Abstract

In this research, we first check that the abstract cone -metric is discontinuous in the case of normal cone by a counter example. We obtain several meaningful results about generalized -quasicontraction and Ćirić-type -quasicontraction in cone -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.

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 ([3], Theorem 2.2). Sequentially, Gajić and Rakoević [4] showed the result holds when in the same spaces. In 2011, the notion of cone -metric space was given by Hussian and Shah [5], which generalized -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 is substituted with the Banach algebra . 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 -metric space over Banach algebra. They obtained some fixed points of contractions in such spaces which were not equivalent to the corresponding work in -metric spaces. Later on, numerous interesting fixed point theorems in these spaces were promoted to be studied by many scholars, see [919] and their references, but most of the results were established under the completeness of the spaces and some even required the continuity of -metric (while cone metric and metric are continuous) (see [14, 68, 1829]).

In 2018, Aleksić et al. [20] proved that -metric is discontinuous in general by some examples, which is a generalization of metric. However, the fixed point conclusions of -quasicontraction in -metric spaces were also discussed under continuous -metric and complete -metric spaces. In order to improve these too strong conditions, we prove that the abstract cone -metric is discontinuous even with a normal cone. Furthermore, we gain some fixed point results in cone -metric spaces over Banach algebras when the cone -metric is discontinuous. Some other conditions are weakened by giving several new concepts, such as -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.

Suppose is a real Banach algebra and is a cone over Banach algebra with , the notation expresses the partial ordering in terms of . For the definitions of Banach algebras and cones, the readers may refer to [24, 31].

Definition 1. (see [5, 8]). Suppose is a nonempty set and is a constant, the mapping is said to be a cone -metric if
(d1) for all and if and only if
(d2) for all
(d3) for all
The pair is called a cone -metric space over Banach algebra .

Definition 2. (see [8]). Suppose is a cone -metric space over Banach algebra , , and is a sequence in , we say (i) converges to if for each with , there is an integer such that for all (ii) is a Cauchy sequence if for each with , there is an integer such that for all (iii) is complete if each Cauchy sequence in is convergent

It is significant to note that different from the usual metric and cone metric with a normal cone, cone -metric is generally discontinuous even with a normal cone. Let us show an example.

Example 3. Take with a norm For any and in , set the multiplication as Let . Then, is a real Banach algebra owing the unit element , and the cone is normal. Set and be defined as In the above definition, is odd if both and are odd; is if both and are . We can check that is a cone -metric space over Banach algebra where .
Let , . We have which indicates but So, we have showed that the cone -metric is discontinuous in the case of normal cone.

Definition 4. (see [25]). Suppose is a solid cone, . A sequence is a -sequence if for any , there is such that for all .

Lemma 5. (see [8]). Suppose is a solid cone, . If , , and are -sequences in , then is a -sequence in .

Lemma 6. (see [31]). Suppose is a Banach algebra with a unit and , if the spectral radius of satisfies then is invertible. Actually,

Throughout this paper, we always suppose is a cone -metric space over Banach algebra with a unit and .

2. Orbital Completeness

In this section, we give several fixed point theorems of generalized -quasicontraction in orbitally complete cone -metric spaces over Banach algebras. The cone is neither regular nor normal. The cone -metric and the self mapping are not required continuous. At first, encouraged by the concepts of orbital continuity, -orbital completeness [32], and -continuity [21] in usual metric space, we provide the analogous concepts in cone -metric space over Banach algebra , which are important in our proof.

Definition 7. Suppose is a cone -metric space over Banach algebra and , take and , namely, the orbit of under .

The mapping is orbitally continuous at an element if for any sequence (for all ), as implies as . Note that each continuous mapping is orbitally continuous, but not the converse.

The mapping is -continuous for , if implies (). It is clear that is 1-continuous if and only if it is continuous, and -continuity implies -continuity for any but not the converse. Furthermore, continuity of and -continuity of are independent when . See the following examples.

Example 8. Suppose with a norm, For any and in , set the multiplication as Let . It follows that there is a unit element in the real Banach algebra . Let and define by for any . It is obvious that is complete and . Suppose is defined as For any if , then implies while . Clearly, is an orbitally continuous mapping rather than continuous. Moreover, notice that is 2-continuous but not continuous, that is, 2-continuity of does not imply continuity of . Furthermore, for each integer , is discontinuous while is -continuous. This indicates that -continuity of does not imply continuity of in usual situation.

In [12], we have given the following definitions of -orbital completeness.

Definition 9. The space is named -orbitally complete, if each Cauchy sequence included in for some converges in . Each complete space is -orbitally complete for any but not the converse.

For being convenient, we give the notion of generalized -quasicontraction in . The mapping is named a generalized -quasicontraction if there exists with , and one has where

When , 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 mapping is a generalized -quasicontraction in the space , for each , let . Then, for any integers , it holds that

Proof. At first, we show that for any and , If , the result is trivial. Assume , then where Obviously, and ; otherwise, there is a contradiction.
If , then If , then which implies that . Thus, (13) is true. Now, we suppose for all integers and , We have to prove for . Now, we prove By (10), we obtain where If , then If , then .
If , then which yields . That is, .
If , then At last, we only check that (20) is true when . That is, where Clearly, and . If , then If , then where Similarly, we also have . If equals to one of and , then by the assumption (18), we have It remains to check (20) when ; that is, By a similar analysis, we can deduce that where Since and , we know equals to one of and . Therefore, we finally obtain or or Hence, (20) is always true. Moreover, by (18), (20), and (33), we know For , we have while If equals to one of and , then by (18) and (38), If by (18) and (20), (33), and (38), we conclude that Therefore, (19) is true. The proof is finished.

Now, we present and prove our main results without requiring the cone to be normal or to be continuous.

Theorem 11. Suppose is a generalized -quasicontraction mapping in the -orbitally complete space , if , then the mapping possesses one and only one fixed point , and the sequence converges to for each .

Proof. For each , take . If there is some such that , then is the fixed point. Hence, we assume for all . Let us show that is a Cauchy sequence. For any , write . From the concept of generalized -quasicontraction, for any , there exits satisfying . Thus, it follows that where . The last inequality is obtained by Lemma 10. Since as . Therefore, for any with , there exists an integer satisfying That is, is a Cauchy sequence. By the -orbital completeness of , we have such that as . Next, we check . By (10), we see where There are the following three cases:
Case 1. If equals to one of and , then is a -sequence.
Case 2. If , we get that is, . Thus, is a -sequence.
Case 3. If , we gain that is, . Then, is a -sequence.
In summary, is always a -sequence. This gives as . According to the uniqueness of the limit, we know .
It remains to prove the uniqueness of . We assume there exists another fixed point such that , and then where It is a contradiction. In conclusion, the fixed point is unique, and the sequence converges to . The proof is finished.

Taking , we obtain the fixed point results in cone metric spaces over Banach algebras.

Corollary 12. Suppose is a generalized quasicontraction mapping in -orbitally complete cone metric space over Banach algebra with a unit , if , then the mapping possesses one and only one fixed point , and the sequence converges to for every .

Remark 13. Theorem 11 greatly improves Theorem 3.1 in [20], while Theorem 3.1 in [20] depends strongly on the continuity of -metric. It also generalizes the condition (i.e. ) of Theorem 2.13 in [23] to . The assumption of completeness in Theorem 3.1 of [11] and Theorem 2.13 in [23] is relaxed by -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 is continuous.

Turning to the next theorem, we show that another type of -quasicontraction in the space has a unique fixed point when . Before giving the related result, we require an important lemma in [26].

Lemma 14. Suppose is a sequence in satisfying for some with and . Then, is a Cauchy sequence in .

Theorem 15. Suppose the space is -orbitally complete, Assume the mapping satisfies for all and , where If is -continuous for some or orbitally continuous, then the mapping possesses one and only one fixed point , and the sequence converges to for every .

Proof. From Theorem 11, we obtain a sequence by and suppose for all and . In view of (52), we see that where We immediately get and . If , then If , then which gives In fact, So, the conditions of Lemma 14 are satisfied for each case. By Lemma 14, is a Cauchy sequence. In view of -orbital completeness of , we have such that .
We are now in a position to show that . If is -continuous, then by -continuity of and as . According to the uniqueness of the limit, we have .
If is orbitally continuous, then due to the fact that . This yields .
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 .

Theorem 16. Suppose is the same as in Theorem 15, assume the mapping that satisfies for all and , where Then, the mapping possesses one and only one fixed point , and the sequence converges to for every .

Proof. The proof is analogous to Theorem 15. We at first gain a sequence and suppose that . By an analogous analysis with Theorem 15, for some . We proceed to show that . By (60), we see that where There are the following three cases.
Case 1. If equals to one of and , then is a -sequence.
Case 2. If , we have that is, . Thus, is a -sequence.
Case 3. If , we have that is, . Thus, is a -sequence.
In summary, we always deduce that is a -sequence. This gives as . Since the limit is unique, we know . The remaining proof is analogous to Theorem 11.

Remark 17. In Theorem 15, we complement and perfect Theorem 11 in [8], which obtained the conclusions under the condition in complete spaces . Moreover, the conditions in Theorem 15 are much weaker than Theorem 2.1 in [20], since we obtain the results by -continuity for some or orbital continuity of the mapping, without appealing the continuity of cone -metric or the mapping .
According to the proof of Theorem 16 and the symmetry of the cone -metric , we see at once that (61) can be replaced by for all .

Example 18. Set with a norm , for any . The multiplication in is taken as the pointwise multiplication. We conclude that is a Banach algebra owing the unit element . Let . For all , define where . Note that the cone is nonnormal, and the cone -metric is discontinuous. Indeed, let , and then So, . However, that is, Thus, the cone -metric is discontinuous.
The mapping is defined as It suffices to show that is orbitally continuous rather than continuous (one also can check that is -continuous for each integer but is discontinuous for each ). In addition, is -orbitally complete but not complete cone -metric space over Banach algebra with the coefficient . In fact, for , , we get but there exists no satisfying Hence, is not complete. Take . We can calculate that . Now, there are the following three cases to verify the inequality (52).
For all , we get and Then, (52) holds by taking .
For all , we gain and Then, (52) holds by taking .
For all , then . We observe that and So, (52) holds trivially. In the same manner, we can prove that (52) is true for all , . Therefore, possesses a unique fixed point and the sequence converges to for each by Theorem 15.
Furthermore, there is no with satisfying for in Case 2, which means that is not a Banach-type -contraction. The work from -metric space, cone -metric space, or cone -metric space over Banach algebra which requires completeness, continuity, or Banach-type -contraction is not applicable here (see [6, 8, 11, 1820, 23, 26]). Due to the continuity of metric and cone metric, the corresponding theorems from such spaces (see [14, 7, 22, 24, 28]) cannot be used in this example, either.

3. Orbital Compactness

Garai et al. [29] and Haokip and Goswami [33] defined -orbitally compact metric spaces and -orbitally compact -metric spaces, respectively, which extend sequentially compact metric (-metric) spaces. Now, the similar definition of -orbital compactness and a fixed point theorem of Ćirić-type -quasicontraction in cone -metric spaces over Banach algebras is showed.

Definition 19. The mapping is named a Ćirić-type -quasicontraction in , if for all with , we have where

Definition 20. Suppose the mapping , the set is -orbitally compact, if each sequence in has a convergent subsequence for all . Clearly, every -orbitally compact cone -metric space over Banach algebra does not need to be complete.

Example 21. Suppose the Banach algebra and cone are the same as in Example 8, take and be given by . Then, is -orbitally compact rather than complete.

In the last theorem, suppose that owes a regular cone with , where and , the cone -metric is continuous.

Theorem 22. Suppose is -orbitally compact, if is a Ćirić-type -quasicontraction and orbitally continuous, then the mapping possesses one and only one fixed point .

Proof. For any , set Note that for all . In fact, if for some , then is a fixed point of . Let for every . From (77), we know where Indeed, ; so, it remains the following two cases.
Case 1. When , then .
Case 2. When , then which means . Thus, by repeating the above process, we deduce that By the regularity of the cone, there exists such that as . Because is -orbitally compact, we have a convergent subsequence of and a point satisfying , according to orbital continuity of , .
When , we gain Moreover, since the cone is regular, we see that where It is evidence to get a contradiction. So, and , namely, is a fixed point.
Finally, let us prove that the fixed point is unique by (77). Otherwise, if there is another fixed point , then where If or , then this is a contradiction. If then a contradiction too. The result follows.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

The research is partially supported by the Special Basic Cooperative Research Programs of Yunnan Provincial Undergraduate Universities’ Association (No. 202101BA070001-045), Teaching Reform Research Project of Zhaotong University in 2021-2022 Academic Year (Nos. Ztjx202203, Ztjx2022014) and Teaching Team for Advanced Algebra (No. Ztjtd202108).