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Algebraic cone metric spaces and fixed point theorems of contractive mappings
Fixed Point Theory and Applications volume 2014, Article number: 160 (2014)
Abstract
In this paper we introduce algebraic cone metric spaces, prove some fixed point theorems of contractive mappings on algebraic cone metric spaces, and improve some well-known results in the non-normal case.
1 Introduction and preliminary
There exist a number of generalizations of metric spaces. One of them is the cone metric space initiated by Huang and Zhang [1]. They described the convergence in cone metric spaces, introduced the notion of completeness and proved some fixed point theorems of contractive mappings on these spaces. Let us recall some notions and definitions.
Let be a real Banach space and be a subset of . is called a cone if and only if
-
(i)
is closed and nonempty and ,
-
(ii)
if ; ; , then ,
-
(iii)
if and , then .
Given a cone , a partial ordering ⪯ with respect to is defined by if . Furthermore, we write if and , while will stands for , where is the interior of .
The cone is called normal if there is a number , such that for every , implies . The least positive number satisfying the above condition is called the normal constant of .
Lemma 1.1 The following conditions are equivalent for a cone in the Banach space :
-
(a)
.
-
(b)
is a normal cone.
-
(c)
For arbitrary sequences , and in E, if for each n and , then .
-
(d)
There exists a norm on , equivalent to the given norm , such that the cone is monotone with respect to , i.e., if , then .
Definition 1.2 Let X be a nonempty set, and let be a real Banach space and be a cone. Suppose the mapping satisfies:
-
(i)
for every and if and only if ;
-
(ii)
for every ;
-
(iii)
for every .
Then d is called a cone metric on X and is called a cone metric space.
Let be a cone metric space, be a sequence in X, and . is said to be convergent to x, if for every there is , such that for all , . Likewise, is called a Cauchy sequence in X if for every there is , such that for each , . A cone metric space X is said to be complete if every Cauchy sequence in X is convergent in X.
To replace the standard properties of a metric, the properties in the following lemma of cone metrics are often useful while dealing with cone metrics when the cone is not normal.
Lemma 1.3 Let be a cone metric space with corresponding cone . If and and are sequences in X and , respectively, then:
-
(a)
If for all , then .
-
(b)
If and , then for each there exists a natural number , such that for each .
-
(c)
If and , then there exists a natural number , such that for every , we have .
Proof See [3], p.2598. □
It follows from Lemma 1.3(c) that the sequence converges to if , and is a Cauchy sequence if as . The converse is true if is a normal cone. To see that does not necessarily imply at non-normal cones, see example (e) in Section 4.
There are several states of cone metric spaces. For example, after cone metric spaces over topological vector spaces, in [6] the authors introduce the concept of tvs-cone b-metric spaces over a solid cone. In the following section another state of metric spaces is presented: an algebraic cone metric space.
2 Algebraic cones and Banach cone algebras
The concept of cone metric spaces over Banach algebras was previously introduced by Liu and Xu in [7]. They proved some fixed point theorems of generalized Lipschitz mappings with weaker conditions on the generalized Lipschitz constant k by means of the spectral radius (see Theorems 2.1, 2.2, and 2.3 in [7]). Indeed for the mapping the generalized Lipschitz condition is
for all , where with .
The same authors in [8] introduced the concept of quasicontractions on cone metric spaces over Banach algebras and proved the existence and uniqueness of fixed points of such mapping. Afterwards Xu and Radenović deleted the superfluous assumption of normality in [7] (see Theorems 3.1, 3.2, and 3.3 in [9]).
In this section, we will present the definition of the algebraic cones, Banach cone algebras, and some properties related to this concept.
Definition 2.1 (cf. [10, 11] and [12])
Let be a cone, and X be a vector space over ℂ. Suppose the mapping satisfies:
(N1) if and only if ,
(N2) , for every ,
(N3) for every and .
Then is called a cone norm on X, and the pair is called a cone normed space (CNS).
Sometimes, to emphasize the cone , we use instead of .
A real topological algebra is an algebra over furnished with a topology τ such that
-
(i)
the map is continuous,
-
(ii)
the map is continuous,
-
(iii)
the map is continuous.
When , is called a complex topological algebra and if , is called a real topological algebra.
It is obvious that every Banach algebra is a topological algebra.
Definition 2.2 Let be a Banach algebra with identity element and be a cone. is an algebraic cone if , and for each , .
Therefore, for each and if and , then .
For example if is a -algebra and is its positive elements, i.e. the set of all Hermitian element with , then is a cone in , and whenever is commutative, is an algebraic cone.
Definition 2.3 Let X be an algebra, be a real topological algebra (or as a special case be a Banach algebra), and be an algebraic cone in . Furthermore, let be a cone norm, such that is a Banach cone metric space, and for each . Then we say that is a cone Banach cone algebra.
In the following, X is an algebra with identity and is a Banach algebra with identity . Moreover, in Section 4, some examples are presented in the topological algebra case.
Theorem 2.4 Suppose is a normal Banach cone algebra, and , then
-
(a)
is invertible,
-
(b)
,
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(c)
for every homomorphism , such that .
Proof (a) Since and , the elements
form a Cauchy sequence X by the normality of . Since X is complete, there exists , such that . As and , the continuity of multiplication implies that s is the inverse of .
(b) .
(c) Suppose , . By (a) is invertible. So . Hence
This completes the proof. □
Note that if the condition in Theorem 2.5 is replaced by , then (a) is still true. In the next theorem, let be the group of invertible elements of X.
Theorem 2.5 If X is a Banach cone algebra on , then is open in X and the mapping is a homeomorphism.
Proof Let and . If , then by Theorem 2.4, and belong to (notice that ) and is contained in . Also by Theorem 2.4,
Therefore, . Hence the mapping is continuous and homeomorphism. □
Note that for each , the spectrum of x is the set of all complex numbers λ, such that is not invertible. The complement of is the resolvent set of x. It consists of all for which there exists .
The spectral radius of x is the number . For every , and if then lies in , by Theorem 2.4, and so does . Thus, . This proves that is an upper bound for , and . In addition by [13], there exists a norm on , such that in is a normal cone with constant normal 1. Hence we can assume that . So is compact and
(see Theorem 10.13 in [14]). Since the combination with is a real-valued norm, we can present an extension of the Gelfand-Mazur theorem.
Theorem 2.6 (Gelfand-Mazur)
If X is a normal Banach cone algebra in which every nonzero element is invertible, then X is isometrically isomorphic to the complex field.
3 Fixed point theorems in algebraic cone metric spaces
In this section some fixed point theorems of generalized Lipschitz mappings with weaker conditions than the condition , are proved (see Definition 3.1 for the condition). Therefore some theorems in [7] and [9] are improved (see Theorems 3.3 and 3.9 for instance).
Let be an algebraic cone metric space and be a self map. F is called Lipschitzian if and only if there exists a constant , called the Lipschitz coefficient for F, such that
for each . In a similar way, we can say that F is a contraction if . In the following, we will see that this definition of contraction maps is general.
When is a complete metric space, and F is a contraction, we know that the statement ‘ if and only if ’ has an important role in the proof of Banach fixed point theorem. But in a Banach algebra , the statement ‘ if and only if ’ is not true. In fact for , does not imply that (see example (f) in Section 4). For this reason, we have to state a scholastic definition of contraction.
Definition 3.1 Let be an algebraic cone metric space, and let be a Lipschitzian map with Lipschitz coefficient L. F is a contraction map if , i.e. the series is absolutely convergent.
Let be a Lipschitz coefficient for a self-map F, then each of the properties and imply that , but example (f) of Section 4 shows that the converse cannot necessarily be established.
Lemma 3.2 Let be a Banach algebra with a unit , and . If , then is invertible and .
Proof The proof is obvious. □
Before presenting some fixed point theorems we remark that although each cone metric space over a solid cone is metrizable, as was shown by various methods, however, not all fixed point results can be reduced in this way to their standard metric counterparts. So this line of investigation is still open (see, e.g., [15]).
The next theorem is an extension of Theorem 2.1 in [7], Theorem 3.1 in [9] and Theorem 1.1 in [16].
Theorem 3.3 Let be a complete algebraic cone metric space and be a contraction map with Lipschitz coefficient L. Then F has a unique fixed point .
Furthermore, for each , we have with
Proof Since F is a contraction with Lipschitz coefficient L, we have
for each , and is absolutely convergent. For existence, select . We first show that is Cauchy sequence. Let , then
Thus for ,
that is, for , and ,
By Lemma 1.3(b), this shows that is a Cauchy sequence and, since X is complete, there exists a with . Hence by continuity . It is obvious that u is unique. □
For each and , let and . Then the collection forms a basis for the cone metric topology of . Also it is easily seen that, for each , is an open set in cone metric space . As in the real case, the equality does not necessarily hold for all and . In the next lemma we express a necessary and sufficient condition for this equality.
Lemma 3.4 If is a cone metric space, then the following statements are equivalent.
-
(a)
For each and each , .
-
(b)
For each and each , there exists , such that and .
Proof (a) implies (b). If and , then . Now (a) implies that and so for each , there exists such that .
(b) implies (a). For each , if then . If , then let . So there exists such that and . Hence there exists a sequence in such that . This implies . □
Corollary 3.5 Let be an algebraic cone metric space. Then for each and each , .
Theorem 3.6 Let be a complete algebraic cone metric space, and . Suppose is a contraction with Lipschitz coefficient L, and
Then F has a unique fixed point in .
Proof There exists a , such that , with . (Otherwise for each we have and then .) We will show that .
Let , then
By Theorem 3.3, F has a unique fixed point in . Again it is easy to see that F has only one fixed point in . □
Before going to the next theorem, we note that in a Banach algebra for each with we can assume that . Indeed, if , then is invertible and . So . Now by Lemma 1.3(c), there exists a natural number N, such that .
Theorem 3.7 Let C be a nonempty, closed, and convex subset of an algebraic cone norm space X. Also let be a Lipschitzian map with Lipschitz coefficient L such that and be a subset of a compact subset of C. Then F has a fixed point.
Proof Since and for all , let and, for , define
Since C is convex and , is a contraction. Therefore, by Theorem 3.3 each has a unique fixed point . That is,
In addition, since lies in a compact subset of C, there exists a subsequence S of integers and a with in S. Thus,
in S. So by continuity, as in S, and therefore . □
Theorem 3.8 (See Theorem 2.1 in [17])
Let be a complete cone metric space, and be an algebraic cone. Suppose that the mapping f and g are two self-maps of X satisfying
for all , where and
Then f and g have a unique common fixed point in X. Moreover, any fixed point of f is a fixed point of g, and vice versa.
Proof Suppose is an arbitrary point of X and define by , for . Then
which implies that , where
(since , is invertible). Similarly, it can be shown that
Therefore, for all n we have
and, hence, for ,
so that Lemma 1.3(b) implies . Hence is a Cauchy sequence. Since X is complete, there exists , such that . Now by (3.1),
and so
Thus , by Lemma 1.3(a). Now, , and so
implies .
To prove uniqueness, suppose that q is another common fixed point of f and g, then , which gives , and . □
Theorem 3.9 (See Theorem 2.2 in [7] and Theorem 3.2 in [9])
Let be a complete cone metric space and be an algebraic cone. Suppose the mapping satisfies the contraction condition
for all , where , and are absolutely convergent. Then T has a unique fixed point in X, and for any , the iterative sequence converges to the fixed point.
Proof Choose and for each set . Then
So . Since is invertible, so
Let , then h is invertible. For ,
Since converges to , so is a Cauchy sequence. Therefore, is Cauchy. Since X is complete, there is a , such that . Thus,
So and
Since d is continuous, so and hence .
Now, let be another fixed point of T, then
Hence . □
Remark 3.10 In Theorem 3.9 if , then the series and show absolute convergence.
Theorem 3.11 Let be a complete algebraic cone metric space, , be absolutely convergent and be a function. If for each there exists a positive integer , such that
for some
and for all , then T has a unique fixed point . Moreover, for every , .
Proof The proof is similar to Theorem 3.1 in [18]. □
4 Examples
In this section, and , will denote the set of all complex-valued and the set of all continuous complex-valued functions on a topological space X, respectively. There are several natural ways of introducing a topology on and and they all follow the general approach indicated below.
Let S be a family of subsets of X. For each and each U of open subset of ℂ, let consisting the set of all for which . Put . Sets of this form are a subbase of some topology on , or similarity on , called the topology of convergence with respect to elements of S. Under this construction, the space is a subspace of the space . The most important topologies obtained in this way are the following: the topology of pointwise convergence, corresponding to the case when S consists of all finite subsets of X, and the compact-open topology, arising when S consists of all compact subsets of the space X. The topologies of uniform convergence on and are described in slightly different ways. Their bases consist of the sets and , respectively. If X is a completely regular compact space, then the topology of uniform convergence and the compact-open topology are equal on .
-
(a)
Let be equipped with the pointwise topology. So in , a sequence converges to if and only if converges to for each x in X. It is obvious that under the pointwise topology is a topological algebra (see the definitions in Section 2). Define
It is obvious that is an algebraic cone. Now define
by for each . It is obvious that has the following properties:
-
(i)
for each and each .
-
(ii)
for each and each .
-
(iii)
for each .
-
(iv)
if and only if .
is a vector-valued inner product on . Now define . It is obvious that is a cone algebraic norm on . It is obvious that . is a Banach cone algebra. For this purpose, we prove that is complete. Let be a Cauchy sequence, then for each , is a Cauchy sequence in ℂ. Therefore, there exists , such that is pointwise convergent to g. This shows that is complete.
(b) Let be a completely regular non-pseudocompact topological space. Let be equipped with pointwise convergence topology. It is obvious that is a topological subalgebra of . But is not a cone complete subalgebra of , because is not closed.
(c) Let be a completely regular non-pseudocompact topological space. Let be equipped with uniform convergence topology. A sequence converges to if and only if uniformly converges to g on X. under uniform convergence on X is not a topological algebra, because multiplication is not continuous.
(d) Let be a completely regular non-pseudocompact topological space. Let be equipped with compact-open topology. A sequence converges to if and only if uniformly converges to g on every compact subset of X. It is obvious that under the compact-open topology is a topological algebra. Define
It is obvious that is a cone. Now define
by for each . It is obvious that has the following properties:
-
(i)
for each and each .
-
(ii)
for each and each .
-
(iii)
for each .
-
(iv)
if and only if .
is a vector-valued inner product on . Now define . It is obvious that is a cone algebraic norm on . It is obvious that . is a Banach cone algebra. For this we prove that is complete. Let be a Cauchy sequence, then, for each compact subset of X, is uniformly Cauchy, so for each is uniformly convergent. Therefore, there exists such that is compact-open convergent to g. This shows that is complete.
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(e)
Let with the norm and , that is, a non-normal solid cone ([2], p.221). We note that:
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(1)
with pointwise operations is a (cone) Banach algebra.
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(2)
Consider for and and let , where 0 is the constant function zero. Further, define the cone metric by . Then in the cone metric space , however, . Therefore is a Banach cone metric space with an algebraic non-normal cone.
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(3)
Suppose X is as (2) and define by , and . Then is a cone metric space too and for and , since , we have in the cone metric space , but . Therefore is a Banach cone metric space with a non-continuous metric.
-
(f)
This example was first treated by Kaplansky for another purpose. Let be the natural orthonormal basis of . The weighted shift operators defined by
for some bounded sequence of weights (see [19], Example 2.2.16). The norm of such an operator is easily seen to be
For each , define and for each . The norm of is given by
So . This implies that is convergent and .
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Tootkaboni, M.A., Bagheri Salec, A. Algebraic cone metric spaces and fixed point theorems of contractive mappings. Fixed Point Theory Appl 2014, 160 (2014). https://doi.org/10.1186/1687-1812-2014-160
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DOI: https://doi.org/10.1186/1687-1812-2014-160