Some results of (Ćirić, 1974) on a nonunique fixed point theorem on the class of metric spaces are extended to the class of cone metric spaces. Namely, nonunique fixed point theorem is proved in orbitally T complete cone metric spaces under the assumption that the cone is strongly minihedral. Regarding the scalar weight of cone metric, we are able to remove the assumption of strongly minihedral.
1. Introduction and Preliminaries
In 1980, Rzepecki [1] introduced a generalized metric dE on a set X in a way that dE:X×X→S where E is a Banach space and S is a normal cone in E with partial order ⪯. In that paper, the author generalized the fixed point theorems of Maia type [2].
In 1987, Lin [3] considered the notion of K-metric spaces by replacing real numbers with cone K in the metric function, that is, d:X×X→K. In that manuscript, some results of Khan and Imdad [4] on fixed point theorems were considered for K-metric spaces. Without mentioning the papers of Lin and Rzepecki, in 2007, Huang and Zhang [5] announced the notion of cone metric spaces (CMSs) by replacing real numbers with an ordering Banach space. In that paper, they also discussed some properties of convergence of sequences and proved the fixed point theorems of contractive mapping for cone metric spaces.
Recently, many results on fixed point theory have been extended to cone metric spaces (see, e.g., [5–13]).
Ćirić type nonunique fixed point theorems were considered by many authors (see, e.g., [14–20]). In this paper, some of the known results (see, e.g., [2, 14, 15]) are extended to cone metric spaces.
Throughout this paper E:=(E,∥·∥) stands for a real Banach space. Let P:=PE always be a closed nonempty subset of E. P is called cone if ax+by∈P for all x,y∈P and non-negative real numbers a,b where P∩(-P)={0} and P≠{0}.
For a given cone P, one can define a partial ordering (denoted by ≤ or ≤P) with respect to P by x≤y if and only if y-x∈P. The notation x<y indicates that x≤y and x≠y while x≪y will show y-x∈intP, where intP denotes the interior of P. From now on, it is assumed that intP≠∅.
The cone P is called normal if there is a number K≥1 for which 0≤x≤y⇒∥x∥≤K∥y∥ holds for all x,y∈E. The least positive integer K, satisfying this equation, is called the normal constant of P. The cone P is said to be regular if every increasing sequence which is bounded from above is convergent, that is, if {xn}n≥1 is a sequence such that x1≤x2≤⋯≤y for some y∈E, then there is x∈E such that limn→∞∥xn-x∥=0.
Lemma 1.1.
(i) Every regular cone is normal.
(ii) For each k>1, there is a normal cone with normal constant K>k.
(iii) The cone P is regular if every decreasing sequence which is bounded from below is convergent.
Proof of (i) and (ii) are given in [6] and the last one follows from definition.
Definition 1.2.
Let X be a nonempty set. Suppose that the mapping d:X×X→E satisfies
0≤d(x,y) for all x,y∈X,
d(x,y)=0 if and only if x=y,
d(x,y)≤d(x,z)+d(z,y), for all x,y∈X,
d(x,y)=d(y,x) for all x,y∈X,
then d is called cone metric on X, and the pair (X,d) is called a cone metric space (CMS).
Example 1.3.
Let E=ℝ3,P={(x,y,z)∈E:x,y,z≥0}, and X=ℝ. Define d:X×X→E by d(x,x̃)=(α|x-x̃|,β|x-x̃|,γ|x-x̃|), where α,β,andγ are positive constants. Then (X,d) is a CMS. Note that the cone P is normal with the normal constant K=1.
Definition 1.4.
Let (X,d) be a CMS, x∈X, and {xn}n≥1 a sequence in X. Then
{xn}n≥1 converges to x whenever for every c∈E with 0≪c there is a natural number N, such that d(xn,x)≪c for all n≥N. It is denoted by limn→∞xn=x or xn→x.
{xn}n≥1 is a Cauchy sequence whenever for every c∈E with 0≪c there is a natural number N, such that d(xn,xm)≪c for all n,m≥N.
(X,d) is a complete cone metric space if every Cauchy sequence is convergent.
Lemma 1.5 (see [5]).
Let (X,d) be a CMS, P a normal cone with normal constant K, and {xn} a sequence in X. Then,
the sequence {xn} converges to x if and only if d(xn,x)→0 (or equivalently ∥d(xn,x)∥→0),
the sequence {xn} is Cauchy if and only if d(xn,xm)→0 (or equivalently ∥d(xn,xm)∥→0asm,n→∞),
the sequence {xn} converges to x and the sequence {yn} converges to y, then d(xn,yn)→d(x,y).
Lemma 1.6 (see [8]).
Let (X,d) be a CMS over a cone P in E. Then
int(P)+int(P)⊆int(P) and λint(P)⊆int(P),λ>0.
If c≫0, then there exists δ>0 such that ∥b∥<δ implies that b≪c.
For any given c≫0 and c0≫0 there exists n0∈ℕ such that c0/n0≪c.
If an,bn are sequences in E such that an→a, bn→b, and an≤bn,for alln, then a≤b.
Definition 1.7 (see [21]).
P is called minihedral cone if sup{x,y} exists for all x,y∈E and strongly minihedral if every subset of E which is bounded from above has a supremum. (equivalently, if every subset of E which is bounded from below has an infimum.)
Lemma 1.8.
(i) Every strongly minihedral normal (not necessarily closed) cone is regular.
(ii) Every strongly minihedral (closed) cone is normal.
The proof of (i) is straightforward, and for (ii) see, for example, [22].
Example 1.9.
Let E=C[0,1] with the supremum norm and P={f∈E:f≥0}. Then P is a cone with normal constant M=1 which is not regular. This is clear, since the sequence xn is monotonically decreasing but not uniformly convergent to 0. This cone, by Lemma 1.8, is not strongly minihedral. However, it is easy to see that the cone mentioned in Example 1.3 is strongly minihedral.
2. Non unique Fixed Points on Cone Metric SpacesDefinition 2.1.
A mapping T on CMS (X,d) is said to be orbitally continuous if limi→∞Tni(x)=z implies that limi→∞T(Tni(x))=Tz. A CMS (X,d) is called T orbitally complete if every Cauchy sequence of the form {Tni(x)}i=1∞,x∈X, converges in (X,d).
Remark 2.2.
It is clear that orbital continuity of T implies orbital continuity of Tm for any m∈ℕ.
Theorem 2.3.
Let T:X→X be an orbitally continuous mapping on CMS (X,d) over strongly minihedral normal cone P. Suppose that CMS (X,d) is T orbitally complete and that T satisfies the condition
u(x,y)-inf{d(x,T(y)),d(T(x),y)}≤kd(x,y)
for all x,y∈X and for some 0≤k<1, where u(x,y)∈{d(x,T(x)),d(T(x),T(y)),d(T(y),y)}. Then, for each x∈X, the iterated sequence {Tn(x)} converges to a fixed point of T.
Proof.
Fix x0∈X. For n≥1 set x1=T(x0) and recursively xn+1=T(xn)=Tn+1(x0). It is clear that the sequence xn is Cauchy when the equation xn+1=xn holds for some n∈ℕ. Consider the case xn+1≠xn for all n∈ℕ. By replacing x and y with xn-1 and xn, respectively, in (2.1), one can get
u(xn-1,xn)-inf{d(xn-1,T(xn)),d(T(xn-1),xn)}={d(xn,xn+1),d(xn-1,xn)}≤kd(xn-1,xn),
where u(xn-1,xn)∈{d(xn-1,T(xn-1)),d(T(xn-1),T(xn)),d(T(xn),xn)}. Since k<1, the case d(xn-1,xn)≤kd(xn-1,xn) yields contradiction. Thus, d(xn,xn+1)≤kd(xn-1,xn). Recursively, one can observe that
d(xn,xn+1)≤kd(xn-1,xn)≤k2d(xn-2,xn-1)≤⋯≤knd(x0,T(x0)).
By using the triangle inequality, for any p∈ℕ, one can get
d(xn,xn+p)≤d(xn,xn+1)+d(xn+1,xn+2)+⋯+d(xn+p-1,T(xn+p))≤(kn+kn+1+⋯+kn+p-1)d(x0,T(x0))=kn(1+k+⋯+kp-1)d(x0,T(x0))≤kn1-kd(x0,T(x0)).
Let c∈int(P). Choose a natural number M0 such that (kn/(1-k))d(T(x0),x0)≪c for all n>M0. Thus, for any p∈ℕ, d(xn+p,xn)≪c for all n>M0. So {xn} is a Cauchy sequence in (X,d). Since (X,d) is T orbitally complete, there is some z∈X such that limn→∞xn=limn→∞Tn(x0)=z. Regarding the orbital continuity of T, T(z)=limn→∞T(Tn(x0))=limn→∞Tn+1(x0)=z, that is, z is a fixed point of T.
A point z is said to be a periodic point of a function T of period m if Tm(z)=z, where T0(x)=x and Tm(x) is defined recursively by Tm(x)=T(Tm-1(x)).
Theorem 2.4.
Let T:X→X be an orbitally continuous mapping on T orbitally complete CMS (X,d) over strongly minihedral normal cone P and c∈int(P). Suppose that there exists a point x0∈X such that d(x0,Tn(x0))≪c for some n∈ℕ and that T satisfies the condition
0<d(x,y)≪c⇒u(x,y)≤kd(x,y)
for all x,y∈X and for some k<1, where u(x,y)∈{d(x,T(x)),d(T(x),T(y)),d(T(y),y)}. Then, T has a periodic point.
Proof.
Set M={n∈ℕ:d(x,Tn(x))≪cforsomex∈X}. By assumption of theorem M≠∅. Set m=minM and let x∈X such that d(x,Tm(x))≪c which is equivalent to saying that c-d(x,Tm(x))∈int(P).
Suppose that m=1. By replacing y=T(x) in (2.5), one can get
u(x,T(x))≤kd(x,T(x)),
where u(x,T(x))∈{d(x,T(x)),d(T(x),T(T(x))),d(T(T(x)),T(x))}. There are two cases. Consider the first case, d(x,T(x))≤kd(x,T(x)), which is a contraction by, regarding k<1. Thus, one has d(T(x),T(T(x)))=d(T(x),T2(x))≤kd(x,T(x)).
As in the proof of Theorem 2.3, one can consider the iterative sequence xn+1=T(xn),x=x0 and observe that Tz=z for some z∈X.
Suppose that m≥2. It is equivalent to saying that for each y∈X, the condition
c-d(T(y),y)∉int(P).
Taking account of d(x,Tm(x))≪c and applying into (2.5), one can get
u(x,Tm(x))≤kd(x,Tm(x)),
where u(x,Tm(x))∈{d(x,T(x)),d(T(x),T(Tm(x))),d(T(Tm(x)),Tm(x))}.
Recall that Tm(x)∈X and say that Tm(x)=z. Then, d(T(Tm(x)),Tm(x))=d(T(z),z) is observed. Regarding (2.7), c-d(T(z),z)=c-d(T(Tm(x)),Tm(x))∉int(P) and also c-d(T(x),x)∉int(P). Thus,
min{d(x,T(x)),d(T(x),T(Tm(x))),d(T(Tm(x)),Tm(x))}=d(T(x),Tm+1(x)),
and hence, (2.8) turns into
d(T(x),Tm+1(x))≤kd(x,Tm(x)).
Recursively, one can get
d(T2(x),Tm+2(x))≤kd(T(x),Tm+1(x))≤k2d(x,Tm(x)).
Continuing in this way, for each p∈ℕ, one can obtain
d(Tp(x),Tm+p(x))≤kd(Tp-1(x),Tm+p-1(x))≤⋯≤kpd(x,Tm(x)).
Thus, for the recursive sequence xn+1=Tm(xn) where x0=x,
d(xn,xn+1)=d(Tnm(x0),T(n+1)m(x0))=d(Tnm(x0),Tm+nm(x0))≤knmd(x0,Tm(x0)).
By using the triangle inequality, for any p∈ℕ, one can get
d(xn,xn+p)≤d(xn,xn+1)+d(xn+1,xn+2)+⋯+d(xn+p-1,xn+p)=knm(1+km+⋯+k(p-1)m)d(x0,T(x0))≤knm1-kmd(x0,Tm(x0)).
Let c∈int(P). Choose a natural number M0 such that ((knm)/(1-km))d(T(x0),x0)≪c for all n>M0. Thus, for any p∈ℕ, d(xn+p,xn)≪c for all n>M0. So {xn} is a Cauchy sequence in (X,d). Since (X,d) is T orbitally complete, there is some z∈X such that limn→∞Tn(x0)=z. Regarding Remark 2.2, the orbital continuity of T implies that
Tm(z)=limn→∞Tm(Tnm(x0))=limn→∞T(n+1)m(x0)=z,
that is, z is a periodic point of T.
Theorem 2.5.
Let T:X→X be an orbitally continuous mapping on CMS (X,d) over strongly minihedral normal cone P. Suppose that T satisfies the condition
u(x,y)-inf{d(x,T(y)),d(T(x),y)}<d(x,y)
for all x,y∈X,x≠y where u(x,y)∈{d(x,T(x)),d(T(x),T(y)),d(T(y),y)}. Suppose that the sequence {Tn(x0)} has a cluster point z∈X, for some x0∈X. Then, z is a fixed point of T.
Proof.
Suppose that Tm(x0)=Tm-1(x0) for some m∈ℕ, then Tn(x0)=Tm(x0)=z for all n≥m. It is clear that z is a required point.
Suppose that Tm(x0)≠Tm-1(x0) for all m∈ℕ. Since {Tn(x0)} has a cluster point z∈X, one can write limi→∞Tni(x0)=z. By replacing x and y with Tn-1(x0) and Tn(x0), respectively, in (2.16),
u(Tn-1(x0),Tn(x0))-inf{d(Tn-1(x0),T(Tn(x0))),d(T(Tn-1(x0)),Tn(x0))}<d(Tn-1(x0),Tn(x0)),
where u(Tn-1(x0),Tn(x0)) lies in {d(Tn-1(x0),T(Tn-1(x0))),d(T(Tn-1(x0)),T(Tn(x0))), d(T(Tn(x0)),Tn(x0))}. The case d(Tn-1(x0),Tn(x0))<d(Tn-1(x0),Tn(x0)) is impossible. Thus, (2.17) is equivalent to d(Tn(x0),Tn+1(x0))<d(Tn-1(x0),Tn(x0)). It shows that
{d(Tn(x0),Tn+1(x0))}1∞
is decreasing. Since the cone P is strongly minihedral, then by Lemma 1.1 (iii) and Lemma 1.8 (i), {d(Tn(x0),Tn+1(x0))}1∞ is convergent. Due to Lemma 1.5, and T orbital continuity,
limi→∞d(Tni(x0),Tni+1(x0))=d(z,Tz).
By {d(Tni(x0),Tni+1(x0))}1∞⊂{d(Tn(x0),Tn+1(x0))}1∞ and (2.19),
limn→∞d(Tn(x0),Tn+1(x0))=d(z,Tz).
Regarding limi→∞Tni+1(x0)=Tz, limi→∞Tni+2(x0)=T2z{d(Tni+1(x0),Tni+2(x0))}1∞⊂{d(Tn(x0),Tn+1(x0))}1∞, and (2.20),
d(Tz,T2z)=d(z,Tz).
Assume that Tz≠z, that is, d(z,Tz)>0. So, one can replace x and y with z and Tz, respectively, in(2.16)
u(z,Tz)-inf{d(z,T(T(z))),d(T(z),T(z))}<d(z,T(z)),
where u(z,Tz)∈{d(z,T(z)),d(T(z),T(T(z))),d(T(T(z)),T(z))}.
It yields that d(Tz,T2z)<d(z,Tz). But it contradicts (2.21). Thus, Tz=z.
3. Non unique Fixed Points on Scalar Weighted Cone Metric SpacesDefinition 3.1.
Let (X,d) be a CMS. The scalar weight of the cone metric d is defined by ds(x,y):=∥d(x,y)∥.
Notice that for normal cone P with the normal constant K=1, the scalar weight of the cone metric ds behaves as a metric on X. In the following theorems normal constant K has no restriction.
Theorem 3.2.
Let T:X→X be an orbitally continuous mapping on T orbitally complete CMS (X,ds) over normal cone P with normal constant K. Suppose that T satisfies the condition
min{ds(x,T(x)),ds(T(x),T(y)),ds(T(y),y)}-min{ds(x,T(y)),ds(T(x),y)}≤kds(x,y)
for all x,y∈X and for some k<1. Then, for each x∈X, the iterated sequence {Tn(x)} converges to a fixed point of T.
Proof.
Fix x0∈X. For n≥1 set x1=T(x0) and recursively xn+1=T(xn)=Tn+1(x0). It is clear that the sequence xn is Cauchy when xn+1=xn hold for some n∈ℕ. Consider the case xn+1≠xn for all n∈ℕ. By replacing x and y with xn-1 and xn, respectively, in (3.1), one can get
min{ds(xn-1,T(xn-1)),ds(T(xn-1),xn),ds(xn,T(xn))}-min{ds(xn-1,T(xn)),ds(T(xn-1),xn)}=min{ds(xn,xn+1),ds(xn-1,xn)≤kds(xn-1,xn).
Since k<1, the case ds(xn-1,xn)≤kds(xn-1,xn) yields contradiction. Thus, ds(xn,xn+1)≤kds(xn-1,xn). Recursively, one can observe that
ds(xn,xn+1)≤kds(xn-1,xn)≤k2ds(xn-2,xn-1)≤⋯≤knds(x0,T(x0)).
By using the triangle inequality, for any p∈ℕ, one can get
ds(xn,xn+p)≤K(ds(xn,xn+1)+ds(xn+1,xn+2)+⋯+ds(xn+p-1,T(xn+p)))≤K((kn+kn+1+⋯+kn+p-1)ds(x0,T(x0)))=Kkn(1+k+⋯+kp-1)ds(x0,T(x0))≤Kkn1-kds(x0,T(x0)).
By routine calculation, one can obtain that {xn} is a Cauchy sequence in (X,d). Since (X,d) is T orbitally complete, there is some z∈X such that
limn→∞xn=limn→∞Tn(x0)=z.
Regarding the orbital continuity of T,
T(z)=limn→∞T(Tn(x0))=limn→∞Tn+1(x0)=z,
that is, z is a fixed point of T.
Theorem 3.3.
Let T:X→X be an orbitally continuous mapping on T orbitally complete CMS (X,d) over normal cone P with normal constant K and ɛ>0. Suppose that there exists a point x0∈X such that ds(x0,Tn(x0))<ɛ for some n∈ℕ and that T satisfies the condition
0<ds(x,y)<ɛ⇒min{ds(x,T(x)),ds(T(x),T(y)),ds(T(y),y)}≤kds(x,y)
for all x,y∈X and for some k<1. Then, T has a periodic point.
Proof.
Set M={n∈ℕ:ds(x,Tn(x))<ɛ:forx∈X}. By assumption of the theorem M≠∅. Let m=minM and x∈X such that ds(x,Tm(x))<ɛ.
Suppose that m=1, that is, ds(x,T(x))<ɛ. By replacing y=T(x) in (3.7), one can get
min{ds(x,T(x)),ds(T(x),T(T(x))),ds(T(T(x)),T(x))}≤kds(x,T(x)).
The case ds(x,T(x))≤kds(x,T(x)) implies a contraction due to the fact that k<1. Thus, ds(T(x),T(T(x)))=ds(T(x),T2(x))≤kds(x,T(x)).
As in the proof of Theorem 3.2, one can consider the iterative sequence xn+1=T(xn),x=x0 and observe that Tz=z for some z∈X.
Suppose that m≥2. It is equivalent to saying that the condition
ds(T(y),y)≥ɛ
holds for each y∈X. Then, from ds(x,Tm(x))<ɛ and (3.7) it follows that
min{ds(x,T(x)),ds(T(x),T(Tm(x))),ds(T(Tm(x)),Tm(x))}≤kds(x,Tm(x)).
Considering Tm(x)∈X, say Tm(x)=z, one has ds(T(Tm(x)),Tm(x))=ds(T(z),z). Regarding (3.9), ds(T(z),z)=ds(T(Tm(x)),Tm(x))≥ɛ and ds(T(x),x)≥ɛ. Thus,
min{ds(x,T(x)),ds(T(x),T(Tm(x))),ds(T(Tm(x)),Tm(x))}=ds(T(x),Tm+1(x)).
and hence
ds(T(x),Tm+1(x))≤kds(x,Tm(x)).
Recursively, one can get
ds(T2(x),Tm+2(x))≤ds(T(x),Tm+1(x))≤k2ds(x,Tm(x)).
Continuing in this way, for each p∈ℕ, one can obtain
ds(Tp(x),Tm+p(x))≤ds(Tp-1(x),Tm+p-1(x))≤⋯≤kpds(x,Tm(x)).
Thus, for the recursive sequence xn+1=Tm(xn) where x0=x,
ds(xn,xn+1)=ds(Tnm(x0),T(n+1)m(x0))=ds(Tnm(x0),Tm+nm(x0))≤knmds(x0,Tm(x0)).
By using the triangle inequality and regarding the normality of the cone, for any p∈ℕ, one can get
ds(xn,xn+p)≤K[ds(xn,xn+1)+ds(xn+1,xn+2)+⋯+ds(xn+p-1,xn+p)]=Kknm[1+km+⋯+k(p-1)m]ds(x0,Tm(x0))≤Kknm1-kmds(x0,Tm(x0)).
Let ɛ>0. Choose a natural number M0 such that (Kknm/(1-km))ds(Tm(x0),x0)<ɛ for all n>M0. Thus, for any p∈ℕ, ds(xn+p,xn)<ɛ for all n>M0. So {xn} is a Cauchy sequence in X. Since X is T orbitally complete, there is some z∈X such that limn→∞Tn(x0)=z. Regarding Remark 2.2, the orbital continuity of T implies that
Tm(z)=Tm(limn→∞Tnm(x0))=limn→∞Tm(Tnm(x0))=limn→∞T(n+1)m(x0)=z,
that is, z is a periodic point of T.
Theorem 3.4.
Let T:X→X be an orbitally continuous mapping on CMS (X,d) over normal cone P with normal constant K. Suppose that T satisfies the condition
min{ds(x,T(x)),ds(T(x),T(y)),ds(T(y),y)}-min{ds(x,T(y)),ds(T(x),y)}<ds(x,y)
for all x,y∈X,x≠y. If the sequence {Tn(x0)} has a cluster point z∈X, for some x0∈X, then z is a fixed point of T.
The proof of Theorem 3.4 is omitted by regarding the analogy with the proof of Theorem 2.5. In the proof of Theorem 2.5, to conclude that the decreasing sequence (2.18) is convergent, we need to use the assumption of strong minihedrality of the cone P. Since we use the scalar weight of cone metric in the proof of Theorem 3.4, we can conclude that the corresponding decreasing sequence of (2.18) is convergent without the assumption of strong minihedrality of the cone P.
Theorem 3.5.
Let T:X→X be an orbitally continuous mapping on T orbitally complete CMS (X,d) over normal cone P with normal constant K and ɛ>0. Suppose that T satisfies the condition
0<ds(x,y)<ɛ⇒min{ds(x,T(x)),ds(T(x),T(y)),ds(T(y),y)}<ds(x,y)
for all x,y∈X. If for some x0∈X, the sequence {Tn(x0)}n=1∞ has a cluster point of z∈X, then z is a periodic point of T.
Proof.
Set limi→∞Tni(x0)=z, that is, for any ɛ>0 there exists N0∈ℕ such that ds(Tni(x0),z)<ɛ/2K for all i>N0. Hence, by triangle inequality and normality of the cone it yields that
ds(Tni(x0),Tni+1(x0))≤ds(Tni(x0),z)+ds(z,Tni+1(x0))<ɛ.
Define a set
M={j∈ℕ:ds(Tn(x0),Tn+j(x0))<ɛforsomen∈ℕ}
which is nonempty by assumption of the theorem. Let m=minM. Consider two cases. Suppose ds(Tn(x0),Tn+m(x0))=0 for some n∈ℕ. Then, z=Tn(x0)=Tn+m(x0)=Tm(Tn(x0))=Tm(z) and the assertion of theorem follows.
Suppose that ds(Tn(x0),Tn+m(x0))>0 for all n∈ℕ. Let r∈ℕ be such that ds(Tr(x0),Tr+m(x0))<ɛ.
If m=1, then replacing x and y with Tn(x0) and Tn+1(x0), respectively, in (3.19) one can obtain that
min{ds(Tn(x0),T(Tn(x0))),ds(T(Tn(x0)),T(Tn+1(x0))),ds(T(Tn+1(x0)),Tn+1(x0))}<ds(Tn(x0),Tn+1(x0)).
Since the case ds(Tn(x0),Tn+1(x0))<ds(Tn(x0),Tn+1(x0)) is impossible, (3.22) turns into ds(Tn+1(x0),Tn+2(x0))<ds(Tn(x0),Tn+1(x0)), that is, the sequence {ds(Tn(x0),Tn+1(x0))} is decreasing for n≥r. Thus, by routine calculation, one can conclude that Tz=z.
Assume that m≥2, that is, for every n∈ℕ,
ds(Tn(x0),Tn+1(x0))≥ɛ.
By orbital continuity of T, limi→∞Tni+r(x0)=Tr(z), and by (3.23), one can get
ds(Tr(z),Tr+1(z))=limi→∞ds(Tni+r(x0),Tni+r+1(x0))≥ɛ.
for every r∈ℕ
Regarding (3.19) under the assumption 0<ds(Tj(x0),Tj+m(x0))<ɛ one can obtain
min{ds(Tj(x0),Tj+1(x0)),ds(Tj+1(x0),Tj+m+1(x0)),ds(Tj+m(x0),Tj+m+1(x0))}<ds(Tj(x0),Tj+m(x0)).
Thus, due to (3.23), ds(Tj+1(x0),Tj+m+1(x0))<ds(Tj(x0),Tj+m(x0))<ɛ.
By continuing this process, it yields that
⋯<ds(Tj+2(x0),Tj+m+2(x0))<ds(Tj+1(x0),Tj+m+1(x0))<ds(Tj(x0),Tj+m(x0))<ɛ.
Hence, the sequence {ds(Tn(x0),Tn+m(x0)):n≥j} is decreasing and thus is convergent. Notice that the subsequences {ds(Tni(x0),Tni+m(x0)):i∈ℕ} and {ds(Tni+1(x0),Tni+1+m(x0)):i∈ℕ} are convergent to d(z,Tmz) and d(Tz,Tm+1z), respectively. By orbital continuity of T and limi→∞Tni(x0)=z, one can get
ds(T(z),Tm+1(z))=ds(z,Tm(z))=limn→∞ds(Tn(x0),Tn+m(x0)).
One can conclude that ds(z,Tmz)<ɛ from (3.26) and (3.27). If ds(z,Tmz)=0, then Tmz=z. Thus, the desired result is obtained. Suppose that ds(z,Tmz)>0. Applying (3.19),
min{ds(z,T(z)),ds(T(z),T(Tm(z))),ds(T(Tm(z)),Tm(z))}<ds(z,Tmz)<ɛ.
Taking account of (3.24), (3.28) yields that ds(T(z),Tm+1(z))<ds(z,Tmz) which contradicts (3.27). Thus, ds(z,Tmz)=0, and so Tmz=z.
Theorem 3.6.
Let T:X→X be an orbitally continuous mapping on T orbitally complete CMS (X,ds) over normal cone P with normal constant K. Suppose that T satisfies the condition
min{[ds(x,T(x))]2,ds(x,y)ds(T(x),T(y)),[ds(T(y),y)]2}-min{ds(x,T(x))ds(T(y),y),ds(x,T(y))ds(T(x),y)}≤kds(x,T(x))ds(T(y),y)
for all x,y∈X and for some k<1. Then, for each x∈X, the iterated sequence {Tn(x)} converges to a fixed point of T.
Proof.
As in the proof of Theorem 3.2, fix x0∈X and define the sequence {xn} in the following way. For n≥1 set x1=T(x0) and recursively xn+1=T(xn)=Tn+1(x0). It is clear that the sequence xn is Cauchy when xn+1=xn hold for some n∈ℕ. Consider the case xn+1≠xn for all n∈ℕ. By replacing x and y with xn-1 and xn, respectively, in (3.29), one can get
min{[ds(xn-1,T(xn-1))]2,ds(xn-1,xn)ds(T(xn-1),T(xn)),[ds(T(xn),xn)]2}-min{ds(xn-1,T(xn-1))ds(T(xn),xn),ds(xn-1,T(xn))ds(T(xn-1),xn)}≤kds(xn-1,T(xn-1))ds(T(xn),xn).
Since k<1, the case ds(xn-1,xn)ds(xn,xn+1)≤kds(xn-1,xn)ds(xn,xn+1) yields contradiction. Thus, one gets
ds(xn,xn+1)≤kds(xn-1,xn).
Recursively, one can observe that
ds(xn,xn+1)≤kds(xn-1,xn)≤k2ds(xn-2,xn-1)≤⋯≤knds(x0,T(x0)).
By routine calculation as in the proof of Theorem 3.2, one can show that T has a fixed point.
Theorem 3.7.
Let X be a nonempty set endowed in two cone metrics d,ρ, and let T be a mapping of X into itself. Suppose that
X is orbitally complete space with respect to ds,
ds(x,y)≤ρs(x,y) for all x,y∈X,
T is orbitally continuous with respect to ds,
T satisfies
min{[ρs(T(x),T(y))]2,ρs(x,y)ρs(T(x),T(y)),[ρs(y,T(y))]2}-min{ρs(x,T(x))ρs(y,T(y)),ρs(x,Ty)ρs(y,T(x))}≤kρs(x,T(x)),ρs(y,Ty)
for all x,y∈X, where 0≤k<1.
Then T has a fixed point in X.
Proof.
As in the proof of Theorem 3.2, fix x0∈X and define the sequence {xn} in the following way. For n≥1 set x1=T(x0) and recursively xn+1=T(xn)=Tn+1(x0). Replacing x,y with xn-1,xn, respectively, in (3.33), one can get
min{[ρs(T(xn-1),T(xn))]2,ρs(xn-1,xn)ρs(T(xn-1),T(xn)),[ρs(xn,T(xn))]2}-min{ρs(xn-1,T(xn-1))ρs(xn,T(xn)),ρs(xn-1,T(xn))ρs(xn,T(xn-1))}≤kρs(xn-1,T(xn-1)),ρs(xn,T(xn)).
Since the case kρs(xn-1,T(xn-1)),ρs(xn,T(xn))≤kρs(xn-1,T(xn-1)),ρs(xn,T(xn)), (3.34) is equivalent to ρs(xn,xn+1)≤kρs(xn-1,xn). Recursively one can obtain
ρs(xn,xn+1)≤kρs(xn-1,xn)≤⋯≤knρs(x0,x1).
Regarding the triangle inequality and the normality of the cone, (3.35) implies that
ρs(xn,xn+p)≤Kkn1-kρs(x0,x1),
for any p∈ℕ. Taking account of assumption (ii) of the theorem, one can get
ds(xn,xn+p)≤Kkn1-kρs(x0,x1).
Thus, {xn} is a Cauchy sequence with respect to ds. Since X is T orbitally complete, there exists z∈X such that limn→∞Tn(x)=z. From orbital continuity of T, one can get the desired result, that is, Tz=limn→∞T(Tn(x))=z.
Remark 3.8.
Theorem 3.6 can be restated by replacing (3.29) with
min{[ds(T(x),T(y))]2,ds(x,y)ds(T(x),T(y)),[ds(T(y),y)]2}-min{ds(x,T(y)),ds(y,T(x))}≤kds(x,T(x))ds(T(y),y).
Note also that, Theorem 3.7 remains valid by replacing (3.33) with
min{[ρs(T(x),T(y))]2,ρs(x,y)ρs(T(x),T(y)),[ρs(T(y),y)]2}-min{vs(x,T(y)),ρs(y,T(x))}≤kρs(x,T(x))ρs(T(y),y).
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