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Research article Special Issues

Nonlocal scalar conservation laws with discontinuous flux

  • Received: 04 October 2022 Revised: 14 December 2022 Accepted: 19 December 2022 Published: 29 December 2022
  • We prove the well-posedness of entropy weak solutions for a class of space-discontinuous scalar conservation laws with nonlocal flux. We approximate the problem adding a viscosity term and we provide L and BV estimates for the approximate solutions. We use the doubling of variable technique to prove the stability with respect to the initial data from the entropy condition.

    Citation: Felisia Angela Chiarello, Giuseppe Maria Coclite. Nonlocal scalar conservation laws with discontinuous flux[J]. Networks and Heterogeneous Media, 2023, 18(1): 380-398. doi: 10.3934/nhm.2023015

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  • We prove the well-posedness of entropy weak solutions for a class of space-discontinuous scalar conservation laws with nonlocal flux. We approximate the problem adding a viscosity term and we provide L and BV estimates for the approximate solutions. We use the doubling of variable technique to prove the stability with respect to the initial data from the entropy condition.



    The (conventional) vector space is based on two basic operations, which are vector addition and scalar multiplication. Under these two operations, some required axioms are needed to extend the concepts of finite-dimensional Euclidean space Rn. Usually, there are eight axioms are provided in the real vector space. However, some spaces cannot satisfy all of the axioms in vector space. In this case, the weak concept of a so-called near vector space will be studied in this paper. We provide three well-known spaces that cannot be the (conventional) vector spaces. However, they can be checked to be near vector spaces.

    ● Let I be the space of all bounded and closed intervals in R. The interval addition and scalar multiplication of intervals can be treated as the vector addition and scalar multiplication. An interval which subtracts itself may not provide a zero element in I. In this case, we cannot consider the concept of an inverse element in I. This says that I cannot be a (conventional) vector space by referring to Example 2.1. However, it is not difficult to check that I is a near vector space, which will be presented in the context of this paper.

    ● Let Fcc(R) be the space of all fuzzy numbers in R. The fuzzy number addition and scalar multiplication of fuzzy numbers can be treated as the vector addition and scalar multiplication. A fuzzy number which subtracts itself may not provide a zero element in Fcc(R). In this case, we cannot consider the concept of an inverse element in Fcc(R). This says that Fcc(R) cannot be a (conventional) vector space as according to Subsection 2.1.

    ● Let U be a (conventional) vector space, and let P(U) be a collection of all subsets of U. The collection P(U) is also called a hyperspace. The set addition and scalar multiplication of sets can be treated as the vector addition and scalar multiplication. A set which subtracts itself may not provide a zero element in P(U). In this case, we cannot consider the concept of an inverse element in P(U). This says that P(U) cannot be a (conventional) vector space as according to Subsection 2.2.

    The main issue of the above three spaces is that the concept of an inverse element is not available. Therefore, in this paper, a concept of a so-called null set will be adopted for the purpose of playing the role of a zero element in the so-called near vector space that can include the space consisting of all bounded and closed intervals in R, the space consisting of all fuzzy numbers in R, and the hyperspace consisting of all subsets of R.

    We can also attach a norm to this near vector space to form a so-called near normed space, which is a completely new concept with no available related references for this topic. The readers may just refer to the monographs [1,2,3,4,5] on topological vector spaces and the monographs [6,7,8] on functional analysis.

    Based on the concept of a null set, we can define the concept of almost identical elements in near vector space. The norm which is defined in near vector space is completely different from the conventional norm defined in vector space, since the so-called near normed space involves the null set and almost identical concept. The triangle inequality is still considered in near normed space. The concepts of limit and class limit of a sequence in near normed space will be defined. For this setting, we can similarly define the concept of a Cauchy sequence, which can be used to define the completeness of a near normed space. A near normed space that is also complete is called a near Banach space. The main purpose of this paper is to establish the so-called near fixed point in near Banach space, where the near fixed point is based on the almost identical concept. The near fixed point theorems in the normed interval space, the space of fuzzy numbers and the hyperspace have been studied by Wu [9,10,11]. This work will consider the general near normed space such that the near fixed point theorems established in this paper will extend the results obtained by Wu [9,10,11].

    In Sections 2 and 3, the concepts of a near vector space and near normed space are proposed, where some interesting properties are derived in order to study the near fixed point theorem. In Section 4, the concept of a Cauchy sequence in near normed space will be defined. Also, the so-called near Banach space will be defined based on the concept of a Cauchy sequence. In Section 5, we present the near fixed point theorem and the Meir-Keeler type of near fixed point theorem that are established using the almost identical concept in near normed space. In Section 6, we present the near fixed point in the space of fuzzy numbers in R. In Section 7, we present the near fixed point in the hyperspace.

    Let U be a universal set such that it is endowed with the vector addition and scalar multiplication as follows:

    ● (Vector addition). Given any x,yU, the vector addition xy is in U.

    ● (Scalar multiplication). Given any αR and xU, the scalar multiplication αx is in U.

    In this case, we also say that U is a universal set over R. It is clear that the (conventional) vector space V over R is a universal set over R satisfying eight axioms. In the conventional vector space over R, the additive inverse element of vV is denoted by v, and it can also be shown that v=(1)v, which means that the inverse element v is equal to the scalar multiplication (1)v. In this paper, we are not going to consider the concept of an inverse element. However, we still adopt x=(1)x for convenience. In other words, when we write x, it just means that x is multiplied by the scalar 1, since we are not going to consider the concept of an inverse element in the universal set U over R.

    For any x and y in the universal set U over R, the substraction xy is defined by

    xy=x(y).

    Recall that y means the scalar multiplication (1)y. On the other hand, given any xU and αR, we remark that

    (α)xαxandα(x)αx

    in general, unless this law α(βx)=(αβ)x holds true for any α,βR. However, in this paper, this law will not be assumed to be true, since U is not a vector space over R. Next, we present a space that cannot have the concept of an inverse element.

    Example 2.1. Let I be the family of all bounded and closed intervals in R. The vector addition and scalar multiplication are given below.

    ● (Vector addition). Given any two bounded closed intervals [A,B] and [C,D], their vector addition is given by

    [A,B][C,D]=[A+C,B+D]I.

    ● (Scalar multiplication). Given any kR and [A,B]I, the scalar multiplication is given by

    k[A,B]={[kA,kB],ifk0[kB,kA],ifk<0.

    Before introducing the inverse element, we need to point out the zero element. It is clear to see that [0,0] is the zero element of I, since we have

    [0,0][A,B]=[A,B][0,0]=[A,B].

    However, the problem is that [A,B] cannot have an inverse element. The main reason is that we cannot find an interval II satisfying

    [A,B]I=I[A,B]=[0,0].

    This also says that the family I is not a (conventional) vector space. On the other hand, we cannot have the following equality

    (α+β)I=αIβI

    for any II and α,βR. This shows another reason why the family I is not a (conventional) vector space.

    Definition 2.2. Let U be a universal set over R. The following set

    Ψ={xx:xU}

    is called the null set of U. Many other terminologies are also given below:

    ● The null set Ψ is said to satisfy the neutral condition when

    ψΨimpliesψΨ,

    where ψ means (1)ψ, since the concept of an inverse element is not considered in U.

    ● The null set Ψ is said to be closed under the condition of the vector addition when we have

    ψ1ψ2Ψforanyψ1,ψ2Ψ.

    ● The element θU is said to be a zero element when we have

    x=xθ=θxforanyxU.

    Example 2.3. Continued from Example 2.1, we are going to present the null set of the family I. Given any [A,B]I, we have

    [A,B][A,B]=[A,B]([A,B])=[A,B][B,A]=[AB,BA]=[(BA),BA].

    It is clear to see that the null set Ψ of I is given by

    Ψ={[C,C]:C0}={C[1,1]:C0}.

    Definition 2.4. Let U be a universal set over R. We say that U is a near vector space over R when the following conditions are satisfied.

    ● For any xU, the equality 1x=x holds true.

    ● For any x,y,zU and αR, the identity x=y implies the following identities

    xz=yzandαx=αy.

    ● (Commutative law). For any x,yU, the following equality

    xy=yx

    holds true.

    ● (Associative Law). For any x,y,zU, the following equality

    (xy)z=x(yz)

    holds true.

    It is clear to see that any (conventional) vector space over R is also a near vector space over R. However, the converse is not true. Although the family I of all bounded and closed intervals as shown in Example 2.1 is not a (conventional) vector space, it is easy to check that I is a near vector space over R. Next, we define the concept of an almost identical element.

    Definition 2.5. Let U be a near vector space over R with the null set Ψ. Any two elements x and y in U are said to be almost identical when any one of the following conditions is satisfied:

    ● We have x=y;

    ● There exists ψΨ such that

    x=yψorxψ=y;

    ● There exist ψ1,ψ2Ψ such that

    xψ1=yψ2.

    We also write xΨ=y to indicate that x and y are almost identical.

    Remark 2.6. In this paper, when we plan to study some properties using the concept xΨ=y, it is enough to just consider the third case, i.e.,

    xψ1=yψ2,

    since the same arguments are still valid for the first and second cases.

    Regarding the binary relation Ψ= in Definition 2.5, given any xU, we consider the following set

    [x]={yU:xΨ=y}. (2.1)

    We also define the following family

    [U]={[x]:xU}.

    The proof of the following proposition is left for the readers.

    Proposition 2.7. Let U be a near vector space over R with the null set Ψ. Suppose that the null set Ψ is closed under the condition of the vector addition. In other words, we have

    ψ1ψ2Ψforanyψ1,ψ2Ψ.

    Then, the binary relation Ψ= in Definition 2.5 is an equivalence relation.

    The above proposition also says that, when the null set Ψ is not closed under the condition of the vector addition, the binary relation Ψ= is not necessarily an equivalence relation. Therefore, given any y[x], we may not have [y]=[x], unless the binary relation Ψ= is an equivalence relation.

    Suppose that the null set Ψ is closed under the condition of the vector addition. Then Proposition 2.7 says that the sets defined in (2.1) form the equivalence classes. We also have that y[x] implies [x]=[y], which says that the family of all equivalence classes forms a partition of the whole universal set U. Even though in this situation, the space [U] is still not a (conventional) vector space, since not all of the axioms taken in the (conventional) vector space are not necessarily to be satisfied in [U]. For example, we consider the near vector space I over R from Example 2.1. The quotient space [I] cannot be a (conventional) vector space. The reason is that

    (α+β)[x]α[x]+β[x]forαβ<0,

    since

    (α+β)xαx+βxforxIandαβ<0.

    Therefore, we need to seriously study the so-called near vector space.

    Let U be a topological space. The fuzzy subset ˜A of U is defined by a membership function ξ˜A:U[0,1]. The α-level set of ˜A is denoted and defined by

    ˜Aα={xU:ξ˜A(x)α}

    for all α(0,1]. The 0-level set ˜A0 is defined as the closure of the set {xU:ξ˜A(x)>0}.

    Now, we take U=R. Let denote any of the four basic arithmetic operations ,,, between two fuzzy subsets ˜A and ˜B in R. The membership function of ˜A˜B is defined by

    ξ˜A˜B(z)=sup{(x,y):z=xy}min{ξ˜A(x),ξ˜B(y)}

    for all zR. More precisely, the membership functions are given by

    ξ˜A˜B(z)=sup{(x,y):z=x+y}min{ξ˜A(x),ξ˜B(y)};ξ˜A˜B(z)=sup{(x,y):z=xy}min{ξ˜A(x),ξ˜B(y)};ξ˜A˜B(z)=sup{(x,y):z=xy}min{ξ˜A(x),ξ˜B(y)};ξ˜A˜B(z)=sup{(x,y):z=x/y,y0}min{ξ˜A(x),ξ˜B(y)},

    where ˜A˜B˜A(˜B).

    Definition 2.8. Let U be a real topological vector space. We denote by Fcc(U) the set of all fuzzy subsets of U such that each ˜AFcc(U) satisfies the the following conditions:

    ξ˜A(x)=1 for some xU;

    ● The membership function ξ˜A(x) is upper semicontinuous and quasi-concave;

    ● The 0-level set ˜A0 is a compact subset of U.

    In particular, if U=R then each element of Fcc(R) is called a fuzzy number.

    For ˜AFcc(R), it is well-known that, for each α[0,1], the α-level set ˜Aα is a bounded closed interval in R, and it is also denoted by

    ˜Aα=[˜ALα,˜AUα].

    We say that ˜1a is a crisp number with a value of a when the membership function of ˜1a is given by

    ξ˜1a(r)={1ifr=a0ifra.

    It is clear that each α-level set of ˜1a is a singleton {a} for α[0,1]. Therefore, the crisp number ˜1a can be identified with the real number a. In this case, we can identify the inclusion RFcc(R). For convenience, we also write λ˜A˜1λ˜A.

    Let ˜A and ˜B be two fuzzy numbers with

    ˜Aα=[˜ALα,˜AUα]and˜Bα=[˜BLα,˜BUα]forα[0,1].

    It is well known that

    (˜A˜B)α=[˜ALα+˜BLα,˜AUα+˜BUα]forα[0,1] (2.2)

    and

    (˜A˜B)α=[˜ALα˜BUα,˜AUα˜BLα]forα[0,1]. (2.3)

    For λR, we also have

    (λ˜A)α={[λ˜ALα,λ˜AUα],ifλ0[λ˜AUα,λ˜ALα],ifλ<0}forα[0,1]. (2.4)

    Given any ˜AFcc(R), we have

    (˜A˜A)α=[˜ALα˜AUα,˜AUα˜ALα]=[(˜AUα˜ALα),˜AUα˜ALα]forα[0,1]. (2.5)

    This says that each α-level set (˜A˜A)α can be treated as an "approximated real zero number" with symmetric uncertainty ˜AUα˜ALα. We can also see that the real zero number has the highest membership degree of 1 given by ξ˜A˜A(0)=1. In this case, we can say that ˜A˜A is a fuzzy zero number.

    The space Fcc(R) cannot be a (conventional) vector space over R since we cannot identify the inverse elements of any elements in Fcc(R). It is not hard to check that the space Fcc(R) of fuzzy numbers is a near vector space over R by treating the vector addition as the fuzzy addition ˜A˜B and the scalar multiplication as λ˜A=˜1λ˜A. Then, the null set Ψ of the near vector space Fcc(R) is given by

    Ψ={˜A˜A:˜AFcc(R)}.

    Therefore, given any ˜ψΨ, there exists ˜AFcc(R) satisfying ˜ψ=˜A˜A. Equivalently, from (2.5), we see that ˜ψΨ if and only if ˜ψUα0 and ˜ψLα=˜ψUα for all α[0,1], i.e.,

    ˜ψα=[˜ψLα,˜ψUα]=[˜ψUα,˜ψUα],

    where the bounded closed interval ˜ψα is an "approximated real zero number" with symmetric uncertainty ˜ψUα. In other words, each ˜ψΨ is a fuzzy zero number. It is also clear that the crisp number ˜1{0} with a value of 0 is in Ψ. We also see that ˜1{0} is a zero element of Fcc(R), since we have

    ˜A˜1{0}=˜1{0}˜A=˜A

    for any ˜AFcc(R).

    Remark 2.9. It is not hard to check that the null set Ψ is closed under the condition of the vector addition (i.e., fuzzy addition) and satisfies the neutral condition.

    Given any two ˜A and ˜B in Fcc(R), the definition says

    ˜AΨ=˜Bifandonlyif˜A˜ψ(1)=˜B˜ψ(2)forsome˜ψ(1),˜ψ(2)Ψ.

    By considering the α-level sets, for any α[0,1], we have

    [˜ALα,˜AUα][(˜ψ(1))Uα,(˜ψ(1))Uα]=[˜BLα,˜BUα][(˜ψ(2))Uα,(˜ψ(2))Uα],

    which says

    ˜ALα(˜ψ(1))Uα=˜BLα(˜ψ(2))Uαand˜AUα+(˜ψ(1))Uα=˜BUα+(˜ψ(2))Uα.

    Let Kα=(˜ψ(2))Uα(˜ψ(1))Uα. Then, we have

    ˜ALα=˜BLαKαand˜AUα=˜BUα+Kα.

    Therefore, we obtain

    ˜Aα=˜Bα[Kα,Kα]forallα[0,1].

    Therefore ˜AΨ=˜B means that the α-level sets ˜Aα and ˜Bα are essentially identical differing with symmetric uncertainty Kα for α[0,1].

    Let U be a (conventional) vector space, and let P(U) be a collection of all subsets of U. Given any A,BP(U), the set addition of A and B in R is defined by

    AB={a+b:aAandbB}

    and the scalar multiplication in P(U) is defined by

    λA={λa:aA},

    where λ is a constant in R. The substraction between A and B is denoted and defined by

    ABA(B)={ab:aAandbB}.

    Then, we have the following properties:

    λ(AB)=λAλB for λR;

    λ1(λ2A)=(λ1λ2)A for λ1,λ2R;

    ● Let A be a convex subset of U. If λ1 and λ2 have the same sign, then we have

    (λ1λ2)A=λ1Aλ2A.

    In P(U), the set addition AB can be treated as the vector addition and λA can be treated as the scalar multiplication. Let θU be the zero element of U. It is clear that the singleton {θU} can be regarded as the zero element of P(U), since we have

    A{θU}={θU}A=A.

    Since AA{θU}, it means that AA is not the zero element of P(U). In other words, the additive inverse element of A in P(U) does not exist. Therefore, the space P(U) cannot be a (conventional) vector space over R, since we cannot identify the inverse elements of any elements in P(U). It is not hard to check that P(U) is a near vector space over R. In this case, the null set of P(U) is given by

    Ψ={AA:AP(U)}.

    Remark 2.10. It is not hard to check that the null set Ψ is closed under the condition of the vector addition (i.e., set addition) and satisfies the neutral condition.

    Let U be a near vector space over R with the null set Ψ. We are going to endow a norm to the space U. Because we do not have an elegant structure in U like in the (conventional) vector space, many kinds of so-called near normed spaces are proposed below.

    Definition 3.1. Let U be a near vector space over R with the null set Ψ, and let ∥:UR+ be a nonnegative real-valued function defined on U.

    ● The function is said to satisfy the null condition when

    x∥=0ifandonlyifxΨ.

    ● The function is said to satisfy the null super-inequality when

    xψ∥≥∥xforanyxUandψΨ.

    ● The function is said to satisfy the null sub-inequality when

    xψ∥≤∥xforanyxUandψΨ.

    ● The function is said to satisfy the null equality when

    xψ∥=∥xforanyxUandψΨ.

    Definition 3.2. Let U be a near vector space over R with the null set Ψ. Given a nonnegative real-valued function ∥:UR+ defined on U, we consider the following conditions:

    (i) αx∥=|α|x for any xU and αR;

    (i') αx∥=|α|x for any xU and αR with α0.

    (ii) xy∥≤∥x+y for any x,yU.

    (iii) x∥=0 implies xΨ.

    Many kinds of near normed spaces are defined below.

    (U,) is said to be a near pseudo-seminormed space when conditions (i) and (ii) are satisfied.

    (U,) is said to be a near seminormed space when conditions (i) and (ii) are satisfied.

    (U,) is said to be a near pseudo-normed space when conditions (i), (ii) and (iii) are satisfied.

    (U,) is said to be a near normed space when conditions (i)–(iii) are satisfied.

    Remark 3.3. Suppose that the norm satisfies the null condition. Then, we have the following observations.

    ● We want to claim that the norm also satisfies the null sub-inequality. Indeed, using the triangle inequality, it follows that

    xψ∥≤∥x+ψ∥=∥xforanyψΨ.

    ● We want to claim that Ψ is closed under the condition of the scalar multiplication. Indeed, using conditions (i) and (iii) in Definition 3.2, it follows that

    αψ∥=|α|ψ∥=0impliesαψΨ.

    ● We want to claim that Ψ is closed under the condition of the vector addition. Indeed, using condition (ii) in Definition 3.2, it follows that

    0≤∥ψ1ψ2∥≤∥ψ1+ψ2∥=0.

    which implies ψ1ψ2Ψ.

    Example 3.4. Continued from Examples 2.1 and 2.3, we define the norm ∥:IR+ on I by

    [A,B]∥=|A+B|.

    It is easy to check that the family (I,) is a near normed space such that the norm satisfies the null equality and null condition.

    Let (U,) be a near pseudo-seminormed space. In general, we cannot have the following equality

    xy=(yx).

    The reason is that we do not assume the laws

    α(xy)=αxαyandα(βx)=(αβ)y

    for x,yU and α,βR. It also says that, in general, we cannot obtain the following equality

    xy∥=∥yx.

    Therefore, we propose the following definition.

    Definition 3.5. Let (U,) be a near pseudo-seminormed space. The norm is said to satisfy the symmetric condition when

    xy∥=∥yxforanyx,yU.

    Next, we are going to provide the sufficient conditions to guarantee that the norm is able to satisfy the symmetric condition.

    Proposition 3.6. Let (U,) be a near pseudo-seminormed space. Suppose that the norm satisfies the null equality. Then, this norm satisfies the symmetric condition.

    Proof. We first have

    [y(x)]([y(x)])=[y(x)][y(x)])ψΨ.

    Using the null equality, we obtain

    x(y)∥=∥x(y)ψ∥=∥[x(y)][y(x)]([y(x)]).

    Using the associative and commutative laws, we also have

    [x(y)][y(x)]([y(x)])=∥[x(x)][y(y)]([y(x)])=∥ψ1ψ2([y(x)]),

    where ψ1=x(x)Ψ and ψ1=y(y)Ψ. Using the null equality two times, we obtain

    ψ1ψ2([y(x)])∥=∥ψ2([y(x)])∥=∥[y(x)].

    Finally, using condition (i) in Definition 3.2, it follows that

    [y(x)]∥=∥y(x).

    Combining the above equalities, we obtain

    x(y)∥=∥y(x).

    This completes the proof.

    Let U be a near vector space over R. Then, the following equality

    (xy)=(x)(y)

    does not hold true in general, since U does not have the elegant structure like the conventional vector space. However, if (U,) is a near pseudo-seminormed space, we can have the following interesting results.

    Proposition 3.7. Let (U,) be a near pseudo-seminormed space. Suppose that the norm satisfies the null equality. Given any x,y,zU, we have

    z(xy)∥=∥zxy∥=∥z(x)(y)

    and

    z(xy)∥=∥zxy∥=∥z(x)y.

    However, in general without using the norm, we have

    z(xy)zxyz(x)(y)

    and

    z(xy)zxyz(x)y.

    Proof. Let ψ1=xxΨ and ψ2=yyΨ. Then, we have

    xyxy=ψ1ψ2.

    By adding z(xy) on both sides, we obtain

    z(xy)xyxy=ψ1ψ2z(xy).

    Let ψ3=(xy)(xy)Ψ. Then, we obtain

    ψ3zxy=ψ1ψ2z(xy).

    Since the norm satisfies the null equality, we have

    zxy=∥ψ3zxy∥=∥ψ1ψ2z(xy)=∥ψ2z(xy)∥=∥z(xy).

    Now, we also have

    (x)yxy=ψ1ψ2.

    Let ψ4=(xy)(xy)Ψ, and add z(xy) on both sides. Then, we obtain

    ψ4z(x)y=ψ1ψ2z(xy).

    Using the null equality, we can similarly obtain the desired result by taking the norm on both sides. This completes the proof.

    Proposition 3.8. Let (U,) be a near pseudo-normed space. Suppose that the norm satisfies the null super-inequality. Given any x,z,y1,,ymU, we have

    xz∥≤∥xy1+y1y2++yjyj+1++ymz.

    Proof. Since yj(yj)=yjyj=ψjΨ for j=1,,m, using the null super-inequality for m times, we have

    xz≤∥x(z)ψ1ψm=∥x(z)y1(y1)ym(ym). (3.1)

    Using the commutative and associative laws, we also have

    x(z)y1(y1)ym(ym)=∥[x(y1)][y1(y2)]++[yj(yj+1)]++[ym(z)]≤∥xy1+y1y2++yjyj+1++ymz(usingthetriangleinequality).

    Using (3.1), the proof is complete.

    Proposition 3.9. We have the following properties.

    (i) Let (U,) be a near pseudo-normed space. Given any x,yU,

    xy∥=0impliesxΨ=y.

    (ii) Let (U,) be a near pseudo-seminormed space. Suppose that the norm satisfies the null equality. Given any x,yU,

    xΨ=yimpliesx∥=∥y.

    (iii) Let (U,) be a near pseudo-seminormed space. Suppose that the norm satisfies the null super-inequality and null condition. Given any x,yU,

    xΨ=yimpliesxy∥=0.

    Proof. To prove Part (i), for xy∥=0, we have xyΨ, i.e., xy=ψ1 for some ψ1Ψ. Let ψ2=yyΨ, and add y on both sides. Then, we obtain

    xψ2=xyy=yψ1,

    which says that xΨ=y.

    To prove Part (ii), the definition says that xΨ=y implies

    xψ1=yψ2forsomeψ1,ψ2Ψ.

    Using the null equality, it follows that

    x∥=∥xψ1∥=∥yψ2∥=∥y.

    To prove Part (iii), we first note that the null set Ψ is closed under the condition of the vector addition from Remark 3.3. For xΨ=y, we have

    xψ1=yψ2forsomeψ1,ψ2Ψ.

    Let ψ3=yyΨ, and add y on both sides. Then, we obtain

    xyψ1=y(y)ψ2=ψ3ψ2ψ4Ψ. (3.2)

    Using the null super-inequality and (3.2), it follows that

    xy∥≤∥xyψ1∥=∥ψ4∥=0.

    This completes the proof.

    Example 3.10. We are going to define a norm in Fcc(R). Given any ˜AFcc(R), we define

    ˜A∥=supα[0,1]|˜ALα+˜AUα|.

    We first claim that satisfies the null condition. In other words, we want to check that

    ˜A∥=0ifandonlyif˜AΨ.

    Suppose that ˜A∥=0. Then, we have |˜ALα+˜AUα|=0 for all α[0,1], which also says that ˜ALα=˜AUα for all α[0,1]. This shows that ˜AΨ. On the other hand, suppose that ˜AΨ. Then, we have ˜ALα=˜AUα for all α[0,1], which says that ˜A∥=0.

    Using (2.4), we have

    λ˜A∥=supα[0,1]|(λ˜A)Lα+(λ˜A)Uα|=|λ|supα[0,1]|˜ALα+˜AUα|=|λ|˜A.

    Using (2.2), we also have

    ˜A˜B=supα[0,1]|(˜A˜B)Lα+(˜A˜B)Uα|=supα[0,1]|˜ALα+˜BLα+˜AUα+˜BUα|supα[0,1](|˜ALα+˜AUα|+|˜BLα+˜BUα|)supα[0,1]|˜ALα+˜AUα|+supα[0,1]|˜BLα+˜BUα|=∥˜A+˜B.

    This shows that (Fcc(R),) is a near normed space such that the null condition is satisfied.

    Furthermore, we can show that the null equality is also satisfied. Given any ˜ψΨ, it means that ˜ψLα=˜ψUα for all α[0,1]. Therefore, we have

    ˜A˜ψ=supα[0,1]|(˜A˜ψ)Lα+(˜A˜ψ)Uα|=supα[0,1]|˜ALα+˜ψLα+˜AUα+˜ψUα|=supα[0,1]|˜ALα˜ψUα+˜AUα+˜ψUα|=supα[0,1]|˜ALα+˜AUα|=∥˜A,

    which shows that the null equality is indeed satisfied.

    Example 3.11. Let (U,U) be a (conventional) normed space. We consider the hyperspace P(U) in the normed space (U,U). We want to define a norm in P(U) such that (P(U),) is a near normed space. Given any AP(U), we define

    A∥=supaAaU.

    Let θU be the zero element in the normed space (U,U). We want to claim that A∥=0 if and only if A={θU}Ψ. Suppose that A={θU}. Then, we haveA∥=0. On the other hand, suppose that A∥=0. Then, we have aU=0 for all aA, which says that A={θU}.

    Now, we have

    λA∥=supaλAaU=supbAλbU=|λ|supbAbU=|λ|A.

    and

    AB=supcABcU=sup{(a,b):aA,bB}a+bUsup{(a,b):aA,bB}(aU+bU)(usingthetriangleinequalityin(U,U))supaAaU+supbBbU=∥A+B.

    This shows that (P(U),) is indeed a near normed space.

    Let (U,) be a near pseudo-seminormed space. Since the symmetric condition is not necessarily satisfied, we can define many concepts of limits based on . For a sequence {xn}n=1 in U, since

    xnx∥≠∥xxn

    in general, many concepts of limits are proposed below.

    Definition 4.1. Let (U,) be a near pseudo-seminormed space.

    ● A sequence {xn}n=1 in U is said to -converge to xU when we have the following limit

    limnxnx∥=0.

    ● A sequence {xn}n=1 in U is said to -converge to xU when we have the following limit

    limnxxn∥=0.

    ● A sequence {xn}n=1 in U is said to converge to xU when we have the following limit

    limnxnx∥=limnxxn∥=0.

    Remark 4.2. We have the following observations.

    ● Suppose that the norm satisfies the null equality. Then, Proposition 3.6 says that the symmetric condition is satisfied, i.e.,

    xnx∥=∥xxnforalln.

    This also means that the above three concepts of convergence are equivalent.

    ● Suppose that the sequence {xn}n=1 is simultaneously -convergent and -convergent. It says that there exists x,yU such that we have the following limits

    limnxnx∥=limnyxn∥=0.

    However, in this situation, x is not necessarily equal to y.

    Let U be a near vector space over R with the null set Ψ. Suppose that the null set Ψ is closed under the condition of the vector addition. Then, Proposition 2.7 says that the binary relation Ψ= in Definition 2.5 is an equivalence relation, which also says that the classes defined in (2.1) form the equivalence classes. In this case, we have many interesting results as follows.

    Proposition 4.3. Let (U,) be a near pseudo-normed space with the null set Ψ such that Ψ is closed under the condition of the vector addition.

    (i) Suppose that the norm satisfies the null super-inequality. Then, we have the following results.

    If the sequence {xn}n=1 in (U,) -converges to x and -converges to y, then [x]=[y].

    If the sequence {xn}n=1 in (U,) converges to x and y simultaneously, then [x]=[y].

    (ii) Suppose that the norm satisfies the null equality. If the sequence {xn}n=1 in (U,) converges to xU, then, given any y[x], the sequence {xn}n=1 also converges to y.

    Proof. To prove Part (i), suppose that the sequence {xn}n=1 -converges to x and -converges to y. Then, we have the following limits

    limnxxn∥=limnxny∥=0.

    Using Proposition 3.8, we obtain

    0≤∥xy∥≤∥xxn+xnyforalln,

    which implies

    0≤∥xy∥≤limnxxn+limnxny∥=0+0=0. (4.1)

    This shows that xy∥=0. By Definition 3.2, it follows that xyΨ, i.e., xΨ=y. Since the binary relation Ψ= is an equivalence relation by Proposition 2.7, we obtain [x]=[y], which shows the first case. The second case by assuming that {xn}n=1 converges to x and y simultaneously can be similarly obtained.

    To prove Part (ii), given any y[x], we have

    xψ1=yψ2forsomeψ1,ψ2Ψ.

    Using Proposition 3.6, the symmetric condition is satisfied. Therefore, we obtain

    0≤∥xny∥=∥yxn∥=∥ψ2yxn∥=∥ψ1xxn=∥xxn∥=∥xnxforalln,

    which says that

    limnxxn∥=limnxnx∥=0implieslimnxny∥=limnyxn∥=0.

    This completes the proof.

    Inspired by Part (ii) of Proposition 4.3, we propose the following concept of a limit.

    Definition 4.4. Let (U,) be a near pseudo-seminormed space. When a sequence {xn}n=1 in U converges to some xU, the equivalence class [x] is called the class limit of the sequence {xn}n=1. In this case, we also write

    limnxn=[x].

    Remark 4.5. Suppose that [x] is a class limit of the sequence {xn}n=1. Then, for y[x], it is not necessarily that the sequence {xn}n=1 converges to y, unless the norm satisfies the null equality as given by Part (ii) of Proposition 4.3.

    The uniqueness of the class limit is shown below.

    Proposition 4.6. Let (U,) be a near pseudo-normed space. Suppose that the norm satisfies the null super-inequality. Then, the class limit is unique.

    Proof. Suppose that the sequence {xn}n=1 is convergent with two class limits [x] and [y]. It means

    limnxxn∥=limnxnx∥=0=limnyxn∥=limnxny.

    Using (4.1), we have

    0≤∥xy∥≤limnxxn+limnxny∥=0+0=0,

    which says that xy∥=0. Using Part (i) of Proposition 3.9, we obtain xΨ=y, i.e., [x]=[y]. This completes the proof.

    In the near pseudo-seminormed space (U,), the symmetric condition is not necessarily satisfied. Therefore, we can propose many different concepts of a Cauchy sequence and completeness as follows.

    Definition 4.7. Let (U,) be a near pseudo-seminormed space, and let {xn}n=1 be a sequence in U.

    {xn}n=1 is called a -Cauchy sequence when, given any ϵ>0, there exists an integer N such that n>m>N implies xnxm∥<ϵ.

    a. If every -Cauchy sequence in U is convergent, we say that U is -complete.

    b. If every -Cauchy sequence in U is -convergent, we say that U is (,)-complete.

    c. If every -Cauchy sequence in U is -convergent, we say that U is (,)-complete.

    {xn}n=1 is called a -Cauchy sequence when, given any ϵ>0, there exists an integer N such that n>m>N implies xmxn∥<ϵ.

    a. If every -Cauchy sequence in U is convergent, we say that U is -complete.

    b. If every -Cauchy sequence in U is -convergent, we say that U is (,)-complete.

    c. If every -Cauchy sequence in U is -convergent, we say that U is (,)-complete.

    {xn}n=1 is called a Cauchy sequence when, given any ϵ>0, there exists an integer N such that m,n>N with mn implies xnxm∥<ϵ and xmxn∥<ϵ.

    a. If every Cauchy sequence in U is convergent, we say that U is complete.

    b. If every Cauchy sequence in U is -convergent, we say that U is -complete.

    c. If every Cauchy sequence in U is -convergent, we say that U is -complete.

    Remark 4.8. Suppose that satisfies the symmetric condition, i.e.,

    xnxm∥=∥xmxn∥<ϵ.

    Then all of the concepts of a Cauchy sequence are equivalent, and all of the concepts of completeness are equivalent.

    Remark 4.9. It is clear to see that if {xn}n=1 is a Cauchy sequence then {xn}n=1 is both a -Cauchy sequence and -Cauchy sequence.

    Remark 4.10. From Remark 4.9, we have the following observations.

    ● If U is complete, then it is also -complete and -complete.

    ● If U -complete, then it is also (,)-complete and (,)-complete.

    ● If U is -complete, then it is also (,)-complete and (,)-complete.

    Proposition 4.11. Let (U,) be a near pseudo-seminormed space. Suppose that the norm satisfies the null super-inequality. Then, we have the following properties.

    (i) Every convergent sequence is a Cauchy sequence.

    (ii) Suppose that the norm satisfies the null condition. Then, every simultaneously -convergent and -convergent sequence is a Cauchy sequence.

    (iii) Given any fixed xU, suppose that the following conditions are satisfied:

    The sequence {xn}n=1 -converges to x.

    The sequence {yn}n=1 -converges to x.

    Then, the sequence {xnyn}n=1 is a Cauchy sequence satisfying

    limnxnyn∥=0.

    Proof. To prove Part (i), let {xn}n=1 be a convergent sequence. Therefore, by the definition of convergence, given any ϵ>0, we have

    xnx∥<ϵ2andxxn∥<ϵ2

    for a sufficiently large n. Using Proposition 3.8, we obtain

    xmxn∥≤∥xmx+xxn∥<ϵ2+ϵ2=ϵ

    and

    xnxm∥≤∥xnx+xxm∥<ϵ2+ϵ2=ϵ

    for sufficiently large n and m. This shows that {xn}n=1 is a Cauchy sequence.

    To prove Part (ii), we first note that the null set Ψ is closed under the condition of the vector addition from Remark 3.3. Assume that the sequence {xn}n=1 simultaneously -converges to x and -converges to y. Then, Part (i) of Proposition 4.3 says that xΨ=y, which also implies xy∥=0 by Part (iii) of Proposition 3.9. Given any ϵ>0, we also have

    xnx∥<ϵ2andyxn∥<ϵ2

    for a sufficiently large n. Since xy∥=0, using Proposition 3.8, we obtain

    xmxn∥≤∥xmx+xy+yxn∥<ϵ2+ϵ2=ϵ

    and

    xnxm∥≤∥xnx+xy+yxm∥<ϵ2+ϵ2=ϵ

    for a sufficiently large n and m. This shows that {xn}n=1 is a Cauchy sequence.

    To prove Part (iii), the assumption says that

    limnxnx∥=0andlimnxyn∥=0. (4.2)

    Therefore, given any ϵ>0, we have

    xnx∥<ϵ4andxyn∥<ϵ4 (4.3)

    for a sufficiently large n. Let zn=xnyn. Then, we have

    znzm=∥(xnyn)[(xmym)]≤∥xnyn+(xmym)=∥xnyn+xmym(bytheconditionofnorm)≤∥xnx+xyn+xmx+xym(byProposition3.8),

    which shows that

    znzm∥<ϵ4+ϵ4+ϵ4+ϵ4=ϵ

    for a sufficiently large n and m by using (4.3). We can similarly obtain zmzn∥<ϵ. This shows that the sequence {zn}n=1 is a Cauchy sequence. Using Proposition 3.8 and (4.2), we obtain

    xnyn∥≤∥xnx+xyn∥→0.

    This completes the proof.

    Many different kinds of near Banach spaces can also be proposed as follows.

    Definition 4.12. Let (U,) be a near pseudo-seminormed space. Different kinds of near Banach spaces are defined below.

    ● If U is complete, then it is called a near pseudo-semi-Banach space.

    ● If U is -complete, then it is called a near -pseudo-semi-Banach space.

    ● If U is -complete, then it is called a near -pseudo-semi-Banach space.

    ● If U is -complete, then it is called a near -pseudo-semi-Banach space.

    ● If U is (,)-complete, then it is called a near (,)-pseudo-semi-Banach space.

    ● If U is (,)-complete, then it is called a near (,)-pseudo-semi-Banach space.

    ● If U is -complete, then it is called a near -pseudo-semi-Banach space.

    ● If U is (,)-complete, then it is called a near (,)-pseudo-semi-Banach space.

    ● If U is (,)-complete, then it is called a near (,)-pseudo-semi-Banach space.

    Definition 4.13. Different kinds of near Banach spaces are defined below.

    ● Let (U,) be a near seminormed space. If U is complete, then it is called a near semi-Banach space. According to the different kinds of completeness in Definition 4.12, the other kinds of near Banach spaces can be similarly defined.

    ● Let (U,) be a near pseudo-normed space. If U is complete, then it is called a near pseudo-Banach space. According to the different kinds of completeness in Definition 4.12, the other kinds of near Banach spaces can be similarly defined.

    ● Let (U,) be a near normed space. If U is complete, then it is called a near Banach space. According to the different kinds of completeness in Definition 4.12, the other kinds of near Banach spaces can be similarly defined.

    Example 4.14. Continued from Example 3.4, we want to claim that the near normed space (I,) is complete. In other words, we want to claim that (I,) is a near Banach space. We first have

    [A,B][C,D]=∥[A,B][D,C]∥=∥[AD,BC]∥=|(AD)+(BC)|=|(A+B)(C+D)|=∥[C,D][A,B].

    Therefore, the norm satisfies the symmetric condition. Let {[An,Bn]}n=1 be a Cauchy sequence in the space (I,). Therefore, given any ϵ>0, for a sufficiently large n and m, we have

    ϵ>∥[An,Bn][Am,Bm]=∥[An,Bn][Bm,Am]=∥[AnBm,BnAm]∥=|(An+Bn)(Am+Bm)|. (4.4)

    Let Cn=An+Bn. Then, the expression (4.4) says that {Cn}n=1 is a Cauchy sequence in R. The completeness of R says that there exists CR satisfying |CnC|<ϵ for a sufficiently large n. In this case, we can define a closed interval [A,B] satisfying A+B=C. Therefore, we obtain

    [An,Bn][A,B]=∥[An,Bn][B,A]∥=∥[AnB,BnA]=|(An+Bn)(A+B)|=|CnC|<ϵ

    for a sufficiently large n. This says that the sequence {[An,Bn]}n=1 is convergent, since the norm satisfies the symmetric condition. This shows that (I,) is a near Banach space.

    Example 4.15. Continued from Example 3.10, we want to claim that the near normed space of fuzzy numbers (Fcc(R),) is complete. Suppose that {˜A(n)}n=1 is a Cauchy sequence in (Fcc(R),). By definition, we have

    ˜A(n)˜A(m)∥<ϵform,n>Nwithmn.

    We write

    (˜A(n))Lα=˜A(n,L)αand(˜A(n))Uα=˜A(n,U)α.

    Then, we have

    ϵ>∥˜A(n)˜A(m)∥=supα[0,1]|(˜A(n)˜A(m))Lα+(˜A(n)˜A(m))Uα|=supα[0,1]|˜A(n,L)α˜A(m,U)α+˜A(n,U)α˜A(m,L)α|(using(2.3))=supα[0,1]|(˜A(n,L)α+˜A(n,U)α)(˜A(m,L)α+˜A(m,U)α)|. (4.5)

    For each fixed α[0,1], we define

    C(n)α=˜A(n,L)α+˜A(n,U)αandC(m)α=˜A(m,L)α+˜A(m,U)α.

    Let fn(α)=C(n)α. Using (4.5), we have

    |fn(α)fm(α)|supα[0,1]|fn(α)fm(α)|=supα[0,1]|C(n)αC(m)α|<ϵforallα[0,1]. (4.6)

    According to the properties of fuzzy numbers, the function fn is continuous on [0,1]. In this case, we consider a sequence of continuous functions {fn}n=1 on [0,1]. Then (4.6) says that the sequence of functions {fn}n=1 satisfies the Cauchy condition for uniform convergence. By referring to Apostol [12,Theorem 9.3], it follows that {fn}n=1 converges uniformly to a limit function f(α)Cα on [0,1]. In other words, for sufficiently large n, we have

    |C(n)αCα|<ϵ2forallα[0,1]. (4.7)

    Since each fn is continuous on [0,1], it follows that the limit function f(α)Cα is continuous on [0,1] according to Apostol [12,Theorem 9.2]. The continuity of Cα on [0,1] allows us to find a fuzzy number ˜A satisfying

    ˜ALα+˜AUα=Cαforallα[0,1].

    Therefore, for a sufficiently large n, we have

    ˜A(n)˜A=supα[0,1]|(˜A(n)˜A)Lα+(˜A(n)˜A)Lα|=supα[0,1]|˜A(n,L)α˜AUα+˜A(n,U)α˜ALα|=supα[0,1]|(˜A(n,L)α+˜A(n,U)α)(˜ALα+˜AUα)|=supα[0,1]|C(n)αCα|ϵ2<ϵ(using(4.7)).

    This shows that the sequence {˜A(n)}n=1 is convergent. Therefore, we conclude that (Fcc(R),) is a near Banach space of fuzzy numbers.

    Example 4.16. Continued from Example 3.11, we further assume that (U,U) is a (conventional) Banach space. Then, we want to claim that the near normed space (P(U),) is complete. Suppose that {An}n=1 is a Cauchy sequence in (P(U),). Let A be a collection of all sequences generated by the sequence {An}n=1. More precisely, each element in A is a sequence {an}n=1 with anAn for all n. We want to claim that each sequence (each element) in A is convergent. Since {An}n=1 is a Cauchy sequence, by definition, we have

    AnAm∥<ϵform,n>Nwithmn,

    which says that

    ϵ>∥AnAm∥=supxAnAmxU=sup{(an,am):anAn,amAm}anamU, (4.8)

    which says anamU<ϵ for any sequence {an}n=1 with anAn for all n in the uniform sense; that is to say, ϵ is independent of an and am. Using the completeness of (U,U), we see that each sequence {an}n=1 is convergent to some aU such that

    anaU0asnintheuniformsense, (4.9)

    where the uniform sense meas that anaU<ϵ such that ϵ is independent of anAn and a for a sufficiently large n. Indeed, if ϵ is dependent on an and a, then

    anamU≤∥anaU+aamU

    says that ϵ is dependent on an and am, which is a contradiction.

    We can define a subset A of U that collects all of the limit points of each sequence in A. Then, we want to show AnA∥→0 as n. Given any xAnA, we have x=ana for some anAn and aA. Since a is a limit point of some sequence {ˆan}n=1 in A, for m>n>N, using (4.8), we have

    anaU≤∥anˆamU+ˆamaU<ϵ+ˆamaU,

    where ϵ is independent of an and ˆam according to (4.8). Since ˆamaU0 as m in the unform sense according to (4.9), it follows that anaU0 as n in the uniform sense. Therefore, we obtain

    AnA∥=supxAnAxU=sup{(an,a):anAn,aA}anaU0asn.

    This shows that the sequence {An}n=1 is convergent. Therefore, we conclude that (P(U),) is a near Banach space.

    Let T:UU be a function from a universal set U into itself. Any point xU is called a fixed point when we have T(x)=x. Recall that (I,) presented in Example 3.4 is not a (conventional) normed space. Therefore, we are not able to study the fixed point of contractive mappings defined on (I,) into itself. In this paper, we shall study the so-called near fixed point that is defined below.

    Definition 5.1. Let U be a near vector space over R with a null set Ψ, and let T:UU be a function defined on U into itself. A point xU is called a near fixed point of T when we have T(x)Ψ=x.

    In the sequel, we shall consider three different contractions to study the near fixed point theorem.

    We are going to propose the concept of a contraction on a near pseudo-seminormed space. Under some suitable conditions, we can obtain the near fixed point theorem based on near Banach space.

    Definition 5.2. Let (U,) be a near pseudo-seminormed space, and let T:(U,)(U,) be a function from (U,) into itself. The function T is called a contraction on U when there exists a real number 0<α<1 such that the inequality

    T(x)T(y)∥≤αxy

    is satisfied for any x,yU.

    Given any initial element x0U, we can generate an iterative sequence {xn}n=1 using the composition of function T given by

    xn=Tn(x0). (5.1)

    The main goal of this paper is to show that the sequence {xn}n=1 in U can converge to a near fixed point.

    Theorem 5.3. (Near fixed point theorem). Let (U,) be a near pseudo-Banach space with the null set Ψ. Suppose that the following conditions are satisfied.

    The norm satisfies the null equality.

    The null sets Ψ is closed under the condition of the vector addition and satisfies the neutral condition.

    The function T:(U,)(U,) is a contraction on U.

    Then T has a near fixed point xU satisfying T(x)Ψ=x. Moreover, the near fixed point x is obtained by the following limit

    limnxxn∥=limnxnx∥=0,

    where the sequence {xn}n=1 is generated according to (5.1). We also have the following properties.

    (a) There is a unique equivalence class [x] such that, for any x[x], x cannot be a near fixed point. If x is a near fixed point of T, then we have [x]=[x], i.e., xΨ=x.

    (b) Every point x in the equivalent class [x] is also a near fixed point of T such that the following equalities

    T(x)Ψ=xandxΨ=x

    are satisfied.

    Proof. Proposition 3.6 says that the norm satisfies the symmetric condition. Using Proposition 2.7, we also see that the family of all sets [x] forms the equivalence classes. Given any initial element x0U, we are going to show that the sequence {xn}n=1 generated by (5.1) is a Cauchy sequence. Now, we have

    xm+1xm=∥T(xm)T(xm1)αxmxm1(sinceTisacontractiononU)=αT(xm1)T(xm2)α2xm1xm2(sinceTisacontractiononU)αmx1x0(sinceTisacontractiononU). (5.2)

    Given any two integers n and m satisfying n<m, we obtain

    xmxn≤∥xmxm1+xm1xm2++xn+1xn(usingProposition3.8)(αm1+αm2++αn)x1x0(using(5.2))=αn1αmn1αx1x0.

    Since 0<α<1, it follows that

    xmxn∥≤αn1αx1x0∥→0asn,

    which shows that {xn}n=1 is a Cauchy sequence. Since the norm satisfies the symmetric condition and the near normed space (U,) is complete, there exists xU satisfying

    limnxxn∥=limnxnx∥=0. (5.3)

    Next, we want to claim that any point x in the equivalence class [x] is a near fixed point. We first have

    xψ1=xψ2forsomeψ1,ψ2Ψ. (5.4)

    Since the norm satisfies the null equality, we have

    xT(x)∥=∥(xψ1)T(x). (5.5)

    Using Proposition 3.8, we also have

    (xψ1)T(x)∥≤∥(xψ1)xm+xmT(x)=∥(xψ1)xm+T(xm1)T(x)≤∥(xψ1)xm+αxm1x(sinceTisacontractiononU). (5.6)

    The the neutral condition says that ψ1Ψ. Since the norm satisfies the null equality, we have

    xm1x∥=∥xm1x(ψ1).

    Using Proposition 3.7, we also have

    xm1x(ψ1)∥=∥xm1(xψ1),

    which says that

    xm1x∥=∥xm1(xψ1). (5.7)

    Combining (5.5)–(5.7), we obtain

    xT(x)∥≤∥(xψ1)xm+αxm1(xψ1).

    Using (5.4), we also obtain

    xT(x)∥≤∥(xψ2)xm+αxm1(xψ2). (5.8)

    Now, we have

    (xψ2)xm=∥xxmψ2=∥xxm(usingthenullequality) (5.9)

    and

    xm1(xψ2)∥=∥xm1x(ψ2)(usingProposition3.7)=∥xm1xψ3)(usingtheneutralconditionbysettingψ3=ψ2Ψ)=∥xm1x(usingthenullequality). (5.10)

    Combining (5.8)–(5.10), we obtain

    xT(x)∥≤∥xxm+xm1x,

    which implies xT(x)∥=0 as m by using (5.3). Using Part (i) of Proposition 3.9, we obtain

    T(x)Ψ=xforanypointx[x].

    Now, we assume that x[x] is another near fixed point x of T, i.e., xΨ=T(x). Since xΨ=T(x), we have

    xψ1=T(x)ψ2andxψ3=T(x)ψ4 (5.11)

    for some ψiΨ, i=1,,4. Now, we have

    (xψ1)(xψ3)∥=∥(xψ1)x(ψ3)(usingProposition3.7)=∥(xψ1)xψ5)(usingtheneutralconditionbysettingψ5=ψ3Ψ)=∥(xψ1)x(usingthenullequality)=∥xx(usingthenullequalityagain). (5.12)

    We can similarly obtain

    (T(x)ψ2)(T(x)ψ4)∥=∥T(x)T(x). (5.13)

    Therefore, we obtain

    xx=∥(xψ1)(xψ3)(using(5.12))=∥(T(x)ψ2)(T(x)ψ4)(using(5.11))=∥T(x)T(x)(using(5.13))αxx(sinceTisacontractiononU).

    Since 0<α<1, it forces xx∥=0, i.e., xΨ=x by Part (i) of Proposition 3.9, which contradicts x[x]. This shows that any x[x] cannot be the near fixed point of T. Equivalently, if x is a near fixed point of T, then we must have x[x]. This completes the proof.

    We are going to propose the concept of a weakly strict contraction on a near pseudo-seminormed space. Under some suitable conditions, we can obtain the near fixed point theorem based on a near -Banach space and near -Banach space.

    Definition 5.4. Let (U,) be a near pseudo-normed space. A function

    T:(U,)(U,)

    is called a weakly strict contraction on U when the following conditions are satisfied:

    xΨ=y implies T(x)T(y)∥=0;

    xΨy implies T(x)T(y)∥<∥xy.

    Part (i) of Proposition 3.9 says that xΨy implies xy∥≠0. Therefore, the second condition of the weakly strict contraction is well-defined. In other words, when we consider the weakly strict contraction, the space (U,) should be assumed to be a near pseudo-normed space rather than a near pseudo-seminormed space.

    Proposition 5.5. Let (U,) be a near pseudo-normed space. Suppose that the norm satisfies the null super-inequality and null condition. If T is a contraction on U, then it is also a weakly strict contraction on U.

    Proof. Since T is a contraction on U, we have

    T(x)T(y)∥≤αxy

    for 0<α<1. We consider the following two cases.

    ● Suppose that xΨ=y. Part (iii) of Proposition 3.9 says that xy∥=0, which implies

    0≤∥T(x)T(y)∥≤αxy∥=0,

    i.e., T(x)T(y)∥=0.

    ● Suppose that xΨy. Then, we have

    T(x)T(y)∥≤αxy∥<∥xy,

    since 0<α<1.

    This completes the proof.

    Theorem 5.6. (Near fixed point theorem). Let (U,) be a near pseudo -Banach space (resp. near pseudo -Banach space) with the null set Ψ. Suppose that the following conditions are satisfied.

    The null set Ψ satisfies the neutral condition.

    The norm satisfies the null super-inequality and null condition.

    The function T:(U,)(U,) is a weakly strict contraction on U.

    If {Tn(x0)}n=1 forms a -Cauchy sequence (resp. -Cauchy sequence) for some x0U, then T has a near fixed point xU satisfying T(x)Ψ=x. Moreover, the near fixed point x is obtained by the following limit

    limnTn(x0)x∥=limnxTn(x0)∥=0.

    We further assume that the norm satisfies the null equality. Then, we also have the following properties.

    (a) There is a unique equivalence class [x] such that, for any x[x], x cannot be a near fixed point. If x is a near fixed point of T, then we have [x]=[x], i.e., xΨ=x.

    (b) Every point x in the equivalent class [x] is also a near fixed point of T such that the following equalities

    T(x)Ψ=xandxΨ=x

    are satisfied.

    Proof. Since (U,) is a near -Banach space, using the -completeness, the -Cauchy sequence {Tn(x0)}n=1 says that there exists xU satisfying

    limnTn(x0)x∥=0=limnxTn(x0). (5.14)

    Therefore, given any ϵ>0, there exists an integer N such that

    nNimpliesTn(x0)x∥<ϵ. (5.15)

    Since T is a weakly strict contraction on U, we consider the following two cases.

    ● For Tn(x0)Ψ=x, using Part (iii) of Proposition 3.9, the weakly strict contraction says that

    Tn+1(x0)T(x)∥=0<ϵ.

    ● For Tn(x0)Ψx, using (5.15), the weakly strict contraction says that

    Tn+1(x0)T(x)∥<∥Tn(x0)x∥<ϵfornN.

    The above two cases show that

    limnTn+1(x0)T(x)∥=0. (5.16)

    Using Proposition 3.8, we obtain

    xT(x)∥≤∥xTn+1(x0)+Tn+1(x0)T(x)

    Using (5.14) and (5.16), we also obtain

    0≤∥xT(x)∥≤limnxTn+1(x0)+limnTn+1(x0)T(x)∥=0+0=0,

    which says that xT(x)∥=0, i.e., T(x)Ψ=x by using Part (i) of Proposition 3.9. This shows that x is a near fixed point.

    Similarly, when (U,) is assumed to be a near -Banach space and {Tn(x0)}n=1 is assumed to be a -Cauchy sequence, the above arguments are still valid to show that x is a near fixed point.

    Now, we further assume that the norm satisfies the null equality. Proposition 3.6 says that the norm satisfies the symmetric condition. Since x is a near fixed point, in the sequel, we shall claim that each point x in the class [x] is also a near fixed point of T. For any x[x], we first note xΨ=x, which says that

    xψ1=xψ2forsomeψ1,ψ2Ψ. (5.17)

    Then, we have

    Tn(x0)x=∥xTn(x0)(sincethesymmetricconditionissatisfied)=∥(xψ1)Tn(x0)(usingthenullequality)=∥(xψ2)Tn(x0)(using(5.17))=∥xTn(x0)(usingthenullequalityagain).

    Using (5.14), we obtain

    limnTn(x0)x∥=limnxTn(x0)∥=0. (5.18)

    Therefore, given any ϵ>0, there exists an integer N such that

    nNimpliesTn(x0)x∥<ϵ. (5.19)

    Since T is a weakly strict contraction on U, we consider the following two cases.

    ● For Tn(x0)Ψ=x, using Part (iii) of Proposition 3.9, the weakly strict contraction says that

    Tn+1(x0)T(x)∥=0<ϵ.

    ● For Tn(x0)Ψx, using (5.19), the weakly strict contraction says that

    Tn+1(x0)T(x)∥<∥Tn(x0)x∥<ϵfornN.

    The above two cases show that

    limnTn+1(x0)T(x)∥=0. (5.20)

    Using Proposition 3.8, we have

    xT(x)∥≤∥xTn+1(x0)+Tn+1(x0)T(x)=∥Tn+1(x0)x+Tn+1(x0)T(x)(sincethesymmetricconditionissatisfied).

    Using (5.18) and (5.20), we also obtain

    0≤∥xT(x)∥≤limnTn+1(x0)x+limnTn+1(x0)T(x)∥=0+0=0,

    which says that xT(x)∥=0, i.e., T(x)Ψ=x by using Part (i) of Proposition 3.9. This shows that x is a near fixed point for any point x[x].

    Suppose that x[x] is another near fixed point of T. Then, we have

    T(x)Ψ=xandxΨx.

    Since we also have T(x)Ψ=x, it follows that

    T(x)ψ1=xψ2andT(x)ψ3=xψ4 (5.21)

    for some ψiΨ for i=1,2,3,4. On the other hand, we have

    (xψ2)(xψ4)∥=∥(xψ2)x(ψ4)(usingProposition3.7)=∥(xψ2)xψ5(usingtheneutralconditionbysettingψ5=ψ4Ψ)=∥xx(usingthenullequalityfortwice). (5.22)

    We can similarly obtain

    (T(x)ψ1)(T(x)ψ3)∥=∥T(x)T(x). (5.23)

    Therefore, we have

    xx∥=∥(xψ2)(xψ4)(using(5.22))=∥(T(x)ψ1)(T(x)ψ3)(using(5.21))=∥T(x)T(x)(using(5.23))<∥xx(sincexΨxandTisaweaklystrictcontraction).

    This contradiction shows that any x[x] cannot be the near fixed point of T. Equivalently, if x is a near fixed point of T, then we must have x[x]. This completes the proof.

    We are going to propose the concept of a weakly uniformly strict contraction on a near pseudo-normed space. Under some suitable conditions, we can obtain the near fixed point theorem based on the near -Banach space and near -Banach space. The concept of a weakly uniformly strict contraction was proposed by Meir and Keeler [13].

    Definition 5.7. Let (U,) be a near pseudo-normed space with the null set Ψ. A function

    T:(U,)(U,)

    is called a weakly uniformly strict contraction on U when the following conditions are satisfied

    xΨ=y implies T(x)T(y)∥=0;

    ● given any ϵ>0, there exists δ>0 such that

    ϵ≤∥xy∥<ϵ+δimpliesT(x)T(y)∥<ϵ

    for any xΨy.

    Part (i) of Proposition 3.9 says that xΨy implies xy∥≠0. Therefore, the second condition of the weakly uniformly strict contraction is well-defined. In other words, when we consider the weakly uniformly strict contraction, the space (U,) should be assumed to be a near pseudo-normed space rather than a near pseudo-seminormed space.

    Remark 5.8. Since the second condition implies

    T(x)T(y)∥<ϵ≤∥xy,

    it says that if T is a weakly uniformly strict contraction on U, then T is also a weakly strict contraction on U.

    Lemma 5.9. Let (U,) be a near pseudo-normed space with the null set Ψ, and let

    T:(U,)(U,)

    be a weakly uniformly strict contraction on U. Suppose that the norm satisfies the null super-inequality and null condition. Then, for any xU, the sequence {Tn(x)Tn+1(x)}n=1 is decreasing and satisfying

    limnTn(x)Tn+1(x)∥=0.

    Proof. Let Tn(x)=an and bn=∥anan+1 for all n. We consider the following two cases.

    ● For [an1][an], we have

    bn=∥anan+1∥=∥Tn(x)Tn+1(x)<∥Tn1(x)Tn(x)(usingRemark5.8)=∥an1an∥=bn1.

    ● For [an1]=[an], we have

    bn=∥anan+1∥=∥Tn(x)Tn+1(x)=∥T(Tn1(x))T(Tn(x))∥=∥T(an1)T(an)=0(usingthefirstconditionofDefinition5.7)bn1.

    The above two cases show that the sequence {bn}n=1 is indeed decreasing.

    Suppose that akΨak+1 for all k1. Since the sequence {bn}n=1 has been proven to be decreasing, we assume bnϵ>0, i.e., bnϵ>0 for all n. Then, there exists δ>0 satisfying ϵbk<ϵ+δ for some k, i.e.,

    ϵ≤∥akak+1∥<ϵ+δ.

    Therefore, we have

    bk+1=∥ak+1ak+2∥=∥Tk+1(x)Tk+2(x)∥=∥T(Tk(x))T(Tk+1(x))=∥T(ak)T(ak+1)∥<ϵ(usingthesecondconditionofDefinition5.7),

    which contradicts bk+1ϵ. This contradiction says that akΨ=ak+1 for some k1. Therefore, we can now assume that k is the first index in the sequence {an}n=1 satisfying ak1Ψ=ak. Then, we want to claim that bk1=bk=bk+1==0. Since ak1Ψ=ak, Part (iii) of Proposition 3.9 says that

    bk1=∥ak1ak∥=0.

    We also have

    0=∥T(ak1)T(ak)(usingthefirstconditionofDefinition5.7)=∥T(Tk1(x))T(Tk(x))∥=∥Tk(x)Tk+1(x)=∥akak+1∥=bk,

    which says that akΨ=ak+1 by Part (ii) of Proposition 3.9. Using the similar arguments, we can also obtain bk+1=0 and ak+1Ψ=ak+2. Therefore, the sequence {bn}n=1 is decreasing to zero. This completes the proof.

    Theorem 5.10. (Meir-Keeler type of near fixed point theorem). Let (U,) be a near pseudo -Banach space or near pseudo -Banach space with the null set Ψ. Suppose that the following conditions are satisfied.

    The null sets Ψ satisfies the neutral condition.

    The norm satisfies the null super-inequality and null condition.

    The function T:(U,)(U,) is a weakly uniformly strict contraction on U.

    There exists x0U satisfying

    Tn(x0)Tn+1(x0)∥=∥Tn+1(x0)Tn(x0)foralln. (5.24)

    Then T has a near fixed point satisfying T(x)Ψ=x. Moreover, the near fixed point x is obtained by the following limit

    limnTn(x0)x∥=limnxTn(x0)∥=0.

    We further assume that the norm satisfies the null equality. Then, we also have the following properties.

    (a) There is a unique equivalence class [x] such that, for any x[x], x cannot be a near fixed point. If x is a near fixed point of T, then we have [x]=[x], i.e., xΨ=x.

    (b) Every point x in the equivalent class [x] is also a near fixed point of T such that the following equalities

    T(x)Ψ=xandxΨ=x

    are satisfied.

    Proof. Using Theorem 5.6 and Remark 5.8, we remain to show that if T is a weakly uniformly strict contraction, then {Tn(x0)}n=1 forms both a -Cauchy sequence and -Cauchy sequence.

    Let an=Tn(x0) and bn=∥anan+1. Suppose that {Tn(x0)}n=1={an}n=1 is not a -Cauchy sequence. Then, there exists ϵ>0 such that, given an integer N, there exist n>mN satisfying aman∥>2ϵ. Since T is a weakly uniformly strict contraction on U, the definition says that there exists δ>0 such that

    ϵ≤∥xy∥<ϵ+δimpliesT(x)T(y)∥<ϵforanyxΨy. (5.25)

    Let η=min{δ,ϵ}. We are going to claim that

    ϵ≤∥xy∥<ϵ+ηimpliesT(x)T(y)∥<ϵforanyxΨy. (5.26)

    We consider the following two cases.

    ● If η=δ, it is clear to see that (5.25) implies (5.26).

    ● If η=ϵ, it means that ϵ<δ. Therefore, we have

    ϵ+η=ϵ+ϵ<ϵ+δ,

    which also says that (5.25) implies (5.26)

    For n>mN, we have

    aman∥>2ϵϵ+η. (5.27)

    Therefore, using Part (iii) of Proposition 3.9, we obtain

    amΨanforn>mN.

    From (5.24), we have

    bn=∥anan+1∥=∥an+1an. (5.28)

    Lemma 5.9 says that the sequence {bn}n=1 is decreasing to zero. Therefore, we can find an integer N satisfying bN<η/3. Then, for mN, we obtain

    amam+1∥=bmbN<η3ϵ3<ϵ. (5.29)

    For k with m<kn, Proposition 3.8 says that

    amak+1∥≤∥amak+akak+1. (5.30)

    We want to show that there exists k with m<kn satisfying

    amΨakandϵ+2η3<∥amak∥<ϵ+η. (5.31)

    Let γk=∥amak for k=m+1,,n. Then, (5.27) implies

    γn>ϵ+η, (5.32)

    and (5.29) implies

    γm+1<ϵ. (5.33)

    Let k0 be an index satisfying

    k0=max{k{m+1,,n}:γkϵ+2η3}. (5.34)

    Then, we have

    γk0ϵ+2η3. (5.35)

    From (5.32) and (5.33), we also see that m+1k0<n, which says that k0 is well defined. Since k0 is an integer, we also have k0+1n. By the definition of k0, it means that k0+1 does not satisfy (5.34), i.e.,

    amak0+1∥=γk0+1>ϵ+2η3, (5.36)

    which also says that amΨak0+1 by using Part (iii) of Proposition 3.9. Therefore, from (5.36), the expression (5.31) will be sound if we can show that γk0+1<ϵ+η. Suppose that this is not true, i.e.,

    γk0+1ϵ+η. (5.37)

    From (5.30), we have

    γk0+1γk0+ak0ak0+1. (5.38)

    Therefore, we have

    η3>bNbk0=∥ak0ak0+1(using(5.29),sincek0m+1>N)γk0+1γk0(using(5.38))ϵ+ηϵ2η3(using(5.35)and(5.37))=η3.

    This contradiction says that (5.31) is sound.

    Using (5.26), we see that (5.31) implies

    am+1ak+1∥=∥T(am)T(ak)∥<ϵforamΨak. (5.39)

    Therefore, we obtain

    amak≤∥amam+1+am+1ak+1+ak+1ak(usingProposition3.8)<bm+ϵ+bk(using(5.39)and(5.28))<η3+ϵ+η3(using(5.29),sinceNm<k)=ϵ+2η3,

    which contradicts (5.31). This contradiction shows that the sequence {Tn(x)}n=1 is a -Cauchy sequence.

    We can similarly prove that the sequence {Tn(x)}n=1 is a -Cauchy sequence. This completes the proof.

    Example 5.11. Continued from Example 4.15, suppose that the function

    T:(Fcc(R),)(Fcc(R),)

    is a contraction on Fcc(R). Using Theorem 5.3, the contraction T has a near fixed point ˜xFcc(R) satisfying T(˜x)Ψ=˜x. Moreover, the near fixed point ˜x is obtained by the following limit

    limn˜xn˜x∥=0

    in which the sequence {˜xn}n=1 is generated according to (5.1). We also have the following properties.

    (a) There is a unique equivalence class [˜x] such that, for any ˜x[˜x], ˜x cannot be a near fixed point. If ˜x is a near fixed point of T, then we have [˜x]=[˜x], i.e., ˜xΨ=˜x.

    (b) Every point ˜x in the equivalent class [˜x] is also a near fixed point of T such that the following equalities

    T(˜x)Ψ=˜xand˜xΨ=˜x

    are satisfied.

    Example 5.12. Continued from Example 4.15, suppose that the function

    T:(Fcc(R),)(Fcc(R),)

    is a weakly uniformly strict contraction on Fcc(R). Theorem 5.10 says that T has a near fixed point satisfying T(˜x)Ψ=˜x. Moreover, the near fixed point ˜x is obtained by the following limit

    limnTn(˜x)˜x∥=0forsome˜xFcc(R).

    We also have the same properties (a) and (b) given in Example 5.11.

    The so-called near vector space, which extends the concept of a vector space, is considered in this paper. The main issue of a near vector space is that it lacks the additive inverse elements. Three well-known space near vector spaces are the space of all bounded and closed intervals in R, the space of all compact and convex subsets of a topological space, and the space of all fuzzy numbers in R. We see that these three spaces do not own the concept of additive inverse elements. Therefore, the so-called null set is introduced to play the role of zero element.

    Based on the null set, a so-called near normed space is proposed, which just endows a norm to a near vector space. Furthermore, the concept of completeness is then introduced to propose the so-called near Banach space. Therefore, under these settings, we are able to establish the near fixed point theorems in near Banach space. In the future research, many advanced topics regarding the fixed point theorems can be hopefully established in near vector space.

    The author declares that he has no conflict of interest.



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