩾N. We denote this by limn→∞xn=x or xn→x;(2)
{xn}is a Cauchy sequence whenever for each c∈A with 0<c there is a natural number N such that d(xn,xm)<c for all n,m⩾N;(3)
(X,d)is a complete cone metric space if every Cauchy sequence is convergent.The following facts are often used.Proposition 5 (see [8]).
Let(X,d)be a cone metric space, let Pbe a normal cone with normal constantM, and let{xn}be a sequence inX. Then, {xn}converges toxif and only ifd(xn,x)→0(n→∞).Proposition 6 (see [8]).
Let(X,d)be a cone metric space, let Pbe a normal cone with normal constantM, and let{xn}be a sequence inX. Then, {xn}is a Cauchy sequence if and only if d(xn,xm)→0(n,m→∞).


## 2. Main Results

In this section we will define quasicontractions in the setting of cone metric spaces with Banach algebras and prove the fixed point theorem of such mappings.Definition 7.
Let(X,d)be a cone metric space with Banach algebraA. A mappingT:X→Xis called a quasicontraction if for somek∈Pwithρ(k)<1and for allx,y∈X, one has
(5)d(Tx,Ty)⩽ku,
where
(6)u∈{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)}.Remark 8.
In Definition7, we only suppose the spectral radius ofkis less than 1, while neitherk<enor∥k∥<1is assumed. In fact, the conditionρ(k)<1is weaker than that∥k∥<1. See the example in [6].Theorem 9.
Let(X,d)be a complete cone metric space with a Banach algebraA, and let Pbe a normal cone with normal constantM. If the mappingT:X→Xis a quasicontraction, thenThas a unique fixed point inX. And for anyx∈X, iterative sequence{Tnx}converges to the fixed point.In the rest of the paper, we choosex0∈X and denotexn=Tnx0. For the sake of clarity, we divide the proof into several steps.Lemma 10.
Assume that the hypotheses in Theorem9 are satisfied. Then, for eachn⩾1, and for alli,jsuch that1⩽i,j⩽n, one has
(7)d(xi,xj)⩽k(e-k)-1d(x0,x1).Proof.
We present the proof by induction.
Whenn=1, which impliesi=j=1, the conclusion is trivial.
Assume that the statement is true forn=m; that is,
(8)d(xi,xj)⩽k(e-k)-1d(x0,x1),for1⩽i,j⩽m.
Now, we will prove that the statement is true forn=m+1. Note that in this case, if1⩽i,j⩽m, then the statement is just (8). Thus, without loss of generality, we suppose thatj=m+1and1⩽i⩽m and denotei=i0.
By the definition of quasicontraction, we have(9)d(xi0,xm+1)⩽ku,
where
(10)u∈{d(xi0-1,xm),d(xi0-1,xi0),d(xm,xm+1),u∈d(xi0-1,xm+1),d(xi0,xm)}.
Firstly, we consider the case thati0=1; that is,
(11)u∈{d(x0,xm),d(x0,x1),d(xm,xm+1),hhhhd(x0,xm+1),d(x1,xm)}.
Ifu=d(x0,xm), then
(12)d(xi0,xm+1)⩽kd(x0,xm)⩽k(d(x0,x1)+d(x1,xm))⩽k(d(x0,x1)+k(e-k)-1d(x0,x1))=k(e+k(e-k)-1)d(x0,x1)=k(e+∑t=1∞‍kt)d(x0,x1)=k(e-k)-1d(x0,x1),
and the statement follows.
Ifu=d(x0,x1), then
(13)d(xi0,xm+1)⩽kd(x0,x1)⩽(∑t=1∞‍kt)d(x0,x1)=k(e-k)-1d(x0,x1),
and the statement also follows.
Ifu=d(xm,xm+1), then we seti1=mand we have
(14)d(xi0,xm+1)⩽kd(xi1,xm+1).
Ifu=d(x0,xm+1), then
(15)d(xi0,xm+1)⩽kd(x0,xm+1)⩽k(d(x0,x1)+d(x1,xm+1))=k(d(x0,x1)+d(xi0,xm+1)),
which implies
(16)(e-k)d(xi0,xm+1)⩽kd(x0,x1).
Note that(e-k)-1=∑t=0∞‍kt⩾0 and thatkand(e-k)-1commute. Multiplying both sides by(e-k)-1, we have
(17)d(xi0,xm+1)⩽k(e-k)-1d(x0,x1),
and the statement also follows.
Ifu=d(xi0,xm), then
(18)d(xi0,xm+1)⩽kd(xi0,xm)⩽k2(e-k)-1d(x0,x1)=(∑t=2∞‍kt)d(x0,x1)⩽(∑t=1∞‍kt)d(x0,x1)=k(e-k)-1d(x0,x1),
and the statement also follows.
Secondly, we consider the case that2⩽i0⩽m.
Ifu=d(xi0-1,xm) oru=d(xi0-1,xi0) oru=d(xi0,xm), then, by (8), we have
(19)d(xi0,xm+1)⩽ku⩽k2(e-k)-1d(x0,x1)=(∑t=2∞‍kt)d(x0,x1)⩽k(e-k)-1d(x0,x1),
and the statement follows.
Ifu=d(xm,xm+1) oru=d(xi0-1,xm+1), then we seti1=mori1=i0-1⩾1, respectively. And we have
(20)d(xi0,xm+1)⩽ku=kd(xi1,xm+1).
In conclusion from discussions of both cases, it results that either the proof is complete, that is,(21)d(xi0,xm+1)⩽k(e-k)-1d(x0,x1),
or there exists an integeri1such that
(22)d(xi0,xm+1)⩽kd(xi1,xm+1),1⩽i1⩽m.
As for the latter situation, we continue in a similar way, and come to the result that either(23)d(xii,xm+1)⩽k(e-k)-1d(x0,x1),
which implies that
(24)d(xi0,xm+1)⩽kd(xi1,xm+1)⩽k2(e-k)-1d(x0,x1)⩽k(e-k)-1d(x0,x1),
and the proof is complete, or there exists an integeri2such that
(25)d(xi1,xm+1)⩽kd(xi2,xm+1),1⩽i2⩽m,
which implies that
(26)d(xi0,xm+1)⩽k2d(xi2,xm+1),1⩽i2⩽m.
Generally, if the procedure ends by theℓ-th step with ℓ⩽m-1, that is, there exist ℓ+1integers
(27)i0,i1,…,iℓ∈{1,…,m},
such that
(28)d(xi0,xm+1)⩽kd(xi1,xm+1)⩽⋯⩽kℓd(xil,xm+1),
and such that
(29)d(xil,xm+1)⩽k(e-k)-1d(x0,x1),
then
(30)d(xi0,xm+1)⩽kℓ+1(e-k)-1d(x0,x1)=(∑t=ℓ+1∞‍kt)d(x0,x1)⩽k(e-k)-1d(x0,x1).
Hence, the proof is complete.
Finally, if the procedure continues more thanmsteps, then there existm+1integers
(31)i0,i1,…,im∈{1,…,m},
such that
(32)d(xi0,xm+1)⩽kd(xi1,xm+1)⩽⋯⩽kmd(xim,xm+1).
Thus, there must exist two integers,pandq, say, such that
(33)0⩽p<q⩽m,ip=iq.
From (32), one sees that
(34)d(xip,xim+1)⩽kq-pd(xiq,xm+1)=kq-pd(xip,xm+1),
and therefore
(35)(e-kq-p)d(xip,xm+1)⩽0.
Note that
(36)ρ(kq-p)⩽ρ(k)q-p<1,
which impliese-kq-pis invertible. And since that
(37)(e-kq-p)-1=∑t=0∞‍k(q-p)t⩾0,
we have
(38)d(xip,xm+1)⩽0.
So,
(39)d(xip,xm+1)=0,(40)d(xi0,xm+1)⩽kpd(xip,xm+1)=0⩽k(e-k)-1d(x0,x1).
Therefore, by induction, the statement is proved.Remark 11.
Lemma10 simply says that
(41)d(xi,xj)⩽k(e-k)-1d(x0,x1),∀i,j⩾1.Lemma 12.
Assume that the hypotheses in Theorem9 are satisfied. Then, {xn}is a Cauchy sequence.Proof.
For1<m<n, denote that
(42)C(m,n)={d(xi,xj)∣m⩽i,j⩽n}.
By the definition of quasicontraction, it follows that, for eachu∈C(m,n), there existsv∈C(m-1,n), such that
(43)u⩽kv.
Consequently,
(44)d(xm,xn)⩽ku1⩽k2u2⩽⋯⩽km-1um-1⩽km(e-k)-1d(x0,x1),
where
(45)u1∈C(m-1,n),(46)u2∈C(m-2,n),…,um-1∈C(1,n),
and the last inequality comes from Lemma 10.
By the normality ofP, and noting that∥km∥→0(m→∞), we have
(47)∥d(xm,xn)∥⩽M∥km∥∥(e-k)-1∥×∥d(x0,x1)∥⟶0(n>m⟶∞).
The proof is complete.Now, we finish the remaining part of the proof of Theorem9.Proof.
By Lemma12 and the completeness of(X,d), there isx*∈Xsuch thatxn→x*(n→∞). Then,
(48)d(x*,Tx*)⩽d(x*,xn)+d(xn,Tx*)⩽d(x*,xn)+ku,
where
(49)u∈{d(xn-1,x*),d(xn-1,xn),d(x*,Tx*),ui∈d(xn-1,Tx*),d(x*,xn)}.
Ifu=d(xn-1,x*) oru=d(xn-1,xn) oru=d(x*,xn), then∥u∥→0(n→∞). Hence,
(50)∥d(x*,Tx*)∥⩽M∥d(x*,xn)∥+∥k∥∥u∥⟶0∥d(x*,Tx*)∥iM∥d(x*,xn)∥(n⟶∞).
Ifu=d(x*,Tx*), then
(51)(e-k)d(x*,Tx*)⩽d(x*,xn).
Hence,
(52)∥d(x*,Tx*)∥⩽M∥(e-k)-1∥∥d(x*,xn)∥⟶0∥d(x*,Tx*)∥iiM∥(e-k)-1∥(n⟶∞).
Ifu=d(xn-1,Tx*), then
(53)d(x*,Tx*)⩽d(x*,xn)+kd(xn-1,Tx*)⩽d(x*,xn)+kd(xn-1,x*)+kd(x*,Tx*).
Hence,
(54)∥d(x*,Tx*)∥⩽M∥(e-k)-1∥×(∥d(x*,xn)∥+∥k∥∥d(xn-1,x*)∥)⟶0,
asn→∞.
In each case, we have∥d(x*,Tx*)∥=0. Thus, Tx*=x*.
Now, ify*is another fixed point, then
(55)d(x*,y*)=d(Tx*,Ty*)⩽ku,
where
(56)u∈{d(x*,y*),d(x*,Tx*),d(y*,Ty*),ui∈d(x*,Ty*),d(y*,Tx*)}.
Ifu=d(x*,Tx*)=d(y*,Ty*)=0, thend(x*,y*)=0.
Ifu=d(x*,y*)=d(x*,Ty*)=d(y*,Tx*), then
(57)(e-k)d(x*,y*)⩽0,
which implies
(58)d(x*,y*)=0.
Thus, the fixed point is unique. And we obtain Theorem9.



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