- Research Article
- Open Access
Best Approximations Theorem for a Couple in Cone Banach Space
© Erdal Karapınar and Duran Türkoğlu. 2010
- Received: 23 March 2010
- Accepted: 8 June 2010
- Published: 29 June 2010
The notion of coupled fixed point is introduced by Bhaskar and Lakshmikantham, (2006). In this manuscript, some result of Mitrović, (2010) extended to the class of cone Banach spaces.
- Fixed Point Theorem
- Normal Constant
- Finite Subset
- Unique Fixed Point
- Couple Fixed Point
for all , has a unique fixed point. Recently, many results on fixed point theorems have been extended to cone metric spaces (see, e.g., [3, 5–11]). In , the authors extends to cone metric spaces over regular cones. In this manuscript, some results of some result of Mitrović in  are extended to the class of cone metric spaces.
Throughout this paper stands for real Banach space. Let always be closed subset of . is called cone if the following conditions are satisfied:
for all and nonnegative real numbers ,
For a given cone , one can define a partial ordering (denoted by or ) with respect to by if and only if . The notation indicates that and while will show , where denotes the interior of . It can be easily shown that and where . Throughout this manuscript .
The cone is called
regular if every increasing sequence which is bounded from above is convergent. That is, if is a sequence such that for some , then there is such that .
In , the least positive integer satisfying (1.2) is called the normal constant of . Note that, in [3, 5], normal constant is considered a positive real number, ( ), although it is proved that there is no normal cone for in (see e.g., Lemma , ).
Lemma 1.1 (see e.g., ).
One has the following.
(i)Every regular cone is normal.
(ii)For each , there is a normal cone with normal constant .
(iii)The cone is regular if every decreasing sequence which is bounded from below is convergent.
Definition 1.2 (see ).
is called minihedral cone if exists for all ; and strongly minihedral if every subset of which is bounded from above has a supremum.
Let with the supremum norm and Since the sequence is monotonically decreasing, but not uniformly convergent to , thus, is not strongly minihedral.
Let be nonempty set. Suppose that the mapping satisfies the following:
for all ,
if and only if ,
, for all .
Then is called cone metric on , and the pair is called a cone metric space (CMS).
Let and and . Define by , where are positive constants. Then is a CMS. Note that the cone is normal with the normal constant
It is quite natural to consider Cone Normed Spaces (CNSs).
Let be a vector space over . Suppose that the mapping satisfies the following:
for all ,
if and only if ,
, for all .
for all .
Then is called cone norm on , and the pair is called a cone normed space (CNS).
Note that each CNS is CMS. Indeed, .
Let be a CNS, and a sequence in . Then one has the following.
(i) converges to whenever for every with there is a natural number , such that for all . It is denoted by or .
(ii) is a Cauchy sequence whenever for every with there is a natural number , such that for all .
(iii) is a complete cone normed space if every Cauchy sequence is convergent.
Complete cone-normed spaces will be called cone Banach spaces.
Let be a CNS, let be a normal cone with normal constant , and let be a sequence in . Then, one has the following:
(i)the sequence converges to if and only if , as ,
(ii)the sequence is Cauchy if and only if as ,
(iii)the sequence converges to and the sequence converges to and then .
The proof is direct by applying Lemmas 1, 4, and 5 in  to the cone metric space , where , for all .
Let be a CNS over a cone in . Then (1) and . (2) If then there exists such that implies . (3) For any given and there exists such that . (4) If are sequences in such that , and for all then .
holds for all . A CNS together with a convex structure is said to be convex CNS. A subset is convex, if holds for all and .
where for all and .
Let be a CMS and . Then the mapping such that forms a cone metric on . A sequence is said to be a double sequence of . A sequence is convergent to if, for every , there exists a natural number such that for all .
Let and . Then, if and only if and .
Suppose . Thus, for any , there exist such that for all . Hence, and for all , that is, and .
Conversely, assume and . Thus, for any , there exist such that for all , and also for all . Hence, for all , where .
Let be a CMS. A function is said to be sequentially continuous if implies that . Analogously, a function is sequentially continuous if implies that .
Lemma 2.3 (see ).
Let be a CNS. Then is continuous if and only if is sequentially continuous.
Note that this definition reduces the notion of mixed monotone function on where represents usual total order in .
Definition 2.6 (see ).
Let be a CMS and . is said to be sequentially compact if for any sequence in there is a subsequence of such that is convergent in .
Remark 2.7 (see ).
Every cone metric space is a topological space which is denoted by . Moreover, a subset is sequentially compact if and only if is compact.
where denotes the convex hull.
Let be a topological vector space, let be a nonempty subset of and let be called KKM map with closed values. If is compact for at least one then .
for each . Since , then . Regarding that the mappings and are continuous, is closed for each . Since is compact, then is compact for each . Thus, is a KKM map.
Let be a CNS over strongly minidhedral cone , and let be a nonempty convex compact subset of . If is continuous mapping and is continuous almost quasiconvex mapping with respect to such that , then and have a coupled coincidence point.
Thus, and .
If we take as an identity, , in Theorem 2.11, then we get the following result.
Let be a CNS over strongly minidhedral cone , and let be a nonempty convex compact subset of . If is continuous mapping, then has a coupled fixed point.
for all .
Take which implies (2.14). It is sufficient to show that . The inequality (2.14) implies that either or .
This is a contradiction. Analogously one can get the contradiction from the case . Thus, .
Let be a CNS over strongly minidhedral cone , and let be a nonempty convex compact subset of . Suppose that is continuous mapping. Then has a coupled fixed point if one of the following conditions is satisfied for all such that :
It is clear that (iii) (ii) (i). To finalize proof, it is sufficient to show that is satisfied. Suppose that holds but has no coupled fixed point. Take Theorem 2.13 into account; then there exist such that (2.14) holds which contradicts .
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