Open Access

Fixed Point Theorems in Cone Banach Spaces

Fixed Point Theory and Applications20092009:609281

DOI: 10.1155/2009/609281

Received: 23 October 2009

Accepted: 15 December 2009

Published: 15 December 2009

Abstract

In this manuscript, a class of self-mappings on cone Banach spaces which have at least one fixed point is considered. More precisely, for a closed and convex subset of a cone Banach space with the norm , if there exist , , and satisfies the conditions and for all , then has at least one Fixed point.

1. Introduction and Preliminaries

In 1980, Rzepecki [1] introduced a generalized metric on a set in a way that , where is Banach space and is a normal cone in with partial order . In that paper, the author generalized the fixed point theorems of Maia type [2].

Let be a nonempty set endowed in two metrics , and a mapping of into itself. Suppose that for all , and is complete space with respect to , and is continuous with respect to , and is contraction with respect to , that is, for all , where . Then has a unique fixed point in .

Seven years later, Lin [3] considered the notion of -metric spaces by replacing real numbers with cone in the metric function, that is, . In that manuscript, some results of Khan and Imdad [4] on fixed point theorems were considered for -metric spaces. Without mentioning the papers of Lin and Rzepecki, in 2007, Huang and Zhang [5] announced the notion of cone metric spaces (CMS) 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: any mapping of a complete cone metric space into itself that satisfies, for some , the inequality

(1.1)

for all , has a unique fixed point.

Recently, many results on fixed point theorems have been extended to cone metric spaces (see, e.g., [59]). Notice also that in ordered abstract spaces, existence of some fixed point theorems is presented and applied the resolution of matrix equations (see, e.g., [1012]).

In this manuscript, some of known results (see, e.g., [13, 14]) are extended to cone Banach spaces which were defined and used in [15, 16] where the existence of fixed points for self-mappings on cone Banach spaces is investigated.

Throughout this paper stands for real Banach space. Let always be a closed nonempty subset of . is called cone if for all and nonnegative real numbers where and .

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 . From now on, it is assumed that

The cone is called

normal if there is a number such that for all :

(1.2)

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 .

Lemma 1.1 (see [6, 17]).
  1. (i)

    Every regular cone is normal.

     
(ii)For each , there is a normal cone with normal constant .
  1. (iii)

    The cone is regular if every decreasing sequence which is bounded from below is convergent.

     

Proofs of (i) and (ii) are given in [6] and the last one follows from definition.

Definition 1.2 (see [5]).

Let be a nonempty set. Suppose the mapping satisfies

for all ,

if and only if ,

for all ,

for all ,

then is called cone metric on , and the pair is called a cone metric space (CMS).

Example 1.3.

Let , , 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 (CNS).

Definition 1.4 (see [15, 16]).

Let be a vector space over . Suppose the mapping satisfies

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, .

Definition 1.5.

Let be a CNS, and a sequence in . Then

(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.

Lemma 1.6.

Let be a CNS, a normal cone with normal constant , and a sequence in . Then,

(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 then .

The proof is direct by applying [5, Lemmas , , and ] to the cone metric space , where , for all .

Lemma 1.7 (see [7, 8]).

Let be a CNS over a cone in . Then and , . If then there exists such that implies . For any given and , there exists such that . If are sequences in such that , , and , for all then .

The proofs of the first two parts followed from the definition of . The third part is obtained by the second part. Namely, if is given then find such that implies . Then find such that and hence . Since is closed, the proof of fourth part is achieved.

Definition 1.8 (see [17]).

is called minihedral cone if exists for all , and strongly minihedral if every subset of which is bounded from above has a supremum.

Lemma 1.9 (see [18]).

Every strongly minihedral normal cone is regular.

Example 1.10.

Let with the supremum norm and Then is a cone with normal constant which is not regular. This is clear, since the sequence is monotonically decreasing, but not uniformly convergent to . This cone, by Lemma 1.9, is not strongly minihedral. However, it is easy to see that the cone mentioned in Example 1.3 is strongly minihedral.

Definition 1.11.

Let be a closed and convex subset of a cone Banach space with the norm and a mapping which satisfies the condition
(1.3)

for all . Then, is said to satisfy the condition .

For , the set of fixed points of is denoted by .

Definition 1.12 (see [14]).

Let be a closed and convex subset of a cone Banach space with the norm and a mapping. Consider the conditions
(1.4)
(1.5)

Then is called nonexpansive (resp., quasi-nonexpansive) if it satisfies the condition (1.4) (resp., (1.5)).

2. Main Results

From now on, will be a cone Banach space, a normal cone with normal constant and a self-mapping operator defined on a subset of .

Theorem 2.1.

Let with and let be a complete cone metric space an onto mapping which satisfies the condition
(2.1)

Then, has a unique fixed point.

Proof.

Let and , then by (2.1), one can observe which is a contradiction. Thus, is one-to-one and it has an inverse, say . Hence,
(2.2)

By [5, Theorem ], has a unique fixed point which is equivalent to saying that has a unique fixed point.

The following statement is consequence of Definition 1.11.

Proposition 2.2.

Every nonexpansive mapping satisfies the condition .

Proposition 2.3.

Let satisfy the condition and , then is a quasi-nonexpansive.

Proof.

Let and . Since and satisfies the condition ,
(2.3)

Theorem 2.4.

Let be a closed and convex subset of a cone Banach space with the norm and a mapping which satisfies the condition
(2.4)

for all , where . Then, has at least one fixed point.

Proof.

Let be arbitrary. Define a sequence in the following way:
(2.5)
Notice that
(2.6)
which yields that
(2.7)
for Combining this observation with the condition (2.4), one can obtain
(2.8)
Thus, , where . Hence, is a Cauchy sequence in and thus converges to some Regarding the inequality
(2.9)
and by the help of Lemma 1.6(iii), one can obtain
(2.10)
Taking into account (2.6) and (2.4), substituting and implies that
(2.11)

Thus, when , one can get , that is, .

Notice that identity map, , satisfies the condition (2.4). Thus, maps that satisfy the condition (2.4) may have fixed points.

From the triangle inequality,

(2.12)

By (2.4),

(2.13)

Thus, letting implies that

(2.14)

Hence we have the following conclusion.

Theorem 2.5.

Let be a closed and convex subset of a cone Banach space with the norm and a mapping which satisfies the condition
(2.15)

for all , where . Then has a fixed point.

Theorem 2.6.

Let be a closed and convex subset of a cone Banach space with the norm and a mapping which satisfies the condition
(2.16)

for all , where . Then has at least one fixed point.

Proof.

Construct a sequence as in the proof of Theorem 2.4, that is, (2.5), (2.6) and also
(2.17)
hold. Thus the triangle inequality implies
(2.18)
Then, by (2.17) and (2.7) we obtain
(2.19)
Replacing and in (2.16) and regarding (2.7) and (2.19), one can obtain
(2.20)

and thus, . Since , the sequence is a Cauchy sequence that converges to some . Since also converges to as in the proof of Theorem 2.4, the inequality (2.16) (under the assumption and ) by the help of Lemma 1.6(iii) yields that which is equivalent to saying that

Theorem 2.7.

Let be a closed and convex subset of a cone Banach space with the norm . If there exist and satisfies the conditions
(2.21)
(2.22)

for all . Then, has at least one fixed point.

Proof.

Construct a sequence as in the proof of Theorem 2.4. We claim that the inequality (2.22) for and implies that
(2.23)
for all that satisfy (2.21). For the proof of the claim, first recall from (2.7) that
(2.24)
The case is trivially true. Indeed, taking into account (2.22) with and together with (2.24) and (2.19), one can get
(2.25)
which is equivalent to (2.23) since . For the case , consider the inequality which is equivalent to
(2.26)
By substituting and in (2.22) together with (2.24), (2.26) and (2.17), one can get
(2.27)

which is equivalent to (2.23) since . Hence, the claim is proved.

By (2.23), one can obtain
(2.28)
Due to (2.21), we have . Thus, the sequence is a Cauchy sequence that converges to some . By substituting with and with in (2.22), one can obtain
(2.29)

as . This last condition is equivalent to saying that as

Authors’ Affiliations

(1)
Department of Mathematics, Atılım University

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Copyright

© Erdal Karapınar. 2009

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