# Fixed Points and Continuity Conditions of Generalizedb-Quasicontractions

Authors: Yan Han; Shaoyuan Xu
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this research, we first check that the abstract coneb-metric is discontinuous in the case of normal cone by a counter example. We obtain several meaningful results about generalized b-quasicontraction and Ćirić-type b-quasicontraction in cone b-metric spaces over Banach algebras, weakening certain important conditions of the spaces and the mappings. Meanwhile, several valid examples are given to demonstrate the new notions and fixed point results, when the existing theorems in the literature are not applicable.


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## Body

## 1. Introduction and Preliminaries

Since the concept of cone metric space was reintroduced by Huang and Zhang [1], a large number of fixed point results were gained in such spaces. In 2009, the authors in [2] defined quasicontraction in cone metric space with a normal cone. Subsequently, by removing the normality of the cone, Kadelburg et al. [3] established a fixed point theorem with a quasicontractive constant k∈0,1/2 ([3], Theorem 2.2). Sequentially, Gajić and Rakoc˘ević [4] showed the result holds when k∈0,1 in the same spaces. In 2011, the notion of cone b-metric space was given by Hussian and Shah [5], which generalized b-metric space and cone metric space. Afterwards, Huang and Xu [6] gave several fixed point results of different classes contraction in this spaces. In 2013, the scholars in [7] claimed the cone metric space over Banach algebra while the Banach space E is substituted with the Banach algebra A. In their paper, the most important work was to verify that the fixed point conclusions in cone metric spaces over Banach algebras were not equivalent to those in metric spaces by a nontrivial example. In [8], the authors redefined cone b-metric space over Banach algebra. They obtained some fixed points of contractions in such spaces which were not equivalent to the corresponding work in b-metric spaces. Later on, numerous interesting fixed point theorems in these spaces were promoted to be studied by many scholars, see [9–19] and their references, but most of the results were established under the completeness of the spaces and some even required the continuity of b-metric (while cone metric and metric are continuous) (see [1–4, 6–8, 18–29]).In 2018, Aleksić et al. [20] proved that b-metric is discontinuous in general by some examples, which is a generalization of metric. However, the fixed point conclusions of b-quasicontraction in b-metric spaces were also discussed under continuous b-metric and complete b-metric spaces. In order to improve these too strong conditions, we prove that the abstract cone b-metric is discontinuous even with a normal cone. Furthermore, we gain some fixed point results in cone b-metric spaces over Banach algebras when the cone b-metric is discontinuous. Some other conditions are weakened by giving several new concepts, such as T-orbital completeness, orbital continuity, and orbital compactness in these spaces. Our work develops and broadens some significant well-known theorems in the literature [8, 11, 20, 23, 28, 30]. Furthermore, some nontrivial examples are provided to demonstrate that the new concepts and main theorems in this paper are genuine developments and generalizations of some existing ones in the literature.Now, we start our paper with some preliminary definitions in the literature.SupposeA is a real Banach algebra and P is a cone over Banach algebra A with intP≠∅, the notation ≤ expresses the partial ordering in terms of P. For the definitions of Banach algebras and cones, the readers may refer to [24, 31].Definition 1. (see [5, 8]).
SupposeX is a nonempty set and s≥1 is a constant, the mapping d:X×X⟶A is said to be a cone b-metric if
(d1)θ≤dϖ,η for all ϖ,η∈X and dϖ,η=θ if and only if ϖ=η
(d2)dϖ,η=dη,ϖ for all ϖ,η∈X
(d3)dϖ,η≤sdϖ,ζ+dζ,η for all ϖ,η,ζ∈X
The pairX,d is called a cone b-metric space over Banach algebra A.Definition 2. (see [8]).
SupposeX,d is a cone b-metric space over Banach algebra A, ϖ∈X, and ϖn is a sequence in X, we say
(i)
ϖn converges to x if for each c∈A with c≫θ, there is an integer N≥1 such that dϖn,ϖ≪c for all n>N(ii)
ϖn is a Cauchy sequence if for each c∈A with c≫θ, there is an integer N≥1 such that dϖn,ϖm≪c for all n,m>N(iii)
X,d is complete if each Cauchy sequence in X is convergentIt is significant to note that different from the usual metric and cone metric with a normal cone, coneb-metric is generally discontinuous even with a normal cone. Let us show an example.Example 3.
TakeA=ℝ2 with a norm ϖ1,ϖ2=ϖ1+ϖ2. For any ϖ=ϖ1,ϖ2 and η=η1,η2 in A, set the multiplication as
(1)ϖη=ϖ1,ϖ2η1,η2=ϖ1η1,ϖ1η2+ϖ2η1.
LetP=ϖ1,ϖ2∈ℝ2:ϖ1,ϖ2≥0. Then, A is a real Banach algebra owing the unit element e=1,0, and the cone P⊆A is normal. Set X=ℕ∪∞×ℕ∪∞ and d:X×X⟶A be defined as
(2)dϖ1,ϖ2,η1,η2=0,0,ϖ1=η1,ϖ2=η2;1ϖ1−1η1,1ϖ2−1η2,oneofϖ1,ϖ2andη1,η2isoddandtheotherisoddor∞;5,5,oneofϖ1,ϖ2andη1,η2isevenandtheotherisevenor∞;2,2,otherwise.
In the above definition,ϖ1,ϖ2 is odd if both ϖ1 and ϖ2 are odd; ϖ1,ϖ2 is ∞ if both ϖ1 and ϖ2 are ∞. We can check that X,d is a cone b-metric space over Banach algebra A where s=5/2.
Letζm=4m−1,4m+1, m∈ℕ+. We have
(3)dζm,∞=d4m−1,4m+1,∞,∞=14m−1−1∞,14m+1−1∞=14m−1,14m+1⟶0,0,which indicates ζm⟶∞ but
(4)dζm,2=d4m−1,4m+1,2,2=2,2↛5,5=d∞,∞,2,2=d∞,2.
So, we have showed that the coneb-metric is discontinuous in the case of normal cone.Definition 4. (see [25]).
SupposeP is a solid cone, P⊆A. A sequence σn⊂P is a c-sequence if for any c≫θ, there is n0∈ℕ such that σn≪c for all n≥n0.Lemma 5. (see [8]).
SupposeP is a solid cone, P⊆A. If α,β∈P, ϖn, and ηn are c-sequences in A, then αϖn+βηn is a c-sequence in A.Lemma 6. (see [31]).
SupposeA is a Banach algebra with a unit e and ϖ∈A, if the spectral radius ρϖ of ϖ satisfies
(5)ρϖ=limn⟶∞ϖn1/n=infn≥1ϖn1/n<1,then e−ϖ is invertible. Actually, e−ϖ−1=∑i=0∞ϖi.Throughout this paper, we always supposeX,d is a cone b-metric space over Banach algebra A with a unit e and s≥1.


## 2. Orbital Completeness

In this section, we give several fixed point theorems of generalizedb-quasicontraction in orbitally complete cone b-metric spaces over Banach algebras. The cone is neither regular nor normal. The cone b-metric and the self mapping are not required continuous. At first, encouraged by the concepts of orbital continuity, Φ-orbital completeness [32], and k-continuity [21] in usual metric space, we provide the analogous concepts in cone b-metric space over Banach algebra A, which are important in our proof.Definition 7.
SupposeX,d is a cone b-metric space over Banach algebra A and Φ:X⟶X, take ϖ∈X and OΦϖ=ϖ,Φϖ,Φ2ϖ,Φ3ϖ,⋯, namely, the orbit of ϖ under Φ.The mappingΦ is orbitally continuous at an element ζ∈X if for any sequence ϖn⊂OΦϖ (for all ϖ∈X), ϖn⟶ζ as n⟶∞ implies Φϖn⟶Φζ as n⟶∞. Note that each continuous mapping is orbitally continuous, but not the converse.The mappingΦ is k-continuous for k=1,2,⋯, if Φk−1ϖn⟶ζ implies Φkϖn⟶Φζ (n⟶∞). It is clear that Φ is 1-continuous if and only if it is continuous, and k-continuity implies k+1-continuity for any k=1,2,⋯ but not the converse. Furthermore, continuity of Φk and k-continuity of Φ are independent when k>1. See the following examples.Example 8.
SupposeA=Cℝ10,1×Cℝ10,1 with a norm,
(6)ϖ1,ϖ2=ϖ1∞+ϖ2∞+ϖ′1∞+ϖ′2∞.
For anyϖ=ϖ1,ϖ2 and η=η1,η2 in A, set the multiplication as
(7)ϖη=ϖ1,ϖ2η1,η2=ϖ1η1,ϖ1η2+ϖ2η1.
LetP=ϖ1z,ϖ2z∈A:ϖ1z≥0,ϖ2z≥0,z∈0,1. It follows that there is a unit element e=1,0 in the real Banach algebra A. Let X=0,4×0,4 and define d:X×X⟶A by
(8)dϖ1,ϖ2,η1,η2z=ϖ1−η12expz,ϖ2−η22expz,for any ϖ=ϖ1,ϖ2,η=η1,η2∈X. It is obvious that X,d is complete and s=2. Suppose Φ:X⟶X is defined as
(9)Φϖ1,ϖ2=sinϖ12,sinϖ23,ϖ1,ϖ2∈0,2×0,2;log1+ϖ12,log1+ϖ23,otherwise.
For anyζ0=ϖ1,ϖ2∈X, if ζn=Φζn−1,n∈ℕ, then ζn⟶θ implies Φζn⟶Φθ=θ while θ=0,0. Clearly, Φ is an orbitally continuous mapping rather than continuous. Moreover, notice that Φ is 2-continuous but not continuous, that is, 2-continuity of Φ does not imply continuity of Φ2. Furthermore, for each integer k≥2, Φk is discontinuous while Φ is k-continuous. This indicates that k-continuity of Φ does not imply continuity of Φk in usual situation.In [12], we have given the following definitions of T-orbital completeness.Definition 9.
The spaceX,d is named Φ-orbitally complete, if each Cauchy sequence included in OΦϖ for some ϖ∈X converges in X. Each complete space X,d is Φ-orbitally complete for any Φ but not the converse.For being convenient, we give the notion of generalizedb-quasicontraction in X,d. The mapping Φ:X⟶X is named a generalized b-quasicontraction if there exists r∈P with ρr<1/s, and one has
(10)dΦϖ,Φη≤ruϖ,η,ϖ,η∈X,where
(11)uϖ,η∈dϖ,η,dϖ,Φϖ,dη,Φη,dϖ,Φη,dη,Φϖ.Whens=1, we call it a generalized quasicontraction in cone metric space over Banach algebra. Before showing our main results, we give an important lemma without the assumptions of completeness and normality.Lemma 10.
Assume the mappingΦ:X⟶X is a generalized b-quasicontraction in the space X,d, for each ϖ0∈X, let ϖn=Φϖn−1. Then, for any integers i,j≥1, it holds that
(12)dϖi,ϖj≤sre−sr−1dϖ0,ϖ1.Proof.
At first, we show that for anyn≥1 and 1≤i≤n,
(13)dϖi,ϖn≤sre−sr−1dϖ0,ϖ1.
Ifn=1, the result is trivial. Assume n=2,i=1, then
(14)dϖ1,ϖ2=dΦϖ0,Φϖ1≤ruϖ0,ϖ1,where
(15)uϖ0,ϖ1∈dϖ0,ϖ1,dϖ0,ϖ1,dϖ1,ϖ2,dϖ0,ϖ2,dϖ1,ϖ1.
Obviously,uϖ0,ϖ1≠dϖ1,ϖ2 and uϖ0,ϖ1≠dϖ1,ϖ1; otherwise, there is a contradiction.
Ifuϖ0,ϖ1=dϖ0,ϖ1, then
(16)dϖ1,ϖ2≤rdϖ0,ϖ1≤sre−sr−1dϖ0,ϖ1.
Ifuϖ0,ϖ1=dϖ0,ϖ2, then
(17)dϖ1,ϖ2≤rdϖ0,ϖ2≤srdϖ0,ϖ1+dϖ1,ϖ2,which implies that dϖ1,ϖ2≤sre−sr−1dϖ0,ϖ1. Thus, (13) is true. Now, we suppose for all integers n≥2 and 1≤i≤n,
(18)dϖi,ϖn≤sre−sr−1dϖ0,ϖ1.
We have to prove(19)dϖi,ϖn+1≤sre−sr−1dϖ0,ϖ1,for n+1≥2,1≤i≤n+1. Now, we prove
(20)dϖ1,ϖn+1≤sre−sr−1dϖ0,ϖ1.
By (10), we obtain
(21)dϖ1,ϖn+1=dΦϖ0,Φϖn≤ruϖ0,ϖn,where
(22)uϖ0,ϖn∈dϖ0,ϖn,dϖ0,ϖ1,dϖn,ϖn+1,dϖ0,ϖn+1,dϖn,ϖ1.
Ifuϖ0,ϖn=dϖ0,ϖn, then
(23)dϖ1,ϖn+1≤rdϖ0,ϖn≤srdϖ0,ϖ1+dϖ1,ϖn=srdϖ0,ϖ1+sre−sr−1dϖ0,ϖ1=sre+∑i=1+∞sridϖ0,ϖ1=sre−sr−1dϖ0,ϖ1.
Ifuϖ0,ϖn=dϖ0,ϖ1, then dϖ1,ϖn+1≤rdϖ0,ϖ1≤sre−sr−1dϖ0,ϖ1.
Ifuϖ0,ϖn=dϖ0,ϖn+1, then
(24)dϖ1,ϖn+1≤rdϖ0,ϖn+1≤srdϖ0,ϖ1+dϖ1,ϖn+1,which yields e−srdϖ1,ϖn+1≤srdϖ0,ϖ1. That is, dϖ1,ϖn+1≤sre−sr−1dϖ0,ϖ1.
Ifuϖ0,ϖn=dϖ1,ϖn, then
(25)dϖ1,ϖn+1≤rdϖ1,ϖn≤r·sre−sr−1dϖ0,ϖ1≤sre−sr−1dϖ0,ϖ1.
At last, we only check that (20) is true when uϖ0,ϖn=dϖn,ϖn+1. That is,
(26)dϖ1,ϖn+1≤rdϖn,ϖn+1=rdΦϖn−1,Φϖn≤r2uϖn−1,ϖn,where
(27)uϖn−1,ϖn∈dϖn−1,ϖn,dϖn−1,ϖn,dϖn,ϖn+1,dϖn−1,ϖn+1,dϖn,ϖn.
Clearly,uϖn−1,ϖn≠dϖn,ϖn+1 and uϖn−1,ϖn≠dϖn,ϖn. If uϖn−1,ϖn=dϖn−1,ϖn, then
(28)dϖ1,ϖn+1≤r2dϖn−1,ϖn≤r2·sre−sr−1dϖ0,ϖ1≤sre−sr−1dϖ0,ϖ1.
Ifuϖn−1,ϖn=dϖn−1,ϖn+1, then
(29)dϖ1,ϖn+1≤rdϖn,ϖn+1≤r2dϖn−1,ϖn+1≤r3uϖn−2,ϖn,where
(30)uϖn−2,ϖn∈dϖn−2,ϖn,dϖn−2,ϖn−1,dϖn,ϖn+1,dϖn−2,ϖn+1,dϖn−1,ϖn.
Similarly, we also haveuϖn−2,ϖn≠dϖn,ϖn+1. If uϖn−2,ϖn equals to one of dϖn−2,ϖn,dϖn−2,ϖn−1 and dϖn−1,ϖn, then by the assumption (18), we have
(31)dϖ1,ϖn+1≤r3·sre−sr−1dϖ0,ϖ1≤sre−sr−1dϖ0,ϖ1.
It remains to check (20) when uϖn−1,ϖn=dϖn−2,ϖn+1; that is,
(32)dϖ1,ϖn+1≤rdϖn,ϖn+1≤r2dϖn−1,ϖn+1≤r3dϖn−2,ϖn+1.
By a similar analysis, we can deduce that(33)dϖ1,ϖn+1≤rdϖn,ϖn+1≤r2dϖn−1,ϖn+1≤⋯≤rn−1dϖ2,ϖn+1≤rnuϖ1,ϖn,where
(34)uϖ1,ϖn∈dϖ1,ϖn,dϖ1,ϖ2,dϖn,ϖn+1,dϖ1,ϖn+1,dϖn,ϖ2.
Sinceuϖ1,ϖn≠dϖn,ϖn+1 and uϖ1,ϖn≠dϖ1,ϖn+1, we know uϖ1,ϖn equals to one of dϖ1,ϖn,dϖ1,ϖ2 and dϖn,ϖ2. Therefore, we finally obtain
(35)dϖ1,ϖn+1≤rndϖ1,ϖn≤rn·sre−sr−1dϖ0,ϖ1≤sre−sr−1dϖ0,ϖ1or
(36)dϖ1,ϖn+1≤rndϖ1,ϖ2≤rn·sre−sr−1dϖ0,ϖ1≤sre−sr−1dϖ0,ϖ1or
(37)dϖ1,ϖn+1≤rndϖn,ϖ2≤rn·sre−sr−1dϖ0,ϖ1≤sre−sr−1dϖ0,ϖ1.
Hence, (20) is always true. Moreover, by (18), (20), and (33), we know
(38)dϖn,ϖn+1≤rn−1uϖ1,ϖn≤rn−1·sre−sr−1dϖ0,ϖ1≤sre−sr−1dϖ0,ϖ1.
For2≤i≤n+1, we have
(39)dϖi,ϖn+1=dΦϖi−1,Φϖn≤ruϖi−1,ϖn,while
(40)uϖi−1,ϖn∈dϖi−1,ϖn,dϖi−1,ϖi,dϖn,ϖn+1,dϖi−1,ϖn+1,dϖn,ϖi.
Ifuϖi−1,ϖn equals to one of dϖi−1,ϖn,dϖi−1,ϖi,dϖn,ϖi and dϖn,ϖn+1, then by (18) and (38),
(41)dϖi,ϖn+1≤r·sre−sr−1dϖ0,ϖ1≤sre−sr−1dϖ0,ϖ1.
Ifuϖi−1,ϖn=dϖi−1,ϖn+1, by (18) and (20), (33), and (38), we conclude that
(42)dϖi,ϖn+1≤rdϖi−1,ϖn+1≤ri−1uϖ1,ϖn≤ri−1·sre−sr−1dϖ0,ϖ1≤sre−sr−1dϖ0,ϖ1.
Therefore, (19) is true. The proof is finished.Now, we present and prove our main results without requiring the cone to be normal ord to be continuous.Theorem 11.
SupposeΦ:X⟶X is a generalized b-quasicontraction mapping in the Φ-orbitally complete space X,d, if ρr<1/s, then the mapping Φ possesses one and only one fixed point ζ∈X, and the sequence Φnϖ0 converges to ζ for each ϖ0∈X.Proof.
For eachϖ0∈X, take ϖn=Φϖn−1. If there is some n∈ℕ such that ϖn=ϖn+1=Φϖn, then ϖn is the fixed point. Hence, we assume ϖn≠ϖn+1 for all n∈ℕ. Let us show that ϖn is a Cauchy sequence. For any n>m>1, write Em,n=dϖi,ϖj:m≤i<j≤n. From the concept of generalized b-quasicontraction, for any u∈Em,n, there exits v∈Em−1,n satisfying u≤rv. Thus, it follows that
(43)dϖm,ϖn≤rum−1≤r2um−2≤⋯≤rm−1u1≤rm−1·sre−sr−1dϖ0,ϖ1,where um−1∈Em−1,n,um−2∈Em−2,n,⋯,u1∈E1,n. The last inequality is obtained by Lemma 10. Since srme−sr−1dϖ0,ϖ1⟶0 as m⟶∞rm⟶0asm⟶∞. Therefore, for any c∈A with c≫θ, there exists an integer N≥1 satisfying
(44)dϖm,ϖn≤srme−sr−1dϖ0,ϖ1≪c,n>m≥N.
That is,ϖn is a Cauchy sequence. By the Φ-orbital completeness of X,d, we have ζ∈X such that ϖn⟶ζ as n⟶∞. Next, we check Φζ=ζ. By (10), we see
(45)dϖn+1,Φζ=dΦϖn,Φζ≤ruϖn,ζ,where
(46)uϖn,ζ∈dϖn,ζ,dϖn,ϖn+1,dζ,Φζ,dϖn,Φζ,dζ,Φϖn.
There are the following three cases:
Case 1. Ifuϖn,ζ equals to one of dϖn,ζ,dϖn,ϖn+1 and dζ,Φϖn, then dϖn+1,Φζ is a c-sequence.
Case 2. Ifuϖn,ζ=dζ,Φζ, we get
(47)dϖn+1,Φζ≤rdζ,Φζ≤srdζ,ϖn+1+dϖn+1,Φζ,that is, dϖn+1,Φζ≤sre−sr−1dϖn+1,ζ. Thus, dϖn+1,Φζ is a c-sequence.
Case 3. Ifuϖn,ζ=dϖn,Φζ, we gain
(48)dϖn+1,Φζ≤rdϖn,Φζ≤srdϖn,ϖn+1+dϖn+1,Φζ,that is, dϖn+1,Φζ≤sre−sr−1dϖn,ϖn+1. Then, dϖn+1,Φζ is a c-sequence.
In summary,dϖn+1,Φζ is always a c-sequence. This gives ϖn⟶Φζ as n⟶∞. According to the uniqueness of the limit, we know ζ=Φζ.
It remains to prove the uniqueness ofζ. We assume there exists another fixed point ζ^ such that Φζ^=ζ^, and then
(49)dζ,ζ^=dΦζ,Φζ^≤ruζ,ζ^,where
(50)uζ,ζ^∈dζ,ζ^,dζ,Φζ,dζ^,Φζ^,dζ,Φζ^,dζ^,Φζ=dζ,ζ^,θ.
It is a contradiction. In conclusion, the fixed pointζ is unique, and the sequence Φnϖ0 converges to ζ. The proof is finished.Takings=1, we obtain the fixed point results in cone metric spaces over Banach algebras.Corollary 12.
SupposeT:X⟶X is a generalized quasicontraction mapping in Φ-orbitally complete cone metric space over Banach algebra A with a unit e, if ρr<1, then the mapping Φ possesses one and only one fixed point ζ∈X, and the sequence Φnϖ0 converges to ζ for every ϖ0∈X.Remark 13.
Theorem11 greatly improves Theorem 3.1 in [20], while Theorem 3.1 in [20] depends strongly on the continuity of b-metric. It also generalizes the condition sh<1 (i.e. h<1/s) of Theorem 2.13 in [23] to ρr<1/s. The assumption of completeness in Theorem 3.1 of [11] and Theorem 2.13 in [23] is relaxed by Φ-orbital completeness. Corollary 12 mainly improves and generalizes Theorem 9 in [28], while the results rely on the conditions that the cone is normal and the d is continuous.Turning to the next theorem, we show that another type ofb-quasicontraction in the space X,d has a unique fixed point when 1/s<ρr<1. Before giving the related result, we require an important lemma in [26].Lemma 14.
Supposeϖn is a sequence in X,d satisfying
(51)dϖn,ϖn+1≤rdϖn−1,ϖn,for some r∈P with ρr<1 and n∈ℕ. Then, ϖn is a Cauchy sequence in X,d.Theorem 15.
Suppose the spaceX,d is Φ-orbitally complete, Assume the mapping Φ:X⟶X satisfies
(52)dΦϖ,Φη≤ruϖ,η,for all ϖ,η∈X and ρr∈1/s,1, where
(53)uϖ,η∈dϖ,η,dϖ,Φϖ,dη,Φη,dϖ,Φη2s,dη,Φϖ2s.
IfΦ is k-continuous for some k≥1 or orbitally continuous, then the mapping Φ possesses one and only one fixed point ζ∈X, and the sequence Φnϖ0 converges to ζ for every ϖ0∈X.Proof.
From Theorem11, we obtain a sequence ϖn by ϖn+1=Φϖn and suppose ϖn≠ϖn+1 for all n∈ℕ and n≥0. In view of (52), we see that
(54)dϖn,ϖn+1=dΦϖn−1,Φϖn≤ruϖn−1,ϖn,where
(55)uϖn−1,ϖn∈dϖn−1,ϖn,dϖn−1,ϖn,dϖn,ϖn+1,dϖn−1,ϖn+12s,dϖn,ϖn2s=dϖn−1,ϖn,dϖn,ϖn+1,dϖn−1,ϖn+12s,θ.
We immediately getuϖn−1,ϖn≠dϖn,ϖn+1 and uϖn−1,ϖn≠θ. If uϖn−1,ϖn=dϖn−1,ϖn, then
(56)dϖn,ϖn+1≤rdϖn−1,ϖn.
Ifuϖn−1,ϖn=dϖn−1,ϖn+1/2s, then
(57)dϖn,ϖn+1≤r·sdϖn−1,ϖn+dϖn,ϖn+12s=rdϖn−1,ϖn+dϖn,ϖn+12,which gives
(58)dϖn,ϖn+1≤r2e−r−1dϖn−1,ϖn.
In fact,(59)ρr2e−r−1≤ρrρ2e−r−1≤ρr2−ρr<1.So, the conditions of Lemma 14 are satisfied for each case. By Lemma 14, ϖn is a Cauchy sequence. In view of Φ-orbital completeness of X,d, we have ζ∈X such that ϖn⟶ζ.
We are now in a position to show thatΦζ=ζ. If Φ is k-continuous, then Φkϖn⟶Φζ by k-continuity of Φ and Φk−1ϖn⟶ζ as n⟶∞. According to the uniqueness of the limit, we have Φζ=ζ.
IfΦ is orbitally continuous, then Φϖn⟶Φζ due to the fact that ϖn⟶ζ. This yields ζ=Φζ.
Uniqueness of the fixed point follows immediately from (52).Once we replace the set (53) by the following set (61), then the result is true without any continuity of the mapping Φ.Theorem 16.
SupposeX,d is the same as in Theorem 15, assume the mapping Φ:X⟶X that satisfies
(60)dΦϖ,Φη≤ruϖ,η,for all ϖ,η∈X and ρr∈1/s,1, where
(61)uϖ,η∈dϖ,η,dϖ,Φϖ,dη,Φηs,dϖ,Φη2s,dη,Φϖ.
Then, the mappingΦ possesses one and only one fixed point ζ∈X, and the sequence Φnϖ0 converges to ζ for every ϖ0∈X.Proof.
The proof is analogous to Theorem15. We at first gain a sequence ϖn=Φϖn−1=Φnϖ0,n≥1 and suppose that ϖn≠ϖn+1,∀n∈ℕ. By an analogous analysis with Theorem 15, ϖn⟶ζ for some ζ∈X. We proceed to show that ζ=Φζ. By (60), we see that
(62)dϖn,Φζ=dΦϖn−1,Φζ≤ruϖn−1,ζ,where
(63)uϖn−1,ζ∈dϖn−1,ζ,dϖn−1,ϖn,dζ,Φζs,dϖn−1,Φζ2s,dζ,ϖn.
There are the following three cases.
Case 1. Ifuϖn−1,ζ equals to one of dϖn−1,ζ,dϖn−1,ϖn and dζ,ϖn, then dϖn,Φζ is a c-sequence.
Case 2. Ifuϖn−1,ζ=dζ,Φζ/s, we have
(64)dϖn,Φζ≤rdζ,Φζs≤srdζ,ϖn+dϖn,Φζs,that is, dϖn,Φζ≤re−r−1dϖn,ζ. Thus, dϖn,Φζ is a c-sequence.
Case 3. Ifuϖn−1,ζ=dϖn−1,Φζ/2s, we have
(65)dϖn,Φζ≤rdϖn−1,Φζ2s≤srdϖn−1,ϖn+dϖn,Φζ2s,that is, dϖn,Φζ≤r2e−r−1dϖn−1,ϖn. Thus, dϖn,Φζ is a c-sequence.
In summary, we always deduce thatdϖn,Φζ is a c-sequence. This gives ϖn⟶Φζ as n⟶∞. Since the limit is unique, we know ζ=Φζ. The remaining proof is analogous to Theorem 11.Remark 17.
In Theorem15, we complement and perfect Theorem 11 in [8], which obtained the conclusions under the condition ρr<1/s in complete spaces X,d. Moreover, the conditions in Theorem 15 are much weaker than Theorem 2.1 in [20], since we obtain the results by k-continuity for some k≥1 or orbital continuity of the mapping, without appealing the continuity of cone b-metric or the mapping Φ.
According to the proof of Theorem16 and the symmetry of the cone b-metric d, we see at once that (61) can be replaced by
(66)uϖ,η∈dϖ,η,dϖ,Φϖs,dη,Φη,dϖ,Φη,dη,Φϖ2s,for all ϖ,η∈X.Example 18.
SetA=Cℝ10,2 with a norm ϖ=ϖ∞+ϖ′∞, for any ϖ∈A. The multiplication in A is taken as the pointwise multiplication. We conclude that A is a Banach algebra owing the unit element e=1. Let X=0,2. For all ϖ,η∈X, define
(67)dϖ,ηz=ϖ−η2ψ,ϖ,η∈0,1;2ϖ−η2ψ,otherwise,where ψ∈P=φz∈A:φz≥0,z∈0,2. Note that the cone P is nonnormal, and the cone b-metric d is discontinuous. Indeed, let ϖn=1+1/n,n∈ℕ, and then
(68)dϖn,1z=d1+1n,1z=21+1n−12ψ⟶0.
So,ϖn⟶1. However,
(69)dϖn,12z=d1+1n,12z=21+1n−122ψ⟶12ψ,d1,12z=1−122ψ=14ψ,that is,
(70)dϖn,12z↛d1,12z.
Thus, the coneb-metric d is discontinuous.
The mappingΦ:X⟶X is defined as
(71)Φϖ=ϖ3,ϖ∈0,1,0,ϖ∈1,2.
It suffices to show thatΦ is orbitally continuous rather than continuous (one also can check that Φ is k-continuous for each integer k≥2 but Φk is discontinuous for each k≥1). In addition, X,d is Φ-orbitally complete but not complete cone b-metric space over Banach algebra A with the coefficient s=4. In fact, for ϖn=2−1/n, n∈ℕ, we get
(72)dϖn,ϖm=21n−1m2ψ⟶0,n,m⟶∞,but there exists no ϖ∈X satisfying dϖn,ϖ⟶0. Hence, X,d is not complete. Take rz=z/4+1/2. We can calculate that ρr=3/4∈1/s,1. Now, there are the following three cases to verify the inequality (52).
For allϖ,η∈0,1, we get dΦϖ,Φηz=ϖ/3−η/32ψ and
(73)uϖ,ηz∈ϖ−η2ψ,ϖ−ϖ32ψ,η−η32ψ,18ϖ−η32ψ,18η−ϖ32ψ.Then, (52) holds by taking uϖ,ηz=ϖ−η2ψ.
For allϖ∈0,1,η∈1,2, we gain dΦϖ,Φηz=ϖ/3−02ψ=ϖ2/9ψ and
(74)uϖ,ηz∈2ϖ−η2ψ,ϖ−ϖ32ψ,2η−02ψ,18ϖ−02ψ,18·2η−ϖ32ψ=2ϖ−η2ψ,4ϖ29ψ,2η2ψ,ϖ28ψ,14η−ϖ32ψ.Then, (52) holds by taking uϖ,ηz=4ϖ2/9ψ.
For allϖ,η∈1,2, then Φϖ=Φη=0. We observe that dΦϖ,Φηz=0 and
(75)uϖ,ηz∈2ϖ−η2ψ,2ϖ−02ψ,2η−02ψ,18·2ϖ−02ψ,18·2η−02ψ=2ϖ−η2ψ,2ϖ2ψ,2η2ψ,ϖ24ψ,η24ψ.
So, (52) holds trivially. In the same manner, we can prove that (52) is true for all ϖ∈1,2, η∈0,1. Therefore, Φ possesses a unique fixed point 0∈X, and the sequence Φnϖ converges to 0 for each ϖ∈X by Theorem 15.
Furthermore, there is nor∈P with ρr∈0,1 satisfying ϖ2/9ψ≤2rϖ−η2ψ for
(76)ϖ=1∈0,1,η=1+19∈1,2andψz>0,in Case 2, which means that Φ is not a Banach-type b-contraction. The work from b-metric space, cone b-metric space, or cone b-metric space over Banach algebra which requires completeness, continuity, or Banach-type b-contraction is not applicable here (see [6, 8, 11, 18–20, 23, 26]). Due to the continuity of metric and cone metric, the corresponding theorems from such spaces (see [1–4, 7, 22, 24, 28]) cannot be used in this example, either.


## 3. Orbital Compactness

Garai et al. [29] and Haokip and Goswami [33] defined Φ-orbitally compact metric spaces and Φ-orbitally compact b-metric spaces, respectively, which extend sequentially compact metric (b-metric) spaces. Now, the similar definition of Φ-orbital compactness and a fixed point theorem of Ćirić-type b-quasicontraction in cone b-metric spaces over Banach algebras is showed.Definition 19.
The mappingΦ:X⟶X is named a Ćirić-type b-quasicontraction in X,d, if for all ϖ,η∈X with ϖ≠η, we have
(77)dΦϖ,Φη<uϖ,η,where
(78)uϖ,η∈dϖ,η,dϖ,Φϖ,dη,Φη,dϖ,Φη+dη,Φϖ2s.Definition 20.
Suppose the mappingΦ:X⟶X, the set X is Φ-orbitally compact, if each sequence in OΦϖ has a convergent subsequence for all ϖ∈X. Clearly, every Φ-orbitally compact cone b-metric space over Banach algebra does not need to be complete.Example 21.
Suppose the Banach algebraA and cone P are the same as in Example 8, take X=0,2×0,2 and Φ:X⟶X be given by Φϖ1,ϖ2=ϖ1/3,ϖ2/5. Then, X is Φ-orbitally compact rather than complete.In the last theorem, suppose thatX,d owes a regular cone P with dϖ,η∈intP, where ϖ,η∈X and ϖ≠η, the cone b-metric d is continuous.Theorem 22.
SupposeX,d is Φ-orbitally compact, if Φ:X⟶X is a Ćirić-type b-quasicontraction and orbitally continuous, then the mapping Φ possesses one and only one fixed point ζ∈X.Proof.
For anyϖ0∈X, set ϖn=Φϖn−1=Tnϖ0,n≥1. Note that ϖn≠ϖn+1 for all n∈ℕ. In fact, if ϖn=ϖn+1=Φϖn for some n∈ℕ, then ϖn is a fixed point of Φ. Let ln=dϖn,ϖn+1 for every n∈ℕ. From (77), we know
(79)ln=dϖn,ϖn+1=dΦϖn−1,Φϖn<uϖn−1,ϖn,where
(80)uϖn−1,ϖn∈dϖn−1,ϖn,dϖn−1,ϖn,dϖn,ϖn+1,dϖn−1,ϖn+1+dϖn,ϖn2s.
Indeed,uϖn−1,ϖn≠dϖn,ϖn+1; so, it remains the following two cases.
Case 1. Whenuϖn−1,ϖn=dϖn−1,ϖn, then dϖn,ϖn+1<dϖn−1,ϖn.
Case 2. Whenuϖn−1,ϖn=dϖn−1,ϖn+1+dϖn,ϖn/2s, then
(81)dϖn,ϖn+1<dϖn−1,ϖn+1+dϖn,ϖn2s≤sdϖn−1,ϖn+dϖn,ϖn+12s,which means dϖn,ϖn+1<dϖn−1,ϖn. Thus, by repeating the above process, we deduce that
(82)θ<ln=dϖn,ϖn+1<ln−1<⋯<l0=dϖ1,ϖ0.
By the regularity of the cone, there existsc0∈A,c0≥θ such that ln⟶c0 as n⟶∞. Because X is Φ-orbitally compact, we have a convergent subsequence ϖni of ϖn and a point ζ∈X satisfying ϖni⟶ζasn⟶∞, according to orbital continuity of Φ, Φϖni⟶Φζ.
Whenc0>θ, we gain
(83)θ<c0=limi⟶∞dϖni,Φϖni=dζ,Φζ.
Moreover, since the cone is regular, we see that(84)θ<c0=limi⟶∞lni=limi⟶∞dΦϖni,Φ2ϖni=dΦζ,Φ2ζ<uζ,Φζ,where
(85)uζ,Φζ∈dζ,Φζ,dζ,Φζ,dΦζ,Φ2ζ,dζ,Φ2ζ+dΦζ,Φζ2s.
It is evidence to get(86)c0=dΦζ,Φ2ζ<dζ,Φζ=c0,a contradiction. So, c0=θ and ζ=Φζ, namely, ζ is a fixed point.
Finally, let us prove that the fixed point is unique by (77). Otherwise, if there is another fixed point η, then
(87)dη,ζ=dΦη,Φζ<uη,ζ,where
(88)uη,ζ∈dη,ζ,dη,Φη,dζ,Φζ,dη,Φζ+dζ,Φη2s=dη,ζ,θ,dη,Φζ+dζ,Φη2s.
Ifuη,ζ=dη,ζ or uη,ζ=θ, then this is a contradiction. If uη,ζ=dη,Φζ+dζ,Φη/2s, then
(89)dη,ζ<dη,Φζ+dζ,Φη2s≤sdη,ζ+dζ,Φζ+sdζ,η+dη,Φη2s=dη,ζ,a contradiction too. The result follows.



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*Source: 7861865-2022-07-04.xml*