Near Fixed Point Theorems in Hyperspaces

The hyperspace consists of all the subsets of a vector space. It is well-known that the hyperspace is not a vector space because it lacks the concept of inverse element. This also says that we cannot consider its normed structure, and some kinds of fixed point theorems cannot be established in this space. In this paper, we shall propose the concept of null set that will be used to endow a norm to the hyperspace. This normed hyperspace is clearly not a conventional normed space. Based on this norm, the concept of Cauchy sequence can be similarly defined. In addition, a Banach hyperspace can be defined according to the concept of Cauchy sequence. The main aim of this paper is to study and establish the so-called near fixed point theorems in Banach hyperspace.


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 K(X) be the collection of all nonempty compact and convex subsets of X.Given A, B ∈ K(X), the set addition is defined by A ⊕ B = {a + b : a ∈ A and b ∈ B} and the scalar multiplication in K(X) is defined by λA = {λa : a ∈ A} , where λ is a constant in R. It is clear to see that K(X) 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 K(X).
The substraction between A and B is denoted and defined by Let θ X be the zero element of the vector space X.It is clear to see that the singleton set {θ X }, denoted by θ K(X) , can be regarded as the zero element of K(X) since A ⊕ θ K(X) = A ⊕ {θ X } = A. (1) On the other hand, since A A = {θ X }, it means that A A is not the zero element of K(X).In other words, the additive inverse element of A in K(X) does not exist.This says that K(X) cannot form a vector space under the above set addition and scalar multiplication.The following set Ω = {A A : A ∈ K(X)} is called the null set of K(X), which can be regarded as a kind of "zero element" of K(X).We also recall that the true zero element of K(X) is θ K(X) ≡ {θ X }, 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 K(X) is not a vector space, we cannot consider the (conventional) normed space (K(X), • ).Therefore we cannot study the fixed point theorem in (K(X), • ) using the conventional way.In this paper, although K(X) is not a vector space, we still can endow a norm to K(X) 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 (K(X), • ).
Let T : K(X) → K(X) be a function from K(X) into itself.We say that A ∈ K(X) is a fixed point if and only if T(A) = A. Since K(X) lacks the vector structure, we cannot expect to obtain the fixed point of the mapping T using conventional methods.In this paper, we shall try to construct a subset A of X satisfying T(A) ⊕ ω 1 = A ⊕ ω 2 for some ω 1 , ω 2 ∈ Ω.Since the null set Ω can play the role of zero element in K(X), i.e., the elements ω 1 and ω 2 can be ignored in some sense, this kind of subset A is said to be a near fixed point of the mapping T.
In Sections 2 and 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.

Hyperspaces
Let X be a vector space with zero element θ X , and let K(X) be the collection of all subsets of X.Under the set addition and scalar multiplication in K(X), it is clear to see that K(X) cannot form a vector space.One of the reasons is that, given any A ∈ K(X), the difference A A is not a zero element of K(X).It is clear to see that the singleton set {θ X } is a zero element, since In this section, we shall present some properties involving the null set Ω, which will be used for establishing the so-called near fixed point theorems in K(X).

Remark 1.
For further discussion, we first recall some well-known properties given below: A is a convex subset of X and λ 1 and λ 2 have the same sign, then (λ We also recall that the following family is called the null set of K(X).For further discussion, we present some useful properties.
Proposition 1.The following statements hold true: Since the null set Ω is treated as a zero element, we can propose the almost identical concept for elements in K(X).
Definition 1.Given any A, B ∈ K(X), we say that A and B are almost identical if and only if there exist which shows A Ω = C, since Ω is closed under the set addition.This completes the proof.
According to the equivalence relation Ω =, for any A ∈ K(X), we define the equivalence class The family of all classes [A] for A ∈ K(X) is denoted by [K(X)].In this case, the family [K(X)] is called the quotient set of K(X).We also have that In other words, the family of all equivalence classes form a partition of the whole set K(X).We also remark that the quotient set [K(X)] is still not a vector space.The reason is for αβ < 0, since (α + β)A = αA + βA for A ∈ K(X) with αβ < 0.

Normed Hyperspaces
Notice that K(X) is not a vector space.Therefore we cannot consider the normed space (K(X), • ).However, we can propose the so-called normed hyperspace involving the null set Ω as follows.
Definition 2. Given the nonnegative real-valued function • : K(X) → R + , we consider the following conditions: We say that • satisfies the null condition when condition (iii) is replaced by A = 0 if and only if A ∈ Ω. Different kinds of normed hyperspaces are defined below.

•
We say that (K(X), • ) is a pseudo-seminormed hyperspace if and only if conditions (i ) and (ii) are satisfied.

•
We say that (K(X), • ) is a seminormed hyperspace if and only if conditions (i) and (ii) are satisfied.

•
We say that (K(X), • ) is a pseudo-normed hyperspace if and only if conditions (i ), (ii) and (iii) are satisfied.

•
We say that (K(X), • ) 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 A ⊕ ω ≥ A for any A ∈ K(X) and ω ∈ Ω.

•
We say that • satisfies the null sub-inequality if and only if A ⊕ ω ≤ A for any A ∈ K(X) and ω ∈ Ω.

•
We say that • satisfies the null equality if and only if A ⊕ ω = A for any A ∈ K(X) and ω ∈ Ω.
For any A, B ∈ K(X), since −(B A) = A B, we have Example 1.We consider the (conventional) normed space (X, • X ).For any A ∈ K(X), we define Then we have the following properties.
Therefore we obtain sup

Now we have
We conclude that (K(X), • ) is a normed hyperspace.For ω ∈ Ω, it means that ω = B B for some B ∈ K(X).Then we have Since ω is not equal to zero in general, it means that the null condition is not satisfied.
Proof.We have (using the triangle inequality).
This completes the proof.
Proposition 4. The following statements hold true.
(i) Let (K(X), • ) be a pseudo-seminormed hyperspace such that • satisfies the null equality.For any Let (K(X), • ) be a pseudo-seminormed hyperspace such that • satisfies the null super-inequality and null condition.For any A, B ∈ K(X), Proof.To prove part (i), we see that Therefore, using the null equality, we have To prove part (ii), suppose that A B = 0. Then A B ∈ Ω, i.e., A B = ω 1 for some ω 1 ∈ Ω.Then, by adding B on both sides, we have for some ω 3 ∈ Ω.Using the null super-inequality, null condition and (2), we have This completes the proof.

Cauchy Sequences
We have the following interesting results.Proposition 5. Let (K(X), • ) be a pseudo-normed hyperspace with the null set Ω.
Proof.To prove the first case of part (i), we have By Proposition 3, we have which says that A B = 0.By Definition 2, we see that A B ∈ Ω, i.e.A Ω = B, which also says that B is in the equivalence class [A].
Inspired by part (ii) of Proposition 5, we propose the following concept of limit.Definition 4. Let (K(X), • ) be a pseudo-seminormed hyperspace.If the sequence {A n } ∞ n=1 in K(X) converges to some A ∈ K(X), then the equivalence class [A] is called the class limit of {A n } ∞ n=1 .We also write lim We need to remark that if [A] is a class limit and B ∈ [A] then it is not necessarily that the sequence {A n } ∞ n=1 converges to y unless • satisfies the null equality.In other words, for the class limit [A], if • satisfies the null equality, then part (ii) of Proposition 5 says that sequence {A n } ∞ n=1 converges to B for any B ∈ [A].Proposition 6.Let (K(X), • ) be a pseudo-normed hyperspace such that • satisfies the null super-inequality.Then the class limit is unique. .This shows the uniqueness in the sense of class limit.Definition 5. Let (K(X), • ) be a pseudo-seminormed hyperspace.A sequence {A n } ∞ n=1 in K(X) is called a Cauchy sequence if and only if, given any > 0, there exists N ∈ N such that • ) be a pseudo-seminormed hyperspace such that • satisfies the null super-inequality.Then every convergent sequence is a Cauchy sequence.
n=1 is a convergent sequence, then, given any > 0, A n A = A A n < /2 for sufficiently large n.Therefore, by Proposition 3, we have for sufficiently large n and m, which says that {A n } ∞ n=1 is a Cauchy sequence.This completes the proof.Definition 6. Different kinds of Banach hyperspaces are defined below.

•
Let (K(X), • ) be a pseudo-seminormed hyperspace.If K(X) is complete, then it is called a pseudo-semi-Banach hyperspace.
Example 2. Continued from Example 1, we further assume that (X, • X ) is a (conventional) Banach space.Then we want to show that the normed hyperspace (K(X), • ) is complete.Suppose that {A n } ∞ n=1 is a Cauchy sequence in (K(X), • ).Let A be the collection of all sequences induced from the sequence {A n } ∞ n=1 .More precisely, each element in A is a sequence {a n } ∞ n=1 with a n ∈ A n for all n.Firstly, we need to claim that each sequence in A is convergent.Since {A n } ∞ n=1 is a Cauchy sequence, i.e., A n A m < for m, n > N with m = n, we have which says that a n − a m X < for any sequence {a n } ∞ n=1 with a n ∈ A n for all n, where is independent of a n and a m .By the completeness of (X, • X ), it follows that each sequence {a n } ∞ n=1 is convergent to some a ∈ X, i.e., a n − a X → 0 as n → ∞ in the uniform sense, which means a n − a X < such that is independent of a n and a.Indeed, if is dependent of a n and a, then says that is dependent of a n and a m , which is a contradiction.We can define a subset A of X that collects all of the limit points of each sequence in A. Then, finally, we want to claim A n A → 0 as n → ∞.For any x ∈ A n − A, we have x = a n − a for some a n ∈ A n and a ∈ A. Since a is a limit point of some sequence {a • n } ∞ n=1 , for m > n > N, using (4), we have where is independent of a n and a • m by referring to (4) again.Since a • m − a X → 0 as m → ∞ in the uniform sense, it follows that a n − a X → 0 as n → ∞ in the uniform sense.Therefore we obtain This shows that the sequence {A n } ∞ n=1 is convergent, i.e., (K(X), • ) is a Banach hyperspace.

Near Fixed Point Theorems
Let T : K(X) → K(X) be a function from K(X) into itself.We say that A ∈ K(X) is a fixed point if and only if T(A) = A.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 (K(X), • ) is not a vector space, we cannot study the corresponding fixed point theorems based on (K(X), • ).However, we can study the so-called near fixed point that is defined below.Definition 7. Let T : K(X) → K(X) be a function defined on K(X) into itself.A point A ∈ K(X) is called a near fixed point of T if and only if T(A)

By definition, we see that T(A)
Given any initial element A 0 ∈ K(X), we define the iterative sequence {A n } ∞ n=1 using the function T as follows: Under some suitable conditions, we are going to show that the sequence {A n } ∞ n=1 can converge to near a fixed point.Theorem 1.Let (K(X), • ) be a Banach hyperspace with the null set Ω such that • satisfies the null equality.Suppose that the function T : (K(X), • ) → (K(X), • ) is a contraction on K(X).Then T has a near fixed point A ∈ K(X) satisfying T(A) Ω = A.Moreover, the near fixed point A is obtained by the limit n=1 is generated according to (5).We also have the following properties.

•
The uniqueness is in the sense that there is a unique equivalence class [A] such that any Ā ∈ [A] cannot be near a fixed point.
Proof.Given any initial element A 0 ∈ K(X), we are going to show that {A n } ∞ n=1 is a Cauchy sequence.Since T is a contraction on K(X), we have For n < m, using Proposition 3, we obtain Since 0 < α < 1, we have 1 − α m−n < 1 in the numerator, which says that We are going to show that any point Ā ∈ [A] is a near fixed point.Now we have Ā ⊕ ω 1 = A ⊕ ω 2 for some ω 1 , ω 2 ∈ Ω.Using the triangle inequality and the fact of contraction on K(X), we have (using −ω 2 ∈ Ω, the null equality and Remark 1), which implies Ā T( Ā) = 0 as m → ∞.We conclude that T( Ā) Now assume that there is another near fixed point Ā of T with Ā ∈ [A], i.e., Ā Ω = T( Ā).Then Since T is a contraction on K(X) and • satisfies the null equality, we obtain (using −ω 3 ∈ Ω, the null equality and Remark 1) (using −ω 4 ∈ Ω, the null equality and Remark 1) . Therefore, any Ā ∈ [A] cannot be the near fixed point.Equivalently, if Ā is a near fixed point of T, then Ā ∈ [A].This completes the proof.
) is called a weakly strict contraction on K(X) if and only if the following conditions are satisfied: By part (ii) of Proposition 4, we see that if A Ω = B, then A B = 0, which says that the weakly strict contraction is well-defined.In other words, (K(X), • ) 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 K(X), then it is also a weakly strict contraction on K(X).
Theorem 2. Let (K(X), • ) be a Banach hyperspace with the null set Ω Suppose that • satisfies the null super-inequality and null condition, and that the function T : (K(X), • ) → (K(X), • ) is a weakly strict contraction on K(X).If {T n (A 0 )} ∞ n=1 forms a Cauchy sequence for some A 0 ∈ K(X), then T has a near fixed point A ∈ K(X) satisfying T(A) Ω = A.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 [A] such that any Ā ∈ [A] cannot be a near fixed point.Proof.Since {T n (A 0 )} ∞ n=1 is a Cauchy sequence, the completeness says that there exists A ∈ K(X) such that T n (A 0 ) A = A T n (A 0 ) → 0.
Definition 10.Let (K(X), • ) be a pseudo-normed hyperspace with the null set Ω. A function T : (K(X), • ) → (K(X), • ) is called a weakly uniformly strict contraction on K(X) if and only if the following conditions are satisfied: By part (ii) of Proposition 4, we see that if A Ω = B, then A B = 0, which says that the weakly uniformly strict contraction is well-defined.In other words, (K(X), • ) should be assumed to be a pseudo-normed hyperspace rather than pseudo-seminormed hyperspace.
Remark 2. We observe that if T is a weakly uniformly strict contraction on K(X), then T is also a weakly strict contraction on K(X).
Then the sequence { T n (A) T n+1 (A) } ∞ n=1 is decreasing to zero for any A ∈ K(X).
Proof.For convenience, we write Then, by the first condition of Definition 10, The above two cases say that the sequence {c n } ∞ n=1 is decreasing.We consider the following cases.

•
Let m be the first index in the sequence Using the first condition of Definition 10, we also have Since the sequence {c n } ∞ n=1 is decreasing, we assume that c n ↓ > 0, i.e., c n ≥ > 0 for all n.There exists δ > 0 such that ≤ c m < + δ for some m, i.e., ≤ A m A m+1 < + δ.
By the second condition of Definition 10, we have This completes the proof.
Theorem 3. Let (K(X), • ) be a Banach hyperspace with the null set Ω. Suppose that • satisfies the null super-inequality, and that the function T : (K(X), • ) → (K(X), • ) is a weakly uniformly strict contraction on K(X).Then T has a near fixed point satisfying T(A) Ω = A.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 [A] such that any Ā ∈ [A] cannot be a near fixed point.
Indeed, if δ = , i.e., < δ, then + δ = + < + δ.Let c n = A n A n+1 .Since the sequence {c n } ∞ n=1 is decreasing to zero by Lemma 1, we can find N such that c N < δ /3.For n > m ≥ N, we have which implicitly says that A m Ω = A n .Since the sequence {c n } ∞ n=1 is decreasing by Lemma 1 again, we obtain For j with m < j ≤ n, using Proposition 3, we have We want to show that there exists j with m < j ≤ n such that A m Ω = A j and Let γ j = A m A j for j = m + 1, • • • , n.Then ( 7) and ( 8) says that γ m+1 < and γ n > + δ .Let j 0 be an index such that j 0 = max j ∈ [m + 1, n] : γ j ≤ + 2δ 3 .
Then we see that j 0 < n, since γ n > + δ .By the definition of j 0 , we also see that j 0 + 1 ≤ n and γ j 0 +1 > + 2δ 3 , which also says that A m Ω = A j 0 +1 .Therefore expression (10) will be obtained if we can show that Suppose that this is not true, i.e., γ j 0 +1 ≥ + δ .From (9), we have This contradiction says that (10) is sound.Since A m Ω = A j , using (6), we see that (10) implies Therefore we obtain A m A j ≤ A m A m+1 + A m+1 A j+1 + A j+1 A j (by Proposition 3) < c m + + c j (by ( 11)) which contradicts (10).This contradiction says that the sequence {T n (A)} ∞ n=1 = {A n } ∞ n=1 is a Cauchy sequence, and the proof is complete.

Proposition 2 .
When A, B and C are not singleton sets, if A B = C, then we cannot have A = B ⊕ C.However, we can have A Ω = B ⊕ C. Indeed, since A B = C, by adding B on both sides, we obtain A ⊕ ω = B ⊕ C, where ω = B B ∈ Ω.This says that A Ω = B ⊕ C. The binary relation Ω = is an equivalence relation.Proof.For any A ∈ K(X), A = A implies A Ω = A, which shows the reflexivity.The symmetry is obvious by the definition of the binary relation Ω =.Regarding the transitivity, for A Ω = B and B Ω = C, we want to claim A Ω = C.By definition, we have

Proof.
Suppose that the sequence {A n } ∞ n=1 is convergent with the class limits [A] and [B].Then, by definition, we have lim n→∞ A A n = lim n→∞ A n A = lim n→∞ B A n = lim n→∞ A n B = 0, which says that A B = 0 by referring to (3).By part (ii) of Proposition 4, we have A Ω = B, i.e., [A] = [B] is also a near fixed point of T satisfying T( Ā)Ω = Ā and [ Ā] = [A].• If Ā is a near fixed point of T, then Ā ∈ [A], i.e., [ Ā] = [A].Equivalently, if A and Ā are the near fixed points of T, then A Ω = Ā.
also a near fixed point of T satisfying T( Ā)Ω = Ā and [ Ā] = [A].• If Ā is a near fixed point of T, then Ā ∈ [A], i.e., [ Ā] = [A].Equivalently, if A and Ā are the near fixed points of T, then A Ω = Ā.Proof.According to Theorem 2 and Remark 2, we just need to claim that if T is a weakly uniformly strict contraction, then {T n(A 0 )} ∞ n=1 = {A n } ∞ n=1 forms a Cauchy sequence.Suppose that {A n } ∞ n=1is not a Cauchy sequence.Then there exists 2 > 0 such that, given any N, there exist n > m ≥ N satisfying A m A n > 2 .Since T is a weakly uniformly strict contraction on K(X), there exists δ > 0 such that≤ A B < + δ implies T(A) T(B) < for any A Ω = B.Let δ = min{δ, }.We are going to claim≤ A B < + δ implies T(A) T(B) < for any A Ω = B.