1. Introduction
Given a universal set
X, the collection of all nonempty subsets of
X is called the hyperspace. The topic of set-valued analysis is to study the mathematical properties on hyperspace by referring to the monographs [
1,
2,
3,
4,
5,
6,
7]. Especially, the set-valued optimization and differential inclusion have been well-studied. The purpose of this paper is to study the near fixed point theorem in hyperspace. The concept of near fixed point is the first attempt at studying the fixed point theory in hyperspace. The conventional fixed point theorem in normed space can be used to study the existence of solutions for some mathematical problems. Especially, the application of fixed point theorems in economics can refer to the articles [
8,
9,
10]. Therefore, the potential applications of near fixed point theorems established in this paper can be used to study the existence of solutions for the mathematical problems that involve the set-valued mappings. This could be future research.
Let
X be a topological vector space, and let
be the collection of all nonempty compact and convex subsets of
X. Given
, the set addition is defined by
and the scalar multiplication in
is defined by
where
is a constant in
. It is clear to see that
cannot be a vector space under the above set addition and scalar multiplication. The main reason is that there is no additive inverse element for each element in
.
The
substraction between
A and
B is denoted and defined by
Let
be the zero element of the vector space
X. It is clear to see that the singleton set
, denoted by
, can be regarded as the zero element of
since
On the other hand, since
, it means that
is not the zero element of
. In other words, the additive inverse element of
A in
does not exist. This says that
cannot form a vector space under the above set addition and scalar multiplication. The following set
is called the
null set of
, which can be regarded as a kind of “zero element” of
. We also recall that the true zero element of
is
, since (
1) is satisfied.
Recall that the (conventional) normed space is based on the vector space by referring to the monographs [
11,
12,
13,
14,
15,
16,
17]. Since
is not a vector space, we cannot consider the (conventional) normed space
. Therefore we cannot study the fixed point theorem in
using the conventional way. In this paper, although
is not a vector space, we still can endow a norm to
in which the axioms are almost the same as the axioms of conventional norm. The only difference is that the concept of null set is involved in the axioms. Under these settings, we shall study the so-called near fixed point theorem in the normed hyperspace
.
Let be a function from into itself. We say that is a fixed point if and only if . Since lacks the vector structure, we cannot expect to obtain the fixed point of the mapping using conventional methods. In this paper, we shall try to construct a subset A of X satisfying for some . Since the null set can play the role of zero element in , i.e., the elements and can be ignored in some sense, this kind of subset A is said to be a near fixed point of the mapping .
In
Section 2 and
Section 3, the concept of normed hyperspace is proposed, where some interesting properties are derived in order to study the near fixed point theorem. In
Section 4, according to the norm, the concept of Cauchy sequence can be similarly defined. In addition, the Banach hyperspace is defined based on the Cauchy sequence. In
Section 5, we present many near fixed point theorems that are established using the almost identical concept in normed hyperspace.
2. Hyperspaces
Let
X be a vector space with zero element
, and let
be the collection of all subsets of
X. Under the set addition and scalar multiplication in
, it is clear to see that
cannot form a vector space. One of the reasons is that, given any
, the difference
is not a zero element of
. It is clear to see that the singleton set
is a zero element, since
for any
. However, when
is not a singleton set, we cannot have
. In this section, we shall present some properties involving the null set
, which will be used for establishing the so-called near fixed point theorems in
.
Remark 1. For further discussion, we first recall some well-known properties given below:
;
for ;
for ;
if A is a convex subset of X and and have the same sign, then .
We also recall that the following family
is called the null set of
. For further discussion, we present some useful properties.
Proposition 1. The following statements hold true:
The singleton set is in the null set Ω;
implies ;
for with ;
Ω is closed under the set addition; that is, for any .
Since the null set is treated as a zero element, we can propose the almost identical concept for elements in .
Definition 1. Given any , we say that A and B are almost identical if and only if there exist such that . In this case, we write .
When A, B and C are not singleton sets, if , then we cannot have . However, we can have . Indeed, since , by adding B on both sides, we obtain , where . This says that .
Proposition 2. The binary relation is an equivalence relation.
Proof. For any
,
implies
, which shows the reflexivity. The symmetry is obvious by the definition of the binary relation
. Regarding the transitivity, for
and
, we want to claim
. By definition, we have
for some
for
. Then
which shows
, since
is closed under the set addition. This completes the proof. ☐
According to the equivalence relation
, for any
, we define the equivalence class
The family of all classes
for
is denoted by
. In this case, the family
is called the quotient set of
. We also have that
implies
. In other words, the family of all equivalence classes form a partition of the whole set
. We also remark that the quotient set
is still not a vector space. The reason is
for
, since
for
with
.
3. Normed Hyperspaces
Notice that is not a vector space. Therefore we cannot consider the normed space . However, we can propose the so-called normed hyperspace involving the null set as follows.
Definition 2. Given the nonnegative real-valued function , we consider the following conditions:
- (i)
for any and ;
- ()
for any and with ;
- (ii)
for any ;
- (iii)
implies .
We say that satisfies the null condition when condition (iii) is replaced by if and only if . Different kinds of normed hyperspaces are defined below.
We say that is a pseudo-seminormed hyperspace if and only if conditions () and (ii) are satisfied.
We say that is a seminormed hyperspace if and only if conditions (i) and (ii) are satisfied.
We say that is a pseudo-normed hyperspace if and only if conditions (), (ii) and (iii) are satisfied.
We say that is a normed hyperspace if and only if conditions (i), (ii) and (iii) are satisfied.
Now we consider the following definitions:
We say that satisfies the null super-inequality if and only if for any and .
We say that satisfies the null sub-inequality if and only if for any and .
We say that satisfies the null equality if and only if for any and .
For any
, since
, we have
Example 1. We consider the (conventional) normed space
. For any
, we define
Then we have the following properties.
if and only if . Indeed, if , then it is obvious that . For the converse, if , then we have for all , i.e., .
We want to claim
. We denote by
Then we see that for all and .
We conclude that
is a normed hyperspace. For
, it means that
for some
. Then we have
Since is not equal to zero in general, it means that the null condition is not satisfied.
Proposition 3. Let be a pseudo-seminormed hyperspace such that satisfies the null super-inequality. For any , we have Proof. This completes the proof. ☐
Proposition 4. The following statements hold true.
- (i)
Let be a pseudo-seminormed hyperspace such that satisfies the null equality. For any , if , then .
- (ii)
Let be a pseudo-normed hyperspace. For any , implies .
- (iii)
Let be a pseudo-seminormed hyperspace such that satisfies the null super-inequality and null condition. For any , implies .
Proof. To prove part (i), we see that
implies
for some
. Therefore, using the null equality, we have
. To prove part (ii), suppose that
. Then
, i.e.,
for some
. Then, by adding
B on both sides, we have
for some
, which says that
. To prove part (iii), for
, we have
for some
. Since
is closed under the vector addition, it follows that
for some
. Using the null super-inequality, null condition and (
2), we have
This completes the proof. ☐
4. Cauchy Sequences
Let be a pseudo-seminormed hyperspace. Given a sequence in , it is clear that . The concept of limit is defined below.
Definition 3. Let
be a pseudo-seminormed hyperspace. A sequence
in
is said to
converge to
if and only if
We have the following interesting results.
Proposition 5. Let be a pseudo-normed hyperspace with the null set Ω.
- (i)
If the sequence in converges to A and B simultaneously, then .
- (ii)
Suppose that satisfies the null equality. If the sequence in converges to , then, give any , the sequence converges to B.
Proof. To prove the first case of part (i), we have
By Proposition 3, we have
which says that
. By Definition 2, we see that
, i.e.
, which also says that
B is in the equivalence class
.
To prove part (ii), for any
, i.e.,
for some
, using the null equality, we have
This completes the proof. ☐
Inspired by part (ii) of Proposition 5, we propose the following concept of limit.
Definition 4. Let
be a pseudo-seminormed hyperspace. If the sequence
in
converges to some
, then the equivalence class
is called the
class limit of
. We also write
We need to remark that if is a class limit and then it is not necessarily that the sequence converges to y unless satisfies the null equality. In other words, for the class limit , if satisfies the null equality, then part (ii) of Proposition 5 says that sequence converges to B for any .
Proposition 6. Let be a pseudo-normed hyperspace such that satisfies the null super-inequality. Then the class limit is unique.
Proof. Suppose that the sequence
is convergent with the class limits
and
. Then, by definition, we have
which says that
by referring to (
3). By part (ii) of Proposition 4, we have
, i.e.,
. This shows the uniqueness in the sense of class limit. ☐
Definition 5. Let
be a pseudo-seminormed hyperspace. A sequence
in
is called a
Cauchy sequence if and only if, given any
, there exists
such that
for
with
. If every Cauchy sequence in
is convergent, then we say that
is
complete.
Proposition 7. Let be a pseudo-seminormed hyperspace such that satisfies the null super-inequality. Then every convergent sequence is a Cauchy sequence.
Proof. If
is a convergent sequence, then, given any
,
for sufficiently large
n. Therefore, by Proposition 3, we have
for sufficiently large
n and
m, which says that
is a Cauchy sequence. This completes the proof. ☐
Definition 6. Different kinds of Banach hyperspaces are defined below.
Let be a pseudo-seminormed hyperspace. If is complete, then it is called a pseudo-semi-Banach hyperspace.
Let be a seminormed hyperspace. If is complete, then it is called a semi-Banach hyperspace.
Let be a pseudo-normed hyperspace. If is complete, then it is called a pseudo-Banach hyperspace.
Let be a normed hyperspace. If is complete, then it is called a Banach hyperspace.
Example 2. Continued from Example 1, we further assume that
is a (conventional) Banach space. Then we want to show that the normed hyperspace
is complete. Suppose that
is a Cauchy sequence in
. Let
be the collection of all sequences induced from the sequence
. More precisely, each element in
is a sequence
with
for all
n. Firstly, we need to claim that each sequence in
is convergent. Since
is a Cauchy sequence, i.e.,
for
with
, we have
which says that
for any sequence
with
for all
n, where
is independent of
and
. By the completeness of
, it follows that each sequence
is convergent to some
, i.e.,
as
in the uniform sense, which means
such that
is independent of
and
a. Indeed, if
is dependent of
and
a, then
says that
is dependent of
and
, which is a contradiction. We can define a subset
A of
X that collects all of the limit points of each sequence in
. Then, finally, we want to claim
as
. For any
, we have
for some
and
. Since
a is a limit point of some sequence
, for
, using (
4), we have
where
is independent of
and
by referring to (
4) again. Since
as
in the uniform sense, it follows that
as
in the uniform sense. Therefore we obtain
This shows that the sequence is convergent, i.e., is a Banach hyperspace.
5. Near Fixed Point Theorems
Let be a function from into itself. We say that is a fixed point if and only if . This concept is completely different from the concept of fixed point in set-valued functions. Some conventional fixed point theorems are based on the normed space that is also a vector space. Since is not a vector space, we cannot study the corresponding fixed point theorems based on . However, we can study the so-called near fixed point that is defined below.
Definition 7. Let be a function defined on into itself. A point is called a near fixed point of if and only if .
By definition, we see that if and only if there exist such that , , or or .
Definition 8. Let
be a pseudo-seminormed hyperspace. A function
is called a
contraction on if and only if there is a real number
such that
for any
.
Given any initial element
, we define the iterative sequence
using the function
as follows:
Under some suitable conditions, we are going to show that the sequence can converge to near a fixed point.
Theorem 1. Let be a Banach hyperspace with the null set Ω
such that satisfies the null equality. Suppose that the function is a contraction on . Then has a near fixed point satisfying . Moreover, the near fixed point A is obtained by the limitin which the sequence is generated according to (
5).
We also have the following properties. The uniqueness is in the sense that there is a unique equivalence class such that any cannot be near a fixed point.
Each point is also a near fixed point of satisfying and .
If is a near fixed point of , then , i.e., . Equivalently, if A and are the near fixed points of , then .
Proof. Given any initial element
, we are going to show that
is a Cauchy sequence. Since
is a contraction on
, we have
For
, using Proposition 3, we obtain
Since
, we have
in the numerator, which says that
This proves that
is a Cauchy sequence. Since
is complete, there exists
such that
We are going to show that any point
is a near fixed point. Now we have
for some
. Using the triangle inequality and the fact of contraction on
, we have
which implies
as
. We conclude that
for any point
by part (ii) of Proposition 4.
Now assume that there is another near fixed point
of
with
, i.e.,
. Then
for some
,
. Since
is a contraction on
and
satisfies the null equality, we obtain
Since , we conclude that , i.e., , which contradicts . Therefore, any cannot be the near fixed point. Equivalently, if is a near fixed point of , then . This completes the proof. ☐
Definition 9. Let be a pseudo-normed hyperspace. A function is called a weakly strict contraction on if and only if the following conditions are satisfied:
, i.e., implies .
, i.e., implies .
By part (ii) of Proposition 4, we see that if , then , which says that the weakly strict contraction is well-defined. In other words, should be assumed to be a pseudo-normed hyperspace rather than pseudo-seminormed hyperspace. We further assume that satisfies the null super-inequality and null condition. Part (iii) of Proposition 4 says that if T is a contraction on , then it is also a weakly strict contraction on .
Theorem 2. Let be a Banach hyperspace with the null set Ω
Suppose that satisfies the null super-inequality and null condition, and that the function is a weakly strict contraction on . If forms a Cauchy sequence for some , then has a near fixed point satisfying . Moreover, the near fixed point A is obtained by the limitAssume further that satisfies the null equality. Then we also have the following properties. The uniqueness is in the sense that there is a unique equivalence class such that any cannot be a near fixed point.
Each point is also a near fixed point of satisfying and .
If is a near fixed point of , then , i.e., . Equivalently, if A and are the near fixed points of , then .
Proof. Since
is a Cauchy sequence, the completeness says that there exists
such that
Therefore, given any , there exists an integer N such that for . We consider the following two cases.
Suppose that
. Since
is a weakly strict contraction on
, it follows that
by part (iii) of Proposition 4.
Suppose that
. Since
is a weakly strict contraction on
, we have
The above two cases say that
. Using Proposition 3, we obtain
which says that
, i.e.,
by part (ii) of Proposition 4. This shows that
A is a near fixed point.
Assume that
satisfies the null equality. We are going to claim that each point
is also a near fixed point of
. Since
, we have
for some
. Then, using the null equality for
, we obtain
Using the above argument, we can also obtain
as
. Using Proposition 3, we have
which says that
. Therefore we conclude that
for any point
by part (ii) of Proposition 4.
Suppose that
is another near fixed point of
. Then
and
, i.e.,
. Then
and
, where
for
. Therefore we obtain
This contradiction says that cannot be a near fixed point of . Equivalently, if is a near fixed point of , then . This completes the proof. ☐
Now we consider another fixed point theorem based on the concept of weakly uniformly strict contraction which was proposed by Meir and Keeler [
18].
Definition 10. Let be a pseudo-normed hyperspace with the null set . A function is called a weakly uniformly strict contraction on if and only if the following conditions are satisfied:
for , i.e., , ;
given any , there exists such that implies for any , i.e., .
By part (ii) of Proposition 4, we see that if , then , which says that the weakly uniformly strict contraction is well-defined. In other words, should be assumed to be a pseudo-normed hyperspace rather than pseudo-seminormed hyperspace.
Remark 2. We observe that if is a weakly uniformly strict contraction on , then is also a weakly strict contraction on .
Lemma 1. Let be a pseudo-normed hyperspace with the null set Ω, and let be a weakly uniformly strict contraction on . Then the sequence is decreasing to zero for any .
Proof. For convenience, we write for all n. Let .
Suppose that
. By Remark 2, we have
Suppose that .
Then, by the first condition of Definition 10,
The above two cases say that the sequence is decreasing. We consider the following cases.
Let
m be the first index in the sequence
such that
. Then we want to claim
. Since
, we have
Using the first condition of Definition 10, we also have
which says that
, i.e.,
. Using the similar arguments, we can obtain
and
. Therefore the sequence
is decreasing to zero.
Suppose that for all .
Since the sequence
is decreasing, we assume that
, i.e.,
for all
n. There exists
such that
for some
m, i.e.,
By the second condition of Definition 10, we have
which contradicts
.
This completes the proof. ☐
Theorem 3. Let be a Banach hyperspace with the null set Ω.
Suppose that satisfies the null super-inequality, and that the function is a weakly uniformly strict contraction on . Then has a near fixed point satisfying . Moreover, the near fixed point A is obtained by the limit Assume further that satisfies the null equality. Then we also have the following properties.
The uniqueness is in the sense that there is a unique equivalence class such that any cannot be a near fixed point.
Each point is also a near fixed point of satisfying and .
If is a near fixed point of , then , i.e., . Equivalently, if A and are the near fixed points of , then .
Proof. According to Theorem 2 and Remark 2, we just need to claim that if
is a weakly uniformly strict contraction, then
forms a Cauchy sequence. Suppose that
is not a Cauchy sequence. Then there exists
such that, given any
N, there exist
satisfying
. Since
is a weakly uniformly strict contraction on
, there exists
such that
Let
. We are going to claim
Indeed, if , i.e., , then .
Let
. Since the sequence
is decreasing to zero by Lemma 1, we can find
N such that
. For
, we have
which implicitly says that
. Since the sequence
is decreasing by Lemma 1 again, we obtain
For
j with
, using Proposition 3, we have
We want to show that there exists
j with
such that
and
Let
for
. Then (
7) and (
8) says that
and
. Let
be an index such that
Then we see that
, since
. By the definition of
, we also see that
and
, which also says that
. Therefore expression (
10) will be obtained if we can show that
Suppose that this is not true, i.e.,
. From (
9), we have
This contradiction says that (
10) is sound. Since
, using (
6), we see that (
10) implies
Therefore we obtain
which contradicts (
10). This contradiction says that the sequence
is a Cauchy sequence, and the proof is complete. ☐